MECHANICAL ENGINEERING
REVIEW MANUAL
By : ENGR. YURI G. MELLIZA
1
PART 1:THERMODYNAMICS
DEFINITION OF TERMS
Mass: It is the absolute quantity of matter in it.
m – mass, kg
Conversion
1 kg = 2.205 lbs
1 Metric Ton = 1,000 kg
1 Ton = 2000 Lbs
Velocity: It is the distance per unit time.
d m
t sec
Where
d – distance, m
t – time, sec
v
Conversion
1 meter = 3.28 ft = 100 cm = 1000 mm
1 ft. = 12 inches
Acceleration: Rate of change of velocity with respect to time.
dv m
a
dt sec 2
Gravitational Acceleration: It is the acceleration due to gravity.
At standard condition or Sea level condition:
g = 9.81 m/sec2 = 32.2 ft/sec2
Force: It is the mass multiplied by the acceleration.
F  ma N (Newton)
F
ma
KN (Kilo Newton)
1000
Newton: It is the force required to accelerate 1 kg mass at the rate of 1 m/sec per second.
Where:
m – mass in kg
a – acceleration in m/sec2
Weight: It is the forge due to gravity.
W  mg N
mg
KN
1000
where
W
g - gravitatio nal accelerati on,
g  9.81
m
sec2
m
 Standard gravitatio nal accelerati on 
sec2
2
Force of Attraction: From Newton’s Law of Gravitation the force of attraction between two masses m1 and m2 is given by the
equation:
Fg 
Gm1m 2
Newton
r2
Where:
m1 and m2 - masses in kg
r - distance apart in meters
G - Gravitational constant in N-m2/kg2
G  6.670 x 10-11
N - m2
kg 2
PROPERTIES OF FLUIDS
Density () - it is the mass per unit volume.
ρ
m kg
V m3
Where:
m - mass, kg
V - volume , m3
Conversion:
1
1
lb
ft 3
kg
 16.0185
3
kg
m3
 62,427.9606
cm
g
kg
1
1 3
liter
m
lb
ft 3
Specific volume () - it is the volume per unit mass or the reciprocal of its density.
υ
V m3
m kg
υ
1 m3
ρ kg
Specific weight () - it is the weight per unit volume.
W KN
V m3
mg
KN
γ
1000V m 3
ρg KN
γ
1000 m 3
1000γ kg
ρ
g m3
γ
Specific Gravity Or Relative Density (S):
FOR LIQUIDS: Its specific gravity or relative density is equal to the ratio of its density to that of water at standard temperature and
pressure.
SL 
ρ
γ
 L
ρw γw
3
At standard condition:
water = 1000 kg/m3
water = 9.81 KN/m3
FOR GASES: Its specific gravity or relative density is equal to the ratio of its density to that of either air or hydrogen at some
specified temperature and pressure
PV  mRT
P kg
RT m 3
8.3143 KJ
R
M
kg - K
ρ
Where:
ah - density of either air or hydrogen at some value of P and T.
Equation of State for Gases
Conversion
F  32
1 .8
F  1.8(C)  32
C 
Where:
P – absolute pressure in KPa
V – volume in m3
m – mass in kg
R – Gas constant in KJ/kg-K
T – absolute temperature in K.
M – molecular weight of gas, kg/kgmol
Temperature: It is the measure of the intensity of heat.
FAHRENHEIT SCALE
Freezing Point = 32F
Boiling Point = 212F
CENTIGRADE SCALE
Freezing Point = 0C
Boiling Point = 100C
ABSOLUTE SCALE
ρG
SG 
ρ AH
R = F + 460
K = C + 273
4
Pressure: - is defined as the normal component of a force per unit area.
F KN
or KPa (Kilo Pascal)
A m2
KN
1 2  1 KPa
m
P
If a force dF acts on infinitesimal area dA the intensity of pressure P is
P
dF
dA
PASCAL`S LAW
At any point in a homogeneous fluid at rest the pressures are the same in all directions.
Atmospheric Pressure (Pa): It is the average pressure exerted by the atmosphere.
At sea level (Standard Condition)
Pa
1Bar
1MPa
=
=
=
=
=
=
=
=
=
=
101.325
0.101325
760
10.33
1.033
14.7
29.921
33.88
100
1000
KPa
MPa
mm Hg
m of H2O
Kg/cm2
Lb/in2
in Hg
Ft of H2O
KPa
KPa
ABSOLUTE AND GAGE PRESSURE
Absolute Pressure - is the pressure measured referred to absolute zero and using absolute zero as the base.
Gage Pressure - is the pressure measured referred to atmospheric pressure and using atmospheric pressure as the base.
5
VARIATION OF PRESSURE WITH DEPTH OR ELEVATION
Viscosity: It is the property of a fluid that determines the amount of its resistance to shearing stress.
6
Kinematic Viscosity: It is the ratio of the absolute or dynamic viscosity to the mass density.
μ m2
ν
ρ sec
Elasticity: If a pressure is applied to a fluid, it contracts; if the pressure is released, it expands, the elasticity of a fluid is related to
the amount of deformation(expansion or contraction) for a given pressure change. Quantitatively, the degree of elasticity is equal
to:
dP
Ev  
dV
V
Where negative sign is used because dV/V is negative for a positive dP.
Ev 
dP
dV dρ
; because 
ρ
dρ
V
ρ
Where:
Ev - bulk modulus of elasticity
dV - is the incremental volume change
V - is the original volume
dP - is the incremental pressure change
Surface Tension: Capillarity
Surface tension is the tendency of fluid surfaces to shrink into the minimum surface area possible. Surface tension allows insects,
usually denser than water, to float and slide on a water surface.
Capillarity is the result of surface, or interfacial, forces. The rise of water in a thin tube inserted in water is caused by forces of
attraction between the molecules of water and the glass walls and among the molecules of water themselves.


h


r
2σ cos θ
γr
h
7
Where:
 - surface tension, N/m
 - specific weight of liquid, N/m3
r – radius, m
h – capillary rise, m
Table 1.Surface Tension of Water
C

0
0.0756
10
0.0742
20
0.0728
30
0.0712
40
0.0696
60
0.0662
80
0.0626
100
0.0589
Entropy (S): It is a property of a fluid that determines the amount of its randomness and disorder of a substance. If during a process
an amount of heat is taken at certain instant and is divided by the absolute temperature at which it is taken the result is called the
“Change of Entropy”
T
Area  Heat
1
dQ  TdS
T
ΔS 
2
dS
dQ
T
dQ  TdS
dS 
T
dQ
S
MANOMETERS
Manometer is an instrument used in measuring gage pressure in length of some liquid column.
1. Open Type Manometer : It has an atmospheric surface and is capable in measuring gage pressure.
2. Differential Type Manometer : It has no atmospheric surface and is capable in measuring differences of pressure.
Open Type
Differential Type
8
Problem No. 1 (Force of Attraction)
How far from the earth must a body be along a line toward the sun so that the gravitational pull of the sun balances that of the earth?
Earth to sun distance is 9.3 x 107 mi; mass of sun is 3.24 x105 times mass of earth. (1.63 x 105 mi.)
Problem No. 2 (Measuring Temperature)
If the F scale is twice the C scale, what will be the corresponding reading in each scale? (160 ; 320)
F  32
C 
1.8
F  1.8C  32
F  2C
2C  1.8C  32
32
C 
 160
2  1 .8
F  320
9
Problem No. 3 (Density, Specific gravity)
A cylindrical tank 2 m diameter, 3 m high is full of oil. If the specific gravity of oil is 0.9, what is the mass of oil in the tank? (8482.3
kg)
Vcylinder 
S
π 2
π
D H  (2 2 )3  3π m 3
4
4
ρ
ρ water
ρ
1000
kg
ρ  900 3
m
m
ρ
V
m  900(3π)  8482.3 kg
0 .9 
Problem No. 4 (Variation in Pressure)
An open tank contains 5 m of water covered with 2 m of oil ( = 8 KN/m3). Find the pressure at the interface and at the bottom of
the tank.
Problem No. 5 (Force)
10 liters of an incompressible liquid exert a force of 20 N at the earth’s surface. What force would 2.3 Liters of this liquid exert on
the surface of the moon? The gravitational acceleration on the surface of the moon is 1.67 m/sec2.
F  ma
ρ
m
V
 1 m3 
 kg
m  ρV  ρ(10.0)

 1000 L 
ρ(10.0)
9.81 N
20 
1000
kg
ρ  203.874 3
m
On the surface of the moon
 2 .3 
F  203.874
(1.67)  0.783 N
 1000 
10
Problem No. 6 (Manometer)
An open manometer is used to measure the pressure in the tank. The tank is half filled with 50,000 kg of a liquid chemical that is
not miscible in water. The manometer tube is filled with liquid chemical. What is the pressure in the tank relative to atmospheric
pressure?
4
 1 2

2

V   πr 3  πr 2 (6)   πr 3  πr 2 (3)  πr 2  r  (3)  54.45 m 3
3
 2 3

3

m 50,000
kg
ρ

 918.2 3
V
54.45
m
ρg
918.29.81
KN
γ

 9.008 3
1000
1000
m
P  9.008(0.225)  0.350(9.81)  0
P  1.4067 KPa
Pa bs  101.325  1.4067  102.73 KPa
Problem No. 7 (Temperature)
If the temperature inside a furnace is 700 K, what is the corresponding reading in F? (800.6)
Solution:
t = 700 – 273 = 427C
F = (427)(1.8) + 32
F = 800.6F
Problem No. 8 (Absolute Pressure)
The suction pressure of a pump reads 540 mm Hg vacuum. What is the absolute pressure in KPa? ( 29.33)
Solution:
P = 760 – 540 = 220 mm Hg absolute
P
220
(101.325)  29.33 mm Hg
760
Problem No. 9 (Pressure Variation)
A storage tank contains oil with a specific gravity of 0.88 and depth of 20 m. What is the hydrostatic pressure at the bottom of the
tank in kg/cm2.(1.7)
Solution:
Using: g = 9.81 m/sec2
0.88(1000)(9.81)(20)
P 0
 172.656 KPa
1000
P = 1.7 kg/cm2
Problem No. 10 (Variation in Pressure Measuring Altitude of Mountain)
A hiker carrying a barometer that measures 101.3 KPa at the base of the mountain. The barometer reads 85 KPa at the top of the
mountain. The average air density is 1.21 kg/m3. Determine the height of the mountain.
dP   γdh
P2  P1   γ(h 2  h1 )
h 2  h1  h
(P1  P2 )1000
ρg
(101.3  85)1000
h
1.21(9.81)
h  1373 m
h
11
MANOMETER PROBLEMS
1.
The closed tank in the figure is filled with water. The pressure gage on the tank reads 50 KPa. Determine
a. The height h in mm in the open water column
b. The gage pressure acting on the bottom of the tank surface AB
c. The absolute pressure of the air in the top of the tank if the local atmospheric pressure is 101 KPa absolute.
By applying principles in variati on of pressure
50  9.81(0.60) - 9.81(h)  0
h  5.7 m
PBottom  50  9.81(1.2)  61.8 KPa
Pabsolute  50  101  151 KPa absolute
2.
The mercury manometer in the figure indicates a differential reading of 30 m when the pressure in pipe A is 30 mm Hg
vacuum. Determine the pressure in pipe B.
30(101.325)
 4 KPa
760
- 4  (9.81)1.5  13.6(9.81)(0.30) - 0.9(9.81)(0.45)  PB
PA 
PB  46.8 KPa
3.
A closed tank contains compressed air and oil (S = 0.90) as shown in the figure.a U – tube manometer using mercury (S =
13.6) is connected to the tank as shown. For column heights h1 = 90 cm; h2 = 15 cm and h3 = 22 cm, determine the pressure
reading of the gage.
12
4.
The pressure of gas in a pipeline is measured with a mercury manometer having one limb open to the atmosphere. If the
difference in the height of mercury in the limbs is 562 mm, calculate the absolute gas pressure. The barometer reads 761
mm Hg, the acceleration due to gravity is 9.79 m/sec2 and SHg = 13.64.
5.
A turbine is supplied with steam at a gauge pressure of 1.4 MPa, after expansion in the turbine the steam flows into a
condenser which is maintained at a vacuum of 710 mm Hg. The barometric pressure is 772mm Hg. Express the inlet and
exhaust pressure in kg/cm2 . Take the S of mercury is 13.6.
772(101.325)
Patm 
 103 KPa
760
1.033 kg 2
cm  1.05 kg
PInlet  1400  103  1503 KPa (absolute) x
cm 2
101.325 KPa
1.033 kg 2
 710(101.325)

cm  0.09 kg
PExhaust  
 103 x
(Absolute)
cm 2
760

 101.325 KPa
6.
The pressure of steam flowing in a pipe line is measured with a mercury manometer. Some steam condenses in to water.
Estimate the steam pressure in KPa. Take the density of mercury as 13,600 kg/m3, the barometer reading as 76.1 cm Hg
and g = 9.806 m/sec2.
0  13.6(9.806)(0.50)  9.806(0.03)  PSteam
PSteam  66.4 KPag 
76.1(101.325)
 167.85 KPa (Absolute)
76
LAW OF CONSERVATION OF MASS
Mass is indestructible. In applying this law we must except nuclear processes during which mass is converted into energy.
b
1
m1
c
a
d
m
2
m2
The verbal Form of the law is:
Mass Entering - Mass Leaving = Change of Mass Stored in the system
In equation form:
m1  m 2  Δm
13
For a Closed System, a system of fixed mass, no equation is necessary. But for a steady-state, steady flow (SSSF) system (OPEN
SYSTEM) m = 0. Therefore:
m1  m 2  0 ; therefore
m1  m 2
For one dimensional flow, the mass flow rate, m in kg/sec is equal to
m  ρAv 
Av
υ
From m1 = m2 = m
ρ1A1v1  ρ 2 A 2 v 2  ρAv
A1v1 A 2 v 2 Av


υ1
υ2
υ
Where:
m – mass flow rate in kg/sec
A - cross sectional area in m2
v - velocity in m/sec
 - specific volume in m3/kg
 - density in kg/m3
Problem No. 1
A counter flowing heat exchanger is used to cool air at 540 K, 400 KPa to 360 K by using 0.05 kg/sec supply of water at 20C, 200
KPa. The airflow is 0.5 kg/sec in a 10 cm diameter pipe. Find the inlet velocity and the water exit temperature.
Problem No.2
Problem No. 3
FORMS OF ENERGY
Work: It is the force multiplied by the displacement in the direction of the force.

2
W  F  dx
1
By convention:
-W - indicates that work is done on the system
+W - indicates that work is done by the system.
Heat: It is a form of energy that crosses a system's boundary, because of a temperature difference between the system and the
surrounding.
Q - Heat
By convention:
+Q - indicates that heat is added to the system
-Q - indicates that heat is rejected from the system.
14
Internal Energy: It is the energy acquired due to the overall molecular interaction, or the total energy that a molecule has.
U = mu KJ
Where
U - total internal energy KJ, KW
u - specific internal energy KJ/kg
U- change of internal energy
m - mass kg, kg/sec
Flow Eneregy Or Flow Work: It is the energy required in pushing a fluid usually into the system or out from the system.
Ef  Flow work or Flow energy
E f1  F1L1  P1A1L1  P1V1
E f2  F2 L 2  P2 A 2 L 2  P2 V2
ΔEf  E f2  E f1
ma
1000
dv
a
dt
F
m 2 dv
 dx
1000 1 dt
m 2
dx
ΔKE 
dv 
1000 1
dt
dx
v
dt
ΔKE 
ΔEf  P2 V2  P1V1
ΔEf  ΔPV 
ΔPV   P2 V2  P1V1 KJ; KW
ΔPυ   P2 υ 2  P1υ1

ΔKE  F  dx


KJ
kg
Kinetic Energy: It is the energy acquired due to the motion of a body or a system.
ΔKE 
ΔKE 
ΔKE 
ΔKE 
m 2
v  dv
1000 1

 2
m v
 
1000  2 

2(1000)
KJ; KW
v  v  KJ
2
2
m3
V  m ;
sec
3
2
1
m v 2  v12
2
P  KPa
2
1
2(1000) kg
m  kg; kg / sec
m3
υ 
kg
1
υ 
ρ
15
Potential Energy: It is the energy required by virtue of its configuration or elevation.
m
2
z
m
1
Reference
Datum
ΔPE  W  dz

mg
W
1000
mg 2
ΔPE 
dz
1000 1
mg (z 2  z1 )
ΔPE 
KJ; KW
1000
g (z 2  z1 ) KJ
ΔPE 
1000
kg
m  kg; kg / sec

Where:
Z - elevation in m, (+) if above datum and (-) if below datum
g - gravitational acceleration, m/sec2
Enthalpy (h): It is the sum of the Internal Energy and the Flow Energy.
h  U  PV KJ
Δh  ΔU  Δ(PV) KJ
Δh  Δu  Δ(Pυ)
KJ
kg
ZEROTH LAW OF THERMODYNAMICS
If two bodies are in thermal equilibrium with a third body, they are in thermal equilibrium with each other, and hence their
temperatures are equal.
SPECIFIC HEAT: It is the amount of heat required to raise the temperature of a 1 kg mass, 1 K or 1 C.
C
dQ dQ KJ
KJ

or
dt dT kg - C
kg - K
then dQ = Cdt
by integration, where C is constant
Q  C ( Δt )  C ( ΔT )
Considerin g m (mass)
Q  mC (Δt )  mC (ΔT )
where :
Δt  Δ T
t - temperatu re in C
T - absolute temperatu re in K
16
Sensible Heat: The amount of heat per unit mass that must be transferred (added or remove) when a substance undergoes a
change in temperature without a change in phase.
Q  mC (Δt )  mC (ΔT)
Where:
m - mass , kg
C - heat capacity or specific heat, KJ/kg-C or KJ/kg-K
t - temperature in C
T - temperature in K
HEAT OF TRANSFORMATION (LATENT HEAT): The amount of heat per unit mass that must be transferred when a substance
completely undergoes a phase change without a change in temperature.
Q  mh
A. Heat of Vaporization: Amount of heat that must be added to vaporize a liquid or that must be removed to condense a gas.
Q  mh v
Where
hv- latent heat of vaporization, KJ/kg
B. Heat of Fusion : Amount of heat that must be added to melt a solid or that must be removed to freeze a liquid.
Q  mh F
Where
hF – latent heat of fusion, KJ/kg
WATER EQUIVALENT: The water equivalent of a substance is the mass of water that would require the same heat transfer as
the mass of that substance to cause the same change of temperature.
mw 
ms C ps
C pw
PROPERTIES OF PURE SUBSTANCE
A pure substance is one that is uniform and invariable in chemical composition. A pure substance can exist in more than one phase,
but its chemical composition must be the same in each phase. For example, if liquid water and water vapor form a system with two
phases, the system can be regarded as a pure substance because each phase has the same composition. A uniform mixture of gases
can be regarded as a pure substance provided it remains a gas and does not react chemically.
a - sub-cooled liquid
b - saturated liquid
c - saturated mixture
d - saturated vapor
e - superheated vapor
Considering that the system is heated at constant pressure where P = 101.325 KPa, the 100C is the saturation temperature
corresponding to 101.325 KPa, and 101.325 KPa pressure is the saturation pressure at 100C.
Saturation Temperature (tsat) - is the highest temperature at a given pressure in which vaporization takes place.
Saturation Pressure (Psat) - is the pressure corresponding to the temperature.
Sub-cooled Liquid - is one whose temperature is less than the saturation temperature corresponding to the pressure.
17
Compressed Liquid - is one whose pressure is greater than the saturation pressure corresponding to the temperature.
Saturated Mixture - a mixture of liquid and vapor at the saturation temperature.
Superheated Vapor - a vapor whose temperature is greater than the saturation temperature.
Temperature - Specific volume Diagram (T- diagram)
Region I – Sub-cooled Liquid Compressed Liquid region
Region II -Saturated Mixture region
Region III - Superheated Vapor region
F (critical point)- at the critical point the temperature and pressure is unique.
For Steam: At Critical Point, P = 22.09 MPa; t = 374.136C
Temperature-Entropy Diagram (T-S Diagram)
Entropy (S): Is that property that determines the randomness and disorder of a substance. If during a process, an amount of heat is
taken and is divided by the absolute temperature at which it is taken the result is called the “Change of Entropy”.
dS 
dQ
T
ΔS 
T
dQ
18
Enthalpy-Entropy Diagram (h versus S Diagram or Mollier Chart)
The properties h,S,U,and  at saturated liquid, saturated vapor, sub-cooled or compressed liquid and superheated vapor condition,
can be determined using the Steam Table.
For the properties at the saturated mixture condition, its properties is equal to
r  rf  x (rfg )
rfg  rg  rf
where r stands for any property, such as h, S, U,and , where subscript f refers to saturated liquid condition and fg refers to the
difference in property between saturated vapor and saturated liquid and x is called the quality.
QUALITY
x
m vapor
m vapor  mliquid

mv
m v  ml
Where:
m - mass
v - refers to vapor
l - refers to liquid
Note: For sub-cooled liquid, its properties are approximately equal to the properties at saturated liquid which corresponds to the
sub-cooled temperature.
THROTTLING CALORIMETER:
An apparatus that is used to determine the quality of a de-superheated steam flowing in a steam line.
A throttling process is a Steady State, Steady Flow Process in which Q = 0; W = 0; KE = 0; PE = 0 but the Enthalpy (h) remains
constant (h = C).
From the figure, steam from the main steam line expands in the calorimeter to the calorimeter pressure and temperature at h =
constant.. A throttling calorimeter is an instrument used to determine the quality of steam.
19
IDEAL OR PERFECT GAS
Ideal Gas: A hypothetical gas that obeys the gas laws perfectly at all temperatures and pressures.
1. CHARACTERISTIC EQUATION
PV  mRT
Pυ  RT
RT
P
P
ρ
RT
PV
Pυ
 C or
C
T
T
υ
Where:
P - absolute pressure in KPa
V - volume in m3
m -mass in kg
R -Gas constant in KJ/kg-K
T - absolute temperature in K
 - specific volume in m3/kg
 - density in kg/m3
2. GAS CONSTANT
R
R 8.3143 KJ

M
M kg - K
R = 8.3143 KJ/kgm-K
R - universal gas constant, KJ/kgm-K
M - molecular weight, kg/kg mol
m kg
M
n kg mol
m  mass , kg
n - no. of moles
3. BOYLE`S LAW (T = C) Robert Boyle (1627-1691) :If the temperature of a certain quantity of gas is held constant, the volume
V is inversely proportional to the absolute pressure P, during a quasi-static change of state.
1
1
or V  C
P
P
PV  C or
Vα
P1 V1  P2 V2  C
4. CHARLE`S LAW ( P = C and V = C) Jacques Charles (1746-1823) and Joseph Louis Gay-Lussac (1778-1850)
 At constant pressure (P = C), the volume V of a certain quantity of gas is directly proportional to the absolute temperature
T, during a quasi static change of state.
V  T or V  CT
V
V V
 C or 1  2
T
T1 T2

At constant volume (V = C), the pressure P of a certain quantity of gas is directly proportional to the absolute temperature
T, during a quasi-static change of state.
P  T or P = CT
P
C
T
or
P1 P2

T1 T2
20
5. AVOGADRO`S LAW: Amedeo Avogadro (1776-1856) :All gases at the same temperature and pressure, under the action of a
given value of g, have the same number of molecules per unit of volume. From which it follows that the the specific weight is
directly proportional to its molecular weight.
At the same temperatu re and pressure for all gases
n1 n 2
n


V1 V2 V
or
γ1 M 1 R 2


γ 2 M 2 R1
6. SPECIFIC HEATS
Specific Heat At Constant Pressure (Cp)
Q  mC p (ΔT)  mΔh
considerin g m
Δh  h 2  h1
From
Q  m( Δh )  ΔH
dQ  du  Pdυ
dQ
dT
dQ dh
Cp 

 (C p - Specific heat at P  C)
dT dT
dQ  C p dT
dh  dQ  υdP
Q  C p (ΔT)  C p (Δt)
At P  C (Constant Pressure)
Q  Δh  mC p (ΔT)  mC p (Δt)
h  u  Pυ
dh  du  Pdυ  υdP
but
From C 
dP  0
dh  dQ
Q  Δh
Specific Heat At Constant Volume (Cv)
Q  mC V ΔT  mΔu
ΔU  m(u 2  u1 )
From
dQ  du  Pdυ
at υ  C; dυ  0
therefore
From
dQ
dT
dQ du
Cv 

 (C v  specific heat at V  C)
dT dT
dQ  C v (dT)
C
Q  C v (ΔT)  C v (Δt )
Q  ΔU  mC v (ΔT)  mC v (Δt )
dQ  du
From
Q  Δu  (u 2 - u1 )
h  u  Pυ
Q  ΔU  m(u 2 - u1 )
Pυ  RT
h  u  RT
dh  du  Rdt
C p dT  C v dT  Rdt
dividing the equation by dT
Cp  Cv  R
21
7. RATIO OF SPECIFIC HEATS
C p dh ΔH
k


C V du ΔU
C p  kCv
Cv 
Cp
k
Cp  Cv  R
kCv  C v  R
R
k 1
Cp  Cv  R
Cv 
Cp 
Cp
R
k
Rk
R
k 1
8. ENTROPY CHANGE (S)
Entropy is that property of a substance that determines the amount of randomness and disorder of a substance. If during a process,
an amount of heat is taken and is by divided by the absolute temperature at which it is taken, the result is called the “Entropy
Change”
ENTROPY CHANGE.
dQ
T
dQ  T (dS)
dS 
by integratio n
ΔS 
2 dQ
 T
1
From :
h  U  Pυ
h  U  RT
dQ  dU  Pdυ
RT
and dQ  TdS
υ
RT
TdS  C v dT 
dυ
υ
dividing by T
P
dT
dυ
R
T
υ
By integratio n
dS  C v
T
υ
ΔS  C vln 2  Rln 2
T1
υ1
From
dh  dQ  υdP
dQ  dh 
RT
dP
P
TdS  C p dT  RT
dP
P
dividing by T
dT
dP
R
T
P
by integratio n
dS  C p
T
P
ΔS  C pln 2  Rln 2
T1
P1
Actual – Gas equation of State
In actual gases the molecular collision are inelastic; at high densities in particular there are intermolecular forces that the simplified
equation of the state do not account for. There are many gas equations of state that attempt to correct for the non - ideal behavior of
gases. The disadvantage of all methods is that the equations are more complex and require the use of experimental coefficients.
a. Van der Waals Equation
22




 P  a  υ  b  RT
2

υ 

where :
a and b a coefficien ts that compensate for the nonideal behavior of the gas.
m3
kg mol
b. Beattie-Bridgeman Equation
υ in
2


υ
P  A 
 RT
2

(
1

ε
)(
υ  B)
υ 

where :
 a
A  A 0 1  
 υ
 b
B  B0 1  
 υ
c
ε 3
υT
Compressibility Factor
Pυ
1
RT
For nonideal behavior of gases
Pυ
Z
RT
Where: Z – compressibility factor
IDEAL GAS MIXTURE
Mixture of gases are common in many applications. Our most common example is air - mainly consisting of nitrogen, oxygen and
water vapor - as moist air. A combustion gas with nitrogen, water vapor and carbon dioxide is an other example.

Total Mass Of A Mixture
m
m

i
MASS FRACTION
xi 

Total Moles Of A Mixture
n

n
i
Mole Fraction
yi 

mi
m
ni
n
Equation Of State
Mass Basis
For the Mixture
PV  mRT
Fort the Components
Pi Vi  mi R i Ti
23
Mole Basis
For the Mixture
PV  n RT
Fort the Components
Pi Vi  n i RTi

AMAGAT’S LAW
The total volume of a mixture of gases is equal to the sum of the volume occupied by each component at the mixture
pressure P and temperature T.
2
V2
1
V1
3
V3
P, T
P = P1 = P2 = P3
T = T1 = T2 = T3
n  n1  n 2  n 3
n
PV
PV

RT
RT
; n1 
PV1
RT

PV1
RT
PV2
RT

; n2 
PV2
RT
; n3 
PV3
RT
PV3
RT
 PV PV1 PV2 PV3  RT 





RT  P 
 RT RT RT
V  V1  V2  V3

V
V
yi 
n i Vi

n V
i
DALTON’S LAW
The total pressure of a mixture of gases P is equal to the sum of the partial pressure that each gas would exert at the mixture
volume V and temperature T.
V  V1  V2  V3
T  T1  T2  T3
n  n1  n 2  n 3
n
PV
RT
; n1 
P1V
RT
; n2 
P2V
RT
; n3 
P3V
RT
24
PV

RT
P1V
RT
P2 V

RT

P3V
RT
 PV P1V P2 V P3V  RT 





 RT RT RT RT  V 
P  P1  P2  P3
P
P
i
n i Pi

n
P
yi 



Molecular Weight Of A Mixture (M)
M
y M
M
R 8.3143 KJ

R
R
kg - K
i
i
Gas Constant (R)
R
x R
R
R 8.3143 KJ

M
M
kg - K
i
i
Specific Heat Of A Mixture
At Constant Volume
Cv 
x C
i
vi
At Constant Pressure
CP 
x C
i
Pi
Cp  C v  R

Ratio Of Specific Heat
k
CP
CV
Rk
k 1
R
CV 
k 1
Gravimetric And Volumetric Analysis
Gravimetric analysis gives the mass fractions of the components in the mixture.
Volumetric analysis gives the volumetric or molal fractions of the components in the mixture.
CP 

CONVERSION
xi 
yi M i
y M
i
yi 

i
yi M i
M
xi
Mi
xi
Mi

25
THE FIRST LAW OF THERMODYNAMICS
(The Law of Conservation of Energy)
“Energy can neither be created nor destroyed but can only be converted from one form to another.”
The Verbal form of the Law is:
Energy Entering - Energy Leaving = Change of Energy Stored within the system
In equation Form:
E1  E 2  ΔEs
Corollary Laws of the First Law:

First Corollary: Is the application of the Law of Conservation of Energy principle to a Closed System. A system of Fixed
mass.
W
U
Q
For a Closed system (Nonflow system) the KE and PE are negligible and (PV) doesn't exists.
Q - W = Es
where
Es = U
Q - W = U  1
Q = U + W  2
By differentiation:
dQ = dU + dW  3
Work for a Moving Boundary of a Closed System
Piston

2
W  P  dV  4
1
dW  P  dV  5
System
dQ  dU  dW
Boundary
dQ  dU  dW
P
dQ  dU  P  dV  6
Area = W
2
W   P  dV
P
1
dV
V
26

Second Corollary: Is the application of the Law of Conservation of Energy principle to an Open System (Steady-state,
Steady Flow System) a system of Fixed space.
For an OPEN SYSTEM (Steady-State, Steady-Flow system)
ES = 0, therefore
Energy Entering = Energy Leaving,
In equation form
E1  E 2  0
E1  E 2
U1  P1V1  KE1  PE1  Q  U 2  P2 V2  KE2  PE 2  W  1
Q  U 2  U1  P2 V2  U1  KE2  KE1  PE 2  PE1  W  2
Q  ΔU  Δ(PV)  ΔKE  ΔPE  W  3
Enthalpy (h):It is the sum of internal energy and flow energy.
h  U  PV  5
Δh  ΔU  Δ (PV)  6
Q  Δh  ΔKE  ΔPE  W  7
W  Q  Δh  ΔKE  ΔPE  8
By differenti ation
dh  dU  PdV  VdP  9
but
dU  PdV  dQ
dh  dQ  VdP
dW  dQ - dh - dKE - dPE  10
dW  dQ - dQ  VdP - dKE - dPE
dW   VdP - dKE - dPE  11

W  - VdP - ΔKE - ΔPE  12

- VdP  Q - Δh  13
If KE = 0 and PE = 0 or negligible;

W   VdP  14
W  Q  Δh  15
27
PROCESSES OF FLUIDS
o
ISOBARIC PROCESS (P = C):
An Isobaric Process is an internally reversible Constant Pressure process.
CLOSED SYSTEM
For any substance
For Ideal Gas
Q  ΔU  W  1
V1 V2

1
T1 T2

W  P  dV
W  P(V2  V1 )  mR (T2  T1 )  5
At P  C
ΔU  mC v (T2  T1 )  3
W  P(V2 - V1 )  2
Q  Δh  mC p (T2  T1 )  6
ΔU  m(U 2 - U1 )  3
ENTROPY CHANGE
from
For any substance
h  U  PV
ΔS 
dh  dU  PdV  VdP
 T  T  S  S  7
dQ
dh
2
dP  0 at P  C
For Ideal Gas
dU  PdV  dQ
dQ  dh  mC p dT
dQ  dh
ΔS 
Q  Δh
1
2 dT
 T  T  mC  T
dQ
dh
p
1
T
ΔS  mC p ln 2  8
T1
Q  m ( h 2  h1 )  4
OPEN SYSTEM
Q  Δh  ΔKE  ΔPE  W  9

W   VdP  ΔKE  ΔPE  10

dP  0 at P  C and Q  Δh;  V  dP  0
W  -ΔKE - ΔPE  11
If ΔKE  0 and ΔPE  0
W  0  12
28
o
ISOMETRIC PROCESS OR ISOCHORIC PROCESS(V = C):
An Isometric Process is an internally reversible “Constant Volume” process.
CLOSED SYSTEM
For any substance
Q  ΔU  W  1

W  P  dV
At V  C
For Ideal Gas
P1 P2

4
T1 T2
Q  ΔU  mCv (T2  T1 )  5
dV  0
ENTROPY CHANGE
W0 2
ΔS 
dQ  dU  PdV
dV  0
dQ  dU
 T  mC  T
dQ
ΔS  mC v ln
dT
v
T2
6
T1
Q  ΔU
Q  m(U 2 - U1 )  3
OPEN SYSTEM
Q  Δh  ΔKE  ΔPE  W  7

W   VdP  ΔKE  ΔPE  8

 V  dP  V(P1  P2 )  9
W  -ΔKE - ΔPE  11
If ΔKE  0 and ΔPE  0
W  0  12
29
o
ISOTHERMAL PROCESS (T = C or PV = C):
An Isothermal Process is an internally reversible “Constant Temperature” Process
CLOSED SYSTEM
For any substance
ENTROPY CHANGE
Q  ΔU  W  1
For any substance

W  P  dV  2
ΔS  S2 - S1  8
ΔU  m(U 2 - U1 )  3
dQ  Tds
From
For Ideal Gas
At T  C
C
PV  C or P 
V
P1V1  P2 V2  C
Q  T ΔS
Q
9
T
For ideal gas
ΔS 
ΔU  mC v (T2  T1 )
ΔU  0, therefor e
But T1  T2
W  Q  10
ΔU  0  4

V
W  P1V1 ln
V2
5
V1
W  PdV  C
W  mRT1 ln
dV
V2
6
V1
V2 P1

V1 P2
W  mRT1 ln
OPEN SYSTEM
P1
7
P2
Q  Δh  ΔKE  ΔPE  W  1

W  - VdP - ΔKE - ΔPE  2
Δh  m(h 2  h1 )  3
For ideal gas
and applying laws of logarithm
 mRT ln
6

P
P
V
 VdP  mRT ln
7

V
 VdP  P1V1 ln
P1
P1
1
2
2
2
1
1
Δh  mC p (T2 - T1 )
If ΔKE  0 and ΔPE  0
but T1  T2

W   VdP   P1V1 ln
Δh  0  4
P2
8
P1
From
PV  C or V 
C
P
P1V1  P2 V2  C

P
P
 VdP   P V ln
5

P
 VdP  C
dP
1 1
2
1
30
o
ISENTROPIC PROCESS (S = C):
An Isentropic Process is an internally “reversible adiabatic” process in which the entropy remains constant where S = C
(for any substance) or PVk = C (for an ideal or perfect gas)
dQ  dU  PdV
dQ  0, for adiabatic
dU  PdV  1
dh  dQ  VdP
dQ  0
dh  VdP  2
dh
VdP

3
dU  PdV
but
Cp
Cv
2 dP
2 dV
1
1
 P  k  V
V 
P
V
V
ln 2  k ln 2  k ln 1  ln  1 
P1
V1
V2
 V2 
k
taking antilog
k

dh
 k , hence
dU
P2  V1 
Vk
  1
 
P1  V2 
V2 k
or P1V1k  P2 V2 k  1
By int egration
U sin g
PV
PV k  C and
C
T
PV P V
P1V1k  P2 V2 k and 1 1  2 2
T1
T2
k 1
V 
T2  P2  k
 
  1 
T1  P1 
 V2 
k 1
2
C
and V 
Vk
1
Ck
1
Pk
P2 V2  P1V1
2
 PdV  1  k  4
k P V  P V 
 VdP 
5

1 k
 VdP  k PdV  6


1
2
2 2
1
2
2
1
1
1 1
P1V1  mRT1  7
From PV k  C
P
VdP
PdV
dP
dV
 k
P
V
by integratio n
k
From
P2 V2  mRT 2  8
3
31
CLOSED SYSTEM
For any substance
Q  ΔU  W
Q  0 1
W  ΔU  2
For Ideal Gas
W  -ΔU  -mC v (T2 - T1 )  3

W  PdV 
P2 V2  P1V1 mR (T2  T1 )

4
1 k
1 k
k 1
T2  P2  k
 
T1  P1 
From
k 1
k 1




mR (T2  T1 ) mRT1  P2  k
P1V1  P2  k


 
 
W  PdV 

 1 
 1  5




1 k
1  k  P1 
1  k  P1 








ENTROPY CHANGE

ΔS  0
OPEN SYSTEM
Q  Δh  ΔKE  ΔPE  W

W   VdP  ΔKE  ΔPE
Q  0 1
W  Δh  ΔKE  ΔPE  2

 VdP  Δh  3
For Ideal Gas
Δh  mC p (T2 - T1 )  4
If ΔKE  0 and ΔPE  0

W  - VdP  -Δh


 VdP  k PdV
k 1
k 1




k (P2 V2  P1V1 ) kmR (T2  T1 ) kmRT1  P2  k
kP1V1  P2  k






 VdP 



1


1
 1  k  P 
 5
1 k
1 k
1  k  P1 
1











o
POLYTROPIC PROCESS (PVn = C):
A Polytropic Process is an internally reversible process of an ideal or perfect gas in which PVn = C, where n stands for
any constants.
U sin g
By int egration
PV
C
T
PV P V
P1V1n  P2 V2 n and 1 1  2 2  1
T1
T2
 PdV  1  n  4
n P V  P V 
 VdP 
5

1 n
 VdP  n PdV  6


PV n  C and
n 1
T2  P2  n  V1 
 
  
T1  P1 
 V2 
n 1
2
C
Vn
and V 
1
2
2 2
1
2
2
1
1
1 1
P1V1  mRT1  7
P2 V2  mRT 2  8
From PV n  C
P
P2 V2  P1V1
2
1
Cn
1
Pn
3
32
From
Q  ΔU  W  9

W  Pdυ 
P2 υ 2  P1υ1 R (T2  T1 )

1 n
1 n
RdT
 10
1 n
dQ  dU  dW
dW 
dQ  C v dT 
 k 1
1 n  k 1
dQ  C v dT1 
  C v dT

 1 n 
 1 n

kn
kn
dQ  C v dT
  Cv 
dT
 1 n 
 1 n 
kn
let : C n  C v 

 1 n 
dQ  C n dT
RdT
1 n
Q  C n (T2  T1 )  10
R  Cp  Cv
Considerin g m
Q  mC n (T2 - T1 )  11
C p  kC v
kn
Cn  Cv 
  Polytropic specific heat
 1 n 
kC v dT  C v dT
1 n
 k 1
dQ  CvdT  CvdT

1 n 
dQ  C v dT 
CLOSED SYSTEM
Q  ΔU  W
Q  ΔU  12
Q  mC n (T2  T1 )  13
ΔU  -mC v (T2 - T1 )  14

W  PdV 
P2 V2  P1V1 mR (T2  T1 )

 15
1 n
1 n
n 1
From
T2  P2  n
 
T1  P1 
n 1
n 1




mR (T2  T1 ) mRT1  P2  n
 P1V1  P2  n





1


1
 1  n  P 
  16
1 n
1  n  P1 
1










ENTROPY CHANGE

W  PdV 
ΔS 
 T  mC  T
dQ
ΔS  mC n ln
dT
n
T2 KJ
 17
T1 K
33
OPEN SYSTEM
Q  Δh  ΔKE  ΔPE  W  18

W   VdP  ΔKE  ΔPE  19
W  Q  Δh  ΔKE  ΔPE  20
h  U  PV
Δh  ΔU  Δ(PV)
Δ(PV)  Δh  ΔU
Δh  mC p (T2 - T1 )  21
Q  mC n (T2  T1 )  22
n 1
n 1




n (P2 V2  P1V1 ) nmR (T2  T1 ) nmRT1  P2  n
nP1V1  P2  n






 VdP 



1


1
 1  n  P 

1 n
1 n
1  n  P1 
1










If ΔKE  0 and ΔPE  0


W  - VdP
o
ISOENTHALPIC PROCESS or THROTTLING PROCESS(h = C):
An Isoenthalpic Process is a steady state, steady flow, process in which W = 0, KE = 0, PE = 0, and Q = 0, where the
enthalpy h remains constant.
h1  h 2
or
hC
IRREVERSIBLE OR PADDLE WORK
W
U
WP
Q
m
Q  ΔU  W  WP
Where:
WP – irreversible or Paddle Work
34
SAMPLE EXERCISES
Processes (Ideal Gas)
A steady flow compressor handles 113.3 m3/min of nitrogen ( M = 28; k = 1.399) measured at intake where P1= 97 KPa and T1=
27 C. Discharge is at 311 KPa. The changes in KE and PE are negligible. For each of the following cases, determine the final
temperature and the work if the process is:
a) PVk= C
b) PV = C
Given: V1 = 113.3/60 = 1.89 m3sec ; P1 = 97 KPa ; T1 = 27 + 273 = 300 K
P2 = 311 KPa
For Isentropic
k 1
P  k
T2  T1  2 
 418 K
 P1 
W
kP1V1  T2  
   1  - 253 KW
1  k  T1  
For Isothermal
T1 = T2 = 300K
P
W  P1V1 ln 1  - 213.6 KW
P2
Processes (Ideal Gas)
Air is contained in a cylinder fitted with a frictionless piston. Initially the cylinder contains 500 L of air at 150 KPa and 20  C. The
air is then compressed in a polytropic process ( PVn = C) until the final pressure is 600 KPa, at which point the temperature is
120 C. Determine the work W and the heat transfer Q. (R = 0.287 KJ/kg-K ; k = 1.4)
Given: V1 = 0.500 m3 ; P1 = 150 KPa; T1 = 293K
P2 = 600 KPa; T2 = 393K
T
ln 2
n 1
T1

P
n
ln 2
P1
n  1.27
m
P1 V1
 0.892 kg
RT1
W
P1V1  T2  
   1  95 KJ
1  n  T1  
∆U = mCv(T2 – T1) = 64 KJ
Q = ∆U + W = 64 – 95 = - 31 KJ
Ideal Gas (Open System; Compressor)
An air compressor receives 20 m3/min of air at 101.325 KPa and 20C and compresses it to 10 000 KPa in an isentropic process.
Calculate the power of the compressor.
For isentropic process; ΔKE  0 and ΔPE  0
k 1


kP1V1  P2  k



W

1

1 - k  P1 




Ideal Gas (Closed System)
A closed gaseous system undergoes a reversible process in which 40 KJ of heat are rejected and the volume changes 0.15 m 3 to
0.60 m3. The pressure is constant at 200 KPa. Determine the change of internal energy U in KJ.
Given:
Q = -40 KJ;
V1 = 0.15 m3
V2 = 0.60 m3; P = 200 KPa
Solution:
Q = U + W ; W = P(V2 – V1)
W = 200(0.60 – 0.15) = 90 KJ
U = -40 – 90 = -130 KJ
35
Ideal Gas (Isentropic Process)
Air in a piston - cylinder occupies 0.12 m3 at 550 KPa. The air expands isentropically doing work on the piston until the volume is
0.25 m3. Determine the work W in KJ. For Air R = 0.287 KJ/kg-K and k = 1.4.
Given: V1 = 0.12 m3; P1 = 550 KPa; V2 = 0.25 m3
k 1

P V  V 
W  I 1  1   1  42 KJ
(1  k )  V2 



Ideal Gas (Paddle Work)
A piston cylinder arrangement contains 0.02 m3/min of air at 50 C and 400 KPa. Heat is added in the amount of 50 KJ and work
is done by a paddle wheel until the temperature reaches 700 C. If the pressure is held constant, how much paddle wheel work
must be added to the air. (R = 0.287 KJ/kg-K ; k = 1.4)
m3
min
T1  50  273  323 K; T2  993 K
V1  0.02
P  400 KPa
Q  50 KJ
Process : Isobaric
PV  mRT
m  0.086 kg
V2  0.06 m 3
W  P(V2  V1 )  16.6 KJ
Q  ΔU  W - WP
WP  8.1 KJ
Ideal Gas (Polytropic Process)
Air at 0.07 m3 and 4000 KPa is expanded in an engine cylinder and the pressure at the end of expansion is 320 KPa. If the
expansion is polytropic with PVn = C where n = 1.35, find the final volume.
Given: V1 = 0.07 m3 ; P1 = 4000 KPa ; P2 = 320 KPa ; PV1.35 = C
P1V1n  P2 V2 n
1
 P n
V2  V1  1   0.455
 P2 
Ideal Gas
Gas at a pressure of 100 KPa, volume 0.20 m3 and temperature 300K, is compressed until the pressure is 320 KPa and the
volume is 0.09 cu.m.. Calculate the final temperature in K.
Given: P1 = 100 KPa; V1 = 0.20 m3; T1 = 300K; P2 = 320 KPa; V2 = 0.09 m3
P1V1 P2 V2

T1
T2
T 2  432K
Ideal Gas
Helium gas ( R=2.077 KJ/kg-K; k= 1.667) at 800 KPa and 300K occupies a volume of 0.30 m3. Determine the mass of helium
in kg.
PV  mRT
m  0.39kg
Ideal Gas (Isometric Process)
A rigid container containing 25 kg of nitrogen gas at 298K receives heat and its temperature is increased to 370K. Determine
the amount of heat added in KJ.
For N2: R = 0.297 KJ/kgK; k = 1.399
Cv = 0.7444KJ/kg-K
Q = mCvT = 1334 KJ
36
Ideal Gas Mixture
A 3 m3 drum contains a mixture at 101 KPa and 35C of 60% Methane (CH4) and 40% oxygen (O2) on a volumetric basis.
Determine the amount of oxygen that must be added at 35C to change the volumetric analysis to 50% of each component.
Determine also the new mixture pressure.
For CH 4: M = 16; k = 1.321
O2: R = 32 ; k = 1.395
GIVEN :
V  3 m 3 ; P  101 KPa ; T  308K ; y CH 4  60% ; y O 2  40%
M  0.60(16)  0.40(32)  22.4
8.3143
KJ
 0.371
22.4
kg - K
PV  mRT
R
101(3)
0.371(308)
m  2.65 kg
m
yi M i
M
0.60(16)
x CH 4 
 0.429
22.4
0.40(32)
xO2 
 0.571
22.4
m CH 4
x CH 4 
2.65
m CH 4  1.137 kg
xi 
m O 2  2.65 - 1.137  1.513 kg
a)
at y CH 4  0.50 ; y O 2  0.50
M  0.50(16)  0.50(32)  24
8.3143
KJ
 0.346
24
kg - K
0.50(16)
x CH 4 
 0.33
24
x O 2  0.67
R
x CH 4 
0.33 
1.137  m CH 4
2.65  m CH 4
1.137  m CH 4
2.65  m CH 4
m CH $  - 0.39 kg ; therefore some CH 4 is removed from the mixture because it is negative.
m  2.65 - .39  2.26 kg
b) y CH 4  0.50 ; y O 2  0.50
M  0.50(16)  0.50(32)  24
8.3143
KJ
 0.346
24
kg - K
0.50(16)
x CH 4 
 0.33
24
x O 2  0.67
R
0.67 
1.513  m O 2
2.65  m O 2
m O 2  0.796
m  2.65  0.796  3.446 kg
c) for condition a;
mRT
V
2.26(0.346)(308)
P
3
P  80.28 KPa
P
For condition b :
3.446(0.346)(308)
3
P  122.41 KPa
P
37
Ideal Gas (Closed System)
Air from the discharge of a compressor enters a 1 m3 storage tank. The initial air pressure in the tank is 500 KPa and the
temperature is 600K. The tank cools, and the internal energy decreases 213 KJ/kg. Determine
a) the work done
b) the heat loss
c) the change of enthalpy
d) the final temperature
For Air
k = 1.4
R = 0.287
V  1 m3
P  500 KPa
T  600 K
KJ
kg
W  0  Cons tan t Volume
ΔU  -213
PV  mRT
m  2.9 kg
Q  mΔU  2.9(-213)  618.5 KJ (Heat is rejected)
Δh
ΔU
Δh  1.4(-618.5)  866 KJ
k
Δh  mC p (T2  T1 )
T2  303 K
Pure Substance (Isometric Process)
A 2 kg steam-water mixture at 1000 KPa and 90% quality (U = 2401.41 KJ/kg) is contained in a rigid tank. Heat is added until the
pressure rises to 3500 KPa and the temperature is 400C (U = 2926.4 KJ/kg). Determine the heat added in KJ.
Q = m(U2 – U1) = 1045 KJ
Ideal Gas
Calculate the pressure of 2 moles of air at 400K, with a total volume of 0.5 m3.
PV  mRT  nMRT  n RT
P  13.3MPa
Ideal Gas
A 100 Liters oxygen tank use in a hospital has a pressure of 1 atmosphere and a temperature of 16C. For a molecular weight of
32 kg/kgmol, determine the mass of oxygen in kilograms.
PV = mRT
T = 289K ; P = 101.325 KPa
m = 0.135 kg
Ideal Gas
An unknown gas has a mass of 1.5 kg and occupies 2.5 m3 while at a temperature of 300K and a pressure of 200 KPa. Determine
the gas constant for the gas in KJ/kg-K.
PV
R
mT = 1.111 KJ/kg-K
Open System (Polytropic Process)
A gas turbine expands 50 kg/sec of helium (M = 4; k = 1.666) polytropically, PV 1.8 = C, from 1000K and 500 KPa to 350K.
Determine;
a. The final pressure in KPa
b. The power produced in KW
c. The heat loss in KW
d. The entropy change in KW/K
38
Given :
m  50 kg/sec
R
8.3143
KJ
 2.08
M
kg - K
Pr ocess : PV1.8  C (Polytropi c)
System : Gas Turbine (Open)
T1  1000 K; P1  500 KPa
T2  350 K
n 1
T2  P2  n
 
T1  P1 
n
P2  T2  n 1
 
P1  T1 
P2  627.14 KPa
nmR (T2  T1 )
 151,995.80 KW
1 n
Q  mC n (T2  T1 )  -16,989.85KW
W  Q  Δh 
kn
Cn  Cv 
  0.523
 1 n 
Δh  mC p (T2  T1 )  -168,985.65 kw
Open System
At one point in a pipeline the water speed is 3 m/sec and the gage pressure is 50 KPa. Find the gage pressure at a second point in
the line, 11 m lower than the first , if the pipe diameter at the second point is one half the first.
Ideal Gas (Open system; Nozzle)
Oxygen expands in a reversible adiabatic manner through a nozzle from an initial pressure and initial temperature and with an
initial velocity of 50 m/sec. There is a decrease of 38K in temperature across the nozzle. Determine
a. the exit velocity
b. for inlet conditions of 410 KPa and 320 K, find the exit pressure.
39
Q  Δh  ΔKE  ΔPE  W
Q  0; W  0; ΔPE  0
ΔKE  Δh  C p (T2  T1 )
v 2 2  v12
 C p (T1  T2 )
2(1000)
v 2  2(1000)C p (T1  T2 )  v12
Rk
KJ
8.3143
KJ
 0.918
;R 
 0.26
; k  1.395
k 1
kg - K
32
kg - K
m
v 2  268.8
sec
Cp 
k 1
T2  P2  k
 
T1  P1 
k
k
 T  k 1
P2  T2  k 1
   ; P2  P1  2 
P1  T1 
 T1 
(T1  T2 )  38
(320  T2 )  38
T2  282 K
P2  262.14
Ideal Gas
A pressure in the cylinder in the figure varies in the following manner with volume, P = C/V2. If the initial pressure is 500 KPa,
initial volume is 0.05 m3 and the final pressure is 200 KPa, find the work done by the system.
P1  500 KPa; V1  0.05 m3 ; P2  200 KPa
PV 2  C
P1V12  P2 V2 2  C
1
 P 2
V2   1  V1  0.08 m3
 P2 
Hydrostatic Pressure
A storage tank contains oil with a specific gravity of 0.88 and depth of 20 m. What is the hydrostatic pressure at the bottom of the
tank in kg/cm2.
1.033 kg 2
cm  1.76 kg
PBottom  0  0.88(9.81)20  172.656 KPa x
cm2
101.325 KPa
Variation of Pressure
A hiker carrying a barometer that measures 101.3 KPa at the base of the mountain. The barometer reads 85 KPa at the top of the
mountain. The average air density is 1.21 kg/m3. Determine the height of the mountain.
ΔP   γΔh
γ
ρg
1000
P2  P1  
ρg
h 2  h1 
1000
h  h 2  h1 
h
1000P1  P2 
 1373.2 m
ρg
40
Closed System
During the execution of a non-flow process by a system, the work per degree temperature increase is dW/dT = 170.9 J/K and the
internal energy may be expressed as U = 27 + 0.68T Joules, a function of the temperature T in Kelvin. Determine the heat in KJ if
the temperature changes from 10C to 38C.
dW  170.9dT
U  27  0.68T
dU  0.68dT
dQ  dU  dW
dQ  0.68dT  170.9dT
dQ  171.58dT
1

Q  171.58 dT
1
Q  171.58(T 2  T1)  171.58(38  10)  4804.24 Joules  48.0424 KJ
Pure Substance (Nozzle)
A nozzle receives 10 kg/sec of steam at 4 MPa and 260C (h = 2836.3 KJ/kg ;  = 51.74 x10-3 m3/kg) and discharges it at 1.4
MPa (h =2634.07 KJ/kg;  = 129.72 x 10-3 m3/kg). If the velocity at inlet is negligible, find the diameter at exit in cm.
Q  Δh  ΔKE  ΔPE  W
Q  0; W  0; ΔPE  0
ΔKE  Δh
v 2 2  v12
  Δh
2(1000)
v 2  2(1000)( Δh )  v12
v1  0
v 2  2(1000)( Δh )
Δh  2634.07  2836.3  -202.23
v 2  636
m
sec
m
A2v2
kg
 10
υ2
sec
m
πd 2 v 2 ρ 2 πd 2 v 2

4υ2
4
d
4m
x 100 cm  5.10 cm
πρ 2 v 2
Properties of fluids
A 3 m diameter by 4.5 m height vertical tank is receiving water ( = 978 kg/m3) at the rate of 1.13 m3/min and is discharging
through a 150 mm  with a constant velocity of 1.5 m/sec. At a given instant, the tank is half full. Find the water level and the
mass change in the tank 15 minutes later.
m1  1.13(978)15  16,577.1 kg  entering
π (0.150) 2 (1.5)(60)(15)(978)
 23331.63 kg  leaving
4
π(3) 2 (4.5)(978)
mi 
 15,554.42 kg  initial mass of water
4( 2)
Δm  m1  m i  m 2  8,799.9 kg
m2 
Volume of water in the tank 
Height of water 
8,799.9
 9 .0 m 3
978
9 .0
 1.27 m
π 2
(3)
4
41
Zeroth Law
An engineering student wants to cool 0.25 kg of Omni Cola (mostly water) initially at 20C by adding ice that is initially at -20C.
How much ice should be added so that the final temperature will be 0C with all the ice melted, if the heat capacity of the
container neglected.
QCola  Qice
0.25(4.187)( 20  0)  mice 2.0935(0  20)  334.9
mice  0.06 kg
Zeroth Law
2.5 kg of brass of specific heat 0.39 KJ/kg-C at a temperature of 176C is dropped into a 1.2 liters of water at 14C. Find the
resulting temperature of the mixture. (At 14C density of water is 999 kg/m3)
Q Brass  Q water
2.5(0.39)(176  t ) 
1.2
999(4.187)t  14
1000
t  40.35C
Open system (Turbine)
Steam enters a turbine with a velocity of 1.5 m/sec and an enthalpy of 2093 KJ/kg and leaves with an enthalpy of 1977 KJ/kg and
a velocity of 91.5 m/sec. Heat losses are 8 KCal/min and the steam flow rate is 27 kg/min. The inlet of the turbine is 3.5 m higher
than its outlet. What is the work output of the turbine if the mechanical losses is 15%.
(1 KCal = 4.187 KJ)
Q  Δh  ΔKE  ΔPE  W
 8(4.187)
Q
 0.56 KW  Rejected
60
27
1977  2093  -52.2 KW
Δh 
60
27  (91.5)2 - (1.5)2 
ΔKE 

  1.9 KW
60 
2(1000)

27(9.81)
 3.5  0  15.45 KW
ΔPE 
60
W  0.56  52.2 - 1.9  15.45  65.2 KW
Woutput  (65.2)(1 - .15)  55.41 KW (work done by the system)
Open System (Compressor)
An air compressor handles 8.5 m3/min of air with a density of 1.26 kg/m3 and a pressure of 101 KPaa and discharges at 546 KPaa
with a density of 4.86 kg/m3. The changes in specific internal energy across the compressor is 82 KJ/kg and the heat loss by
cooling is 24 KJ/kg. Neglecting changes in kinetic and potential energies, find the work in KW.
System : Compressor (0pen system)
8.5
1.26  0.18 kg
m
60
sec
Q  Δh  ΔKE  ΔPE  W
Δh  Δ(Pυ)  ΔU
KJ
ΔU  82
kg
ΔPE  0
 P  546 101
KJ
Δ(Pυ)  Δ  

 32.19
kg
 ρ  4.86 1.26
KJ
Δh  32.19  82  114.19
kg
KJ
Q  24
(heat loss)
kg
KJ
W  Q - Δh  ΔKE  ΔPE  -138.2
kg
42
Closed system (Piston in cylinder)
Four kilograms of a certain gas is contained within a piston–cylinder assembly. The gas undergoes a process for which the
pressure - volume relationship is PV1.5 = C. The initial pressure is 300 KPa, the initial volume is 0.10 m 3, and the final volume is
0.2 m3. The change in specific internal energy of the gas in the process is U = - 4.6 kJ/kg. There are no significant changes in
kinetic or potential energy. Determine the net heat transfer for the process, in kJ.
System : Piston - Cylinder (Closed)
KJ
P1  300 KPa; V1  0.10 m 3 ; V2  0.20 m3 ; ΔU  4.6
kg
Q  ΔU  W
P V -PV
W  PdV  2 2 1 1
1 n

P2 V2  P1V1n
n
P2  106.07 KPa
W  17.57 KJ
ΔU  4.6(4)  18.4 KJ
Q  35.97 KJ
Entropy Change
Calculate the change of entropy per kg of air (R = 0.287 KJ/kg-K; k = 1.4) when heated from 300K to 600K while the pressure
drops from 400 KPa to 300 KPa.
Rk
KJ
 1.0045
k 1
kg - K
KJ
ΔS  0.78
kg - K
CP 
Law of Conservation of Mass
Steam at 1 MPa, 300°C flows through a 30 cm diameter pipe with an average velocity of 10 m/s. Determine mass flow rate of this
steam. (Density at 1 MPa and 300C = 3.875 kg/m3)
m  ρAv
kg
π
m  3.875 0.302 10  2.74
sec
4
Closed System (Piston in Cylinder)
Air is compressed in a piston-cylinder device. Using constant specific heats and treating the process as internally reversible,
determine the amount of work required to compress this air from 100 KPa, 27°C to 2000 KPa, 706°C.
Ideal Gas
A certain perfect gas of mass 0.1 kg occupies a volume of 0.03 m 3 at a pressure of 700 KPa and a temperature of 131C. The gas
is allowed to expand until the pressure is 100 KPa and the final volume is 0.02 m3. Calculate;
a) the molecular weight of the gas
b) the final temperature
PV  mRT
700(0.03)  0.1R (131  273)
KJ
R  0.52
kg - K
8.3143
kg
M
 16
0.52
kg mol
100(0.02)  0.1(0.52)T
T  38.5 K
43
System : Piston - Cylinder (closed)
Process : PV n  C
Q  ΔU  W

W  PdV 
P2 V2 - P1V1 R (T2  T1 )

1 n
1 n
P1V1n  P2 V2 n
n
 V2 
P

  1
V
P
2
 1
n 1
T2  P2  n
 
T1  P1 
U sin g Laws of logarithm
P1
P2
n
V
ln 2
V1
ln
n 1

n
ln
T2
T1
P 
ln  2 
 P1 
n  1.652
KJ
W  266.8
kg
Open System (Nozzle)
Air enters the after burner nozzle of a jet fighter at 427°C with a velocity of 100 m/s. It leaves this adiabatic nozzle at 377°C.
Assuming that the air specific heats do not change with temperature, determine the velocity at the nozzle exit.
Q  Δh  ΔKE  ΔPE  W
Q  0; W  0; ΔPE  0
ΔKE  Δh  C p (T2  T1 )
v 2 2  v12
 C p (T1  T2 )
2(1000)
v 2  2(1000)C p (T1  T2 )  v12
C p  1.0045
T1  427C
T2  377C
v1  100
m
sec
v 2  332.3
m
sec
Gas Mixture
One mole of a gaseous mixture has the following gravimetric analysis: O2 = 16% ; CO2 = 44% ; N2 = 40%. Find
a) the molecular weight of the mixture
b) the mass of each constituents
c) the moles of each constituent in the mixture
44
Formulas
xi
Mi
xi
Σ
Mi
xi 
mi
m
M  ΣyiMi
M
m
n
yi 
8.3143
M
ni
yi 
n
R
Open System (Nozzle)
Air enter the nozzle as shown at a pressure of 2700 KPa at a velocity of 30 m/sec and with an enthalpy of 923 KJ/kg, and leaves
with a pressure of 700 KPa and enthalpy of 660 KJ/kg. If the heat loss is 0.96 KJ/kg, find the exit velocity in m/sec if the mass
flow rate is 0.2 kg/sec.
System : Nozzle (Open)
Q  Δh  ΔKE  ΔPE  W
KJ
Q  -0.96
kg
W  0; ΔPE  0
ΔKE  Q - Δh
v 2 2  v12
 Q - Δh
2000
v 2  2000Q - Δh   v12
Δh  (660 - 923)  -263
v1  30
KJ
kg
m
sec
v 2  724.6
m
sec
Force
10 liters of an incompressible liquid exert a force of 20 N at the earth’s surface. What force would 2.3 Liters of this liquid exert on
the surface of the moon? The gravitational acceleration on the surface of the moon is 1.67 m/sec 2.
V  0.01 m 3
m
ρ  ; m  Vρ
V
m  0.01ρ
F  ma
20  0.01ρ(9.81)
kg
ρ  203.9 3
m
FMoon  0.0023(203.9)1.67  0.80 N
45
Manometers
A closed tank contains compressed air and oil (S = 0.90) as shown in the figure.a U – tube manometer using mercury (S = 13.6) is
connected to the tank as shown. For column heights h1 = 90 cm; h2 = 15 cm and h3 = 22 cm, determine the pressure reading of
the gage.
ΔP  γΔh
0  13 .6(9.81)(0.22 )  0.90 (9.81)(0.15 )  0.90 (9.81)(0.90 )  Pgage
Pgage  20.08 KPa
Manometer
In the figure pipe A contains carbon tetrachloride (S = 1.60) and the closed storage tank B contains a salt brine
(S =1.15). Determine the air pressure in tank B in KPa if the gage pressure in pipe A is 1.75 kg/cm 2.
(101.325)
 171.65 KPa
1.033
3 ft  0.914 m
PA  1.75
171.65 - (0.914)(1.6)(9.81)  (0.914)(1.15)(9.81) - 1.2(1.15)(9.81)  PB
PB  154.08 KPa gage
Ideal Gas (Isothermal Process)
Air which is initially at 120 KPa and 320K occupies 0.11 m3. It is compressed isothermally until the volume is halved and then
compressed it at constant pressure until the volume decreases to ¼ of the initial volume. Sketch the process on the PV and TS
diagrams. Then determine the pressure, the volume and temperature in each state.
46
Pr ocess Isothermal : P1V1  P2 V2
Fluid : Air (R  0.287 KJ/kg - K) k  1.4
P1  120 KPa; T1  320 K; V1  0.11 m 3
PV  mRT
120(0.11)
 0.144 kg
0.287(320)
At 1 to 2 (Isorherma l)
m
0.11
 0.055 m 3
2
PV
P2  1 1  240 KPa;T2  320 K
V2
V2 
At 2 to 3 (Isobaric)
P 2  P3  240 KPa
0.11
 0.0275 m 3
4
T3 V3

T2 V2
V3 
T3  160 K
Gas Mixture
One mole of a gaseous mixture has the following gravimetric analysis; O2 = 16% ; CO2 = 44%; N2 = 40%.Find
a) the molecular weight of the mixture
b) the mass of each constituent
c) the moles of each constituent in the mixture
d) the gas constant R
e) the partial pressure for P = 207 KPa
47
PRACTICE PROBLEMS
1. Determine the molecular weight of a gas if its specific heats at constant pressure and volume are Cp = 2.286 kJ/kg K and Cv =
1.768 kJ/kg K.
Cp  Cv  R
R  Cp  Cv
8.3143
R
M
8.3143
 Cp  Cv
M
8.3143
kg
M
 16.05
Cp  Cv
kgm
2. A perfect gas at pressure of 750 kPa and 600 K is expanded to 2 bar pressure. Determine final temperature of gas if initial and
final volume of gas are 0.2 m3 and 0.5 m3 respectively.
P1  750 KPa
T1  600 K
V1  0.2 m 3
P2  2 Bar  200 KPa
V2  0.5 m 3
T2 
P1V1 P2 V2

T1
T2
T2  400 K
3. A vessel of 5 m3 capacity contains air at 100 kPa and temperature of 300K. Some air is removed from vessel so as to reduce
pressure and temperature to 50 kPa and 7ºC respectively. Find the amount of air removed and volume of this mass of air at initial
states of air. Take R = 287 J/kg.K for air.(Answer: 2.696 kg; 2.32 m3)
4. A cylindrical vessel of 1 m diameter and 4 m length has hydrogen gas at pressure of 100 kPa and 27ºC. Determine the amount of
heat to be supplied so as to increase gas pressure to 125 kPa. For hydrogen take Cp = 14.307 kJ/kg.K, Cv = 10.183 kJ/kg K.
(Answer: Q = 194 KJ)
5. A spherical balloon of 5 m diameter is filled with Hydrogen at 27ºC and atmospheric pressure of 1.013 bar. It is supposed to lift
some load if the surrounding air is at 17ºC. Estimate the maximum load that can be lifted. (Answer: m = 74.344 kg)
6. A pump draws air from large air vessel of 20 m 3 at the rate of 0.25 m3 /min. If air is initially at atmospheric pressure and
temperature inside receiver remains constant then determine time required to reduce the receiver pressure to 1/4 of its original
value. (t = 110.9 minutes)
48
in this case T is constant and V is constant, and mass and pressure changes with time t
PV  mRT
 dP 
 dm 
V   RT 

dt
 
 dt 
dm
 is the mass change wrt time t
dt
dm
Pν

(- sign is used because mass decreases with time
dt
RT
 dP 
 Pν 
V   RT  

dt
 
 RT 
 dP 
V     Pν
 dt 
dt ( ν )
dP

V
P
V  dP 
dt    
ν P 
by integratio n
t-
V  P2 
 ln 
ν  P1 
P1
4
P2 1

P1 4
P2 
t
20  1 
 ln   110.9 minutes
0.25  4 
7. A vessel of volume 0.2 m3 contains nitrogen at 101.3 KPa and 15ºC. If 0.2 kg of nitrogen is now pumped
into the vessel,
calculate the new pressure when the vessel has returned to its initial temperature. For nitrogen: M = 28 and k = 1.399. (187
KPa)
8. A certain perfect gas of mass 0.01 kg occupies a volume of 0.003 m3 at a pressure of 700 KPa and a temperature of 131ºC. The
gas is allowed to expand until the pressure is 100 KPa and the final volume is 0.02 m3. Calculate:
a) the molecular weight of the gas (16)
b) the final temperature (111.5ºC)
9. A perfect gas has a molecular weight of 26 kg/kgmol and a value of k = 1.26. Calculate the heat rejected
a) when 1 kg of the gas in contained in a rigid vessel at 300 KPa and 315ºC, and is then cooled until the pressure falls to
150 KPa. (-361 KJ)
b) when 1 kg/sec mass flow rate of the gas enter a pipeline at 280ºC and flows steadily to the end of the pipe where the
temperature is 20ºC. Neglect changes in kinetic and otential energies.(-403 KW)
10. Calculate the internal energy and enthalpy of 1 kg of air occupying 0.05 m3 2000 KPa. If the internal
120 KJ as the air is compressed to 5000 KPa, calculate the new volume occupied by 1 kg of the air.
For air: R = 0.287 KJ/kg-ºK and k = 1.4.( 250.1 KJ/kg; 350.1 KJ/kg; 0.0296 m 3)
energy is increased by
11. When a certain perfect gas is heated at constant pressure from 15ºC to 95ºC, the heat required is 1136 KJ/kg. When the same
gas is heated at constant volume between the same temperatures the heat required is 808 KJ/kg. Calculate Cp, Cv, k, and M
of the gas. (14.2 KJ/kg; 10.1 KJ/kg; 1.405; 4.1 and 2.208)
12. A quantity of a certain perfect gas is compressed from an initial state of 0.085 m3, 100 KPa to a final state of 0.034 m3, 390
KPa. the Cv = 0.724 KJ/kg-ºC and Cp = 1.020 KJ/kg-ºC. The observed temperature rise is 146ºK. Calculate R, the mass
present, and U of the gas.(0.296 KJ/kg-K; 0.11 kg; 11.63 KJ)
13. A mass of 0.05 kg of air is heated at constant pressure of 200 KPa until the volume occupied is 0.0658 m3. Calculate the heat
supplied, the work and the change in entropy for the process if the initial temperature is 130ºC. (Q = 25.83 KJ; W = 7.38 KJ)
14. A 1 kg of nitrogen is compressed reversibly and isothermally from 101 KPa, 20ºC to 420 KPa. Calculate
and the heat flow during the process assuming nitrogen to be a perfect gas. ( Q = W = 124 kJ/KG)
the nonflow work
49
15. Air at 102 KPa, 22ºC, initially occupying a cylinder volume of 0.015 m3 is compressed isentropically by a piston to a pressure
of 680 KPa. Calculate the final temperature, the final volume, the work done on the mass of air in the cylinder.
(234.3 ºC; .00387 m3; 2.76 KJ)
16. 1 kg of air is compressed from 110 KPa, 27 ºC in a polytropic process where n = 1.3 until the final pressure is 660 KPa.
Calculate:
a) ∫PdV
b) - ∫VdP
c) S
17. There are 1.36 kg of air at 138 KPa stirred with internal paddles in an insulated rigid container, whose volume is 0.142 m3 until
the pressure becomes 689.5 KPa. Determine the work input and PV. ( 196.2 KJ; 78.3 KJ)
18. During an isentropic process of 1.36 kg/sec of air, the temperature increases from 4.44ºC to 115.6 ºC. for a nonflow process
and for a steady flow process (KE = 0 and PE = 0) Find:
a) U in KW
b) H in KW
c) W in KW
d) S in KW/ºK
e) Q in KW
19. A certain perfect gas is compressed reversibly from 100 KPa, 17 ºC to a pressure of 500 KPa in a perfectly thermally insulated
cylinder, the final temperature being 77 ºC. The work done on the gas during the compression is 45 KJ/kg. Calculate, k , Cv,
R and M of the gas.( 1.132; 0.75 KJ/kg-ºK; 0.099 KJ/kg-ºK; 84)
20. 1 kg of air at 102 KPa, 20 ºC is compressed reversibly according to a law PV1.3 = C to a pressure of 550 KPa. Calculate the work
done on the air and the heat supplied during the compression. (133.46 KJ/kg; -33.3 KJ/kg)
21. Oxygen (M = 32) is compressed polytropically in a cylinder from 105 KPa, 15ºC to 420 KPa in such a way that one third of the
work input is rejected as heat to the cylinder walls. Calculate the final temperature of the oxygen. Assume oxygen to be perfect
gas and take Cv = 0.649 KJ/kg-K. (113 ºC)
22. Air at 690 KPa, 260ºC is throttled to 550 KPa before expanding through the nozzle to a pressure of 110 KPa. Assuming that the
air flows reversibly in steady flow through the nozzle and that no heat is rejected, calculate the velocity of the air at exit from
the nozzle when the inlet velocity is 100 m/sec. ( 636 m/sec)
23. Air at 40ºC enters a mixing chamber at a rate of 225 kg/sec where it mixes with air at 15ºC entering at a rate of 540 kg/sec.
Calculate The temperature of the air leaving the chamber, assuming steady flow conditions. Assume that the heat loss is
negligible. (22.4ºC)
24. A gaseous mixture composed of 25 kg of N2, 3.6 kg of H2, and 60 kg of CO2 is at 200 KPa, 50C. Find the respective partial
pressures and compute the volume of each component at its own partial pressure and 50C.
Given: mN2 = 25 kg ; mH2 = 3.6 kg ; mCO2 = 60 kg
m = 25 + 3.6 + 60 = 88.6 kg
P = 200 KPa ; T = 323 K
25. Assume 2 kg of O2 are mixed with 3 kg of an unknown gas. The resulting mixture occupies a volume of 1.2 m3 at 276 KPa and
65C. Determine
a) R and M of the unknown gas constituent
b) the volumetric analysis
c) the partial pressures
Given; mO2 = 2 kg; mx = 3 kg
V = 1.2 m3 ; P = 276 KPa; T = 338 K
a)
m = 5 kg
xO2 = 0.40 ; xx = 0.60
PV  mRT
R = 0.1361 KJ/kg-K
R = .40(0.26) + 0.60(Rx)
Rx = 0.535 KJ/kg-K
Mx = 15.54 kg/kgm
x
.40
.60
 i 

 0.0511
b)
Mi 32 15.54
yO2 = 0.245 ; yx = 0.755
c) PO2 = .245(276) = 67.62 KPa ;
50
Px = 0.755(276) = 208.38 KPa
26. Assume 28 m3 of a gaseous mixture whose gravimetric analysis is 20% CO2, 15% O2, 65% N2, are at 103.4 KPa and 150C.
Find
a) the volumetric analysis
b) the respective partial pressures
c) R and M
d) the moles of mixture and of each constituent
e) the heat transferred with no change in pressure to reduce the temperature to 75C
f) the volume the mixture occupies after the cooling
27. Consider 2 kg of CO and 1 kg of CH4 at 32C that are in a 0.6 m3 rigid drum. Find:
a) the mixture pressure P in KPa
the volumetric analysis
c) the partial pressures in KPa
d) the heat to cause a temperature rise of 50C.
28. A gaseous mixture has the following volumetric analysisO2, 30%; CO2, 40% N2, 30%. Determine
a) the analysis on a mass basis
b) the partial pressure of each component if the total pressure is 100 KPa and the temperature is 32C
c) the molecular weight and gas constant of the mixture
29. A gaseous mixture has the following analysis on a mass basis, CO2, 30%; SO2, 30%; He, 20% and N2, 20%. For a total pressure
and temperature of 101 KPa and 300 K, Determine
a) the volumetric or molal analysis
b) the component partial pressure
c) the mixture gas constant
d) the mixture specific heats
30. A cubical tank 1 m on a side, contains a mixture of 1.8 kg of nitrogen and 2.8 kg of an unknown gas. The mixture pressure and
temperature are 290 KPa and 340 K. Determine
a) Molecular weight and gas constant of the unknown gas
b) the volumetric analysis
31. A mixture of ideal gases at 30C and 200 KPa is composed of 0.20 kg CO2, 0.75 kg N2, and 0.05 kg He. Determine the mixture
volume.10. An ideal gas with R = 2.077 KJ/kg-K and a constant k= 1.659 undergoes a constant pressure process during which
527.5 KJ are added to 2.27 kg of the gas. The initial temperature is 38C. Find the S in KJ/K.
a) 2.687
b) 1.586
c) 8.64
d) 3.861
32.Calculate the volume occupied by 13.6 kg of chlorine at a pressure of 73 mm Hg and a temperature of 21C. molecular
weight of chlorine may be taken as 71. (4.7 m 3)
33.A 3.8 m3 tank filled with acetylene at 103.4 KPa is supercharged with 1.95 kg of the gas. What would be the final
pressure, assuming no change in temperature (16C) that takes place during the charging period. (155 KPa)
34. A volume of gas having initial entropy of 5317.2 KJ/K is heated at constant temperature of 540C until the entropy
is 8165.7 KJ/K. How much heat is added and how much work is done during the process.
35. An ideal gas with R = 2.077 KJ/kg-K and a constant k= 1.659 undergoes a constant pressure process during which
527.5 KJ are added to 2.27 kg of the gas. The initial temperature is 38C. Find the S in KJ/K.
a) 2.687
b) 1.586
c) 8.64
d) 3.861
36. Calculate the volume occupied by 13.6 kg of chlorine at a pressure of 73 mm Hg and a temperature of 21C. molecular
weight of chlorine may be taken as 71. (4.7 m 3)
37. A volume of gas having initial entropy of 5317.2 KJ/K is heated at constant temperature of 540C until the entropy
is 8165.7 KJ/K. How much heat is added and how much work is done during the process.
51
SAMPLE PROBLEMS
Problem No. 1
Consider the design of a nozzle in which nitrogen gas flowing in a pipe at 500 KPa, 200C, and a velocity of 10 m/sec, is to be
expand to produce a velocity of 300 m/sec. Determine the exit pressure and cross-sectional area of the nozzle if the mass flow rate
is 0.15 kg/sec, and the expansion is reversible and adiabatic. For N2: R = 0.2968 KJ/kg-K ; k = 1.399; Cp = 1.04 KJ/kg-K
Given
P1  500 KPa
T1  200  273  473 K
m
sec
m
v 2  300
sec
kg
m  0.15
sec
v1  10
Pr ocess : PV k  C
R  0.2968
KJ
kg - K
k  1.399
Problem No. 2
A counterflowing heat exchanger is used to cool air at 540 K, 400 KPa to 360 K by using 0.05 kg/sec supply of water at 20C, 200
KPa. The airflow is 0.5 kg/sec in a 10 cm diameter pipe. Find the inlet velocity and the water exit temperature.
Given
Ta  540 K
Pa  400 KPa
kg
ma  o.5
sec
kg
mw  0.05
sec
tw  20C
Pw  200 KPa
d  0.10 m (pipe diameter)
Problem No. 3
Consider that 1 kg of air has a decrease in internal energy of 22 KJ while its centigrade temperature is reduced to one-third of the
initial temperature during a reversible nonflow constant pressure process. Determine
a) the initial temperature inC
b) the final temperature inC
c) the heat in KJ
d) the work in KJ
e) the entropy change in KJ/K
For air: R = 0.287 KJ/kg-K; k = 1.4
m  1kg
ΔU  22 KJ
t
t2  1
3
t

  2 t1 
Q  ΔU  22  1(0.7175) 1  t1   0.7175

3

 3 
3(22)
t1 
 46C
 2(0.7175)
T1 319 K
t 2  15.33C
T2  288.33 K
Q  22KJ
ΔS  mC v ln
T2
KJ
 -0.0726
T1
K
52
Problem No. 4
Nitrogen gas enters a turbine operating at steady state through a 5 cm duct with a velocity of 60 m/sec, a pressure of 345 KPa and a
temperature of 550 K. At the exit, the velocity is 0.6 m/sec, the pressure is 138 KPa and the temperature is 389 K. Heat transfer
from the surface is 37.2 KJ/kg of nitrogen flowing. Neglecting PE effects and using the ideal gas model, determine the power
developed by the turbine in KW. (For N2, M = 28 kg/kg mol ; k = 1.399)
Problem No. 5
Air (R = 0.287 KJ/kg-K; k =1.4) is contained in a cylinder fitted with a frictionless piston. Initially the cylinder contains 500 liters
of air at 150 KPa and 20C. The air is then compressed in a polytropic process until the final pressure is 600 KPa and the temperature
is 120C. Calculate
a. The polytropic exponent n
b. The final volume V 2 in li
c. The work W
d. The heat transfer Q
Problem No. 6
Consider 2 kg of CO and 1 kg of CH4 at 32C that are in a 0.6 m3 rigid drum. Find:
a. the mixture pressure P in KPa
b. the volumetric analysis
c. the partial pressures in KPa
d. the heat to cause a temperature rise of 50C.
Given :
m CO  2 kg
m CH4  1 kg
T  32  273  305 K
V  0.6 m3
53
Problem No. 7
A gaseous mixture has the following volumetric analysisO2, 30%; CO2, 40% N2, 30%. Determine
a. the analysis on a mass basis
b. the partial pressure of each component if the total pressure is 100 KPa and the temperature is 32C
c. the molecular weight and gas constant of the mixture
Given
Volumetric Analysis
y O2  30%
y CO2  40%
y N2  30%
Problem No. 8
A gaseous mixture has the following analysis on a mass basis, CO2, 30%; SO2, 30%; He, 20% and N2, 20%. For a total pressure and
temperature of 101 KPa and 300 K, Determine
a. the volumetric or molal analysis
b. the component partial pressure
c. the mixture gas constant
d. the mixture specific heats
Given
Gravimetri c Analysis
y CO2  30%
ySO2  30%
yHe  20%
yN2  20%
P  101 KPa
T  300 K
Problem No. 9
A cubical tank 1 m on a side, contains a mixture of 1.8 kg of nitrogen and 2.8 kg of an unknown gas. The mixture pressure and
temperature are 290 KPa and 340 K. Determine
a. Molecular weight and gas constant of the unknown gas
b. the volumetric analysis
Problem No. 10
A mixture of ideal gases at 30C and 200 KPa is composed of 0.20 kg CO2, 0.75 kg N2, and 0.05 kg He. Determine the mixture
volume.
Given
T  30  273  303 K
m CO2  20 kg
m N2  0.75 kg
m He  0.05 kg
Problem No. 11
A gaseous mixture composed of 25 kg of N2, 3.6 kg of H2, and 60 kg of CO2 is at 200 KPa, 50C. Find the
respective partial
pressures and compute the volume of each component at its own partial pressure and
50C.
Given: mN2 = 25 kg ; mH2 = 3.6
kg ; mCO2 = 60 kg
m = 25 + 3.6 + 60 = 88.6 kg
xN2 = 0.282 ; xH2 = 0.041 ; xCO2 = 0.678
P = 200 KPa ; T = 323 K
xi
Mi
Pi
yi 
xi yi 

P
Mi
xi 0.282 .041 .678




 0.046
Mi
28
2
44
y N 2  0.219
y H 2  0.446
y CO 2  0.335
Pi Vi  mi R i Ti
25(0.297(323)
 54.76 m3
43.8
3.6(4.16)(323)
VH 2 
 54.23 m3
89.2
60(0.189)(323)
VCO 2 
 54.67 m3
67
VN 2 
54
PN2 = .219(200) = 43.8 KPa
PH2 = .446(200) = 89.2 KPa
PCO2 = 0.335(200) = 67 KPa
Problem 12
Assume 2 kg of O2 are mixed with 3 kg of an unknown gas. The resulting mixture occupies a volume of 1.2 m3 at 276 KPa and
65C. Determine
a) R and M of the unknown gas constituent
b) the volumetric analysis
c) the partial pressures
Given; mO2 = 2 kg; mx = 3 kg
V = 1.2 m3 ; P = 276 KPa; T = 338 K
a) m = 5 kg
xO2 = 0.40 ; xx = 0.60
PV  mRT
R = 0.1361 KJ/kg-K
R = .40(0.26) + 0.60(Rx)
Rx = 0.535 KJ/kg-K
Mx = 15.54 kg/kgm
x
.40
.60
 i 

 0.0511
b)
M i 32 15.54
yO2 = 0.245 ; yx = 0.755
c) PO2 = .245(276) = 67.62 KPa ;
Px = 0.755(276) = 208.38 KPa
Problem 13
A piston cylinder contains air at 600 KPa, 290 K and a volume of 0.01 m 3. A constant pressure process gives 54 KJ of work out.
Determine the heat transfer of the process.
Given:
W  P(V2  V1 )  mR (T2  T1 )
P1 = P2 = 600 KPa
P
T1 = 290 K
1V1  mRT1
3
PV
V1 = 0.01 m
m  1 1  0.07 kg
RT1
W = 54 KJ (work out)
W
 T1  478.4 K
mR
Q  mC p (T2 - T1 )
T2 
Rk
KJ
 1.0045
k -1
kg - K
Q  13.24 KJ
Cp 
Problem 14
A piston cylinder device, whose piston is resting on a set stops, initially contains 3 kg of air at 200 KPa and 27C. The mass of the
piston is such that a pressure of 400 KPa is required to move it. Heat is now transferred to the air until its volume doubles. Determine
the work done by the air and the total heat transferred to the air during this process. Also show the process on a P-V diagram.
Given:
m = 3 kg
P
P1 = 200 KPa
T1 = 27 + 273 = 300 K
3
P = 400 KPa
V2 = 2V1
2
1
Q
V
55
At V  C
V1  V2
P2 T2

P1 T1
T2  600 K
Q v  3(0.7175)(600 - 300)  645.75 KJ
At P  C
V3 T3

V2 T2
V3  2V1  2.6 m 3
T
2.6
 3
1.3 600
T3  1200 K
Q p  mC p (T3 - T2 )  3(1.0045)(1200 - 600)
Q p  1808.1 KJ
W  P(V3 - V2 )  400(2.6 - 1.3)
W  520 KJ
Q T  645.75  1808.1  2453.85 KJ
Problem No. 15
During some actual expansion and compression processes in piston cylinder devices, the gases have been observed to satisfy the
relationship PVn = C, where n and C are constants. Calculate the work done when a gas expands from a state of 150 KPa and 0.03
m3 to a final volume of 0.2 m3 for the case of n = 1.3. Also show the process on the PV diagram.
Given:
P1 = 150 KPa ; V1= 0.03 m3
V2 = 0.2 m3
PV n  C
P1V1n  P2 V2 n  C
For a Closed system

W  P  dV
P
C
Vn
By integratio n

2
W  PdV 
1
P2 V2  P1V1
1 n
n
V 
 0.03 
 P1  1   150

V
V2
 0.2 
 2
P2  12.74 KPa
P2 
P1V1n
1.3
n
P2 V2  P1V1 12.74(0.2)  150(0.03)

1 n
1  1.3
W  6.533 KJ

2
W  PdV 
1
Problem No. 16
1. A. How many kilograms of nitrogen must be mixed with 3.6 kg of CO2 in order to produce a gaseous mixture that is 50% by
volume of each constituent?
B. For the resulting mixture, determine M and R, and the partial pressure of the N2 if that of the CO2 is 138 KPa.
mCO2 = 3.6 kg ; yCO2 = 0.50 ; yN2 = 0.50
56
y CO 2 M CO

y N2M N 2
2 ; x
;M 
yi M i
N2 
M
M
M  0.5(44)  0.50(28)  22  14  36M  0.5(44)  0.50(28)  22  14  36
x CO 2 
22
14
 0.61; x N2 
 0.39
36
36
m CO 2
x CO 2 
m CO 2  m N 2
x CO 2 
0.61 
3.6
3 .6  m N 2
m N2 
3.6  2.196
 2.3 kg
0.61
M  36
8.3143
 0.231 KJ/kg - K
36
Pi
yi 
P
138
y CO 2 
 0.50
P
P  276 KPa
R
PN2  0.50(276)  138 KPa
Problem No. 17
Assume 2 kg of O2 are mixed with 3 kg of an unknown gas. The resulting mixture occupies a volume of 1.2 m3 at 276 KPa and
65C. Determine
a) R and M of the unknown gas constituent
b) the volumetric analysis
c) the partial pressures
Given; mO2 = 2 kg; mx = 3 kg
V = 1.2 m3 ; P = 276 KPa; T = 338 K
a)
m = 5 kg
xO2 = 0.40 ; xx = 0.60
PV  mRT
R = 0.1361 KJ/kg-K
R = .40(0.26) + 0.60(Rx)
Rx = 0.535 KJ/kg-K
Mx = 15.54 kg/kgm

b)
xi
.40
.60


 0.0511
M i 32 15.54
yO2 = 0.245 ; yx = 0.755
c) PO2 = .245(276) = 67.62 KPa ; Px = 0.755(276) = 208.38 KPa
Problem No. 18
One mole of a gaseous mixture has the following gravimetric analysis: O2 = 16%, CO2 = 44%, N2 = 40%. Find :
a) the molecular weight of the mixture
b) the mass of each constituent,
c) the moles of each constituent in the mixture,
d) R and
e) Partial pressure for P = 207 KPa.
Given: xO2 = 0.16 ; xCO2 = .44 ; xN2 = 0.40
57
xi
Mi
yi 
xi

Mi
xi
0.16 .44 .40




 0.0293
Mi
32
44 28
y O 2  0.171
y CO 2  0.341
y N 2  0.488
M = .171(32) + .341(44) + .488(28) = 34.14 kg/kgm
n
 ni
ni
n
nO2 = .171 moles; nCO2 = .341; nN2 = .488
yi 
M
m
n
mO2 = .171(32) = 5.472 kg
mCO2 =.341(44) = 15.004 kg
mN2 = .488(28) = 13.664 kg
m = 5.472 + 15.004 + 13.664 = 34.14 kg
8.3143
M = 0.244 KJ/kg-K
Pi
yi 
P
R
PO2= .171(207) = 35.4 KPa
PCO2 = .341(207) = 70.6 KPa
PN2 = .488(207) = 101.016 KPa
Problem No. 19
A gaseous mixture composed of 25 kg of N2, 3.6 kg of H2, and 60 kg of CO2 is at 200 KPa, 50C. Find the respective partial
pressures and compute the volume of each component at its own partial pressure and
50C.
Given: mN2 = 25 kg ; mH2 = 3.6 kg ; mCO2 = 60 kg
m = 25 + 3.6 + 60 = 88.6 kg
xN2 = 0.282 ; xH2 = 0.041 ; xCO2 = 0.678
P = 200 KPa ; T = 323 K
xi
yi  Mi
xi

Mi
xi 0.282 .041 .678




 0.046
Mi
28
2
44
y N 2  0.219
y H 2  0.446
y CO 2  0.335
yi 
Pi
P
PN2 = .219(200) = 43.8 KPa
PH2 = .446(200) = 89.2 KPa
PCO2 = 0.335(200) = 67 KPa
58
Pi Vi  m i R i Ti
25(0.297 (323)
 54.76 m 3
43.8
3.6(4.16)(323)
VH 2 
 54.23 m 3
89.2
60(0.189 )(323)
VCO 2 
 54.67 m 3
67
VN 2 
Problem No. 20
A 0.23 m3 drum contains a gaseous mixture of CO2 and CH4 each 50% by mass at P = 689 KPa, 38C; 1 kg of O2 are added to the
drum with the mixture temperature remaining at 38C. For the final mixture, find;
a) the gravimetric analysis
b) the volumetric analysis
c) the Cp (For CO2: k = 1.288; CH4: k = 1.321; O2: k = 1.395)
d) the total pressure P
Problem No. 21
Assume 28 m3 of a gaseous mixture whose gravimetric analysis is 20% CO2, 15% O2, 65% N2, are at 103.4 KPa and 150C. Find
a) the volumetric analysis
b) the respective partial pressures
c) R and M
d) the moles of mixture and of each constituent
e) the heat transferred with no change in pressure to reduce the
temperature to 75C
f) the volume the mixture occupies after the cooling
Problem 22
Consider 2 kg of CO and 1 kg of CH4 at 32C that are in a 0.6 m3 rigid drum. Find:
a) the mixture pressure P in KPa
b) the volumetric analysis
c) the partial pressures in KPa
d) the heat to cause a temperature rise of 50C.
Problem 23
A gaseous mixture has the following volumetric analysisO2, 30%; CO2, 40% N2, 30%. Determine
a) the analysis on a mass basis
b) the partial pressure of each component if the total pressure is 100 KPa and the temperature is 32C
c) the molecular weight and gas constant of the mixture
Problem 24
A gaseous mixture has the following analysis on a mass basis, CO2, 30%; SO2, 30%; He, 20% and N2, 20%.
For a total pressure and temperature of 101 KPa and 300 K, Determine
a) the volumetric or molal analysis
b) the component partial pressure
c) the mixture gas constant
d) the mixture specific heats
Problem 25
A cubical tank 1 m on a side, contains a mixture of 1.8 kg of nitrogen and 2.8 kg of an unknown gas. The
mixture pressure and temperature are 290 KPa and 340 K. Determine
a) Molecular weight and gas constant of the unknown gas
b) the volumetric analysis
Problem 26
A mixture of ideal gases at 30C and 200 KPa is composed of 0.20 kg CO2, 0.75 kg N2, and 0.05 kg He.
Determine the mixture volume.
59
FUELS AND COMBUSTION
Fuel - a substance composed of chemical elements, which in rapid chemical union with oxygen produced combustion.
Combustion - is that rapid chemical union with oxygen of an element, whose exothermic heat of reaction is sufficiently great and
whose rate of reaction is sufficiently fast, whereby useful quantities of heat are liberated at elevated temperatures. It is the burning
or oxidation of the combustible elements.
TYPES OF FUEL
1) Solid Fuels
Example: a. coal
b. charcoal
c. coke
d. woods
2) Liquid Fuels (obtained by the distillation of petroleum)
Example: a. Gasoline
b. kerosene
c. diesoline
d. Fuel oil
e. alcohol (these are not true hydrocarbons, since it contains oxygen in the molecule)
3) Gaseous Fuels (a mixture of various constituents hydrocarbons, its combustion products do not have sulfur
components)
Example:
a. Natural Gas (example: methane, ethane, propane)
b. Coke oven gas -obtained as a byproduct of making coke
c. Blast furnace gas - a byproduct of melting iron ore
d. LPG
e. Producer Gas - fuel used for gas engines
4) Nuclear Fuels
Example:
a. Uranium
b. Plutonium
COMBUSTIBLE ELEMENTS
1. Carbon (C)
2. Hydrogen (H2)
3. Sulfur (S)
TYPES OF HYDROCARBONS
1) Paraffin - all ends in "ane"
Formula: CnH2n+2
Structure: Chain (saturated)
Example:
GAS
a. Methane(CH4)
b. Ethane (C2H6)
LPG
a. Propane (C3H8)
b. Butane (C4H10)
c. Pentane ( C5H12)
GASOLINE
a. n-Heptane (C7H16)
b. Triptane (C7H16)
c. Iso- octane (C8H18)
FUEL OIL
a. Decane (C10H22)
b. Dodecane (C12H26)
c. Hexadecane (C16H34)
d. Octadecane (C18H38)
60
2) Olefins - ends in "ylene" or "ene"
Formula: CnH2n
Structure: Chain (unsaturated)
Example:
a. Propene (C3H6)
b. Butene (C4H8)
c. Hexene ( C6H12)
d. Octene ( C8H16)
3) DIOLEFIN - ends in "diene"
Formula: CnH2n-2
Structure: Chain (unsaturated)
Example:
a. Butadiene (C4H6)
b. Hexadiene (C6H10)
4) NAPHTHENE - named by adding the prefix "cyclo"
Formula: CnH2n
Structure: Ring (saturated)
Example:
a. Cyclopentane (C5H10)
b. Cyclohexane (C6H12)
5) AROMATICS - this hydrocarbon includes the;
A. Benzene Series (CnH2n-6)
B. Naphthalene Series (CnH2n-12)
Structure: Ring (unsaturated)
Example:
a. Benzene (C6H6)
b. Toluene (C7H8)
c. Xylene (C8H10)
6) ALCOHOLS - These are not true hydrocarbon, but sometimes used as fuel in an internal combustion engine. The characteristic
feature is that one of the hydrogen atom is replaced by an OH radical.
Example:
a. Methanol (CH4O or CH3OH)
b. Ethanol (C2H6O or C2H5OH)
Saturated Hydrocarbon - all the carbon atoms are joined by a single bond.
Unsaturated Hydrocarbon - it has two or more carbon atoms joined by a double or triple bond.
Isomers - two hydrocarbons with the same number of carbon and hydrogen atoms, but at different structure.
61
STRUCTURE OF CnHm
Chain Structure (saturated)
Chain Structure (unsaturated)
c. Ring Structure (saturated)
62
THE COMBUSTION CHEMISTRY
A. Oxidation of Carbon
C + O2  CO2
Mole Basis:
1 + 11
Mass Basis:
1(12) + 1(32)  1(12 + 32)
12 + 32  44
3 + 8  11
B. Oxidation of Hydrogen
H2 + ½O2  H2O
Mole Basis:
1 + ½ 1
Mass Basis:
1(2) + ½(32)  1(2 + 16)
2 + 16  18
1 + 8 9
C. Oxidation of Sulfur
S + O2  SO2
Mole Basis:
1 + 11
Mass Basis:
1(32) + 1(32)  1(32+32)
32 + 32  64
1 + 12
Complete Combustion: Occurs when all the combustible elements has been fully oxidized.
Example: C + O2  CO2
Incomplete Combustion: Occurs when some of the combustible elements have not been fully oxidized and it may result from;
a. Insufficient oxygen
b. Poor mixing of fuel and oxygen
c. the temperature is too low to support combustion.
Result: Soot or black smoke that sometimes pours out from chimney or smokestack.
Example: C + ½O2  CO
Composition of Air
a) Volumetric or Molal analysis
O2 = 21%
N2 = 79%
b) Gravimetric Analysis
O2 = 23.3%
N2 = 76.7%
moles of N2 79

 3.76
mole of O2
21
63
COMBUSTION WITH AIR
A) Combustion of Carbon with air
C + O2 + (3.76)N2  CO2 + (3.76)N2
Mole Basis
1 + 1 + 3.76  1 + 3.76
Mass Basis
1(12) + 1(32) + (3.76)(28)  1(44) + (3.76)(28)
12 + 32 + 3.76(28)  44 + 3.76(28)
3 + 8 + 3.76(7)  11 + 3.76(7)
B) Combustion of Hydrogen with Air
H2 + ½O2 +½3.76)N2  H2O + ½(3.76)N2
Mole Basis:
1 + ½ + ½(3.76)  1 + ½(3.76)
Mass Basis:
1(2) + ½(32) + ½(3.76)(28)  1(18) + ½(3.76)(28)
2 + 16 + 3.76(14)  18 + 3.76(14)
C) Combustion of Sulfur with air
S + O2 + 3.76N2  SO2 + 3.76N2
Mole Basis:
1 + 1 + 3.76  1 + 3.76
Mass Basis:
1(32) + 1(32) + (3.76)(28)  1(64) + (3.76)(28)
8 + 8 + (3.76)(7)  16 + 3.76(7)
THEORETICAL AIR
It is the minimum amount of air required to oxidized the reactants. With theoretical air no O2 is found in the products.
EXCESS AIR
It is an amount of air in excess of the theoretical air required to influence complete combustion. With excess air, O2 is present in the
products. Excess air is usually expressed as a percentage of the theoretical air. But in actual combustion, although there is an
amount of excess air, the presence of CO and other hydrocarbon in the products cannot be avoided.
Example: 25% excess air is the same as 125% theoretical air.
64
COMBUSTION OF HYDROCARBON FUEL (CnHm)
Combustion of CnHm with 100% theoretical air
CnHm + aO2+ a(3.76)N2  bCO2 + cH2O + a(3.76)N2
kg of air
 A  a (32)  a (3.76)(28)
  
F
12
n

m
kg
of CnHm
 t
In theoret ical combustion , for the combustion of CnHm,
a  n  0.25m
bn
c  0.5m
Combustion with excess air
CnHm + (1+e)aO2 + (1+e)a(3.76)N2  bCO2 + cH2O +dO2 +(1+e)a(3.76)N2
A
A
   (1  e) 
F
 a
 F t
a (32)  a (3.76)(28)
kg of air
A
   (1  e)
F
12
n

m
kg
of CnHm
 a
d  e(n  0.25m)
where:
e - excess air in decimal
(A/F)t - theoretical air-fuel ratio
(A/F)a - actual air-fuel ratio
H

100  4.76  11.28  CO2
C

e
H

4.76  14.28 C  CO2


H

11.44(1  e) 1  3 
C  kg o air
A

  
H
F
kg of Fuel


 a
1  C 


where:
H and C are in % (by mass)
CO2 is in % by volume from the exhaust gas analysis
Actual or Real world Combustion Process
Combustion with CO in the products due to incomplete combustion (100% theoretical air)
CnHm  aO2  a (3.76 ) N 2  bCO 2  cH2O  dCO  a (3.76 ) N 2
Combustion with CO in the products due to incomplete combustion (with excess air)
CnHm  (1  e)aO2  (1  e)a (3.76 ) N 2  bCO 2  cH 2O  dCO  fO 2  (1  e)a (3.76 ) N 2
65
Typical Real-World Engine Combustion Process:
Fuel (CnHm)  Air (O 2 and N 2 )  CO2  H 2O  O 2  N 2  CH(VOC's)  CO  NOx
CH(VOC's) - Volatile Organic Compounds
CO - Carbon Monoxide
NOx - Nitrogen Oxides
Note: Values of a, b, c, and d in terms of n and m in theoretical combustion, cannot be applied in actual in an actual combustion
process.
EQUIVALENCE RATIO
ER 
(F A) actual
(F A)Stoichiometric
Stoichiometric (or chemically correct) mixture of air and fuel, is one that contains just sufficient oxygen for complete combustion
of the fuel
A “Weak Mixture”, is one which has an excess of air
A “Rich Mixture”, is one which has a deficiency of air
Percentage Excess Air 
A F
 F
 A
A F
Actual
T heoretical
T heoretical
Common Combustion Gases
Gases
Molecular
Weight
C
12
H
1
H2
2
O
16
O2
32
N
14
N2
28
S
32
COMBUSTION OF SOLID FUELS
Components of Solid Fuels: C, H2, O2, N2, S, and Moisture
A) Combustion with 100% theoretical air
aC + bH2 + cO2 + dN2 + eS + fH2O + gO2 + g(3.76)N2  hCO2 + iH2O + jSO2 + kN2
B) Combustion with excess air x
Where:
x - excess air in decimal
aC + bH2 + cO2 + dN2 + eS +fH2O + (1+x)gO2 + (1+x)g(3.76)N2  hCO2 + iH2O + jSO2 + lO2 + mN2
The theoretical and actual air-fuel ratio of solid fuels can be computed based on their balance combustion equation above.
32g  3.76(28)g
A
  
 F  t 12a  2b  32c  28d  32e  18f
A
A
   (1  x ) 
F
 a
 F t
66
DEW POINT TEMPERATURE
The Dew Point Temperature (tdp) is the saturation temperature corresponding the partial pressure of the water vapor in the mixture
(products of combustion).
ULTIMATE ANALYSIS
Ultimate Analysis gives the amount of C, H2, O2, N2, S and moisture in percentages by mass, sometimes the percentage amount of
Ash is given.
100% = %C + %H2 + %O2 + %N2 + %S + %H2O + %Ash
O 
kg of air

A
   11 .44 C  34 .32  H  2   4.29S
8 
kg of Fuel
 F t

where: C, H, O and S are in decimals obtained from the Ultimate Analysis
PROXIMATE ANALYSIS
Proximate Analysis gives the percentage amount of Fixed Carbon, Volatiles, Ash and Moisture.
100% = % Fixed Carbon + % Volatiles + % Ash + % Moisture
ORSAT ANALYSIS
Orsat Analysis gives the volumetric or molal analysis of the products of combustion or exhaust gases on a Dry Basis.
ORSAT ANALYSIS
Orsat Analysis gives the volumetric or molal analysis of the products of combustion or exhaust gases on a Dry Basis.
100% = % CO2 + % CO + % O2 + % SO2 + % N2 + % CH + % NO + NOx
H

100  4.76  11.28  CO2
C

e
H

4.76  14.28 C  CO2


A
  
 F a
H

11.44(1  e) 1  3 
C  kg o air

kg of Fuel
 H
1  C 


where:
H and C are in % (by mass)
CO2 is in % by volume from the exhaust gas analysis
MASS FLOW RATE OF FLUE GAS
a) Without considering Ash loss:
A
m g  m F 
F

 1

b) Considering Ash loss
A
m g  m F 
F

 1- Ash loss 

where ash loss in decimal
67
Properties of Fuels and Lubricants
a) Viscosity - a measure of the resistance to flow that a lubricant offers when it is subjected to shear stress.
b) Absolute Viscosity - viscosity which is determined by direct measurement of shear resistance.
c) Kinematics Viscosity - the ratio of the absolute viscosity to the density
d) Viscosity Index - the rate at which viscosity changes with temperature.
e) Flash Point - the temperature at which the vapor above a volatile liquid forms a combustible mixture with air.
f) Fire Point - The temperature at which oil gives off vapor that burns continuously when ignited.
g) Pour Point - the temperature at which oil will no longer pour freely.
h) Dropping Point - the temperature at which grease melts.
i) Condradson Number(carbon residue) - the percentage amount by mass of the carbonaceous residue remaining after destructive
distillation.
j) Octane Number - a number that provides a measure of the ability of a fuel to resist knocking when it is burnt in a gasoline
engine. It is the percentage by volume of iso-octane in a blend with normal heptane that matches the knocking behavior of the fuel.
k) Cetane Number - a number that provides a measure of the ignition characteristics of a diesel fuel when it is burnt in a standard
diesel engine. It is the percentage of cetane in the standard fuel.
PROPERTIES OF COMMON GASES
GAS
M
R
k
Cp
Cv
k
O2
32
0.260
1.395
0.918
0.658
1.395
CO2
44
0.189
1.288
0.845
0.656
1.288
N2
28
0.297
1.399
1.041
0.744
1.399
C2H2
26
0.320
1.232
1.698
1.378
1.232
AIR
28.97
0.287
1.4
1.004
0.717
1.4
ARGON
39.95
0.208
1.666
0.521
0.312
1.666
CO
28
0.297
1.399
1.041
0.744
1.399
Cl2
70.914
0.117
1.324
0.479
0.362
1.324
C2H6
30
0.277
1.187
1.759
1.482
1.187
C2H4
28
0.297
1.24
1.534
1.237
1.24
He
4
2.079
1.666
5.200
3.121
1.666
H2
2
4.157
1.4
14.550
10.393
1.4
N2H4
32
0.260
1.195
1.592
1.332
1.195
CH4
16
0.520
1.321
2.138
1.619
1.321
Ne
20.183
0.412
1.666
1.030
0.619
1.666
C3H8
44
0.189
1.127
1.677
1.488
1.127
SO2
64
0.130
1.263
0.624
0.494
1.263
H2O
18
0.462
1.329
1.866
1.404
1.329
Xe
131.3
0.063
1.666
0.158
0.095
1.666
Problem No. 1
An automotive internal combustion engine burns liquid Octane (C3 H8 ) at the rate of 0.005 kg/sec, and uses 20% excess air. The
air and fuel enters the engine at 25C and the combustion products leaves the engine at 900 K. It may be assumed that 90% of the
carbon in the fuel burns to form CO2 and the remaining 10% burns to form CO. Determine
a. The actual air – fuel ratio
b. The kg/s of actual air
c. The M and R of the products
Combustion with 100% theoretical air
Combustion Equation w ith 20% excess air and 10% of C burns to CO
C3H8  5.82O 2  21.88N 2  2.7CO2  4H 2O  0.97O 2  0.3CO  21.88N 2
68
Problem No. 2
A hydrocarbon fuel represented by C12H26 is used as fuel in an IC engine and requires 25 % excess air for complete combustion.
Determine
a. The combustion equation
b. The theoretical air – fuel ratio
c. The actual air – fuel ratio
d. The volumetric and gravimetric analysis of the products
e. The molecular weight M and gas constant R of the products
f. The kg of CO2 formed per kg of fuel
g. % C and %H in the fuel
Solution
Fuel: C12H26
Combustion with 100% theoretical air
C12H 26  aO 2  a (3.76) N 2  bCO2  cH 2 O  a(3.76)N 2
Carbon balance
12  b
Hydrogen balance
26  2c
c  13
Oxygen balance
2a  2b  c
a  b
c
 18.5
2
Combustion with excess air e = 0.25
C12 H 26  (1.25)aO 2  (1.25)a (3.76) N 2  bCO2  cH 2 O  dO 2  (1.25)a(3.76)N 2
By oxygen balance
1.25(2)a  2b  c  2d
1.25(2)a - 2b - c
d
 4.625
2
C12 H 26  23.125O 2  86.95N 2  12CO2  13H 2 O  4.625O2  86.95N 2
Gases
Mi
ni
yi
mi
xi
yiMi
yi(%)
xi(%)
CO2
44
12
0.103
528
0.158
4.529
10.3
15.8
H2O
18
13
0.112
234
0.070
2.007
11.2
7.0
O2
32
4.625
0.040
148
0.044
1.270
4.0
4.4
N2
total
28
86.95
0.746
2434.6
0.728
20.884
74.6
72.8
116.575
1.00
3344.6
1.00
28.691
100.0
100.0
The combustion equation
C12H 26  23 .125 O 2  86 .95 N 2  12CO 2  13H 2 O  4.625O 2  86 .95 N 2
The theoretical A/F ratio
a (32 )  a (3.76 )(28)
kg of air
A
 14 .94
  
12 (12 )  26
kg of C12H 26
 F t
The actual A/F ratio
kg of air
A
A
   (1  e)   14 .94 (1.25)  18 .674
kg of C12H 26
 F a
 F t
69
The Volumetric and gravimetric Analysis
Gases
yi(%)
xi(%)
CO2
10.3
15.8
H2O
11.2
7.0
O2
4.0
4.4
N2
total
74.6
72.8
100.0
100.0
Molecular weight and Gas constant
kg
M  ΣyiMi  28.691
kg m
R
 ΣxiRi
M
8.3143
KJ
R
 0.2898
28.691
kg - K
R
Kg of CO2 per kg of fuel
kg of CO2
528

 3.106
kg of C12H 26 12(12)  26
% C and % H in the fuel
12n
x 100%
12n  m
12(12)
%C 
x 100  84.7%
12(12)  26
m
%H 
x 100%
12n  m
26
%H 
x 100  15.3%
12(12)  26
%C 
Problem No. 3
A fuel represented by C7H16 is oxidized with 20% excess air and the mass of fuel required for combustion is 50 kg/hr. Determine
the mass flow rate of the products in kg/hr.
A 137.28(1.20)(7  4)
kg of air

 20.592
F
(84  4)
kg fuel
m p  m F (20.592)  m F
m p  50(21.592)  1079.6 kg/hr
The mass analysis of hydrocarbon fuel A is 88.5% Carbon and 11.5% Hydrogen. Another hydrocarbon fuel B requires 6% more air
than fuel A for complete combustion. Calculate the mass analysis of Fuel B.
a) C = 84.8%; H = 15.2% c) C = 74.8%; H = 25.2%
b) C = 15.2% ; H = 84.8% d) C = 25.2% ; H = 74.8%
Solution:
Fuel A: C = 0.885 ; H = 0.115
Fuel B: C = ; H =
(A/F)B = 1.06(A/F)A
(A/F)A = 11.44(0.885) + 34.32(0.115) = 14.0712 kg/kg
(A/F)B = 1.06(14.0712)= 14.9155 kg/kg
For fuel B: H + C = 1
H = (1 – C)
14.9155 = 11.44C + 34.32(1-C)
34.32  14.9155
C
 0.8481
34.32  11.44
H  0.1519
70
C = 84.8%
H = 15.2%
The analysis of the natural gas showed the following percentages by volume: C2H6 = 9%; CH4 = 90%; CO2 = 0.2 % and N2 = 0.8
%. Find the volume of air required per cu,m. of gas if the gas and air are at temperature of 16C and a pressure of 101.6 KPa.
a) 10.07
b) 14.9
c) 17.1
d) 19.08
solution: basis 100 moles of fuel)
9C2H6 + 90CH4 + 0.2CO2 + 0.8N2 + aO2 + a(3.76)N2 
bCO2 + cH2O + dN2
By Carbon balance:
2(9) + 90 + 0.20 = b
b = 108.20
By Hydrogen Balance:
6(9) + 4(90) = 2c
c = 207
By O2 balance:
0.2 + a = 108.20 + (207/2)
a = 211.5
na = a(1 + 3.76)
nF = 100
na/nF = 10.07
A diesel engine uses a hydrocarbon fuel represented by C12H26 and is burned with 30% excess air. The air and fuel is supplied at 1
atm and 25C. Determine
a. the actual air-fuel Ratio
b. the m3 of CO2 formed per kg of fuel if the product temp. is 400C and a pressure of 1 atm.
c. The M and R of the Products
d. The M and R of the dry flue gas
Combustion with 100% theoretical air (basis 1 mole of fuel)
C12H26 + aO2 + a(3.76)N2  bCO2 + cH2O + a(3.76)N2
a = n + 0.25m = 18.5
b = n = 12
c = 0.5m = 13
Combustion with 30% excess air
C12H26 + 1.30aO2 + 1.30a(3.76)N2  bCO2 + cH2O + dO2 + 1.30a(3.76)N2
d = ea = 5.55
137 .28 (1  e)(n  0.25 m)
kg of air
A
 19 .42
 a 
F
12
n

m
kg
of fuel
 
VCO 2 
n RT 12(8.3143)( 400  273)

 662.7 m3 of CO2
P
101.325
m3 of CO2
662.7

 3.9
kg of Fuel 12(12)  26
nP = b + c + d + 1.3(18.5)(3.76) = 120.978
1
1
12(44)  13(18)  5.55(32)  90.428(28)
M
niM 
n
120.978
M  28.7
R  0.2897
n d  n p  c  107.978


1
1
12(44)  5.55(32)  90.428(28)
niM 
n
107.978
M  29.983
M
R  0.2773
71
A fuel consisting 80% C12H26 and 20% C14H30 is burned with 30% excess air. The flue gas is at atmospheric pressure. Find the
minimum exhaust temperature to avoid condensation. (Answer: 50.5C; 47.45C)
Assuming that the given fuel analysis is percentages by mass, converting it to molal analysis;
Solution:
Assuming that the given fuel analysis is percentages by mass, converting it to molal analysis;
.8
170
C12H 26 
 .823
.8 0 . 2

170 198
.2
198
C14H 30 
 0.177
.8 0 . 2

170 198
Combustion with 100% theoretical air (basis; 100 moles of fuel)
82.3C12H26 + 17.7C14H30 + aO2 + a(3.76)N2  bCO2 + cH2O + a(3.76)N2
b = 1235.4 ; c = 1335.4; a = 1903.1
Combustion with 30% excess air
82.3C12H26 + 17.7C14H30 + (1.3)aO2 + (1.3)a(3.76)N2 
bCO2 + cH2O + dO2 + (1.30)a(3.76)N2
By O2 balance
d = 570.93
nP = b + c + d + 1.30a(3.76) = 12,444.0828
yH2O = c/nP = 0.107
PH2O = 0.107(101.325) = 10.9 KPa
From steam table tsat at 10.9 KPa = 47.45C
Assuming that the given analysis is molal analysis
80C12H26 + 20C14H30 + aO2 + a(3.76)N2  bCO2 + cH2O + a(3.76)N2
b = 1240; c = 1340; a = 1910
with excess air:
80C12H26 + 20C14H30 + (1.3)aO2 + (1.3)a(3.76)N2 
bCO2 + cH2O + dO2 + (1.30)a(3.76)N2
d = 573
nP = 12489.08
yH2O = c/nP = .1073
PH2O = 10.9 KPa ; tsat = 47.45C
A fuel has the following volumetric analysis: CH4 = 68% C2H6 = 32%. Assume complete combustion with 15% excess air at
101.325 KPa, 21C wet bulb and 27C dry bulb. What is the partial pressure of the water vapor in KPa?
72
Problem Natural Gas
A natural gas fuel showed the following percentages by volume: C2H6 = 9%; CH4 = 90%; CO2 = 0.2 % and N2 = 0.8 %. If this gas
is used as fuel and is burned with 20% excess air for complete combustion, Determine
The Combustion Equation
The theoretical and actual air fuel ratio
The molecular weight and gas constant of the products
The volume of air required per m3 of natural gas if the gas and air are at temperature of 16C and a pressure of 101.6 KPa.
Combustion with 100% theoretic al air
9C 2 H 6  90CH4  0.2CO2  0.8N 2  aO 2  a(3.76)N 2  bCO2  cH 2 O  dN 2
Carbon Balance
2(9)  90  0.2  b
b  108.2
Hydrogen Balance
9(6)  90(4)  2c
c  207
Oxygen Balance
2(0.2)  2a  2b  c
2(108.2)  207 - 2(0.2)
2
a  211.5
a
Nitrogen Balance
2(0.8)  2(a)(3.76)  2d
2(0.8)  2(211.5)(3.76)
2
d  796.04
d
Combustion with excess air
9C2 H 6  90CH4  0.2CO2  0.8N2  (1.20)aO2  (1.20)a(3.76)N2  bCO2  cH 2O  eO2  fN 2
By oxygen balance
2(0.2)  2(1.20)(211.5)  2(108.2)  207  2e
2(0.2)  2(1.20)(211.5)  2(108.2)  207
2
e  42.3
e
By Nitrogen balance
2(0.8)  2(1.20)(211.5)(3.76)  2f
2(0.8)  2(1.20)(211.5)(3.76)
f
2
f  955.088
Combustion Equation
9C2 H 6  90CH4  0.2CO2  0.8N2  253.8O2  954.288N2  108.2CO2  207H 2O  42.3O2  955.088N2
Mass of Fuel
9(30)  90(16)  0.2(44)  0.8(28)  1741.2 kg
Mass of air
253.8(32)  954.288(28)  34841.664 kg
34841.664
kg of air
A

 20.01
 
F
1741.2
kg
of fuel
 actual
A
A
 (1  e) 
 
 F actual
 F  theoretical
20.01
kg of air
A

 16.68
 
kg of fuel
 F  theoretical 1.20
73
Moles of Products
108.2  207  42.3  955.088  1312.588
 yiMi  M 
M  27.9
R
108.2(44)  207(18)  42.3(32)  955.088(28)
1312.588
kg
kg mol
8.3143
KJ
 0.298
27.9
kg - K
PVa  n a RT
Va 
(253.8  954.288)(8.3143)(16  273)
 28571.2 m 3
101.6
PVF  n F RT
(9  90  .2  0.8)(8.143)(16  273)
 2365 m 3
101.6
Va 28571.2

 12.08
VF
2365
VF 
Sample Problem (Known Orsat analysis and Fuel type)
A fuel oil C12H26 is used in an internal combustion engine and the Orsat analysis are as follows: CO 2 = 12.8% ; O2 = 3.5%; CO =
0.2% and N2 = 83.5%. Determine the actual air-fuel ratio and the percent excess air.
Solution: (Basis 100 moles of dry flue gas)
aC12H26 + bO2 + b(3.76)N2  12.8CO2 + cH2O + 0.2CO + 3.5O2 + 83.5N2
By C balance
12a = 12.8 + 0.2
a = 1.0833
By N2 Balance
b(3.76) = 83.5
b = 22.207
By H balance
26a = 2c
c = 26(1.0833)/2
c = 14.083
Dividing the equation by a
C12H26 + 20.5O2 + 77.08N2  11.816CO2 + 13H2O + 0.185CO + 3.23O2 + 77.08N2
20 .5(32 )  77 .08(28)
kg of air
A
 16 .56
  
F
12
(
12
)

26
kg
of fuel
 a
Combustion of C12H26 with 100% theoretical air
C12H 26  aO 2  a (3.76) N 2  bCO2  cH 2 O  a (3.76) N 2
12  b
26  2c
c  13
2a  2b  c
a  18.5
kg of air
 A  18.5(32)  (18.5)(3.76)( 28
 14.94
  
12(12)  26
kg of C12H 26
 F t
74
A
A
   (1  e) 
 F a
 F t
A
 
 F a
1 e 
A
 
 F t
A
 
 F a
e
 1  0.108  10.8%
A
 
 F t
Sample Problem (Unknown Fuel – Known Orsat analysis)
An unknown hydrocarbon is used as fuel in a diesel engine, and after an emission test the orsat analysis shows,
CO2 = 12.5% ; CO = 0.3% ; O2 = 3.1% ; N2 = 84.1%.Determine
a. the actual air-fuel ratio
b. the percent excess air
c. the fuel analysis by mass
CnHm + aO2 + a(3.76)N2  12.5CO2 + bH2O + 0.3CO + 3.1O2 + 84.1N2
By Carbon balance
n = 12.5 + 0.3
n = 12.8
By Hydrogen balance
m = 2b  eq. 1
By Oxygen balance
2a = 2(12.5) + b + 0.3 + 2(3.1)  eq. 2
By Nitrogen balance
a(3.76) = 84.1
a = 22.367
substituting a to eq. 2
b = 13.234
substituting b to eq. 1
m = 26.47
C12.8 H 26.47  22.367O 2  84.1N 2  12.5CO2  13.234H 2 O  0.3CO  3.1O2  84.1N 2
22.367(32)  84.1(28)
kg of air
A
 17.05
  
12(12.8)  26.47
kg of fuel
 F a
Combustion with 100% theoretical air
n = 12.8 ; m = 26.47
C12.8 H 26.47  aO 2  a (3.76) N 2  bCO2  cH 2 O  a (3.76) N 2
12.8  b
26.47  2c
c  13.235
2a  2 b  c
a  19.4175
kg of air
 A  19.4175(32)  19.4175(3.76)( 28
 14.8
  
12(12.8)  26.47
kg of C12H 26
 F t
75
A
A
   (1  e) 
 F a
 F t
e  0.152  15.2 %
12n
12(12.8)

 85.3%
12n  m 12(12.8)  26.47
m
26.47
%H 

 14.7%
12n  m 12(12.8)  26.47
%C 
Problem (Combustion of Coal Fuel)
A coal fired steam power plant uses that has the following Ultimate Analysis;
C = 74% ; H2 = 5% ; O2 = 6%; S = 1%; N2 = 1.2%; H2O = 3.8% and Ash = 9%. If this coal is burned with 30% excess air, determine
a. The combustion equation
b. The actual air – fuel ratio
c. The M and R of the products
d. The Orsat analysis of the products
e. The HHV of coal in KJ/kg
76
ENTHALPY OF FORMATION
The “Enthalpy of Formation of a compound is the enthalpy at the Arbitrary Reference State (t = 25C and P = 1 Atm).
1 mole C
25C, 1 atm
Combustion
Chamber
1 mole CO2
25C, 1 atm
1 mole O2
25C, 1 atm
Q
Let;
HR – total enthalpy of Reactants
HP – total enthalpy of products
From 1st Law
Q  HR  HP
or
Q  Σ ni hi  Σ ni hi
R
P
but the enthalpy of all the reactants (at the reference state) is Zero (for they are all elements)
Q  H P  -393,757 KJ
therefore
(h f ) CO 2  393,757
M Carbon  12
KJ
 Enthalpy of formation of CO 2
kg mol
kg
kg m
(h f ) CO 2  32,813
KJ
kg C
Note : Negative sign is due to the reaction' s being " Exothermic "
For Hydrogen (Vapor)
(h f ) H 2O  241,971
KJ
kg mol
(h f ) H 2O  120,985.5
KJ
kg H 2
For Hydrogen (Liquid)
(h f ) H 2O  286,010
KJ
kg mol
(h f ) H 2O  143,005
KJ
kg H 2
For Sulfur
(h f ) S  296,840
KJ
kg mol
(h f ) S  9,276.25
KJ
kg Sulfur
77
ENTHALPY OF COMBUSTION OR HEATING VALUE
It is the difference between the enthalpies of the products and the reactants at the same temperature T and Pressure P.
h RP  H P  H R

KJ
kg mol


 kgKJ
h RP  Σ n j h f  (h  h298)  Σn i h f  (h  h 298) i
P
R
mol
(H  H R ) KJ
h RP  P
M
kg
where
M - molecular weight
Example:
Determine the heating value of C12H26 (liquid) when burned with 25% excess air. Air and fuel enters at 25C and products leaves at
900 K. Assume P =101 KPa.
Combustion Equation
C12H 26  23.13O 2  86.95N 2  12CO2  13H 2 O  4.63O 2  86.95N 2
LOWER HEATING VALUE (H 2 O in the products is VAPOR
H R  1(394,199)
KJ
kg mol
H P  12(393,757  28,041)  13(241,971  21,924)  4.63(0  19,246)  86.95(0  18,221)
H P  -5,575,874.30
KJ
kg mol
LHV  -5,181,675.30
LHV 
KJ
kg mol
- 5,181,675.30
KJ
 30,480.44
)
170
kg C12 H 26
HIGHER HEATING VALUE (H 2 O in the products is LIQUID)
H R  1(394,199)
KJ
kg mol
H P  12(393,757  28,041)  13(-286,010.0  21,924)  4.63(0  19,246)  86.95(0  18,221)
H P  -6,148,381.30
KJ
kg mol
HHV  -5,754,182.30
HHV 
KJ
kg mol
- 5,754,182.30
KJ
 33,848.13
170
kg C12 H 26
78
FROM BUREAU OF STANDARD
HEATING VALUE
Heating Value - is the energy released by fuel when it is completely burned and the products of combustion are cooled to the original
fuel temperature.
Higher Heating Value (HHV) - is the heating value obtained when the water in the products is liquid.
Lower Heating Value (LHV) - is the heating value obtained when the water in the products is vapor.
For Solid Fuels
O 
KJ

HHV  33,820C  144,212 H 2  2   9,304S
8 
kg

where: C, H2, O2, and S are in decimals from the ultimate analysis
For Coal and Oils with the absence of Ultimate Analysis
 A  HHV
kg of air/kg of fuel
  
 F  t 3,041
HHV = 31,405C + 141 647H KJ/kg
For Liquid Fuels
HHV = 43,385 + 93(Be - 10) KJ/kg
Be - degrees Baume
For Gasoline
HHV = 41,160 + 93 (API) KJ/kg
LHV = 38,639 + 93 (API) KJ/kg
For Kerosene
HHV = 41,943 + 93 (API) KJ/kg
LHV = 39,035 + 93 (API) KJKkg
For Fuel Oils
HHV = 41,130 + 139.6(API) KJ/kg
LHV = 38,105 + 139.6(API) KJ/kg
API - American Petroleum Institute
For Fuel Oils (From Bureau of Standard Formula)
HHV = 51,716 - 8793.8 (S)2 KJ/kg
LHV = HHV - QL KJ/kg
QL = 2,442.7(9H2) KJ/kg
H2 = 0.26 - 0.15(S) kg of H2/ kg of fuel
141.5
131.5  API
140
S
130  Be
S
Where:
S @ t = S - 0.0007(t-15.56)
S - specific gravity of fuel oil at 15.56 C
H2 - hydrogen content of fuel oil, Kg H/Kg fuel
79
QL - heat required to evaporate and superheat the water vapor formed by the combustion of hydrogen in the fuel,KJ/kg
S @ t - specific gravity of fuel oil at any temperature t
Oxygen Bomb Calorimeter - instrument used in measuring heating value of solid and liquid fuels.
Gas Calorimeter - instrument used for measuring heating value of gaseous fuels.
SECOND LAW OF THERMODYNAMICS
Whenever energy is transferred, the level of energy cannot be conserved and some energy must be permanently reduced to a lower
level.
When this is combined with the first law of thermodynamics, the law of energy conservation, the statement becomes:
Whenever energy is transferred, energy must be conserved, but the level of energy cannot be conserved and some energy must be
permanently reduced to a lower level.
Kelvin-Planck statement of the second law:
No cyclic process is possible whose sole result is the flow of heat from a single heat reservoir and the performance of an equivalent
amount of work.
For a system undergoing a cycle: The net heat is equal to the network.
 dW   dQ or W = Q
W - net work
Q - net heat
CARNOT CYCLE
Nicolas Léonard Sadi Carnot (1796-1832), French physicist and military engineer, son of Lazare Nicolas Marguerite Carnot, born
in Paris, and educated at the École Polytechnique. In 1824 he described his conception of the perfect engine, the so-called Carnot
engine, in which all available energy is utilized. He discovered that heat cannot pass from a colder to a warmer body, and that the
efficiency of an engine depends upon the amount of heat it is able to utilize. These discoveries led to the development of the Carnot
Cycle, which later became the basis for the second law of thermodynamics.
A. Carnot Engine Cycle
Processes:
1 to 2 - Heat Addition (T = C)
2 to 3 - Expansion (S = C)
3 to 4 - Heat Rejection (T = C)
4 to 1 - Compression (S = C)
P
T
1
QA
T=C
S=C
1
2
4
3
2
4
S=C
T=C
3
QR
V
S
Heat Added (T = C)
Q A = TH ( ΔS)  1
Heat Rejected (T = C)
Q R = TL ( ΔS)  2
ΔS = S2 - S1 = S4 - S3
3
Net Work
W = ΣQ = Q A - Q R
4
W = (TH - TL )( ΔS)  5
80
Thermal Efficiency
e
W
x 100%  6
QA
e
QA  QR
x 100%  7
QA
 Q

e  1  R x  100%  8
 QA 
Eq. 5 to and Eq. 1 to Eq. 6
e
TH  TL
x 100%  9
TH
B. Carnot Refrigeration Cycle (Reversed Carnot Cycle)
Processes:
1 to 2 - Compression (S =C)
2 to 3 - Heat Rejection (T = C)
3 to 4 - Expansion (S = C)
4 to 1 - Heat Addition (T = C)
T
3
2
QR
4
1
QA
S
COEFFICIENT OF PERFORMANCE
QA
6
W
QA
COP 
7
QR - QA
COP 
COP 
TL
8
TH - TL
81
Carnot Heat Pump:
A heat pump uses the same components as the refrigerator but its purpose is to reject heat at high energy level.
Performance Factor
PF =
PF =
PF =
QR
W
 11
QR
QR - Q A
TH
TH - TL
 12
 13
PF = COP + 1  14
TON OF REFRIGERATION
It is the heat equivalent to the melting of 1 ton (2000 lb) of water ice at 0C into liquid at
0C in 24 hours.
2000(144)
BTU
TR 
 12,000
24
hr
KJ
TR  211
min
where 144 BTU/lb is the latent heat of fusion
Problem No. 1
A Carnot engine operates between temperature levels of 397C and 7C and rejects 20 KJ/min to the environment. The total network
output of the engine is used to drive heat pump which is supplied with heat from the environment at 7C and rejects heat to a home
at 40C. Determine:
a) the network delivered by the engine in KJ/min
b) the heat supplied to the heat pump in KJ/min
c) the overall COP for the combined devices which is defined as the energy rejected to the home divided by the initial
energy supplied to the engine.
Problem No. 2
A reversed Carnot cycle is used for refrigeration and rejects 1000 KW of heat at 340 K while receiving heat at 250 K. Determine:
a) the COP
b) the power required in KW
c) the refrigerating capacity in Tons
82
SAMPLE PROBLEMS
1. A Carnot engine operating between 775 K and 305K produces 54 KJ of work. Determine the change of entropy during heat
addition. (S12= -0.12 KJ/K)
2. A Carnot engine operates between temperature reservoirs of 817C and 25C and rejects 25 KW to the low temperature reservoir.
The Carnot engine drives the compressor of an ideal vapor compression refrigerator, which operates within pressure limits of 190
KPa and 1200 Kpa. The refrigerant is ammonia. Determine the COP and the refrigerant flow rate.(4; 14.64 kg/min)
3. A Carnot engine operates between temperatures of 1000 K and 300K The engine operates at 200 RPM and develops 200 KW.
The total engine displacement is such that the mean effective pressure is 300 Kpa. Determine:
a) the thermal efficiency (71%)
b) the heat added ( 286 KW)
c) the total engine displacement in m3/cycle ( 0.02 m3/cycle)
4. A Carnot engine operating between 775 K and 305K produces 54 KJ of work. Determine the change of entropy during heat
addition. (S12= -0.12 KJ/K)
5. A Carnot engine produces 25 KW while operating between temperature limits of 1000 K and 300 K. Determine (a) the heat
supplied in KW (b) the heat rejected in KW
6. A Carnot heat engine rejects 230 KJ of heat at 25C. The net cycle work is 375 KJ. Determine the cycle thermal
efficiency and the cycle high temperature.
83
7. A Carnot refrigerator operates between temperature limits of -5C and 30C. The power consumed is 4 KW and the heat absorbed
is 30 KJ/kg. Determine; (a) the COP (b) the refrigerant flow rate
8. A non polluting power plant can be constructed using the temperature difference in the ocean. At the surface of the ocean in
tropical climates, the average water temperature year round is 30C. At a depth of 305 m, the temperature is 4.5C. Determine the
maximum thermal efficiency of such a power plant. (8.4%)
9. A heat is used to heat a house in the winter months. When the average outside temperature is 0C and the indoor
temperature is 23C, the heat loss from the house is 20 KW. Determine the minimum power required to heat the
heat pump.
10. A Carnot heat pump is being considered for home heating in a location where the outside temperature may as low
as -35C. The expected COP for the heat pump is 1.50. To what temperature could this unit provide?
INTERNAL COMBUSTION ENGINE CYCLE
 Air Standard Cycle
 Otto Cycle
 Diesel Cycle
 Dual Cycle
AIR STANDARD OTTO CYCLE
Otto, Nikolaus August
Born:June 10, 1832, Holzhausen, Nassau
Died: Jan. 26, 1891, Cologne
German engineer who developed the four-stroke internal-combustion engine, which offered the first practical alternative to the
steam engine as a power source.
Otto built his first gasoline-powered engine in 1861. Three years later he formed a partnership with the German industrialist Eugen
Langen, and together they developed an improved engine that won a gold medal at the Paris Exposition of 1867.
In 1876 Otto built an internal-combustion engine utilizing the four-stroke cycle (four strokes of the piston for each ignition). The
four-stroke cycle was patented in 1862 by the French engineer Alphonse Beau de Rochas, but since Otto was the first to build an
engine based upon this principle, it is commonly known as the Otto cycle. Because of its reliability, its efficiency, and its relative
quietness, Otto's engine was an immediate success. More than 30,000 of them were built during the next 10 years, but in 1886 Otto's
patent was revoked when Beau de Rochas' earlier patent was brought to light.
Processes
1 to 2 – Isentropic Compression (S = C)
2 to 3 – Constant Volume Heat Addition (V = C)
3 to 4 – Isentropic Expansion (S = C)
4 to 1 – Constant volume Heat Rejection (V = C)
84
Compression Ratio (r)
V1 V4

1
V2 V3
r
V1 = V4 and V2 = V3
V1 - volume at bottom dead center (BDC)
V2 – volume at top dead center (TDC)(clearance volume)
Displacement Volume (VD)
VD =V1 – V2
 Eq. 2
Percent Clearance (C)
C
r=
V2
VD  Eq. 3
1C
 Eq. 3’
C
Heat Added (QA)
At V = C ; Q = mCv(T)
QA = mCv(T3 – T2)
 Eq. 4
Heat Rejected (QR)
 Eq. 5
QR = mCv(T4 – T1)
Net Work (W)
W = Q
W = QA – QR
 Eq. 6
P,V and T Relations
At point 1 to 2 (S = C)
k 1
T2  V1 
 
T1  V2 
k 1
 r 
k 1
P  k
  2 
 P1 
T2  T1 (r)k1  Eq. 7
85
At point 2 to 3 (V = C)
T3 P3

T2 P2  Eq. 8
At point 3 to 4 (S = C)
k 1
k 1
T3  V4 
 
T4  V3 
P  k
  3   (r)k1
 P4 
k 1
T3  T4 (r)  Eq.9
At point 4 to 1 (V = C)
T4 P4
 Eq. 10

T1 P1
Entropy Change
a) S during Heat Addition
S3 – S2 = mCvln
T3
T2
 Eq. 11
b) S during Heat Rejection
T1
T4
 Eq. 12
S1 - S4 = -( S3 – S2)
 Eq. 13
S1 - S4 = mCvln
Thermal Efficiency
e
W
x 100%  Eq. 14
QA
e
Q A  QR
x 100%  Eq. 15
QA
 Q 
e  1  R  x 100%  Eq. 16
 QA 
 (T  T ) 
e  1  4 1  x 100%  Eq.17
 (T3  T2 ) 
1 

e  1  k1  x 100%  Eq. 18
 (r) 
Mean Effective Pressure
Pm 
W
KPa
VD
 Eq. 19
where:
Pm – mean effective pressure, KPa
W – Net Work KJ, KJ/kg, KW
VD – Displacement Volume m3, m3/kg, m3/sec
86
Displacement Volume
VD = V1 – V2 m3  Eq. 20
VD = 1 - 2 m3/kg  Eq.21
Note:
For Cold Air Standar: k = 1.4
For Hot Air Standard: k = 1.3
AIR STANDARD DIESEL CYCLE
Diesel, Rudolf (Christian Karl)
Born: March 18, 1858, Paris, France
Died: September 29, 1913, at sea in the English Channel
German thermal engineer who invented the internal-combustion engine that bears his name. He was also a distinguished connoisseur
of the arts, a linguist, and a social theorist.
Diesel, the son of German-born parents, grew up in Paris until the family was deported to England in 1870 following the outbreak
of the Franco-German War. From London Diesel was sent to Augsburg, his father's native town, to continue his schooling. There
and later at the Technische Hochschule (Technical High School) in Munich he established a brilliant scholastic record in fields of
engineering. At Munich he was a protégé of the refrigeration engineer Carl von Linde, whose Paris firm he joined in 1880.
Diesel devoted much of his time to the self-imposed task of developing an internal combustion engine that would approach the
theoretical efficiency of the Carnot cycle. For a time he experimented with an expansion engine using ammonia. About 1890, in
which year he moved to a new post with the Linde firm in Berlin, he conceived the idea for the diesel engine . He obtained a German
development patent in 1892 and the following year published a description of his engine under the title Theorie und Konstruktion
eines rationellen Wäremotors (Theory and Construction of a Rational Heat Motor). With support from the Maschinenfabrik
Augsburg and the Krupp firms, he produced a series of increasingly successful models, culminating in his demonstration in 1897 of
a 25-horsepower, four-stroke, single vertical cylinder compression engine. The high efficiency of Diesel's engine, together with its
comparative simplicity of design, made it an immediate commercial success, and royalty fees brought great wealth to its inventor.
Diesel disappeared from the deck of the mail steamer Dresden en route to London and was assumed to have drowned.
Processes:
1 to 2 – Isentropic Compression (S = C)
2 to 3 – Constant Pressure Heat Addition (P = C)
3 to 4 – Isentropic Expansion (S = C)
4 to 1 – Constant Volume Heat Rejection (V = C)
T
P
2
3
QA
S=C
P=C
4
S=C
2
QR
W
CVD
VD
4
V=C
1
Pm
3
1
V
S
87
Compression Ratio
r
V1 V4

V2 V2
 1
V1 V4

V2 V3
Cut-Off Ratio(rc)
rc 
V3
V2  2
Percent Clearance
V2
VD  3
1C
r
C  4
C
Displacement Volume (VD)
VD =V1 – V2
Heat Added (QA)
 Eq. 5
At P = C ; Q = mCp(T)
QA = mCp(T3 – T2)
 Eq. 6
QA = mkCv(T3 – T2)  Eq. 7
Heat Rejected (QR)
QR = mCv(T4 – T1)
 Eq. 8
Net Work (W)
W = Q
W = QA – QR
 Eq. 9
W = mkCv(T3 – T2) - mCv(T4 – T1)
W = mCv[k (T3 – T2) - (T4 – T1)]  Eq. 10
P,V and T Relations
At point 1 to 2 (S = C)
k 1
T2  V1 
 
T1  V2 
P 
 r k1   2 
 P1 
T2  T1 (r)k1  Eq. 11
k 1
k
At point 2 to 3 (P = C)
T3 V3
  rc  Eq. 12
T2 V2
T3  T1 (r)k 1 (rc )  Eq. 13
At point 3 to 4 (S = C)
88
k 1
T4  V3 
 
T3  V4 
k 1
P  k
  4 
 P3 
k 1
V 
T4  T1 (r) (rc ) 3 
 V4 
V k1 V V k1
T4  T1 1k1 3 3k1
V2 V2 V4
k 1
T4  T1 (rc )k  Eq.14
At point 4 to 1 (V = C)
T4 P4
 Eq. 15

T1 P1
Entropy Change
a) S during Heat Addition
T3
T2
S3 – S2 = mCpln
 Eq. 16
b) S during Heat Rejection
T1
T4
 Eq. 17
S1 - S4 = -( S3 – S2)
 Eq. 18
S1 - S4 = mCvln
Thermal Efficiency
e=
W
x 100%
QA
e=
Q A  QR
x 100%  Eq. 20
QA
 Eq. 19
 Q 
e  1  R  x 100%
 QA 
 Eq. 21

(T4  T1 ) 
e  1 
 x 100%
 k (T3  T2 ) 
 Eq.22
Substituting Eq. 11, Eq. 13 and Eq. 14 to Eq. 22

1  r k  1 
e  1  k 1  c
 x 100%  Eq. 23
 (r )  k (rc  1) 
Mean Effective Pressure
Pm 
W
KPa
VD
 Eq. 24
where:
Pm – mean effective pressure, KPa
W – Net Work KJ, KJ/kg, KW
VD – Displacement Volume m3, m3/kg, m3/sec
Displacement Volume
VD = V1 – V2 m3  Eq. 25
VD = 1 - 2 m3/kg  Eq.26
89
AIR STANDARD DUAL CYCLE
Processes:
1 to 2 – Isentropic Compression (S = C)
2 to 3 – Constant Volume Heat Addition Q23 ( V = C)
3 to 4 – Constant Pressure Heat Addition Q34 (P = C)
4 to 5 – Isentropic Expansion (S = C)
5 to 1 - Constant Volume Heat Rejection (V = C)
P
T
3
QA34
4
QA23
4
P=C
3
2
S=C
2
V=C
5
S=C
5
V=C
1
1
CVD
QR
V
VD
S
Compression Ratio
V
r 1
V2  Eq. 1
V1 = V5 and V2 = V3
Cut-Off Ratio
rc 
V4
V3  Eq. 2
Pressure Ratio
P
rp  3
P2  Eq. 3
P3 = P4
Percent Clearance
V2
VD  Eq. 4
1C
r
 Eq. 5
C
C
Displacement Volume
VD = V1 – V2 m3  Eq. 6
VD = 1 - 2  Eq. 7
P, V, and T Relations
At point 1 to 2 (S = C)
T2  V1 


T1  V2 
k 1
k 1
 r 
k 1
P  k
  2 
 P1 
T2  T1 (r ) k 1  Eq. 8
90
At point 2 to 3 (V = C)
T3 P3

 rP  Eq. 9
T2 P2
T3  T1 ( r ) k 1 ( rP )  Eq. 10
At point 3 to 4 (P = C)
T4  V4 
  rc

T3  V3 
T4  T1 (r ) k 1 (rP )( rc )  Eq. 11
At point 4 to 5 (S = C)
T5  V4 


T4  V5 
T5  T1
k 1
V1k 1

V2
V2  V1
k 1 
r  V4  V4k 1 
k 1 P

 Eq. 12
T5  T1 (rc ) k (rP )  Eq. 13
At 5 to 1 (V = C)
T5 P5

 Eq. 14
T1 P1
Entropy change
a) At 2 to 3 (V = C)
S3 – S2 = mCvln
T3
T2
 Eq. 15
b) At 3 to 4 (P = C)
S4 – S3 = mCPln
T4
 Eq. 16
T3
S4 – S2 = (S3 – S2) + (S4 – S3)  Eq. 17
c) At 5 to 1 (V = C)
S1 – S5 = mCvln
T1
 Eq. 18
T5
Heat Added
QA = QA23 + QA34  Eq. 19
QA23 = mCv(T3 - T2)
 Eq. 20
QA34 = mCp(T4 - T3) = mkCv(T4 - T3)
 Eq. 21
QA = mCv[(T3 - T2)+ k(T4 - T3)
 Eq. 22
Heat Rejected
QR = mCv(T5 - T1)
 Eq. 23
Net Work
W = (QA - QR)  Eq. 24
W = mCv[(T3 - T2) + k(T4 - T3) - (T5 - T1)]
 Eq. 25
Thermal Efficiency
e=
W
x 100%
QA
 Eq. 26
91
e=
Q A  QR
x 100%  Eq. 27
QA
 Q 
e  1  R  x 100%  Eq. 28
 QA 


(T5  T1 )
e  1 
 x 100%  Eq.29
 (T3  T2 )  k(T4  T3 ) 
Substituting Eq. 8, Eq. 10 and Eq. 11and Eq. 13 to Eq. 29


1 
rPrck  1

 x 100%  Eq. 30
e  1  k1 
 (r)  (rP  1)  krP (rc  1) 
BRAYTON CYCLE
Basic Components
Air Compressor
Heater
Gas Turbine
Cooler
Processes
1 to 2 – Isentropic Compression
2 to 3 – Isobaric Heat Addition
3 to 4 – Isentropic Expansion
4 to 1 – Isobaric Heat Rejection
Figure 1: Closed Gas Turbine Cycle
92
Pr essure Ratio
rp 
P2 P3

 Eq.1
P1 P4
Compressor Work
Wc  mC p (T2 - T1 )  Eq. 2
  k  Eq.3
k 1
T3  T4 rp  k  Eq.4
k 1
T2 T3

 rp  k
T
T
T2  T1 rp
1
k 1
4
Heater
Q A  mC p (T3  T2 )  Eq.5
Turbine
Wt  mC p (T3 - T4 )  Eq. 6
Cooler
Q R  mC p (T4  T1 )  Eq.7
Net Work
W  Q A  Q R  Eq.8
Thermal Efficiency
e
W
x100%  Eq.9
QA
e
QA  QR
x100%  Eq.10
QA
 Q 
e  1  R  x100%  Eq.11
 QA 
 (T  T1 ) 
e  1  4
 x100%  Eq.12
 (T3  T2 ) 
Substituti ng Eq. 3 and Eq.4 to Eq. 12


1 

e  1 
x100%  Eq.13
k 1 
rp k 

 
Work Ratio
WR 
Net Work
Turbine Work
93
Figure 2: Open Gas Turbine Cycle
W  WT urbine  WCompressor
ACTUAL BRAYTON CYCLE
Turbine Efficiency
Actual Turbine Work
x100%
Ideal Turbine Work
Wt '  mC p (T3 - T4' )  Actual Work
ηT urbine 
Wt  mC p (T3 - T4 )  Ideal Work
ηT urbine 
(T3 - T4' )
x100%
(T3  T4 )
Compressor Efficiency
Ideal Compressor Work
x100%
Actual Compressor Work
Wc  mC p (T2 - T1 )  Actual Work
ηCompressor 
Wc '  mC p (T2' - T1 )  Ideal Work
ηCompressor 
(T2 - T1 )
x100%
(T2' - T1 )
Actual Thermal efficiency
e t' 
Net Work
x100%
Heat added
94
Problem
mum
temperature is 2750K; the maximum pressure is 6894 KPa. Determine:
a. the work in KJ/kg
b. the heat added in KJ/kg
c. the thermal efficiency
d. the mean effective pressure in KPa
e. the entropy change during the heat addition processes
f. the percent clearance c
Solution:
Problem No. 1
In the air standard diesel cycle, the air is compressed isentropically from 26C and 105 KPa to 3700 KPa. The entropy change
during heat rejection is -0.6939 KJ/kg-K. Determine
a) the heat added per kg
b) the thermal efficiency
c) the maximum temperature
d) the temperature at the start of heat rejection
P
T
2
3
QA
3
P=C
S=C
2
S=C
4
V=C
4
1
1
QR
S
V
Given:
T1 = 26 + 273 = 299 K
P1 = 105 KPa
P2 = P3 = 3700 KPa
S1 – S4 = -0.6939 KJ/kg-K
S3 – S2 = 0.6939 KJ/kg-K
Q R  C V T4  T1 
Q R  349.76 KJ/kg
W  Q A - Q R  477.34 KJ/kg
e
W
x 100%  57.71 %
QA
95
k -1
T2  P2  k
 
T1  P1 
T 2  827.3 K
S3  S 2  ΔS  mC p ln
0.6939  1.0045 ln
T3
T2
T3
827.3
0.6939
T3
827.3
T3  1650.71 K
e 1.0045
Q A  C p T3 - T2   827.1 KJ/kg
S4 - S1  C v ln
0.6939
e 0.7175
T4
T1
T4
299
T4  786.47 K
96
Problem No. 2
For an ideal Otto engine with 17% clearance and an initial pressure of 93 Kpa, determine the pressure at the end of compression. If
the pressure at the end of constant volume heating is 3448 KPa, what is the mean effective pressure.
P
T
QA
3
3
V=C
S=C
2
4
2
V=C
S=C
4
1
1
QR
V
S
Given:
c= 0.17
P1 = 93 KPa
P3 = 3448 KPa
e
W
QA
W  eQ A  eq.9
1  c 1.17
r

c
.17
r  6.88
Q A  C v (T3  T2 )
V
V
r 1  4
V2 V3
Q A  2.315T1  eq.10
P2
 rk
P1
Q A  0.71755.39T1  2.163T1 
Q A  0.7175T15.39  2.163
W  0.542.315T1 
W  1.25T1  eq.11
W
1.25T1

VD 0.0026T1
P2  936.881.4  1383.91 KPa
Pm 
1 

e  1  k 1   0.54  54%
 r 
Pm  480.9 KPa
T2  T1 r k 1  2.163T1  eq.1
P 
T3  T2  3   5.39T1  eq.2
 P2 
T
T4  k31  T3 r 1 k  2.49T1  eq.3
r 
VD  υ1  υ 2  eq.4
r
υ1
υ2
υ1  rυ 2  eq.5
VD  rυ 2  υ 2
VD  υ 2 r  1  eq.6
υ2 
RT2 0.2872.163T1 

 0.00045T1  eq.7
P2
1383.91
VD  0.00045T1 6.88  1
VD  0.0026T1  eq.8
97
Problem No. 3
Air at 300 K and 100 kPa enters into the compressor of a gas turbine engine that operates on a Brayton cycle. The air mass fl ow
rate is 5 kg/s and the maximum air temperature in the engine is 1200 K. If the pressure ratio of the cycle is rp = 4,
a. Find the efficiency of the engine, assuming ideal compression and expansion.
b. Find the net power
c. What are the exit temperatures from the turbine and the compressor if Compressor = Turbine = 60.95%
given
T1  300 K
P1  100 KPa
T3  1200 K
rp  4
rp 
P2 P3

P1 P 4
Problem No. 4
An air standard diesel cycle has a compression ratio of 18 and the heat transferred is 1800 KJ/kg. At the beginning of the compression
process the pressure is 100 KPa and the temperature is 15C. For 1kg of air, determine
The entropy change during isobaric heat addition
The temperature at the start of heat rejection
For Air
The work in KJ
k = 1.4
The thermal efficiency
R = 0.287 KJ/kg-K
The mean effective pressure in KPa
k 1
T2  T1 r   915  K
P2  P1 r   100 18   5719 .8 KPa
k
1 .4
Q A  mC p (T3 - T2 )
QA
 T2  2707 .1 K
mC p
T3 
S 3  S 2  mC p ln
T3
KJ
 1.0894
T2
K
S1  S 4  ( S 3  S 2 )  mCv ln
T1
T4
T4  1314 .61 K
QR  mCv (T4  T1 )  736 .6 KJ
e
QA - QR
 59.1%
QA
W  Q A - Q R  1063 .4 KJ
Pm 
W
VD
VD V 1V2
V1 
mRT1
mRT2
; V2 
P1
P2
VD  0.781 m 3
Pm  1362 .2 KPa
98
Problem No. 5
An ideal dual combustion cycle operates on 0.454 kg of air. At the beginning of compression air is at 97 KPa, 316K. Let rp = 1.5,
rc= 1.6 and r = 11. Determine
the percent clearance (C = 10%)
the heat added, heat rejected and the net cycle work
the thermal efficiency
1 C
r
C
a)
1
C
 10 %
r 1
b) QA = QA1 + QA2
QA1 = mCv(T3 – T2)
T2  T1 (r ) k 1  825 K
P2  P1 (r ) k  2,784 KPa
T3  T2 (rp )  1238 K
QA1 = 0.454(0.7175)(1238 – 825) = 135 KJ
QA2 = mCp(T4 – T3)
T
rc  4  1.6
T3
T4  1.6T3
T4  1981K
QA2 = 0.454(1.0045(1981 – 1238) = 339 KJ
Q = QA1 + QA2 = 474 KJ


1 
rP rc k  1
 x 100%
e  1  k 1 
 (r )  (rP  1)  krP (rc  1) 
e = 58.7 %
Problem No. 6
An air standard Otto cycle has a compression ratio of 8 and has air conditions at the beginning of compression of 100 KPa and 25C.
The heat added is 1400 KJ/kg. Determine:
the thermal efficiency (56.5%)
the mean effective pressure (1057 KPa)
the percent clearance c (14.3%)
Given:
r = 8; P1 = 100 KPa ; T1 = 298K; QA = 1400 KJ/kg
Solution:

1 
e  1  k 1  x 100%
a)
 (r ) 
e = 56.5%
b
)W = eQA = 0.565(1400) = 791 KJ/kg
Pm =
W
VD
VD = 1 - 2
 1
VD = 1 1  
 r
RT1  1 
VD =
1   = 0.748 m3/kg
P1  r 
Pm = 1057 KPa
c)
1 C
C
1
C=
r 1
r=
C = 14.3%
99
Problem No. 7
An engine operates on the air standard Otto cycle. The cycle work is 900 KJ/kg, the maximum cycle temperature is 3000C and the
temperature at the end of isentropic compression is 600C. Determine the engine's compression ratio and the thermal efficiency.(r
= 6.35; e = 52.26%)
Given: W = 900 KJ/kg; T3 = 3273K; T2 = 873K
QA = Cv(T3 – T2) = 0.7175(3273 – 873) = 1722 KJ/kg
e = W/QA
e = 52.26%
1
 1  k 1
r

 1 e 
r = 6.36
Problem No. 8
An air standard diesel cycle has a compression ratio of 20 and a cut-off ratio of 3. Inlet pressure and temperature are 100 KPa and
27C. Determine:
the heat added in KJ/kg (1998 KJ/kg)
the net work in KJ/kg (1178 KJ/kg)
the thermal efficiency (61%)
Given: r = 20; rc = 3 ; P1 = 100 KPa ; T1 = 300C
Solution:
r
V1
V2
rk 
P2
P1
P2  100(20)1.4  6229 KPa
P2  P3  6229 KPa
T2
 (r ) k 1
T1
T2  300(20)1.4 1  994.3 K
T3 V3

 rc
T2 V2
T3  994.3(3)  2983 K
υ3 
RT3 0.287(2983)

 0.14 m 3
P3
6229
υ1  υ 4 
RT1 0.287(300)

 0.861 m 3
P1
100
T4  V3 


T3  V4 
k 1
1.4 1
 0.14 
T4  2983
 1442.45 K

 0.861 
QA = Cp(T3 – T2) = 1.0045(2983 – 994.3)
QA = 1997.65 KJ/kg
QR = Cv(T4 – T1) = 0.7175(1442.45 – 300)
QR = 820 KJ/kg
W = QA - QR = (1997.65 – 820)
W = 1177.65 KJ/kg
e
W
x 100%
QA
1177.65
x 100%
1997.65
e  59 %

1  r k  1 
e  1  k 1  c

 (r )  k (rc  1) 
e
100

 (3)1.4  1 
1

 x 100%
e  1 
1.4 1 

1
.
4
(
3

1
)
 (20)


e = 60.6 %
VAPOR POWER CYCLE
Rankine Cycle
Processes:
1 to 2 - Constant entropy (Isentropic) expansion (Turbine)
2 to 3 - Constant pressure (Isobaric) heat rejection (Condenser)
3 to 4 - Constant entropy (Isentropic) compression (Pump)
4 to 1 - Constant pressure (Isobaric) heat addition (Boiler)
Q A  Q E  Q Ev  Q s
Q E - heat required by the Economizer
Q Ev - heat required by the Evaporator
Q s - heat required by the Superheate r
Q R - heat rejected by steam in the condenser
101
Ideal Cycle
Actual Cycle
A. System: Turbine
Ideal Cycle
Wt = (h1 - h2)KJ/kg
Wt = ms(h1 - h2) KW
Actual Cycle
Wt' = (h1 - h2')KJ/kg
Wt' = ms(h1 - h2') KW
where:
ms is the steam flow rate in kg/sec
Wt - ideal turbine work
Wt' - actual turbine work
Turbine Efficiency
e Turbine 
Wt '
x 100%
Wt
B. System: Condenser
mw - cooling water flow rate that enters and leaves the condenser in kg/sec
Ideal Cycle
QR = (h2 - h3) KJ/kg
QR = ms(h2 - h3) KW
QR = mw Cpw (twB-twA)KW
Actual Cycle
102
QR = (h2'-h3) KJ/kg
QR = ms(h2'-h3) KW
QR = mw Cpw (twB-twA)KW
where:QR - heat rejected in the condenser
mw - cooling water flow rate in kg/sec
Cpw - specific heat of cooling water KJ/kg-C or KJ/kg-K
Cpw = 4.187 KJ/kg-C or KJ/kg-K (average value)
twA - inlet cooling water temperature,C
twB - outlet cooling water temperature,C
C. System: Pump
Ideal Cycle
WP = (h4-h3) KJ/kg
WP = ms(h4-h3) KW
Actual Cycle
WP' = (h4'-h3) KJ/kg
WP' = ms(h4'-h3) KW
Pump Efficiency
epump 
Wp
Wp'
x 100%
D. System: Boiler or Steam Generator
Ideal Cycle
QA =(h1-h4) KJ/kg
QA =ms(h1-h4) KW
Actual Cycle
QA =(h1-h4') KJ/kg
QA = ms(h1-h4') KW
Boiler or Steam Generator Efficiency
eBoiler 
QA
x 100%
Qs
QA = Heat absorbed by boiler or steam generator, KW
QS = Heat supplied to boiler or steam generator, KW
QA = ms(h1-h4) KW (for the ideal cycle)
QA = ms(h1-h4') KW (for the actual cycle)
If the heat supplied to boiler came from the burning of fuel or products of combustion, the heat supplied Q S is:
QS = mf(HV) KW
where: mf - is the fuel consumption in kg/sec
HV - heating value of fuel in KJ/kg
Heating Value (HV) - is the energy released by fuel when it is completely burned and the products of combustion are cooled to the
original fuel temperature.
STEAM RATE
SR 
Steam Flow Rate
kg
Power (KW) Produced KW - hr
HR 
Heat Supplied
KJ
Power (KW) Produced KW - hr
HEAT RATE
103
Reheat Cycle
For a steam power plant operating on a reheat cycle, after partial expansion of the steam in the turbine, the steam flows back to a
heat exchanger called the "re-heater" to heat the steam almost the same to its initial temperature.
104
Regenerative Cycle
For a regenerative cycle steam power plant some of the steam were taken from the turbine before it reached the condenser and used
to heat the feed-water with no extra energy input and the latent heat of vaporization would not be lost from the system in the
condenser. Ideally the water leaving the feed-water heater is saturated liquid at the heater pressure and the heater is a direct-contact
type. If the system uses a shell-and-tube heat exchanger the extracted steam doesn't mixed with the feed-water and the drains is send
back to the previous heat exchanger, also a separate drain pump may be employed.
m - fraction of steam bleed for feed-water heating, kg of bled steam/kg of throttled steam
ms - steam flow rate, kg/sec
Reheat-regenerative Cycle
For a reheat-regenerative cycle, further increase in thermal efficiency will occur because this cycle uses the combined
effect of reheating the steam after partial expansion and extracting some part of the circulating steam for feed-water heating, which
result to an increase in turbine work and reduces heat required by the boiler.
105
Steam Generator (boiler) – an integrated assembly of several components which converts water into steam at a pre-determined
pressure and temperature.
Steam turbine – is a device used to convert the kinetic energy of the substance into mechanical energy/work.
A.C. Generator – an electrical device used to convert the mechanical energy into electrical energy.
Condenser – a heat exchanger used to condense the exhaust steam from the turbine for recycling.
Pump – a mechanical device used to raise the condensate pressure to that of steam generator pressure.
Heat exchangers – are auxiliary equipment used to assist in steam plant operation.
Super heater (closed) – use to superheat the saturated steam from steam drum.
De-superheater (closed or open) –to de-superheat the superheated steam.to final state.
Reheater (closed) - used to reheat the steam from turbine after expansion.
Condenser (closed or open) – used to condemns the exhaust stream.
Feedwater heater (closed or open) – used to preheat the condensate/feedwater.
Economizer (closed) – used to convert the sub-cooled liquid to saturated liquid.
Air pre-heater (closed) - used to preheat the air for combustion.
106
Example no. 1
A 15,000 KW turbine has a guaranteed steam rate of 5.4 kg/KW-hr at rated load. Steam enters the turbine at 4.15 MPa and 400C,
with exhaust at 25.4 mm Hg absolute. Condensate is sub-cooled 3C, radiation and frictional losses in the turbine is estimated to be
3% of the turbine output. Calculate:
a) the enthalpy of exhaust steam
b) the total heat removed by the condenser in KJ/hr
c) the thermal efficiency
d) m3/min of condenser cooling water based on 6C temperature rise
P1 = 4150 KPa
P2 = P3 = 25.4 mm Hg absolute
P2 = P3 = 3.39 KPa (tsat at 3.39 KPa = 26.16C)
At P1 = 4150 KPa and t1 = 400C
h1= 3211.5 KJ/kg ; S1 = 6.7526 Kj/kg-K
h3 = h at 3.39 KPa and t3 = (26.16 – 3) = 23.16C
h3 = 97.016C ; S3 = 0.3408 KJ/kg-K
h4 = h at 4150 KPa and S4 = 0.3408 KJ/kg-K
h4 = 101.18 KJ/kg
T
1
4
26.16
C
23.16
C
P1 = P4 = 4150
KPa
P2 = P3 = 3.39
KPa
2
2’
3
S
Steam flow rate
KW produced
m s  5.4(15,000)  81,000 kg/hr
SR 
m s  22.5 kg/sec
Wt'  15,000(1- .03)  14,550 KW
Wt'  m s h1 - h 2' 
h 2'  h1 
WT '
 2564.83 KJ/kg
ms
Q R  m s (h 2' - h 3 )  22.5(2564.83 - 97.016)  55,525.82 KW
Q R  m w (C pw )(t wB - t wA )
55,525.82  m w (4.187)(6)
m w  2,210.3 kg/sec
Q w - volume flow rate of water in m 3 /min
 m3 
m3
(60)  132.6
Q w  2210.3 kg/sec 
min
 1000 kg 
WP  m s (h 4  h 3 )  93.7 KW
Q A  m s (h1  h 4 )  69,982.2 KW
e
WT ' - Wp
QA
x 100%
e  20.7%
107
ExampleNo. 2
A 2500 KW steam turbo generator set has a combined steam rate of 7 kg/KW-hr with a throttle steam at 3.4 MPa (tsat = 240.87C)
and 400C and a condenser pressure 8.5 KPa absolute(tsat = 42.69C). Feedwater enters the boiler at 138C. The steam generator
supplying the steam is coal fired, with a coal heating value of 24,000 KJ/kg. Overall steam generator efficiency is 74%. Assuming
no losses, Determine:
a) the Rankine cycle efficiency
b) the brake thermal efficiency if generator efficiency is 94%
c) the combined thermal efficiency
d) the fuel consumption in kg/hr
e) the enthalpy of exhaust steam if brake power is equal to actual turbine work
T
1
400C
240.87C
138C
4’
P1 = P4 = 3400 KPa
4
42.69C
3
P2 = P3 = 8.5 KPa
2
2’
S
At 3400 KPa and 400C
h1 = 3224.6 KJ/kg
S1 = 6.8620 KJ/kg-K
At 8.5 KPa and S2 = S1 = 6.8620 KJ/kg-K
h2 = 2153.4 KJ/kg ; x2 = 82.302%
At 8.5 KPa; saturated liquid,
h3 = 178.63 KJ/kg ; S3 = 0.6078 KJ/kg-K
At S3 = S4 to 3400 KPa
h4 = 182.04 KJ/kg
At 3400 KPa and 138C
h4’ = 582.22 KJ/kg
2,659.6
 x100%
17,496.84
e B  15.2%
eB 
2,500
x100%  14.3%
17,496.84
Q
17,496.84
kg
mF  S 
(3600)  2624.526
HV
24,000
hr
Wt '  WB  ms (h1  h 2' )
e combined 
h 2'  3224.6 
2,659.6
KJ
 2681.824
4. 9
kg
108
kg
kg
 4. 9
hr
sec
(h1  h 2)  (h 4  h 3)
e
x100%
(h1  h 4)
e  35.07%
m s  7(2500)  17,500
eg 
Wo
x 100%
WB
0.94 
2500
WB
2500
 2,659.6 KW
0.94
W
e B  B x100%
QS
WB 
Q s  m F (HV )
e Boiler 
QA
x100%
QS
Q A  m s (h1  h 4' )  4.9(3224.6  582.22)
Q A  12,947.662 KW
QS 
12,947.662
 17,496.84 KW
0.74
Example No. 3
An ideal steam power plant operating on single stage reheat cycle has steam conditions at throttle of 7.6 MPa and 500C and
expands to the turbine to 2 MPa after which the steam is withdrawn and is reheated in the reheater to 450C and re-expands again
in the turbine to the condenser at a pressure of 7 KPa. For a steam flow rate of 10 kg/sec. determine
The ideal turbine work in KW
The ideal pump work in KW
The net cycle work in KW
The ideal thermal efficiency
The ideal steam rate in kg/KW-hr
The heat rate in KJ/KW-hr
T
1
3
P1 = P6
6
P2 = P3
2
P4 = P5
5
4
S
Steam Properties
At P1 = 7600 KPa and t1 = 500C
h1 = 3402.7 KJ/kg ; S1 = 6.7482 KJ/kg-K
At S1 = S2 to P2 = 2000 KPa
h2 = 3012.7 KJ/kg
At P3 = 2000 KPa and t3 = 450C
h3 = 3357.9 KJ/kg ; S3 = 7.2900 KJ/kg-K
At S3 = S4 to P4 = 7 KPa
h4 = 2264.0 KJ/kg ; x4 = 87.2311%
At P5 = P4 saturated liquid
h5 = 163.33 KJ/kg; S5 = 0.5590 KJ/kg-K
At S5 = S6 to P6 = P1
a. Wt  m s (h 1 - h 2 )  (h 3 - h 4 ) KW
b. Wp  m s (h 6 - h 5 ) KW
c. W  Wt - Wp KW
d. e 
W
x 100%
QA
Q A  m s (h 1 - h 6 ) KW
3600ms
kg
W
KW - HR
3600QA
KJ
f . HR 
W
KW - hr
e. SR 
109
h6 = 170.97 KJ/kg
mS = 10 kg/sec
T
Additional work
S
Example No. 4
A steam power plant operates on an ideal reheat– regenerative Rankine cycle and has a net power output of 80 MW. Steam enters
the high-pressure turbine at 10 MPa and 550°C and leaves at 0.8 MPa. Some steam is extracted at this pressure to heat the feedwater
in an open feed-water heater. The rest of the steam is reheated to 500°C and is expanded in the low-pressure turbine to the condenser
pressure of 10 KPa. Show the cycle on a T-s diagram with respect to saturation lines, and determine (a) the mass flow rate of steam
through the boiler and (b) the thermal efficiency of the cycle.
Example No. 5
A steam power plant is operating on a reheat – regenerative cycle with 2 – stages of regenerative heating and 1 – extraction for
reheating. The turbine receives steam at 7000 KPa and 550C, the steam expands to 2000 KPa isentropically where a fraction is
extracted for feedwater heating and the remainder is reheated at constant pressure until the temperature is 540C. The steam expands
isentropically to 400 KPa where a fraction is withdrawn for LP feedwater heating on a closed type heater as shown on figure; the
remainder expands isentropically in the turbine to 7.5 KPa, at which point it enters the condenser. For a mass flow rate of 10
kg/secDetermine
a) The mass fractions of extracted steam
b) The turbine work in KW
c) The pump work in KW
d) The net cycle work in KWThe heat added in KJ/hr
f) The heat rejected from the condenser in KJ/hr
110
Wt  ms (h1  h 2 )  (1  m1 )(h 3  h 4 )  (1  m1  m 2 )(h 4  h 5  KW
Wp1  ms ( h12  h11)
Wp 2  m s (1  m1 )( h 8  h 6 )
ΣWp  Wp1  Wp 2
W  Wt  ΣWp
QA  ms (h1  h10 )  (1  m1 )( h 3  h 2 ) kw
e
W
W
x 100%e 
x 100%
QA
QA
QR  ms (1 - m1 - m 2 )(h 5 - h 6 ) KW
SAMPLE PROBLEMS
Rerheat Cycle
Steam power plant that operates on a reheat Rakine cycle with 80 MW of net output. Steam enters the high pressure turbine at 10
MPa and 500 °C, and is reheated to and enters the low pressure turbine at 1 MPa and 500 °C. Steam leaves the condenser as a
saturated liquid at 10 kPa. Isentropic efficiencies of the turbine and compressor are 80 percent and 95 percent, respectively. Find:
a) quality or temperature of steam at turbine exit
b) thermal efficiency of the cycle
c) mass flow rate of the steam
d) condenser cooling watert flow rate allowing a water temperature rise of 15C
e) cycle steam rate kg/KW-hr
f} cycle heat rate KJ/KW - hr
111
Reheat – Regenerative Cycle
Ideal reheat-regenerative Rankine cycle with one open feedwater heater. The boiler pressure is 10 MPa, condenser pressure is 15
kPa, reheater pressure is 1 MPa, and feedwater pressure is 0.6 MPa. Steam enters the high and low pressure turbines at 500 °C.
Find:
a) fraction of steam extracted for regeneration
b) thermal efficiency of the cycle.
Example 3
Calculate the power plant efficiency and the net work for a steam power plant that has turbine inlet conditions of 6 MPa and 500C,
bleed steam to the heater occurring at 800 KPa and exhausted to the condenser at 15 KPa. The turbine and pump have an internal
efficiencies of 90%. The flow rate is 63 kg/sec. (e = 35.6% ; W = 60,511 KW)
A steam power plant operates using the reheat Rankine cycle. Steam enters the high pressure turbine at 12.5 MPa and 550ºC at a
rate of 7.7 kg/s and leaves at 2 MPa. Steam is then reheated at a constant pressure to 450ºC before it expands in the low pressure
turbine. The isentropic efficiencies of the turbine and the pump are 85% and 90%, respectively. Steam leaves the condenser as a
saturated liquid at 150 KPa. Determine
(a) the net power output, and
(b) the thermal efficiency.
1
T
(ms-m)
3
8'
8
ms
(ms-m)
2'
7
m
2
6'
6
(ms-m)
(ms-m)
5
4'
4
S
P = 12.5 Mpa ; t = 550 C
h1
s1
3473.7
6.6250
P = 2000 Kpa ; S = 6.6250
h2
t2
tsat
2936.9
263.9
212.36
P = 2000 Kpa ; t = 450 C
h3
s3
3357
7.2900
P = 96 Kpa ; S = 7.2900
h4
tsat
2631
96.712
P =96 Kpa ; Sat. Liquid
h5
s5
412.52
1.2896
P = 2000 Kpa ; s = 1.2896
112
h6
t6
414.51
98.629
P = 2000 Kpa ; Saturatted liq.
h7
s7
908.02
2.4465
P = 12500 Kpa ; s = 2.4465
h8
tsat
920.35
214.14
0.85
0.9
h2'
3017.42
h4'
2739.9
h6'
414.7311
h8'
921.72
A coal fired steam power plant produces 150 MW of electric power with a thermal efficiency of 35%. If the energetic efficiency of
the boiler is 75%, heating value of coal is 30 MJ/kg, and the temperature rise of the cooling water in the condenser is 10oC, determine
(a) the fuel consumption rate in kg/h, and (b) mass flow rate of cooling water.
A steam power plant operates on an ideal Rankine cycle. Superheated steam flows into the turbine at 2 MPa and 500oC with a flow
rate of 100 kg/s and exits the condenser at 50oC as saturated water. Determine (a) the net power output, (b) the thermal efficiency,
and (c) the quality of steam at the turbine exit. (d) What would the efficiency be if the condenser temperature can be lowered to
30oC.
A power plant operates on a simple Rankine cycle producing a net power of 100 MW. The turbine inlet conditions are 15 MPa and
600oC and the condenser pressure is 10 kPa. If the turbine and pump each has an isentropic efficiency of 85% and there is a 5%
pressure drop in the boiler, determine (a) the thermal efficiency, (b) the mass flow rate of steam in kg/h, and (c) the back work
ratio. How sensitive is the thermal efficiency on (d) boiler pressure drop, (e) pump efficiency, and (f) turbine efficiency?
A steam power plant operates on an ideal reheat-regenerative Rankine cycle with two feedwater heaters, one open and one closed.
Steam enters the first turbine at 15 MPa and 500oC, expands to 1 MPa, and is reheated to 450oC before it enters the second turbine
where it expands to 10 kPa. Steam is extracted from the first turbine at 3 MPa and sent to the closed FWH, where the feedwater
leaves at 5oC below the temperature at which the saturated condensate leaves. The condensate is fed through the trap to the open
FWH, which operates at 0.5 MPa. Steam is extracted at 0.5 MPa from the second turbine and is fed to the open FWH. The flow out
of the open FWH is saturated liquid at 0.3 MPa. If the power output of the cycle is 150 MW, determine (a) the thermal efficiency,
(b) the mass flow rate through the boiler, (c) the bleeding rate from the first turbine, and (d) the bleeding rate from the second
turbine. What would the thermal efficiency be if (e) the closed FWH were eliminated, (f) both the FWH's were eliminated?
(a) 44.8%, (b) 116.13 kg/s, (c) ,(d) , (e) 43.8%, (f) 41.4%
The cogeneration plant has a peak capacity of 25 MW for process heating. Steam leaves the boiler at 8 MPa and 600oC. The process
heater and the condenser operate at a pressure of 400 kPa and 8 kPa respectively. Assuming ideal behavior of each component,
determine (a) the mass flow rate of steam through the boiler and (b) the maximum turbine output. To meet a certain load condition,
10% of the steam is diverted to the expansion valve, and half the steam flowing through the turbine is extracted. Determine (c) the
process heat load, and (d) the utilization factor.
ANSWER: (a) 8.23 kg/s, (b) 11.9 MW, (c) 10.61 MW, (d) 71.8%,
113
A combined gas turbine-steam power plant has a net power output of 500 MW. Air enters the compressor of the gas turbine at 100
kPa, 300 K, and has a compression ratio of 12 and an isentropic efficiency of 85%. The turbine has an isentropic efficiency of 90%,
inlet conditions of 1200 kPa and 1400 K, and an exit pressure of 100 kPa. The air from the turbine exhaust passes through a heat
exchanger and exits at 400 K. On the steam turbine side, steam at 8 MPa, 400oC enters the turbine, which has an isentropic efficiency
of 85%, and expands to the condenser pressure of 8 kPa. Saturated water at 8 kPa is circulated back to the heat exchanger by a pump
with an isentropic efficiency of 80%. Determine (a) the ratio of mass flow rates in the two cycles, (b) the mass flow rate of air, and
(c) the thermal efficiency. (d) What would the thermal efficiency be if the turbine inlet temperature increased to 1600 K?
ANSWER: (a) 6.673, (b) 1107 kg/s, (c) 53.1%, (d) 54.8%
YURI G. MELLIZA
114
PRACTICE EXERCISES
Problem 1
Consider a steam power plant that operates on a simple ideal Rankine cycle and has a net power output of 45MW. Steam enters the
turbine at 7 MPa and 500°C and is cooled in the condenser at a pressure of 10 kPa by running cooling water from a lake through the
tubes of the condenser at a rate of 2000 kg/s. Show the cycle on a T-s diagram with respect to saturation lines, and determine (a) the
thermal efficiency of the cycle, (b) the mass flow rate of the steam, and (c) the temperature rise of the cooling water.
State
1
2
3
4
P (KPa)
7000
10
10
7000
T(°C)
500
45.83
45.999
S
6.7940
6.7940
0.6492
0.6492
h
3409.90
2151.30
191.76
198.81
x
(Quality)
81.924
Problem 2
In a regenerative steam cycle as shown below employing two closed feedwater heaters, the steam is supplied to the turbine at 4000
KPa and 500C and is exhausted to the condenser at 3.5 KPa. The intermediate bleed pressures are obtained such that the saturation
pressure intervals are approximately equal, giving pressures of 1000 KPa and 110 KPa, respectively. If the plant to produces a net
power output of 10 MW, Calculate the following
The hourly amount of bleed steam at each stage (m s1 and ms2)
The cycle thermal efficiency
Note: Draw the matching TS Diagram & assume all liquids leaving bleed steam heaters is saturated liquid at the heater pressure.
Schematic Diagram
Answer
ms1
4,653.06 Kg/hr
ms2
4,041.48 Kg/hr
e
42.5 %
115
Problem 3
A steam power plant operates using the reheat Rankine cycle. Steam enters the high pressure turbine at 12.5 MPa and 550ºC at a
rate of 7.7 kg/s and leaves at 2 MPa. Steam is then reheated at a constant pressure to 450ºC before it expands in the low pressure
turbine. The isentropic efficiencies of the turbine and the pump are 85% and 90%, respectively. Steam leaves the condenser as a
saturated liquid. If the moisture content of the steam at the exit of the turbine is not to exceed 5% determine (a) the condenser
pressure, (b) the net power output, and (c) the thermal efficiency.
Problem 4
A steam power plant operates on an ideal Rankine cycle. Superheated steam flows into the turbine at 2 MPa and 500 oC with a flow
rate of 100 kg/s and exits the condenser at 50oC as saturated water. Determine
the net power output,
the thermal efficiency, and
the quality of steam at the turbine exit.
What would the efficiency be if the condenser temperature can be lowered to 30 oC?
Problem 5
A steam power plant operates on an ideal reheat-regenerative Rankine cycle with two feedwater heaters, one open and one closed.
Steam enters the first turbine at 15 MPa and 500oC, expands to 1 MPa, and is reheated to 450oC before it enters the second turbine
where it expands to 10 kPa. Steam is extracted from the first turbine at 3 MPa and sent to the closed FWH, where the feedwater
leaves at 5oC below the temperature at which the saturated condensate leaves. The condensate is fed through the trap to the open
FWH, which operates at 0.5 MPa. Steam is extracted at 0.5 MPa from the second turbine and is fed to the open FWH. The flow out
of the open FWH is saturated liquid at 0.3 MPa. If the power output of the cycle is 150 MW, determine
the thermal efficiency,
the mass flow rate through the boiler,
the bleeding rate from the first turbine, and
the bleeding rate from the second turbine.
Problem 6
Consider a regenerative vapor power cycle with one open feedwater heater. Steam enters the turbine at 8.0 MPa, 480?C and
expands to 0.7 MPa, where some of the steam is extracted and diverted to the open feedwater heater operating at 0.7 MPa.
The remaining steam expands through the second-stage turbine to the condenser pressure of 0.008 MPa. Saturated liquid exits
the open feedwater heater at 0.7 MPa. The isentropic efficiency of each turbine stage is 85% and each pump operates isentropically. If the net power output of the cycle is 100 MW, determine
the thermal efficiency and
the mass flow rate of steam entering the first turbine stage, in kg/h.
Problem no. 7
A coal fired steam power plant produces 150 MW of electric power with a thermal efficiency of 35%. If the energetic efficiency of
the boiler is 75%, heating value of coal is 30 MJ/kg, and the temperature rise of the cooling water in the condenser is 10oC, determine
(a) the fuel consumption rate in kg/h, and (b) mass flow rate of cooling water.
(a) 68.57 kg/h , (b) 23,969 kg/h
116
Multiple Choice Problems
In a Rankine cycle, saturated liquid water at 1 bar is compressed isentropically to 150 bar. First by reheating in a boiler and then by
superheating at constant pressure of 150 bar, the water substance is brought to 750K. After adiabatic reversible expansion in a
turbine to 1 bar, it is then cooled in a condenser to a saturated liquid. How much work is generated in the turbine? (Steam properties
h, kJ/kg, s, kJ/kg-K: @ 150 bar&750 K, h = 3240.5, s1 = 6.2549; @ 1 bar, hf=417.46, hfg=2258, sf=1.3026, sfg=6.0568)
769.9 b. 796.9
c .967.9
d.976.9
A reheat steam has 13850 kPa throttle pressure at the turbine inlet and 2800 kPa reheat pressure. The throttle and reheat temperature
of the steam is 540oC, condenser pressure is 3.4 kPa, engine efficiency of high pressure and low pressure is 75%. Find the cycle
thermal efficiency.
34.46% b. 35.56
c. 36.66
d. 37.76
In a Rankine cycle, steam enters the turbine at 2.5 MPa and condenser of 50 kPa. What is the thermal efficiency of the cycle in
percent?
25.55 b. 28.87
c. 30.12
d. 31.79
A supercritical steam Rankine cycle has turbine inlet conditions of 17.5 MPa and 530oC expands in a turbine to 7 kPa. The turbine
and pump polytropic efficiencies are 0.9 and 0.7, respectively. Pressure losses between pump and turbine inlet are 1.5 MPa. What
should be the pump work in kJ/kg.
27.13 b. 29.87
c. 32.47
d. 33.25
Steam enters the superheater of a boiler at a pressure of 2.5 MPa and dryness of 0.98 and leaves at the same pressure at a temperature
of 370oC. Calculate the heat energy supplied per kg of steam supplied in the superheater.
405.51 b. 504.15
c. 154.15
d. 245.25.
A back pressure steam turbine of 100 MW capacity serves as a prime mover in a cogeneration system. The boiler admits the return
water at a temperature of 66oC and produces the steam at 6.5 MPa and 455oC. Steam then enters a back pressure turbine and expands
to the pressure of the process, which is 0.52 MPa. Assuming a boiler efficiency of 80% and neglecting the effect of pumping and
the pressure drops at various location, what is the incremental heat rate for electric? The following enthalpies have been found: at
turbine entrance = 3306.8 kJ/kg, exit = 2700.8 kJ/kg; boiler entrance = 276.23 kJ/kg, exit = 3306.8 kJ/kg)
22,504.23 kJ/kW-hr
b. 52,244.32 kJ/kW-hr
c. 12,435.72kJ/kW-hr
d. 32,234.82 kJ/kW-hr
In an open feedwater for a steam power plant, saturated steam at 7 bar is mixed with sub-cooled liquid at 7 bar and 25oC. Just enough
steam is supplied to ensure that the mixed steam leaving the heater will be saturated liquid at 7 bar when heater efficiency is 90%.
Calculate the mass flow rate of sub cooled liquid if steam flow rate is 0.865 kg/s. (Steam properties h, kJ/kg, @ 7 bar, h g = 2763.5,
hf = 697.22; @ 7 bar & 25oC, hf= 105.5)
2.725 b. 2.286
c. 3.356
d. 3.948
A steam plant operates with an initial pressure of 1.7 MPa and 370oC temperature and exhaust to a heating system at 0.17 MPa. The
condensate from the heating system is returned to the boiler at 65.5oC and the heating system utilizes from its intended purpose 90%
of the energy transferred from the steam it receives. The turbine efficiency is 70%. If the boiler efficiency is 80%, what is the
cogeneration efficiency of the system in percent? Neglect pump work. (Steam properties h, kJ/kg, s, kJ/kg-K: @ 1.7 MPa & 370oC;
h = 3787.1, s = 7.1081; @ 1.7 MPa, h f= 483.20, hfg= 2216.0, sf= 1.4752, sfg= 5.7062; @ 65oC, h f=274.14)
69
b. 78
c. 91.24
d. 102.10
In a cogeneration plant, steam enters the turbine at 4 MPa and 400oC. One fourth of the steam is extracted from the turbine at 600kPa
pressure for process heating. The remaining steam continues to expand to 10 kPa. The extracted steam is then condensed and mixed
with feedwater at constant pressure and the mixture is pumped to the boiler pressure of 4 MPa. The mass flow rate of the steam
through the boiler is 30 kg/s. Disregarding any pressure drops and heat losses in the piping, and assuming the turbine and pump to
be isentropic, how much process heat is required in kW? (Steam properties h, kJ/kg, s, kJ/kg-K: @ 4 MPa & 400oC, h = 3213.6 s =
6.7690; @ 600 kPa, h f= 670.56, hfg= 2086.3, sf= 1.9312, sfg= 4.8288)
1,026.90 b. 2,468.2
c. 3,578.5
d. 15,646.8
A 23.5 kg/s at 5 MPa and 400oC is produced by a steam generator. The feedwater enters economizer at 145 oC and leaves at 205oC.
The steam leaves the boiler drum with a quality of 98%. The unit consumes 2.75 kg of coal per second as received having an heating
value of 25,102 kJ/kg. What would be the overall efficiency of the unit in percent? (Steam properties h, kJ/kg, s, kJ/kg-K: @ 5 MPa
& 400oC, h=3195.7; @ 0 MPa, h f= 1154.23, h fg= 1640.1; @ 205oC , hf= 610.63)
65
b. 78
c. 88
d. 95
A coal-fired power plant has a turbine-generator rated at 1000 MW gross. The plant required about 9% of this power for its internal
operations. It uses 9800 tons of coal per day. The coal has a heating value of 6,388.9 kCal/kg, and the steam generator efficiency
is 86%. What is the net station efficiency of the plant in percent?
117
30.12
b. 33.07
c. 36.74
d. 40.01
Steam enters the turbine of a cogeneration plant at 7 MPa and 500C. Steam at a flow rate of 7.6 kg/s is extracted from the turbine
at 600 kPa pressure for process heating. The remaining steam continues to expand to 10 kPa. The recovered condensates are pumped
back to the boiler. The mass flow rate of steam that enters the turbine is 30 kg/s. Calculate the cogeneration efficiency in percent.
(Steam properties h, kJ/kg, s, kJ/kg-K: @ 7 MPa & 500C, h = 3410.3 s = 6.7975; @ 600 kPa, h f= 670.56, hfg= 2086.3, sf= 1.9312,
sfg= 4.8228; @ 10 kPa, h f= 191.83, h fg= 2392.8, sf= 0.6493, sfg= 7.5009)
50
b. 55
c. 60
d. 65
A 60 MW turbine generator running at 3600 rpm receives steam at 4.0 MPa and 450oC with back pressure of 10 kPa. Engine
efficiency is 78% and the combined mechanical and electrical efficiency is 95%. What would be the exhaust enthalpy of the steam
in kJ/kg.
2,400.12 kJ/kg
b. 20,432.10 kJ/kg
c. 28,124.20 kJ/kg
d. 30,101.15 kJ/kg
Steam enters a throttling calorimeter at a pressure of 1.03 MPa. The calorimeter downstream pressure and temperature are
respectively 0.100 MPa and 125oC. What is the percentage moisture of the supply steam? (Steam properties h, kJ/kg, s, kJ/kg-K:
@1.03 MPa, hfg = 2010.7, hg = 2779.25; @ 0.1 MPa & 125oC, h=2726.6)
1.98
b. 2.62
c.3.15
d. 5.21
Steam expands adiabatically in a turbine from 2 MPa, 400oC to 400 kPa, 250oC. What is the effectiveness of the process in percent
assuming an atmospheric temperature of 15oC. Neglect changes in kinetic and potential energy. (Steam properties h, kJ/kg, s, kJ/kgK: @ 2.0 MPa and 400oC; h = 3247.6 s = 7.1271; @ 400 kPa & 250oC, h= 2964.2, s= 7.3789)
79.62 b. 84.52
c. 82.45
d. 74.57
A drum containing steam with 2.5 m in diameter is 7.5 m long. Of the total volume, 1/3 contains saturated steam at 800 kPa and the
other 2/3 contains saturated water. If this tank should explode, how much water would evaporate? Consider the process to be of
constant enthalpy. (Steam properties h, kJ/kg, v, m3/kg @0.8 MPa, hf = 721.11, hg = 2769.1, vf= 0.0011148, vg=0.2404; @
0.101325 MPa & 100oC, hf=419.04, hg=2676.1, vf=0.0010435, vg=2769.1)
a. 2,123.76 kg
b. 2,424.62 kg
c. 2,651.24 kg
d. 2,948.11 kg
A Batangas base industrial company operates a steam power plant with reheat and regeneration. The steam enters a turbine at 300
bar and 900 K and expands to 1 bar. Steam leaves the first stage at 30 bar and part of it entering a closed heater while the rest
reheated to 800K. Both section of the turbine have adiabatic efficiency of 93%. A condensate pump exists between the main
condenser and the heater. Another pump lies between the heater and condensate outlet line from the heater (condensed extracted
steam). Compute for the extracted fraction of the total mass flow to the heater.
a. 0.234
b. 0.543
c. 0.765
d. 0.485
BOILER OR STEAM GENERATOR
A steam generator is a complex combination of economizer, evaporator(Boiler), superheater, reheater and air pre-heater. In addition
it has various auxiliaries, such as stokers, pulverizers, burners, fans, emission control equipment, stack and ash handling equipment.
A boiler is that portion in the steam generator where saturated liquid is converted to saturated steam, although it may be difficult to
separate it, physically, from the economizer. The term “Boiler” is often used to mean the whole steam generator in the literature,
however. Steam generators are classified in different ways. They may, for example, be classified as either (1) Utility or (2) Industrial
steam generators.
Utility Steam Generators are those used by utilities for electric – power generating plants. Modern utility steam generators essentially
are of two basic kinds: (1) the subcritical water-tube drum type and (2) the supercritical once-through type. The supercritical units
operate at about 24 Mpa and higher, above the steam critical pressure of 22.11 Mpa. The subcritical drum group usually operate at
either 13 Mpa gage or 18 Mpag. The steam capacities of modern utility steam generators are high ranging from 125 to 1250 kg/sec.
They power electric power plants with an output ranging from 125 to 1300 Megawatts.
Industrial Steam Generators, on the other hand are those used by industrial and institutional concerns and are of many types. These
include the water-tube pulverized-coal units similar to those used by utilities. Some are heat recovery types that use the waste heat
from other industrial processes. They may be of the fire-tube variety . Industrial steam generators usually do not produced
superheated steam. Rather, they usually produced saturated steam, or even only hot water (in which case they should not be called
as steam generators). They operate at pressures ranging from a few Mpa to as much as 10.5 Mpag and steam or hot water capacities
ranging from a few kg/sec to 125 kg/sec.
Fossil-fueled steam generators are more broadly classified [12,13] as those having the following components or characteristics.
 Fire-tube boilers
 Water-tube boilers
 Natural-circulation boilers
 Controlled-circulation boilers
118



Once-through flow
Subcritical pressure
Supercritical pressure
FIRE-TUBE BOILER
Fire tube boilers have been used in various early forms to produce steam for industrial purposes at the upper limits of 1.8 Mpag
pressure and 6.3 kg/sec capacity. Although their size has increased, their general design has not change appreciably in the past 25
years.
Fire-tube boiler is a special form of the shell-type boiler. A shell-type boiler is a closed, usually cylindrical, vessel or shell that
contains water. A portion of the shell such as its underside is simply exposed to heat, such as gases from an externally fired flame.
The shell boiler evolve into more modern forms such as the electric boiler, in which heat is supplied by electrodes embedded in the
water, or the accumulator, in which heat is supplied by steam from an outside source passing through tubes within the shell. In both
cases the shell itself is no longer exposed to heat.
The shell boiler evolves into the fire-tube boiler. Hot gases, instead of steam, were now made to pass through the tubes. Because of
improved heat transfer the fire-tube boiler is more efficient than the original shell boiler and can reach efficiencies of about 70
percent. The fire tubes were placed in horizontal, vertical or inclined positions. The most common was the horizontal-tube boiler,
in which the furnace and grates are located underneath the front end of the shell. The gases pass horizontally along its underside to
the rear, reverse direction, and pass through the horizontal tubes to the stack at the front.
There are two types of fire-tube boilers: (1) the fir-box and (2) the scotch marine. In the fire-box boiler, the furnace or the fire box,
is located within the shell, together with the fire tubes. In the scotch marine boiler, combustion takes place within one or more
cylindrical chambers that are usually situated inside and near the bottom of the main shell. The gases leave these chambers at the
rear, reverse direction, and return through the fire tubes to the front and out through the stack. Scotch marine boilers are usually
specified with liquid or gas fuels.
Because boiling occurs in the same compartment where water is, fire-tube boilers are limited to saturated-steam production. They
are presently confined to relatively small capacities and low steam pressures, such as supplying steam for space heating and in
decreasing numbers, for railroad locomotive service. The largest scotch marine offered in the United States today is rated at 2000
Boiler HP, containing two combustion chambers within a 13-ft diameter, 30-ft long shell.
WATER TUBE BOILER
Demands for increased capacity and pressure over that economically obtained from the fire-tube unit led to the development of the
water-tube boiler. Although available for smaller capacities, the field of the water-tube boiler begins at about 2 kg/sec of steam. The
water tube boiler puts the pressure instead in tubes and relatively small diameter drums that are capable of withstanding the extreme
pressures of the modern steam generators. In general appearance, the early water-tube boiler looks much like the fire tube-boiler
except that the higher pressure water were inside the tubes and the combustion gases were on the outside. The water-tube boiler
went through several stages of development.
119
Straight-Tube Boiler: The 1st water-tube boiler was the straight-tube boiler, in which straight tubes, 3 to 4 in OD, inclined at about
15 and staggered with 7 to 8 in spacing, connected two vertical headers. One header was a downcomer, or downtake, which is
supplied nearly saturated water to the tubes. The water partially boiled in the tubes. The other header was a riser, or uptake, which
received the water-steam mixture. The water density in the downcomer was larger than the two-phase density in the riser, which
caused natural circulation in a clockwise direction. As capacity increased, more than one header each and more than one tube “deck”
were used. The two-phase mixture went into the upper drum that was arranged either parallel to the tubes (the longitudinal drum)
or perpendicular to them (the cross drum). These drums received the feedwater from the last feedwater heater and supplied saturated
steam to the super-heater through a steam separator within the drum which separated steam from the bubbling water. The lower end
of the downcomer was connected to the mud drum, which collected sediments from the circulating water.
A single longitudinal drum, usually 1.2 m in diameter, can allow only a limited number of tubes and hence a limited heating surface.
Longitudinal-drum boilers were built with one more or more than one parallel drums, depending upon capacity. They were built
with heating surfaces of 93 to 930 m2 and were limited to low pressures of 1.2 Mpa to 2.3 Mpa and steam capacities from 0.63 to
10 kg/sec.
Cross-drum boilers, because of geometry, could accommodate many more tubes than longitudinal-drum boilers and were built with
heating surfaces of 93 to 2300 m2, pressures of 1.2 Mpa to 10 Mpa, and steam capacities of 0.63 to 63 kg/sec.
Baffles were installed across the tubes in both kinds to allow for up to three gas passes to ensure maximum exposure of the tubes to
the hot combustion gases and minimal gas dead spots.
The Bent-Tube Boiler: There were many versions of the bent-tube boiler. In general, a bent-tube boiler used bent, rather than straight
tubes between several drums or drum and headers. The tubes were bent so that they entered and left the drums radially. The number
of drums usually varied from two to four. Gas baffles were installed to allow for one or more gas passages.
Water-Tube Boilers: Recent Developments
The advent of the water-cooled furnace walls, called water walls, eventually led to the integration of furnace, economizer, boiler,
superheater, reheater, and air preheater into the modern steam generator. Water-cooling is also used for superheater and economizer
compartment walls and various other components, such as screens, dividing walls, etc. The use of a large number of feedwater
heaters (up to seven or eight) means a smaller economizer, and the high pressure means a smaller boiler surface because of the latent
heat of vaporization decreases rapidly with pressure. Thus, a modern high-pressure steam generator requires more superheating and
reheating surface and less boiler surface than older units. Beyond about 10 Mpa, the water tubes represent the entire boiler surface
and no other tubes.
Water at 230C to 260C from the plant high-pressure feedwater heater enters the economizer and leaves saturated or as a twophase mixture of low quality. It then enters the steam drum at midpoint. Water from the steam drum flows through insulated
downcomers, which are situated outside the furnace, to a header. The header connects to the water tubes that line the furnace walls
and act as risers. Thw water in the tubes receives heat from the combustion gases and boils further. The density differential between
the water inn the downncomer and that in the water tubes helps circulation. Steam is separated from the bubbling water in the drum
and goes to the superheater and the high pressure section of the turbine. The exhaust from the turbine returns to the reheater, after
which it goes to the low-pressure section of the turbine.
Atmospheric air from a FD fan is preheated by the flue gases just before they exhausted to the atmosphere. From there it flows into
the furnace, where it mixes with the fuel and burn to some 1650C. The combustion gases impart portion of some of their energy to
the water tubes and then the superheater, reheater, and economizer, and leave latter at about 320C. From there they reheat the
incoming atmospheric air in the air preheater, leaving it about 150C. An ID fan draws the flue gases from the system and sends
them up the stack. The temperature of about 150C of the exiting flue gases represents an availability loss to the plant. This, however,
is deem acceptable because (1) the gas temperature should be kept well above the dew point of the water vapor in the gases to
prevent condensation which should form acids that will corrode metal components in its path and (2) the flue gases must have
enough buoyancy to rise in a high plume above the stack for proper atmospheric dispersion.
120
Boiler Walls
The water tubes that cool the water walls are closely spaced for maximum heat absorption. Tube construction has varied over the
years from bare tubes (a) tangent to or (2) embedded in the refractory, to (c) studded tubes, to the now-common membrane design
(d). The membrane design consists of tubes spaced on centers slightly wider than their diameter, connected by bars or membranes
welded to the tubes at their centerlines. The membranes acts as fins to increase the heat transfer as well as to afford a continuous
rigid and pressure-tight construction for the furnace. No additional inner casing is required to contain the combustion gases.
Insulation and metal lagging to protect it are provided on the outer side of the wall. One manufacturer has standardized its design
on 3-in diameter tubes on 3.75 in between centers, another on 3 in on 4 1n, and yet a third on 2.75 in on 3.75 in.
Radiant Boiler
Heat is transferred from the combustion gas to the water walls by both radiation and convection. A radiant boiler, as the name
implies, receives most of its heat by radiation.
The combustion gases have characteristics that depend upon the fuel used, the combustion process and the air-fuel ratio. They may
be luminous, i.e., emit all wavelengths and hence strong visible radiation if there are particulates such as soot particles during the
combustion process. This is the case with coal and oil. They may be nonluminous, in which case they burn cleanly without
particulates, as in the case of gaseous fuels. No combustion gases are truly nonluminous because the heavier gases in the combustion
products, in particular the triatomic CO2 nd H2O, but also SO2, ammonia and sulfur dioxide, are selective radiators that emit (and
absorbs) radiation in certain wavelengths, mostly outside the visible range. The portion of radiation within the visible range is small
but gives the combustion gases a green-blue appearance. Lighter gases such as monatomic or diatomic gases are poor radiators.
The radiant energy emitted by the combustion gases depends upon the gas temperature (to the fourth power), the partial pressure
of the individual constituents radiating gases, the shape and size of the gases, their proximity to the absorbing body, and the
temperature of that body (to the fourth power).
The convective portion of the heat transfer follows the usual Nusselt-Reynolds turbulent forced-convection relationship. It is smaller
than the radiant portion because a thick body of gas causes radiation, whereas convection is localized near the tube surface.
Heat received by the water walls is conducted through the membranes and tube walls and is then convected to the two-phase mixture
inside the tubes by forced convection nucleate-boiling heat transfer. The heat transfer resistance of the latter is much smaller than
the others so that it may be neglected in design calculations with little error.
Radiant boilers are designed for electric-generating stations to use coal or lignite for pulverized or cyclone furnace applications, oil
or natural gas. They are built to supply a wide range of steam pressures and temperatures, but usually around 540C and steam
capacities up to 1260 kg/sec. They are limited to subcritical pressures, usually 12.5 Mpag to 17 Mpag.
Boiler Auxiliaries and Accessories
Superheater – a heat exchanger that is used to increase the temperature of the water vapor greater than the saturation temperature
corresponding the boiler pressure.
Evaporator – a heat exchanger that changes saturated liquid to saturated vapor.
Economizers – is the heat exchanger that raises the temperature of the water leaving the highest pressure feedwater heater to the
saturation temperature corresponding to the boiler pressure.
Air Preheater – is a heat exchanger use to preheat air that utilizes some of the energy left in the flue gases before exhausting them
to the atmosphere.
Fans – a mechanical machine that assist to push the air in, pull the gas out or both.
Stoker – combustion equipment for firing solid fuels (used in water tube boilers)
Burners – combustion equipment for firing liquid and gaseous fuels.
Feedwater pump – a pump that delivers water into the boiler.
Pressure Gauge – indicates the pressure of steam in the boiler.
Safety Valve – A safety device which automatically releases the steam in case of over pressure.
Temperature Gauge – indicates the temperature of steam in the boiler.
Fusible Plug – a metal plug with a definite melting point through which the steam is released in case of excessive temperature
which is usually caused by low water level.
Water Walls – water tubes installed in the furnace to protect the furnace against high
temperature and also serve as extension
of heat transfer area for the feed-water.
Gage Glass (Water column) – indicates the water level existing in the boiler.
Baffles – direct the flow of the hot gases to effect efficient heat transfer between the hot gases and the heated water.
Furnace – encloses the combustion equipment so that the heat generated will be utilized effectively.
Soot blower – device which uses steam or compressed air to remove the soot that has accumulated in the boiler tubes and drums.
Blowdown Valve – valve through which the impurities that settle in the mud drum are remove. Sometimes called blow 0ff valve.
Breeching – the duct that connects the boiler and the chimney.
Chimney or Smokestack – a structure usually built of steel or concrete that is used to dispose the exhaust gases at suitable height
to avoid pollution in the vicinity of the plant.
121
BOILER PERFORMANCE
1.Heat Generated by Fuel
Q S  m f ( HHV )
KJ
HR
Where: mf – fuel consumption, kg/hr
HHV – higher heating value of fuel KJ/kg
2. Rated Boiler Horsepower(R.BHp)
For Water Tube Type
R.BHp 
HS
0.91
For Fire Tube Type
HS
R.BHp 
1.1
Where: HS – required heating surface, m2
3. Developed Boiler Horsepower (D.BHp)
m (h  h f )
D.BHp  s s
15.65(2257)
m (h  h f )
D.BHp  s s
35,322
One Boiler Horsepower is equivalent to the generation of 15.65 kg/hr of steam from water at 100C to saturated steam at
100C. The latent heat of vaporization of water at 100C was taken at 2257 KJ/kg.
4. Percentage Rating
%R 
D.BHp
x 100%
R.BHp
5. ASME Evaporation Units
ASME Evap. Units  ms (h s  h f )
6.Factor of Evaporation (FE)
(h  h f )
FE  s
2257
7. Boiler Efficiency
m (h  h f )
ηBoiler  s s
x 100%
mF (HHV )
8. Net Boiler Efficiency
m (h  h f )  Auxiliarie s
ηNet  s s
x 100%
mF (HHV )
9. Actual Specific Evaporation
m kg of steam
Actual Sp. Evap.  s
m F kg of fuel
10. Equivalent Evaporation
Equiv . Evap.  m s (FE)
11. Equivalent Specific Evaporation
Equiv . Sp. Evap. 
ms
(FE)
mF
122
Economizer, Evaporator, Superheater, Reheater and Air Pre–Heater of a thermal Power Plant
T
d
b
ts
c
a
QS
QE
QEv
S
a - subcooledliquid
b - saturated liquid
c - saturated vapor
d - superheated vapor
QE - heat required by the economizer
QEv - heat required by the evaporator
Qs - heat required by the superheater
For Shell and Tube type Heat Exchanger
Q  F(UA)(LMTD )
LMTD 
θ 2 - θ1
θ
ln 2
θ1
where
Q - Total heat Transfer, KW
F - Correction Factor for complex design
U - Overall Coefficien t of heat trans fer,
KW
m2 - K
or
KW
m 2 - C
A - total heat trans fer surface area, m 2
(LMTD) - Log Mean Temperatur e Difference , C or K
θ - Terminal Temperatur e Difference
A  πDL(N t )  For U based on Outside surface
A  πdL (N t )  For U based on inside surface
N t - number of tubes per Pass
L - length of tubes for single pass
For multiple tube passes
L
Number of passes
Total Number of tubes is N t ( Number of passes)
Tube length is
Actual tube length
L'  Tube Length  2t
where t  tube sheet thic kness
Volume Flow rate of Tube Fluid
π 2
d (Velocity )( N t )
4
Mass flow rate of tube fluid in kg/sec m 3
QV 
sec
Density of tube fluid in kg/m 3
QV 
123
Economizer
Q E  mg (Cpg )( t g1  t g 2 )  ms (h c  h b )  F( UA )( LMTD )
mg  mass flow rate of flue gas, kg/sec
ms - mass flow rate of feed water or steam produced, kg/sec
Cpg  specific heat of flue gas,
KJ
KJ
or
kg - C kg - K
θ 2  ( t g1  t a )
θ1  ( t g2  t b )
θ 2  ( t g1  t b)
θ1  ( t g 2  t a )
LMTD 
θ 2  θ1
θ
ln 2
θ1
Evaporator (Boiler)
QEv  mg (Cpg )( t g 2  t g 3 )  ms (h b  h a )  F( UA )( LMTD )
mg  mass flow rate of flue gas, kg/sec
ms - mass flow rate of feed water or steam produced, kg/sec
Cpg  specific heat of flue gas,
KJ
KJ
or
kg - C kg - K
θ 2  ( t g 2  t sat )
θ1  ( t g 3  t sat )
124
Superheater
QS  m g (C pg )( t g 3  t g 4 )  m s (h d  h c )  F( UA )(LMTD )
QS  m s (C psteam)( t d  t sat )
m g  mass flow rate of flue gas, kg/sec
m s - mass flow rate of feed water or steam produced, kg/sec
C pg  specific heat of flue gas,
C psteam  1.86
KJ
KJ
or
kg - C kg - K
KJ
KJ
or
kg - C kg - K
θ 2  (t g3  t c )
θ1  ( t g 4  t d )
θ 2  (t g3  t d )
θ1  ( t g 4  t c )
LMTD 
θ 2  θ1
θ
ln 2
θ1
Re - Heater
Hot Fluid – Flue Gas
Cold fluid – Reheated Steam
125
Air Preheater
Hot Fluid – Flue Gas
Cold fluid – Combustion Air
126
127
BOILER HEAT BALANCE
Energy supplied to the boiler by 1 kg of fuel is distributed among the following items in the ASME short-form heat balance, all
expressed in units of KJ/kg of fuel.
128
129
GEOTHERMAL POWER PLANT
Geothermal energy is the power obtained by using heat from the Earth's interior. Most geothermal resources are in regions of active
volcanism. Hot springs, geysers, pools of boiling mud, and fumaroles (vents of volcanic gases and heated groundwater) are the most
easily exploited sources of such energy
The most useful geothermal resources are hot water and steam trapped in subsurface formations or reservoirs and having
temperatures ranging from 176° to 662° F (80° to 350° C). Water and steam hotter than 356° F (180° C) are the most easily exploited
for electric-power generation and are utilized by most existing geothermal power plants. In these plants hot underground water is
drilled from wells and passes through a separator-collector where the hot water is flashed to steam, which is then used to drive a
steam turbine whose mechanical energy is then converted to electricity by a generator.
130
IDEAL TURBINE WORK
Wt = ms(h1 – h2) KW
ACTUAL TURBINE WORK
Wt’ = Tms(h1 – h2) KW
GENERATOR POWER OUTPUT
W0 = GTms(h1 – h2) KW
131
where
ms – steam flow rate in kg/sec
T - turbine efficiency
G – generator efficiency
o– underground water
H – well head condition
1 – saturated vapor condition leaving flasher – separator
B – saturated liquid condition leaving flasher – separator
2 – turbine exhaust
3 – saturated liquid leaving condenser
DRY STEAM POWER PLANT (Vapour Dominated System)
132
FLASH STEAM GEOTHERMAL POWER PLANT (Liquid Domain System)
133
134
Example
A geothermal power plant has an output of 16,000 KW and mechanical - electrical efficiency of 80%. The pressurized groundwater
at 17.0 MPa, 280C leaves the well to enter the flash chamber maintained at 1.4 MPa. the flash vapor passes through the separator
- collector to enter the turbine as saturated vapor at 1.4 MPa. the turbine exhaust at 0.1 MPa. The unflashed water runs to waste. If
one well discharges 195,000 kg/hr of hot water, how many wells are required. ( 4 wells)
From Steam Table
At 17,000 KPa and 280C
ho = 1231.7 KJ/kg
At 1400 KPa Saturated Vapor
h1 = 2786.4 KJ/kg ; S1 = 6.4642 KJ/kg-K
At S1 = S2 to 100 KPa
h2 = 2341.1 KJ/kg
At 1400 KPa saturated Liquid
hB = 829.6 KJ/kg
W0  0.80m s (h1  h 2 )
16000  m s (0.80)(2786.4  2341.1)
m s  45 kg/sec
By mass and energy balance on the flasher - collector
mo  ms  m B
m B  m o  m s  eq. 1
m o h o  m s h1  (m o - m s )h B
mo 
m s (h1  h B )
 219 kg/sec  788,400 kg/hr
ho  hB)
N wells 
788,400
 4 wells
195,000
HYDRO ELECTRIC POWER PLANT
A. IMPULSE TYPE (Pelton type)
headwater
Dam or Reservoir
Penstock
Y - Gross head
Turbine
tailwater
135
B. REACTION TYPE (Francis Type)
h  Y  HL
PA VA2

 ZA
γ
2g
Q
V
A
π
A  D2
4
h
D - penstock diameter, m
Y - Gross head, m
VB - velocity at inlet, m/sec
A - area of penstock, m2
HL - head loss, m
ZA - turbine setting above tail - water level, m
PUMP STORAGE HYDRO-ELECTRIC PLANT
136
How turbines work
The rotor is the rotating part of a turbine. In a simple turbine, it consists of a disc or wheel mounted on an axle. The axle sits either
horizontally or vertically. The wheel has curved blades or buckets around the edges. Nozzles or movable gates called guide vanes
aim the fluid at the blades or buckets and adjust its speed. In many turbines, a casing encloses the rotor. The casing holds the fluid
against the rotor so that none of the fluid's energy is lost.
As a fluid passes through a turbine, it hits or pushes against the blades or buckets and causes the wheel to turn. When the wheel
rotates, the axle turns with it. The axle is connected directly or through a series of gears to an electric generator, air compressor, or
other machine. Thus, the circular motion of the spinning rotor drives a machine.
The rotors of some turbines have only one wheel. However, the rotors of others have as many as 50 or more. Multiple wheels
increase the efficiency of turbines, because each wheel extracts additional energy from the moving fluid. In a turbine with more
than one wheel, the wheels are mounted on a common axle, one behind the other. A stationary ring of curved blades is attached to
the inside of the casing in front of each wheel. These stationary blades direct the flow of the fluid toward the wheels. A wheel and
a set of stationary blades is called a stage. Multistage turbines are those that have many stages.
Kinds of turbines
Turbines are sometimes classified according to their principle of operation. All turbines operate by impulse or reaction, or by a
combination of these principles. In an impulse turbine, the force of a fast-moving fluid striking the blades makes the rotor spin. In
a reaction turbine, the rotor turns primarily as a result of the weight or pressure of a fluid on the blades.
Turbines are more commonly classified by the type of fluid that turns them. According to this method, there are four main kinds of
turbines: (1) water turbines, (2) steam turbines, (3) gas turbines, and (4) wind turbines.
Water turbines are also called hydraulic turbines. Most water turbines are driven by water from waterfalls or by water that is stored
behind dams. The turbines are used primarily to power electric generators at hydroelectric power plants. There are three main kinds
of water turbines: (1) the Pelton wheel, (2) the Francis turbine, and (3) the Kaplan turbine. The type of water turbine used at a plant
depends on the head available. A head is the distance the water falls before it strikes the turbine. Heads range from about 2.5 meters
to more than 300 meters.
The Pelton wheel is an impulse turbine. It is used with heads of more than 300 meters. A Pelton's rotor consists of a single wheel
mounted on a horizontal axle. The wheel has cup-shaped buckets around its perimeter. Water from a lake or reservoir drops toward
the turbine through a long pipe called a penstock. One to six nozzles at the end of the penstock increase the water's velocity and
aim the water toward the buckets. The force of these high-speed jets of water against the buckets turns the wheel.
The Francis turbine is used when the head is between about 30 meters and 300 meters. A Francis turbine's rotor is enclosed in a
casing. Its wheel has as many as 24 curved blades. Its axle is vertical. The wheel of a Francis turbine operates underwater. It is
encircled by a ring of guide vanes, which can be opened or closed to control the amount of water flowing past the wheel. The spaces
between the vanes act as nozzles to direct the water toward the center of the wheel. The rotor is turned chiefly by the weight or
pressure of the flowing water.
The Kaplan turbine is used for heads of less than 30 meters. The Kaplan rotor resembles a ship's propeller. It has from three to
eight blades on a vertical axle. It works in a manner similar to that of a Francis turbine. Both the Kaplan turbine and the Francis
turbine are reaction turbines.
FUNDAMENTAL EQUATIONS
1. Total dynamic head or Net effective head
a. For an Impulse type
h = Y - HL
Y - Gross head at plant
Gross head - difference in elevation between head water level and tail water level.
b. For a Reaction type
h  Y  HL
PA VA2

 ZA
γ
2g
Q
V
A
π
A  D2
4
h
where:
PB - pressure at turbine inlet in KPa
VB - velocity of water at penstock, m/sec
2. Discharge or Rate of Flow (Q)
Q = AV m3/sec
A
π 2
D
4
where: D - diameter of penstock
137
3. Water Power (WP)
WP = Qh
4. Brake Power (BP)
BP 
KW
2πTN
KW
60,000
where:
T - brake torque N-m
N - no. of RPM
5. Head loss
HL 
fLV 2
meters
2gD
f - Moody friction factor
L - length of penstock
6. Turbine Efficiency (e)
e = ehemev
e = BP x 100%
WP
where: eh - hydraulic efficiency
em - mechanical efficiency
ev - volumetric efficiency
7. Generator Efficiency (g)
g = GP x 100%
BP
where: GP - electrical output of the generator, KW
8. Rotative Speed (N)
N
120f
RPM
n
where:
n - number of generator poles (usually divisible by 4)
9. Turbine Specific Speed
Ns 
N BP
RPM
5
3.813h  4
10. Wheel Diameter
D
60φ 2gh
meters
πN
where
φ  peripheral velocity factor
TERMS AND DEFINITION
Reservoir - stores the water coming from the upper river or waterfalls.
Headwater - the water in the reservoir or upper pool.
Spillway - a weir in the reservoir which discharges excess water so that the head of the plant will be
maintained.
Dam - the concrete structure that encloses the reservoir to impound water.
Silt Sluice - a chamber which collects the mud and through which the mud is discharged.
Trash Rack - a screen which prevents the leaves, branches and other water contaminants to enter into the
penstock.
Valve - opens or closes the entrance of the water into the penstock.
Surge Chamber - a standpipe connected to the atmosphere and attached to the penstock so that the water will
be at atmospheric pressure.
Penstock - a channel or a large pipe that conducts the water from the reservoir to the turbine.
Turbine - a device or a machine that converts the energy of the water to mechanical energy.
138
Generator - a device or a machine that converts mechanical energy of the turbine into electrical energy.
Draft Tube - a pipe that conducts the water from the turbine to the tailrace so that the turbine can be set above
the tail water level.
Tailrace - is the canal that is used to carry the water away from the plant.
Undershot wheel - water enters at the bottom of the wheel tangential to its periphery and impinges on the
buckets or vanes.
Breast shot wheel - a wheel used for heads up to 16 ft, where water enters between the bottom and top of the
wheel at an angle and is prevented from leaving the wheel by a breast wall on the side of the wheel.
Over shot wheel - a wheel used for high heads, where water enters the wheel at the top by being discharged
from a flume.
Gross head - is the difference between the headwater and tail water elevation.
Spiral case - it conducts the water around a reaction type turbine.
SAMPLE PROBLEMS
1. A pelton type turbine was installed 30 m below the head gate of the penstock. The head loss due to friction is
15% of the given elevation. The length of the penstock is 80 m and the coefficient of friction is 0.00093.
Determine
a) the diameter of the penstock in mm. (421.6 mm)
b) the power output in KW (781.234 KW)
2. What power in KW can be developed by the impulse turbine shown if the turbine efficiency is 85%. Assume
that the resistance coefficient f of the penstock is 0.015 and the head loss in the nozzle itself is negligible.
What will be the speed of the wheel , assuming ideal conditions where VJET = 2VBUCKET and what torque will
be exerted on the turbine shaft.
3. A hydroelectric plant has a 20 MW generator with an efficiency of 96%. The generator is directly coupled to
a vertical Francis type hydraulic turbine having an efficiency of 80%.The total gross head of the turbine is
150 m while the loss of head due to friction at the penstock up to the turbine inlet flange is 4% of gross head.
the runaway speed is not to exceed 750 RPM, determine:
a) Brake horsepower rating of the turbine (27 927 hp)
b) the flow of water through the turbine in cfs (650.53 cfs)
c) check if the specific speed falls under that of Francis type turbine. (Ns = 29)
d) the rated speed of the turbine (N = 400 RPM)
4. A Francis turbine is installed with a vertical draft tube. The pressure gauge located at the penstock leading to
the turbine casing reads 372.6 KPa and velocity of water at inlet is 6 m/sec. The discharge is 2.5 m 3/sec. The
hydraulic efficiency is 85%, and the overall efficiency is 82%. The top of the draft tube is 1.5 m below the
centerline of the spiral casing, while the tailrace level is 2.5 m from the top of the draft tube. There is no
velocity of whirl at the top or bottom of the draft tube and leakage losses are negligible. Calculate,
a) the net effective head in meters (43.817 m)
b) the brake power in kw. (881.2 kw)
c) the plant output for a generator efficiency of 92%. (810.7 kw)
d) the mechanical efficiency (96.550)
5. A hydroelectric power plant using a Francis type turbine has the following data:
Headwater elevation - 190 m
Tailwater elevation - 50 m
Head loss due to friction - 3.5% of gross head
Turbine discharge at full gate opening - 6 m3/sec
Turbine – Generator Speed – 600 RPM
Turbine efficiency at rated capacity - 90%
Turbine is to be direct connected to a 60 hertz a-c generator
Determine:
a) the brake power in kw (7156.8 kw)
b) the number of generator poles (12 poles)
c) the electrical power output of the generator if the efficiency is 94% (6727.4 kw)
d) the torque developed in N-m (102 925.31 N-m)
e) the approximate length of the penstock, if the diameter is 1.5 m and friction factor
f is 0.018. (693 m)
6. The flow of a river is 21.25 m3/sec and the head on the site is 30.5 m. It is proposed to developed the
maximum capacity at the site with the installation of two turbines, one of which is twice the capacity of the
other. The efficiency of both units is assumed to be 85%. Determine:
a) Rotative speed of each unit in rpm if the specific of both is 70 rpm.
b) Brake power of each unit in kw.
c) Number of poles of the generator for 60 cycle current
139
7. The flow of a river is 21 m3/sec and the head at the power site is 24.4 m (80 ft). It is proposed to developed
this site with an installation of 3-turbines, 2 similar units and another of half their size, all having the same
efficiency of 85%. Find the rotative speed of all these units. (348 rpm; 492 rpm)
Ns at 80 ft head = 70 rpm
8. A hydroelectric power plant discharging water at the rate of 0.75 m3/sec and entering the turbine at 0.35
m/sec with a pressure of 275 KPag has a runner of 55 cm internal diameter. Speed is 514 rpm at 260 BHP.
The casing is 2 m above the tailwater level. Calculate:
a) the net effective head in m (30.039 m)
b) the peripheral coefficient (0.61)
c) the efficiency (87.8%)
9. A Mindanao province where a mini-hydroelectric plant is to be constructed has an average annual rainfall of
139 cm. The catchment's area is 206 km2 with an available head of 23 m.Only 82% of the rainfall can be
collected and 75% of the impounded water is available for power. Hydraulic friction loss is 6%, turbine
efficiency is 78% and generator efficiency is 93%. Determine the average kw power that could be generated
for continuous operation.
POWER = 1.39(206)(1000)2(0.82)(0.75)(9.81)(23)(1-.06)(0.78)(0.93)
8760(3600)
= 859 kw
10. A Mindanao province where a hydroelectric power station is to be constructed has an average annual
rainfall of 1.93 m. The catchment's area 260 km 2 with an available head of 32 m. Only 82% of the rainfall
can be collected and 5/8 of the impounded water is available for power.The load factor of the station is 80%,
the penstock efficiency is 94%, turbine efficiency 78% and the generator efficiency is 88%.
a) Calculate the average power that could be generated
b) Calculate the installed capacity of the station
c) Assuming no standby unit, give the possible number and size of the units
Average power = load factor(peak power)
11. The difference in elevation between the surfaces of water in the storage reservoir and the intake to a turbine
was 40.4 m. During a test the pressure head at the later point was 38.6 m and the discharge was1.25 m/sec.
The inside diameter of the penstock is762 mm.
a) What is the efficiency of the pipeline
b) What was the power delivered to the turbine in KW
12. The difference in elevation between the source of the water supply and the centerline at the base of the
nozzle of a pelton wheel is 373 m. During a test the pressure at the end of the nozzle was 3,520 KPa when the
flow was 1.3 m3/sec. Inside diameter of the penstock is 762 mm. Compute:
a) What power at the base of the nozzle.
b) Shaft power developed by the pelton wheel if efficiency is 80%.
13. In a hydroelectric power plant, the head water surface on the dam is at elevation 75.4 m while the
water surface just at the outlet of the head gate is at elevation 70.4 m. The head gate has 5 gates of 1 m x
1 m leading to the penstock and are fully opened. Turbine is 122 m below the entrance of the penstock with
an efficiency of 90%. Assuming 60% as coefficient of discharge, and Generator efficiency of 87% ,determine
the Megawatt output of electrical power produced by the plant.
140
14. A hydro - storage plant has a 20,000 KW rated capacity, with a utilization factor of 76%. For a 1 1/2 hour
peak, determine the hydraulic impoundment in cubic meters of water required with friction factor loss of
6%. Dam elevation at 40 m and hydraulic turbine elevation of 14 m and tailrace at elevation of 7 m.
Generator efficiency is 92% and turbine efficiency of 85%, evaporation factor of 20%. If the water is
pumped from the lower reservoir with friction factor of 8%, pump efficiency of 76% and motor efficiency
of 85%.
a) How many KW-hr of power is required to pump the water required to carry the peak
b) What is the overall thermal efficiency of the hydro plant.
15. A remote community in the mountain province plans to put up a small hydroelectric plant to service six
closely located barangays estimated to consume 52,650,000 KW-hrs per annum. Expected flow of water is
28 m3/sec. The most favorable location for the plant fixes the tail water level at 480 m. The manufacturer of
the turbine generator set have indicated the following performance data:
Turbine efficiency - 87&
Generator Efficiency - 92%
Loss in headwork will not exceed 3.8% of the available head
In order to pinpoint the most suitable area for the dam, determine
a. Headwater elevation
b. Type of turbine to be used
c. Synchronous speed of generator if number of poles is 6, and frequency is 60 hertz.
16. A proposed hydro electric power plant has the following data:
Elevation of normal headwater surface - 194 m
Elevation of normal tailwater surface - 60 m
Loss of head due to friction - 6.5 m
Turbine discharge at full gate opening - 5 m3/sec
Turbine efficiency at rated capacity - 90%
Turbine is direct connected to a 60 cycle AC generator
Required:
a. What type of hydraulic turbine would you specify? (Francis)
b. Find the Brake Power of the turbine in KW (5628.5 KW)
c. Find the number of poles of the generator if Ns = 33 RPM (10 poles)
d. Find the KW output of the direct-connected generator if the efficiency is 94%
(5,291 KW)
17. The flow of a river of 22 m3/sec produces a total brake power of 5400 KW. It is proposed to install two
turbines one of which is twice the capacity of the other. The efficiency and specific speed of both units are
assumed to be 85% and 70 RPM, respectively. Determine
a. Head in meters (30.5 m)
b. Rotative speed of each unit (450 RPM; 320 RPM)
c. Number of poles of generator if f = 60 Hz. (16 Poles ; 24 Poles)
18. At a proposed hydro-electric power plant site, the average elevation of the headwater is 600 m, the tailwater
elevation is 480 m. Average annual water flow is determined to be equal to that volume flowing through a
rectangular channel 4 m wide and0.5 m deep and average velocity of 5.5 m/sec. Assuming that the plant will
operate 350 days per year, find the annual energy in KW-HR that the power site can develop if the hydraulic
turbine that will be used has an efficiency 80% and generator efficiency of 92%. Consider a headwork loss
of 4% of the available head. (76,854,851 KW-hr)
I
A Francis turbine is installed with a vertical draft tube. The pressure gauge located at the penstock leading to the turbine casing
reads 372.6 KPa and velocity of water at inlet is 6 m/sec. The discharge is 2.5 m 3/sec. The hydraulic efficiency is 85%, and the
overall efficiency is 82%. The top of the draft tube is 1.5 m below the centerline of the spiral casing, while the tailrace level is 2.5
m from the top of the draft tube. There is no velocity of whirl at the top or bottom of the draft tube and leakage losses are negligible.
Calculate,
a) the net effective head in meters
b) the brake power in KW
c) the plant output for a generator efficiency of 92%.
d) the mechanical efficiency
II
The discharge gauge of a centrifugal pump handling potable water for a class A subdivision reads 175 KPa, while the suction gauge
indicates a vacuum of 305 mm Hg. The discharge pressure gauge is 10 m above the pump centerline and the point of attachment of
the suction gauge is 3 m below the centerline. The diameters of the suction and discharge pipes are 76 mm and 63.5 mm, respectively.
Assuming a head loss of 6 m, and a pump-motor efficiency of 75, Calculate the power required if the flow is 12.5 L/sec of water. (
= 9.81 KN/m3)
141
DIESEL POWER PLANT
Diesel engine is a type of internal combustion engine that uses low grade fuel oil and which burns this fuel inside the cylinder by
heat of compression. It is used chiefly for heavy-duty work. Diesel engines drive huge freight trucks, large buses, tractors, and
heavy road-building equipment. They are also used to power submarines and ships, and the generators of electric-power stations in
small cities. Some motor cars are powered by diesel engines.
Gasoline engine - is a type of internal combustion engine, which uses high grade of oil. It uses electricity and spark plugs to ignite
the fuel in the engine's cylinders.
Kinds of diesel engines. There are two main types of diesel engines. They differ according to the number of piston strokes required
to complete a cycle of air compression, exhaust, and intake of fresh air. A stroke is an up or down movement of a piston. These
engines are (1) the four-stroke cycle engine and (2) the two-stroke cycle engine.
Four Stroke Cycle Engine
1. Intake
Compression
Power
Exhaust
Intake
2. Compression
3. Power
4. Exhaust
In a four-stroke engine, each piston moves down, up, down, and up to complete a cycle. The first down stroke draws air into the
cylinder. The first upstroke compresses the air. The second down stroke is the power stroke. The second upstroke exhausts the
gases produced by combustion. A four-stroke engine requires exhaust and air-intake valves.
It completes one cycle in two revolutions of the crankshaft.
Two Stroke Cycle Engine
Intake-compression
Power-Exhaust
1. Intake-Compression stroke
2. Power-exhaust stroke
Intake port
Exhaust port
In a two-stroke engine, the exhaust and intake of fresh air occur through openings in the cylinder near the end of the down stroke,
or power stroke. The one upstroke is the compression stroke. A two-stroke engine does not need valves. These engines have twice
as many power strokes per cycle as four-stroke engines, and are used where high power is needed in a small engine. It completes
one cycle in one revolution of the crankshaft.
142
INTERNAL COMBUSTION ENGINE POWER PLANT
Two stroke cycle engine: An engine that completes one cycle in one revolution of the crankshaft.
Four stroke cycle engine: An engine that completes one cycle in two revolution of the crankshaft.
ENGINE PERFORMANCE
HEAT SUPPLIED BY FUEL
KJ
Qs  m F (HV )
hr
where:
mf - fuel consumption in kg/hr
HV - heating value of fuel in KJ/kg
INDICATED POWER
IP 
PmiπLD 2 Nn'
KW
4(60)
where:
Pmi - indicated mean effective pressure in KPa
L - length of stroke in m
D - diameter of bore in m
n' - no. of cylinders
IP - indicated power in KW
N  (RPM )  (For 2 Stroke - Single acting)
N  2(RPM)  (For 2 Stroke - Double acting)
(RPM)
 (For 4 Stroke - Single acting)
2
N  (RPM )  (For 4 Stroke - Double acting)
N
BRAKE POWER
IP 
PmbπLD 2 Nn'
KW
4(60)
where:
Pmb - brake mean effective pressure in KPa
N  (RPM )  (For 2 Stroke - Single acting)
N  2(RPM)  (For 2 Stroke - Double acting)
(RPM)
 (For 4 Stroke - Single acting)
2
N  (RPM )  (For 4 Stroke - Double acting)
N
143
BP 
2πTN
KW
60,000
where:
T - brake torque in N-m
N = RPM
4. FRICTION POWER
FP  IP  BP
5. INDICATED MEAN EFFECTIVE PRESSURE
AS'
Pmi 
KPa
L'
where:
A - area of indicator card, m2
S - spring scale, KPa/m
L' - length of indicator card, m
6. BRAKE TORQUE
T  (P  Tare )R N - m
where:
P - Gross load on scales, N
Tare - tare weight, N
R - length of brake arm, m
7. PISTON SPEED
m
min
8. DISPLACEMENT VOLUME
πLD 2 Nn'
VD 
KPa
4(60)
IP
VD 
KPa
Pmi
PS  2LN
VD 
BP
KPa
Pmb
Where
N  (RPM )  (For 2 Stroke - Single acting)
N  2(RPM)  (For 2 Stroke - Double acting)
(RPM)
 (For 4 Stroke - Single acting)
2
N  (RPM )  (For 4 Stroke - Double acting)
N
9. SPECIFIC FUEL CONSUMPTION
a. Indicated specific fuel consumption
m
kg
mfi  f
IP KW - hr
b. Brake specific fuel consumption
m
kg
mfb  f
BP KW - hr
c. Combined specific fuel consumption
m
kg
mfc  f
GP KW - hr
where: GP - Generator output
144
10. HEAT RATE (HR): Heat supplied divided by the KW produced.
a. Indicated heat rate
Q
KJ
HRi  s
IP KW - hr
b. Brake heat rate
Q
KJ
HRb  s
BP KW - hr
c. Combined heat rate
Q
KJ
HRc  s
GP KW - hr
11. GENERATOR SPEED
120f
RPM
N
n
WHERE: n - number of generator poles (usually divisible by 4)
12. MECHANICAL EFFICIENCY
BP
ηm 
x 100%
IP
13. GENERATOR EFFICIENCY
GP
ηg 
x 100%
BP
14. Indicated Thermal Efficiency
3600(IP)
ei 
x 100%
Qs
15. Brake Thermal Efficiency
3600(BP)
eb 
x 100%
Qs
16. Combined Thermal efficiency
3600(GP)
ec 
x 100%
Qs
17. Indicated Engine Efficiency
e
ηi  i x 100%
e
18. Brake Engine Efficiency
e
ηb  b x 100%
e
where:
e - cycle thermal efficiency
19. Volumetric Efficiency
ηv 
m3
sec x 100%
3
Actual volume of air entering,
Displacement Volume,
m
sec
20. Correction Factor for Nonstandard condition
a. Considering temperature and pressure effect
 B  T 
Ph  Ps  S  h 


 Bh  Ts 
b. Considering temperature effect alone
 T 
Ph  Ps  h 
 T 
 s
c. Considering pressure effect alone
145
B 
Ph  Ps  S 
 Bh 
Note: From US Standard atmosphere:
83.312h 

Bh   Bs 
 mm Hg
1000 

6.5h 

Th   Ts  K
1000


where: B - barometric pressure, mm Hg
T - absolute temperature, K
h - at elevation h condition
s - at sea level condition
21. ENGINE HEAT BALANCE: The total heat supplied to th engine was broken down into four heat items.
Q2
Q1
QS
engine
Q3
Q4
QS = Q1 + Q2 + Q3 + Q4
Q1 - heat converted to useful work
Q2 - heat lost to cooling water
Q3 - heat lost to exhaust gases
Q4 - heat lost due to friction, radiation and unaccounted for
Q1 = 3600(BP) KJ/hr
Q2 = mwCpw(two - twi) KJ/hr
Q3 = Qa + Qb KJ/hr
Qa = mgCpg(t g - ta) KJ/hr
Qb = mf(9H2)(2442.7) KJ/hr
Q4 = QS - (Q1 + Q2 + Q3) KJ/hr
H2 = 0.26 - 0.15S kgH/kgfuel
141.5
S
131.5  API
140
S
130  BeI
where:
Qa - sensible heat of products of combustion
Qb - heat required to evaporate and superheat moisture formed from the
combustion of hydrogen in the fuel
tg - temperature of flue gas, C
ta - temperature of air, C
H2 - amount of hydrogen in the fuel kg H/kg fuel
146
TERMS AND DEFINITIONS
Diesel engine is a type of internal combustion engine that uses low grade fuel oil and which burns this fuel inside the cylinder by
heat of compression. It is used chiefly for heavy-duty work. Diesel engines drive huge freight trucks, large buses, tractors, and
heavy road-building equipment. They are also used to power submarines and ships, and the generators of electric-power stations in
small cities. Some motor cars are powered by diesel engines.
Gasoline engine - is a type of internal combustion engine, which uses high grade of oil. It uses electricity and spark plugs to ignite
the fuel in the engine's cylinders.
Kinds of diesel engines. There are two main types of diesel engines. They differ according to the number of piston strokes required
to complete a cycle of air compression, exhaust, and intake of fresh air. A stroke is an up or down movement of a piston.These
engines are (1) the four-stroke cycle engine and (2) the two-stroke cycle engine.
Four Stroke Cycle Engine
1. Intake
2. Compression
3. Power
4. Exhaust
In a four-stroke engine, each piston moves down, up, down, and up to complete a cycle. The first down stroke draws air into the
cylinder. The first upstroke compresses the air. The second down stroke is the power stroke. The second upstroke exhausts the
gases produced by combustion. A four-stroke engine requires exhaust and air-intake valves.
It completes one cycle in two revolutions of the crankshaft.
Two Stroke Cycle Engine
1. Intake-Compression stroke
2. Power-exhaust stroke
In a two-stroke engine, the exhaust and intake of fresh air occur through openings in the cylinder near the end of the down stroke,
or power stroke. The one upstroke is the compression stroke. A two-stroke engine does not need valves. These engines have twice
as many power strokes per cycle as four-stroke engines, and are used where high power is needed in a small engine. It completes
one cycle in one revolution of the crankshaft.
Governor - is a device used to govern or control the speed of an engine under varying load conditions.
Purifier - a device used to purify fuel oil and lube oil.
Generator - a device used to convert mechanical energy.
Crank scavenging - is one that the crankcase is used as compressor.
Thermocouple - is made of rods of different metal that are welded together at one end.
Centrifuge - is the purification of oil for separation of water.
Unloader - is a device for automatically keeping pressure constant by controlling the suction valve.
Planimeter - is a measuring device that traces the area of actual P-V diagram.
Tachometer - measures the speed of the engine.
Engine indicator - traces the actual P-V diagram.
Dynamometer - measures the torque of the engine.
Supercharging - admittance into the cylinder of an air charge with density higher than that of the surrounding air.
Bridge Gauge - is an instrument used to find the radial position of crankshaft motor shaft.
Piston - is made of cast iron or aluminum alloy having a cylinder form.
Atomizer - is used to atomize the fuel into tiny spray which completely fill the furnace in the form of hollow cone.
Scavenging - is the process of cleaning the engine cylinder of exhaust gases by forcing through it a pressure of m
fresh air.
Flare back - is due the explosion of a maximum fuel oil vapor and air in the furnace.
Single acting engine - is one in which work is done on one side of the piston.
Double acting engine - is an engine in which work is done on both sides of the piston.
Triple-expansion engine - is a three-cylinder engine in which there are three stages of expansion.
The working pressure in power cylinder is from 50 psi to 500 psi.
The working temperature in the cylinder is from 800F to 1000F.
Air pressure used in air injection fuel system is from 600 psi to 1000 psi.
Effect of over lubricating a diesel engine is:
Carbonization of oil on valve seats and possible explosive mixture is produced.
The average compression ratio of diesel engine is from 14:1 to 16:1.
Three types of piston:
1. barrel type
2. trunk type
3. closed head type
Three types of cam follower:
1. flat type
2. pivot type
3. roller type
147
Methods of mechanically operated starting valve:
1. the poppet
2. the disc type
Three classes of fuel pump:
1. continuous pressure
2. constant stroke
c. variable stroke
Type of pump used in transferring oil from the storage to the service tanks:
1. rotary pump
2. plunger pump
3. piston pump
4. centrifugal pump
Valve that is found in the cylinder head of a 4-stroke cycle engine:
1. fuel valve
2. air starting valve
3. relief valve
4. test valve
5. intake valve
6. exhaust valve
Four common type of governors used on a diesel engine:
1. constant speed governor
2. variable speed governor
3. speed limiting governor
4. load limiting governor
Kinds of piston rings used in an internal combustion engines:
1. compression ring
2. oil ring
3. firing ring
4. oil scraper ring
Reasons of smoky engine:
1. overload
2. injection not working
3. choked exhaust pipe
4. fuel or water and leaky things
Methods of reversing diesel engines:
1. sliding camshaft
2. shifting roller
c. rotating camshaft
148
Arrangements of cylinders:
1. in-line
2. radial
3. opposed cylinder
4. V
5. opposed piston
Position of cylinders:
1. vertical
2. horizontal
3. inclined
Methods of starting:
1. manual, crank, rope, and kick
2. electric (battery)
3. compressed air
4. using another engine
Applications:
1. automotive
2. marine
3. industrial
4. stationary power
5. locomotive
6. aircraft
Types of internal combustion engine:
1. Gasoline engine
2. Diesel engine
3. Kerosene engine
4. Gas engine
5. Oil-diesel engine
Methods of ignition:
1. Spark
2. Heat of compression
Reasons for supercharging:
1. to reduce the weight to power ratio
2. to compensate the power loss due to high altitude
Types of superchargers:
1. engine-driven compressor
2. exhaust-driven compressor
3. separately-driven compressor
Auxiliary systems of a diesel engine:
1. Fuel system
a. fuel storage tank
b. fuel filter
c. transfer pump
d. day tank
e. fuel pump
2. Cooling system
a. cooling water pump
b. heat exchanger
c. surge tank
d. cooling tower
e. raw water pump
3. Lubricating system:
a. lub oil tank
b. lub oil pump
c. oil filter
d. oil cooler
e. lubricators
149
4. Intake and exhaust system
a. air filter
b. intake pipe
c. exhaust pipe
d. silencer
5. Starting system
a. air compressor
b. air storage tank
Advantages of diesel engine over other internal combustion engines:
1. low fuel cost
2. high efficiency
3. needs no large water supply
4. no long warm-up period
5. simple plant layout
Types of scavenging:
1. direct scavenging
2. loop scavenging
3. uniflow scavenging
Color of the smoke:
1. efficient combustion - light brown
baze
2. insufficient air - black smoke
3. excess air - white smoke
Causes of black smoke:
1. fuel valve open too long
2. too low compression pressure
3. carbon in exhaust pipe
4. overload on engine
Causes of white smoke:
1. one or more cylinders not getting
enough fuel
2. too low compression pressure
3. water inside the cylinder
ENGINE FOUNDATION
Functions of a Foundation:
Support the weight of the engine.
Maintain proper alignment with the machinery and
Absorb the vibration produced by unbalanced forces created by reciprocating revolving masses.
Materials:
Mixture: 1 : 2 : 4
1 part Cement
2 parts sand
4 parts broken stone or Gravel
For good firm soil, reinforced concrete foundations for large engines may use a leaner mixture down to
1:3:6
Soil Bearing Pressure:
The safe loads vary from about 4,890 kg/m2 for alluvial soil or wet clay, to 19,560 kg/m 2 for gravel, coarse sand and dry clay.
(12,225 kg/m2 is assumed to be safe load average).
In computation 2,406 kg/m2 may be used as weight of concrete.
Depth:
The foundation depth maybe taken as a good practical rule, to be 3.2 to 4.2 times the engine stroke; the lower factor for wellbalanced multi-cylinder engines and increased factor for engines with fewer cylinders or on less firm soil.
Weight:
The minimum weight required to absorb vibration could be expressed as a function of the reciprocating masses and the speed
of the engine. However, for practical purposes it is simpler to use the empirical formula.
150
Volume:
If the weight and speed of the engine is not known, the volume of concrete for the foundation may be estimated from the data
in the table.
Anchor Bolts:
To prevent pulling out of the bolts when the nuts are tightened, the length embedded in concrete should be equal to at least
thirty (30) times the bolt diameter. The upper ends are surrounded by a 50 mm or 75 mm sheet metal pipe, 460 mm to 610 mm
long to permit them to be bent slightly to fit the holes of the bedplate.
FORMULA:
WF = e (WE) N
Where:
WF – weight of foundation in kgs
WE – weight of engine in kgs
e – empirical coefficient
N – engine speed in RPM
Values of e
Type of Engine
Single Acting
Single Acting
Single Acting
Single Acting
Single Acting
Single Acting
Single Acting
Double acting
Double acting
Cylinder Arrangement
Vertical
Vertical
Vertical
Vertical
Horizontal
Horizontal Duplex
Horizontal Twin Duplex
Horizontal
Horizontal Twin Tandem
No. of Cylinders
1
2
3
4, 6, 8
1
2
4
1, 2
4
Volume of Concrete Foundation (Cubic Feet per HP)
No. of Cylinders
1
2
3
High speed engine
4.0
2.5
2.0
Medium speed engine
5.0
3.1
2.5
Low speed engine
6.0
4.0
3.0
Note: 1 cubic meter (m3) = 35.315 ft3
1 Horsepower = 0.746 KW
4
1.7
2.1
2.6
e
0.15
0.14
0.12
0.11
0.25
0.24
0.23
0.32
0.20
5-8
1.5
1.9
2.3
General Requirements
1. All heavy machinery shall be supported on solid foundation of sufficient mass and base area to prevent or
minimize the transmission of objectionable vibration to the building and occupied space and to maintain the
supported machine at its proper elevation and alignment.
2. Foundation mass should be from 3 to 5 times the weight of the machinery it is supposed to support. If the
unbalanced inertial forces produced by the machine can be calculated, a mass of weight equal to 10 to 20 times
the forces should be used to dampen vibration.
For stability, the total combined engine, driven equipment and foundation center of gravity must be kept below
the foundation’s top.
3. the weight of the machine plus the weight of the foundation should be distributed over a sufficient soil area which
is large enough to cause a bearing stress within the safe bearing capacity of the soil with a factor of safety of 5.
4. Foundation should be isolated from floor slabs and building footings by at least 25 mm around its perimeter to
eliminate transmission of vibration. Fill opening with water tight-mastic.
When installing machinery above grade level of a building, additional stiffness must be provided on its structural
members of the building to dampen machine vibration.
5. Foundations are preferably built of concrete in the proportions of one (1) measure of Portland Cement to two
(2) measures of sand and four (4) measures of screened crushed stones. The machine should not be placed
on thefoundation until ten (10) days have elapsed or operated until another ten (10) days have passed.
151
6. Concrete foundation should have steel bar reinforcements placed both vertically and horizontally, to avoid
thermal cracking. Weight of reinforcing steel should be from ½ % to 1 % of the weight of the foundation.
7. Foundation bolts of specified size should be used and surrounded by a pipe sleeve with an inside diameter of
at least three (3) times the diameter of the anchor bolt and a length of at least eighteen (18) times the diameter
of the bolt. No foundation bolts shall be less than 12 mm diameter.
8. Machine should be leveled by driving wedges between the machine’s base and concrete foundation and with
The aid of a spirit level. Grout all spaces under the machine bed with a thin mixture of one (1) part cement and
one part sand. The level wedges should be removed after grout has thoroughly set and fill wedges holes with
grout.
152
SAMPLE PROBLEMS
No. 1
A 2 - stroke, 4 - cylinders, 38 cm x 53 cm diesel engine is guaranteed to deliver 522 KW at 300 rpm.The fuel rate is 0.26
kg/KW-hr. If the heating value of the fuel is 44,320 KJ/kg Calculate:
a) the brake thermal efficiency (32%)
b) the brake mean effective pressure (422 KPa)
c) suction displacement in m3/min-KW of shaft power (0.15)
d) heat supplied to cylinder per Liter of displacement
Given: 2-stroke, n’ = 4; D = 0.38 m; L = 0.53 m; BP = 522 KW; N = 300 RPM
mFB = 0.26 kg/KW-hr; HV = 44,320 KJ/kg
3600BP 3600BP
3600
a. eb 


 31.2%
.
Qs
mF (HV) mFB (HV)
PmBπLD 2 Nn'
4(60)
N  RPM (for 2 - stroke, single acting)
b. BP 
PmB 
240(BP)
 434 KPa
πLD 2 Nn'
.
c. BP  PmB (VD )
VD 1(60)
m3

 0.14
BP PmB min - KW
d.
QS
m (HV ) KJ m FB (BP )( HV )PmB m FB (HV )PmB
 F


KJ/L  1.4 KJ/L
VD BP (3600 ) m3
BP (3600 )
(3600 )(1000 )
PmB
No. 2
Find the volume in Liters needed for a two weeks supply of 26API fuel oil to operate a 750 KW engine 70 % of the time at
full load, 10 % at 3/4 load and idle 20% of the time. Fuel rate is 0.25 kg/KW-hr at full load and 0.24 kg/KW-hr at 3/4 load.
Temperature of oil is 21C.
T  2weeks(7 days)(24 hours)  336 hrs
m F  336(0.70)0.25(750)  0.10(336)(0.75)(750)(0.24)  48,636 kg
141.5
 0.898
131.5  26
S @ t  0.898  0.0007(21  15.56)  0.895
S
kg
kg
 0.895
L
m3
48,636
VF 
 54,342 Liters
0.895
ρ  895
No. 3
A diesel power plant is to have 3- supercharged, 7 cylinders, 590 KW, 720 RPM, 4-stroke cycle diesel engine. The full load
brake specific fuel consumption is 0.24 kg/KW-hr. Determine the most economical size and dimensions of a day tank you
would install to contain enough fuel for the three units to operate for 24 hours.
Assume D = 0.75H and fuel oil is at 26API.
MF = 3(0.24)(590)(24) = 10,195.2 kg (total mass of fuel consumed)
S of fuel at 26API = 0.898;  = 898 kg/m3
10,195.2
 11.4 m3 (volume of fuel)
VF =
898
π 2
V D H
4
D  0.75H
H  3 m ; D  2.25 M
No.4
It is claimed that 1 KW is developed for each 9200 KJ supplied per hour based on a lower heating value, bya supercharged
Dorman diesel engine when operated at a brake mean effective pressure of 900 KPa. The lower heating value of the fuel is
41,900 KJ/kg and the engine is a 4 - stroke cycle with 16 cylinders, 39 cm x 56 cm. If the engine requires 20 kg of combustion
air per kg of fuel, Determine:
a) the brake specific fuel consumption in kg/KW-hr
b) the brake power in KW at N = 360 RPM
c) the volumetric efficiency assuming combustion air is at P =101 KPa and T = 298K.
a.
153
QS
m (HV )
 F
9200
9200
9200
m FB 
 0.22 kg/KW - hr
41,900
BP 
b.
PmBπLD 2 Nn'
BP 
4(60)
RPM
N
for 4 - stroke single acting
2
BP  2,890 KW
c.
mF = 0.22(2,890) = 635.8 kg/hr = 0.177 kg/sec
ma = 0.177(20) = 3.54 kg/sec
mRT
V
P
Va = 3 m3/sec
BP
VD 
PmB
VD = 3.211 m3/sec
Va
ev 
x100%
VD
ev = 93.5 %
No. 5
A single acting, 4-cycle diesel engine uses 11 kg/hr of 24API fuel when running at 420 RPM. Engine specifications: 23 cm x
36 cm. The prony brake used to determine the brake power has 1 m arm and registers on the scale 130 kg gross. If the tare mass
is 12 kg, calculate the brake thermal efficiency based on the lower heating value of fuel.
T = (P – tare)R = (130 – 12)1 = 118 kg-m (9.81 N/kg) = 1,157.58 N-m
2πTN
BP 
60,000
BP = 51 KW
LHV = 38,105 + 139.6(API) = 41,455.4 KJ/kg
3600BP
eB 
x 100%
mF (LHV)
eB = 40.3 %
SAMPLE PROBLEMS
1. A 2 - stroke, 4 - cylinders, 38 cm x 53 cm diesel engine is guaranteed to deliver
522 KW at 300 rpm. The fuel rate is 0.26 kg/KW-hr. If the heating value of the
fuel is 44,320 KJ/kg Calculate:
a) the brake thermal efficiency (32%)
b) the brake mean effective pressure (422 KPa)
c) suction displacement in m3/min-KW of shaft power (0.15)
d) heat supplied to cylinder per Liter of displacement
2. Find the volume in Liters needed for a two weeks supply of 26API fuel oil to
operate a 750 KW engine 70 % of the time at full load, 10 % at 3/4 load and idle
20% of the time. Fuel rate is 0.25 kg/KW-hr at full load and 0.24 kg/KW-hr at 3/4
load. Temperature of oil is 21C.
3. A diesel power plant is to have 3- supercharged, 7 cylinders, 590 KW, 720 RPM, 4stroke cycle diesel engine. The full load brake specific fuel consumption is 0.24
kg/KW-hr. Determine the most economical size and dimensions of a day tank you
would install to contain enough fuel for the three units to operate for 24 hours.
Assume D = 0.75H.
4. It is claimed that 1 KW is developed for each 9200 KJ supplied per hour based on
a lower heating value, by a supercharged Dorman diesel engine when operated at a
brake mean effective pressure of 900 KPa. The lower heating value of the fuel is
41,900 KJ/kg and the engine is a 4 - stroke cycle with 16 cylinders, 39 cm x 56
cm. If the engine requires 20 kg of combustion air per kg of fuel, Determine:
a) the brake specific fuel consumption in kg/KW-hr
b) the brake power in KW
c) the volumetric efficiency
5. A single acting, 4-cycle diesel engine uses 11 kg/hr of 24API fuel when running at
420 RPM. Engine specifications: 23 cm x 36 cm. The prony brake used to
determine the brake power has 1 m arm and registers on the scale 130 kg gross. If
the tare mass is 12 kg, calculate the brake thermal efficiency based on the lower
154
heating value of fuel.
6. A 4 cylinders, 4 stroke cycle diesel engine, 30.5 cm x 46 cm, 260 RPM, single
acting diesel engine is rated at 200 KW. The fuel rate at rated load is 0.26
kg/KW-hr, and the fuel used has a heating value of 44,000 KJ/kg. Determine
a) the brake mean effective pressure
b) displacement in m3/min-KW of shaft power
c) the brake thermal efficiency
7. A torque of 28 kg-m is developed by a diesel engine when running at 1200 RPM
and using 10 kg of fuel per hour. the heating value of the fuel is 44,200 KJ/kg.
The engine is 4-stroke cycle and has 4 cylinders and the bore is equal the stroke.
The engine takes in 25 kg of air per kg of fuel and the volumetric efficiency is 80%.
Calculate the brake thermal efficiency and the bore in cm.
8. Performance values of a 3000 KW Diesel power plant unit are as follows:
Fuel rate: 1.5 barrels for 900 KW-hr of 25API fuel
Generator Efficiency: 92%
Mechanical efficiency: 82%
Determine:
a) Engine fuel rate in kg/KW-hr
b) Engine-Generator fuel rate in kg/KW-hr
c) Indicated thermal efficiency
d) Brake thermal efficiency
e) Overall thermal efficiency
9. A twin tandem, 4-stroke cycle, double acting Blast furnace gas engine is to
developed 2700 KW of brake power at 90 RPM. Expected operating data are:
PmB = 450 KPa
m = 83%
b = 65%
HHV = 31,700 KJ/m3
L/D = 1.35
r = 7.1
k = 1.32
155
Determine:
a) the cylinder dimensions in cm
b) fuel consumption in L/hr if the specific gravity of fuel is 0.88.
10. An 8 cylinders,2-stroke cycle, single acting diesel engine rated at 940 KW at
standard condition (P = 760 mm Hg; t = 15.56C) is directly coupled to a 24 poles
alternator; 3-phase; 60 Hertz. Assuming brake mean effective pressure of 520
KPa; m = 85% & g = 94%
a) Find the diameter of the cylinder and the length of stroke of the piston if the
average piston speed is 310 m/min.
b) What will be the actual thermal efficiency for a combined fuel rate of 0.20 kg
of diesel fuel per KW hour with a HV = 46,530 KJ/kg
c) What will be the KW output if the above diesel engine is operated in Baguio
City whose elevation is 1525 m above sea level and the actual sea level
temperature is 29C.
11. A diesel engine under test gave the following performance data;
Brake Power - 3360 KW
Jacket loss - 23%
Mechanical Efficiency - 83%
Indicated thermal efficiency - 34%
A waste heat recovery boiler recovers 35% of the exhaust loss. If cooling water
leaving the engine is at 66C and is used as feedwater to boiler, determine the
quantity of 0.14 MPa steam that can be produced in kg/hr.
12. When the pressure is 101.3 KPa and the temperature is 27C, a diesel engine has
the full throttle characteristics listed:
Brake power - 275 KW
Brake specific fuel consumption - 0.25 kg/KW-hr
Air - Fuel ratio - 22
Mechanical efficiency - 88%
If the engine is operated at a pressure of 84.5 KPa and temperature of 16C, Find:
a) Brake power in KW
b) Mechanical efficiency
c) Brake specific fuel consumption
d) air-Fuel ratio
13. A 40 KW blast furnace engine shows by test a gas consumption of 4.2 m 3/KW-hr.
Heating value of gas is 33,500 KJ/m3. Mechanical efficiency of the gas engine is
86%, Calculate:
a) The brake thermal efficiency
b) Indicated thermal efficiency
c) Heat rate in KJ/KW-hr
d) If the heat rejected to cooling water is 28% of the heat generated in engine
cylinder, determine the quantity of cooling water in L/hr to be circulated
if the allowable rise in temperature of the cooling water is 10C
14. On the test bed, a 16 cylinders, V type, 4-stroke cycle, turbo charge diesel engine
developed 7350 KW at 514 RPM. Heat rejection to lubricating oil was 460
KJ/KW-hr. Specific gravity of lubricating oil was 0.90 and its specific heat was
1.9 KJ/kg-C. Direct connected generator is rated at 9,150 KVA at 80% power
factor and efficiency of 95%, Calculate
a) Liters of lubricating oil per hour to be circulated through engine if
lubricating oil temperature entering engine is 54C and the allowable rise
in temperature of oil is 6C.
b) Liters of make up new lubricating oil per 24 hours operated at rated load
to maintain the level of oil in the sump tank constant if manufacturers
guarantees a lubricating oil consumption of 1.36 gram/KW-hr.
15. A diesel generating set has the following specifications:
Mitsubishi, 16 cylinders, Model V-280, 4 stroke cycle, supercharge
Bore and stroke - 280 mm x 290 mm
Brake power and RPM - 225 kw per cylinder at 900 RPM
Fuel type and heating value - Bunker C, 10 000 KCal/kg
Brake specific fuel consumption - 0.2064 kg/kw-hr
If the engine is operating at full load with an ambient room temperature of 32C
db and 24 C wb and 760 mm Hg barometric pressure, determine
a) the fuel consumption in L/hr (743.04 kg/hr)
b) the brake thermal efficiency (42%;39%)
c) the amount of cooling water in L/min, if the cooling loss is 30% and the
temperature rise of the cooling water is 22 C. (1689 L/min)
d) the cubic meter per minute of air entering the engine at ambient
conditions if the air-fuel ratio is 22:1 (241 m3/min)
at 32 C db and 24 C wb
v = 0.885 m3/kg
16. The following data were obtained during a full load test on a 2 cylinder, single
acting, 4 stroke cycle diesel engine, 280 mm bore by 381 mm stroke; average
RPM, 385; length of prony brake arm, 92 cm; net weight on scales (due to
torque), 227 kg; total; fuel used in 15 min. run, 5.2 kg; heating value of fuel oil,
45 400 KJ/kg; Mean effective pressure from indicator cards (average of each
cylinder), 620 KPa; Find
a) Brake power (82.6 KW
b) Brake specific fuel consumption (0.252 kg/KW-hr)
c) Brake thermal efficiency (32%)
d) Indicated power (93.33 KW)
e) Friction Power (10.7 KW)
f)Mechanical efficiency (89%)
17. A diesel engine under test gave the following performance data:
Brake power - 3357 KW
Jacket loss - 23%
Mechanical efficiency - 83%
Indicated thermal efficiency - 34%
A waste heat recovery boiler recovers 35% of the exhaust loss. If cooling water
leaving the engine is 66C and used as feed water to boiler, determine the
quantity of 34.5 KPa steam that can be produced in kg/hr.
m = BP x 100%
IP
IP = 4044.6 KW
ei = 3600(IP)_ x 100%
mf(HV)
Qs = mf(HV) = 42 825 176 KJ/hr
BP = 3600(3357) = 12 085 200 KJ/hr
FP = 3600(IP - BP) = 2 475 360 KJ/hr
Qc = cooling loss = 0.23Qs = 9 849 791 KJ/hr
QG = exhaust gas loss = Qs - (BP + Qc + FP) = 18 414 825 KJ/hr
QB = heat absorbed by waste heat recovery boiler
QB = 0.35QG = 6 445 189 KJ/hr
QB = ms (hs - hf)
hs = 2689 KJ/kg
hf = 276.23 KJ/kg
ms = 2672 kg/hr
18. A spark ignition engine produces 224 KW while using 0.0169 kg/sec of fuel. The
fuel has a higher heating value of 44 186 KJ/kg and the engine has a
compression ratio of 8:1. The friction power is found to be 22.4 KW. Determine
the indicated engine efficiency. (58%)
19. A six cylinder automotive engine with a 9 cm x 9 cm has a fuel consumption of
0.306 kg/KW-hr at 300 RPM. The brake power developed is 86 KW and the
indicated power is 105 KW. The thermal efficiency of the ideal cycle is 47%, and
the fuel has a heating value of 44 186 KJ/kg. Compute the brake and indicated
engine efficiency. (56.6%;69.2%)
20. A 6 cylinder, 4 stroke cycle, single acting, spark ignition engine with a
compression ratio of 9.5 is required to develop 67.14 KW with a torque of
194 N-m. Under the conditions the m = 78% and the Pmb = 552 KPa. For the
ideal cycle, P1 = 101.35 KPa, t1 = 35C and the hot air standard k = 1.32. If
D/L =1.1 and the mfi = 0.353 kg/KW-hr and HV = 43 967 KJ/kg, Determine
a) the bore and stroke in cm (10.1 cm; 9.2 cm)
b) the indicated thermal efficiency (23.3%)
c) the brake engine efficiency ( 34.4%)
d) the percentage clearance (11.8%)
21. A 6 cylinder, 700 mm x 900 mm, single acting, 2 stroke cycle diesel engine
develops 22.35 KW at 128 RPM. compression ratio is 13, cut off ratio is 2.45 and
overall k = 1.33. Suction pressure is 97 KPa, suction temperature 54C. the
average Pmi = 620 KPa. The fuel consumed during a 30 minutes test was 286 kg
with a LHV = 42 563 KJ/kg. Determine
a) the brake thermal efficiency
b) the mechanical efficiency
c) the indicated power
d) the brake engine efficiency
22. A single cylinder,4 stroke, compression ignition oil engine gives 15 KW at 60 RPM
and uses fuel having the composition by mass of 84% carbon and 16% hydrogen.
The air supply is 100% in excess of that required for perfect combustion. The
fuel has a calorific value of 45 000KJ/kg and the brake thermal efficiency is 30%.
Find:
a) the mass of fuel used per cycle ( 0.0022)
b) the actual mass of air taken in per cycle
c) the volume of air taken in per cycle at 100 KPa and 15C.Take
R = 0.29 KJ/kg-C
1. A 6 cylinder, 4 stroke cycle, single acting, spark ignition engine with a
compression ratio of 9.5 is required to develop 67.14 KW with a torque of
194 N-m. Under the conditions the m = 78% and the Pmb = 552 KPa. For the
ideal cycle, P1 = 101.35 KPa, t1 = 35C and the hot air standard k = 1.32. If
D/L =1.1 and the mfi = 0.353 kg/KW-hr and HV = 43 967 KJ/kg, Determine
a) the bore and stroke in cm
b) the indicated thermal efficiency
c) the brake engine efficiency
d) the percentage clearance
2. A spark ignition engine produces 224 KW while using 0.0169 kg/sec of fuel. The
fuel has a higher heating value of 44 186 KJ/kg and the engine has a
compression ratio of 8:1. The friction power is found to be 22.4 KW. Determine
the indicated engine efficiency. (58%)
3. Performance values of a 3000 KW Diesel power plant unit are as follows:
Fuel rate: 1.5 barrels for 900 KW-hr of 25API fuel
Generator Efficiency: 92%
Mechanical efficiency: 82%
Determine:
a) Engine fuel rate in kg/KW-hr
b) Engine-Generator fuel rate in kg/KW-hr
c) Indicated thermal efficiency
d) Brake thermal efficiency
e) Overall thermal efficiency
1. A 40 KW blast furnace engine shows by test a gas consumption of 4.2 m3/KW-hr. Heating value of
gas is 33,500 KJ/m3. Mechanical efficiency of the gas engine is 86%, Calculate:
a) The brake thermal efficiency
b) Indicated thermal efficiency
c) Heat rate in KJ/KW-hr
2. An 8 cylinders,2-stroke cycle, single acting diesel engine rated at 940 KW at standard condition
(P = 760 mm Hg; t = 15.56C) is directly coupled to a 24 poles alternator; 3-phase; 60 Hertz.
Assuming brake mean effective pressure of 520 KPa; m = 85% & g = 94%
a) Find the diameter of the cylinder and the length of stroke of the piston if the average piston
speed is 310 m/min.
b) What will be the actual thermal efficiency for a combined fuel rate of 0.20 kg of diesel fuel per
KW hour with a HV = 46,530 KJ/kg
3. A single cylinder,4 stroke, compression ignition oil engine gives 15 KW at 60 RPM and uses fuel
having the composition by mass of 84% carbon and 16% hydrogen. The air supply is 100% in excess
of that required for perfect combustion. The fuel has a calorific value of 45 000 KJ/kg and the brake
thermal efficiency is 30%. Find
a) the mass of fuel used per cycle
b) the actual mass of air taken in per cycle
c) the volume of air taken in per cycle at 100 KPa and 15C. Take
R = 0.29 KJ/kg-C
PUMPS AND PIPING
BERNOULLI’S ENERGY EQUATION
From 1st Law (OPEN SYSTEM)
(U  P  KE  PE)2
Q
W
(U  P  KE  PE)1
GENERAL ENERGY EQUATION
(P  P1 ) v 2  v 1
0 2

 ( z 2  z1 )
g
2g
1000
but
2
Q  U  (P)  KE  PE  W
Considering
Q  0; U  0 and W  0
0  (P)  KE  PE
  g
v2  v
g( z 2  z1 )

2(1000)
1000
For one dimensional flow
2
2
1
0  P2 2  P11 
2
2
2
2
P1 v 1
P
v

 z1  2  2  z 2

2g

2g

1
 2  1  

0
(P2  P1 ) v 2  v 1
g( z 2  z1 )



2(1000)
1000
2
Multiplyin g both sides of the equation by
1000(P2  P1 ) v 2  v 1

 ( z 2  z1 )
g
2g
2
0
1000
(P  P1 ) v 2  v 1
0 2

 ( z 2  z1 )

2g
1   2  
2
2
2
1000
g
Bernoulli’s Energy Equation
P
 Pr essure head , meters

v2
 Velocity head, meters
2g
z - Elevation head, meters
P - Pressure,KPa
v - velocity, m/sec
 - specific weight, KN/m 3
z  elevation, m
g - gravitatio nal acceleration,
m
sec 2
Note
z is positive if meassured above datum
z is negative if measured below datum
Considering Head Loss and W = 0
U  Q  HL in meters
With Energy Head added to the fluid (Work done on the system; let –W = ht in meters)
With Energy Head given up by the fluid (Work done on the system; let W = h in meters)
APPLICATION OF BERNOULLI’S ENERGY THEOREM
Nozzle
JET POWER
2
v
PJet  Q 2
2g
If a nozzle makes an angle  with the horizontal,
v  velocity of water at tip of nozzle
at 1 to 0
vo 2  v12 - 2gd
0  v12 - 2gd
(vsin θ) 2
1
2g
v0  v1  gt
d
v sin θ
2
g
at 0 to 2
t
d  v1t  1 2 gt 2
d  0  1 2 gt 2
d  1 2 gt 2  3
(vsin θ) 2 1 2
 2 gt
2g
(vsin θ) 2
g2
(vsin θ)
t
4
g
R  v cos θ(2 t )
t2 
2(vsin θ)( v cos θ) v 2 (2sin θ cos θ)

g
g
(2sin θ cos θ)  sin( 2θ)
R
R
v 2 (sin2 θ)
5
g
Venturi Meter
Orifice
By applying Bernoulli's Energy theorem:
2
2
P1 v 1
P
v

 Z1  2  2  Z2

2g

2g
But P1 = P2 = Pa and v1 is negligible, then
2
v2
 Z1  Z2
2g
and from figure: Z1 - Z2 = h, therefore
2
v2
h
2g
v 2  2gh
let v2 = vt
v t  2gh
where:
vt - theoretical velocity, m/sec
h - head producing the flow, meters
g - gravitational acceleration, m/sec2
COEFFICIENT OF VELOCITY (Cv)
actual velocity
theoretica l velocity
v'
Cv 
vt
Cv 
COEFFICIENT OF CONTRACTION
area of jet @ vena contracta
area of the orifice
a
Cc 
A
Cc 
where: a - area of jet at vena contracta, m2
A - area of the orifice, m2
COEFFICIENT OF DISCHARGE
actual flow
theoretica l flow
Q'
Cd 
Q
Cd  C v Cc
Cd 
where:
v' - actual velocity
vt - theoretical velocity
a - area of jet at vena contracta
A - area of orifice
Q' - actual flow
Q - theoretical flow
Cv - coefficient of velocity
Cc - coefficient of contraction
Cd - coefficient of discharge
Sample Problem no. 1
2
2
P1 v 1
P
v

 z1  2  2  z 2

2g

2g
400
 0  0  0  0  z2

400
z2 
 40.8 m
9.81
Sample Problem no. 2
Sample Problem No. 3
PUMPS
By Engr. Yuri G. Melliza
Pump is a device that moves or compresses liquids and gases. Pumps are used in a variety of machines and other devices, including
home heating systems, refrigerators, oil wells and water wells, and turbojet and car engines. The fluids (gases or liquids) moved by
pumps range from air for inflating bicycle tires to liquid sodium and liquid potassium for cooling nuclear reactors. Most pumps are
made of steel, but some are made of glass or plastic. Gas pumps are also called compressors, fans, or blowers.
Types of Pumps
Dynamic Pump: Dynamic pumps maintain a steady flow of fluid.
Positive Displacement Pump: Positive displacement pumps, on the other hand, trap individual portions of fluid that are in an enclosed
area before moving them along.
Dynamic pumps
Centrifugal pumps consist of a motor-driven propeller like device, called an impeller, which is contained within a circular housing.
The impeller is a wheel of curved blades that rotates on an axis. Before most centrifugal pumps can start pumping liquid, they must
be primed (filled with liquid). As the impeller rotates, it creates suction that draws a continuous flow of fluid through an inlet pipe.
Fluid enters the pump at the center of the impeller and travels out along the blades due to centrifugal (outward) force. The curved
ends of the blades sweep the fluid to an outlet port. Centrifugal pumps are inexpensive and can handle large amounts of fluid. They
are widely used in chemical processing plants and oil refineries.
Axial-flow pumps have a motor-driven rotor that directs fluid along a path parallel to its axis. The fluid thus travels in a relatively
straight path from the inlet pipe through the pump to the outlet pipe. Axial-flow pumps are most often used as compressors in turbojet
engines. Centrifugal pumps are also used for this purpose, but axial-flow pumps are more efficient. Axial-flow compressors consist
of alternating rows of rotors and stationary blades. The blades and rotors produce a pressure rise in the air as it moves through the
axial-flow compressor. Air then leaves the compressor under high pressure.
Jet pumps get their name from the way they move fluid. They operate on the principle that a high-velocity fluid will carry along any
other fluid it passes through. Most jet pumps send a jet of steam or water through the fluid that needs to be moved. The jet carries
the fluid with it directly into the outlet pipe and, at the same time, creates a vacuum that draws more fluid into the pump. The amount
of fluid carried out of most jet pumps is several times the amount in the jet itself. Jet pumps can be used to raise water from wells
deeper than 60 meters. In such cases, a centrifugal pump at ground level supplies water for a jet at the bottom of the well. The jet
carries well water with it back up to ground level. Jet pumps are also used in high vacuum diffusion pumps to create a vacuum in
an enclosed area. In high vacuum diffusion pumps, a high-velocity jet of mercury or oil vapor is sent into the enclosed area. The
vapor molecules collide with the molecules of air and force them out the outlet port.
Electromagnetic pumps are used chiefly to move liquid sodium and liquid potassium, which serve as coolants in nuclear reactors.
These pumps consist of electrical conductors and magnetized pipes. The conductors send current through the fluid, which thereby
becomes an electromagnet. The fluid is then moved by the magnetic attraction and repulsion (pushing away) between the fluid's
magnetic field and that of the pipes. The fluid is therefore moved in an electromagnetic pump in much the same way as an armature
is moved in an electric motor.
Positive displacement pumps
Rotary pumps are the most widely used positive displacement pumps. They are often used to pump such viscous (sticky) liquids as
motor oil, syrup, and paint. There are three main types of rotary pumps. These types are: (1) gear pumps, (2) lobe pumps, and (3)
sliding vane pumps.
Gear pumps consist of two gears that rotate against the walls of a circular housing. The inlet and outlet ports are at opposite sides
of the housing, on line with the point where the teeth of the gears are fitted together. Fluid that enters the pump is trapped by the
rotating gear teeth, which sweep the fluid along the pump wall to the outlet port.
Lobe pumps operate in a manner similar to gear pumps. However, instead of gears, lobe pumps are equipped with impellers that
have lobes (rounded projections) fitted together. Lobe pumps can discharge large amounts of fluid at low pressure.
Sliding vane pumps consist of a slotted impeller mounted off-center in a circular housing.
Sliding vanes (blades) move in and out of the slots. As the vanes rotate by the inlet port,
they sweep up fluid and trap it against the pump wall. The distance between the impeller
and the pump wall narrows near the outlet port. As the fluid is carried around to this port, the
vanes are pushed in and the fluid is compressed. The pressurized fluid then rushes out the
outlet port.
Reciprocating pumps consist of a piston that moves back and forth within a cylinder. One end of the cylinder has an opening through
which the connecting rod of the piston passes. The other end of the cylinder, called the closed
end, has an inlet valve or an
outlet valve, or both, depending on the type of pump. In some reciprocating pumps, the inlet valve or the outlet valve is on the
piston. Common reciprocating pumps include lift pumps, force pumps, and bicycle tire pumps.
Lift pumps draw water from wells. In a lift pump, the inlet valve is at the closed end of the cylinder and the outlet valve is on the
piston. As the piston is raised, water is drawn up through the inlet valve. As the piston moves down, the inlet valve closes, forcing
water through the outlet valve and up above the piston. As the piston is raised again, the outlet valve closes and the water is lifted
to an opening, where it leaves the pump. At the same time, more water is drawn through the inlet valve. It is theoretically possible
for a lift pump to raise water in a well almost 10 meters. However, because of leakage and resistance, a lift pump cannot raise water
that is deeper than about 7.5 meters.
Force pumps are similar to lift pumps. However, in force pumps, both the inlet valve and the outlet valve are at the closed end of
the cylinder. As the piston moves away from the closed end, fluid enters the cylinder. When the piston moves toward the closed
end, the fluid is forced out the outlet valve.
Bicycle tire pumps differ in the number and location of the valves they have and in the way air enters the cylinder. Some simple
bicycle tire pumps have the inlet valve on the piston and the outlet valve at the closed end of the cylinder. Air enters the pump near
the point where the connecting rod passes through the cylinder. As the rod is pulled out, air passes through the piston and fills the
areas between the piston and the outlet valve. As the rod is pushed in, the inlet valve closes and the piston forces air through the
outlet valve.
History
Pumping devices have been an important means of moving fluids for thousands of years. The ancient Egyptians used water wheels
with buckets mounted on them to move water for irrigation. The buckets scooped water from wells and streams and deposited it in
ditches that carried it to fields. In the 200's B.C., Ctesibius, a Greek inventor, made a reciprocating pump for pumping water. About
the same time, Archimedes, a Greek mathematician, invented a screw pump that was made up of a screw rotating in a cylinder. This
type of pump was used to drain and to irrigate the Nile Valley.
True centrifugal pumps were not developed until the late 1600's, when Denis Papin, a French-born inventor, made one with straight
vanes. The British inventor John Appold introduced a curved-vane centrifugal pump in 1851. Axial-flow compressors were first
used on turbojet engines in the 1940's.
CLASSIFICATION OF PUMPS
Centrifugal: It consist essentially of an impeller arranged to rotate within a casing so that the liquid will enter at the center or eye of
the impeller and be thrown outward by centrifugal force to the outer periphery of the impeller and discharge into the outer case. It
operates at high discharge pressure, low head, high speed and they are not self priming.
a. Centrifugal
b. Mixed Flow
1. single stage
2. multi stage
3. Propeller or axial flow
4. Peripheral
Rotary: It is a positive displacement pump consisting of a fixed casing containing gears, cams, screws, vanes, plungers or similar
element actuated by the rotation of the drive shaft. A rotary pump traps a quantity of liquid and moves it along toward the discharge
point. For a gear type rotary pump the unmeshed gears at the pump provides a space for the liquid to fill as the gears rotate. The
liquid trapped between the teeth and the pump casing is eventually released at the discharge line. It operates at low heads, low
discharge and is used for pumping viscous liquids like oil.
a. cam
c. gear
b. screw
d. vane
Reciprocating: It is a positive displacement unit wherein the pumping action is accomplished by the forward and backward movement
of a piston or a plunger inside a cylinder usually provided with valves.
a. Piston
b. Direct Acting
1. single
2. duplex
c. Crank and Flywheel
d. Plunger
e. Power Driven
1. simplex
2. duplex
3. triplex
Deepwell pumps: It is used when pumping water from deep wells. The pump is lowered into the well and operated close to water
level. They are usually motor driven with the motor being at the ground level and
connected to the pump by a long vertical line shaft.
a. Turbine
b. Ejector or centrifugal
c. reciprocating
d. Airlift
For a final choice of a pump for a particular operation the following data is needed.
Number of units required
Nature of liquid
Capacity
Suction conditions
Discharge conditions
Intermittent or continuous service
Total dynamic head
Position of pump, vertical or horizontal
Location, geographical, indoor, outdoor, elevation
Type of power drive
Closed
Tank
Open Tank
PUMP
FUNDAMENTAL EQUATIONS
TOTAL DYNAMIC HEAD
2
2
P  P1 v 2  v 1
Ht  2

 Z 2  Z 1  HL
γ
2g
FLUID POWER or WATER POWER
meters
WP = QHt KW
DISCHARGE or CAPACITY
Q = Asvs = Advd
s – refers to suction
d – refers to discharge
BRAKE or SHAFT POWER
BP 
2 πTN
KW
60,000
PUMP EFFICIENCY
ηP 
WP
x 100%
BP
MOTOR EFFICIENCY
ηm 
BP
x 100%
MP
COMBINED PUMP-MOTOR EFFICIENCY
WP
x 100%
MP
MOTOR POWER
ηC  ηP ηm
ηC 
For Single Phase Motor
MP 
EI(cosθ)
KW
1000
For 3 Phase Motor
MP 
3 EI(cosθ)
KW
1000
where: P - pressure in KPa
v - velocity, m/sec
T - brake torque, N-m
N - no. of RPM
 - specific weight of liquid, KN/m3
FP - fluid power, KW
Z - elevation, meters
WP or BP - brake power, KW
g - gravitational acceleration, m/sec2
MP - power input to motor
HL - total head loss, meters
E - energy, Volts
s – suction
I - current, amperes
d - discharge
(cos) - power factor
1 & 2 – Reference Points (@ suction and
discharge)
Sample Problem no. 1
A pump handles brine having a temperature of 0C and S = 1.2 when the discharge pressure is 173 KPag and the suction gauge
indicates a vacuum of 305 mm Hg. The discharge gauge is 61 cm above the pump centerline and the point of attachment of the
suction gauge is 25 cm below the pump centerline. Suction and discharge pipes inside diameters are 100 mm and 80 mm respectively.
Find the total head of the pump in meters of the fluid handled. If the efficiency of the motor-driven pump is 88%, find the brake
power load of the motor when the pump handles 32 L/sec of brine. (8.5 KW)
0.61 m
2
0.25 m
Datum line
(PCL)
PUMP
1
γ  1.2(9.81)  11.772
KN
m
3
z1  0.25 m
z 2  0.61 m
Reference Datum : Pump Centerline
 101.325 
P1  -305
  40.7 KPa
 760 
P2  173 KPa
Q
m3
; Q  0.032
A
sec
π
2
As  (0.100)  0.00785 m 2
4
π
Ad  (0.08) 2  0.005 m 2
4
m
v1  4.08
sec
m
v 2  6.4
sec
v
 P  P   v 2  v12 
h   2 1    2
 z 2  z1   H L L

2g
 γ  

2
2
 172  40.7   6.4  4.08 
h 

 0.61  0.25  0

 11.772   2(9.81) 
h  18.07  0.85  0.86  19.78 m
Power 
0.032(11.772)(19.78)
 8.5 KW
0.88
HEAD LOSSES
HL = Major loss + Minor losses
Major Loss: Head loss due to friction and turbulence in pipes
Minor Losses: Minor losses include, losses due to valves and fittings, enlargement, contraction, pipe entrance and pipe exit. Minor
losses are most easily obtained in terms of equivalent length of pipe "Le". The advantage of this approach is that both pipe and
fittings are expressed in terms of "Equivalent Length" of pipe of the same relative roughness.
Considering Major loss only
Using the Darcy-Weisbach Equation
(Henry Darcy, a French Engineer and Julius Weisbach, a German engineer)
hf 
fLv 2
meters
2gD
v2
hf  K
2g
fL
K
D
Considering Major and Minor losses
hf 
f(L  ΣL e )v 2
2gD
meters
Where;
f - friction factor from Moody's Chart
L - length of pipe, m
Le - equivalent length in straight pipe of valves and fittings, m
v - velocity, m/sec
D - pipe inside diameter, m
g - gravitational acceleration, m/sec2
REYNOLD'S NUMBER: Reynold's Number is a non dimensional one which combines the physical quantities which describes the
flow either Laminar or Turbulent flow.
The friction loss in a pipeline is also dependent upon this dimensionless factor.
NR 
where;
ρvD vD

μ
ν
μ m2
ν
ρ sec
 - absolute or dynamic viscosity, Pa-sec
 - kinematic viscosity, m2/sec
For a Reynold's Number of less 2100 flow is said to Laminar
For a Reynold's Number of greater than 3000 the flow is Turbulent
FRICTION FACTOR (Resistance Coefficient)
For Laminar Flow
f
64
NR
For Turbulent Flow
Prandtl Equation for smooth pipes for NR > 3000
1


 2 log N R f  0.8
f
Cyril F. Colebrook Equation (1910 – 1997)

2.51 
 - 2 log  D 
 3.7 N f 
f
R


0.25
f
2
 

5
.
74

log  D 
  3.7 N R 0.9 

 
1
5
Q  2.22D 2
gh f



 m3
1.78ν
D

 log 

L
3
 3.7
 sec
gh
2
f


D
L

Zigrang-Sylvester Equation (From Fluid Mechanics by F.M. White)


5.02   D 13 
 4.0 log  D 
log

 3.7 N 
 3.7 N R
f
R



1
f 


5.02   D 13 
 4.0 log  D 
log

 3.7 N R 
 3.7 N R



S. E. Haaland Equation (1983)
1
1.11

  
 6.9  D  
 1.8 log 

N R  3.7  
f


 
1
In 1944, Lewis Ferry Moody plotted the Darcy–Weisbach friction factor against Reynolds number NR for various values of relative
roughness

This chart became commonly known as the Moody Chart or Moody Diagram.
D
Moody's Chart
 - absolute roughness
D - inside diameter
/D - relative roughness
Absolute Pipe Roughness () is a measure of pipe wall irregularities of commercial pipes. Other than pipes, absolute roughness is
also used for representing the irregularities of other equipment walls, for example, walls of heat exchanger shell. The absolute
roughness has dimensions of length and is usually expressed in millimeter (mm).
Relative Roughness' or 'Roughness factor of a pipe wall can be defined as the ratio of absolute roughness to the pipe nominal
diameter. Relative roughness factor is often used for pressure drop calculations for pipes and other equipment. The relative roughness
factor is an important parameter for determining friction factor based on Reynold's number for flow in a pipe.
Relative roughness = /D
Absolute Roughness is usually defined for a material and can be measured experimentally.
Following table gives typical roughness values in millimeters for commonly used piping materials.
ABSOLUTE ROUGHNESS () OF PIPE MATERIAL
 (mm)
MATERIAL
Aluminum
Aluminum, Lead
Drawn Brass, Drawn Copper
0.0015
0.001 - 0.002
0.0015
Drawn Tubing, Glass, Plastic
0.0015-0.01
Drawn Brass, Copper, Stainless Steel (New)
0.0015-0.01
Flexible Rubber Tubing - Smooth
Flexible Rubber Tubing - Wire Reinforced
0.006-0.07
0.3-4
Wrought Iron (New)
0.045
Carbon Steel (New)
0.02-0.05
Carbon Steel (Slightly Corroded)
0.05-0.15
Carbon Steel (Moderately Corroded)
0.15-1
Carbon Steel (Badly Corroded)
1-3
Carbon Steel (Cement-lined)
1.5
Asphalted Cast Iron
0.1-1
Cast Iron (new)
New cast iron
0.25
Cast Iron (old, sandblasted)
Fiberglass
1
Worn cast iron
0.8 - 1.5
Corroding cast iron
1.5 - 2.5
Asphalted cast iron
0.012
Sheet Metal Ducts (with smooth joints)
0.02-0.1
Galvanized Iron
Galvanized steel
0.025-0.15
Wood Stave
0.18-0.91
Wood Stave, used
0.25-1
Smooth Cement
0.5
Concrete – Very Smooth
0.025-0.2
Concrete – Fine (Floated, Brushed)
0.2-0.8
Concrete – Rough, Form Marks
PVC, Plastic Pipes
0.8-3
Galvanized iron
0.015
Riveted Steel
Rusted steel
0.91-9.1
Stainless steel
0.015 – 0.03
Steel commercial pipe
0.045 - 0.09
Stretched steel
0.015
Water Mains with Tuberculations
Weld steel
1.2
Brickwork, Mature Foul Sewers
3
0.25 - 0.8
0.005
0.15
0.0015
0.15 - 4
0.045
Smoothed cement
0.3
Ordinary concrete
0.3 - 3
Well planed wood
0.18 - 0.9
Ordinary wood
5
PVC
0.0015
PIPE, TUBING AND FITTINGS
Nominal Pipe Diameter: Pipe sizes are based on the approximate diameter and are reported as nominal pipe sizes. Regardless of wall
thickness, pipes of the same nominal diameter have the same outside diameter. This permits interchange of fittings. Pipe may be
manufactured with different and various wall thickness, so some standardization is necessary. A method of identifying pipe sizes
has been established by ANSI (American National Standard Institute). By convention, pipe size and fittings are characterized in
terms of Nominal Diameter and wall thickness.
For steel pipes, nominal diameter is approximately the same as the inside diameter for 12" and smaller. For sizes of 14" and larger,
the nominal diameter is exactly the outside diameter.
SCHEDULE NUMBER: The wall thickness of pipe is indicated by a schedule number, which is a function of internal pressure and
allowable stress.
Schedule Number  1000P/S
where P - internal working pressure, KPa
S - allowable stress, KPa
Schedule number in use: 10,20,30, 40,60, 80, 100, 120, 140, and 160.
Schedule 40 "Standard Pipe"
Schedule 80 " Extra Strong Pipe"
TUBINGS: Tubing specifications are based on the actual outside diameter with a designated wall thickness. Conventional systems
such as the Birmingham Wire Gauge "BWG" are used to indicate the wall thickness.
FITTING: The term fitting refers to a piece of pipe that can:
1. Join two pieces of pipe
ex. couplings and unions
2. Change pipeline directions
ex. elbows and tees
3. Change pipeline diameters
ex. reducers
4. Terminate a pipeline
ex. plugs and valves
5. Join two streams to form a third
ex. tees, wyes, and crosses
6. Control the flow
ex. valves
VALVES: A valve is also a fitting, but it has more important uses than simply to connect pipe. Valves are used either to control the
flow rate or to shut off the flow of fluid.
DESIGN OF A PIPING SYSTEM
The engineer should consider the following items when he is developing the design of a piping system.
1. Choice of material and sizes
2. Effects of temperature level and temperature changes.
a. insulation
b. thermal expansion
c. freezing
3. Flexibility of the system for physical and thermal shocks.
4. Adequate support and anchorage
5. Alteration in the system and the service.
6. Maintenance and inspection.
7. Ease of installation
8. Auxiliary and standby pumps and lines
9. Safety
a. Design factors
b. Relief valves and flare systems
TABLE FOR DN and NPS
Nominal Pipe Size NPS, Nominal Bore NB, Outside Diameter OD
What is a relation between NPS, NB & OD of a pipe?
What is NPT & what are its applications?
What is a relation between NPS, NB & OD of a pipe?
Nominal Pipe Size (NPS) is a North American set of standard sizes for pipes used for high or low pressures and temperatures. Pipe
size is specified with two non-dimensional numbers: a nominal pipe size (NPS) for diameter based on inches, and a schedule (Sched.
or Sch.) for wall thickness. NPS is often incorrectly called National Pipe Size, due to confusion with national pipe thread (NPT)
NB (nominal bore) is the European designation equivalent to NPS is DN (diamètre nominal/nominal diameter/Durchmesser nach
Norm), in which sizes are measured in millimeters. NB (nominal bore) is also frequently used interchangeably with NPS.
OD is the outside diameter of the pipe and is fixed for a given size.
The NPS is very loosely related to the inside diameter in inches, but only for NPS 1/8 to NPS 12. For NPS 14 and larger, the NPS is
equal to the outside diameter (OD) in inches. For a given NPS, the OD stays constant and the wall thickness increases with larger
SCH. For a given SCH, the OD increases with increasing NPS while the wall thickness increases or stays constant. Pipe sizes are
documented by a number of standards, including API 5L, ANSI/ASME B36.10M in the US, BS 1600 and BS EN 10255 in the
United Kingdom and Europe, and ISO 65 internationally.
For NPS of 5 and larger, the DN is equal to the NPS multiplied by 25 (not 25.4).
Table below shows the relation between Nominal pipe size, Nominal diameter & outside diameter for pipes:
Nominal
NPS
[inches]
Pipe
size
Nominal
DN
[mm]
Diameter
Outside
OD
[mm]
1/2
15
21.3
3/4
20
26.7
1
25
33.4
1 1/4
32
42.16
1 1/2
40
48.26
2
50
60.3
2 1/2
65
73.03
3
80
88.9
4
100
114.3
5
125
141.3
6
150
168.28
Diameter
8
200
219.08
10
250
273.05
12
300
323.85
14
350
355.6
16
400
406.4
18
450
457.2
20
500
508
24
600
609.6
28
700
711.2
32
800
812.8
36
900
914.4
40
1000
1016
42
1050
1066.8
44
1100
1117.6
48
1200
1219.2
52
1300
1320.8
56
1400
1422.4
60
1500
1524
64
1600
1625.6
68
1700
1727.2
72
1800
1828.8
76
1900
1930.4
80
2000
2032
88
2200
2235.2
96
2400
2438.4
104
2600
2641.6
112
2800
2844.8
120
3000
3048
128
3200
3251.2
Definition and Details of Nominal Pipe Size
NOMINAL PIPE SIZE
Nominal Pipe Size (NPS) is a North American set of standard sizes for pipes used for high or low pressures and temperatures. The
name NPS is based on the earlier "Iron Pipe Size" (IPS) system.
That IPS system was established to designate the pipe size. The size represented the approximate inside diameter of the pipe in
inches. An IPS 6" pipe is one whose inside diameter is approximately 6 inches. Users started to call the pipe as 2inch, 4inch, 6inch
pipe and so on. To begin, each pipe size was produced to have one thickness, which later was termed as standard (STD) or standard
weight (STD.WT.). The outside diameter of the pipe was standardized.
As the industrial requirements handling higher pressure fluids, pipes were manufactured with thicker walls, which has become known
as an extra strong (XS) or extra heavy (XH). The higher pressure requirements increased further, with thicker wall pipes.
Accordingly, pipes were made with double extra strong (XXS) or double extra heavy (XXH) walls, while the standardized outside
diameters are unchanged. Note that on this website only terms XS and XXS are used.
PIPE SCHEDULE
So, at the IPS time only three wall thickness were in use. In March 1927, the American Standards Association surveyed industry and
created a system that designated wall thicknesses based on smaller steps between sizes. The designation known as nominal pipe size
replaced iron pipe size, and the term schedule (SCH) was invented to specify the nominal wall thickness of pipe. By adding schedule
numbers to the IPS standards, today we know a range of wall thicknesses, namely:
SCH 5, 5S, 10, 10S, 20, 30, 40, 40S, 60, 80, 80S, 100, 120, 140, 160, STD, XS AND XXS.
Nominal pipe size (NPS) is a dimensionless designator of pipe size. It indicates standard pipe size when followed by the specific
size designation number without an inch symbol. For example, NPS 6 indicates a pipe whose outside diameter is 168.3 mm.
The NPS is very loosely related to the inside diameter in inches, and NPS 12 and smaller pipe has outside diameter greater than the
size designator. For NPS 14 and larger, the NPS is equal to 14inch.
For a given NPS, the outside diameter stays constant and the wall thickness increases with larger schedule number. The inside
diameter will depend upon the pipe wall thickness specified by the schedule number.
SUMMARY:
Pipe size is specified with two non-dimensional numbers,
nominal pipe size (NPS)
schedule number (SCH)
and the relationship between these numbers determine the inside diameter of a pipe.
Stainless Steel Pipe dimensions determined by ASME B36.19 covering the outside diameter and the Schedule wall thickness. Note
that stainless wall thicknesses to ASME B36.19 all have an "S" suffix. Sizes without an "S" suffix are to ASME B36.10 which is
intended for carbon steel pipes.
The International Standards Organization (ISO) also employs a system with a dimensionless designator.
Diameter nominal (DN) is used in the metric unit system. It indicates standard pipe size when followed by the specific size
designation number without a millimeter symbol. For example, DN 80 is the equivalent designation of NPS 3. Below a table with
equivalents for NPS and DN pipe sizes.
NPS
1/2
3/4
1
1¼
1½
2
2½
3
3½
4
DN
15
20
25
32
40
50
65
80
90
100
Note: For NPS ≥ 4, the related DN = 25 multiplied by the NPS number.
Do you now what is "ein zweihunderter Rohr"?. Germans means with that a pipe NPS 8 or DN 200. In this case, the Dutch talking
about a "8 duimer". I'm really curious how people in other countries indicates a pipe.
EXAMPLES OF ACTUAL O.D. AND I.D.
ACTUAL OUTSIDE DIAMETERS
NPS 1 actual O.D. = 1.5/16" (33.4 mm)
NPS 2 actual O.D. = 2.3/8" (60.3 mm)
NPS 3 actual O.D. = 3½" (88.9 mm)
NPS 4 actual O.D. = 4½" (114.3 mm)
NPS 12 actual O.D. = 12.3/4" (323.9 mm)
NPS 14 ACTUAL O.D. = 14"(355.6 MM)
ACTUAL INSIDE DIAMETERS OF A 1 INCH PIPE.
NPS 1-SCH 40 = O.D.33,4 mm - WT. 3,38 mm - I.D. 26,64 mm
NPS 1-SCH 80 = O.D.33,4 mm - WT. 4,55 mm - I.D. 24,30 mm
NPS 1-SCH 160 = O.D.33,4 mm - WT. 6,35 mm - I.D. 20,70 mm
Such
as
above
defined,
no
inside
diameter
corresponds
to
the
truth
1"
(25,4
mm).
The inside diameter is determined by the wall thickness (WT).
ANSI Standard Steel Pipe Chart ASME B36.10 ASME B36.19
Gauge Sizes | Sheet Metal Gauge Sizes | Pipe Sizes | Pipe Specification | Pipe Schdule | ANSI Pipe Chart | NPS
According to ASME B36.10 and ASME B 36.19.
NPS
1/8
1/4
3/8
1/2
3/4
1
1 1/4
1 1/2
OD
Schedule Designations
Wall Thickness
Inside Diameter
Weight
(Inches)
(ANSI/ASME)
(Inches)
(Inches)
(lbs./ft.)
10/10S
0.049
0.307
0.1863
Std./40/40S
0.068
0.269
0.2447
XS/80/80S
0.095
0.215
0.3145
10/10S
0.065
0.41
0.3297
Std./40/40S
0.088
0.364
0.4248
XS/80/80S
0.119
0.302
0.5351
10/10S
0.065
0.545
0.4235
Std./40/40S
0.091
0.493
0.5676
XS/80/80S
0.126
0.423
0.7388
5/5S
0.065
0.71
0.5383
10/10S
0.083
0.674
0.671
Std./40/40S
0.119
0.622
0.851
XS/80/80S
0.147
0.546
1.088
160
0.188
0.466
1.309
XX
0.294
0.252
1.714
5/5S
0.065
0.92
0.6838
10/10S
0.083
0.884
0.8572
Std./40/40S
0.113
0.824
1.131
XS/80/80S
0.154
0.742
1.474
160
0.219
0.618
1.944
XX
0.308
0.434
2.441
5/5S
0.065
1.185
0.8678
10/10S
0.109
1.097
1.404
Std./40/40S
0.133
1.049
1.679
XS/80/80S
0.179
0.957
2.172
160
0.25
0.815
2.844
XX
0.358
0.599
3.659
5/5S
0.065
1.53
1.107
10/10S
0.109
1.442
1.806
Std./40/40S
0.14
1.38
2.273
XS/80/80S
0.191
1.278
2.997
160
0.25
1.16
3.765
XX
0.382
0.896
5.214
5/5S
0.065
1.77
1.274
0.405
0.54
0.675
0.84
1.05
1.315
1.66
1.9
2
2 1/2
3
3 1/2
4
4 1/2
5
2.375
2.875
3.5
4
4.5
5
5.563
10/10S
0.109
1.682
2.085
Std./40/40S
0.145
1.61
2.718
XS/80/80S
0.2
1.5
3.631
160
0.281
1.338
4.859
XX
0.4
1.1
6.408
5/5S
0.065
2.245
1.604
10/10S
0.109
2.157
2.638
Std./40/40S
0.154
2.067
3.653
XS/80/80S
0.218
1.939
5.022
160
0.344
1.689
7.462
XX
0.436
1.503
9.029
5/5S
0.083
2.709
2.475
10/10S
0.12
2.635
3.531
Std./40/40S
0.203
2.469
5.793
XS/80/80S
0.276
2.323
7.661
160
0.375
2.125
10.01
XX
0.552
1.771
13.69
5/5S
0.083
3.334
3.029
10/10S
0.12
3.26
4.332
Std./40/40S
0.216
3.068
7.576
XS/80/80S
0.3
2.9
10.25
160
0.438
2.624
14.32
XX
0.6
2.3
18.58
5/5S
0.083
3.834
3.472
10/10S
0.12
3.76
4.973
Std./40/40S
0.226
3.548
9.109
XS/80/80S
0.318
3.364
12.5
XX
0.636
2.728
22.85
5/5S
0.083
4.334
3.915
10/10S
0.12
4.26
5.613
Std./40/40S
0.237
4.026
10.79
XS/80/80S
0.337
3.826
14.98
120
0.438
3.624
19
160
0.531
3.438
22.51
XX
0.674
3.152
27.54
Std./40/40S
0.247
4.506
12.53
XS/80/80S
0.355
4.29
17.61
XX
0.71
3.58
32.43
5/5S
0.109
5.345
6.349
10/10S
0.134
5.295
7.77
6
7
8
9
10
6.625
7.625
8.625
9.625
10.75
Std./40/40S
0.258
5.047
14.62
XS/80/80S
0.375
4.813
20.78
120
0.5
4.563
27.04
160
0.625
4.313
32.96
XX
0.75
4.063
38.55
5/5S
0.109
6.407
7.585
10/10S
0.134
6.357
9.289
Std./40/40S
0.28
6.065
18.97
XS/80/80S
0.432
5.761
28.57
120
0.562
5.491
36.39
160
0.719
5.189
45.35
XX
0.864
4.897
53.16
Std./40/40S
0.301
7.023
23.57
XS/80/80S
0.5
6.625
38.05
XX
0.875
5.875
63.08
5S
0.109
8.407
9.914
10/10S
0.148
8.329
13.4
20
0.25
8.125
22.36
30
0.277
8.071
24.7
Std./40/40S
0.322
7.981
28.55
60
0.406
7.813
35.64
XS/80/80S
0.5
7.625
43.39
100
0.594
7.439
50.95
120
0.719
7.189
60.71
140
0.812
7.001
67.76
XX
0.875
6.875
72.42
160
0.906
6.813
74.69
Std./40/40S
0.342
8.941
33.9
XS/80/80S
0.5
8.625
48.72
XX
0.875
7.875
81.77
5S
0.134
10.482
15.19
10S
0.165
10.42
18.7
20
0.25
10.25
28.04
30
0.307
10.136
34.24
Std./40/40S
0.365
10.02
40.48
XS/60/80S
0.5
9.75
54.74
80
0.594
9.564
64.43
100
0.719
9.314
77.03
120
0.844
9.064
89.29
140
1
8.75
104.13
11
12
12
14
16
11.75
12.75
12.75
14
16
160
1.125
8.5
115.64
Std./40/40S
0.375
11
45.55
XS/80/80S
0.5
10.75
60.07
XX
0.875
10
101.63
5S
0.165
12.42
22.18
10S
0.18
12.39
24.2
20
0.25
12.25
33.38
30
0.33
12.09
43.77
Std./40S
0.375
12
49.56
40
0.406
11.938
53.53
XS/80S
0.5
11.75
65.42
60
0.562
11.626
73.15
80
0.688
11.376
88.63
100
0.844
11.064
107.32
120
1
10.75
125.49
140
1.125
10.5
139.67
160
1.312
10.126
160.27
10S
0.188
13.624
27.73
10
0.25
13.5
36.71
20
0.312
13.376
45.61
Std./30/40S
0.375
13.25
54.57
40
0.438
13.124
63.44
XS/80S
0.5
13
72.09
60
0.594
12.814
85.05
80
0.75
12.15
106.13
100
0.938
12.126
130.85
120
1.094
11.814
150.9
140
1.25
11.5
170.21
160
1.406
11.188
189.1
10S
0.188
15.624
31.75
10
0.25
15.5
42.05
20
0.312
15.376
52.27
Std./30/40S
0.375
15.25
62.58
XS/40/80S
0.5
15
82.77
60
0.656
14.688
107.5
80
0.844
14.314
136.61
100
1.031
13.938
164.82
120
1.219
13.564
192.43
140
1.438
13.124
223.64
160
1.594
12.814
245.25
18
20
NPS
22
24
10S
0.188
17.624
35.76
10
0.25
17.5
47.39
20
0.312
17.376
58.94
Std./40S
0.375
17.25
70.59
30
0.438
17.124
82.15
XS/80S
0.5
17
93.45
40
0.562
16.876
104.67
60
0.75
16.5
138.17
80
0.938
16.126
170.92
100
1.156
15.688
207.96
120
1.375
15.25
244.14
140
1.562
14.876
274.22
160
1.781
14.438
308.5
10
0.25
19.5
52.73
20
0.375
19.25
78.6
30
0.5
19
104.13
40
5.94
18.814
123.11
60
8.12
18.376
166.4
80
1.031
17.938
208.87
100
1.281
17.438
256.1
120
1.5
17
296.37
140
1.75
16.5
341.09
160
1.969
16.064
379.17
OD
Schedule Designations
Wall Thickness
Inside Diameter
Weight
(Inches)
(ANSI/ASME)
(Inches)
(Inches)
(lbs./ft.)
10/10S
0.25
21.5
58.07
Std./20/40S
0.375
21.25
86.61
XS/30/80S
0.5
21
114.81
60
0.875
20.25
197.41
80
1.125
19.75
250.81
100
1.375
19.25
302.88
120
1.625
18.75
353.61
140
1.875
18.25
403
160
2.125
17.75
451.06
10/10S
0.25
23.5
63.41
Std./20/40S
0.375
23.25
94.62
XS/80S
0.5
23
125.49
30
0.562
22.876
140.68
40
0.688
22.626
171.29
60
0.969
22.064
238.35
18
20
22
24
26
28
30
32
34
36
42
48
26
28
30
32
34
36
42
48
Related References:
80
1.219
21.564
296.58
100
1.531
20.938
367.39
120
1.812
20.376
429.39
140
2.062
19.876
483.1
160
2.344
19.314
542.13
10
0.312
25.376
85.6
Std./40S
0.375
25.25
102.63
XS/80S
0.5
25
136.17
10
0.312
27.376
92.26
Std./40S
0.375
27.25
110.64
20/80S
0.5
27
146.25
30
0.625
26.75
182.73
10
0.312
29.376
98.93
Std./40S
0.375
29.25
118.65
XS/20/80S
0.5
29
157.53
30
0.625
28.75
196.08
10
0.312
31.376
105.59
Std.
0.375
31.25
126.66
20
0.5
31
168.21
30
0.625
30.75
109.43
40
0.688
30.624
230.08
10
0.312
33.376
112.25
Std.
0.375
33.25
134.67
20
0.5
33
178.89
30
0.625
32.75
222.78
40
0.688
32.624
244.77
10
0.312
35.375
118.92
Std./40S
0.375
35.25
142.68
XS/80S
0.5
35
189.57
Std./40S
0.375
41.25
166.71
XS/80S
0.5
41
221.61
30
0.625
40.75
276.18
40
0.75
40.5
330.41
Std./40S
0.375
47.25
XS/80S
0.5
47
190.74
PVC Pipe Reference
Select Mechanical or Inflatable Pipe Plugs based on the below pipe ID and along with the pressure to be blocked
PVC and CPVC Pipes - Schedule 40
Nominal Pipe Size Outside
(inches)
(inches)
Diameter Minimum
Thickness
(inches)
Wall Inside
(inches)
Diameter *) Weight
(lb/ft)
PVC
CPVC
1/2
0.840
0.109
0.622
0.16
0.17
3/4
1.050
0.113
0.824
0.21
0.23
1
1.315
0.133
1.049
0.32
0.34
1 1/4
1.660
0.140
1.380
0.43
0.46
1 1/2
1.900
0.145
1.610
0.51
0.55
2
2.375
0.154
2.067
0.68
0.74
2 1/2
2.875
0.203
2.469
1.07
1.18
3
3.500
0.216
3.068
1.41
1.54
4
4.500
0.237
4.026
2.01
2.20
5
5.563
0.258
5.047
2.73
6
6.625
0.280
6.065
3.53
3.86
8
8.625
0.322
7.981
5.39
5.81
10
10.750
0.365
10.020
7.55
8.24
12
12.750
0.406
11.938
10.01
10.89
14
14.000
0.438
13.124
11.80
16
16.000
0.500
15.000
15.43
PVC and CPVC Pipes - Schedule 80
1/2
0.840
0.147
0.546
0.20
0.22
3/4
1.050
0.154
0.742
0.27
0.30
1
1.315
0.179
0.957
0.41
0.44
1 1/4
1.660
0.191
1.278
0.52
0.61
1 1/2
1.900
0.200
1.500
0.67
0.74
2
2.375
0.218
1.939
0.95
1.02
2 1/2
2.875
0.276
2.323
1.45
1.56
3
3.500
0.300
2.900
1.94
2.09
4
4.500
0.337
3.826
2.75
3.05
5
5.563
0.375
4.813
3.87
6
6.625
0.432
5.761
5.42
5.82
8
8.625
0.500
7.625
8.05
8.83
10
10.750
0.593
9.564
12.00
13.09
PVC and CPVC Pipes - Schedule 40
Nominal Pipe Size Outside
(inches)
(inches)
Diameter Minimum
Thickness
(inches)
Wall Inside
(inches)
Diameter *) Weight
(lb/ft)
PVC
CPVC
18.0
12
12.750
0.687
11.376
16.50
14
14.000
0.750
12.500
19.30
16
16.000
0.843
14.314
25.44
Inside Diameter = Outside Diameter - 2 x Minimum Wall Thickness
PVC - Polyvinyl Chloride
strong and rigid
resistant to a variety of acids and bases
may be damaged by some solvents and chlorinated hydrocarbons
maximum usable temperature 140oF (60oC)
usable for water, gas and drainage systems
not useable in hot water systems
Operating Temp (°F) Operating Temp (°C) De-Rating Factor
73
22.8
1.00
80
26.7
0.88
90
32.2
0.75
100
37.8
0.62
110
43.3
0.51
120
48.9
0.40
130
54.4
0.31
140
60
0.22
CPVC - Chlorinated Polyvinyl Chloride
similar to PVC - but designed for water up to approx. 180oF (82oC)
Operating Temp (°F) Operating Temp (°C) De-Rating Factor
73 - 80
22.8 - 26.7
1.00
90
32.2
0.91
100
37.8
0.82
110
43.3
0.72
120
48.9
0.65
130
54.4
0.57
140
60
0.50
150
65.6
0.42
160
71.1
0.40
170
76.7
0.29
180
82.2
0.25
200
93.3
0.20
SAMPLE PROBLEMS
Water at 15C is flowing in a 5 cm diameter horizontal pipe made of stainless steel at a rate of 0.0057 m 3/sec. Determine the pressure
drop, the head loss over a 60 m long section of the pipe.
For water at 15C
kg
ρ  999.14 3
m
μ  0.001140 Pa - sec
m2
sec
for stainless steel
ν  1.141 x 10-6
 0.015 mm
 0.015

 0.0003
D
50
u sin g Colebrook Eq.
f  0.015
5m
60 m
fLv 2
hL 

2gD
Example:
Water at 20C the rate of 30 L/sec is flowing in a 70 m long pipeline, (ND = 100 mm; OD = 114.23 mm; ID = 102.29 mm) containing
1-Open Gate valve; 1 – Check valve and 15 90-Elbows. Determine the pipe material with minimal head losses if pipe material
available are Stainless Steel, Cast Iron and Galvanized Iron.
Pipe Material
Stainless Steel
Cast Iron
Galvanized Iron
Absolute Roughness ()
0.015
0.25
0.025
At 20C Water
Density
Absolute Viscosity
Kinematic Viscosity
Properties
998.29
0.001003
1.005 x 10-6
Fitting
Gate Valve
Check Valve
90Elbow
Total
Qty.
1
1
15
L/D
8
135
30
Le
0.8
13.5
45
59.3
Unit
Kg/m3
Pa-sec
m2/sec
Unit
m
m
m
m
Formulas
Q
A
π
A  D2
4
f ( L  ΣLe) v 2
hf 
2gD
f
v
NR 
vD
ν
For Stainless Steel
HEAD LOSS CALCULATION
Q (Capacity)
30
L/sec
Nominal Diameter
100
mm
Outside Diameter (OD)
114.33
mm
Inside Diameter (ID)
102.29
mm
Wall Thickness (t)
6.02
mm
Area (A)
0.0082
m^2
Velocity (v)
3.65
m/sec
Velocity Head (V^2/2g)
0.679
m
998.29
kg/m^3
Absolute Viscocity (m)
0.001003
Pa-sec
Kinematic Viscosity (n)
1.005E-06
m^2/sec
Reynold's Number(NR)
69,161.354
x
L (Length of Pipeline)
70
m
59.3
m
m
Pipe Roughness 
0.015
mm
Relative Roughness
0.00015
x
Friction Factor (f)
0.0200
Head Loss (hf)
m
For Cast Iron Steel
HEAD LOSS CALCULATION
Q (Capacity)
30
L/sec
Nominal Diameter
100
mm
Outside Diameter (OD)
114.33
mm
Inside Diameter (ID)
102.29
mm
Wall Thickness (t)
6.02
mm
0.25
 

log  D  5.74 
  3.7 N R 0.9 

 
2
Area (A)
0.0082
m^2
Velocity (v)
3.65
m/sec
Velocity Head (V^2/2g)
0.679
m
(Density) r
998.29
kg/m^3
Absolute Viscocity (m)
0.001003
Pa-sec
Kinematic Viscosity (n)
1.005E-06
m^2/sec
Reynold's Number(NR)
69,161.354
x
L (Length of Pipeline)
70
m
59.3
m
m
Pipe Roughness (Î)
0.25
mm
Relative Roughness
0.00244
x
Friction Factor (f)
0.0271
Head Loss (hf)
m
For Galvanized Iron
HEAD LOSS CALCULATION
Q (Capacity)
30
L/sec
Nominal Diameter
100
mm
Outside Diameter (OD)
114.33
mm
Inside Diameter (ID)
102.29
mm
Wall Thickness (t)
6.02
mm
Area (A)
0.0082
m^2
Velocity (v)
3.65
m/sec
Velocity Head (V^2/2g)
0.679
m
(Density) r
998.29
kg/m^3
Absolute Viscocity (m)
0.001003
Pa-sec
Kinematic Viscosity (n)
1.005E-06
m^2/sec
Reynold's Number(NR)
69,161.354
x
L (Length of Pipeline)
70
m
59.3
m
m
0.025
mm
Relative Roughness
0.00024
Friction Factor (f)
0.0205
Head Loss (hf)
x
m
FUNDAMENTAL EQUATIONS OF CENTRIFUGAL PUMPS
1. Total Head
HT = nH
n - number of stages
H - head per stage
2. Specific Speed: Is the speed in RPM at which a theoretical pump geometrically
similar to the actual pump would run at its best efficiency if proportion to
deliver 1 m3/sec against a total head of 1 m. It serves as a convenient index of
the actual pump type.
NS 
N Q
3
0.0194H 4
where: Q - flow for a single suction pump in m3/sec and H is the head per stage
3. Suction Specific Speed (S)01
S
N Q
3
0.0194NPSH  4
3
N S  NPSH  4


S  H 
NPSH - net positive suction head, meters
4. Net Positive Suction Head: The amount of pressure in excess of the vapor
pressure of the liquid to prevent cavitation.
NPSH = Hp  Hz - Hvp - hfs , meters
where: Hp - absolute pressure head at liquid surface at suction, m
Hz - elevation of liquid surface at suction, above or below the pump
centerline, m
(+) if above PCL
(-) if below PCL
Hvp - vapor pressure head corresponding the temperature of the liquid, m
hfs - friction head loss from liquid surface at suction to PCL.
5. Cavitation: The formation of cavities of water vapor in the suction side of the pump due to low suction pressure.
CAUSES OF CAVITATION
 Sharp bends.
 High temperature
 High velocity
 Rough surface
 Low atmospheric pressure
EFFECTS OF CAVITATION
 Noise
 Vibration
 Corrosion
 Decreased capacity
 6. Cavitation Parameter:
4
NPSH  NS  3
δ
 
H
 S 
D
60 2gH
meter
πN
7. Impeller Diameter;
where:  - peripheral velocity factor whose value ranges from 0.95 to1.09
8. Affinity Laws or Similarity Laws for Centrifugal Machines
a. For Geometrically similar pumps
Q  ND3
Power   N3D5
2 2
HN D
T   N2D5
b. For pumps with Variable Speed and Constant impeller diameter
QN
Power  N3
2
HN
c. For pumps at Constant Speed with Variable impeller diameter
QD
Power  D3
2
HD
FUNDAMENTAL EQUATIONS FOR RECIPROCATING PUMPS
Specification: Ds x Dw x L
where: Ds - diameter of steam cylinder
Dw - diameter of water cylinder
L - length of stroke
1. Volumetric Efficiency
ηV 
Q
x 100%
VD
where: Q - capacity, m3/sec
VD - displacement volume, m3/sec
Fw  e m Fs
(Dw) 2 (Pd  Psu )
KPa
4
Ds
(Pd  Psu )

Dw
e m (Ps  Pe )
Fw 
2. Displacement Volume
a. For Single acting
VD 
L(D W )2 Nn' m3
4(60)
sec
b. For Double acting without considering piston rod
c. For Double acting considering piston rod


VD 
LNn'
m3
2DW 2 - d2
4(60)
sec
VD 
L(D W )2 Nn' m3
4(60)
sec
where: N - no. of strokes per minute
L - length of stroke, m
D - diameter of bore, .
d - diameter of piston rod, m
n' - no. of cylinders
n' = 1 (For Simplex)
n' = 2 (For Duplex)
n' = 3 (For Triplex)
3. Percent Slip
%Slip = 100 - v
4. Slip
Slip = VD - Q
5. Thermal Efficiency
3600(WP)
x 100%
ms (hs  he )
where: hs - enthalpy of supply steam, KJ/kg
he - enthalpy of exhaust steam, KJ/kg
ms - steam flow rate, kg/hr
FP - fluid power, KW
e
DS 2 (Ps  Pe )
KPa
4
6. Force produced and acting on the piston rod
where: Ps - supply steam pressure, KPa
Pe - exhaust steam pressure, KPa
Ds - diameter of steam cylinder, m
(Ps - Pe) - mean effective pressure
7. Force transmitted to the liquid piston
where: em - mechanical efficiency
Fw  e mFs
Fs 
w 
F
(Dw) 2 (Pd  Psu )
KPa
4
Ds

Dw
(Pd  Psu )
e m (Ps  Pe )
Psu - suction pressure of water cylinder, KPa
Pd - discharge pressure of water cylinder, KPa
8. Pump Duty: Work done on the water cylinder expressed in Newton-meter per
Million Joules
Pump Duty 
9.81mw (Hd  Hsu ) x 106
1000ms (hs - he )
where: mw - water flow rate, kg/hr
Hd - discharge head of pump, m
Hsu - suction head of pump, m
9. Pump Speed
V = 43.64(L)1/2(ft), m/min
(D W ) 2 Vn' m 3
VD 
4(60) sec
N-m
Million Joules
where: ft - temperature correction factor
L - length of stroke, m
10. Temperature Correction Factor
ft
= 1 For cold water
= 0.85 for 32.2C water
= 0.71 for 65.5C water
= 0.55 for 204.4C water
11. For Indirect Acting pumps
N
907f t
L
EXAMPLE
A mechanical engineer of an industrial plant wishes to install a pump to lift 15 L/sec of water at 20C from a sump to a tank on a
tower. The water is to be delivered into a tank 200 KPag. The water level in the tank is 20 m above the pump centerline and the
pump is 4 m above the water level in the sump. The suction pipe is 100 mm in diameter, 7.0 m long and will contain 2 - standard
elbows and 1 - Foot valve. The discharge pipe to the tank is 75 mm in diameter and is 120 m long and contains 5 - 90elbows, 1 check valve, and 1 - gate valve. Pipe material is Cast iron. Assuming a motor-pump efficiency of 75%, calculate the KW power
required and the current drawn by the 3 – Phase motor drive if E = 220 Volts and 0.90 Power Factor.
From table
At 20C
 = 998 kg/m3
 = 9.79 KN/m3
= 0.25 mm
 = 1.002 x 10-3 Pa-sec
 = 1.004 x 10-6 m2/sec
DISCHARGE LINE
L =120 m
dd = 75 mm
5 – Elbows
1 – Check valve
1 – Gate Valve
120 m

Q = 15 L/sec
P2 = 200 Kpag
P1 = 0 gage
20 m
4m
SUCTION LINE
L=7m
ds = 100 mm
2 – Elbows
1 – foot valve
1
Suction Line
L
ID

/D
NR
vd
fd
7m
0.100 m
0.25 mm
0.0025
190,224
1.91 m/sec
0.026
 0.25

 0.0025
D 100

A  (0.100)2  0.00785 m2
4
0.015
m
vs 
 1.91
0.00785
sec
1.91(0.100)
NR 
 190,224
1.004x106
From Moody diagram
f  0.026
0.0267  13.51.91
 0.99 m
2(9.81)(0.100)
2
hL Suction 
Fitting
L/D
30
90Standard
Elbow
Foot Valve
75
Total (Le)
Discharge Line
L
120 m
ID
0.075 m
0.25 mm

0.0033
/D
NR
253,631
vd
3.40 m/sec
fd
0.028
Quantity
2
D
0.100
Le
6
1
0.100
7.5
13.5
 0.25

 0.0033
D
75

A d  (0.075) 2  0.00442 m 2
4
0.015
m
vd 
 3 .4
0.00442
sec
3.4(0.100)
NR 
 253,631
1.004x10 6
From Moody diagram
fd  0.028
0.028120  25.033.40
 32 m
2(9.81)(0.075)
2
hL Discharge 
For Pipe Exit
K 1
fL
D
L K
1
 
 35.7
D f 0.028
K
Fitting
90Standard Elbow
Check Valve
Gate Valve
Pipe Exit
Total (Le)
L/D
30
135
13
35.7
Quantity
5
1
1
1
D
0.075
0.075
0.075
0.075
Le
11.25
10.125
0.975
2.68
25.03
HL  hLs  hLd
HL  32  0.99  32.99 m
P2 - P1 v 2  v 1

 ( z 2  z1 )  HL

2g
2
ht 
2
v1  0
v2  0
P1  0 gage
z1  20 m
z 2  -4 m
200
 0  (20  4)  32.99  77.42 m
9.79
Water Power  Qh t
ht 
Water Power
x100%
Motor Power
0.015(9.79)(77.42)
MP 
 15.2 KW
0.75
Efficiency 
3 EI(Power Factor)
1000
1000(15.2)
I
 44.3 Amperes
220(0.90) 3
MP 
EXAMPLE NO. 1
The diameters of the suction and discharge pipes of a pump are 152 mm and 102 mm, respectively. The discharge pressure is read
by a gauge at a point 1.5 m above the pump centerline and the suction pressure is read by a gauge 61 cm below the pump centerline.
If the discharge pressure gauge reads 137.9 KPa and the suction gauge reads a vacuum of 254 mm Hg, when gasoline with s = 0.75
is pumped at the rate of 34 L/sec, find the power delivered to the fluid. If the pump is 75% efficient and is driven by a 3-phase, 440
V, 0.9 power factor and 92% efficiency induction motor, determine the cost of power for 24 hours operation, if power costs P
0.30/KW-hr, and the line current drawn by the motor. (FP = 6.54 KW; C = P51/day; I = 10.37 amp; ht = 26.16 m)
EXAMPLE NO. 2
A boiler feed pump delivers water at 100C ( = 958.31 kg/m3; = 9.40 KN/m3) which it draws from an open hot-well with a friction
loss of 610 mm in the intake pipe between it and the hot-well. The barometric pressure is 737 m Hg and the value of the cavitation
parameter for the pump is 0.10. What must be the elevation of the water surface in the hot-well relative to that of the pump intake?
The total pumping head is 73 m. (Hz = 8.24 m)
EXAMPLE 3
A centrifugal pump design for a 1800 RPM operation and a head of 61 m has a capacity of 190 L/sec with a power input of 132 KW.
What effect will a speed reduction to 1200 RPM have on the head, capacity and power input of the pump? What will be the change
in H, Q and BP if the impeller diameter is reduced from 305 mm to 254 mm while the speed is held constant at 180 RPM. Neglect
effects of fluid viscosity.
EXAMPLE 4
A plant has installed a single suction centrifugal pump with a discharge of 68 m3/hr under a 60 m head and running at 1200 RPM. It
is proposed to install another pump with double suction but of the same type to operate at 30 m head and deliver 90 m 3/hr.
a) Determine the speed of the proposed pump (877 RPM)
b) What must be the impeller diameter of the proposed pump if the diameter of the existing pump is 150 mm.
(145 mm)
Design Problem No. 1
Water (@ t = 20C) from an open reservoir A at 8 m elevation is drawn by a motor driven centrifugal pump to an open reservoir B
at 70 m elevation. The suction line is GI pipe; ND = 200 mm (NPS = 8”); L = 25 m containing 1 –check valve; 1 – Gate valve; 5
standard 90 elbows. Discharge line is GI pipe; ND = 150 mm (NPS = 6”); L = 120 m containing 1-Gate valve, 1-Check valve, and
8 standard 90 elbows. Assume shard edge for pipe entrance and pipe exit. Pump elevation is at 4 m. Overall efficiency of the system
is 78%. For a discharge rate of 10 L/sec and power cost of P 7.00 per KW-hr. Determine
The total dynamic head
The water power
The power required by the motor
The power cost for 10 hrs pumping time
The line current drawn by the motor for E = 220 Volts and power factor PF = 0.92
The pressure gauge reading installed just at the inlet and outlet of the pump
Note: Draw the sketch of the problem

B
Open Tank
Disch arg e Line (GI Pipe)
ND  150 mm
L  120 m
120 m
1 - Check valve
1 - Gate valve
8 - 90Elbows
Suction Line (GI Pipe)
A
ND  200 mm
Open Tank
L  25 m
1 - Check valve
1 - Gate valve
5 - 90 Elbows
CV
GV
GV
Centrifugal Pump
Datum Line
CV
4m
8. 0 m
1
HYDRAULIC MACHINERY: Activity 2 (Pump Design)
Determine the Motor power required of the pumping system shown below.
P2 = 500 KPag
Avialable Material
1. Stainless steel
2. Galvanized Iron (new)
3. Cast Iron (new)

t  20C
Powered Valve
Storage Tank
KN
m3
kg
  998.2 3
m
  1.002 x 10-3 Pa - sec
  9.789
Discharge Line
100 mm (DN)line : Sch. 40
5 - 90 Elbow
ID  102.29 mm
1- GateValve
L  15 m
Pipe: Sch.40
2
m
sec
60 m
  1.003 x 10-6
SuctionLine
L  20m
DN 100mm
ID  102.29mm
L
sec
OverallEfficioency  72%
Q  10
3 - 90  Elbow
1 - Gate Valve
1 - Check Valve
75 mm (DN) Line; Sch. 40
ID  77.95 mm
L  85 m
8 - 90  Elbow
1 - Check Valve
1 - Reducer
Gate Valve
Pressure
Gauge
Pressure
Gauge
3m
Pump
1
200 Kpag Water
Main
Gate Valve Check Valve
Reducer
Pump
Cenbterline
To Sprinkler
System
HYDRAULIC MACHINERY(QUIZ NO. 2)
Gasoline at 38C being drawn from a closed tank having a pressure of 69 KPag in a plant located 900 m above sea level. The level
of gasoline in the tank is 2.5 m above the pump centerline. The suction line friction and turbulence head losses amounts to 0.6 m.
The vapor pressure of the gasoline is 48 KPa absolute and the relative density is 0.72.
a) What is the NPSH of the system
b) If the cavitation parameter is 0.10 and the discharge is 400 L/sec, what is the size of the drive motor
required? Combined pump-motor efficiency is 55%.
At its optimum point of operation a given centrifugal pump with an impeller diameter of 50 cm delivers 3.2 m 3/sec of water against
a head of 25 m when operating at 1450 RPM.
a. If the efficiency is 82%, what is the brake power of the driving shaft.
b. If a homologous pump with an impeller diameter of 80 cm is rotating at 1200 RPM, what would be the Q, H,
and BP.
c. Compute the specific speed of both pumps.
A submersible, multistage centrifugal deep-well pump 984 L/min capacity is installed in a well 8 m below the static water level.
Draw-down when pumping at rated capacity is 3 m. The pump delivers the water into a 95 Liters capacity overhead storage tank.
Total discharge head developed including friction in piping is 74 m. Calculate:
a) Brake power required to drive the pump if pump efficiency is 70%
b) Determine the number of stages in the pump if each impeller develops a head of 12 m at 3450 RPM
c) Determine the diameter of the impeller of this pump in mm.
HYDRAULIC MACHINERY
ACTIVITY NO. 2 (May 06, 2017)
NAME _______________________________________
No. 1
A 4 m3/hr pump delivers water to a pressure tank. At the start, the gauge reads 138 KPa until it reads 276 KPa and then the pump
was shut off. The volume of the tank is 160 Liters. At 276 KPa, the water occupied 2/3 of the tank volume.
a) Determine the volume of water that can be taken out until the gauge reads 138 KPa
b) If 1 m3/hr of water is constantly used, in how many minutes from 138 KPa will the pump run until
the gauge reads 276 KPa
At P  276 KPa
Pair  276  101.325  377.325 KPa
Vair  13 (0.160)  0.0533 m 3
at P  138 KPa
Pair  138  101.325  239.325 KPa
assumng isothermal conditions for air in the tank
377.325(0.0533)
 0.084 m 3 , and the volume of water remaining in the tank is
239.325
Vw ater at 138 KPa  0.160  0.084  0.076 m 3
Vair at 138 KPa 
Volume of water taken out  2 3 (0.160)  0.076  0.0307 m 3
t  pumping time if 1 m 3 /hr is constantly used
4t - 1t  0.0307
t
0.0307
hrs  0.01022 hrs  0.6133 min  36.8 sec
3
No. 2
Water from a reservoir is pumped over a hill through a pipe 900 mm diameter and a pressure 1 kg/cm 2 is maintained at the pipe
discharge where the pipe is 85 m from the pump centerline. the pump has a positive suction head of 5 m. Pumping rate of the pump
at 1000 RPM is 1.5 m3/sec. Friction loss is equivalent to 3 m of head loss.
a) What amount of energy must be furnished by the pump in KW
b) If the speed of the pump is 1200 RPM, what are the new values of Q, h, and power
2. A double suction single stage centrifugal pump is delivering 900 m3/hr of sea water (S = 1.03) from a source where the water level
varies 2 m from a high tide to low tide level. The pump centerline is located 2.6 m above the surface of the water at high tide level.
The pump discharges into a surface condenser 3 m above the pump centerline. Loss of head due to friction in suction pipe is 0.8 m
and that in the discharge side is 3 m. Pump is directly connected to a 1750 RPM,460 Volts, 3-phase, 60 Hertz motor.
a) Make a sketch of the installation in accordance with good practice.
b) Calculate the total suction head in m
c) Calculate the total discharge head in m
d) Calculate the specific speed of the pump in RPM
e) Calculate the BP of the pump if efficiency is 82%
f) Calculate the impeller diameter in mm.
SAMPLE PROBLEMS (INDUSTRIAL PLANT)
1. A centrifugal pump with a 30.5 cm. diameter impeller requires a power input of 45 KW when the flowrate is 200 L/sec
against a 18 m head. The impeller is changed to one with a 25.5 cm. diameter. Determine the expected flowrate, head, and
input power if the pump speed remains the same.
QαD
H α D2
Power α D
3
Q1
H1
P1
D1
200
18
45
30.5
Q2
H2
P1
D2
167.21
12.6
26.30
25.5
2. A small centrifugal pump, when tested at 2875 rpm with water, delivered a flow rate of 16 L/sec and a head of 42 m at its
best efficiency point (efficiency is 76%). Determine the specific speed of the pump at this test condition. Compute the
required power input to the pump.
N Q
NS 
3
Ns 
0.0194H 4
Ns
N
Q
H
2875 0.016
1136.21
2875
0.016
42
3
0.0194 42 4
 
3. A plant has installed a single suction centrifugal pump with a discharge of 68 m 3/hr under a 60 m head and running at 1200
RPM. It is proposed to install another pump with double suction but of the same type to operate at 30 m head and deliver
90 m3/hr.
a) Determine the speed of the proposed pump (877 RPM)
b) What must be the impeller diameter of the proposed pump if the diameter of the existing
pump is 150 mm. (145 mm)
Given:
Q1 = 68 m3/hr = 0.0189 m3/sec
H1 = 60 m
N Q
N1 = 1200 RPM
NS 
3
For the proposed pump
0.0194 H 4
H2 = 30 m
Q2 = 90 m3/hr
Q2 = 45 m3/hr = 0.0125 m3/sec→ For single suction
 
NS1
N
Q
H
394.45
1200
0.0189
60
N
Ns
Q
H
877
394.5
0.0125
30
For Geometrically similar pumps
Q  ND3
Power   N3D5
2 2
HN D
T   N2D5
Q1
H1
D1
N1
0.0189
60
0.15
1200
D2
Q2
H2
N2
0.145
0.0125
30
877
Q2
N D 
 2  2 
Q1
N1  D1 
2
3
H2  N 2   D 2 
 


H1  N1   D1 
2
By: Yuri G. Melliza
COURSE DESCRIPTION
This class provides students with an introduction to principal concepts and methods of fluid mechanics. Topics covered in the course
include pressure, hydrostatics, and buoyancy; open systems and control volume analysis; mass conservation and momentum
conservation for moving the models necessary to study, analyze, and design fluid systems through the application of these concepts,
and to develop the problem-solving skills essential to good engineering practice of fluid mechanics in practical applications.
Students will have the opportunity to demonstrate a familiarity and ability to work on fluid mechanics. These outcomes will be
demonstrated through an assessment of quizzes, assignments and major examinations.
GENERAL OBJECTIVES
By the end of this course, you should expect to be able to calculate:
 fundamental fluid properties for different fluids and flows
 forces on objects submerged in both static and flowing fluids
 pressures in both static and flowing fluids, and the velocities associated with different flows
 forces in complicated momentum balance problems
 energy loss and the flow rates associated with different flow networks in channels and pipes
 dimensionless numbers important for design of experiments and practical engineering work
 numerical solutions for simple fluid flow problems
 properties of a boundary layer, both turbulent and laminar problems with the application of Bernoulli’s Principle
 Fluid Mechanics - it is a science that deals with the mechanics of fluids (liquid or gas) and is based on the same fundamental
principles employed in the mechanics of solids.
STUDENT COURSE OUTCOMES
This is an introductory course on fluid motion, the forces that fluids exert, and the forces that are exerted on them. The study of fluid
mechanics has numerous engineering applications. Fluids interact with structures such as high-rise buildings, dams, and bridges and
the static and dynamic loads imposed by the fluids must be considered in the design and construction of these structures. Cars,
aircraft, and ships all move through fluids, and frictional (fluid drag) forces represent a major energy sink. Water is an important
resource to California and fluid mechanical problems abound in the complex system of dams, aqueducts, treatment plants, pipes,
and valves used to deliver water to urban and agricultural consumers. Finally, the motions of contaminants in water and air are
governed by the mechanics of fluid flow. After this course, the students are expected to:
Branches of Fluid Mechanics
1. Fluid Static - deals with the forces exerted by or upon the fluid at rest.
2. Kinematics - deals with velocities and accelerations without considering forces and
energy.
3. Hydrodynamics - deals with the relation of velocities and accelerations and the forces exerted by or upon
the fluid in motion.
Fluid - is a substance capable of flowing and having particles which easily moved and changed
their relative position without the separation of mass.
Distinction between Liquid and Gas
Liquid
1. It has a free surface
2. A given quantity of liquid occupies a given volume of a container
3. It is incompressible
Gas
1. It has no free surface
2. A given quantity of gas occupies all portions of the container regardless of its shape and
size.
3. It is compressible
Free Surface - a surface in which all pressures can be removed except its own vapor
pressure.
Vapor - is a gas in which its pressure and temperature are such that very near to its liquid state.
PROPERTIES OF FLUIDS
1. Density (): It is the mass on a unit volume.
ρ  m kg3
V m
2. Specific Volume (): It is the volume per unit mass, or the reciprocal of its density.
υ V m
3
m kg
3. Specific Weight or Weight Density (): It is the weight on a unit volume.
γ  W KN3
V m
mg
ρg KN
γ

1000V 1000 m3
4. Specific Gravity or Relative Density (S):
For Liquids
It is the ratio of its density to density of water at standard temperature and pressure.
SL 
L

 L
w  w
For Gases
It is the ratio of its density to the density of either air or hydrogen at some specified
temperature and pressure.
SG 
G
 AH
For Gases
ρ
P Kg
RT m 3
Where:
P - Absolute pressure in KPaa
R -Gas constant in KJ/kg-K
T -Absolute temperature in K
8.3143 KJ
M kg - K
M - molecular weight in kg/kgmole
R
where:
w = 1000 kg/m3 (Standard condition)
w = 9.81 KN/m3 (Standard condition)
AH = is the density of either air or hydrogen at some value of P and T.
5. Temperature: It is the measure of the intensity of heat of a fluid.
Celsius and Fahrenheit Scale
F  32
1.8
F  1.8(C)  32
C 
Absolute Scale
RANKINE:
R  F  460
KELVIN:
K  C  273
6. Pressure (P): It is the normal component of a force per unit area.
F KN
or KPa
A m2
KN
1 2  1 KPa (Kilo Pascal)
m
P
If a force dF acts on an infinitesimal area dA, the intensity of pressure is:
P
dF KN
or KPa
A m2
where:
F - force, KN
A - area , m2
PASCAL`S LAW
At any point in a homogeneous fluid at rest the pressures are the same in all directions.
Fx  0
P1 A 1  P3 A 3 sin   0
Fy  0
P2 A 2  P3 A 3 cos 
P1 A 1  P3 A 3 sin   eq. 1
P2 A 2  P3 A 3 cos   eq.3
From Figure
cos  
A1
;
A3
A 1  A 3 sin   eq. 2
sin 
Equation 2 to equation 1
P1  P3
A2
A3
A 2  A 3 cos   eq.4
Equation 4 to equation 3
P2  P3
therefore
P1  P2  P3
Atmospheric Pressure (Pa): It is the absolute pressure exerted by the atmosphere.
At sea level
Pa = 101.325 KPa
= 0.101325 MPa
= 1.01325 Bar
= 760 mm Hg
= 10.33 m of H2O
= 1.033 kg/cm2
= 14.7 psi
= 29.921 in Hg
= 33.878 ft of H2O
Absolute and Gage Pressure
Absolute Pressure - is the pressure measured referred to absolute zero and using absolute zero as the base.
Gage Pressure - is the pressure measured referred to the existing atmospheric pressure and using atmospheric pressure as the base.
Pgage - if above atmospheric
Pvac- vacuum or negative gage pressure, if below atmospheric
Pabs  Pa  PGage
Pabs  Pa  Pvacuum
7. Viscosity: It is the property of a fluid that determines the amount of its resistance to
shearing stress.
Assumptions:
1. Fluid particles in contact with the moving surface moves with the same velocity of that surface.
2. The rate of change of velocity dv/dx is constant in the direction perpendicular to the direction of motion.
3. The shearing is directly proportional to the rate of change of velocity.
let S - shearing stress in Pa or N/m2
dv
dx
dv
dv v
Sμ
but

dx
dx x
v
Sμ
x
x
N - sec
μ  S Pa - sec or
v
m2
S
Where
 - Absolute or dynamic viscosity in Pa-sec or N-sec/m2
S - shearing stress in Pascal (Pa)
x - distance apart in meters
v - velocity in m/sec
8. Kinematics Viscosity: It is the ratio of the absolute or dynamic viscosity to the mass
density.
ν
μ m2
ρ sec
Conversion
1 Pa-sec = 1 N-sec/m2
1 Pa-sec = 10 Poise
1 Pa-sec = 0.020885 lb-sec/ft2
1 Pa-sec = 0.10197 kg-sec/m2
1 Pa-sec = 1000 Centipoise
1 m2/sec = 10,000 stokes
1 m2/sec = 10.764 ft2/sec
1 m2/sec = 1,000,000 Centistokes
Stoke = 1 cm2/sec
1 kg/m3 = 0.062428 llb/ft3
1 lb/ft3 = 16.018 kg/m3
9. Elasticity: If a pressure is applied to a fluid, it contracts; if the pressure is released, it expands, the
elasticity of a fluid is related to the amount of deformation(expansion or contraction) for a given pressure
change. Quantitatively, the degree of elasticity is equal to:
P
P
Ev 
dP
KPa
dV

V
dP
Ev 
KPa
dρ
ρ
dV dρ

V
ρ
Where negative sign is used because dV/V is negative for a positive dP.
where:
Ev - bulk modulus of elasticity
dV - is the incremental volume change, m3
V - is the original volume, m3
dP - is the incremental pressure change in Kpa
10. Surface Tension
The attractive force exerted upon the surface molecules of a liquid by the molecules beneath that tends to draw the surface molecules
into the bulk of the liquid and makes the liquid assume the shape having the least surface area.
Surface tension is the elastic tendency of a fluid surface which makes it acquire the least surface area possible. Surface tension
allows insects (e.g. water spiders), usually denser than water, to float and stride on a water surface.
At liquid–air interfaces, surface tension results from the greater attraction of liquid molecules to each other (due to cohesion) than
to the molecules in the air (due to adhesion). The net effect is an inward force at its surface that causes the liquid to behave as if its
surface were covered with a stretched elastic membrane. Thus, the surface becomes under tension from the imbalanced forces, which
is probably where the term "surface tension" came from. [1] Because of the relatively high attraction of water molecules for each other
through a web of hydrogen bonds, water has a higher surface tension (72.8 millinewtons per meter at 20 °C) compared to that of
most other liquids. Surface tension is an important factor in the phenomenon of capillarity.
Surface tension has the dimension of force per unit length, or of energy per unit area. The two are equivalent, but when referring to
energy per unit of area, it is common to use the term surface energy, which is a more general term in the sense that it applies also to
solids.
Floating objects
Cross-section of a needle floating on the surface of water. Fw is the weight and Fs are surface tension resultant forces.
When an object is placed on a liquid, its weight Fw depresses the surface, and if surface tension and downward force becomes equal
than is balanced by the surface tension forces on either side Fs, which are each parallel to the water's surface at the points where it
contacts the object. Notice that small movement in the body may cause the object to sink. As the angle of contact decreases surface
tension decreases the horizontal components of the two Fs arrows point in opposite directions, so they cancel each other, but the
vertical components point in the same direction and therefore add up [2] to balance Fw. The object's surface must not be wettable for
this to happen, and its weight must be low enough for the surface tension to support it.
Examples of Surface Tension
Drops of water. When using a water dropper, the water does not flow in a continuous stream, but rather in a series of drops.
The shape of the drops is caused by the surface tension of the water. The only reason the drop of water isn't completely spherical is
because of the force of gravity pulling down on it. In the absence of gravity, the drop would minimize the surface area in order to
minimize tension, which would result in a perfectly spherical shape.
Insects walking on water. Several insects are able to walk on water, such as the water strider. Their legs are formed to distribute
their weight, causing the surface of the liquid to become depressed, minimizing the potential energy to create a balance of forces so
that the strider can move across the surface of the water without breaking through the surface. This is similar in concept to wearing
snowshoes to walk across deep snowdrifts without your feet sinking.
Needle (or paper clip) floating on water. Even though the density of these objects is greater than water, the surface tension along
the depression is enough to counteract the force of gravity pulling down on the metal object. Click on the picture to the right, then
click "Next," to view a force diagram of this situation or try out the Floating Needle trick for yourself.
Capillarity
Capillary Action
Capillary action is the result of adhesion and surface tension.
Adhesion of water to the walls of a vessel will cause an upward
force on the liquid at the edges and result in a meniscus which
turns upward. The surface tension acts to hold the surface intact, so
instead of just the edges moving upward, the whole liquid surface
is dragged upward.




r
h
Where:
 - surface tension, N/m
 - specific weight of liquid, N/m3
r – radius, m
h – capillary rise, m
Table 1.Surface Tension of Water
C

0
0.0756
10
0.0742
20
0.0728
30
0.0712
40
0.0696
60
0.0662
80
0.0626
100
0.0589
h
2σ cos θ
γr
Sample Exercises
1. A mercury barometer at the ground floor of Kingston Tower in Chicago reads 735 mm Hg. At the same time another barometer
at the top of the tower reads 590 mmHg. Assuming the air to be constant at 1.21 kg/m 3, what is the approximate height of the
tower using g = 9.7 m/sec2.
2
h
1
735(101.325)
 98 KPa
760
590(101.325)
P2 
 78.7 KPa
760
ρg
1.21(9.7)
KN
γ

 11.64 x 10- 3 3
1000
1000
m
dP   γdh
P2  P1   γ(h 2  h1 )
P1 
h1  0; h 2  h
P2  P1   γ(h )
h
P2  P1
 1658.08 meters
γ
2. A tank contains a mixture of 20 kg of nitrogen and 20 kg of carbon monoxide. The total tank volume is 20 m 3. Determine the
density, specific volume and specific weight of the mixture if local g = 9.81 m/sec2
m  m1  m 2
m  20  20  40 kg
ρ
m 40
kg

2 3
V 20
m
υ
1
m3
 0 .5
ρ
kg
3. A pressure gage registers 345 KPag in a region where the barometric pressure is 736.7 mm hg. Find the absolute in Bars.
Pabs  Pa  Pgage
Bar
 736.7(101.325) 
Pabs  
  345  443.2 KPaa x 100 KPa
760


Pabs  4.432 Bars
4. The level of water in an enclosed water tank is 40 m above ground level. If the pressure of the air space above the water is 120
Kpa, what is the pressure at ground level in KPa.
lb  sec
1 Pa - sec
x
 0.023Pa - sec
2
lb - sec
ft
0.020885 2
ft
kg
16.018 3
lb
m  864.972kg
  54 3 x
lb
m3
ft
1 3
ft
 0.023
 
864.972

  4.8 x 10-4
Air
1
40 m
2
2
  2.66 x 10-5 m sec
5. A liquid has an absolute viscosity of 4.8 x 10-4lb-sec/ft2. It weighs 54 lb/ft3. What are its absolute and kinematic viscosities in
SI.
0.020885lb - sec 2
ft  2.09 x 10-5 lb - sec
  1.002x103 Pa - sec
ft2
Pa - sec
ft2
2 10.764
 1.002x103 Pa - sec
-6 m
-5 ft2
sec
 
 1.004 x 10 x 2  1.08 x 10 sec
kg
 998 3
sec 1m
sec
m
6. The absolute viscosity of water at 20C is 1.002 x 10-3 Pa-sec and the density is 998 kg/m3. What are its absolute and kinematic
viscosity in English units.
d P   d h
By integ ratio n

2
1
dP    
2
1
dh
P2  P1    (h 2  h1 )
P2
 P1   (h 2  h1 )
h1  0
h2
 4 0 m
P2
 1 20  9.8 1
( 4 0)  5 12 .4 KPa
IDEAL GAS PRINCIPLES
1. CHARACTERISTIC EQUATION
PV = mRT 1
P = RT 2
P
3
RT
RT
υ
4
P
PV
C  5
T
P1V1 P2 V2

 6
T1
T2
ρ
where:
P - absolute pressure in KPa
V - volume in m3
m -mass in kg
R -Gas constant in KJ/kg-K
T - absolute temperature in K
 - specific volumein m3/kg
 - density in kg/m3
2. GAS CONSTANT
R
KJ
 7
M kg - K
KJ
R  8.3143
(Universal Gas Constant)
kg mol - K
R
M  molecular weight of a gas in
kg
kg mol
3. BOYLE`S LAW ( T = C ): Robert Boyle (1627-1691)
If the temperature of a certain quantity of gas is held constant, the volume V is inversely proportional to
the pressure P, during a quasi-static change of state.
Vα
1
1
or V  C
P
P
PV  C
P1V1  P2 V2  8
4. CHARLE`S LAW ( P = C and V = C):Jacques Charles (1746-1823) and Joseph Louis Gay- Lussac
(1778-1850)
A) At constant pressure (P=C), the volume V of a certain quantity of gas is directly proportional to
the absolute temperature T, during a quasi static change of state.
V  T or V = CT
V
C
T
V1 V2

 9
T1 T2
B) At constant volume (V = C), the pressure P of a certain quantity of gas Is directly proportional to
the absolute temperature T, during a quasi-static change of state.
P  T or P = CT
Cp  Cv  R
Rk
k 1
R
Cv 
k 1
Cp
k 
Cv
Cp 
5. AVOGADRO`S LAW:Amadeo Avogadro (1776-1856)
All gases at the same temperature and pressure, under the action of a given value of g, have the
same number of molecules per unit of volume. From which it follows that the  M.
γ1 M1 R 2


γ 2 M 2 R1
6. SPECIFIC HEATS
P
C
T
P1 P2
  10
T1 T2
7. ENTROPY CHANGE (S)
Entropy is that property of a substance that determines the amount of randomness and disorder of a substance.
If during a process, an amount of heat is taken and is by divided by the absolute temperature at which it is taken,
the result is called the ENTROPY CHANGE.
dQ
T
dQ
ΔS 
T
dS 

8. Isentropic Process: It is an internally reversible adiabatic process of an ideal or perfect gas in
which S = C or PVk = C.
P, V, & T relationship:
P1V1k  P2 V2 k
k 1
T2  P2  k  V1 
 
  
T1  P1 
 V2 
k 1
9. Polytropic Process: It is an internally reversible process of an ideal or perfect gas in
which PVn = C. Where n stands for any constant (but not equal to infinity).
P, V, & T relationship:
P1V1n  P2 V2 n
n 1
V 
T2  P2  n
  
  1 
T1  P1 
 V2 
n 1
1. Total moles of a mixture
n = ni
2. Mole Fraction
yi = ni/n
3. Total mass of a mixture
m = mi
4. Mass Fraction
xi = mi/m
5. Equation of State
A. Mass Basis
a. For the mixture
PV = mRT
b. For the components
PiVi = miRiTi
B. Mole Basis
a. For the mixture
PV = nRT
b. For the components
PiVi = niRTi
6. Amagat's Law: The total volume V of a mixture is equal to the sum of the volume occupied by each component at
The mixture pressure P, and temperature T.
V = Vi
n1
m1
V1
n2
m2
V2
n3
m3
V3
P,T
P = P1 = P2 = P3
T = T1 = T2 = T3
n = n 1 + n2 + n3
V = V1 + V2 + V3
 PV PV1 PV2 PV3   RT 






 RT RT RT RT   P 
V = V1 + V2 + V3
V = V
yi 
Vi
V
7. Dalton's Law
The total pressure of a mixture P is equal to the sum of the partial pressure that each gas would exert at
the mixture volume V and temperature T.
P = Pi
T = T1 = T2 = T3
V = V1 = V2 = V3
mixture
1
n1
P1
2
n2
P2
n = n 1 + n2 + n 3
 PV P1 V P2 V P3 V   RT 
 RT  RT  RT  RT   P 
 
P = P1 + P2 + P3
P
yi  i
P
8. Molecular Weight of a Mixture
M
y M
M
R 8.3143 kg

R
R kgmol
i
i
9. Gas Constant
R
x R
R
R 8.3143 KJ

M
R kg - K
i i
10. Specific Heat
KJ
Cp 
 x C kg - K
Cv 
 x C kg - K
i Pi
KJ
i Vi
CP  C V  R
Rk
k 1
R
CV 
k 1
CP 
KJ
kg - K
3
n3
P3
n
P
11. Gravimetric and Volumetric Analysis
Gravimetric Analysis gives the mass fractions of the components in the mixture.
Volumetric Analysis gives the volumetric or molal fractions of the components in the mixture.
xi 
y iMi
yM
 i i
M
y iMi

xi
Mi
yi 
xi
Mi

Example 1
The specific weight of water at ordinary pressure and temperature is 9.81 KN/m3. The specific gravity of mercury is 13.55.
Compute the density of water, and the specific weight and density of mercury.
Example 2
A. Calculate the density, specific weight and specific volume of oxygen at 38C and 104 KPa.
B. What would be the temperature and pressure of the gas if it where compressed isentropically to 40 percent of its original
volume.
C. If the process described in (B) had been isothermal, what would be the temperature and pressure have been? (For O 2: M = 32 ;
k = 1.395)
Example 3
A vessel contains 85L of water at 10C ( = 999.7 kg/m3) and atmospheric pressure. If it is heated to 70C ( = 977.8 kg/m3), what
will be the percentage change in its volume? What mas of water must be removed, to maintain the volume at the original volume?
VARIATION OF PRESSURE WITH ELEVATION
By applying equations for equilibrium
 Fy  0
(P  dP)A  PA  W  0
PA  dPA  PA  W  0
dPA  W
But
W  γV
V  A(dh )
dPA  γA(dh )
dP  γ(dh )
General Equation
dP  -γdh
If the specific weight γ or density ρ is constant
ΔP  -γΔh
Note: Negative sign is used because pressure decreases with increasing elevation and increases with decreasing elevation.
h is positive if measured upward
h is negative if measured downward
Where:
P – Pressure in KPa
 - specific weight in KN/m3
h – elevation in meters
a. For isothermal Condition (T = C)
P
RT
Pg
γ
1000RT
dP   γdh
ln
ρ
h 2  h1  h
ln
Pg
dh
1000RT
dP
g

dh
P
1000RT
2 dP
2
g

dh
1000RT 1
1 P
dP  

P2
g

( h 2  h1 )
P1
1000RT

P1
gh

P2 1000RT
h
1000RT  P1 
ln   Elevation
g
 P2 
P2 
P1
gh
e 1000RT
 Final Pressure
b. Isentropic Condition
Pυ k  C
2
 1 1 
P k 
1
 1    1 / k ( h 2  h1 )
C
1  
 k 1
k 1
(h )
k 1
k 1
P2 k  P1 k   k 1 / k
C
k 1
k 1 

 C1 / k P2 k  P1 k 

  Elevation
h
k 1
k
1
υ
ρ
ργ
P
C
γk
P1
C
γ

γ1k
P2
γ2k
P1 / k
C1 / k
dP   γdh
dP  
dP
1/ k
P
P
1/ k
C


dh
1
C
1
1/ k
C
P 1 / k dP  
1
P2  P
dh
1/ k
P 1 / k dP  
2



1/ k
1
k 1
k 
 
k 1
h
k
C1 / k


k
k 1
 Final Pressure
dh
1
C1 / k
2
 dh
1
Pressure Variation in the Atmosphere
TROPOSPHERE: It is the layer in the atmosphere between sea level and 10.769 km and the temperature decreases linearly with
increasing elevation at a lapse rate of 6.5K/km.
g
P  Ts  0.0065h  6.5R


Ps 
Ts

6.5R


g


P


Ts 1   
  Ps 




h
0.0065
meters
STRATOSPHERE: It is the layer that begins at the top of the troposphere and extends to an elevation of 32.3 km, and in this
layer the temperature is constant at -55C.
P

Ps
h
1
gh
e 1000RT
1000RT Ps
ln
meters
g
P
where:
P - pressure at elevation h, KPa
Ps - Pressure at sea level, KPa
h - elevation, meters
Ts - temperature at sea level in K
FROM NASA
1. Atmospheric pressure decreases at the rate of 83.312 mm Hg per 1000 meters rise
in elevation.
2. Atmospheric temperature decreases at the rate of 6.5K per 1000 meters rise in
elevation.
CONDITIONS OF PRESSURE VARIATION WITH ELEVATION
General Formula
dP  dh
a.
Constant Density
1 dP    1 dh
P2  P1  (h 2  h1 )
2
2
(h 2  h1 )  h
P2  P1  h
P1  P2

Note
h
 h  if measured upward
- h - if measured downward
b.
For Isothermal Condition (T = C)
P
RT
Pg

1000RT
dP   dh

Pg
dP  
dh
1000RT
dP
g

dh
P
1000RT
g
2 dP
2

1
 dh
P
1000RT 1
c.
Isentropic Condition
ln
P2
g

(h 2  h1 )
P1
1000RT
h 2  h1  h
ln
P1
gh

P2 1000RT
h
1000RT  P1 
ln   Elevation
g
 P2 
P2 
P1
gh
e 1000RT
 Final Pressure
P k  C
1



P
C
k
P
P
C  1k  2k
1
2
P 1/ k
C1/ k
dP   dh

P1/ k
dh
C1 / k
dP
1
  1/ k dh
1/ k
P
C
1
P 1/ k dP   1/ k dh
C
1 2
2 1/ k
1 P dP   1/ k 1 dh
C
dP  
d.
2
 1 1 
P k 
1
 1    1/ k (h 2  h1 )
C
1  
 k 1
k 1
(h)
k 1
k 1
P2 k  P1 k   k 1/ k
C
k 1
k 1

1/ k 
 C P2 k  P1 k 

  Elevation
h
k 1
k
k

 k  1  k 1
 k 1 h k  
P2  P1 k   1/ k    Final Pressure
C






Temperature decreases at a Standard Lapse Rate (6.5K per Kilometer)
T  (Ts  0.0065h)

P
P

RT R(Ts  0.0065h)

g
Pg

1000 1000R(Ts  0.0065h)
dP  
Pg(dh)
1000R(Ts  0.0065h)

dP
g 
dh



P 1000R  (Ts  0.0065h) 
2
1

dP
 g 2
dh


1 
P 1000R  (Ts  0.0065h) 
Ts is constant
1
dP
g
2  ( 0.0065)dh 


 
P
( 0.0065)1000R 1  (Ts  0.0065h) 
ln
 (T  0.0065h) 
P2
g

ln  s

P1 6.5R 
Ts

2
g
P2  (Ts  0.0065h)  6.5R


P1 
Ts

g
 (T  0.0065h)  6.5R
P2  P1  s

Ts


6 .5 R


 P2  g 

Ts 1   
  P1 



h
0.0065
MANOMETERS
Manometer is an instrument used in measuring gage pressure in length of some liquid column.
1. Open Type Manometer : It has an atmospheric surface and is capable in measuring gage pressure.
2. Differential Type Manometer : It has no atmospheric surface and is capable in measuring differences of pressure.
Determination of S using a U - tube
Specific gravity or relative density of an unknown liquid can be determine using a U - tube with both ends open to the atmosphere,
and applying the Variation of Pressure with Elevation principle, provided that one liquid of known specific gravity is available.
Problem No. 1
A mercury manometer shown is connected to a pipeline carrying water ( = 9.81 KN/m3). If the elevation at point B is 3 m above
A and the mercury reading y = 1200 mm, what is the pressure in the pipe in KPa. ( of mercury = 133.4 KN/m3)
0
P1 = PB
1
1
2
P2 = PA
P0  133.4(1.2)  3(9.81)  PA
P0  0
PA  189.51KPa
Problem no. 2
In the figure shown, fluid A is water ( = 9.81 KN/m3, fluid B is mercury ( = 133.42 KN/m3). Z = 450 mm and Y = 900 mm.
Compute the pressure difference (Pn – Pm) in KPa.
Pn = P1
1
Pm = P5
5
2
4
3
3
Pn  ( y  x )9.81  z(133.42)  z(9.81)  x (9.81)  Pm
Pn  Pm  ( y  x )9.81  z(133.42)  z(9.81)  x (9.81)
Pn  Pm   y(9.81)  x (9.81)  z(133.42)  z(9.81)  x (9.81)
Pn  Pm   y(9.81)  z(133.42)  z(9.81) KPa
Pn  Pm  64.5 KPa
Problem no. 3
A U tube with both ends open to the atmosphere contains mercury in the lower portion. In one leg, water stands 760 mm above the
surface of mercury, in the other leg , oil (S = 0.80) stands 450 mm above the surface of the mercury. What is the difference in
elevation between the surfaces of mercury in contact with oil and water columns?
HYDROSTATIC FORCES ACTING ON PLANE SURFACES
Fig. (a)
Fig. (b)
C.G. - center of gravity
C.P. - center of pressure
General Equation:
F  γhA
Therefore, the total hydrostatic force exerted by the fluid on any plane surfaces submerged in a homogeneous fluid at rest, is equal
to the product of the surface area A and the pressure at its centroids  h .
LOCATION OF THE CENTER OF PRESSURE
yp 
Ig  A y
Ay
2
yp  y  e
ye
Ig
y
Ay
Ig
Ay
Ig
e
Ss
Ss  A y
e
where:
e - is the perpendicular distance between C.G. andC.P.
Ig - moment of inertia with respect to the axis at its centroids and lying on its plane
Ss- statical moment of inertia with respect to the axis SS not lying on its plane
HYDROSTATIC FORCE ACTING ON CURVED SURFACES
C'C = B'B = L
where L - length of the curved surface AB perpendicular to the paper
F  Fh  Fv KN
2
2
Fh  γhA KN
A = BC x L m2
where: A - area of the vertical projection of the curved surface AB
FV  γV KN
V = AABCDEA m2
Fh - horizontal component of F in KN that passes through the center of pressure of the vertical projection of the curved
surface AB
Fv - vertical component of F, KN
F - total hydrostatic force acting on the curved surface AB
Note: If the liquid is underneath the surface the Force acts upward.
HOOP TENSION
D
H
1m
Considering a semi-circular segment of 1 m length
P = h
T
T
F = 0
F = 2T  1
T 
F
22
F
F
T
1m
t
T
On the vertical projection:
D
1m
F
A
P
where
A = 1D
F = P(1D)
let : S - tensile or hoop stress in KPa
S
T
A
where A = 1t (area subjected to tensile stress)
S
F
P(1D)

2(1t) 2(1t)
S
PD
2t
ARCHIMEDES PRINCIPLE
LAWS OF BUOYANCY:Any body partly or wholly submerged in a liquid is subjected to a Buoyant or Upward force, which is equal
to the weight of the liquid displaced. If the buoyant force is less than to the weight of the body, the body sinks, and if the buoyant
force is greater than the weight of the body the body floats.
A.
W = BF
W = BVBKN; W = BVB kg
BF = VsKN; BF = Vs kg
B.
W = BF - T
W = BVBKN; W = BVB kg
BF = VsKN; BF = Vs kg
C.
W = BF + T
W = BVBKN; W = BVB kg
BF = VsKN ; BF = Vs kg
D.
W = BF - T
W = BVBKN ; W = BVB kg
BF = VsKN ; BF = Vs kg
VB = Vs
E.
W = BF + T
W = BVBKN ; W = BVB kg
BF = VsKN ; BF = Vs kg
VB = Vs
where:
W - weight of the body; KN, kg
BF - buoyant force, KN, kg
VB - volume of body, m3
Vs - volume submerged, m3
B - specific gravity of the body, KN/m3
 - specific gravity of the liquid, KN/m3
B - density of the body, kg/m3
 - density of the liquid, kg/m3
SPECIFIC GRAVITY OF SOLIDS HEAVIER THAN WATER
SB 
W1
W1  W2
SPECIFIC GRAVITY OF AN UNKNOWN LIQUIDS
SB 
W1  W3
W1  W2
Where
W1 – weight of the body in air
W2 - weight of the body in water
W3 - weight of the body in an unknown liquid
SAMPLE PROBLEMS
Problem No. 1 (Manometer)
A mercury manometer shown is connected to a pipeline carrying water ( = 9.81 KN/m3). If the elevation at point B is 3 m above
A and the mercury reading y = 1200 mm, what is the pressure in the pipe in KPa. ( of mercury = 133.4 KN/m3)
0
P1 = PB
1
1
2
P2 = PA
P0  133.4(1.2)  3(9.81)  PA
P0  0
PA  189.51 KPa
Problem no. 2 (Manometer)
In the figure shown, fluid A is water ( = 9.81 KN/m3, fluid B is mercury ( = 133.42 KN/m3). Z = 450 mm and Y = 900 mm.
Compute the pressure difference (Pn – Pm) in KPa.
Pn = P1
1
Pm = P5
5
2
4
3
3
Pn  ( y  x )9.81  z(133.42)  z(9.81)  x (9.81)  Pm
Pn  Pm  ( y  x )9.81  z(133.42)  z(9.81)  x (9.81)
Pn  Pm   y(9.81)  x (9.81)  z(133.42)  z(9.81)  x (9.81)
Pn  Pm   y(9.81)  z(133.42)  z(9.81) KPa
Pn  Pm  64.5 KPa
Problem No. 3 (Variation of Pressure)
An airplane barometer reads 762 mm Hg at sea level and 736 mm Hg at some unknown elevation. If the temperature at sea level is
15 C, what is the elevation assuming:
a. air at constant density
b. air under isothermal condition
c. air temperature decreases linearly with elevation at the rate of 6.5 K per kilometer.
Given:
P1 = 762 mm Hg = 101.6 KPa
P2 = 736 mm Hg = 98.23 KPa
T1 = 15 + 273 = 288 K
a. Constant density
P
101.6
kg

 1.23 3
RT 0.287(288)
m
1.23(9.81)
KN

 0.01206 3
1000
m
P2  P1   (h2  h1 )

P P
h  1 2  279.4 meters

b. Constant Temperature
h
1000RT  P1 
ln 
g  P2 
h  284.2 meters
c. Air temperature decreases linearly with elevation at the rate of 6.5 K per kilometer
g
P2  T1  0.0065h  6.5R


P1 
T1

6.5R


g


P
2

T1 1    
  P1  

 meters
h 
0.0065
h = 283.04 meters
Problem No. 4 (Variation of Pressure)
The bottom of a river is 12 m below the water surface. Underneath which is a silt having a specific gravity of 1.75 and a thickness
t. The pressure at the bottom of silt is 450 KPa. Determine the thickness of the silt.
at 0 to 2
0  9.81(12)  1.75(9.81)t  450
450  (9.81)12
t
 19.355 M
1.75(9.81)
Problem No. 5 (Hydrostatic forces)
A rectangular gate 1.6 m wide and 2 m high has water on one side and is inclined at 45with the horizontal. Water is 1.5 m above
the top of the gate.
a) Compute the total force acting on the gate and its location
b) If the gate is hinged at the top(B)what force P is needed at the bottom(A) to open the gate.
h  1.5  1 sin 45
h  1.5  1 sin 45
 M at A
h  2.207 m
2P  F(1.10)  0
69.3(1.10)
P
2
P  38.3 KN
2
A  1.6(2)  3.2 m
F  9.81(2.207)(3. 2)
F  69.3 KN
From Figure
sin 45 
h
y
2.207
y
 3.12 m
sin 45
For a rectangle
Ig 
yp 
B
yp’
bh3 1.6(2)3

 1.0666 m4
12
12
Ig  A y
2
Ay

2m
F
1.0666  3.2(3.12)2
3.2(3.12)
y p  3.23 m
P
1.5
1.5
y p'  y p 
 3.23 
sin 45
sin 45
y p'  1.109 m
A
Problem No. 6 (Forces of Curved Surfaces)
The curve surface represented by AB in the figure is the surface of a quadrant of a circular cylinder 3 m long. Determine the
horizontal and vertical component of the total hydrostatic force on the surface if the liquid is gasoline with S = 0.72.
Fv
L - length
L=3m
R = 2 m (radius)
Liquid surface
Fh  γ hA
3m
A  2(3)  6 m 2
L=3
m
h  3 1  4 m
γ  0.72(9.810)  7.0632
KN
m3
Fh  169.52 KN
Fv  γV
1

V   π 22  2(3) 3  27.425 m 3
4

Fv  7.0632(27.425)  193.71 KN
F
Fh 2  Fv 2  257.41 KN
2m
Fh
2m
h  4m
Problem No. 7 (Buoyancy)
Two spheres each 1.2 m in diameter weighs 4 KN and 12 KN, respectively. They are connected with a short rope and placed in
water. What portion of the lighter sphere protrudes from the water.
W1  4 KN; W2  12 KN; R 
Volume of Sphere 
1.2
 0.60 m
2
4
πR 3
3
W1  W2  BF1  BF2
4

4  12  9.81VS1  9.81 π(0.60) 3 
3

VS1  0.73 m 3
VF1  volume protrudes above liquid surface
4

VF1   π(0.60) 3   0.73  0.175 m 3
3

Problem No. 8 (Buoyancy)
A block of material of an unknown volume is submerged in water and weighs 500 N. The same block weighs 650 N when weigh
in air. Determine the volume of the material in m3.
S
W1
650

 4.33
W1  W2 650  500
ρ  4.33(1000)  4330
kg
m3
W  mg
650
 66.26 kg
9.81
m
ρ
V
66.26
V
 0.0153 m 3
4330
m
Problem No. 9 (Forces on Plane surface)
A square gate (3 m x 3 m) is hinged at the top
= 0.80) stands on the left side of the gate to a height of 1.5 m above point A. Determine the amount of the hydrostatic force exerted
by the oil on the gate and its location of the center of pressure.
Ig for a square:
Ig =
bh 3
12
SOLUTION
h  1.5  1.5 sin 60  2.79 m
F  0.80(9.81)(2.79)(9)  197.06 KN
yp 
Ig  A y
2
Ay
3(3) 3
Ig 
 6.75 m
12
h
y
 3.22 m
sin 60
6.75  9(3.22) 2
yp 
 3.453 m
9(3.22)
Problem No. 10 (Hoop Tension)
A 1.2 m diameter steel pipe, 6 mm thick, carries oil with S = 0.822 under a head of 122 m of oil. Compute
a. The stress in the steel in KPa
b. The thickness of the steel pipe required to carry a pressure of 1724 KPa with an allowable stress of 124 MPa
PD
2t
P  0.822(9.81)(122)  983.8 KPa
S
983.8(1.2)
 98,379 KPa
2(0.006)
1724(1.2)
t
 0.0083 m  8.3 mm
124,000(2)
S
SAMPLE OF MANOMETER PROBLEMS
7.
The closed tank in the figure is filled with water. The pressure gage on the tank reads 48 KPa. Determine
d. The height h in mm in the open water column
e. The gage pressure acting on the bottom of the tank surface AB
f. The absolute pressure of the air in the top of the tank if the local atmospheric pressure is 101 KPa absolute.
8.
The mercury manometer in the figure indicates a differential reading of 30 m when the pressure in pipe A is 30 mm Hg
vacuum. Determine the pressure in pipe B.
9.
In the figure pipe A contains carbon tetrachloride (S = 1.60) and the closed storage tank B contains a salt brine (S = 1.15).
Determine the air pressure in tank B in KPa if the gage pressure in pipe A is 1.75 kg/cm2.
10. A U tube mercury manometer is connected to a closed pressurized tank as shown. If the air pressure is 14 KPa, determine
the differential reading h. The specific weight of the air is negligible.
11. A closed tank contains compressed air and oil (S = 0.90) as shown in the figure.a U – tube manometer using mercury (S =
13.6) is connected to the tank as shown. For column heights h1 = 90 cm; h2 = 15 cm and h3 = 22 cm, determine the
pressure reading of the gage.
12. The pressure of gas in a pipeline is measured with a mercury manometer having one limb open to the atmosphere. If the
difference in the height of mercury in the limbs is 562 mm, calculate the absolute gas pressure. The barometer reads 761
mm Hg, the acceleration due to gravity is 9.79 m/sec2 and SHg = 13.64.
13. A turbine is supplied with steam at a gauge pressure of 1.4 MPa, after expansion in the turbine the steam flows into a
condenser which is maintained at a vacuum of 710 mm Hg. The barometric pressure is 772mm Hg. Express the inlet and
exhaust pressure in kg/cm2 . Take the S of mercury is 13.6.
14. The pressure of steam flowing in a pipe line is measured with a mercury manometer. Some steam condenses in to water.
/estimate the steam pressure in KPa. Take the density of mercury as 13,600 kg/m 3, the barometer reading as 76.1 cm Hg
and g = 9.806 m/sec2.
15. A manometer is attached to a tank containing three different fluids, as ashow in the figure below. What will be the
difference in elevation h of th mercury column in the manometer.
Quiz no. _______
1. The pressure on top of a mountain is 90 KPa. If the pressure and temperature ate sea level are 101 KPa and 288 K,
respectively, Determine the height of the mountain and the temperature on top, Assuming
a. Constant density
b. Isothermal condition
c. Isentropic condition
d. Standard atmospheric lapse rate prevails
2. A manometer is attached to a tank containing different fluids, as shown in the figure below. What will be the
difference in elevation of the mercury column in the manometer.
1.
A differential manometer is shown in the figure below. Calculate the pressure difference (P A – PB) in kg/cm2.
(1 inch = 25.4 mm; 1 m = 1000 mm).
2.
On top of a mountain the pressure is 85 KPa. If the pressure and temperature ate sea level are 101.33 KPa and 21C
respectively, Determine the altitude and temperature on top of the mountain, assuming Constant density
b. Isothermal condition
c. Isentropic condition
d. Standard atmospheric lapse rate prevails
3. For the configuration shown below, calculate the weight of the piston if the gage pressure reading is 70 KPa.
4. On a certain day a PAL plane is flying at an altitude of 3000 m. If the atmospheric pressure and temperature at sea level are
101 KPa and 288 K, respectively. Determine the pressure and temperature in the plane, assuming
a. Constant density from sea level to elevation 3000 m
b. Isothermal condition
c. Isentropic condition
d. Standard atmospheric lapse rate prevails
5. A mountain is 2000 meters above sea level. If the pressure and temperature ate sea level are 101 KPa and 288 K,
respectively, Determine the temperature and pressure on top of the mountain, assuming
a.
b.
c.
Constant density from sea level to the top of the mountain
Isothermal condition
Isentropic condition
Standard atmospheric lapse rate prevails
6. If the atmospheric pressure is 101.03 KPa and the absolute pressure at the bottom of the tank is 231.3 KPa.
What is the specific gravity of olive oil? (S of SAE oil = 0.89 ; S of Hg = 13.6)
101.3  (0.89)(9.81)1.5  2.5(9.81)  γ oil (2.9)  13.6(9.81)(0.4)  231.3
γ oil  13.34
KN
m3
Soil  1.36
SAMPLE PROBLEMS
1. A vertical piston cylinder device contains a gas at an unknown pressure. If the outside pressure is 100 kPa, determine (a)
the pressure of the gas if the piston has an area of 0.2 m2 and a mass of 20 kg. Assume g = 9.81 m/s2. (b) What-if-Scenario:
How would the answer in (a) changes if the orientation of the device is changed upside down. Answers: (a) 101 kPa, (b) 99
kPa.
2. A piston with a diameter of 50 cm and a thickness of 5 cm is made of a composite material with a density of 4000 kg/m3.
(a) If the outside pressure is 101 KPa, determine the pressure inside the piston-cylinder assembly if the cylinder contains
air. (b) How would the answer change if the piston diameter was 100 cm instead? Answers: (a) 104.8 KPa
3. Water flows through a variable-area pipe with a mass flow rate of 10,000 kg/min. Determine the minimum diameter of the
pipe if the flow velocity is not to exceed 5 m/s. Assume density of water to be 1000 kg/m 3. Answers: 0.206 m
4. A bucket of concrete with a mass of 5000 kg is raised without any acceleration by a crane through a height of 20 m. (a)
Determine the work transferred into the bucket. (b) What happens to the energy as it is transferred to the bucket? (c) Also
determine the power delivered to the bucket if it is raised at a constant speed of 1 m/s.
Answers: (a) 981 kJ, (c) 49.05 kW
5. A building in Makati is 84.5 m high above the street level. The required static pressure of the water line at the top of the
building is 2.5 kg/cm2. What must be the pressure in KPa in the main water located 4.75 m below the street level.
P1 = 245.166 KPa
245 .166  9.81(84 .5)  4.75 (9.81)  1120 .71 KPa
6. A cylindrical tank 2 m diameter, 3 m high is full of oil. If the specific gravity of oil is 0.9, what is the mass of oil in the tank?
kg
ρ  0.9(1000)  900 3
m
π 2
V  (2) 3  9.425 m 3
4
m
ρ
V
m  900(9.425)  8482.5 kg
7. A cubical tank 1 m on a side, contains a mixture of 1.8 kg of nitrogen (M = 28; k = 1.399) and 2.8 kg of an unknown gas. The
mixture pressure and temperature are 290 KPa and 340 K. Determine
a) Molecular weight and gas constant of the unknown gas
b) the volumetric analysis
Given:
8.3143
Mx 
 72.93 kg/kg m
mN2 = 1.8 kg ; mx = 2.8 kg
0.114
P = 290 KPa ; T = 340K
xi 0.391 0.609
 Mi  28  72.93  0.014  0.0084
xi
 Mi  0.0223
m  1.8  2.8  4.6 kg
1.8
 0.391  39.1%
4.6
x x  0.609  60.9%
x N2 
PV  mRT
y N 2  0.6278  62.78%
y x  0.3722  37.22%
 mixture
V  1(1)(1)  1 m
3
290(1)
KJ
 0.1854
4.6(340)
kg - K
R  x N2R N2  X x R x
R
8.4143
KJ
 0.297
28
kg  K
0.1854  0.391(0.297)  0.609( Rx )
R N2 
R x  0.114
KJ
kg - K
8. A closed vessel of 0.7 m3 internal volume contains a gas at 58 KPa and 18C and with R = 0.27 KJ/kg- K.If now 0.35 kg of
another gas at 18C and R = 0.29 KJ/kg-K is also admitted into the vessel. Calculate the final pressure of the mixture.
P1V  m1R 1 T1
Given:
V = 0.7 m3
58(0.7)
m1 
 0.517 kg
P1 = 58 KPa; T1 = 18 + 273 = 291 K
0.27(291)
R1 = 0.27 KJ/kg-K
m  m1  m2  0.517  0.35
m2 = 0.35 kg
m  0.867 kg
T2 = 291 K
R2 = 0.29 KJ/kg-K
0.517
x1 
 0.5963
0.867
0.35
x2 
 0.4037
0.867
R  x1R 1  x2R 2
R  0.5963(0.27)  0.4037(0.29)
R  0.278
KJ
kg - K
mRT 0.867(0.278)(291)

V
0.7
P  100.2 KPa
P
9. In the figure shown the diameters of the two cylinders are 75 mm and 600 mm, the face of the piston is 6 m above the face of the
weight “W” and the intervening passages is filled with oil (S = 0.80). What force F is required if W = 35 KN.
F
W = 35 KN
10. If 1 100 N force F1 is applied to the piston with 5 cm diameter, what is the magnitude of the force F2 that can be resisted by the
piston with the 10 cm diameter? Neglect the weights of the piston.
F2
10 cm
oil
2m
F1
5 cm
2.2 m
11. A glass tube 1.5 m long and 25 mm diameter with one end closed is inserted vertically with the open end down into a tank of
water until the open end is submerged to a depth of 1.2 m. If the barometric pressure is 98 KPa, and neglecting vapor pressure,
how high will water rise in the tube. (Assume isothermal conditions for air)
1.5 m
1.2 - x
1.2 m
x
MANOMETER
1. A mercury manometer shown is connected to a pipeline carrying water ( = 9.81 KN/m3). If the elevation at point B is 3 m
above A and the mercury reading y = 1200 mm, what is the pressure in the pipe in KPa.( of mercury = 133.4 KN/m3)
Open
y = 1200 mm
B
3m
A
2. A is water, Fluid B is oil (S = 0.80), z = 350 mm. Compute the pressure difference between m and n in KPa.
Fluid B
z
m
n
Fluid A
Example 2
What force must be exerted through the bolts to hold the dome in place? The metal dome and pipe weighs 6 KN and has no
bottom.
80 cm
ID = 20 cm
400 cm
160 cm
h  1.6  4  5.6 m
V - Volume of imaginary volume of liquid
V  Vcylinder - Vdome
14
V  π(1.6) 2 5.6     π1.6 3
23
2


V  π1.6 2 5.6  1.6   36.46 m3
3


Tension in bolts  9.81(36.46) - 6  351.7 KN
R = 160
cm
water
Example 3
A hemispherical dome surrounds a closed tank as shown. If the tank and dome are filled with gasoline (S = 0.72) and the gage
indicates a pressure of 60 KPa. What is the total tension in the bolts holding the dome in place.
5.5 m
R=2m
Gasoline
S = 0.72
3.0 m
P =60 KPa
By extending the curved surface to the imaginary free surface of the liquid
with 0 gage
0  0.72(9.81)h  60
h  8.5 m
H - height of imaginary cylindrica l prism
H  8.5 - 3  5.5 m
the weight of imaginary fluid prism
V  Vcylinder - Vdome
14
V  π22 5.5    π23  52.4 m 3
23
W  0.72(9.81)(52.4)  370 KN
Example 4
A cubical block of wood 10 cm on a side floats at the interface between oil and water with its lower surface 1.50 cm below the
interface. The density of the oil is 790 kg/m3 . What is the gage pressure at the lower face of the block? What is the mass and
density of the block?
1.
A piece of wood of S = 0.651 is 8 cm square and 150 cm long. How many kilograms of lead weighing 11,200 kg/m 3 must
be fastened at one end of the stick so that it will float upright with 30.5 cm out of water?
S  0.651
kg
m3
V1  0.08(0.08)(1.5)  0.96 m 3
ρ wood  0.651(1000)  651
W1  0.96(651)  6.25 kg
W2  11,200V2
BF1  (1.5  0.305)(0.08) 21000  7.65 kg
BF2  1000V2
W1  W2  BF1  BF2
6.25  11,200V2  7.65  1000V2
V2 
7.65 - 6.25
 1.4 x 10- 4 m 3
10,200
W2  1.4 x 10- 4 (11200)  1.54 kg
2.
A barge is loaded with 150 Metric tons of coal. The weight of the empty barge in air is 35 Metric ton. If the barge is 5.5 m
wide, 16 m long and 3 m high, what is its draft. (Depth below the water surface)
W  BF
W1  W2  ρVs
(150  35)(1000)  1000(5.5)(16)h
h  2.1 m
3.
A prismatic object 20 cm thick by 20 cm wide by 40 cm long is weighed in water at a depth of 50 cm and found to weigh
50 N. What is its weight in air and its specific gravity?
W1 
W2  50 Newton
V1  (0.20)(0.20)(0.4)  0.016 m 3
ρ1 
m1
V1
m1
V
W1
S 1 
1000 W1  W2
W1  m1g Newton
W1
gV1
W1

1000 W1  W2
W1
W1

9810V1 W1  W2
1
1

9810V1 W1  W2
W1  W2  9,810(0.016)
W1  156.96  50
W1  206.96 Newton
S
4.
206.96
 1.312
206.96 - 50
A ship, with vertical sides near the water line, weighs 3630 Metric ton and draws 6.7 m in saltwater (S = 1.025). Discharge
of 181 Metric ton of water ballast decreases the draft to 6.4 m. What would be the draft d of the ship in fresh water (S = 1)
W1  BF1
3630(1000)  A1L(6.7)(1.025)(1000)
3,630
 Eq.1
6.7(1.025)
W2  BF2
A1L 
(3,630  181)(1000)  A 2 L(6.4)(1.025)(1000)
A 2L 
3449
 Eq.2
(6.4)(1.025)
1.
An object weighs 25.95 N when submerged in kerosene (S = 0.81) and weighs 26.6 N when submerged in gasoline (S =
0.68). Determine the specific weight of the object.
2.
Determine the water power and mechanical efficiency of a centrifugal pump which has an input of 3 KW. If the pump has
a 203 mm diameter suction line and a 152 mm diameter discharge line and handles 10 L/sec of water at 66C ( =980
kg/m3;  = 9.6 KN/m3). The suction line gauge shows 102 mm Hg vacuum and the discharge gauge shows 180 KPa. The
Discharge gauge is located 61 cm above the center of the discharge pipeline and the pump inlet and discharge lines are at
the same elevation.
3.
A piece of wood of S = 0.651 is 8 cm square and 150 cm long. How many kilograms of lead weighing 11,200 kg/m 3 must
be fastened at one end of the stick so that it will float upright with 30.5 cm out of water?
4.
At one point in a pipeline the water speed is 3 m/sec and the gage pressure is 50 KPa. Find the gage pressure at a second
point in the line, 11 m lower than the first, if the pipe diameter at the second point is twice at the first.
1. A circular gate 1.5 m in diameter is inclined at an angle of 45. Sea water (S = 1.02) stands on one side of the gate to a
height of 10 m above the center of the gate. Determine the total hydrostatic force F and the location of C.P.
F  hA  1.02(9.81)(10)
h
sin45 
F  hA
A
e
 2
D
4
Ig
y
10
y
 14.142 m
sin 45
A  1.767 m 2
Ig 
Ay
h
y
sin 45
D 4
Ig 
64

(1.5) 2  176.82 KN
4
yp 
D 4
 0.2485 m 4
64
Ig  A y
Ay
2
 14.15 m
2. An airplane barometer reads 762 mm Hg at sea level and 736 mm Hg at some unknown elevation. If the temperature at
sea level is 15 C, what is the elevation assuming:
a. air at constant density (h = 287.8 m)
b. air under isothermal condition(h = 292.8 m)
c. air under isentropic conditions (h = 291.35 m)
d. air condition follows standard atmosphere condition (h = 291.83 m)
SOLUTION:
762
101.325  101.6 KPa
Ps 
760
Ts  15  273  288 K
736
101.325  98.13 KPa
P
760
1.
What force must be exerted through the bolts to hold the dome in place? The metal dome and pipe weighs 6 KN and has no
bottom.
h  1 .6  4  5 .6 m
V - Volume of imaginary volume of liquid
V  Vcylinder - Vdome
V  (1.6) 2 5.6  
1 4
3
 1.6 
23
2

2
V  1.6  5.6  1.6   36.46 m 3
3


Tension in bolts  9.81(36.46) - 6  351.7 KN
2.
A U tube mercury manometer is connected to a closed pressurized tank as shown. If the air pressure is 14 KPa, determine
the differential reading h. The specific weight of the air is negligible.
14  9.81(0.6)  9.81(0.6)  9.81h  13.6(9.81)h  14
19.81(0.6)  9.81(0.6)  9.81h  13.6(9.81)h  0
h
(9.81)(1.2)
 0.095 m  9.5 cm
12.6(9.81)
1. A BOEING 747 flies at an altitude of 2500 m above sea level where Ps = 101.33 KPa and Ts = 288 K. Determine the
pressure and temperature at this altitude considering
a. air under isothermal condition (P2 = 75.31 KPa ; T = 288 K)
b. air under isentropic conditions (P2 = 74.31 KPa ; T = 263.6 K)
c. air condition follows standard atmosphere condition (P2 = 74.66 KPa ; T = 271.75 K)
h  2500 m
Ps  101.33 KPa
Ts  288 K
1. The closed tank in the figure is filled with water. The pressure gage on the tank reads 48 KPa. Determine
a. The height h in mm in the open water column
b. The gage pressure acting on the bottom of the tank surface AB
c. The absolute pressure of the air in the top of the tank if the local atmospheric pressure is 101 KPa absolute.
48  0.60(9.81)  h(9.81)  0
h  5 .5 m
48  0.6(9.81)  (0.6)(9.81)  PAB
PAB  59.772 KPa
Pabs  48  101  149 KPa
2. The air above the liquid is under a pressure of 40 KPa gage, and the specific gravity of the liquid in the tank is 0.80. If the
rectangular gate is 1 m wide and if y1 = 1 m and y2 = 3 m , What force P is required to hold the gate in place?
SOLUTION
F  40  0.8(9.81)( 2.5) (1)(3)  178.86 KN
Ig
e
Ay
Ig 
1(3)3
 2.25 m 4
12
A y  3(2.5)  7.5 m3
e  0.3 m
 M @ Hinge  0
F(1.5  0.3)  3P
p
178.86(1.8)
 107.32 KN
3
FLUID MECHANICS QUIZ NO. 3
1.
A block of wood has a vertical projection of 15.24 cm when placed in water and 10.2 cm when placed in alcohol. If the
specific gravity of alcohol is 0.82, find the specific gravity of wood gravity of wood.
2.
In water
W  BF  1000A(h - 0.1524)  1
In Alcohol
W  BF  0.82(1000)A(h - 0.102)  2
Equating eq.1 and eq. 2
(h - 0.1524)  0.82(h - 0.102)
h - 0.82h  0.1524  0.82(0.102)
h  0.382 m
W  1000(0.2296)A  3
W  ρ B A(0.382)  4
eq.3  eq. 4
1000(0.2296)  0.382ρ B
SB 
3.
ρB
0.2296

 0.601
1000 0.382
A cubical block of wood 10 cm on a side floats at the interface between oil and water with its lower surface 1.50 cm
below the interface. The density of the oil is 790 kg/m3 . What is the gage pressure at the lower face of the block? What
is the mass and density of the block?
4.
Compartments A and B of the tank shown in the figure below are closed and filled with air and a liquid with S = 0.6. If the
atmospheric pressure is 101 KPa (abs) and the pressure gage reads 3.5 KPa (gage), determine the manometer reading h in
cm.
3.5  9.81h  0.6(9.81)h  0.6(9.81)0.02  13.6(9.81)(0.03)  0
h
5.
3.5  0.6(9.81)0.02  13.6(9.81)(0.03)
 1.9 m
9.81(1  0.6)
The gate shown is hinged at A and rests on a smooth floor at B. The gate is 3 m square. Oil (S = 0.80) stands on the left
side of the gate to a height of 1.5 m above A. Above the oil surface is a gas under a gage pressure of -6.9 KPa. Determine
the amount of the vertical force P applied at B that would be required to open the gate.
P = -6.9 KPag
Free Surface
1.5m
h
A
F
P
CG
CP
3
m
45
h  1.5  1.5 sin 45   2.56 m
F   6.9  0.80 (9.81)2.56 (3)(3)  118 .8 KN
6.
A spherical buoy 2 m in diameter floats half submerge in a liquid with S = 1.5. What is the weight of the lead anchor weighing
7000 kg/m3 will completely submerged the buoy in the liquid.
ρ L  1.5(1000)  1500
kg
m3
4 3
πR  4.18 m 3
3
14

Vs   π(1) 3   2.09 m 3  volume submerge
23

W1  BF1  ρ L (Vs )
Vsphere 
W1  1500(2.09)  3135 kg
from figure 2 :
W1  W2  BF1  BF2
BF1  1500(Vsphere )  1500(4.18)  6270 kg
W1  3135 kg
W2  7000V2
BF2  1500V2
3135  7000V2  6270  1500V2
V2  0.57 m 3
W2  3990 kg
7.
A glass tube 1.5 m long and 25 mm diameter with one end closed is inserted vertically with the open end down
into a tank of water until the open end is submerged to a depth of 1.2 m. If the barometric pressure is 98 KPa,
and neglecting vapor pressure, how high will water rise in the tube. (Assume isothermal conditions for air)
fOR isothermal :
P1V1  P2 V2
π
(0.025) 2 (1.5)  0.00074 m 3
4
P2  98  9.81(1.2  x )  1
V1 
π
(0.025) 2 (1.5  x )  2
4
π

π

98 (0.025) 2 (1.5)  98  9.81(1.2  x ) (0.025) 2 (1.5  x )
4
4




98(1.5)  98  9.81(1.2  x )(1.5  x )
V2 
147  (98  11.772  9.81x )(1.5  x )
147  (109.772  9.81x )(1.5  x )
147  164.658  109.772x  14.715x  9.81x 2
9.81x 2  124.487x  17.658  0
x 2  12.7 x  1.8  0
 12.7  12.98
2
x  0.14 m  14 cm
x
1.
A brass cube 152.4 mm on a side weighs 298.2 N. We want to hold this cube in equilibrium under water by attaching a
light foam buoy to it. If the foam weighs 707.3 N/m3, what is the minimum required volume of the buoy?
W1  298.2 N
BF1  (0.1524)3 (9810)  34.72 N
W2  707.3V2  1
BF2  V2 (9810)  2
298.2  707.3V2  34.72  V2 (9810)
298.2  34.72
 0.029 m 3
9810  707.3
A man dives into a lake and tries to lift a large rock weighing 170 kg. If the density of the granite rock is 2700 kg/m3, find
the force that the man needs to apply to lift it from the bottom of the lake. Assume density of lake water to be 1000 kg/m3.
V2 
2.
W  2700V  170 kg
170
 0.063 m 3
2700
BF  1000(0.063)  63 kg
V
W  BF  T
T  170 - 63  107 kg
Example 1
A large pipe called a penstock in hydraulic work is 1.5 m in diameter. Here it is composed of wooden staves bound together by
steel hoops each 3.23 cm2 in area, and is used to conduct water from a reservoir to a powerhouse. If the maximum tensile stress
permitted in the hoops is 130 MPa, what is the maximum spacing between hoops under a head of 30.5 m.
1.5 m
L
P  9.81(30.5)  299.205 KPa
D  1.5 m
On the vertical projection
F
; A  LD
A
F  299.205(1.5)L  448.81L KN
P
F
 224.4L KN
2
Tensile Stress on the hoop
T
T
; A  3.23 cm 2
A
A  0.000323 m 2
S
S  130000 KPa
224.4L
0.000323
L  0.187 m  18.7 cm
130000 
L
Example 2
Compute the wall stress in a 1200 mm diameter steel pipe 6 mm thick under a pressure of 970 KPa.
Given:
D = 1.2 m
t = 0.006 m
P = 970 KPa
S
PD 970(1.2)

 97,000 KPa
2t 2(0.006)
Example 3
What is the minimum allowable thickness of 600 mm diameter steel pipe under an internal pressure of 860
KPa with a working stress in the steel of 70,000KPa.
S
PD
2t
860(0.600)
2t
t  0.004 m  4 mm
70,000 
Example 4
A wood stave pipe is bound by steel rods which take the entire bursting stress. Find the proper spacing for
25 mm steel rods for a 1800 mm diameter wood stave pipe under a pressure of 590 KPa if the working
stress in the steel is 105,000 KPa.
Given
Dr = 0.025 m
D = 1.8 m
P = 590 KPa
S = 105,000 KPa
F
2T

A DL
For the rods
P
S
T

0.0252
4
105,000 
4T
0.025
2
T  51.54 KN
L 
2(51.54)
 0.10 m
590(1.8)
L  10 cm
Example 5
A vertical cylindrical tank, 2 m in diameter and 4 m high, is held together by means of two steel hoops, one at the top and one at
the bottom. When molasses (S = 1.50) stands to a depth of 3 m in the tank, what is the tensile force in each hoop?
A. NOZZLE
2
2
P1 v 1
P
v

 Z 1  2  2  Z 2  HL

2g

2g
From continuity equation: Q = Av ; for 1 = 2 = 
Q  A 1v 1  A 2 v 2
For a nozzle the head loss HL is equal to:
 1
v
 2  1 2
C
 2g
 v

2
where: Cv - velocity coefficient
JET POWER
2
v
PJet  Q 2
2g
B. VENTURI METER
C. ORIFICE
An orifice is an opening with a closed perimeter in which fluid flows.
By applying Bernoulli's Energy theorem:
2
2
P1 v 1
P v

 Z1  2  2  Z2
 2g

2g
But P1 = P2 = Pa and v1 is negligible, then
2
v2
 Z1  Z2
2g
and from figure: Z1 - Z2 = h, therefore
2
v2
h
2g
v 2  2gh
let v2 = vt
v t  2gh
where:
vt - theoretical velocity, m/sec
h - head producing the flow, meters
g - gravitational acceleration, m/sec2
COEFFICIENT OF VELOCITY (Cv)
actual velocity
theoretica l velocity
v'
Cv 
vt
Cv 
COEFFICIENT OF CONTRACTION
area of jet @ vena contracta
area of the orifice
a
Cc 
A
Cc 
where: a - area of jet at vena contracta, m2
A - area of the orifice, m2
COEFFICIENT OF DISCHARGE
actual flow
theoretica l flow
Q'
Cd 
Q
Cd  C v Cc
Cd 
where:
v' - actual velocity
vt - theoretical velocity
a - area of jet at vena contracta
A - area of orifice
Q' - actual flow
Q - theoretical flow
Cv - coefficient of velocity
Cc - coefficient of contraction
Cd - coefficient of discharge
JET TRAJECTORY:
v  velocity of water at tip of nozzle
at 1 to 0
v o 2  v12 - 2gd
0  v12 - 2gd
(vsin θ) 2
1
2g
v 0  v1  gt
d
v sin θ
2
g
at 0 to 2
t
d  v1t  1 2 gt 2
d  0  1 2 gt 2
d  1 2 gt 2  3
(vsin θ) 2 1 2
 2 gt
2g
t2 
(vsin θ) 2
g2
(vsin θ)
t
4
g
R  v cos θ(2 t )
2(vsin θ)( v cos θ) v 2 (2sin θ cos θ)

g
g
(2sin θ cos θ)  sin( 2θ)
R
R
v 2 (sin2 θ)
5
g
If the jet is flowing from a vertical orifice: If the jet is initially horizontal vx = v.
D. PUMP
Pump is a steady - state, steady - flow machine in which mechanical energy is added to the liquid from one point to another point
of higher pressure.

B
z2
TANK
1
z1
A
TANK
Pump Centerline
PUMP
Points 1 & 2 are the reference points at suction and discharge,
respectively.
a) Total Dynamic Head (Ht)
 P  P   v 2  v12 
h t   2 1    2
  Z2  Z1  HL meters
γ  
2g





b) Fluid or Water Power (FP)
FP  Qh t KW
c) Capacity or Discharge (Q)
Q  A1v1  A 2 v 2 m3/sec
d) Brake or Shaft Power (BP)
2TN
KW
60,000
BP 
e) Motor Power (MP) (For motor driven pump)
1. For single - phase motors
MP 
EI(cos )
KW
1000
2. For 3 - phase motors
MP 
3 EI(cos )
KW
1000
f) Pump Efficiency
p 
FP
x 100%
BP
g) Motor Efficiency
m 
BP
x 100%
MP

h) Combined Pump-Motor Efficiency (Overall Efficiency)
c 
MP
x 100 %
IP
where:
ht - total dynamic head, m
FP - Fluid or Water Power, Kw
BP - Brake or Shaft Power, KW
MP - Motor Power. KW
IP - Power input to motor, KW
Q - Capacity or Discharge, m3/sec
 - Specific weight of the liquid pumped, KN/m3
v - velocity, m/sec
P - pressure, KPa
g - gravitational acceleration, m/sec2
Z - elevation, m (Positive if measured above datum, and negative if measured below datum)
HL - head loss (due to fluid friction, fittings and turbulence in pipes)
T - Brake torque, N-m
N - no. of RPM (revolutions per minute)
E - Volts
I - current drawn by the motor, Amperes
cos - Power Factor
E. HYDRAULIC TURBINE
Impulse Type (Pelton Type)
Reaction Type( Francis Turbine)
Dam
1
Head Water
Head Gates
(Fully Open
Penstock
Y – Gross Head
PA
A
Draft Tube
zA
B
2
Tail Water
Fundamental Equations
1. TOTAL DYNAMIC HEAD (h):
a. For an Impulse Type
h = (Z1 - Z2) - HL meters
h = Y - HL meters
Y = Z1 - Z2 meters
b. For a Reaction Type
h = (Z1 - Z2) - HL meters
h = Y - HL meters
Y = Z1 - Z2 meters
Y - gross head at plant, m
at A to 2
h
PA v 2A

 z A meters
 2g
2. WATER POWER (WP):
FP = Qh KW
3. DISCHARGE OR RATE OF FLOW (Q):
Q = Av m3/sec
4. BRAKE POWER (BP):
BP = 2TN
KW
60 000
T - brake torque, N-m
N - rotative speed, RPM
5. TURBINE EFFICIENCY (e)
e = BP x 100%
FP
e = evehem
where:
ev - volumetric efficiency
eh - hydraulic efficiency
em - mechanical efficiency
6. ROTATIVE SPEED (N):
N
120f
n
where: f - frequency, cps or Hertz
n - no. of generator poles (usually divisible by 4)
QUIZ NO. 4
Problem No. 1
The velocity of water in a 10 cm diameter pipe is 3 m/sec. At the end of the pipe is a nozzle whose velocity coefficient is 0.98. If
the pressure in the pipe is 55 KPa, what is the velocity in the jet? What is the diameter of the jet? What is the rate of discharge? What
is the head loss?
d1  0.10 m
m
sec
C v  0.98
v1  3
P1  55 KPa
Applying Bernoulli' s equation
2
2
P1 v 1
P v

 z1  2  2  z 2  HL
 2g

2g
2
2
 1
v 2  1
v
v
HL   2  1 2  
 1 2  0.041 2
2
2g
 2g
 Cv
 2g  (0.98)
2
2
2
P1 v 1
v
v

 0  0  2  0  0.041 2
 2g
2g
2g
2
2
v2
1  0.041  P1  v1
2g
 2g
v2 
P v 2 
2g 1  1 
  2g   10.75 m
1  0.041
sec

m3
(0.10)2 (3)  0.024
4
sec
 2
m3
Q  d2 (10.75)  0.024
4
sec
d2  0.053 m  5.3 cm
Q  Av 
 (10.75)2 
  0.24 m
HL  0.041
 2g 
Problem No. 2
A centrifugal pump draws water from a well at the rate of 142 L/sec of water through a 203 mm ID suction line and a 152 mm ID
discharge line. The suction gauge located on the pump centerline reads 254 mm Hg vacuum, while the discharge gauge is 6 m above
the pump centerline. If the power input to the water is 75 KW, find the reading of the discharge gauge in KPa.
m3
sec
d1  0.203 m ; A 1  0.032 m2
Q  0.142
d2  0.152 m; A 2  0.018 m2
 2
d
4
Q
v
A
A
m
m
; v 2  7 .8
sec
sec
z1  0; z 2  6 m
v 1  4 .4
 101.325 
P1  -254
  33.86 KPa
 760 
WP  Qht
ht 
75
 53.84 m
(0.142)(9.81)
ht 
P2  P1 v 2  v 1

 z 2  z1   HL

2g
2
2
2
2

 P
v  v1
P2   ht  2
 z 2  z1   HL   1
2
g

 
P2  414.45 KPa
Problem No. 3
A 15 KW suction pump draws water from a suction line whose diameter is 200 mm and discharges through a line whose diameter
is 150 mm. The velocity in 150 mm line is 3.6 m/sec. If the pressure at point A in the suction line is 34.5 KPa below the atmosphere
where A is 1.8 m below that of B on the 150 mm line, Determine the maximum elevation above B to which water can be raised
assuming a head loss of 3 m due to friction.
FP
d1
d2
v2
P1
P2
HL
A1
A2
Q
v1
SW
P1/SW
P2/SW
v1^2/2g
v2^2/2g
z1
z2
ht
D(Phead)
D(Vhead)
KW
m
m
m/sec
KPa
KPa
m
m^2
m^2
m^3/sec
m/sec
KN/m^3
m
m
m
m
m
D(Ehead)
15
0.2
0.15
3.6
-34.5
0
3
0.03
0.02
0.064
2.025
9.81
-3.52
0.000
0.209
0.66
0.0
1.8 + h
24.04
3.52
0.45
(1.8 +
h)
h
15.27
m
m
Problem No. 4
A power nozzle throws a jet of water that is 50 mm in diameter. The diameter of the base of the nozzle and of the approach pipe
is150 mm. If the power of the nozzle jet is 42 HP and the pressure head at the base of the nozzle is 54 m, compute the head lost in
the nozzle.
Problem No. 5
A fire pump delivers water through a 150 mm main to a hydrant to which is connected a 75 mm hose, terminating in a 25 mm nozzle.
The nozzle is 1.5 m above the hydrant and 10 m above the pump. Assuming frictional losses of 3 m from the pump to the hydrant,
2 m in the hydrant and, and 12 m from the hydrant to the base of the nozzle, and a loss in the nozzle of 6% of the velocity head in
the jet, to what vertical height can the jet be thrown if the gage pressure at the pump is 550KPa.
Problem No. 6
Water issues from a circular orifice under a head of 12 m. The diameter of the orifice is 10 cm. If the discharge is found to be 75
L/sec, what is the coefficient of discharge? If the diameter at the vena cotracta is measured to be 8 cm, what is the coefficient of
contraction and what is the coefficient of velocity.
v  2gh  theoretical velocity
v  2(9.81)(12)  15.34
Q'  0.075
m
sec
m3
sec

m3
(0.10)2 (15.34)  0.12
4
sec
Q'
Cd 
 0.625
Q
a d' 8
Cc   
 0.67
A D 12
Cd  Cc(Cv )
Q
Cv 
0.625
 0.93
0.67
Problem No. 7
A jet discharges from an orifice in a vertical plane under a head of 3.65 m. The diameter of the orifice is 3.75 cm and the measured
discharge is 6 L/sec. The coordinates of the centerline of the jet are 3.46 m horizontally from the vena contracta and 0.9 m below the
center of the orifice. Find the coefficient of discharge, velocity and contraction.
Problem No. 8
The inside diameters of the suction and discharge pipes of a pump are 20 cm and 15 cm, respectively. The discharge pressure is read
by a gage at a point 2 m above the centerline of the pump, and the suction pressure is read by a gage
1 m below the pump
centerline. If the pressure gage reads 145 KPa and the suction gage reads a vacuum of 250 mm Hg when diesel fuel (S = 0.82) is
pumped at the rate of 30 L/sec, Find the KW power of the driving motor if overall pump efficiency is 75%.
d1
d2
z1
z2
P1
P2
S
SW(water)
SW
0.20
0.15
-1
2
-33.33
145
0.82
9.81
8.0442
m
m
m
m
Kpa
KPa
KN/m^3
KN/m^3
Q
A1
A2
v1
v2
P1/SW
P2/SW
v1^2/2g
v2^2/2g
D(Phead)
D(Vhead)
D(ElHead)
HL
ht
WP
e
BP
0.03
0.031
0.018
0.955
1.70
-4.14
18.03
0.05
0.15
22.17
0.10
3.00
0.00
25.27
6.10
0.75
8.13
m^3/sec
m^2
m^2
m/sec
m/sec
m
m
m
m
m
m
m
m
m
KW
KW
Problem No. 9
A jet of water 7.6 cm in diameter discharges through a nozzle whose velocity coefficient is 0.96. If the pressure in the pipe is 82.7
KPa and the pipe diameter is 20 cm and if it is assumed that there is no contraction of the jet, what is the velocity at the tip of the
nozzle? What is the rate of discharge?
Q  A 1v 1  A 2 v 2
4
d  2
2
v 1   2  v 2
 d1 
4
2
d  v 2
v1
  2  2
2g  d1  2g
2
2
P1 v 1
P v

 z1  2  2  z 2  HL
 2g

2g
2
2
 1
v 2
P1 v 1
P v

 z1  2  2  z 2   2  1 2
 2g

2g
 Cv
 2g
4
2
2
 1
v 2
P1  d2  v 2
P v
  
 z1  2  2  z 2   2  1 2
  d1  2g

2g
 Cv
 2g
d1
d2
Cv
P1
P2
v1
v2
SW
g
v2
A1
A2
Q
0.20
0.076
0.96
82.7
0
1.80
12.467
9.81
9.81
12.47
0.031
0.005
0.057
m
m
Kpa
KPa
m/sec
m/sec
KN/m^3
m/sec^2
m/sec
m^2
m^2
m^3/sec
4
2

 d2   P1  P2
v 2  1




 1  1     
 ( z1  z 2 )
2g  C v 2

 d1  


4
2
v 2  1  d2   P1  P2
 2     
 ( z1  z 2 )
2g  C v

 d1  

 P1  P2

2
   ( z1  z 2 )
v2


2g
 1  d 4 
 2   2  
 C v
 d1  
SAMPLE PROBLEMS
APPLICATION OF BERNOULLI’S EQUATION
ExampleNo. 1
The water in a 10 m diameter, 2 m high aboveground swimming pool is to be emptied by unplugging a 3 cm diameter, 25 m long
horizontal pipe attached to the bottom of the pool. Determine the maximum discharge rate of water through the pipe.
1
h=2
m
2
P1 v12
P
v 2

 Z1  2  2  Z 2
γ 2g
γ
2g
P1  0; v1  0; P2  0; Z1  0; Z 2  -2 m
0000
2
Q  Av
Q
v22
 (2)
2g
π
(0.03) 2 (6.23)  0.0044 m3/sec
4
v22
2g
v 2  2(9.81)( 2 )  6.23 m/sec
Example No. 2
A large tank open to the atmosphere is filled with water to a height of 5 m from the outlet tap. A tap near the bottom of the tank is
now opened and water flows out from the smooth and rounded outlet. Determine the water velocity at the outlet.
2
1
2
P1 v 1
P
v

 Z1  2  2  Z 2

2g

2g
P1  0; v 1  0; P2  0; Z 1  0; Z 2  - 5 m
2
v
0  0  0  0  2  ( 5)
2g
2
5
v2
2g
v 2  2(9.81)5  9.623 m/sec
Example No. 3
The water level of a tank on a building roof is 20 m above the ground. A hose leads from the tank bottom to the ground. The end of
the hose has a nozzle, which is pointed straight up. What is the maximum height to which the water could rise.
P1 v12
P v 2

 Z1  2  2  Z2
γ 2g
γ
2g
P1  0; P2  0; Z1  0; Z2  h; v2  0
0
v12
0  00h
2g
v12
2g
h  max imum height
h
v1  velocity at the tip of the nozzle
Example no. 4
Water flows through a horizontal pipe at the rate of 1 Gal./sec. The pipe consist of two sections of diameter 4 in. and 2 in with a
smooth reducing section. The pressure difference between the two pipe sections is measured by mercury manometer . Neglecting
frictional effects, determine the differential height of mercury between the two pipe sections.
Q = 1 gal/sec = 0.0038 m3/sec
D1 = 4 in. = 0.1016 m
D2 = 2 in = 0.0508 m
2
1
x
h
P1  9.81x  9.81h  13.6(9.81
)h  9.81x  P
Mercury S = 13.6 2
P2  P1  9.81h  13.6(9.81)h
P2  P1  123.606  1
from 1 to 2
P1 v 1 2
P v 2

 Z1  2  2  Z2
γ 2g
γ 2g
Q
0.0038
v1 

 0.47 m/sec
A1 π
(0.1016)2
4
Q
0.0038
v2 

 1.875m/sec
π
A2
(0.0508)2
4
2
v1
(0.47)2

 0.0113m
2g 2(9.81)
v 2 2 (1.875)2

 0.18 m
2g
2(9.81)
P1
P
 0.0113  0  2  0.18  0
γ
γ
P2  P1
 0.0113  0.18  0.1687
γ
P2  P1  9.81(0.1687)  1.655 KPa  2
Equating eq. 1 and eq. 2
- 123.606h  -1.655
- 1.655
- 123.606
h  0.0134 meters
h
h  0.53 inches
Example No. 5
A 300 mm pipe is connected by a reducer to a 100 mm pipe. Points 1 and 2 are at the same elevation. The pressure at point 1 is
200 KPa. Q = 30 L/sec flowing from 1 to 2, and the energy lost between 1 and 2 is equivalent to 20 KPa. Compute the pressure at
2 if the liquid is oil with S = 0.80. (174.2 KPa)
P1 v12
P
v 2

 z1  2  2  z 2  H L
γ 2g
γ
2g
Q  A1v1  A 2 v 2
v
Q
π
; A  D2
A
4
v1 
0.030
m v12 (0.424) 2
 0.424
;

 0.009 m
π
sec 2g
2(9.81)
(0.3) 2
4
0.030
m v 2 2 (3.82) 2
 3.82
;

 0.744 m
π
sec 2g 2(9.81)
(0.1) 2
4
KN
γ  0.8(9.81)  7.848 3
m
20
HL 
 2.55 m
7.848
v2 
P1 v12
P
v 2

 Z1  2  2  Z 2  H L
γ 2g
γ
2g
Z1  Z 2 : (Z 2 - Z1 )  0
P2 P1 v12 v 2 2
200
 

 HL 
 0.009  0.744  2.55
γ
γ 2g 2g
7.848
P2  174.22 KPa
Example No. 6
A venturi meter having a diameter of 150 mm at the throat is installed in a 300 mm water main. In a differential gage partly filled
with mercury (the remainder of the tube being filled with water) and connected with the meter at the inlet and at the throat, what
would be the difference in level of the mercury columns if the discharge is 150 L/sec? Neglect loss of head. (h=273 mm)
L
1m 3
m3
x
 0.150
sec 1000L
sec
S of mercury  13.6
Q  150
γ  13.6(9.81)  133.416
KN
m3
d1  0.30m
d 2  0.15m
Q  0.150 m 3 / sec
HL  0
v
Q
A
v1 
v2 
0.150
π
(0.30) 2
4
0.150
π
(0.15) 2
4
 2.122 m / sec;
v12
 0.230 m
2g
 8.488 m / sec;
v22
 3.672 m
2g
P1 v12
P
v 2

 Z1  2  2  Z 2
γ 2g
γ
2g
Z1  Z 2  0
P1  P2 v 2 2 v12


 3.672  0.230
γ
2g 2g
P1  P2
 3.442 m
γ
P1  P2  3.442(9.81)  33.766 KPa
P1  9.81( x )  9.81h  133.416h  9.81x  P2
P1  9.81h  133.416h  P2
P1  P2  133.416h  9.81h
P1  P2  h (133.416  9.81)
P1  P2
33.766

(133.416  9.81) (133.416  9.81)
h  0.273 m
h
h  273 mm
Example no. 7
A mechanical engineer of an industrial plant is required to install a centrifugal pump to lift 15 L/sec of water from a sump to a
storage tank on a tower. The water is to be delivered into a 105 KPag tank and the water level in the tank is 20 m above the water
level in the sump. Pump centerline is 4 m above the water level in the sump. The suction pipe is 100 mm diameter and discharge
pipe is 65 mm diameter. Head loss at suction is 3 times the velocity head in the suction line and head loss at discharge is 20 times
the velocity head in the discharge pipeline. Other data are as follows:
p = 75 %
m = 80 %
E = 220 Volts
Motor – 3-Phase
Power Factor = 0.92
Requirements:
a. Sketch of the problem
b. Total dynamic head in m
c. Water Power in KW
d. Motor power in KW
e. Line current drawn by the motor in amperes
f. Total power cost per day for 10 hours a day continuous operation and a power costs of P 5.00/KW-hr
105 KPa
2
20 m
65 mm
100 mm
4m
1
Q  Av
Q
A
Q
0.015
m
vs 

 1.91
A s π (0.10) 2
sec
4
Q
0.015
m
vd 

 4.52
A s π (0.065) 2
sec
4
v
vs 2 (1.91)2

 0.186 m
2g 2(9.81)
v d 2 (4.52)2

 1.041 m
2g
2(9.81)
v 2
h Ls  3 s   0.56 m
 2g 


v 2
h Ld  20 d   20.83 m
 2g 


H L  0.56  20.83  21.39 m
P2  P1 v 2 2  v12

 Z 2  Z1  H L
γ
2g
at 1 to 2, datum line through point 1
ht 
105 - 0
 0  (20  0)  20.83
9.81
h t  52.091 m
ht 
WP  0.015(9.81)(52.091)  7.67 KW
7.67
 10.22 KW
0.75
10.22
MP 
 12.8 KW
0.75
BP 
MP 
3 EI(P.F.)
1000
3 (220)I(0.92)
1000
I  36.44 amperes
12.8 
Cost  12.8(10)(5)  P639.00
Example No. 8
Figure below shows a siphon discharging oil (sp gr 0.90). The siphon is composed of 8 cm. pipe from A to B followed by 10 cm.
pipe from B to the open discharge at C. The head losses are from 1 to 2, 0.34 m; from 2 to 3, 0.2 m; from 3 to 4;0.8 m. Compute
the discharge in L/sec
4.5 m
3m
0
Example No. 9
The liquid in the figure has a specific gravity of 1.5. The gas pressure PA is 35 KPa and PBis -15 KPa. The orifice is 100 mm in
diameter with Cd = Cv = 0.95. Determine the velocity in the jet and the discharge when h = 1.2. (9.025 m/sec; 0.071 m 3/sec)
PA
Without considerin g head lost
1
P1 v12
P
v 2

 Z1  2  2  Z 2
γ 2g
γ
2g
Datum at point 1 :
h
PB
Z1  0 ; Z2  - h; v1  0
v 2 2 P1  P2

 Z1  Z 2
2g
γ
2
P  P

v 2  2g  1 2  h   theoreti cal velocity
 γ

KN
γ  1.5(9.81)  14.715 3
m
 35  15

v 2  v t  2(9.81) 
 1.2  9.5 m/sec
 14.715

Example no. 10
A pump draws water from reservoir A and lifts it to reservoir B as shown in the figure. The loss of head from A to 1 is 3 times the
velocity head in the 150 mm pipe and the loss of head from 2 to B is 20 times the velocity head in the 100 mm pipe. Compute the
horsepower output of the pump and the pressure heads at 1 and 2 when the discharge is 20 L/sec. (FP= 20.73 HP; 5.74 m ; 84.3 m)

72 m
100 mm
1
6m
150
mm
A
B
I-1
I-2
V-1
V-2
pump
20
m3
 0.020
1000
sec
Q
v
A
0.020
vs 
 1.13 m/sec
π
(0.150) 2
4
Q
vd 
0.020
Water Power  0.020(9.81)(78.63)
Water Power  15.43 KW  20.7 HP
At 1 to A (Point 1 on datum)
P1 v12
P
v 2

 Z1  A  S  Z B  h Ls
γ 2g
γ
2g
 2.55 m / sec
π
(0.100) 2
4
v 2
h Ls  3 s   0.20 m
 2g 


000 
v 2
h Ld  20 d   6.63 m
 2g 


PA (1.13) 2

 6  0.20
γ 2(9.81)
PA
 5.73 meters ; PA  56.2 KPa
γ
At B to 2 (Point B on datum)
PB v d 2
P v 2

 Z B  2  2  Z2  h Ld
γ
2g
γ 2g
H L  h Ls  h Ld  6.83 m
At point 1 to 2
PB 2.55) 2

 0  0  0  78  6.63
γ 2(9.81)
P1 v12
P v 2

 Z1  ht  2  2  Z 2  H L
γ 2g
γ 2g
v1 and v 2 are negligible
PB
 84.3 meters; PB  827 KPa
γ
P1  P2  0 gage (Atmospher ic)
h t  Z 2  Z1  H L
h t  72  6.83  78.63 m
Example No. 11
The diameters of the suction and discharge pipe of a fuel pump for a day tank of a gasoline engine are 150 mm and 100 mm,
respectively. The discharge pressure gauge located 10 m above the pump centerline reads 320 KPa and the suction pressure gauge
which is 4 m below the pump centerline reads a vacuum of 60 KPa. Head losses due to pipe friction, turbulence and fittings amounts
to 15 m. If gasoline with a relative density S = 0.75 is pumped at the rate of 35 L/sec, find
a. The total dynamic head developed by the pump
b. The fluid power in KW
c. The brake or shaft power delivered to the fluid for a pump efficiency of 75%
d. The brake torque if the pump speed is 1200 RPM
e. The electrical power input to the pump motor for a motor efficiency of 92%
f. The line current drawn by the motor if the motor is 3 – phase, 240 volts, and 0.9 Power Factor
g. The cost of power for 5 hours operation, if electricity costs P 1.50/KW-hr
Q
S
SW(water)
SW
g
d1
d2
P1
P2
z1
z2
HL
FP
A1
A2
0.035
0.75
9.81
7.3575
9.81
0.150
0.100
-60.00
320.00
-4
10
15
75
0.018
0.008
m^3/sec
KN/m^3
KN/m^3
m/sec^2
m
m
KPa
KPa
m
m
m
KW
m^2
m^2
v1
v2
v1^2/2g
v2^2/2g
P1/sw
P2/sw
D(Phead)
D(vhead)
D(Zhead)
ht
Water Power
Pump
Efficiency
Brake Power
Motor
Efficiency
Motor Power
E (Volts)
Power Factor
Phase
N(RPM)
Torque(N-m)
Line Current
PowerCost
Cost
2.0
4.5
0.20
1.01
-8.15
43.49
51.65
0.81
14
81.46
20.98
m/sec
m/sec
m
m
m
m
m
m
m
m
KW
75.00
%
27.97
KW
92.00
%
30.40
220.00
0.90
3.00
1200.00
222.57
88.65
1.50
228.01
KW
Volts
Phase
RPM
N-m
Amperes
Pesos/KW-hr
Pesos
Example No. 12
A jet of liquid is directly vertically upward. At A (Nozzle tip) its diameter is 75 mm and its velocity is 10 m/sec. Neglecting air
friction, determine its diameter at a point 4 m above A.
d1
v1
P1
P2
g
A1
Q
SW
P1/SW
P2/SW
v1^2/2g
z1
z2
HL
v2^2/2g
v2
A2
d2
d2
0.075
10
0
0
9.81
0.0044
0.0442
9.81
0
0
5.097
0
4
0
9.097
13.36
0.0033
0.065
6.5
m
m/sec
Kpa
KPa
m/sec^2
m^2
m^3/sec
KN/m^3
m
m
m
m
m
m
m
m/sec
m^2
m
cm
Example No. 13
A closed vessel contains water up to a height of 2 m and over the water surface there is air having a pressure of 8.829 N/cm2 above
atmospheric pressure. At the bottom of the vessel there is an orifice of diameter 15 cm. Find the rate of flow of water from orifice
if Cd = 0.6.
Example No. 14
The 600 mm pipe shown in the figure conducts water from a reservoir A to a pressure turbine, which discharges through another
600 mm pipe into tailrace B. The loss of head from A to 1 is 5 times the velocity head in the pipe and the loss of head from 2 to B
is 0.2 times the velocity head in the pipe. If the discharge is 700 L/sec , what horsepower is being given up by the water to the
turbine and what are the pressure heads at 1 and 2.(FP = 537.4 HP; 53.628 m; -4.75 m)
Q  700 L/sec  0.70 m 3 /sec
v
Q 4(0.70)

 2.5 m/sec
A π(0.6) 2
v 2 (2.5) 2

 0.32 m
2g 2(9.81)
h LA-1  5(0.320  1.6 m
h L2- B  0.20(0.32)  0.064 m
H L  1.6  0.064  1.664 meters
At A to B
PA v A 2
P
v 2

 ZA  B  B  ZB  H L  h
γ
2g
γ
2g
With datum line through poin B
Z B  0; ZA  60 m
PA  PB  0 gage
v A  VB  0
h  (ZA  ZB )  H L
h  60  1.664  58.336 meters
WP  Qγh  0.70(9.81)(58.336)  400.6 KW  537 HP
h
P1
g
SW
P1/SW
d(orifice)
A(Orifice)
Theoretical
Velocity
Cd
Q(m^3/sec)
Q(L/sec)
2
88.29
9.81
9.81
9.0
0.15
0.018
m
Kpa
m/sec^2
KN/m^3
m
m
m^2
14.69
m/sec
0.60
0.1558
155.8
m^3/sec
L/sec
At A to 1
P v2
PA v A 2
 Z A  1  1  Z1  h LA 1

γ 2g
2g
γ
with datum through point A
ZA  0 ; Z1  -55.5 m
000
P1
 0.32  55.5
γ
P1
 55.5  0.32  55.18 meters
γ
P1  55.18(9.81)  541.3 KPa
At 2 to B, with datum through point 2
Z 2  0 ; Z B  -4.5 m
v 2
P
P2 v 2 2
 Z2  B  B  ZB  h L2  B

2g
γ
2g
γ
P2
 0.32  0  0  0  4.5  0.064
γ
P2
 4.5  0.32  0.064  4.756 meters
γ
P2  -4.756(9.81)  -46.7 KPa  46.7 KPa vacuum
FLUID MECHANICS (PRE – FINAL S4)
March 15, 2017
Name ____________________________________
5. A piece of wood of S = 0.651 is 8 cm square and 150 cm long. How many kilograms of lead weighing 11,200 kg/m 3 must
be fastened at one end of the stick so that it will float upright with 30.5 cm out of water?
S  0.651
kg
m3
V1  0.08(0.08)(1.5)  0.96 m3
 w ood  0.651(1000)  651
W1  0.96(651)  624.96 kg
W2  11,200V2
BF1  (1.5  0.305)1000  1195 kg
BF2  1000V2
W1  W2  BF1  BF2
624.96  11,200V2  1195  1000V2
1195 - 624.96
 0.056 m 3
10,200
W2  625.93 kg
A barge is loaded with 150 Metric tons of coal. The weight of the empty barge in air is 35 Metric ton. If the barge is 5.5 m
wide, 16 m long and 3 m high, what is its draft. (Depth below the water surface)
V2 
6.
W  BF
W1  W2  Vs
(150  35)(1000)  1000(5.5)(16)h
h  2.1 m
7.
A prismatic object 20 cm thick by 20 cm wide by 40 cm long is weighed in water at a depth of 50 cm and found to weigh
50 N. What is its weight in air and its specific gravity?
W1 
W2  50 Newton
V1  (0.20)(0.20)(0.4)  0.016 m3
1 
m1
V1
m1
W1
V1

S
1000 W1  W2
W1  m1g Newton
W1
W1
gV1

1000 W1  W2
W1
W1

9810V1 W1  W2
1
1

9810V1 W1  W2
W1  W2  9,810(0.016)
W1  156.96  50
W1  206.96 Newton
S
8.
206.96
 1.312
206.96 - 50
A ship, with vertical sides near the water line, weighs 3630 Metric ton and draws 6.7 m in saltwater (S = 1.025). Discharge
of 181 Metric ton of water ballast decreases the draft to 6.4 m. What would be the draft d of the ship in fresh water (S = 1)
W1  BF1
3630(1000)  A 1L(6.7)(1.025)(1000)
3,630
 Eq.1
6.7(1.025)
W2  BF2
A 1L 
(3,630  181)(1000)  A 2L(6.4)(1.025)(1000)
A 2L 
3449
 Eq.2
(6.4)(1.025)
In a hydroelectric power plant, the water surface on the crest of the dam is at elevation 75.3 m while the water surface just at the
outlet of the head gate is at elevation 70.4 m. The head gate has 5 gates of 0.91 m x 0.91 m leading to the penstock and are fully
opened. Assume 61% as coefficient of discharge, determine
a. The quantity of water that enters the hydraulic turbine in m3/sec
b. The KW power that the turbine will developed, assuming eturbine = 90% efficiency and the turbine is 122 m below the
entrance of the penstock
c. The brake torque for a speed N = 1800 rpm
d. The number of generator poles if f = 60 Hertz
e.
The electrical power developed by the generator if electrical and windage loses amounts to 18%
h  75.3  70.4  4.9 m  head producing the flow
v  2gh  theoretical velocity
Q  5(Av)  theoretical flow for 5 gates
Q'  C d Q  Actual flow
Q'  0.61(5)(0.91x0.91) 2(9.81)4.9  24.76
m3
sec
h t  total dynamic head
h t  122  4.9  126.9 m
BP  e Turbine Q' h t  24.76(9.81)(126.9)(0.90)  27,746.2 KW
2TN
60,000
T  147,198 N  m
BP 
GP  22, 751KW
FINAL EXAM (March 25, 2017)
SET 2
NAME _____________________________________
1.
An object weighs 25.95 N when submerged in kerosene (S = 0.81) and weighs 26.6 N when submerged in gasoline (S =
0.68). Determine the specific weight of the object.
  6.23(9.81)  61.116
2.
KN
m3
Determine the water power and mechanical efficiency of a centrifugal pump which has an input of 3 KW. If the pump has
a 203 mm diameter suction line and a 152 mm diameter discharge line and handles 10 L/sec of water at 66C ( =980
kg/m3;  = 9.6 KN/m3). The suction line gauge shows 102 mm Hg vacuum and the discharge gauge shows 180 KPa. The
Discharge gauge is located 61 cm above the center of the discharge pipeline and the pump inlet and discharge lines are at
the same elevation.
BP
3
KW
D1
0.203
m
D2
0.152
m
Q
0.01
m^3/sec
A1
0.032
m^2
A2
0.018
m^2
v1
0.309
m/sec
v2
0.551
m/sec
g
9.810
m/sec^2
v1^2/2g
0.005
m
v2^2/2g
0.015
m
SW
9.6
KN/m^3
P1
-13.60
Kpa
P2
180
KPa
z2
0.61
m
z1
0
m
(P2-P1)/sw
20.17
m
(v2^20.01
m
v1^2)/2g
(z2-z1)
0.61
m
HL
0.00
m
Ht
20.79
m
WP
1.996
KW
em
66.52
%
3.
A piece of wood of S = 0.651 is 8 cm square and 150 cm long. How many kilograms of lead weighing 11,200 kg/m 3 must
be fastened at one end of the stick so that it will float upright with 30.5 cm out of water?
S  0.651
kg
m3
V1  0.08(0.08)(1.5)  0.96 m3
 w ood  0.651(1000)  651
W1  0.96(651)  624.96 kg
W2  11,200V2
BF1  (1.5  0.305)1000  1195 kg
BF2  1000V2
W1  W2  BF1  BF2
624.96  11,200V2  1195  1000V2
1195 - 624.96
 0.056 m 3
10,200
W2  625.93 kg
V2 
4.
At one point in a pipeline the water speed is 3 m/sec and the gage pressure is 50 KPa. Find the gage pressure at a second
point in the line, 11 m lower than the first, if the pipe diameter at the second point is twice at the first.
FLUID MECHANICS (PRE – FINAL S2)
March 04, 2017
Name ____________________________________
1.
The diameters of the suction and discharge pipe of a fuel pump for a day tank of a gasoline engine are 150 mm and 100
mm, respectively. The discharge pressure gauge located 10 m above the pump centerline reads 320 KPa and the suction
pressure gauge which is 4 m below the pump centerline reads a vacuum of 60 KPa. Head losses due to pipe friction,
turbulence and fittings amounts to 15 m. If gasoline with a relative density S = 0.75 is pumped at the rate of 35 L/sec, find
a.
b.
c.
d.
e.
f.
g.
The total dynamic head developed by the pump
The fluid power in KW
The brake or shaft power delivered to the fluid for a pump efficiency of 75%
The brake torque if the pump speed is 1200 RPM
The electrical power input to the pump motor for a motor efficiency of 92%
The line current drawn by the motor if the motor is 3 – phase, 240 volts, and 0.9 Power Factor
The cost of power for 5 hours operation, if electricity costs P 1.50/KW-hr
Q
S
SW(water)
SW
g
d1
d2
P1
P2
z1
z2
HL
FP
A1
A2
v1
v2
v1^2/2g
v2^2/2g
P1/sw
P2/sw
D(Phead)
D(vhead)
D(Zhead)
ht
Water Power
Pump
Efficiency
Brake Power
Motor
Efficiency
Motor Power
E (Volts)
Power Factor
Phase
N(RPM)
0.035
0.75
9.81
7.3575
9.81
0.150
0.100
-60.00
320.00
-4
10
15
75
0.018
0.008
2.0
4.5
0.20
1.01
-8.15
43.49
51.65
0.81
14
81.46
20.98
m^3/sec
75.00
%
27.97
KW
92.00
%
30.40
220.00
0.90
3.00
1200.00
KW
Volts
KN/m^3
KN/m^3
m/sec^2
m
m
KPa
KPa
m
m
m
KW
m^2
m^2
m/sec
m/sec
m
m
m
m
m
m
m
m
KW
Phase
RPM
Torque(N-m)
Line Current
PowerCost
Cost
2.
N-m
Amperes
Pesos/KW-hr
Pesos
A jet of liquid is directly vertically upward. At A (Nozzle tip) its diameter is 75 mm and its velocity is 10 m/sec. Neglecting
air friction, determine its diameter at a point 4 m above A.
d1
v1
P1
P2
g
A1
Q
SW
P1/SW
P2/SW
v1^2/2g
z1
z2
HL
v2^2/2g
v2
A2
d2
d2
3.
222.57
88.65
1.50
228.01
A closed vessel contains
the water surface there is air
above atmospheric pressure.
orifice of diameter 15 cm.
orifice if Cd = 0.6.
0.075
10
0
0
9.81
0.0044
0.0442
9.81
0
0
5.097
0
4
0
9.097
13.36
0.0033
0.065
6.5
h
P1
g
SW
P1/SW
d(orifice)
A(Orifice)
Theoretical
Velocity
Cd
Q(m^3/sec)
Q(L/sec)
m
m/sec
Kpa
KPa
m/sec^2
m^2
m^3/sec
KN/m^3
m
m
m
m
m
m
m
m/sec
m^2
m
cm
2
88.29
9.81
9.81
9.0
0.15
0.018
m
Kpa
m/sec^2
KN/m^3
m
m
m^2
14.69
m/sec
0.60
0.1558
155.8
m^3/sec
L/sec
water up to a height of 2 m and over
having a pressure of 8.829 N/cm2
At the bottom of the vessel there is an
Find the rate of flow of water from
Example No. 15
A Francis turbine is installed with a vertical draft tube. The pressure gauge located at the penstock leading to the turbine casing reads
372.6 KPa and velocity of water at inlet is 6 m/sec. The discharge is 2.5 m3/sec. The hydraulic efficiency is 85%, and the overall
efficiency is 82%. The top of the draft tube is 1.5 m below the centerline of the spiral casing, while the tailrace level is 2.5 m from
the top of the draft tube. There is no velocity of whirl at the top or bottom of the draft tube and leakage losses are negligible.
Calculate,
a) the net effective head in meters (43.817 m)
b) the brake power in kw. (881.2 kw)
c) the plant output for a generator efficiency of 92%. (810.7 kw)
d) the mechanical efficiency (96.550)
GIVEN:
P1 = 372.6 KPa
v = 6 m/sec
Q = 2.5 m3/sec
eh = 0.85
e = 0.82
ZB = (1.5 + 2.5) = 4 m
at point b to 2 (datum line through point 2)
2
2
PA v A
P
v

 Z A  2  2  Z 2  hL A  2  h

2g

2g
P2  0
v2  0
Z2  0
hL A  2  0 (negligible)
2
h
PA v A

 ZA

2g
372.6
( 6 )2

4
9.81 2(9.81)
h  43.817 m
h
WP  Qh
WP  2.5(9.81)( 43.817)  1074.6 kw
BP  1074.6(0.82)  881.2 KW
PLANT OUTPUT  881.2(0.92)  810.7 KW
e  em eh e v
0.82  em (0.85)(1)
em 
0.82
 0.965  96.5%
0.85
Example no. 16
A 4 m3/hr pump delivers water to a pressure tank. At the start, the gauge reads 138 KPa until it reads 276 KPa and then the pump
was shut off. The volume of the tank is 160 Liters. At 276 KPa, the water occupied 2/3 of the tank volume.
a) Determine the volume of water that can be taken out until the gauge reads 138 KPa.
b) If 1 m3/hr of water is constantly used, in how many minutes from 138 KPa will the pump run until the
gauge reads 276 KPa.
P1  138  101.325  239.325 KPa
P2  276  101.325  377.325 KPa
1
(0.160)  0.0533 m3
3
P1V1  P2 V2
V2 
V1  0.084 m3
Vw@138  (0.160 - 0.084)  0.076  76 Liters
2
(0.16)  0.1067  106.7 Liters
3
Vtaken out  0.1067  0.076  0.0307  30.7 Liters
Vw @ 276 
4 t  1t  0.0307
t  0.01023 hrs  0.614 min  36.84 sec
HEAD LOSSES
HL = Major loss + Minor losses
Major Loss: Head loss due to friction and turbulence in pipes
Minor Losses: Minor losses include, losses due to valves and fittings, enlargement, contraction, pipe entrance and pipe exit. Minor
losses are most easily obtained in terms of equivalent length of pipe "Le".the advantage of this approach is that both pipe and
fittings are expressed in terms of "Equivalent Length" of pipe of the same relative roughness.
Considering Major loss only
Darcy-Weisbach Equation
hf 
fLv 2
meters
2gD
Considering Major and Minor losses
hf 
f(L  ΣLe )v 2
meters
2gD
where; f - friction factor from Moody's Chart
L - length of pipe, m
Le - equivalent length in straight pipe of valves and fittings, m
V - velocity, m/sec
D - pipe inside diameter, m
g - gravitational acceleration, m/sec20
REYNOLD'S NUMBER: Reynold's Number is a non dimensional one which combines the physical quantities which describes
the flow either Laminar or Turbulent flow.
The friction loss in a pipeline is also dependent upon this dimensionless factor.
NR 
ρvD vD

μ
ν
where;  - absolute or dynamic viscosity, Pa-sec
 - kinematic viscosity, m2/sec
For a Reynold's Number of less 2100 flow is said to Laminar
For a Reynold's Number of greater than 3000 the flow is Turbulent
FRICTION FACTOR: Moody's Chart
f
/D
 - absolute roughness
D - inside diameter
/D - relative roughness
NR
VALUES OF ABSOLUTE ROUGHNESS  FOR NEW PIPES
Type of Material
Feet
Drawn tubing, brass, lead, glass
centrifugally spun cement, bituminous
lining, transite
0.000005
Commercial Steel, Wrought iron
0.00015
Welded steel pipe
0.00015
Asphalt-dipped cast iron
0.0004
Galvanized iron
0.0005
Cast iron, average
0.00085
Wood stave
0.0006 to
0.003
Concrete
0.001 to 0.01
Riveted steel
0.003 to 0.03
Millimeter
0.0015
0.046
0.046
0.12
0.15
0.25
0.18 to
0.9
0.3 to 3.0
0.9 to 9
For Laminar Flow
f
64
NR
PIPE, TUBING AND FITTINGS
Nominal Pipe Diameter: Pipe sizes are based on the approximate diameter and are reported as nominal pipe sizes. Regardless of
wall thickness, pipes of the same nominal diameter have the same outside diameter. This permits interchange of fittings. Pipe may
be manufactured with different and various wall thickness, so some standardization is necessary. A method of identifying pipe
sizes has been established by ANSI (American National Standard Institute). By convention, pipe size and fittings are characterized
in terms of Nominal Diameter and wall thickness.
For steel pipes, nominal diameter is approximately the same as the inside diameter for 12" and smaller. For sizes of 14" and larger,
the nominal diameter is exactly the outside diameter.
SCHEDULE NUMBER: The wall thickness of pipe is indicated by a schedule number, which is a function of internal pressure
and allowable stress.
Schedule Number  1000P/S
where P - internal working pressure, KPa
S - allowable stress, KPa
Schedule number in use: 10,20,30, 40,60, 80, 100, 120, 140, and 160.
Schedule 40 "Standard Pipe"
Schedule 80 " Extra Strong Pipe"
TUBINGS: Tubing specifications are based on the actual outside diameter with a designated wall thickness. Conventional systems
such as the Birmingham WireGauge "BWG" are used to indicate the wall thickness.
FITTING: The term fitting refers to a piece of pipe that can:
1. Join two pieces of pipe
ex. couplings and unions
2. Change pipeline directions
ex. elbows and tees
3. Change pipeline diameters
ex. reducers
4. Terminate a pipeline
ex. plugs and valves
5. Join two streams to form a third
ex. tees, wyes, and crosses
6. Control the flow
ex. valves
VALVES: A valve is also a fitting, but it has more important uses than simply to connect pipe. Valves are used either to control
the flow rate or to shut off the flow of fluid.
DESIGN OF A PIPING SYSTEM
The engineer should consider the following items when he is developing the design of a piping system.
1. Choice of material and sizes
2. Effects of temperature level and temperature changes.
a. insulation
b. thermal expansion
c. freezing
3. Flexibility of the system for physical and thermal shocks.
4. Adequate support and anchorage
5. Alteration in the system and the service.
6. Maintenance and inspection.
7. Ease of installation
8. Auxiliary and standby pumps and lines
9. Safety
a. Design factors
b. Relief valves and flare systems
1. A piston moves inside a cylinder with a velocity of 5 m/sec. The 150 mm diameter piston is centrally
located within the 150.2 mm inside diameter cylinder. The film of oil separating the piston from the
cylinder has an absolute viscosity of 0.40 Pa-sec. Assuming a linear velocity profile, find the:
a. Shear stress in the oil ( 20KPa)
b. Force required to maintained the given motion. (942 N)
2. An airplane barometer reads 762 mm Hg at sea level and 736 mm Hg at some unknown elevation. If the
temperature at sea level is 15 C, what is the elevation assuming:
a. air at constant density
b. air under isothermal condition
c. air temperature decreases linearly with elevation at the rate of 6.5 K per
kilometer.
3. The bottom of a river is 12 m below the water surface. Underneath which is a silt having a specific
gravity of 1.75 and a thickness t. The pressure at the bottom of silt is 450 KPa. Determine the thickness
of the silt.
4. A building in Makati is 84.5 m high above the street level. The required static pressure of the water line
at the top of the building is 2.5 kg/cm2. What must be the pressure in KPa in the main water located 4.75
m below the street level.
5. A cylindrical tank 2 m diameter, 3 m high is full of oil. If the specific gravity of oil is 0.9, what is the
mass of oil in the tank?
6
8. An iceberg in the ocean floats with 15% of its volume above the
surface. What is its specific gravity relative to ocean water?
a) 0.78
b) 0.97
c) 0.67
d) 0.85
9. A storage tank contains oil with a specific gravity of 0.88 and depth of 20 m. What is the hydrostatic
pressure at the bottom of the tank in kg/cm2. (1.76 kg/cm2)
10. The reading on a pressure gage is 1.65 MPa, and the local barometer reading is 94 KPa. Calculate the
absolute pressure that is being measured in kg/cm2. (17.78 kg/cm2)
11. An object weighs 25.95 N when submerged in kerosene (S = 0.81) and weighs 26.6 N when submerged in gasoline (S = 0.68).
Determine the specific weight of the object. (58.86 KN/m3)
12. A pressure gage at elevation 8 m on a side of a tank containing a liquid reads 57.4 KPa. Another gage at
elevation 5 m reads 80 KPa. Determine the density of the liquid. ( 767.6 kg/m3)
13. A mercury barometer at the ground floor of SEARS TOWER in Chicago reads 735 mm Hg. At the
same time another barometer at the top of the building reads 590 mmHg. Assuming air density to be
constant at 1.22 kg/m3, what is the approximate height of the building. (1617 m)
14. A block of ice 30.5 cm in height with specific gravity of 0.917 floats in fresh water. What must be its surface area to support an
82 kg man if the ice is to be submerged so that its top is coincident with the water surface. (A = 3.24 m2)
Sample Problems
2. A block of ice 305mm in height with a specific gravity of 0.917 floats in fresh water. What must be its
surface area to support an 82 kg man if the ice is to be submerged so that its top is coincident with the
water surface? (3.25 m2)
3. Ice in an iceberg has a specific gravity of 0.922. Determine the percent of expose volume of the iceberg
if it floats in sea water with specific gravity of 1.03. (10.485
4. A block of wood has a vertical projection of 15.24 cm when placed in water and 10.2 cm when placed in
alcohol. If the specific gravity of alcohol is 0.82,find the specific gravity of wood.
a) 0.702
b) 0.601
c) 0.568
d) 0.807
SAMPLE PROBLEMS
1. An iceberg in the ocean floats with 15% of its volume above the surface. What is its specific gravity
relative to ocean water?
a) 0.78
b) 0.97
c) 0.67
d) 0.85
2. A block of material of an unknown volume is submerged in water and weighs 500 N. The same block
weighs 650 N when weigh in air. Determine the volume of the material in m 3.
a) 0.0218
b) 0.0812
c) 0.063
d) 0.0153
3. Two spheres each 1.2 m in diameter weighs 4 KN and 12 KN, respectively. They are connected with a
short rope and placed in water. What portion of the lighter sphere protrudes from the water.
a) 0.179 cu.m.
b) 0.378 cu.m. c) 0.970 cu.m.
d) 0.365 cu.m
4. A block of ice 30.5 cm in height with specific gravity of 0.917 floats in fresh water. What must be its
surface area to support an 82 kg man if the ice is to be submerged so that its top is coincident with the
water surface. (A = 3.24 m2)
5. An iceberg of density of 922 kg/m3 floats in sea water with a density of 1024 kg/m 3. What percent of the
volume of the iceberg is submerged below the seawater surface.
6. A block of wood has a vertical projection of 15.24 cm when placed in water and 10.2 cm when placed in
alcohol. If the specific gravity of alcohol is 0.82,find the specific gravity of wood.
a) 0.702
b) 0.601
c) 0.568
d) 0.807
7.A spherical buoy 2 m in diameter floats half submerge in a liquid with S = 1.5. What is the weight of the lead anchor weighing
7000 kg/m3 will completely submerged the buoy in the liquid. (3998.4 kg)
8. A vertical tube 3 m long with open end closed is inserted vertically with the open end own into a tank of
water until the open end is submerged to depth of 1 m. Neglecting vapor pressure , how far will the water
level in the in the tube be below the level in the tank. (answer: 0,787 m)
9. A building in Makati is 84.5 m high above the street level. The required static pressure of the water line
at the top of the building is 2.5 kg/cm2. What must be the pressure in KPa in the main water located 4.75
m below the street level. (1120.8 KPa)
11. The crest gate shown consist of a cylindrical surface of which AB is the base, supported
by a structural frame hinged at C. The length of the gate perpendicular to the paper is 10 m. Compute
the amount and location of the horizontal and vertical components of the total force on AB. (x = 8.59 m
from C.
wsB
60
A
C
2. An iceberg having a specific gravity of 0.92 floats in salt water having a specific gravity of 1.03. If the
volume of the ice above the water surface is 1000 m3, what is the total volume in m3 of the iceberg.
(9369 m3)
13. A rectangular caisson open at the top having length of 15 m, width of6 m and depth of 7 m is to be
made foundation of a bridge in a pier. If the weight of the foundation is 75 tons, Find:
a) How deep it will sink in water (0.756 m)
b) How much more weight is needed so that the base of the foundation will reach the bottom, the
height of water being 6 m. (471,972.75 kg)
14. A block of ice 305mm in height with a specific gravity of 0.917 floats in fresh water. What must be its
surface area to support an 82 kg man if the ice is to be submerged so that its top is coincident with the
water surface? (3.25 m2)
15. Ice in an iceberg has a specific gravity of 0.922. Determine the percent of expose volume of the iceberg
if it floats in sea water with specific gravity of 1.03. (10.485%)
1. A barrel contains a 0.120 m layer of oil floating on water that is 0.250 m deep. The density of the oil is
600 kg/m3 and water at 1000 kg/m3. a) What is the gage pressure at the oil-water interface? b) What is
the gage pressure at the bottom of the barrel? ( P1 = 0.706 KPa; P2 = 3.16 KPa)
2. A spherical tank is full of water that has a mass of 10 000 kg. If the outside diameter of the tank is 2722
mm, how thick is the wall of the tank? (25 mm)
1. A mercury barometer at the ground floor of SEARS TOWER in Chicago reads 735 mm Hg(98 KPa). At
the same time another barometer at the top of the building reads 590 mmHg(78.7 KPa). Assuming air
density to be constant at 1.22 kg/m3, what is the approximate height of the building. (1617 m)
2. An airplane barometer reads 762 mm Hg(101.6 KPa) at sea level and 736 mm Hg(98.12 KPa) at some
unknown elevation. If the temperature at sea level is 15 C(288K), what is the elevation assuming:
a. air at constant density (288.6 m)
b. air under isothermal condition (293.6547 m)
c. air temperature decreases linearly with elevation at the rate of 6.5 K per kilometer. (292.7 m)
----------------------------------------------------------------
PROPERTIES AND VARIATION OF PRESSURE
1. A building in Makati is 84.5 m high above the street level. The required static pressure of the water line at the top of the
building is 2.5 kg/cm2. What must be the pressure in KPa in the main water located 4.75 m below the street level.
2. A cylindrical tank 2 m diameter, 3 m high is full of oil. If the specific gravity of oil is 0.9, what is the mass of oil in the tank?
3. An open tank contains 5 m of water covered with 2 m of oil ( = 8 KN/m3). Find the pressure at the interface and at the bottom
of the tank.
4. A skin diver wants to determine the pressure exerted by the water on her body after a descent of 35 m to a sunken
ship. The specific gravity of seawater is 1.02 times that of pure water. Determine the pressure in KPa.
5. The water pressure in the mains of a city at a particular location is 400 KPa gage. Determine the pressure if this maincan serve
water to neighborhoods that are 50 m above this location.
P1 = 400 KPagae (pressure at the main)
P2 = pressure at 50 m elevation
P2  P1   γ(h2  h1 )
h1  0; h2  50 m
P2  P1 - γ(h2 - h1 )
P2  400  (9.81(50  0)
P2  90.5 KPa gage
6. In the figure shown, determine the pressure at point A in KPa, if x = 1.15 m and y = 0.51 m.
A
x
Water
S=1
y
Mercury
S = 13.6
7. A pressure gage at elevation 8 m on the side of a tank containing a liquid reads 58 KPa. Another gage at elevation 5 m reads 80
KPa. Compute the specific weight and density of the liquid.
8. The suction gage of a pump reads 600 mm Hg vacuum and the discharge gage reads 270 KPa. Determine the pressure difference
between the discharge and suction gage in psi.
9. An object weighs 25.95 N when submerged in kerosene (S = 0.81) and weighs 26.6 N when submerged in
gasoline(S = 0.68). Determine the specific weight of the object.
10. The reading on a pressure gage is 1.65 MPa, and the local barometer reading is 94 KPa. Calculate the
absolute pressure that is being measured in kg/cm2.
11. Two spheres each 1.2 m in diameter weighs 4 KN and 12 KN, respectively. They are connected with a
short rope and placed in water. What portion of the lighter sphere protrudes from the water?
12. In the figure shown if the pressure at B = 200 KPa, find the pressure at A.
Liquid A
SA = 0.85
h = 0.4 m
2m
A
Manometer
Liquid (Hg)
S = 13.6
Liquid B
SB = 1.25
3m
13. A vertical, cylindrical water storage tank of a certain housing project is 3 m in diameter and 5 m high. The tank is
vented to the atmosphere and its sides are held in position by means of two steel hoops, one at the top and one
B stands to a depth of 4.5 m.
at the bottom. What is the tensile force in each hoop if water
ME 413H
FINAL EXAM
The velocity of water in a 10 cm diameter pipe is 3 m/sec. At the end of the pipe is a nozzle whose velocity coefficient is 0.98. If
the pressure in the pipe is 55 Pascal, what is the velocity in the jet? What is the diameter of the jet? What is the rate of discharge?
What is the head loss?
A centrifugal pump draws water from a well at the rate of 142 L/sec of water through a 203 mm ID suction line and a 152 mm ID
discharge line. The suction gauge located on the pump centerline reads 254 mm Hg vacuum, while the discharge gauge is 6 m
above the pump centerline. If the power input to the water is 75 KW, find the reading of the discharge gauge in KPa.
During a flow of 0.405 m3/sec the gauge pressure is 68.6 KPa in the horizontal 30 cm supply line of a water turbine and –41.4
KPa at a 45 cm section of the draft tube 1.5 m below. Assuming an efficiency of 85 %, compute the power output of the turbine.
A 15 KW suction pump draws water from a suction line whose diameter is 200 mm and discharges through a line whose diameter
is 150 mm. The velocity in 150 mm line is 3.6 m/sec. If the pressure at point A in the suction line is 34.5 KPa below the
atmosphere where A is 1.8 m below that of B on the 150 mm line, Determine the maximum elevation above B to which water can
be raised assuming a head loss of 3 m due to friction.
A 60 mm fire hose discharges a 30 mm jet. If the head lost in the nozzle is 2 m, what gauge pressure must be maintained at the
base of a nozzle to throw a stream to a vertical height of 30 m, neglecting air resistance.
(295.5 KPa)
1. A pelton type turbine was installed 30 m below the head gate of the penstock. The head loss
due to friction is 15% of the given elevation. The length of the penstock is 80 m and the
coefficient of friction is 0.00093. Determine
a) the diameter of the penstock in mm. (421.6 mm)
b) the power output in KW (781.234 KW)
2. What power in KW can be developed by the impulse turbine shown if the turbine efficiency is
85%. Assume that the resistance coefficient f of the penstock is 0.015 and the head loss in
the nozzle itself is negligible. What will be the speed of the wheel , assuming ideal conditions
where VJET = 2VBUCKET and what torque will be exerted on the turbine shaft.
1670 m elevation
1 m diameter; L = 6 km
3m
DJET = 18 cm
1000 m elevation
3. A hydroelectric plant has a 20 MW generator with an efficiency of 96%. The generator is
directly coupled to a vertical Francis type hydraulic turbine having an efficiency of 80%. the
total gross head of the turbine is 150 m while the loss of head due to friction at the
penstock up to the turbine inlet flange is 4% of gross head. the runaway speed is not to
exceed 750 RPM, determine:
a) Brake horsepower rating of the turbine (27 927 hp)
b) the flow of water through the turbine in cfs (650.53 cfs)
c) check if the specific speed falls under that of Francis type turbine. (Ns = 29)
d) the rated speed of the turbine (N = 400 RPM)
4. A Francis turbine is installed with a vertical draft tube. The pressure gauge located at the
penstock leading to the turbine casing reads 372.6 KPa and velocity of water at inlet is 6
m/sec. The discharge is 2.5 m3/sec. The hydraulic efficiency is 85%, and the overall
efficiency is 82%. The top of the draft tube is 1.5 m below the centerline of the spiral
casing, while the tailrace level is 2.5 m from the top of the draft tube. There is no velocity of
whirl at the top or bottom of the draft tube and leakage losses are negligible. Calculate the
following:
a) the net effective head in meters (43.817 m)
b) the brake power in kw. (881.2 kw)
c) the plant output for a generator efficiency of 92%. (810.7 kw)
d) the mechanical efficiency (96.550
5. A hydroelectric power plant using a Francis type turbine has the following data:
Headwater elevation - 190 m
Tailwater elevation - 50 m
Head loss due to friction - 3.5% of gross head
Turbine discharge at full gate opening - 6 m3/sec
Specific speed of turbine - 32 RPM
Turbine efficiency at rated capacity - 90%
Turbine is to be direct connected to a 60 hertz a-c generator
Determine:
a) the brake power in kw (7156.8 kw)
b) the number of generator poles (12 poles)
c) the electrical power output of the generator if the efficiency is 94% (6727.4 kw)
d) the torque developed in N-m (102 925.31 N-m)
e) the approximate length of the penstock, if the diameter is 1.5 m and friction factor f
is 0.018. (693 m)
6. The flow of a river is 21.25 m3/sec and the head on the site is 30.5 m. It is proposed to
developed the maximum capacity at the site with the installation of two turbines, one of
which is twice the capacity of the other. The efficiency of both units is assumed to be 85%.
Determine:
a) Rotative speed of each unit in rpm if the specific of both is 70 rpm.
b) Brake power of each unit in kw.
c) Number of poles of the generator for 60 cycle current
7. The flow of a river is 21 m3/sec and the head at the power site is 24.4 m (80 ft). It is
proposed to developed this site with an installation of 3-turbines, 2 similar units and another
of half their size, all having the same efficiency of 85%. Find the rotative speed of all these
units. (348 rpm; 492 rpm)
Ns at 80 ft head = 70 rpm
8. A hydroelectric power plant discharging water at the rate of 0.75 m3/sec and entering the
turbine at 0.35 m/sec with a pressure of 275 KPag has a runner of 55 cm internal diameter.
Speed is 514 rpm at 260 BHP. The casing is 2 m above the tailwater level. Calculate:
a) the net effective head in m (30.039 m)
b) the peripheral coefficient (0.61)
c) the efficiency (87.8%)
9. A Mindanao province where a mini-hydroelectric plant is to be constructed has an average
annual rainfall of 139 cm. The catchment area is 206 km 2 with an available head of 23 m. Only
82% of the rainfall can be collected and 75% of the impounded water is available for power.
Hydraulic friction loss is 6%, turbine efficiency is 78% and generator efficiency is 93%.
Determine the average kw power that could be generated for continuous operation.
POWER = 1.39(206)(1000)2(0.82)(0.75)(9.81)(23)(1-.06)(0.78)(0.93)
8760(3600)
= 859 kw
10. A Mindanao province where a hydroelectric power station is to be consrtructed has an
average annual rainfall of 1.93 m. The catchment area 260 km 2 wwith an available head of 32
m. Only 82% of the rainfall can be collected and 5/8 of the impounded water is available for
power. The load factor of the station is 80%, the penstock efficiency is 94%, turbine
efficiency 78% and the generator efficiency is 88%.
a) Calculate the average power that could be generated
b) Calculate the installed capacity of the station
c) Assuming no standby unit, give the possible number and size of the units
Average power = load factor(peak power)
In a test to determine the discharge coefficient of a 50 mm by 12.5 mm venture meter the total weight of water passing through the
meter in 5 minutes was 340 kg. A mercury – water differential gage connected to the inlet and throat of the meter shown an
average mercury difference during that time of 360 mm. Determine the meter coefficient.
ME 10E/ygm
4. The horizontal orifice in the figure is 75 mm in diameter with Cc =
0.63, Cv = 0.98. When H = 3 m, neglecting air resistance, compute the
height to which the jet will rise above the plane of the orifice. What
will be the diameter of the jet 1.5 m above the vena contracta?
(h = 2.88 m; d = 72 mm)
H
h
7. A vertical draft tube is installed on a Francis turbine and the total
head to the center of the spiral casing at the inlet is 38 m and
velocity of water at the inlet is 5 m/sec. The discharge is 2.1
m3/sec. The hydraulic efficiency is 87% and overall efficiency is 84%.
The velocities at the inlet and exit of the draft tube are 5 m/sec and
1.5 m/sec, respectively. The top of the draft tube is 1 m below the
centerline of the spiral casing while the tailrace (water) level is 3
m. from the top of the draft tube. There is no velocity of whirl at
either top or bottom of the draft tube and leakage losses are
negligible. What is the power output of the turbine? (747 KW)
8. Two spheres each 1.2 m in diameter weighs 4 KN and 12 KN,
respectively. They are connected with a short rope and placed in
water. What portion of the lighter sphere protrudes from the water.
a) 0.179 cu.m.
b) 0.378 cu.m. c) 0.970 cu.m. d) 0.365 cu.m
9. A block of material of an unknown volume is submerged in water and
weighs 500 N. The same block weighs 650 N when weigh in air. Determine
the volume of the material in m3.
a) 0.0218 b) 0.0812 c) 0.063 d) 0.0153
10. During a flow of 0.405 m3/sec the gauge pressure is 68.6 KPa in the
horizontal 30 cm supply line of a water turbine and –41.4 KPa at a 45
cm section of the draft tube 1.5 m below. Assuming an efficiency of
85 %, compute the power output of the turbine.
a) 56 KW
b) 47.5 KW c) 65 KW
d) 74.5 KW
12. A 15 KW suction pump draws water from a suction line whose diameter
is 200 mm and discharges through a line whose diameter is 150 mm. The
velocity in 150 mm line is 3.6 m/sec. If the pressure at point A in
the suction line is 34.5 KPa below the atmosphere where A is 1.8 m
below that of B on the 150 mm line, Determine the maximum elevation
above B to which water can be raised assuming a head loss of 3 m due
to friction. Answer: 7.97 m
13. A jet of water from a nozzle discharging into the air has a diameter
of 150 mm and a mean velocity of 36 m/sec. Compute the jet power in
KW. (415 KW)
14. A 60 mm fire hose discharges a 30 mm jet. If the head lost in the
nozzle is 2 m, what gauge pressure must be maintained at the base of
a nozzle to throw a stream to a vertical height of 30 m, neglecting
air resistance. (295.5 KPa)
15. A free waterfalls is 50 m high. It Discharges at a constant rate of
1.5 m3/sec. A mini hydro electric power plant is to be constructed
below the water fall. If the turbine efficiency is 770% and generator
efficiency is 90%, calculate:
a) the turbine power in KW
b) the KW output of the generator
c) the annual energy available in KW-hr if the discharge rate is
16. A 20 m3 tank can be filled with water by a motor driven pump. pump
efficiency is 90% and motor efficiency is 80%. If the total head is
20 m, Find the required power of the motor, If the electric energy
cost is P0.40 per KW-hr, find the cost of the filling.
17. A power nozzle throws a jet of water that is 50 mm in diameter. The
diameter of the base of the nozzle and of the approach pipe is150
mm. If the power of the nozzle jet is 42 HP and the pressure head at
the base of the nozzle is 54 m, compute the head lost in the nozzle.
18. A jet of liquid is directly vertically upward. At A its diameter is
75 mm and its velocity is 9 m/sec. Neglecting air friction, determine
its diameter at appoint 3 m above A.
B
3m
A
19. A fire pump delivers water through a 150 mm main to a hydrant to
which is connected a 75 mm hose, terminating in a 25 mm nozzle. The
nozzle is 1.5 m above the hydrant and 10 m above the pump. Assuming
frictional losses of 3 m from the pump to the hydrant, 2 m in the
hydrant and, and 12 m from the hydrant to the base of the nozzle, and
a loss in the nozzle of 6% of the velocity head in the jet, to what
vertical height can the jet be thrown if the gage pressure at the
pump is 550KPa.
constant throughout the year.
20. A 60 mm fire hose discharges water through a nozzle having a jet
diameter of 25 mm. The lost head in the nozzle is 4% of the velocity
head in the jet. If the gage pressure at the base of the nozzle is
400 KPa (a) compute the discharge (b) what is the maximum horizontal
range to which the stream can be thrown if the nozzle makes an angle
of 45 degrees, neglect air resistance.
21. Water discharges through an orifice in the side of a large tank as
shown in the figure. The orifice is circular in cross section and 50
mm in diameter. The jet is the same diameter as the orifice. The
liquid is water , and the surface elevation is maintained at a height
"h" of 4 m above the center of the jet. Compute the discharge
a) neglecting loss of head.
b) considering the loss of head to be 10% of "h"
22. Water issues from a circular orifice under a head of 12 m. The
diameter of the orifice is 10 cm. If the discharge is found to be 75
L/sec, what is the coefficient of discharge? If the diameter at the
vena cotracta is measured to be 8 cm, what is the coefficient of
contraction and what is the coefficient of velocity.
23. A jet discharges from an orifice in a vertical plane under a head of
3.65 m. The diameter of the orifice is 3.75 cm and the measured
discharge is 6 L/sec. The coordinates of the centerline of the jet
are 3.46 m horizontally from the vena contracta and 0.9 m below the
center of the orifice. Find the coefficient of discharge, velocity
and contraction.
24. The velocity of water in a 10 cm diameter pipe is 3 m/sec. At the end
of the pipe is a nozzle whose velocity coefficient is 0.98. If the
pressure in the pipe is 55 Pa, what is the velocity in the jet? What
is the diameter of the jet? What is the rate of discharge? What is
the head loss?
25. A jet of water 7.6 cm in diameter discharges through a nozzle whose
velocity coefficient is 0.96. If the pressure in the pipe is 82.7 KPa
and the pipe diameter is 20 cm and if it is assumed that there is no
conraction of the jet, what is the velocity at the tip of the nozzle?
What is the rate of discharge?
26. The nozzle in the figure throws a stream of water vertically upward
such that the power available in the jet at point 2 is 2.55 KW. If
the pressure at the base of the nozzle, point 1 is 145 Pa, Find
a) the theoretical height to which the jet will rise
b) the coefficient of velocity
c) the head loss between points 1 and 2
d) the theoretical diameter of the jet at a point 6 m above point 2.
6m
2
diameter of orifice = 4 cm
30 cm
1
27. The diameters of the suction and discharge pipes of a pump are 15cm
and 10 cm, respectively. The discharge pressure is read by a gage at
a point 1.5 m above the centerline of the pump, and the suction
pressure is read by a gage 0.61 m below the centerline. If the
pressure gage reads 138 KPa and the suction gage reads a vacuum of
254 mm Hg when gasoline (s = 0.75) is pumped at the rate of 35L/sec,
Find the power delivered to the fluid.
28. An iceberg in the ocean floats with 15% of its volume above the
surface. What is its specific gravity relative to ocean water
a) 0.78 b) 0.97 c) 0.67
d) 0.85
29. The mechanical engineer of an industrial plant wishes to install a
pump to lift 13L/sec of water at 21C from a sump to a tank on a
tower. The water is to be delivered into an storage tank at 120 KPa.
The tank is 20 m above the sump and the pump is 2 m above the water
level in the sump. The suction pipe is 4" sch. 40 steel pipe. 10 m
long and will contain 2-90 elbows and a foot valve. The discharge
pipe to the tank is 2" sch. 40 steel pipe; 120 m long and contains 590, 1-Check valve and one gate valve. Recommend the motor power
requiredto drive the pump if p = 75
ES 423
1. A jet discharges from an orifice in a vertical plane under a head of
3.65 m. The diameter of the orifice is 3.75 cm and the measured
discharge is 6 L/sec. The coordinates of the centerline of the jet
are 3.46 m horizontally from the vena contracta and 0.9 m below the
center of the orifice. Find the coefficient of discharge, velocity
and contraction.
2. The velocity of water in a 10 cm diameter pipe is 3 m/sec. At the end
of the pipe is a nozzle whose velocity coefficient is 0.98. If the
pressure in the pipe is 55 Pa, what is the velocity in the jet? What
is the diameter of the jet? What is the rate of discharge? What is
the head loss?
------------------------------------------------------------------------ES 423
1. A jet discharges from an orifice in a vertical plane under a head of
3.65 m. The diameter of the orifice is 3.75 cm and the measured
discharge is 6 L/sec. The coordinates of the centerline of the jet
are 3.46 m horizontally from the vena contracta and 0.9 m below the
center of the orifice. Find the coefficient of discharge, velocity
and contraction.
2. The velocity of water in a 10 cm diameter pipe is 3 m/sec. At the end
of the pipe is a nozzle whose velocity coefficient is 0.98. If the
pressure in the pipe is 55 Pa, what is the velocity in the jet? What
is the diameter of the jet? What is the rate of discharge? What is
the head loss?
------------------------------------------------------------------------ES 423
1. A jet discharges from an orifice in a vertical plane under a head of
3.65 m. The diameter of the orifice is 3.75 cm and the measured
discharge is 6 L/sec. The coordinates of the centerline of the jet
are 3.46 m horizontally from the vena contracta and 0.9 m below the
center of the orifice. Find the coefficient of discharge, velocity
and contraction.
2. The velocity of water in a 10 cm diameter pipe is 3 m/sec. At the end
of the pipe is a nozzle whose velocity coefficient is 0.98. If the
pressure in the pipe is 55 Pa, what is the velocity in the jet? What
is the diameter of the jet? What is the rate of discharge? What is
the head loss?
------------------------------------------------------------------------ES 423
1. A jet discharges from an orifice in a vertical plane under a head of
3.65 m. The diameter of the orifice is 3.75 cm and the measured
discharge is 6 L/sec. The coordinates of the centerline of the jet
are 3.46 m horizontally from the vena contracta and 0.9 m below the
center of the orifice. Find the coefficient of discharge, velocity
and contraction.
2. The velocity of water in a 10 cm diameter pipe is 3 m/sec. At the end
of the pipe is a nozzle whose velocity coefficient is 0.98. If the
pressure in the pipe is 55 Pa, what is the velocity in the jet? What
is the diameter of the jet? What is the rate of discharge? What is
the head loss?
-------------------------------------------------------------------------
HEAT TRANSFER
SCHOOL YEAR 2018 - 2019
MODES OF HEAT TRANSFER
1. Conduction: It is the transfer of heat from one part of a body to another part of the same body, or from one body to another in
physical contact with it, without appreciable displacement of the particles of the body.
2. Convection: It is the transfer of heat from one point to another point within a fluid, gas, vapor, or liquid by
the mixing of some portion of the fluid with another.
A. Natural or Free Convection: the movement of the fluid is entirely caused by differences in density resulting from
temperature differences.
B. Forced Convection: the motion of the fluid is accomplished by mechanical means, such as a fan
or a blower.
3. Radiation: It is the transfer of heat from one body to another, not in contact with it, by means of "wave
motion" through space.
CONDUCTION:
Qdx   kAdT


Q dx   kA dT
Q(Δx )   kAΔT
(Δx )  L
From FOURIER'S LAW:
Q   kA
dT
dx
k
Note:
Negative sign is used from Fourier's Equation because temperature decreases in the direction of heat flow.
From Ohms Law;
Current heat flow for an electric conductor
V
R
I  current in Amperes
V - Energy in Volts (V)
I
R - electrical resistance in Ohms (Ω)
For conductive heat flow
- Δt - ΔT

R
R
L C
K
R
or
kA Watts
Watts
Q
where:
(-T) and(-t) - temperature potential in K or C
R - thermal resistance in C/W or K/W
Q - conductive heat flow in Watts
k - thermal conductivity in W/m-C or W/m-K
A - surface area in m2
L - thickness in m
THERMAL CIRCUIT DIAGRAM
R
1
2
CONVECTION:
From Newton' s Law of Cooling
Q  hA ( t1  t 2 ) Watts
( t1  t 2 )  ( t 2  t1 )  Δt  ΔT



1
1
1
1
hA
hA
hA
hA
 Δt  ΔT
Q

R
R
Q
where
Q - convective heat flow, Watts
h - convective coefficien t (surface conductanc e),
A - Fuid - Surface area, m
W
m - C
2
or
W
2
m -K
2
 Δt &  ΔT  temperatue potential, C or K
R - resistance ,
C K
or
W
W
RADIATION:
A
Surface or
(Radiator)
T1
T1 > T2
Q
T2
T1 – Surface temperature, K
T2 – surrounding temperature, K
From Stefan-Boltzmann Law: The radiant heat transfer of a blackbody is directly proportional to the product of the surface area A
and its absolute temperature to the fourth power.
Q  AT 4
Q  δAT 4
From Fig., the radiant heat flow Q from the surface (or body) to the surrounding (or to other surface or body) is equal to
4
4
Q  δA T1  T2  Watts


where
W
 Stefan - Boltzmann Constant
m2 - K 4
T1 - absolute surface temperatu re, K
δ  5.678 x 10-8 ,
T2 - absolute surroundin g or other surface temperatu re, K
Black Body - a hypothetical body that absorbs the entire radiation incident upon it.
Gray Body - are actual bodies or surfaces that absorbs a portion of the black body radiation, because they are not perfect radiators
and absorbers.
EMISSIVITY
Emissivity - is the ratio of the actual body (or surface) radiation at temperature T to the black body (or black surface) radiation at the
same temperature T.
Actual Surface radiation @ T

Black Surface radiation @ T
Therefore the rate of heat radiated by an Actual Boby (or Actual Surface or Actual Radiator )
is equal to
4
4
Q  δA T1  T2  Watts


where T1  T2
Relating Radiation to Convection
4
4
Q  δA T1  T2   h r AT1 - T2 


4
4
 δ T1  T2   h r T1 - T2 


4
4
2
2
2
2
2
2
 δ T1  T2   δ T1  T2  T1  T2   δ T1  T2 T1  T2 T1 - T2 







hr 
T1 - T2 
T1 - T2 
T1 - T2 
2
2
h r  δ T1  T2 T1  T2   radiation coefficien t


Combined Radiation and Convection heat transfer
Actual surface exposed to the surrounding air involves convection and radiation simultaneously, the total heat transfer Q
Q  h c A( t1  t 2 )
h combined  h Radiation  h Convection
(Answer: Q = 8400 Watts)
Answer: 8125 W
t1 = 300C
h = 250 W/m2-K
QConvection
0.50 x 0.25 m
t2 = 40C
Answer: 726 W; 547 W
t1 = 150C
Q
t2 = 25C
A = 0.5 m2
0.8
QEmitted  AT 4

Q  A T1  T2
4
4

A plate 30 cm long and 10 cm wide with a thickness 0f 12 mm is made of stainless steel (k = 16 W/m-K), the top of the is exposed
to an air stream of temperature 20C. In an experiment, the plate is heated by an electric heater (also 30 cm by 10 cm) positioned on
the underside of the plate and the temperature of the plate adjacent to the heater is maintained at 100C. A voltmeter and ammeter
are connected to the heater and these read 200 Volts and o.25 Ampere, respectively. Assuming that the plate is perfectly insulated
on all sides except the top surface, what is the convective heat transfer coefficient h if the emittance of the top surface  = 1.
A  0.30(0.10)  0.03 m 2
Tair  20  273  293 K
T1  100  273  373 K
Q  200(0.25)  50 W
0.03(373  T1 )
0.012
16
T1  371.75 K
Q
Q  Qh  QR
4
4
Q R  1(5.67 x 10-8 )(0.03)(371.75  293 )
Q R  20 W
Q h  50  20  30 W
Q h  hA (371.75  293)
h  12.7
W
m 2 - C
A cold storage room has walls constructed of a 5.08 cm layer of corkboard insulation contained between double wooden walls, each
1.27 cm thick. Find the rate of heat loss in Watts if the wall surface temperature is -12C outside the room and 21C inside the room.
The thermal conductivity of corkboard is 0.04325 W/m-C and the thermal conductivity of the wood walls is 0.10726 W/m-C.
Q
(21  12)


0.0127 0.0508 0.0127
A


0.10726 0.4325 0.10726
Example 4
The wall of a house 7 m wide and 6 m high is made from o.3 thick brick with k = 0.6 W/m-K. The surface temperature on the inside
of the wall is 16C and that on the outside is 6C. Find the heat flux through the wall and the total heat loss through it.
A = 7 x 6 m2
L
1
k
2
Q
Example 5
Assume :  1
L
20 mm
Example 7
Assume :  1
Q  Qr  Qc

Qr  A T1  T2
4
4

Qc  hA (T1  T2 )


Q
4
4
  T1  T2  h(T1  T2 )
A
Q
4
4
 T1   T2  hT1  hT2
A
Q
4
4
 T1  T2  hT1  hT2
A
Q
4
4
T1  hT1  hT2  T2   0
A
Q
4
hT2  T2 
h
4
A 0
T1  T1 


4
T1  105,820,105.82T1 - 45,077,281,841.56  0
T2  20  273  293
T2  7,370,050,801
4
T   105820105.82T - 45077601410.935  0
2 2
s
by quadratic formula
Ts  426
2
Ts  426  20.64K
Example 8
Answer: 1233 KW/m2
Example 9
s
CONDUCTION THROUGH A COMPOSITE PLANE WALL
Thermal Circuit diagram:
At point 1 to 4
Q
(t1 - t 4 )
(t1 - t 4 )
(t1 - t 4 )


W
L
L1
L
ΣR
R1  R 2  R 3
 2  3
k 1A k 2 A k 3 A
R1 
L1 C
k 1A W
R2 
L 2 C
k 2A W
R3 
L 3 C
k 3A W
(t1 - t 4 )
A(t1 - t 4 )

W



L1 L 2 L 3 
1 L1 L 2 L 3

 





A  k1 k 2 k 3   k1 k 2 k 3 
(t1 - t 4 )
Q
W

A  L1 L 2 L 3  m 2




 k1 k 2 k 3 
Q
Δt
ΣR
 Δt
Q
L
Σ
kA
A (  Δt )
Q
L
Σ
k
Q (Δt )

L
A
Σ
k
where
Q
(-Δt)  (-ΔT)
OVERALL COEFFICIENT OF HEAT TRANSFER
On the basis of Overall Coefficien t of Heat Transfer
Q  UA(-Δt)
From
A(-Δt)
L
Σ 
k
therefore
Q
U
1
L
Σ 
k
W
W
or 2
m - C
m -K
2
where: U - overall coefficient of heat transfer in W/m2-C or W/m2-K
PARALLEL LAYERS OR COMBINED SERIES – PARALLEL LAYERS
HEAT TRANSFER FROM FLUID TO FLUID SEPARATED BY A COMPOSITE PLANE WALL
L2
L1
Fluid
(i)
L3
Fluid
(o)
t1
A
t2
ho
to
t3
hi
ti
k1
k2
k3
t4
Q
Thermal Circuit Diagram:
i
R2
R1
Ri
1
R3
3
2
4
Ro
o
Q
(t i  t o )
(t i  t o )
(t  t )
; Q
 i o
L
1
L
L
1
R i  R1  R 2  R 3  R o
ΣR
 1  2  3 
h i A k 1A k 2 A k 3 A h o A
(t i  t o )
A(t i  t o )
Q
; Q
 1 L1 L 2 L 3
1  1 L1 L 2 L 3 1 
1
 
    
   


A  h i k1 k 2 k 3 h o 
 h i k1 k 2 k 3 h o 
(t i  t o )
Q

;
A  1 L1 L 2 L3 1 
 


 
h
 i k1 k 2 k 3 h o 
L
1
1
L
L
Ri 
; Ro 
; R1  1 ; R 2  2 ; R 3  3
hiA
h oA
k1A
k 2A
k 3A
(Δt)
(Δt)
Q

;
ΣR  R t
 1
L  R


 
 hA kA 
Q
(Δt)

1 L
A
   
h k
Q
(  Δt)
(  Δt)
; Q
 1
 1  1 L
L 
    


 
hA
kA


A h k 
A(  Δt)
Q
1 L
   
h k
(  Δt)
Q

A
1 L
   
h k 
(  Δt)
Q
R
1 L
ΣR  R t     
h k
Q
OVERALL COEFFICIENT OF HEAT TRANSFER COMPOSITE PLANE WALL
A (  Δt )
1 L
Σ  
h k
Q  UA (Δt )
Q
U
1
1 L
Σ  
h k
SAMPLE PROBLEMS
Problem No. 1
A composite plane wall is made up of an external thickness of brickwork 10 cm thick, followed by a layer of fiber glass 8 cm thick
and an insulating board 2.5 cm thick. The thermal conductivity for the three are as follows;
Brickwork - 1.5 W/m-C
Fiber Glass -0.04 W/m-C
Insulating Board - 0.06 W/m-C
The fluid film coefficient of the inside wall is 3.1 Wm 2-C while that of the outside wall is 2.5 Wm2-C. Determine the
overall coefficient of heat transfer and the heat loss through such a wall 3.5 m high and 15 m long. Take the internal ambient
temperature as 15C and the external temperature as 28C.
L1
L3
L2
1
2
3
o
ho
to
4
k1
k2
i
hi
ti
k3
Q
Problem No. 2
A furnace wall is constructed with 25 cm of firebrick, k = 1.36 W/m-K, 8 cm of insulating brick, k = 0.26 W/m-K, and 15 cm of
building brick, k = 0.69 W/m-K. The inside surface temperature is 600C and the outside air temperature is 30 C. The convective
coefficient for the outside air is 3.6 W/m2-K. Determine
a. The total heat transferred W/m2
b. The interface temperature
c. The overall coefficient of heat transfer in W/m2-K
Problem No. 3
A composite wall is made up of an external thickness of brickwork 11 cm thick, inside which is a layer of fibre glass 7.5 cm thick.
The fibre glass is faced internally by an insulating board 2.5 cm thick. The coefficient of thermal conductivity for the three are as
follows;
Brickwork - 1.5 W/m-C
Fiber Glass -0.04 W/m-C
Insulating Board - 0.06 W/m-C
The surface transfer coefficient of the inside wall is 3.1 Wm 2-C while that of the outside wall is 2.5 Wm2-C. Determine the overall
coefficient of heat transfer and the heat loss through such a wall 6 m high and 10 m long. Take the internal ambient temperature as
10C and the external temperature as 27C.
HEAT TRANSFER
QUIZ NO. 1
NAME ________________________________________
RATING ______________________
1. A wall 3.00 m by 2.44 m is made up of a thickness of 10.0 cm of brick (k = 0.649 W/m-C), 10.0 cm of glass wool (k =
0.0414 W/m-C), 1.25 cm of plaster (k = 0.469 W/m-C), and 0.640 cm of oak wood paneling (k = 0.147 W/m-C). If the
inside temperature of the wall is ti =18 C and the outside temperature t0 =−7C, Determine
a. The total Resistance of the wall in C/Watt
b. The overall coefficient of heat transfer in W/m2-C
c. The total conductive heat flow in Watts
d. The interface temperature in C
2.
A composite plane wall consisting of two layers of materials, 38 mm steel and 51 mm aluminum, separates a hot gas at t i
= 93 C; h i = 11.4 W/m2-C, from a cold gas at to = 27C; ho = 28.4 W/m2-C. If the hot fluid is on the aluminum side,
Find
a. the transmittance U in W/m2-C
b. the total resistance R in C -m2/W
c. the interface temperature at the junction of two metals in C
d. the heat through 9.3 m2 of the surface under steady-state conditions.
QUIZ NO. 2
I
An insulated steam pipe run through a dark warehouse room. The pipe outside diameter is 60 mm and its surface temperature and
emissivity are 165C and 0.95, respectively. The warehouse and air is kept at 5C. If the coefficient of heat transfer by natural
convection from the outside surface to the air is 11.4 W/m2-C and the pipe surface may be treated as a gray body, what is the rate
of heat loss from the surface per meter of pipe length.
A flat furnace wall is constructed of a 11 cm layer of sil-o-cel brick with a thermal conductivity of 0.14 W/m-C, backed by a 23 cm
layer of common brick of conductivity 1.4 W/m-C. The temperature of the inner face of the wall is 760 C, and that of the outer
face is 77C.
a. What is the heat loss through the wall
b. What is the temperature of the interface between the refractory brick and common brick
c. Supposing that the contact between the two brick layers is poor and that a contact resistance of 0.95 C/W
is present, what would be the heat loss.
1
L1
L2
k1
k2
Q ( t1  t 3 )

A L1  L 2
k1 k 2
Q t1  t 2

L1
A
k1
2
3
Q
 Q  L 
t 2  t1    1 
 A  k1 
R1 
L1
k1A
R2 
L2
k2A
C
W
Q
t1  t 3

A R1  R x  R2
R x  0.95
A furnace wall consists of 20 mm of refractory fireclay brick, 100 mm of sil-ocel brick, and 6 mm of steel plate. The fire side of the
refractory is at 1150C and the outside of the steel is at 30C. An accurate heat balance over the furnace shows the heat loss from
the wall to be 300 W/m2. It is known that there may be thin layers of air between the layers of brick and steel. To how many mm of
sil-o-cel are these air layers equivalent.
k = 1.52 W/m-C (fireclay)
k = 0.138 W/m-C (sil-ocel)
k = 4.5 W/m-C (steel)
A carpenter builds an outer house wall with a layer of wood (k = 0.080 W/m-K) 2 cm thick on the outside and a layer of Styrofoam
(k = 0.01 w/m-K) insulation 3.5 cm thick as the inside wall surface. What is the temperature at the plane where the wood meets the
Styrofoam? Interior temperature is 19C; exterior temperature is -10C.
L1
L2
t3
t1
k1
k2
Q
A
An industrial freezer is designed to operate with an internal air temperature of -20C when the external air temperature is 25C and
the internal and external heat transfer coefficient are 12 W/m2-K and 8 W/m2-K, respectively. The walls of the freezer are composite
construction, comprising of an inner layer of plastic (k = 1 W/m-K, and thickness 3 mm), and an outer layer of stainless stell (k = 16
W/m-K, and thickness of 1 mm). Sandwiched between these two layers is a layer of insulation material with k = 0.07 W/m-K. Find
the width of the insulation that is required to reduce the convective heat loss to 15 W/m2.
CONDUCTION THROUGH CYLINDRICAL COORDINATES
L
k
t1  t 2
1
2
Q
r1
r2
L - length of the cylinder perpendicular to the paper
From Fourier' s Equation :
dT
, since heat flows in a radial direction, the heat tran sfer Q is;
dx
dT
Q  - kA
, A  2 πrL
dx
dT
dr
Q  - k(2πrL)
,
; Q  - 2 πkLdT
dr
r
2 dr
2
r
Q
 - 2πkL dT ; Qln 2  2 πkL(T2  T1 )
r1
1 r
1
Q  - kA


r
Qln 2  2πkL(T1  T2 )
r1
Q
2 πkL(T1  T2 )
r
ln 2
r1
;
Q
(T1  T2 )
r
ln 2
r1
2 πkL
where T1 - T2  t1 - t 2
(t1  t 2 )
;
ln r2
r1
2 πkL
r
ln 2
r1
R
2πkL
where R - resistance
Q
Q
Δt
Δt

R
ln ro
ri
2πkL
o – refers to outside
i – refers to inside
Q
(t1  t 2 )
R
where: Q in Watts
k in W/m-C or W/m-K
L in meters
r in m
-t in C or K
R in W/C or W/K
CONDUCTION THROUGH A COMPOSITE CURVED WALL
Let
L - length of the cylinder perpendicular to the paper
at 1 to 2
at 3 to 4 :
Q
Q
t1  t 2 
ln r2
t 3  t 4 
r
ln 4
r3
2 πk 3L
r1
2 πk1L
at 2 to 3
at 1 to 4 :
Q
Q
t 2  t 3 
ln
r3
r2
2 πk 2 L
Q
(t1  t 4 )
r
r
ln 4
ln
ln 3
r3
r1
r2


2 πk1L 2πk 2 L 2 πk 3L
r2
- Δt 
 ln ro 

ri 


 2 πkL 


(t1  t 4 )
Q
r
r
r
ln 4
ln 2
ln 3
r3
r1
r2


2 πk1L 2πk 2 L 2 πk 3L
(t i  t 0 )
r
r
 ln r2
ln 4 
ln 3
r3 
r1
r2
 1 




k2
k3 
 2 πL   k1



2 πL(t i  t 0 )
Q
r
r
 ln r2
ln 4 
ln 3

r3 
r1
r2

 k  k
k3 
2
 1



Q
2 π(t i  t 0 )

r
r
L  ln r2
ln 4 
ln 3

r3 
r1
r2

 k  k

k
2
3 
 1


Q
2 π(-Δt )

L
 ln ro 

ri 


k 




Q

HEAT TRANSFER FROM FLUID TO FLUID SEPARATED BY A COMPOSITE CURVED WALL
L - length of the cylinder perpendicular to the paper
at i to 1:
at 1 to 2:
Q = Aihi(ti - t1)
Q
(t i  t1 )
1
hiAi
A i  2r1L
Q
at 3 to 4:
at 2 to 3:
(t 1 - t 2 )
r
ln 2
r1
2k 1 L
at 4 to 0:
(t - t )
Q 3 4
r
ln 4
r3
2k 3 L
Q  A 0 h 0 (t 4  t 0 )
Q
(t 4  t 0 )
1
A0h 0
A 0  2r4 L
at i to 0:
(t i  t 0 )
r
r
r
ln 4
ln 2
ln 3
r3
r1
r2
1
1




A i h i 2k 1 L 2k 2 L 2k 3 L A 0 h 0
(t i  t 0 )
Q
r
r
r
ln 4
ln 2
ln 3
r3
r1
r2
1
1




2r1 Lh i 2k 1 L 2k 2 L 2k 3 L 2r4 Lh 0
Q
Q
(t 2 - t 3 )
r
ln 3
r2
2k 2 L
Q
(t i  t 0 )
r
r


ln 4
ln r2
ln 3
r3
r1
r2
1  1
1 




2L  r1 h i
k1
k2
k3
r4 h 0 


Q
2L( t i  t 0 )
r
r
r


ln 4
ln 2
ln 3
 1
r3
r1
r2
1 






k1
k2
k3
r4 h 0 
 r1h i


2( t i  t 0 )
Q

r
r
r
L 

ln 4
ln 2
ln 3
 1
r3
r1
r2
1 






k1
k2
k3
r4 h 0 
 r1h i


General Equation:
(t )
Q
r
ln o

r
ln o
;Q
ri
(t )
1

; where R  
R
Ah
2kL
ri
1

Ah
2kL
o - refers to outside
i - refers to inside
Q
(t )
r 

ln o 

ri
1
1
  

2L  rh
k 


2L( t )
Q
ro 

ln
 1
ri 
  

k 
 rh


Q
2( t )

r 
L 
ln o 
 1
ri
  

k 
 rh


Cylindrical coordinates sample Problems
1. Water through a cast steel pipe (k = 50 W/m-K) with an outer diameter of 104 mm and 2 mm wall thickness.
a. Calculate the heat loss by convection and conduction per meter length of insulated pipe (OD = 104 mm ; ID =
100 mm when the water temperature is 15C, the outside air temperature is -10C, the water side heat transfer
coefficient is 30,000 W/m2-K and the outside heat transfer coefficient is 20 W/m2-K.
b. Calculate the corresponding heat loss when the pipe is lagged with insulation having an outer diameter 0f 300 mm,
and thermal conductivity of k = 0.05 W/m-K.
ho
to
ti
hi
i
1
t1
2
t2
Q
r1
r2
2. Saturated steam at 500 K flows in a 20 cm ID; 21 cm OD pipe. The pipe is covered with 8 cm of insulation with a thermal
conductivity of 0.10 w/m-C. The pipes conductivity is 52 W/m-C. the ambient temperature is 300 K. The unit convective
coefficient are hi = 18 000 W/m2-C and ho = 12 W/m2-C. Determine the heat loss from 4 m of pipe. Calculate the overall
coefficient of heat transfer base on the outside area.
k1
ti
hi
1 2
3
k2
t0
ho
Q
r1
r2
r3
3. Steam initially saturated at 2.05 MPa, passes through a 10.10 cm (ID =10 cm; OD =12 cm) standard steel pipe (k = 50 W/m – K)
for a total distance of 152 m. The steam line is insulated with a 5.08 cm thickness of 85% magnesia (k = 0.069 W/m-K). For an
ambient temperature of 22C, what is the quality of the steam which arises at its destination if the mass flow rate is 0.125 kg
steam/sec.
hsteam = 550 W/m2 – K
hstill air = 9.36 W/m2 – K
r1
0.0500
k1
50
tsat at 2.05 MPa = 213.61C
r2
0.06
k2
0.069
r3
0.1108
hi
550
r2/r1
1.2
ho
9.36
L
r3/r2
1.85
m
0.125
ln(r2/r1)
0.182
hs1
2796.3
ln(r3/r2)
0.613
L
152
k2
1/hir1
0.036
2PiL
955.044
Q
k1
1/hor3
0.964
ti
213.61
t  22C
ln(r2/r1)/k1
0.0036
to
22
1
2 3
ti  213.61C
ln(r3/r2)/k2
8.8896
(ti -t0)
191.61
Q
18495.94
Watts
hs2
2648.3
Enthalpy
r1
hf
913.75
r2
hg
2796.3
r3
hfg
1882.55
x
92.14
Quality
0
Q
ms
ms
hs1
hs2
m s (h s1 - h s2 )
Watts
1000
2πLti - to 
Q
r
r2
ln
ln 3
1
1
r1
r2



h i r1
k1
k2
h o r3
Q
hs 2  h f  x s 2 (h g  h f )
x s 2  92.14%
4. A steel pipe 100 mm bore and 7 mm wall thickness, carrying steam at 260C is insulated with 40 mm of a moulded hightemperature diatomaceous earth covering, this covering in turn insulated with 60 mm of asbestos felt. The atmospheric temperature
is 15C. The heat transfer coefficients for the inside and outside surfaces are 550 and 15 W/m2-K, respectively and the thermal
conductivities of steel, diatomaceous earth and asbestos felt are 50, 0.09, and 0.07 W/m-K respectively. Calculate:
a. the rate of heat loss by the steam per unit length of pipe
b. the temperature of the outside surface
c. the overall coefficient of heat transfer based on outside surface
k1
k2
k3
1 2 3
ti
hi
4
to, ho
Q
r1
r2
r3
r4
Q

L
Q

L
Q
2πt i - t o 
r
r
ln 4
ln
ln 3
r3
1
r1
r2
1




h i r1
k1
k2
k3
h o r4
r2
2πt i - t 4 
r
r
ln 4
ln
ln 3
r3
1
r1
r2



h i r1
k1
k2
k3
r2
2πLt i - t o 
 U o A o t i - t o   U o 2πr4 L t i - t o 
r
r
ln 4
ln
ln 3
r3
1
r1
r2
1




h i r1
k1
k2
k3
h o r4
U o  0.481
r2
W
m - C
2
r1
0.050
ti
260
ln(r2/r1)
0.1310
r2
0.057
t0
15
ln(r3/r2)
0.5317
r3
0.097
hi
550
ln(r4/r3)
0.4815
r4
0.157
ho
15
ln(r2/r1)/k1
0.0026
k1
50
(ti - to)
245
ln(r3/r2)/k2
5.9073
k2
0.09
2*PI()
6.28
ln(r4/r3)k3
6.8791
k3
0.07
Ai
0.314
1/hir1
0.0364
L
1
Ao
0.986
1/hor4
0.4246
Q
116.18
SUM
13.2500
Uo
0.481
SUM2
12.8254
t4
22.85
5. Steam at 280C flows in a stainless steel pipe (k = 15 W/m-K) whose inner and outer diameters are 5 cm and 5.5 cm, respectively.
The pipe is covered with 3 cm thick glass wool insulation (k = 0.038 W/m-K). Heat is lost to the surroundings at 5C by natural
convection and radiation, with a combined natural convection and radiation heat transfer coefficient of 22 W/m 2-K. Taking the heat
transfer coefficient inside the pipe is 80 W/m2-K. determine the rate of heat loss from the steam per unit length of pipe. Also
determine the temperature drops across the pipe shell and the insulation.
k1
k2
1 2 3
ti
hi
to, ho
Q
r1
r2
r3
6. Hot water (hi = 6895 W/m2 -C) flows in a 2.5 cm ID; 2.66 cm OD smooth copper pipe (k = 400 W/m -C). The pipe is horizontal
in still air ( ho = 3.56 W/m2 -C )and covered with a 1-cm layer of polystyrene foam insulation (k =0.038 W/m-C) . For a 65°C
water temperature and 20°C air temperature, Calculate
a.the heat loss per unit length of pipe
b.the interface temperature in C
c.Overall coefficient of heat transfer U
OVERALL COEFFICIENT OF HEAT TRANSFER
Q = UA(-t)
A. For a composite Plane Wall
Q
A(Δt)
Δt

Rt
1 L
  
h k
L 
 1



 hA kA 
1
1 L
  
h k

Rt  
U

B. For a Composite Curved Wall
Q
- Δt 

r
ln o
1
ri
 
hA 2kL
r
ln o
1
ri
Rt  

hA 2kL
r
ln o
1
ri
UA  

hA 2kL
 t
Rt
UA = UiAi = UoAo
where:
Ui - overall coefficient of heat transfer based on inside surface, W/m2-C or W/m2-K
Uo - overall coefficient of heat transfer based on outside surface, W/m2-C or W/m2-K
HEAT EXCHANGER OR HEAT TRANSFER EQUIPMENT
TYPES OF HEAT EXCHANGERS
1. DIRECT CONTACT TYPE: The same fluid at two different states is mixed.
2. SHELL AND TUBE TYPE: One fluid flows inside the tubes and the other one on the outside.
mh, hh
mc, hc
m, h
By Mass Balance
m h  mc  m
By Energy Balance (ΔKE and ΔPE are negligible )
m h h h  m c h c  mh
m h h h  m c h c  (m h  m c )h
m h h h  mch c  m h h  mch
m h (h h  h )  m c (h  h c )
Q  m h (h h  h )  m c (h  h c )  Total Heat Transfer
Example:
SHELL AND TUBE TYPE
Shell and Tube Type Heat Exchanger
By energy balance:
Heat Rejected by hot fluid = Heat Absorbed by cold fluid
Qh  Qc
Q h  m h C ph ( t h1  t h 2 )  Hot Fluid
Q c  m c C pc ( t c 2  t c1 )  Cold Fluid
OVERALL COEFFICIENT OF HEAT TRANSFER
Q  Q h  Qc
Q  UA (LMTD )  Based on Log Mean Temperatur e difference
Q  UA (AMTD )  Based on Arithmetic Mean Temperatur e difference
Where:
A – total heat transfer area, m2
LMTD - Log Mean Temperature Difference, C or K
AMTD - Arithmetic Mean Temperature Difference, C or K
U – Overall Coefficient of Heat Transfer,
LMTD 
AMTD 
KW
KW
or 2
2
m  C
m K
θ 2  θ1
θ
ln 2
θ1
θ 2  θ1
2
where:  - terminal temperature difference
TERMINAL TEMPERATURE DIFFERENCE
If the design of the heat exchanger is more complex, the LMTD is modified by a correction factor F.
Q  UAF(LMTD )
Where :
F - correction factor
TUBE DESIGN
Vflow  Area x velocit y x Number of tubes
m3
sec
A  πDLn m 2
D - outside diameter of tube ( for U based on outside surface area)
A  πdLn
d - inside tube diameter (For U based on inside surface area)
n - total number of tubes
L - efdfective length of tube, meters
L t - actual length of tube
L t  L  2t
t - thickness of tube sheet, meters
F - correction factor F can be determine from charts or tables
For Multiple tube passes, divide the length by number of passes and multiply
the number of tubes by the number of passes.
Example No. 1
Exhaust gases flowing through a tubular heat exchanger at the rate of 0.3 kg/sec are cooled from 400 to 120C by water initially at
10C. The specific heat capacities of exhaust gases and water may be taken as 1.13 and 4.19 KJ/kg-K respectively, and the overall
heat transfer coefficient from gases to water is 140 W/m 2-K. Calculate the surface area required when the cooling water flow is 0.4
kg/sec;
a. for parallel flow (4.01 m2)
b. for counter flow (3.37 m2)
Given
m c  0.4 kg/sec ; C pc  4.19 KJ/kg - C
m h  0.3 kg/sec ; C ph  1.13 KJ/kg - C
t h1  400C ; t h2  120C
t c1  10C
Qh  Qc
LMTD 
θ 2 - θ1 390  53.3

 169.2C
θ
390
ln 2
ln
θ1
53.3
For Counter Flow
θ 2  t h 1 - t c2  400  66.7  333.3C
0.3(1.13)(400  120)  0.4(4.19)( t c 2  10)
θ1  t h 2  t c1  120  10  110C
t c 2  66.7C
LMTD 
For Parallel Flow
θ 2  t h1 - t c1  400 - 10  390C
θ1  t h2 - t c2  120 - 66.7  53.3C
Q  0.3(1.13)( 400  120)  94.92 KW
Q  94920 Watts
94,920
 4.01 m 2
140(169.2)
94,920
A
 3 .4 m 2
140(201.43)
A
333.3  110
 201.43C
333.3
ln
110
Q  UA (LMTD )
A
Q
U(LMTD )
SAMPLE PROBLEMS
1. A composite wall is made up of an external thickness of brickwork 11 cm thick, inside which
is a layer of fiber glass 7.5 cm thick. The fiber glass is faced internally by an insulating
board 2.5 cm thick. The coefficient of thermal conductivity for the three are as follows;
Brickwork - 1.5 W/m-C
Fiber Glass -0.04 W/m-C
Insulating Board - 0.06 W/m-C
The surface transfer coefficient of the inside wall is 3.1 Wm2-C while that of the outside
wall is 2.5 Wm2-C. Determine the overall coefficient of heat transfer and the heat loss
through such a wall 6 m high and 10 m long. Take the internal ambient temperature as 10C
and the external temperature as 27C.
2. A furnace is constructed with 20 cm of firebrick, k = 1.36 W/m-K, 10 cm of insulating
brick, k = 0.26 W/m-K, and 20 cm of building brick, k = 0.69 W/m-K. The inside surface
temperature is 650C and the outside air temperature is 32 C. The heat loss from the
furnace wall is 0.56 W/m2. Determine
a. the unit convective coefficient for the air W/m 2-K (3.545)
b. the temperature at 25 cm from the outside surface in C. (460 C)
3. A furnace wall consists of 20 mm of refractory fireclay brick, 100 mm of sil-ocel brick, and
6 mm of steel plate. The fire side of the refractory is at 1150C and the outside of the
steel is at 30C. An accurate heat balance over the furnace shows the heat loss from the
wall to be 300 W/m2. It is known that there may be thin layers of air between the layers of
brick and steel. to how many mm of sil-o-cel are these air layers equivalent. (400 mm)
k = 1.52 W/m-C (fireclay)
k = 0.138 W/m-C (sil-ocel)
k = 4.5 W/m-C (steel)
4. A composite plane wall consisting of two layers of materials, 38 mm steel and 51 mm
aluminum, separates a hot gas at ti = 93 C; hi = 11.4 W/m2-C, from a cold gas at to = 27C;
ho = 28.4 W/m2-C. If the hot fluid is on the aluminum side, Find
a. the transmittance U in W/m2-C (8)
b. the total resistance R in C -m2/W (0.124)
c. the interface temperature at the junction of two metals in C (45)
d. the heat through 9.3 m2 of the surface under steady-state conditions. (4937)
5. A heat exchanger is to be designed to the following specifications:
Hot gas temperature = 1145C
Cold gas temperature = 45C
Unit surface conductance on the hot side = 230 W/m2-K
Unit surface conductance on the cold side = 290 W/m2-K
Thermal conductivity of the metal wall = 115 W/m-K
Find the maximum thickness of metal wall between the hot gas and the cold gas, so that the
maximum temperature of the wall does not exceed 545C. (20.115 mm)
6. A composite furnace wall is made up of a 300 mm lining of magnesite refractory brick, a
130 mm thickness of 85% magnesia and a steel 2.54 mm thick. Flue gas temperature is
1205 C and the boiler is at 27 C. Gas side film coefficient is 85 W/m 2-C and the air side
is 23 W/m2-C.
Determine:
a. the thermal current Q/A in W/m2
b. the interface temperatures in C
c. effect on thermal current and the inside refractory wall temperature if the
magnesia insulation were doubled.
k for magnesite = 296 W/m-C
k for 85% magnesia = 0.0692 W/m-C
k for steel = 43.3 W/m-C
7. Determine the thermal conductivity of a wood that is used in a 1.5 m2 test panel, 25 mm
thick, if during a 4 hours test period there are conducted 190 KJ through the panel with a
temperature differential of 6C between the surfaces. Express answer in W/m-C. (0.0244)
8. The walls of a cold storage are composed of an insulating material (k = 0.065 W/m-C)
10.16 cm thick held between two layers of concrete ( k = 1.04 W/m-C) each 10.16 cm thick.
The film coefficients are 22.7 W/m2-C on the outside and 11.4 W/m2-C on the inside. Cold
storage temperature is -7C and the ambient temperature is 32C. Determine the heat
transmitted in KW through an area of 56 m2. (1.2)
9. A 12 in thick furnace wall with a dimensions of 5 m x 2 m has temperature difference of
60C. The wall has a thermal conductivity of 0.75 BTU/hr-ft-F. Calculate the heat
transmitted across the wall. (2554 W)
10. A 15 cm thick wall has a thermal conductivity of 5 W/m-K. If the inside and outside surface
temperature of the wall are 200C and 30C, respectively. Determine the heat transmitted.
( 5.67 W/m-K)
11. Two walls of cold storage plant are composed on an insulating material (k = 0.07 W/m-K),
100 mm thick at the outer layer and material (k = 0.97 W/m-K), 15 cm thick at the inner
layer. If the surface temperature of the cold side is 30C and hot side is 250C, find the
heat transmitted in W/m2. (138)
12. An insulated steam pipe runs through a dark warehouse room.The pipe outside diameter is
60 mm and its surface temperature and emissivity are 165 C and 0.95, respectively. The
warehouse and air is kept at 5C. If the coefficient of heat transfer by natural convection
from the outside surface to the air is 11.4 w/m 2-C and the pipe surface maybe treated as
a gray body, what is the rate of heat loss from the surface per meter of pipe length.
( 657 W/m)
13. Saturated steam at 500 K flows in a 20 cm ID; 21 cm OD pipe. The pipe is covered with 8
cm of insulation with a thermal conductivity of 0.10 w/m-C. The pipes conductivity is 52
W/m-C. the ambient temperature is 300 K. The unit convective coefficient are h i = 18 000
W/m2-C and ho = 12 W/m2-C. Determine the heat loss from 4 m of pipe. Calculate the
overall coefficient of heat transfer base on the outside area. (822 W; 0.9 W/m 2-C)
14. A tube 60 mm OD is lagged with a 50 mm layer of asbestos for which the conductivity is
0.21 W/m-C, followed with a 40 mm layer of cork with a conductivity of 0.05 W/m-C. If
the temperature of the outer surface of the pipe is 150C and the temperature of the
outer surface of the cork is 30C, Calculate the heat loss in Watts per meter length of
pipe.(59)
15. An economizer receives hot gas (Cp = 1.13o6 KJ/kg-K) and water in the ratio 1.5 kg gas/kg
water. The gas enters at 455C and leaves at 180C; the water enters at 50C. Find the
exit temperature of the water and the LMTD:
a. for parallel flow ( 125.425C)
b. for counter flow (200.81C)
Assume no energy losses external to the system.
16. A flat furnace wall is constructed of a 11 cm layer of sil-o-cel brick with a thermal
conductivity of 0.14 W/m-C, backed by a 23 cm layer of common brick of conductivity 1.4
W/m-C. The temperature of the inner face of the wall is 760 C, and that of the outer
face is 77C.
a. What is the heat loss through the wall
b. What is the temperature of the interface between the refractory brick and
common brick
c. Supposing that the contact between the two brick layers is poor and that a
contact resistance of 0.95 C/W is present, what would be the heat loss.
17. Steam initially saturated at 2.05 MPa, passes through a 10.10 cm standard steel pipe for a
total distance of 152 m. the steam line is insulated with a 5.08 cm thickness of 85%
magnesia. For an ambient temperature of 22C, what is the quality of the steam which
arises at its destination if the mass flow rate is 0.125 kg steam/sec. (x = 93%)
At 2.05 MPa: hf = 914.52 KJ/kg; hfg = 1885.5 KJ/kg; hg = 2800 KJ/kg
18. The hot combustion gases of a furnace are separated from the ambient air and its
surrounding which are at 25C, by a brick wall 15 cm thick.The brick has a thermal
conductivity of 1.2 W/m-C and a surface emissivity of 0.8. Under steady-state
conditions and outer surface temperature of 100C is measured. Free convection heat
transfer to the air adjoining this surface is characterized by a convection coefficient of
20 W/m2-C. what is the brick inner surface temperature inC. (352.5C)
19. A counter flow heat exchanger is designed to heat fuel oil from 28 C to 90 C, while the
heating fluid enters at 138C and leaves at 105C. The fuel has a specific gravity of
21API, a specific heat of 2.1 KJ/kg-K and enters the heat at the rate of 3 000 L/hr.
Determine the required heating surface area in m2 if the overall coefficient of heat
transfer is 465.2 W/m2-K. ((3.5 m2)
20. Brine enters a circulating brine cooler at the rate of 5.7 m3/hr at -10C and leaves at
-16C.Specific heat of the brine is 1.072 KJ/kg-C and the specific gravity is 1.10. The
refrigerant evaporates at -25 C. What is the required heat transfer area if U = 454
W/m2-C. (2.1 m2)
21. A steel pipe 100 mm bore and 7 mm wall thickness, carrying steam at 260C is insulated
with 40 mm of a moulded high-temperature diatomaceous earth covering, this covering in
turn insulated with 60 mm of asbestos felt. The atmospheric temperature is 15C. The
heat transfer coefficients for the inside and outside surfaces are 550 and 15 W/m 2-K,
respectively and the thermal conductivities of steel, diatomaceous earth and asbestos felt
are 50, 0.09, and 0.07 W/m-K respectively. Calculate:
a. the rate of heat loss by the steam per unit length of pipe (116 W)
b. the temperature of the outside surface (22.8C)
22. A furnace wall consists of 125 mm wide refractory brick and a25 mm wide insulating
firebrick separated by an air gap. The outside wall is covered with a 12 mm thickness of
plaster. The inner surface of the wall is at 1100C and the room temperature is 25C. The
heat transfer coefficient from the outside wall surface to the air in the room is
17 W/m2-K and the resistance to heat flow of the air gap is 0.16 K/W. The thermal
conductivities of refractory brick, insulating brick, and plaster are 1.6, 0.3, and
0.14 W/m-K, respectively. Calculate:
a. the rate of heat loss per unit area of wall surface (1344 W)
b. the temperature of each interface throughout the wall
(995, 780, 220, 104 C)
c. the temperature at the outside surface of the wall ( 104.1C)
23. Exhaust gases flowing through a tubular heat exchanger at the rate of 0.3 kg/sec are
cooled from 400 to 120C by water initially at 10C. The specific heat capacities of
exhaust gases and water may be taken as 1.13 and 4.19 KJ/kg-K respectively, and the
overall heat transfer coefficient from gases to water is 140 W/m2-K. Calculate the
surface area required when the cooling water flow is 0.4 kg/sec;
a. for parallel flow (4.01 m2)
b. for counter flow (3.37 m2)
24. A properly designed steam heated tubular pre heater is heating 5.7 kg/sec of air from 21C
to 77C when using steam at 0.14 MPa. It is proposed to double the rate of air flow
through the heater and yet heat the air from 21C to 77C; this is to be accomplished by
increasing the steam pressure. Calculate the new steam pressure required to meet the
change condition expressed in KPa. (661.5 KPa)
25. A 404.34 m2 heating surface counter-flow economizer is used in conjunction with a 72,000
kg/hr boiler. the inlet and outlet water temperature are 100C and 155C. The inlet and
outlet gas temperature are 340C and 192C. Find the overall coefficient in W/m2-C.
(85.6 w/M2-C)
26. A boiler tube for steam at 8.2 MPa (ts = 296.79C; h = 2574.82 KJ/kg) is 9 cm OD and 7 cm
ID. assume an internal film coefficient of 11,350 W/m 2-C and a thermal flow of 157,640
W/m2 based on the outside area, calculate
a) Outside tube temperature
b) Allowable boiler scale thickness if the metal is not to exceed 482C. Assume k for
steel = 43.26 W/m-C anf for scale k = 0.52 W/m-C.
27. An 8" steel pipeline (OD = 22 cm; ID = 19 cm) carries steam at 232C. An 85% magnesia
(k = 0.07 W/m-C) pipe covering is to be applied of such a thickness so as to limit the
surface temperature to 50C with room temperature of 16C. Assume inside and outside
coefficients of 1700 W/m2-C and 0.011 W/m2-C and k for steel = 41 W/m-C, calculate
the magnesia thickness in cm,
28.Calculate the energy transfer rate across 6 in wall of firebrick with a temperature difference across the wall
of 50C. The thermal conductivity of the firebrick is 0.65 BTU/hr-ft-F at the temperature interest.
BTU
W
 1.73 

hr - ft - F
m- C
k = 0.65(1.73) = 1.1245 W/m-C
L = 0.6" = 0.01524 m
Q k (  Δt )

 3689 .3 W/m 2
A
L
29. A carpenter builds an outer house wall with a layer of wood (k = 0.080 W/m-K) 2 cm thick on the outside
and a layer of styrofoam (k = 0.01 w/m-K) insulation 3.5 cm thick as the inside wall surface. What is the
temperature at the plane where the wood meets the Styrofoam? Interior temperature is 19C; exterior
temperature is -10C.
L1 L2
1
2
Q k1
3
k2
Q (t 1  t 3 ) (t 1  t 2 )


L1
A L1 L 2

k1 k 2
k1
t1 = 19C
t3 = -10C
k1 = 0.01
k2 = 0.08
L1 = 0.035 m
L2 = 0.02 m
t2 = - 8C
Exams
1. An 8" steel pipeline (OD = 22 cm; ID = 19 cm) carries steam at 232C. An 85% magnesia (k = 0.07 W/m-C) pipe
covering is to be applied of such a thickness so as to limit the surface temperature to 50C with room temperature of
16C. Assume inside and outside coefficients of 1700 W/m2-C and 0.011 W/m2-C and k for steel = 41 W/m-C,
calculate the magnesia thickness in cm,
2. The emissivity of tungsten is 0.35. A tungsten sphere with a radius of 1.5 cm is suspended within a large enclosure
whose walls are at 290K. What power input is required to maintain the sphere at a temperature of 3000K if heat
conduction along the supports is neglected? (Area of sphere = 4r2)
3. A copper cylinder is initially at 20C. At what temperature will be its volume be 0.150%larger than it is at 20C.
Coefficient of linear expansion  of copper is 1.7 x 10-5.
REFRIGERATION
By: ENGR. YURI G. MELLIZA
Refrigeration system is that area in mechanical engineering that deals with the different mechanism involved in maintaining a
temperature below that of the immediate surroundings. It is a process of moving heat from one location to another in controlled
conditions. The work of heat transport is traditionally driven by mechanical work, but can also be driven by heat, magnetism,
electricity, laser, or other means.
Major Uses of Refrigeration:
1. Air Conditioning
2. Food Preservation
3. Removing of heat from substances in chemical, petroleum and petrochemical plants
4. Special applications in the manufacturing and construction industries.
SECOND LAW OF THERMODYNAMICS
Whenever energy is transferred, the level of energy cannot be conserved and some energy must be permanently reduced to a lower
level.
When this is combined with the first law of thermodynamics, the law of energy conservation, the statement becomes:
Whenever energy is transferred, energy must be conserved, but the level of energy cannot be conserved and some energy must be
permanently reduced to a lower level.
KELVIN-PLANCK STATEMENT OF THE SECOND LAW:
No cyclic process is possible whose sole result is the flow of heat from a single heat reservoir and the performance of an
equivalent amount of work.
For a system undergoing a cycle: The net heat is equal to the network.
 dW   dQ
W Q
W - net work
Q - net heat
CARNOT CYCLE
(Nicolas Leonard Sadi Carnot 1796-1832)
A. Carnot Engine Cycle
Processes:
1 to 2 - Heat Addition (T = C)
2 to 3 - Expansion (S = C)
3 to 4 - Heat Rejection (T = C)
4 to 1 - Compression (S = C)
P
T
1
QA
T=C
S=C
4
1
2
4
3
2
S=C
T=C
3
QR
V
S
Heat Added (T = C)
Q A = TH ( S)  1
Heat Rejected (T = C)
QR = TL ( S)  2
S = S2 - S1 = S3 - S4
3
Net Work
W = Q = Q A - QR
4
W = (TH - TL )( S)  5
Thermal Efficiency
e
W
x 100%  6
QA
e
Q A  QR
x 100%  7
QA
 Q 
e  1  R x  100%  8
 QA 
Eq. 5 to and Eq.1 to Eq. 6
e
TH  TL
x 100%  9
TH
B. Carnot Refrigeration Cycle (Reversed Carnot Cycle)
Processes:
1 to 2 - Compression (S =C)
2 to 3 - Heat Rejection (T = C)
3 to 4 - Expansion (S = C)
4 to 1 - Heat Addition (T = C)
T
3
2
QR
4
1
QA
S
C. Carnot Heat Pump:
A heat pump uses the same components as the refrigerator but its purpose is to reject heat at high energy level.
Performance Factor
PF =
PF =
PF =
QR
W
 11
QR
QR - Q A
TH
TH - TL
 12
 13
PF = COP + 1  14
TON OF REFRIGERATION
It is the heat equivalent to the melting of 1 ton (2000 lb) of water ice at 0C into liquid at 0C in 24 hours.
2000(144)
BTU
 12,000
24
hr
KJ
TR  211
min
TR 
where 144 BTU/lb is the latent heat of fusion of ice
CARNOT REFRIGERATION SYSTEM
Processes
1 to 2 – Isentropic compression
2 to 3 – Isothermal Heat rejection
3 to 4 – Isentropic Expansion
4 to 1 – Isothermal Heat Addition
QR
3
S
COMPONENTS
1. Compressor
2. Condenser
3. Expander
4. Evaporator
Condenser
2
TH
QR
Expansion
Valve
1
R
Compressor
HEAT ADDED (T = C)
QA = TL (S)
S = S1 – S4
TL = Evaporator temperature
HEAT REJECTED
QR = TH (S)
S = S2 – S3
TH = Condenser temperature
NET WORK (compressor work)
Wc = QR - QA
COEFFICIENT OF PEFORMANCE
COP 
4
QA
Evaporator
TL
QA
QA
WC
HEAT PUMP
A heat pump uses the same components as the refrigerator but its main purpose is to reject heat at high thermal energy level. Its
performance is obtained on the basis of Performance Factor.
PERFORMANCE FACTOR
Performance Factor
PF =
PF =
PF =
QR
W
 11
QR
QR - Q A
TH
TH - TL
 12
 13
PF = COP + 1  14
SAMPLE PROBLEMS (CARNOT CYCLE)
1. A Carnot engine operating between 775 K and 305K produces 54 KJ of work. Determine the change of entropy during heat
addition. (S12= 0.115 KJ/K)
e
TH  TL 775  305

 60.6 %
TH
775
W  (TH - TL )S  54
S  0.115 KJ/K
2. A Carnot refrigerator operates between temperature limits of -5C and 30C. The power consumed is 4 KW and the heat
absorbed is 30 KJ/kg. Determine; (a) the COP (b) the refrigerant flow rate
TL = -5 + 273 = 268K ; TH = 30 + 273 = 303K; W = 4 KW
268
 7 .7
303  268
Q A  COP( W )  7.7( 4)  30.8 KW
COP 
Q A  m(30)
m  30.8/30  1.03 kg/sec
T
TH
TH
QR
2
3
QR
R
TL
1
4
W
QA
QA
S
TL
3. A Carnot refrigerator rejects 2500 KJ of heat at 353K while using 1100 KJ of work. Find
a) the cycle low temperature
b) the COP
c) the heat absorbed or refrigerating capacity
W  QR  Q A
1100  2500  Q A
Q A  1400 KJ
Q
TL
COP  A 
W
TH  TL
TL
1400

1100 353  TL
1100TL  494,200  1400TL
T
TH
TH
QR
2
3
QR
R
TL
1
4
W
QA
QA
S
TL
TL  197.7K
COP  1.2727
QUIZ NO 1
1. A Carnot engine operates between temperature reservoirs of 817C and 25C and rejects 25 KW to
the low temperature reservoir. The Carnot engine drives the compressor of an ideal vapor compression refrigerator, which
operates within pressure limits of 190 KPa and 1200 KPa. The refrigerant is R 22. Determine the COP and the refrigerant
flow rate.
For the R22 refrigerator
h1 =394.27 KJ/kg
h2 =440.99 KJ/kg
h3 = 279.91 KJ/kg
h4 = 279.91 KJ/kg
2. A heat pump is used to heat a house in the winter months. When the average outside temperature is
0C and the indoor temperature is 23C, the heat loss from the house is 20 KW. Determine the
minimum power required to heat the heat pump.
TL  0  273  273 K
TH  23  273  296 K
Q R  20 KW
Q
TH
296
PF  R 

 12.9
W TH  TL 296  272
20
W
 1.55 KW
12.9
3. A Carnot heat pump is being considered for home heating in a location where the outside temperature may as low as 35C. The expected COP for the heat pump is 1.50. To what temperature could this unit provide?
TL  35  273  238 K
COP  1.5
TL
COP 
 1.5
TH - TL
1.5TH  1.5( 238)  238
TH  397 K
t H  124C
VAPOR COMPRESSION REFRIGERATION SYSTEM
Processes
1 to 2 – Isentropic Compression
2 to 3 – Isobaric Heat Rejection
3 to 4 – Iso-enthalpic Expansion
4 to 1 – Isobaric Heat Addition
COMPONENTS
1. Compressor
2. Condenser
3. Expansion Valve
4. Evaporator
Ph and TS Diagram
T
P
2
tc
3
Pc
Pc
3
2
S=C
h=C
Pe
te
Pe
4
1
S
4
1
h
Schematic Diagram
COMPRESSOR WORK
a.
Compressor
W  m(h 2  h1 ) KW
where
m - mass flow rate in kg/sec
b. Condenser
Q R  m(h 2  h 3 ) KW
For air cooled condenser
Q R  m a C pa (ΔTa )
For water cooled condenser
Q R  m w C pw (ΔTw )
w - refers to water
a - refers to air
c.
Expansion Valve/Coil
h3  h 4
h 4  h f 4  x 4 (h g 4  h f 4 )
g  saturated vapor
f  saturated liquid
x - Quality
x
mv
mv

m
mL  mv
d. Evaporator
Q A  m(h1  h 4 ) KW
REFRIGERATING CAPACITY
Refrigerating Capacity is the amount of heat removed from an object to be cooled per unit time by a refrigerating machine.
Refrigerating capacity is measured in watts or in kilocalories per hour. Other common units include tons, or Tons of Refrigeration
(TR), which describe the amount of water at a given temperature that can be frozen in a given amount of time.
Q A  m(h1  h 4 ) KW
QA 
60m(h1  h 4 )
TR
211
COEFFICIENT OF PERFORMANCE (COP)
A constant that denotes the efficiency of a refrigerator, expressed as the amount of heat removed from a substance being refrigerated
divided by the amount of work necessary to remove the heat.
QA
W
h1  h 4
COP 
h 2  h1
COP 
PRODUCT LOAD
Cpb
Cpa
t1
m
tf
Q1
t2
m
tf
Q3
Q2
Q  Q1  Q 2  Q 3
Q1  mC pa ( t1  tf )
Q 2  mhF
Q 3  mC pb ( tf  t 2)

Q  mC pa ( t 1  t f )  mh F  mC pb ( t f  t 2 )

Q  m C pa ( t1  t f )  h F  C pb ( t f  t 2 )


where
m  mass , kg; kg / min
KJ
KJ
or
kg - C kg - K
KJ
KJ
C pb  sp.heat of product below freezing,
or
kg - C kg - K
h F  heat of fusion (latent heat of freezing) in KJ/kg
C pa  sp.heat of product above frezing,
Considerin g heat gains/loss es
Q  Q1  Q 2  Q 3 (1  Gain / loss )
Q  QA
Refrigeration Load Calculations
The heat load calculation is separated into the following main sources of heat for a given 24 hour period:
1. Transmission load
2. Air change load
3. Miscellaneous load
4. Product load
Transmission Load
Methods of determining the amount of heat flow through walls, floor and ceiling are well established. This heat gain is directly
proportional to the Temperature Difference (T.D.) between the two sides of the wall. The type and thickness of insulation used in
the wall construction, the outside area of the wall and the T.D. between the two sides of the wall are the three factors that establish
the wall load. Tables are provided to simplify the calculations. Some coolers for above freezing temperatures have been constructed
with only a floor slab (no floor insulation).
Air Change Load
Average Air Change- when the door to a refrigerated room is opened, warm outside air will enter the room. This air must be cooled
to the refrigerated room temperature, resulting in an appreciable source of heat gain. This load is sometimes called the infiltration
load. The probable number of air changes per day and the heat that must be removed from each cubic foot of the infiltrated air, are
given in tables based on experience. For heavy usage, the infiltration may be doubled or more.
Infiltration Through a Fixed Opening- As an alternate to the average air change method using Psychrometry, some formulas may
be used to calculate the infiltration resulting from natural ventilation through external door openings.
Miscellaneous Loads
Although most of the heat load in a refrigerated room or freezer is caused by wall heat leakage, air changes and product cooling or
freezing, there are three other heat sources that should not be overlooked prior to the selection of the refrigeration equipment. Since
the equipment has to maintain temperature under design conditions, these loads are generally averaged to a 24 hour period to provide
for capacity during these times.
Lights- typically storage requirements are 1 to 1-1/2 watt per square foot. Cutting or processing rooms can be double the wattage.
Each watt is multiplied by 3.42 BTU/watt to obtain a BTUH figure. This is then multiplied by 24 to obtain a daily figure.
Motors- smaller motors are usually less efficient and tend to generate more heat per horsepower as compared to larger motors. Also,
motors inside the refrigerated area will reject all of their heat losses. However, motors that are located outside but do the work inside,
like a conveyor, will reject less heat into the refrigerated space. If powered material handling equipment is used, such as forklift
trucks, this must be included under Motor Heat Loads. Generally only battery operated lift trucks are used in refrigerated rooms,
which represent a heat gain of 8,000 to 15,000 BTU/hr. or more over the period of operation. If motor or loading conditions are not
known, then calculate one motor horsepower for each 16,000 cubic foot box in a storage cooler and one HP for each 12,500 C.F. in
a storage freezer which allows for fan motors and some forklift operations. These figures can be higher in a heavily used area, i.e.
loading dock or distribution warehouse.
Occupancy- People working in the refrigerated storage area dissipate heat at a rate depending on the room temperature. Multiple
occupancies for short periods should be averaged over a 24 hour period. If occupancy load is not known, allow one person per 24
hour for each 25,000 cubic foot space.
Product Load
Whenever a product having a higher temperature is placed in a refrigerator or freezer room, the product will lose its heat until it
reaches the storage temperature. This heat load consists of three separate components:
Specific Heat- The amount of heat that must be removed from one pound of product to reduce the temperature of this pound by 1ºC.,
is called its specific heat. It has two values: one applies when the product is above freezing; the second is applicable after the product
has reached its freezing point.
Latent Heat- The amount of heat that must be removed from one kg of product to freeze this kg is called the latent heat of fusion.
Most products have a freezing point in the range of -3.3C. to -0.56C. If the exact temperature is unknown, it may be assumed to
be -2.2C.
There is a definite relationship between the latent heat of fusion and the water content of the product and its specific and latent heats.
Estimating specific and latent heats:
Sp. Ht. above freezing = 0.20 + (0.008 X % water)
Sp. Ht. below freezing = 0.20 + (0.008 X % water)
Latent Heat = 143.3 X % water
Respiration- Fresh fruits and vegetables are alive. Even in refrigerated storage they generate heat which is called the heat of
respiration. They continually undergo a change in which energy is released in the form of heat, which varies with the type and
temperature of the product. Tabulated values are usually in KJ/kg/24 hours, and are applied to the total weight of product being
stored and not just the daily turnover.
Pull down Time- When a product load is to be calculated at other than a 24 hour pull down, a correction factor must be multiplied
to the product load.
Safety Factor
When all four of the main sources of heat are calculated, a safety factor of 10% is normally added to the total
refrigeration load to allow for minor omissions and inaccuracies (additional safety or reserve may be available
from the compressor running time and average loading).
PRODUCT LOAD SAMPLE PROBLEMS
1. Compute the heat to be removed from 110 kg of lean beef if it were to be cooled from 20C to 4C, after which it is frozen and
cooled to -18C. Specific heat of beef above freezing is given as 3.23 KJ/kg-C, and below freezing is 1.68 KJ/kg-C. Freezing
point of beef is -2.2C, and latent heat of fusion is 233 KJ/kg.
Given: m = 110 kg Cpa = 3.23 KJ/kg-C
t1 = 20C
Cpb = 1.68 KJ/kg-C
t2 = 4C
hL = 233 KJ/kg
tf =-2.2C
t3 = -18C
Q = m[Cpa(t1 - t2) + Cpa(t2 - tf) + hL + Cpb(tf - t3)]
2.
Q = 36 438 KJ
A refrigeration storage is supplied with 10 000 kg of fish at a temperature of 2C. The fish has to be cooled to -10C for
preserving it for a long period without deterioration. The cooling takes place in 10 hours. The specific heat of fish is 0.7
KCal/kg-C above freezing point and 0.3 KCal/kg-C below freezing point which is -3C. The latent heat of freezing is 55.5
KCal/kg. Find the refrigerating capacity of the plant in tons of refrigeration. (24.35 tons)
3.
4.
A mass of ice at -4C is needed to cool 114 kg of vegetables in a bunker for 24 hours. The initial temperature of the vegetables
is assumed to be 29C. It also assumed that the average temperature in the bunker is 7C, within the 24 hours period. If the heat
gained per hour in the bunker is 30% of the heat removed to cool the vegetables from 29C to 7C, what would be the required
mass of ice? (29.33 kg)
Specific heat of ice = 1.9387 KJ/kg-C
Specific heat of vegetables = 3.35 KJ/kg-C
Specific heat of water = 4.2292 KJ/kg-C
Heat oof fusion of ice = 335 KJ/kg
4. Four hundred kilograms of poultry enter a chiller at 6C and are frozen and chilled to a final temperature of
-16C for storage in 12 hours. Compute the product load.
Cpa - 3.18 KJ/kg-C
Cpb - 1,55 KJ/kg-C
Latent heat - 246 KJ/kg
Freezing temperature - -3C
ANSWER: 2.73 KW
5. Ten thousand kilograms of dressed poultry are blast frozen on hand trucks each day (24 hours) in a freezing tunnel.The poultry is
pre-cooled to 7C before entering the freezer where it is frozen and its temperature lowered to -20C for storage. The hand trucks
carrying the poultry total 700 kg per day and have a specific heat of 0.47 KJ/kg-C. The specific heat of poultry above and below
freezing are 3.18 KJ/kg-C and 1.55 KJ/kg-C, respectively, the latent heat is 246 KJ/kg and the freezing temperature is -2.75C.
Determine the product load. (35.3 KW)
6. Twenty three hundred liters of partially frozen ice cream at -4Center a hardening room each day. Hardening is completed and
the temperature of the ice cream is lowered to -28C in 10 hours. The average density of ice cream is 0.69 kg/L, the average
latent heat per kg is 233 KJ. determine the product load. (10.86 KW)
7. A food freezing plant receives peas from the field at a temperature of 24C, chills, freezes and chills them further to -18C. What
refrigerating capacity in tons is required to perform this operation on 500 kg of peas per hour. (14.2 tons)
Freezing point of peas = -1.1C
Cpa = 3.31 KJ/kg-C
Cpb = 1.76 KJ/kg-C
Latent heat of freezing - 246.6 KJ/kg
8. An ice plant is freezing 3785.4 liters of water per hour from 18C into ice at -7C. Find the output of the
refrigerating machine in tons, assuming 20% additional load due to losses in heat leakage, etc. Specific heat
of water is 4.187 KJ/kg-C, specific heat of ice is 2.1 KJ/kg-C and latent heat of fusion is equal to 334.9
KJ/kg. ( Q = 152.27 tons)
9. A fish weighing 11 000 kg with a temperature of 20C is brought to a cold storage and which shall be cooled to -10C in 11 hours.
Find the required plant refrigerating capacity in tons of refrigeration if the specific heat of peas is 0.7 KCal/kg-C above freezing
and 0.3 KCal/kg-C below freezing point which is -3C. The latent heat of freezing is 55.5 KJ/kg. (24.4 tons)
Wet Compression
P
Pc
3
Pe
4
2
1
h
Sub – Cooling of
Refrigerant
T
tc
t3
P
2
PC
2
Pc
3
3
Pe
Pe
te
4
4
1
1
S
h
Superheating of
Suction Vapor
P
T
2
3
tc
PC
t1
te
1
Pe
4
Pc
3
Pe
4
2
1
h
S
Sub Cooling and
Superheating
T
P
Sub – Cooling &
Superheating
2
tc
t3
t1
te
PC
3
1
4
Pe
S
Pc
3
Pe
4
2
1
h
ACTUAL VAPOR COMPRESSION CYCLE
As the refrigerant flows through the system there will be pressure drop in the condenser, evaporator and piping. Heat losses or
gains will occur depending on the temperature difference between the refrigerant and the surroundings. Compression will be
polytropic with friction and heat transfer instead of isentropic. The actual vapor compression cycle may have some or all of the
following items of deviation from the ideal cycle:
 Superheating of the vapor in the evaporator
 Heat gain in the suction line from the surroundings
 Pressure drop in the suction line due to fluid friction
 Pressure drop due to wiredrawing at the compressor suction valve.
 Polytropic compression with friction and heat transfer
 Pressure drop at the compressor discharge valve.
 Pressure drop in the delivery line.
 Heat loss in the delivery line.
 Pressure drop in the condenser.
 Sub-cooling of the liquid in the condenser or sub-cooler.
 Heat gain in the liquid line.
 Pressure drop in the evaporator.
COMPRESSOR EFFICIENCY
Ideal Power
x100%
Brake or Shaft Power
Wc
ec 
x100%
BP
ec 
MOTOR EFFICIENCY
e Motor 
Brake or Shaft Power
x 100%
Motor Power
where
BP - Power supplied by the motor
MP - Electrical Power supplied to the motor (For Motor driven)
BRAKE POWER (In terms of Torque)
2 πTN
KW
60,000
where
BP 
T - Brake Torque in N - m
N - No. of RPM
MOTOR POWERa) For single phase motor
MP 
EI(cos θ)
KW
1000
MP 
3 EI(cos θ)
KW
1000
where
E  Volts
I  Current drawn by the motor, Amperes
cos θ - Power Factor
RECIPROCATING COMPRESSOR
ISENTROPIC WORK
W  m(h 2  h1 ) KW
k 1


kmRT1  P2  k



W

1
 KW
k - 1  P1 




P1V1'  mRT1
k 1
T2  P2  k
 
T1  P1 
Where :
m - mass flow rate of refrigeran t in
kg
sec
V1'  volume flow rate measured at intake,
m3
sec
KJ
kg - K
P - absolute pressure in KPa
R - Gas constant in
subscripts
1 - suction
2 - discharge
T - absolute temperat ure, K
h - enthalpy,
KJ
kg
DISPLACEMENT VOLUME (For Reciprocating type of compressor)
a. For Single Acting
πLD 2Nn' m3
VD 
4(60)
sec
b. For Double Acting without considering piston rod
VD 
2πLD 2 Nn' m 3
4(60)
sec
c. For Double Acting considering piston rod
πLNn'
m3
VD 
2D 2 - D 2
4(60)
sec


KW PER TON OF REFRIGERATION
KW 211(h 2 - h1 ) KW

Ton (60)(h1 - h4 ) Ton
COEFFICIENT OF PERFORMANCE
Q
COP  A
W
h1 - h 4
COP 
h 2 - h1
CUBIC METER/min PER TON OF REFRIGERATION
A. For Single Acting
m3
211πLD2Nn'
m3

min - Ton 4m(h1 - h4 ) min - Ton
B. For double acting without considering piston rod
m3
211(2)πLD2Nn'
m3

min - Ton
4m(h1 - h4 )
min - Ton
C. For double acting considering piston rod


m3
211πLNn'
m3

2D2 - d 2
min - Ton 4m(h1 - h4 )
min - Ton
where;
L - length of stroke, m
D - diameter of bore, m
d - diameter of piston rod, m
N - no. of RPM
n' - no. of cylinders
VOLUMETRIC EFFICIENCY
v 
V1' '
x 100%
VD
1


k


P

2
v  1  c  c    x 100%

 P1  


where:
V1' - volume flow rate at intake, m3/sec
VD - displacement volume, m3/sec
v - volumetric efficiency
1 - specific volume, m3/kg
PISTON SPEED
PS  2LN m/min
PS 
2LN
m/sec
60
PERCENT CLEARANCE
Clearance Volume
x 100%
Displaceme nt Volume
For reciprocating type compressor design, values of percent clearance C range from 3 to 10 percent
C
EFFICIENCY
A. Compression Efficiency
Ideal Work
ηcn 
x 100%
Indicated Work
B. Mechanical Efficiency
Indicated Work
ηm 
x 100%
Brake or Shaft Work
C. Compressor Efficiency
Ideal Work
ηc 
x 100%
Brake or Shaft Work
ηc  ηcnηm
Example No. 1
A six cylinder, 6.7 cm x 5.7 cm, R-22 compressor operating at 1800 RPM indicate a refrigerating capacity of
96.4 KW and a power requirement of 19.4 KW at an evaporating temperature of 5C and a condensing
temperature of 35C.
Compute:
a) the clearance volumetric efficiency if c = 5%
b) the actual volumetric efficiency
c)the compressor efficiency
QR
3
2
Condenser
Expansion
Valve
L
HE
CE
4
1
Evaporator
D
QA
Example No. 2
An air conditioning system of a high rise building has a capacity of 350 KW of refrigeration, uses R-12. The evaporating and
condensing temperature are 0C and 35C, respectively. Determine the following:
a) mass of flash gas per kg of refrigerant circulated
b) mass of R-12 circulated per second
c) volumetric rate of flow under suction conditions
d) compression work
e) the COP
QR
3
2
Condenser
Expansion
Valve
W
Compressor
4
Evaporator
QA
1
From Chart or Table for R 12 (Saturated vapor and liquid)
t e  0C
Pe  359 KPa  3.59 Bar
h1  355.51
KJ
kg
m3
kg
KJ
S1  1.5591
kg - K
υ1  0.04827
S1  S2  1.5591
KJ
kg - K
t c  35C
Pc  843 KPa  8.43 Bar
At S1  S2 to 843 KPa
t 3  40.45C
S1  S2  1.5591
KJ
kg - K
KJ
kg
h 3  h 4  h f at 35 (saturated liquid)
h 2  370.67
h 3  h 4  234.12
KJ
kg
h1
355.51
KJ/kg
h2
370.67
KJ/kg
h3
234.12
KJ/kg
h4
234.12
KJ/kg
vg1
0.04827
m3/kg
QA
350
KW
hf4
204.7
KJ/kg
hg4
355.51
KJ/kg
hg4-hf4
150.81
KJ/kg
x4
19.51
%
(h1-h4)
121.39
m
2.9
kg/sec
V1'
0.14
m3/sec
mW
43.71
KW
COP
8.007
Example No. 3
A 4 cylinder, R-12 compressor operates between evaporator and condenser temperature of 4C and 43C. It is to carry a load of 20
tons of refrigeration at 1200 RPM. If the average piston speed is 213 m/min and the actual volumetric efficiency is 80%, what should
be the bore and stroke of the compressor?
From R12 table and Chart
At 4C (Psat  3.48 Bar  348 KPa
t e  4C ; Pe  348 KPa
h1  355.08 KJ/kg
S1  S 2  1.5596 KJ/kg - K
1  0.04976
m3
kg
At 43C Psat  10.29 Bar  1029 KPa
tc  43C; Pc  1029 KPa
h1
355.08
KJ/kg
h 2  h at S 2  1.5596 and P2  1029 KPa
h2
374.46
KJ/kg
h3
242.32
KJ/kg
h4
242.32
KJ/kg
vg1
0.04976
m3/kg
QA
10
TR
QA
35.17
KW
N
1200
RPM
PS
213
m/mim
PS
3.55
m/sec
ev
80
%
n'
4.0
cyl.
(h1 - h4)
112.76
KJ/kg
KJ
kg
t 2  49.95C
m
0.31
kg/sec
V1'
0.016
m3/sec
VD
0.019398
m3/sec
L
0.08875
m
L
8.9
cm
D
0.058981
m
D
5.9
cm
h 2  374.46
h 3  h 4  hf at 43C
KJ
h 3  h 4  242.32
kg
Example No. 4
An ammonia compressor operates at an evaporator pressure of 316 KPa and a condenser pressure of 1514.5 KPa. The refrigerant is
sub-cooled 5C and is superheated 8C. A twin cylinder compressor with bore and stroke ratio of 0.85 is to be used at 1200 RPM.
The mechanical efficiency is 77%. For a load of 87.5 KW, determine:
a) the quantity of the cooling water in the condenser if the increase in temperature is 7C.
b) the bore and stroke in cm assuming c = 5%
c) the size of the driving motor
d) the quality of the refrigerant entering the evaporator
P
3
Pc = 1514.5 KPa
QR
2
3
Pe = 316 KPa
4
2
Condenser
S=C
1
Expansion
Valve
L
HE
h
CE
4
QA
T
2
t2 = 117.36 C
tc = 39.04 C
t3 = 34.04 C
t1 = 0.07 C
te = -7.93 C
Evaporator
3
1
4
S
1
D
For Ammonia at Saturated Condition
With 8C Superheating of refrigerant at inlet of compressor
t1 = -7.93 + 8 = 0.07C
With 5C subcooling of condensate at condenser exit
t3 = 39.04 – 5 = 34.04C
Example No. 5
A refrigerant 22, 4-cylinders, 85 mm x 70 mm, compressor running at 28 RPS, a condensing temperature of 38C and an
evaporating temperature of -2C carries a refrigerating load of 112 KW. The motor driving the compressor has an efficiency of
92% and draws 32.5 KW. There is a 5C sub-cooling and 7C superheating of the suction gas entering the compressor. Compute
a) the actual volumetric efficiency
b) the compression efficiency
For R22 at Saturated Condition
With sub – cooling and super – heating of refrigerant
t1  2  7  5
t 3  38  5  33 C
From R22 chart and table (Superheated and Subcooled condition)
Example No. 6
A cold storage warehouse uses a refrigeration system to keep groceries at 2C while the temperature outside the warehouse is
30C. The groceries and the outside air act as thermal reservoirs in this process. Although the warehouse is insulated, it absorbs
heat from the surroundings at a rate of 775 KW. Determine the power requirement and COP for a Carnot refrigeration cycle and
for an ideal Ammonia vapor – compression refrigeration cycle that will maintain the temperature of the groceries under these
conditions. The condenser operates at 1600 KPa and the evaporator operates at 300 KPa.
QR
3
2
Condenser
Expansion
Valve
HE
L
W
CE
4
1
Evaporator
D
QA
Carnot Cycle
T
QR
TH
3
2
TL
4
1
QA
S
TH
QR
W
R
QA
TL
For the Vapor – Compression Cycle (Refrigerant: Ammonia)
P
Pc
2
3
S=C
Pe
4
1
h
T
2
tc
3
Pc
h=C
te
Pe
4
1
S
Saturated Properties (Ammonia)
Superheated Properties at P2 = 1600 KPa and S1 = S2 = 6.2175 KJ/kg-K
Sample Problem No. 7
The catalog for a refrigerant 22, 4-cylinders, hermetic compressor operating at 29 r/s, a condensing temperature of 40C and an
evaporating temperature of -4C shows a refrigeration capacity of 115 KW. At this operating points the motor whose efficiency is
90 percent, draws 34.5 KW. The bore of the cylinders is 87 mm and the piston stroke is 70 mm. The performance data are based on
8C of sub-cooling of the liquid leaving the condenser. Compute
a. the actual volumetric efficiency
b. the compression efficiency
c. the actual COP of the system
From R22 Table and Chart
Pe  Psat at - 4C  436.28 KPa
Pc  Psat at 40C  1533.58 KPa
t 3  40 - 8  32C
t e  t1  t 4
h1  h @ - 4C  403.55
S1  1.7566
KJ
(saturated vapor)
kg
KJ
m3
; υ1  0.0535
kg - K
kg
S1  S 2
h 2  h @ S1  S 2 to Pc  1533.58 KPa (Superheated vapor)
h 2  435.01
KJ
; t2  60.2C
kg
h3  h 4  h@ 1533.58 KPa and 32C
h3  h 4  239.16
KJ
kg
Q A  m(h1  h 4 )  115 KW
m  0.70
kg
sec
V1 '  mυ1  0.70(0.0535)  0.037
m3
sec
πLD 2Nn'
m3
 0.05
4(60)
sec
V
e v  1 ' x100%  77.53%
VD
VD 
W  m(h 2  h1 )  22.01KW
MP  34.5 KW
BP
x100%  90%
MP
BP  0.90(34.5)  31.05 KW
emotor 
W
x100%
BP
22.01
ec 
x100%  70.9%
31.05
115
COP 
 3.33
34.5
ec 
Sample Problem No. 8
Refrigerant-134a enters the compressor of a refrigerator as superheated vapor at 140 KPa and −10°C at a rate of 0.12 kg/s, and it
leaves at 700 KPa and 50°C. The refrigerant is cooled in the condenser to 24°C and 650 KPa, and it is throttled to 150 KPa. Show
the cycle on a Ph diagram and determine
a. the rate of heat removal from the refrigerated space and the power input to the compressor
b. the isentropic efficiency of the compressor, and
c. the COP of the refrigerator.
P1  140 KP1; t1  - 10C
P2  700 KPa ; t 2  50C
P3  650 KPa; t 3  24C
P4  150 KPa
At P1  140 KPa and t1  -10C
KJ
kg
KJ
S1  1.768
kg - K
h1  394.5
υ1  0.1461
At P2  700 KPa and 50C
KJ
kg
KJ
S 2  1.791
kg - K
At P3  650 KPa ; t 3  24C
h 2  436.67
h3  h 4  233.12
KJ
kg
At S1  S 2  1.768
KJ
to P2  700 KPa (Considering Isentropic compression)
kg - K
KJ
; t 2  42.6
kg
Q A  m(h1  h 4 )  1936.56 KW
h 2  429.32
W  m(h 2 - h1 )  502.44 KW
Wisentropic  0.12(429.32 - 394.5)  417.84 KW
e compression 
COP 
Wisentropic
W
x100%  83.2%
QA
 3.82
W
EFFECTS ON CHANGING OPERATING CONDITIONS
A. Effects on Increasing the Vaporizing Temperature
P
1. The refrigerating effect per unit mass increases.
h
2.The mass flow rate per ton decreases
3. The volume flow rate per ton decreases.
4. The COP increases.
5. The work per ton decreases.
6. The heat rejected at the condenser per ton decreases.
T
S
B. Effects on Increasing the Condensing Temperature
T
P
1. The refrigerating effect per unit mass decreases.
2. The mass flow rate per ton increases.
3. the volume flow rate per ton increases.
4. The COP decreases.
5. The work per ton increases.
6. The heat rejected at the condenser per ton increases.
S
h
C. Effects of Superheating the Suction Vapor
T
P
h
When superheating produces useful cooling:
1. The refrigerating effect per unit mass increases.
2. The mass flow rate per ton decreases
3. The volume flow rate per ton decreases.
4. The COP increases.
5. The work per ton decreases.
When superheating occurs without useful cooling:
1. The refrigerating effect per unit mass remains the same.
2. The mass flow rate per ton remains the same.
3. The volume flow rate per ton increases.
4. The COP decreases.
5. The work per ton decreases.
6. The heat rejected at the condenser per ton increases.
S
D. Effects of Sub-cooling the Liquid
1. The refrigerating effect per unit mass increases.
2. The mass flow rate per ton decreases.
3. The volume flow rate per ton decreases.
4. The COP increases.
5. The work per ton decreases.
6. The heat rejected at the condenser per ton decreases.
LIQUID-SUCTION HEAT EXCHANGER
The function of the heat exchanger is:
1. To ensure that no liquid enters the compressor, and
2. To sub-cool the liquid from the condenser to prevent bubbles of vapor from impeding the flow of refrigerant
through the expansion valve.
QR
4
3
Condenser
Heat
Exchanger
2
5
Expansion
Valve
W
6
Evaporator
1
Compressor
ACTUAL VAPOR COMPRESSION SYSTEM
1. An ammonia compression plant is to be designed for a capacity of 50 tons. The system operates with a condensing
temperature of 40C and an evaporating temperature of -20C. The other data are the following:
Temperatures
leaving evaporator
-10C
entering compressor
-5C
leaving compressor
95C
entering condenser
85C
leaving condenser
33C
entering expansion valve
36C
Wiredrawing
suction valve
20 KPa
discharge valve
40 KPa
Compressor Speed
400 RPM
Mechanical Efficiency
80%
Volumetric Efficiency
77%
Stroke to bore ratio
1.3
A twin cylinder, double acting compressor is to be used. The heat absorbed by the jacket water is 30% of
the indicated work of the compressor. determine
a. the bore and stroke in cm
b. the brake work in KW
c. the heat loss or gain between compressor and condenser
d. the heat rejected from the condenser
h5 = hf at 33 C = 356.3 KJ/kg
h6=h7= hf at 36 C = 370.96 KJ/kg
h8=h at 190.74 KPa and -10 C = 1460 KJ/kg
h9=h at 190.74 KPa and -5 C = 1470 KJ/kg
h1 = h9 = 1470 KJ/kg
v1 = v at 190.74 KPa and h1 = 0.75 m3/kg
h3 = h at 1557 KPa and 95 C = 1650 KJ/kg
h4 = h at 1557 KPa and 85 C = 1628 KJ/kg
a) D = 17.94 cm ;L = 23.3 cm
b) BP = 52 KW d) QR = 12,323 KJ/min
c) Q34 = 213.2 KJ/min
2.
An ammonia refrigeration plant is to be designed for a capacity of 30 TR. The cooling water temperature requires a
condenser pressure of 1400 KPa and the brine temperature a pressure of 291.6 KPa in the brine cooler. The following
temperatures will exist at the points designated:
Compressor suction
0C
entering condenser
90C
leaving condenser
32C
leaving evaporator
-5C
Wire drawing through compressor valves
suction
35 KPa
discharge
70 KPa
A two cylinder, vertical compressor is to be used at 120 m/min piston speed.
Mechanical efficiency
80%
Adiabatic compression efficiency
82%
Volumetric efficiency
75%
stroke to bore ratio
1.2
Determine:
a. the rate of circulation of ammonia (5.7 kg/min)
b. the BHP of the compressor (52.3 HP)
c. the bore and stroke and RPM of the compressor (D=19.7 cm;L=23.6 cm;N=254 RPM)
d. the heat rejected at the condenser (7353.5 KJ/min)
e. the COP (3.38)
h5 = h6 = hf at 32C = 351.5 KJ/kg
h7 = h at 291.6 KPa and -5C = 1465 KJ/kg
h1 = h8 = h at 291.6 KPa and 0C = 1475 KJ/kg
v1 = v at 256.6 KPa and h1 = 0.48 m3/kg
h2' = h at 1470 KPa and S1 = S2' = 1745 KJ/kg
h4 = h at 1400 KPa and 90C = 1645 KJ/kg
MULTIPRESSURE SYSTEMS
A multi-pressure system is a refrigeration system that has two or more Low - Side pressures. The low-side pressure is the pressure
of the refrigerant between the expansion valve and the intake of the compressor. A multi-pressure system is distinguished from the
single-pressure system, which has but one low-side pressure.
Removal of Flash Gas
A savings in the power requirement of a refrigeration system results if the flash gas that develops in the throttling process between
the condenser and the evaporator is removed and recompressed before complete expansion. The vapor is separated from the liquid
by an equipment called the Flash Tank. The separation occurs when the upward speed of the vapor is low enough for the liquid
particles to drop back into the tank. Normally, a vapor speed of less than 1 m/sec will provide adequate separation.
Inter-cooling
Inter-cooling between two stages of compression reduces the work of compression per kg of vapor. Inter-cooling in a refrigeration
system can be accomplished with a water-cooled heat exchanger or by using refrigerant. The water-cooled intercooler may be
satisfactory for two-stage air compression, but for refrigerant compression the water is not usually cold enough. The alternate method
uses the liquid refrigerant from the condenser to do the inter-cooling. Discharge gas from the low stage compressor bubbles through
the liquid in the intercooler. Refrigerant leaves the intercooler as saturated vapor. Inter-cooling with liquid refrigerant will usually
decrease the total power requirement when ammonia is the refrigerant but not when R-12 or R-22 is used.
2-Evaporators and 1-Compressor
2-Compressors and 1-Evaporator
2-Compressors and 2-Evaporators
SAMPLE PROBLEMS
1. Calculate the power required by a system of one compressor serving two evaporators. One evaporator carries a load of 35
KW at 10C and the other at a load of 70 KW at -5C. A backpressure valve reduces the pressure in the 10C evaporator to
that of the -5C evaporator. The condensing temperature is 37C. The refrigerant is ammonia. What is the COP.
From ammonia table or chart
h 3  h 4  h 7  h f at 37C  519.02
h 5  h 6  h g at 10C  1615.27
KJ
kg
KJ
kg
KJ
kg
Energy balance in the HP evaporator
h 8  h g at - 5C  1599.82
35  m HP (h 5 - h 4 )
kg
sec
Energy balance in the LP evaporator
m HP  0.032
70  m LP (h 8  h 7 )
kg
sec
at mixing point
m LP  0.065
m HP h 6  m LP h 8  m1h1
m1  0.032  0.065  0.097
KJ
kg
From table by interpolation
h1  1604.92
At S1 = S2 = 6.1781 to P2 = 1430 KPa
KJ
kg
W  m1 (h 2  h1 )  19.7 KW
h 2  1808.04
COP 
35  70
 5.33
19.7
2.
3.
Calculate the power required in an ammonia system that serves a 210 KW evaporator at -20C. The system uses twostage compression with inter-cooling and removal of flash gas. The condensing temperature is 32C.
For minimum work and with perfect inter-cooling, the intermediate pressure P2 is equal to
P2 = (P1P4)1/2
m7 = m8 = m1 = m2
Q = m1(h1 - h8)
m1 = 0.172 kg/sec
by mass and energy balance about the intercooler;
m2h2 + m6h6 = m7h7 + m3h3
m3 = 0.208 kg/sec
WLP = m1(h2 - h1) = 21.6 KW
WHP = m3(h4 - h3) = 31.1 KW
W = WHP + WLP
In ammonia system, one evaporator is to provide 200 KW of refrigeration at -30C and another evaporator is to provide
220 KW at 5C. The system uses two stage compression with inter-cooling and is arranged as shown below. The
condensing temperature is 40C.
a. Calculate the power required by the compressor.
b. What is the COP of the system?
Psat at 40C = 1557 KPa
Psat at 5C = 517 KPa
Psat at -30C = 120 KPa
h1 = hg at -30C = 1422.9 KJ/kg
h2 = h at S1 = S2 to 517 KPa = 1630 KJ/kg
h3 = hg at 5C = 1466.8 KJ/kg
h4 = h at S3 = S4 to 1557 KPa = 1625 KJ/kg
h5 = h6 = hf at 40C = 390.6 KJ/kg
h7 = h8 = hf at 5C = 223.2 KJ/kg
by energy balance at -30 C evaporator
m1 = 200/(h1 - h8) = 0.167 kg/sec
m1 = m2 = m7 = m8
by energy balance at 5C evaporator
m6 = 220/(h3-h6) = 0.204 kg/sec
from the boundary condition above
m5h5 + m2h2 + 220 = m3h3 + m7h7
m5 = m3 = m4
m2 = m 7
m3 = 0.422 kg/sec
WHP = m3(h4-h3) = 66.76 KW
WLP = m2(h2-h1) = 34.59 KW
W = WHP + WLP = 101.4 KW
COP = 220 +200
101.4
COP = 4.142
4.
A two-stage ammonia refrigeration system with a capacity of 50 TR operates with an evaporator temperature of -50C
and a condensing temperature of 28C. The intermediate (intercooler) pressure is 275 KPa. The discharge vapor from the
LP cylinder is bubbled through a flash intercooler which reduces it to saturated state and before entering the condenser is
sub-cooled by 7C, flashed directly into the inter-stage flash inter-cooler and part of the saturated liquid passed through
the main expansion valve to the evaporator. Other data are as follows:
Assume Isentropic Compression
LP cylinder actual volumetric efficiency = 85 %
HP cylinder actual volumetric efficiency = 75 %
Determine
a) Piston displacement in m3/sec
b) Work of the LP and HP cylinders in KW
c) L/min of condenser cooling water for a temperature rise in cooling water temperature of 10C
5
6
su b -cco l er
4
3
2
7
1
8
9
CASCADE SYSTEM
A cascade system combines two vapor compression units, with the condenser of the low temperature system discharging its heat to
the evaporator of the high temperature system. This system can furnish refrigeration for about -100 C. There are two types of a
cascade system, the closed cascade condenser and the direct contact heat exchanger. In the closed cascade condenser fluids in the
low-pressure and high-pressure may be different, but in the direct contact heat exchanger the same fluid is used throughout the
system.
CLOSED CASCADE CONDENSER
DIRECT CONTACT CONDENSER
SAMPLE PROBLEMS
1. A cascade refrigerating system uses R-22 in the low-temperature unit and R-12 in the hightemperature unit. The system develops 28 KW of refrigeration at -40C. the R-12 system
operates at -10C evaporating and 38C condensing temperature. There is a 10C overlap of
temperatures in the cascade condenser. Calculate:
a) the mass flow rate of R-22 (0.1485 kg/sec)
b) the mass flow rate of R-12 (0.305 kg/sec)
c) the power required by the R-22 compressor (5.7 KW)
d) the power required by the R-12compressor (7.6 KW)
For R 12
Pcondenser = 912.533 KPa  Psat at 38C
P1 = P4 = 218.78 KPa
h1 = 348.29 KJ/kg
S1 = 1.5644
At S1 = S2 t0 912.533 KPa
h2 = 373.71 KJ/kg
h3 = h4 = 237.16 KJ/kg
For R22
Pevaporator = 105.231 KPa  Psat at -40C
h5 = 388.13 KJ/kg
S5 = 1.8231 KJ/kg-K
At S5 = S6 to P6 = 497.988 KPa
h6 = 425.84 KJ/kh
h7 = h8 = 200 KJ/kg
Q A  28  mR 22 (h5  h8 )
kg
sec
By energy balance in the cascade condenser
m  0.149
mR12 (h1  h 4 )  mR 22 (h6  h7 )
mR12 
mR 22 (h6  h7 )
kg
 0.3025
(h1  h 4 )
sec
PowerR12  mR12 (h2  h1 )  7.7 KW
PowerR22  mR22 (h6 - h5 )  5.62 KW
COP 
28
 2.10
7.7  5.62
2. In a certain refrigeration system for low temperature application, a two stage operation is
desirable which employs ammonia system that serves a 30 ton evaporator at -30 C. the
system uses a direct contact cascade condenser, and the condenser temperature is 40 C.
find the following:
a) Sketch the schematic diagram of the system and draw the process on the Ph diagram
b) the cascade condenser pressure in KPa for minimum work
c) the mass flow rate in the high pressure and low pressure loops in kg/sec
d) the total work in KW
AIR CYCLE REFRIGERATION
The air cycle refrigeration system in which a gaseous refrigerant is used throughout the cycle. The compression is accomplished by
a reciprocating or a centrifugal compressor as in the vapor-compression system, but condensation and evaporation is replaced by a
sensible cooling and heating of the gas. An air cooler is used in place of a condenser and a refrigerator in place of an evaporator.
The expansion valve is replaced by an expansion engine or turbine.
The air cycle refrigeration system is ideally suited for use in air craft, because it is lighter in weight and requires less space than the
vapor-compression cycle. One disadvantage of the air cycle is that it is not as efficient as the vapor-compression cycle. The air
cycle refrigeration may be designed an operated either as an open or a closed system as shown above. In the closed or dense-airsystem, the air refrigerant is contained within the piping or component parts of the system at all times and with the refrigerator
usually maintained at pressures above atmospheric level. In the open system, the refrigerator is an actual space to be cooled with
the air expanded to atmospheric pressure, circulated through the cold room and then compressed to the cooler pressure
.
IDEAL AIR REFRIGERATION CYCLE
Processes:
1 to 2: Isentropic Compression
2 to 3: Constant Pressure Heat Rejection
3 to 4: Isentropic Expansion
4 to 1: Constant Pressure Heat Absorption
Refrigerating Effect
QA = mCp(T1-T4)
Heat Rejected
QR = mCp(T3-T2)
Compressor Work
k 1 


kP1V1  P2  k
Wc 
 1
 
1  k  P1 



Expander Work
k 1 


kP3 V3  P4  k
WE 
 1
 
1  k  P3 



Net Work
W = WC - WE
Coefficient of Performance
COP 
Refrigerating Effect
Net Work
SAMPLE PROBLEMS
1. An air cycle refrigeration system operating on a closed cycle is required to produced 50 KW of refrigeration with a cooler
pressure of 1550 KPa and a refrigerator pressure of 448 KPa. Leaving air temperature are 25C for cooler and 5C for the
refrigerator. Assuming a theoretical cycle with isentropic compression and expansion, no clearance and no losses. Determine;
a) the mass flow rate (0.720 kg/sec)
b) the compressor displacement (0.1283 cu.m./sec)
c) the expander displacement (0.0964 cu.m./sec
d) the COP (2.35)
2. An open air refrigeration system carries a load of 35 KW with a suction pressure of 103 KPa and a
discharge pressure of 690 KPa. the temperature leaving the refrigerator is 5C and leaving the cooler is
30C. the compression is polytropic with n = 1.35. Determine:
a) the power required (24 KW)
b) the COP (1.47)
3. An air refrigeration system is required to produce 52.5 KW of refrigeration with a cooler pressure of 1448
KPa and a refrigerator pressure of 207 KPa. Leaving air temperature are 29C for cooler and 5C for
refrigerator. Expansion is isentropic and compression is polytropic with n = 1.34. Determine the COP.
(answer: 1.47)
R - 410A
Food and Foodstuff - Specific Heats
By: Engr. Yuri G. Melliza
Specific heats of some common food and foodstuff as apples, bass, beef, pork and much more
The table below can be used to find the specific heat of food and foodstuffs.
Specific Heat above Freezing
Specific Heat below Freezing
Food
(btu/lboF)
(kJ/kgoC)
(kcal/kgoC)
(btu/lboF)
(kJ/kgoC)
(kcal/kgoC)
Apples
0.87
3.64
0.87
0.42
1.76
0.42
Apricots. fresh
0.88
3.68
0.88
0.43
1.8
0.43
Artichokes
0.87
3.64
0.87
0.42
1.76
0.42
Asparagus
0.94
3.94
0.94
0.45
1.88
0.45
Asparagus beans
0.88
3.68
0.88
0.43
1.8
0.43
Avocados
0.72
3.01
0.72
0.37
1.55
0.37
Bacon
2.0
Bananas
0.8
3.35
0.8
0.4
1.67
0.4
Barracuda
0.8
3.35
0.8
0.4
1.67
0.4
Bass
0.82
3.43
0.82
0.41
1.72
0.41
Beef, carcass
0.68
2.85
0.68
0.48
2.01
0.48
Beef, flank
0.56
2.34
0.56
0.32
1.34
0.32
Beef, hamburger
3.52
Beef, loin
0.66
2.76
0.66
0.35
1.47
0.35
Beef, rib
0.67
2.81
0.67
0.36
1.51
0.36
Beef, round
0.74
3.1
0.74
0.38
1.59
0.38
Specific Heat above Freezing
Specific Heat below Freezing
Food
(btu/lboF)
(kJ/kgoC)
(kcal/kgoC)
(btu/lboF)
(kJ/kgoC)
(kcal/kgoC)
Beef, rump
0.62
2.6
0.62
0.34
1.42
0.34
Beef, shanks
0.76
3.18
0.76
0.39
1.63
0.39
Beef, corned
0.63
2.64
0.63
0.34
1.42
0.34
Beets
0.9
3.77
0.9
0.43
1.8
0.43
Blackberry
0.87
3.64
0.87
0.42
1.76
0.42
Blueberries
0.87
3.64
0.87
0.42
1.76
0.42
Brains
0.84
3.52
0.84
0.41
1.72
0.41
Broccoli
0.92
3.85
0.92
0.44
1.84
0.44
Brussels sprouts
0.88
3.68
0.88
0.43
1.8
0.43
Butter
0.65
2.72
0.65
0.34
1.42
0.34
Butter-fish
0.77
3.22
0.77
0.39
1.63
0.39
Cabbage
0.94
3.94
0.94
0.45
1.88
0.45
Candy
0.93
3.89
0.93
0.93
3.89
0.93
Carp
0.82
3.43
0.82
0.41
1.72
0.41
Carrots
0.91
3.81
0.91
0.44
1.84
0.44
Cauliflower
0.93
3.89
0.93
0.44
1.84
0.44
Celery
0.94
3.94
0.94
0.45
1.88
0.45
Chard
0.93
3.89
0.93
0.43
1.8
0.43
Cheese, cottage
3.27
Specific Heat above Freezing
Specific Heat below Freezing
Food
(btu/lboF)
(kJ/kgoC)
(kcal/kgoC)
(btu/lboF)
(kJ/kgoC)
(kcal/kgoC)
Cherries, sour
0.88
3.68
0.88
0.41
1.72
0.41
Cherries, sweet
0.84
3.52
0.84
0.4
1.67
0.4
Chicken, squab
0.8
3.35
0.8
0.39
1.63
0.39
Chicken, broilers
0.77
3.22
0.77
0.38
1.59
0.38
Chicken, fryers
0.74
3.1
0.74
0.35
1.47
0.35
Chicken, hens
0.65
2.72
0.65
0.44
1.84
0.44
Chicken, capons
0.88
3.68
0.88
0.41
1.72
0.41
Chocolate (aprox.)
1.6
Clams, meat only
0.84
3.52
0.84
0.36
1.51
0.36
Coconut, meat and milk
0.68
2.85
0.68
0.45
1.88
0.45
Coconut, milk only
0.95
3.98
0.95
0.42
1.76
0.42
Codfish
0.86
3.6
0.86
0.39
1.63
0.39
Cod Roe
0.76
3.18
0.76
0.39
1.63
0.39
Cow-peas, fresh
0.73
3.06
0.73
0.22
0.92
0.22
Cow-peas, dry
0.28
1.17
0.28
0.41
1.72
0.41
Crabs
0.84
3.52
0.84
0.41
1.72
0.41
Crab apples
0.85
3.56
0.85
0.43
1.8
0.43
Cranberries
0.9
3.77
0.9
0.38
1.59
0.38
Cream
0.9
3.77
0.9
0.45
1.88
0.45
Specific Heat above Freezing
Specific Heat below Freezing
Food
(btu/lboF)
(kJ/kgoC)
(kcal/kgoC)
(btu/lboF)
(kJ/kgoC)
(kcal/kgoC)
Cucumber
0.98
4.1
0.98
0.45
1.88
0.45
Currants
0.97
4.06
0.97
0.45
1.88
0.45
Dandelion greens
0.88
3.68
0.88
0.43
1.8
0.43
Dates
0.2
0.84
0.2
0.01
0.03
0.01
Eels
0.77
3.22
0.77
0.39
1.63
0.39
Eggs
0.76
3.18
0.76
0.4
1.67
0.4
Eggplant
0.94
3.94
0.94
0.45
1.88
0.45
Endive
0.95
3.98
0.95
0.45
1.88
0.45
Figs, fresh
0.82
3.43
0.82
0.41
1.72
0.41
Figs, dried
0.39
1.63
0.39
0.26
1.09
0.26
Figs, candied
0.37
1.55
0.37
0.26
1.09
0.26
Fish, canned
3.35
Fish, fresh
3.6
Flounders
0.86
3.6
0.86
0.42
1.76
0.42
Flour
0.38
1.59
0.38
0.28
1.17
0.28
Frogs, legs
0.88
3.68
0.88
0.44
1.84
0.44
Garlic
0.79
3.31
0.79
0.4
1.67
0.4
Gizzards
0.78
3.27
0.78
0.39
1.63
0.39
Goose
0.61
2.55
0.61
0.34
1.42
0.34
Specific Heat above Freezing
Specific Heat below Freezing
Food
(btu/lboF)
(kJ/kgoC)
(kcal/kgoC)
(btu/lboF)
(kJ/kgoC)
(kcal/kgoC)
Gooseberry
0.86
3.6
0.86
0.42
1.76
0.42
Granadilla
0.84
3.52
0.84
0.41
1.72
0.41
Grapefruit
0.91
3.81
0.91
0.44
1.84
0.44
Grapes
0.86
3.6
0.86
0.42
1.76
0.42
Grape juice
0.82
3.43
0.82
0.41
1.72
0.41
Guavas
0.86
3.6
0.86
0.42
1.76
0.42
Guinea hen
0.75
3.14
0.75
0.38
1.59
0.38
Haddock
0.85
3.56
0.85
0.42
1.76
0.42
Halibut
0.8
3.35
0.8
0.4
1.67
0.4
Herring, smoked
0.71
2.97
0.71
0.37
1.55
0.37
Horseradish, fresh
0.79
3.31
0.79
0.4
1.67
0.4
Horseradish, prepared
0.88
3.68
0.88
0.43
1.8
0.43
Ice cream
0.74
3.1
0.74
0.4
1.67
0.4
Kale
0.89
3.73
0.89
0.43
1.8
0.43
Kidneys
0.81
3.39
0.81
0.4
1.67
0.4
Kidney beans, dried
0.28
1.17
0.28
0.23
0.96
0.23
Kohlrabi
0.92
3.85
0.92
0.44
1.84
0.44
Kumquats
0.85
3.56
0.85
0.41
1.72
0.41
Lamb, carcass
0.73
3.06
0.73
0.38
1.59
0.38
Specific Heat above Freezing
Specific Heat below Freezing
Food
(btu/lboF)
(kJ/kgoC)
(kcal/kgoC)
(btu/lboF)
(kJ/kgoC)
(kcal/kgoC)
Lamb, leg
0.71
2.97
0.71
0.37
1.55
0.37
Lamb, rib cut
0.61
2.55
0.61
0.34
1.42
0.34
Lamb, shoulder
0.67
2.81
0.67
0.35
1.47
0.35
Lard
0.54
2.26
0.54
0.31
1.3
0.31
Leeks
0.91
3.81
0.91
0.44
1.84
0.44
Lemons
0.91
3.81
0.91
0.44
1.84
0.44
Lemon juice
0.92
3.85
0.92
0.44
1.84
0.44
Lettuce
0.96
4.02
0.96
0.45
1.88
0.45
Lima beans
0.73
3.06
0.73
0.38
1.59
0.38
Limes
0.89
3.73
0.89
0.43
1.8
0.43
Lime juice
0.93
3.89
0.93
0.44
1.84
0.44
Litchi fruits, dried
0.39
1.63
0.39
0.26
1.09
0.26
Liver, raw beef
3.5
Lobsters
0.82
3.43
0.82
0.41
1.72
0.41
Loganberries
0.86
3.6
0.86
0.42
1.76
0.42
Loganberry juice
0.91
3.81
0.91
0.44
1.84
0.44
Milk, cow - whole
pasteurized
0.9
3.77
0.9
0.47
1.97
0.47
Milk, dry no fat
1.52
Specific Heat above Freezing
Specific Heat below Freezing
Food
(btu/lboF)
(kJ/kgoC)
(kcal/kgoC)
(btu/lboF)
(kJ/kgoC)
(kcal/kgoC)
Mushrooms, fresh
0.93
3.89
0.93
0.44
1.84
0.44
Mushrooms, dried
0.3
1.26
0.3
0.23
0.96
0.23
Muskmelons
0.94
3.94
0.94
0.45
1.88
0.45
Nectarines
0.86
3.6
0.86
0.42
1.76
0.42
Nuts
0.28
1.17
0.28
0.24
1
0.24
Olives, green
0.8
3.35
0.8
0.4
1.67
0.4
Onions
0.9
3.77
0.9
0.43
1.8
0.43
Onion, Welsh
0.91
3.81
0.91
0.44
1.84
0.44
Oranges, fresh
0.9
3.77
0.9
0.43
1.8
0.43
Orange juice
0.89
3.73
0.89
0.43
1.8
0.43
Oysters
0.84
3.52
0.84
0.41
1.72
0.41
Peaches, Georgina
0.87
3.64
0.87
0.42
1.76
0.42
Peaches, North Carolina
0.89
3.73
0.89
0.43
1.8
0.43
Peaches, Maryland
0.9
3.77
0.9
0.43
1.8
0.43
Peaches, New Jersey
0.91
3.81
0.91
0.44
1.84
0.44
Peach juice fresh
0.89
3.73
0.89
0.43
1.8
0.43
Pears, Bartlet
0.89
3.73
0.89
0.43
1.8
0.43
Pears, Beurre Bosc
0.85
3.56
0.85
0.41
1.72
0.41
Pears. dried
0.39
1.63
0.39
0.26
1.09
0.26
Specific Heat above Freezing
Specific Heat below Freezing
Food
(btu/lboF)
(kJ/kgoC)
(kcal/kgoC)
(btu/lboF)
(kJ/kgoC)
(kcal/kgoC)
Peas, young
0.85
3.56
0.85
0.41
1.72
0.41
Peas, medium
0.81
3.39
0.81
0.4
1.67
0.4
Peas, old
0.88
3.68
0.88
0.43
1.8
0.43
Peas, split
0.28
1.17
0.28
0.23
0.96
0.23
Peppers, ripe
0.91
3.81
0.91
0.44
1.84
0.44
Perch
0.82
3.43
0.82
0.41
1.72
0.41
Persimmons
0.72
3.01
0.72
0.37
1.55
0.37
Pheasant
0.75
3.14
0.75
0.36
1.51
0.36
Pickerel
0.84
3.52
0.84
0.41
1.72
0.41
Pickles, sweet
0.82
3.43
0.82
0.41
1.72
0.41
Pickles, sour and dill
0.96
4.02
0.96
0.45
1.88
0.45
Pickles, sweet mixed
0.78
3.27
0.78
0.29
1.21
0.29
Pickles, sour mixed
0.95
3.98
0.95
0.45
1.88
0.45
Pig's feet, pickled
0.5
2.09
0.5
0.31
1.3
0.31
Pike
0.84
3.52
0.84
0.41
1.72
0.41
Pineapple, fresh
0.88
3.68
0.88
0.43
1.8
0.43
Pineapple, sliced or crushed
0.82
3.43
0.82
0.41
1.72
0.41
Pineapple, juice
0.9
3.77
0.9
0.43
1.8
0.43
Plums
0.89
3.73
0.89
0.43
1.8
0.43
Specific Heat above Freezing
Specific Heat below Freezing
Food
(btu/lboF)
(kJ/kgoC)
(kcal/kgoC)
(btu/lboF)
(kJ/kgoC)
(kcal/kgoC)
Pomegranate
0.85
3.56
0.85
0.41
1.72
0.41
Pompano
0.77
3.22
0.77
0.39
1.63
0.39
Porgy
0.81
3.39
0.81
0.4
1.67
0.4
Pork, bacon
0.36
1.51
0.36
0.25
1.05
0.25
Pork, ham
0.62
2.6
0.62
0.34
1.42
0.34
Pork, loin
0.66
2.76
0.66
0.35
1.47
0.35
Pork, shoulder
0.59
2.47
0.59
0.33
1.38
0.33
Pork, spareribs
0.62
2.6
0.62
0.34
1.42
0.34
Pork, smoked ham
0.65
2.72
0.65
0.35
1.47
0.35
Pork, salted
0.31
1.3
0.31
0.24
1
0.24
Potatoes
0.82
3.43
0.82
0.41
1.72
0.41
Prickly pears
0.91
3.81
0.91
0.43
1.81
0.43
Prunes
0.81
3.39
0.81
0.4
1.67
0.4
Pumpkin
0.92
3.85
0.92
0.44
1.84
0.44
Quinces
0.88
3.68
0.88
0.43
1.8
0.43
Rabbit
0.76
3.18
0.76
0.39
1.63
0.39
Radishes
0.95
3.98
0.95
0.45
1.88
0.45
Raisins
0.39
1.63
0.39
0.26
1.09
0.26
Raspberries, black
0.85
3.56
0.85
0.41
1.72
0.41
Specific Heat above Freezing
Specific Heat below Freezing
Food
(btu/lboF)
(kJ/kgoC)
(kcal/kgoC)
(btu/lboF)
(kJ/kgoC)
(kcal/kgoC)
Raspberries, red
0.89
3.73
0.89
0.43
1.8
0.43
Raspberry juice, black
0.91
3.81
0.91
0.44
1.84
0.44
Raspberry juice, red
0.93
3.89
0.93
0.44
1.84
0.44
Reindeer
0.73
3.06
0.73
0.37
1.55
0.37
Rhubarb
0.96
4.03
0.96
0.45
1.88
0.45
Rose Apple
0.89
3.73
0.89
0.43
1.8
0.43
Rutabagas
0.91
3.81
0.91
0.44
1.84
0.44
Salmon
0.71
2.97
0.71
0.37
1.55
0.37
Sand dab
0.86
3.6
0.86
0.42
1.76
0.42
Sapodilla
0.91
3.81
0.91
0.44
1.84
0.44
Sapote
0.73
3.06
0.73
0.37
1.55
0.37
Sauerkraut
0.93
3.89
0.93
0.44
1.84
0.44
Sausage, beef and pork
0.56
2.34
0.56
0.32
1.34
0.32
Sausage, Bratwurst
0.71
2.97
0.71
0.37
1.55
0.37
Sausage, Bologna
0.71
2.97
0.71
0.37
1.55
0.37
Sausage,FRANKFURTER
0.69
2.89
0.69
0.36
1.51
0.36
Sausage, salami
0.45
1.88
0.45
0.28
1.17
0.28
Sardines
0.77
3.22
0.77
0.39
1.63
0.39
Shad
0.76
3.18
0.76
0.39
1.63
0.39
Specific Heat above Freezing
Specific Heat below Freezing
Food
Shrimp
(btu/lboF)
(kJ/kgoC)
(kcal/kgoC)
(btu/lboF)
(kJ/kgoC)
(kcal/kgoC)
0.83
3.48
0.83
0.41
1.72
0.41
0.73
0.39
1.63
0.39
Skim milk
Spanish mackerel
4.0
0.73
Starch
3.06
1.75
Strawberries
0.95
3.98
0.95
0.45
1.88
0.45
Strawberry juice
0.79
3.31
0.79
0.39
1.63
0.39
String beans
0.91
3.81
0.91
0.44
1.84
0.44
Sturgeon, raw
0.83
3.48
0.83
0.41
1.72
0.41
Sturgeon, smoked
0.71
2.97
0.71
0.37
1.55
0.37
Sugar apple, fresh
0.79
3.31
0.79
0.39
1.63
0.39
Sweet potatoes
0.75
3.14
0.75
0.38
1.59
0.38
Swordfish
0.8
3.35
0.8
0.4
1.67
0.4
Terrapin
0.8
3.35
0.8
0.4
1.67
0.4
Tomatoes, red
0.95
3.98
0.95
0.45
1.88
0.45
Tomatoes, green
0.96
4.02
0.96
0.45
1.88
0.45
Tomato, juice
0.95
3.98
0.95
0.45
1.88
0.45
Tongue, beef
0.74
3.1
0.74
0.38
1.59
0.38
Tongue, calf
0.79
3.31
0.79
0.4
1.67
0.4
Tongue, lamb
0.76
3.18
0.76
0.38
1.59
0.38
Specific Heat above Freezing
Specific Heat below Freezing
Food
(btu/lboF)
(kJ/kgoC)
(kcal/kgoC)
(btu/lboF)
(kJ/kgoC)
(kcal/kgoC)
Tongue, pork
0.74
3.1
0.74
0.39
1.63
0.39
Tongue, sheep
0.69
2.89
0.69
0.36
1.51
0.36
Tomato soup, concentrate
3.67
Tripe, beef
0.83
3.48
0.83
0.41
1.72
0.41
Tripe, pickled
0.89
3.73
0.89
0.43
1.8
0.43
Trout
0.82
3.43
0.82
0.41
1.72
0.41
Tuna
0.76
3.18
0.76
0.39
1.63
0.39
Turkey
0.67
2.81
0.67
0.35
1.47
0.35
Turnips
0.93
3.89
0.93
0.44
1.84
0.44
Turtle
0.84
3.52
0.84
0.41
1.72
0.41
Veal, carcass
0.74
3.1
0.74
0.38
1.59
0.38
Veal, flank
0.65
2.72
0.65
0.35
1.47
0.35
Veal, loin
0.75
3.14
0.75
0.38
1.59
0.38
Veal, rib
0.73
3.06
0.73
0.37
1.54
0.37
Veal, shank
0.77
3.22
0.77
0.39
1.63
0.39
Veal, quarter
0.74
3.1
0.74
0.38
1.59
0.38
Venison
0.78
3.27
0.78
0.39
1.63
0.39
Watercress
0.95
3.98
0.95
0.45
1.88
0.45
Watermelon
0.94
3.94
0.94
0.45
1.88
0.45
Specific Heat above Freezing
Specific Heat below Freezing
Food
(btu/lboF)
(kJ/kgoC)
(kcal/kgoC)
(btu/lboF)
(kJ/kgoC)
(kcal/kgoC)
Whitefish
0.76
3.18
0.76
0.39
1.63
0.39
Yams
0.78
3.27
0.78
0.39
1.63
0.39
Recommended maximum freezing time:
Storage Time (months)
Food
-12 oC
-18 oC
-24 oC
Apricots, raw
4
18
>24
Apricots, in sugar
3
18
>24
Asparagus, with green spears
3
12
>24
Bacon, sliced and vacuum packed
12
12
12
Beans, green
4
15
>24
18
>24
Beans, lima
Beef carcass, unpackaged
8
15
24
Beef steaks/cuts
8
18
24
Breads
3
Broccoli
15
24
Brussels sprouts
6
15
>24
Butter, lactic and unsalted pH 4.7
15
18
20
Butter, lactic and salted pH 4.7
8
12
14
Storage Time (months)
Food
-12 oC
Butter, sweet-cream and unsalted pH
6.6
Butter, sweet-cream and salted pH 6.6
20
Cakes
-18 oC
-24 oC
>24
>24
>24
>24
15
24
Carrots
10
18
>24
Califlower
4
12
24
Cherries, raw
4
18
>24
Cherries, in sugar
3
18
>24
Chicken, whole
9
18
>24
Chicken, parts/cuts
9
18
>24
Clams
4
6
>9
12
18
6
>12
12
15
15
>24
12
18
12
18
12
>24
5
>9
Corn on the cob
Crab
4
Cream
Cut corn
4
Dough, raw
Duck, whole
6
Egg, whole magma
Fish, fat - glazed
3
Storage Time (months)
Food
Fish, lean
-12 oC
-18 oC
-24 oC
4
9
>12
24
>24
Fruit juice concentrate
Geese, whole
6
12
18
Ground beef
6
10
15
Ice cream
1
6
24
Lamb carcass, unpackaged
18
24
>24
Lamb steaks
12
18
24
Leeks, blanched
18
Liver
4
12
18
Lobster
4
6
>12
Mushrooms, cultivated
2
8
>24
10
15
Onions
Oysters
4
6
>9
Peaches, raw
4
18
>24
Peaches, in sugar
3
18
>24
Peas, green
6
24
>24
6
12
Peppers, red and green
Pork carcass, unpackaged
6
10
15
Pork steak/cuts
6
10
15
Storage Time (months)
Food
-12 oC
-18 oC
-24 oC
Potatoes, french fried
9
24
>24
Raspberries/Strawberries, raw
5
24
>24
Raspberries/Strawberries, in sugar
3
24
>24
Shrimps, in shell and cooked
4
6
>12
Shrimps, peeled and cooked
2
5
>9
Spinach, chopped
4
18
>24
Turkey, whole
8
15
>24
Veal carcass, unpackaged
6
12
15
Veal, steaks/cuts
6
12
15
The latent heat of melting for some common solids are indicated below:
Latent Heat of Melting
Product
(kJ/kg)
(Btu/lbm)
Acetic acid
181
77.8
Acetone
98.3
42.3
Alcohol, ethyl (ethanol)
108
46.4
Alcohol, methyl (methanol)
98.8
42.5
Alcohol, propyl
86.5
37.2
Latent Heat of Melting
Product
(kJ/kg)
(Btu/lbm)
Aluminum
321
138.2
Ammonia
339
146
Aniline
113.5
48.8
Antimony
164
70.5
Benzene
126
54.2
Bismuth
52.9
22.7
Brass
168
72.2
Bromine
66.7
28.7
Cadmium
46
19.8
Carbon disulfide
57.6
24.8
Carbon dioxide
184
79
Carbon tetrachloride
174
74.8
Cast iron
126
54.2
Chloroform
77
33.1
Chromium
134
57.6
Cobalt
243
104.5
Copper
207
89
Decane
201
86.4
Dodecane
216
92.8
Latent Heat of Melting
Product
(kJ/kg)
(Btu/lbm)
Ether
96.2
41.4
Ethyl ether
113
48.6
Ethylene glycol
181
77.8
Glycerine
200
75.6
Gold
67
28.8
Heptane
140
60.2
Hexane
152
65.3
Iodine
62.2
26.7
Iron, gray cast
96
41.4
Iron, white cast
138
59.4
Iron, slag
209
90.0
Lead
22.4
9.65
Manganese
155
66.6
Mercury
11.6
5.08
Naphthaline
151
64.9
Nickel
297
127.7
Octane
181
77.8
Paraffin
147
63.2
Ice, see Water
Latent Heat of Melting
Product
(kJ/kg)
(Btu/lbm)
Phenol
121
52
Phosphrus
21.1
9.1
Platinium
113
48.6
Potassium
59
25.4
Propane
79.9
34.3
Propylene
71.4
30.7
Silver
88.0
37.9
Sulphur
39.2
16.87
Tin
58.5
25.2
Toluene
71.8
30.9
Water, Ice
334
144
Wood's alloy
33.5
14.4
Zinc
118
5.63
Introduction
"Open Dating" on a food product is a date stamped on a product's packages to help the store determine how
long to hold the food for sale. It can be helpful to the consumer to identify the time limit to buy the item or
when the product is at its best quality. It is NOT a safety date.
With the exception of infant formula and baby food, product dating is not required by federal regulations. If a
date is used on perishable foods, it must have both the day and month. On shelf-stable and frozen products the
year must also be included, a phrase explaining the meaning of the date such as "sell by" or "use before" must
also be present.
There is no uniform system used for food dating by manufacturers in the U.S. Although dating of some foods
is required by more than 20 states, some states do not require any date codes.
When present "open dating" is found primarily on perishable foods such as meat, poultry, eggs, and dairy
products. "Closed" or "coded" dating might appear on shelf-stable products such as cans and boxes of food.
Types of Dates
A "Sell-By-Date" tells the store how long to hold the food for sale. You should buy the food before the sellby-date expires.
A "Best-if-Used-By (or Before) -Date" is indicates best flavor or quality. It is not a safety date and does not
indicate when a food should be purchased.
A "Use-By-Date" is the last date recommended for peak quality. The date is determined by the manufacturer
of the product.
"Closed or coded dates" are packing numbers for use by the manufacturer so if there is a problem with the
food, it can be recalled.
Safety after dates expire
Except for "use-by-dates," product dates don't always refer to home storage or use after purchase. Even if the
dates expires during home storage, perishable foods should be safe, wholesome, and of good quality - if
handled properly and kept in the frigerator or frozen.
Changing dates
A retailer can legally sell fresh or processed meat and poultry products beyond the expiration date on the
package as long as the product is wholesome. It is also legal for a retailer to change a date on wholesome fresh
meat that has been cut and wrapped in the meat department of the store. However, retailers cannot change
dates on products packaged under federal inspection. If a food is not handled properly, it can become a health
hazard regardless of the date code.
Dating of infant formula and baby food
Formula and baby food dating is for quality as well as nutrient retention. The "use-by-date" is selected by the
manufacturer on the basis of product analysis throughout its shelf life. It is also based on the conditions of
handling, storage, preparation, and use printed on the label. Do not buy or use baby formula or baby food after
its "use-by-date."
Can Codes
Cans must show a packing code so the food can be tracked through distribution. This helps manufacturers
rotate their stock as well as to locate their products in the event of a recall. These codes aren't meant for the
consumer to interpret as "use-by" dates. Cans may also display "open" or calender dates. Usually these are
"best if used by" dates for optimum quality. In general, high-acid foods such as canned tomatoes can be stored
on the shelf for 12 to 18 months; low-acid foods such as canned meat, poultry, fish and most vegetables will
keep 2 to 5 years -- if the can remains in good condition and if it has been stored in a cool, clean, dry place.
Normally food from these cans is safe to eat unless the can is bulging or leaking, but the quality may be
diminished.
Dates on Egg Cartons
If an egg carton has an expiration date on it such as "EXP July 1", be sure that the date has not passed when
you buy the eggs. That is the last day the store may sell the eggs as "fresh." On eggs which use a federal
grading like, Grade AA, the date cannot be more than 30 days from the date the eggs were put in the carton. If
you purchase a carton of eggs before the expiration date, you should be able to use the eggs safely for 3 to 5
weeks from the date of purchase.
Storage Times
Since dates aren't a good guide for safe use of a product, follow these tips to be sure that food is still top
quality.
Purchase the food before the expiration date.
For perishable food, take it home immediately after purchase and refrigerate or freeze it promptly.
Refrigerators should be kept at 40 o F or below, while freezers should be set a 0o F or below.
Once a perishable food is frozen, it doesn't matter if the date expires because foods kept frozen continuously
are safe indefinitely.
Follow handling recommendations on the product.
Consult the following storage chart for fresh or uncooked meat and poultry.
Refrigerated Home Storage (at 40o F or below) of Fresh or Uncooked Products. If product has a "Use-ByDate", follow that date. If the product has a "Sell-By-Date" or no date, cook or freeze by the times on the
following chart. These storage times are general recommendations at refrigeration temperatures.
PRODUCT
STORAGE TIMES
AFTER PURCHASE
Poultry
1 to 2 days
Beef, veal, pork, lamb
3 to 5 days
Ground beef and ground poultry
1 to 2 days
Variety meats (liver, tongue, brain, kidneys, heart, chitterlings)
1 to 2 days
Cured ham, cook-before-eating
5 to 7 days
Sausage from pork, beef or turkey, uncooked
1 to 2 days
Eggs
3 to 5 weeks
Refrigerator Home Storage (at 40o F or below) for Processed Products Sealed at the Plant. If product has a
"Use-By-Date" follow this date. If product has a "Sell-By-Date" or no date, cook or freeze the product by the
times on the following chart.
UNOPENED,
PROCESSED PRODUCT
AFTER OPENING
AFTERPURCHASE
Cooked poultry
3 to 4 days
3 to 4 days
Cooked sausage
3 to 4 days
3 to 4 days
Sausage, hard/ dry, shelf-stable
6 weeks/pantry
3 weeks
Corned beef, uncooked, in pouch
with pickling juices
5 to 7 days
3 to 4 days
Vacuum-packed dinners, commercial
brand with USDA seal
2 weeks
3 to 4 days
Bacon
2 weeks
7 days
Hot Dogs
2 weeks*
7 days
Lunch Meats
2 weeks*
3 to 5 days
Ham, fully cooked
7 days
3 days - slices, 7 days - whole
Ham, canned, labeled “keep refrigerated”
9 months
3 to 4 days
Ham, canned, shelf-stable
2 years/ pantry
3 to 5 days
Canned meat and poultry, shelf-stable
2 to 5 years/ pantry
3 to 4 days
* no longer than 1 week after a "sell-by-date".
Food Safety and Inspection Service
United States Department of Agriculture
Washington, D.C. 20250-3700
Safe Storage of Meat and Poultry:
From the farm to the store, meat and poultry products must be chilled -- and kept chilled, packaged and
handled properly so it will be safe for consumers to buy. Several government agencies have the responsibility
to assure the food's safety. In the home, food caretakers must do their part to store, handle and cook meat and
poultry right so it's safe to eat. Here are some of the scientific principles behind the safe storage of meat and
poultry.
Why is Chilling Meat and
Poultry Important?
Raw meat and poultry products should be maintained at 40° F or below to
greatly reduce the growth rate of any pathogenic bacteria that may be
present on their surfaces. Chilling is required of all raw product unless it
moves directly from the slaughter line to heat processing or cooking (made
into hot dogs or luncheon meats, for example), which destroys pathogens.
How is Meat and Poultry
Chilled and Maintained at
the Plant?
Meat and poultry products are chilled immediately after slaughter to
acceptable internal temperatures which insure the prompt removal of the
animal heat and preserve the wholesomeness of the products. Generally, red
meat carcasses (which are above 90° F at the time of slaughter) are chilled
in a blast cooler with rapidly moving chilled air, and, in some instances, a
cold water shower.
Poultry is required to be chilled to 40° F or less within specified time
frames, depending upon the size of the carcass. Whole birds and parts of
major size are chilled in ice or ice and water media. Poultry parts are chilled
in ice, air or water spray with continuous drainage. Giblets must be chilled
to 40° or below within two hours of slaughtering the birds.
How Does Packaging
Prolong Storage Times?
Packaging is a physical barrier to cross contamination. Microorganisms
exist everywhere in nature. They are in the soil, air, and water. The simple
act of covering food keeps microorganisms from contacting the food.
Covered perishable foods can be stored longer and at better quality than
uncovered foods. Modified atmosphere packaging (MAP) and vacuum
packaging help prolong storage (see following question). Refer to the charts
on pages 5 and 6 for recommended storage times.
What Effect Does
Vacuum Packaging
and MAP Have on
Meat Storage?
Oxygen in the air hastens both the chemical breakdown and microbial spoilage
of many foods. To help preserve foods longer, scientists have developed ways
to help overcome the effects of oxygen. Vacuum packaging, for example,
removes air from packages and produces a vacuum inside. MAP (modified
atmosphere packaging) helps to preserve foods by replacing some or all of the
oxygen in the air inside the package with other gases such as carbon dioxide or
nitrogen. (examples: lunch meat in a blister package; raw beef brisket or filets
in vacuum packaging; fresh turkeys).
At What Temperature
Can Raw Poultry Be
Maintained?
USDA's new rules for labeling raw poultry products as to their storage
temperature will become effective in December 1997. The term "fresh" may
ONLY be placed on raw poultry that has never been below 26° F. Poultry held
at 0° or below must be labeled "frozen" or "previously frozen." No specific
labeling is required on poultry between 0 and 26° F.
This poultry label rule addresses a truth-in-labeling issue, not food safety,
because most pathogenic bacteria do not grow or grow very slowly at normal
refrigerator temperatures. The Agency concluded that the term "fresh" should
not be used on the labeling of raw poultry products that have been chilled to the
point they are hard to the touch.
How are Meat
Products Kept Safe
During
Transportation?
To prevent rapid growth of pathogenic bacteria, perishable meat and poultry
products should be kept cold (40° or below) or frozen (0° or below) during
transport from the plant to a refrigerated warehouse or retail store.
Microorganisms capable of causing foodborne illness either don't grow or grow
very slowly at refrigerated temperatures of 40°F. Freezing keeps food safe by
slowing the movement of molecules, causing any microbes present to enter a
dormant stage. There's also no risk of dripping juices to contaminate nearby
products and storage areas.
How is the
Temperature Inside
Trucks Monitored?
Trucks should have temperature devices which constantly record temperatures
on a running graph for the duration of a trip. A visual temperature device is
located outside the trucks so the drivers can monitor how cold it is inside.
Trucks can be specially sealed to prevent being opened during transit.
How are Meat
Products Kept Safe
During Loading and
Unloading?
One of the extremely important places in handling perishable meat and poultry
products safely is the receiving dock. Employees should verify the products are
at a safe temperature upon arrival. According to the Food and Drug
Administration Food Code (which covers retail establishments, not federal) the
temperature must be at 5° C (41° F) when received. Employees should also
check the conditions of the packaging materials and the sight and smell of the
products..
In the Store, How Should
Meat Products Be Safely
Handled?
To protect perishable food from contamination after receipt at the store,
food employees must wash their hands before handling it. Raw foods must
be kept separate from cooked ready-to-eat foods during storage, preparation,
holding and display. Some local jurisdictions may require food handlers to
wear gloves.
Frozen foods must be maintained frozen. When "thawed for your
convenience," frozen food must be kept under refrigeration that maintains
the food at 5° C (41° F) or below, or completely submerged under running
water following strict guidelines outlined in the Food Code.
How is Meat Kept Safe in
Display Cases?
Meat and poultry products on display must be protected from contamination
by the use of packaging; counter, service line, or salad bar food guards;
display cases or other effective means, according to the Food Code.
Unpackaged, raw animal food, such as beef, lamb, pork, poultry and fish,
may not be offered for consumer self-service.
Ready-to-eat meat and poultry at consumer self-service operations must be
provided with suitable utensils or effective food dispensing methods that
protect the food from contamination. After being served or sold and in the
possession of a consumer, food that is unused or returned by the consumer
may not be re-offered as food for human consumption.
Should Meat and Poultry
Be Stacked Above the
Cooling Level of the
Display Case?
In order for meat and poultry to be stored at the required safe temperatures,
fresh and frozen products should not be stacked above the cooling level of
refrigerator and freezer display cases. The foods should be arranged to
maintain the recommended temperature of 5° C (41° F) in the Food Code
unless the state prescribes a different temperature (for example, many states
allow shell eggs to be stored at 45° F).
Is Food Product Dating
Required by Federal Law?
Except for infant formula and some baby food (see below), product dating is
not required by Federal regulations. However, if a calendar date is used, it
must express both the month and day of the month (and the year, in the case
of shelf-stable and frozen products). If a calendar date is shown,
immediately adjacent to the date must be a phrase explaining the meaning of
that date such as "sell by" or "use before."
There is no uniform or universally accepted system used for food dating in
the United States. Although dating of some foods is required by more than
20 states, there are areas of the country where much of the food supply has
some type of open date and other areas where almost no food is dated.
What Types of Food Are
Dated?
"Open" dating, or dates you can read, are found primarily on perishable
foods such as meat, poultry, eggs and dairy products. "Closed" or "coded"
dating might appear on shelf-stable products such as cans and boxes of
food.
Types of Dates
There are several types of dates:
A "Sell-By" date tells the store how long to display the product for sale.
You should buy the product before the date expires.
A "Best if Used By (or Before)" date is recommended for best flavor or
quality. It is not a purchase or safety date.
A "Use-By" date is the last date recommended for the use of the product
while at peak quality. The date has been determined by the manufacturer of
the product.
"Closed or coded dates" are packing numbers for use by the manufacturer.
Safety After Date Expires
Except for "use-by" dates, product dates don't always refer to home storage
and use after purchase. But even if the date expires during home storage, a
product should be safe, wholesome and of good quality -- if handled
properly and kept at 40° F or below. See the accompanying refrigerator
charts for storage times of dated products.
Foods can develop an off odor, flavor or appearance due to spoilage
bacteria. If a food has developed such characteristics, you should not use it
for quality reasons.
If foods are mishandled, however, foodborne bacteria can grow and cause
foodborne illness -- before or after the date on the package. For example, if
a package of hot dogs is taken to a picnic and left out several hours, the hot
dogs wouldn't be safe to use, even if the date hasn't expired.
Other examples of potential mishandling are products that have been:
defrosted at room temperature more than two hours; cross contaminated; or
handled by people who don't use proper sanitary practices. Make sure to
follow the handling and preparation instructions on the label to ensure top
quality and safety.
How Can Some Products
Have Lengthy Storage
Dates?
Product dating is not required by Federal regulations. But a plant may use
shelf-life studies to determine a sell-by or use-by date for its products. If the
plant has dated a "keep-refrigerated" product, it promises the consumer it
will be safe to use until at least that time, assuming the consumer is storing
the product at 40° F or below.
Some packaging procedures which can lengthen safe storage of perishable
foods are vacuum packaging and MAP. If a fresh meat and poultry product
does not have a date affixed at the plant: refrigerate raw poultry and ground
meat up to 2 days; beef, veal, pork and lamb, 3 to 5 days. Cook or freeze
within those times. Hot dogs and luncheon meats can be refrigerated
unopened up to 2 weeks but no longer than 1 week after a "sell-by" date.
After opening, use or freeze hot dogs within 7 days; luncheon meats, 3 to 5
days.
Canning is another process which can lengthen the storage of meat and
poultry products. Commercially processed cans, jars and flexible types of
retort packages must exhibit a packing code to enable tracking of the
product in interstate commerce. These codes aren't meant for the consumer
to interpret as "use-by" dates. "Best if used by" dates on these products are
quality dates. Canned meat and poultry will keep at best quality 2 to 5 years
if the can remains in good condition and has been stored in a cool, clean,
dry place.
What Ensures the Safe
Storage Dates of
Processed Products with
Lengthy Dates?
Processed meat and poultry products are either cooked, heat treated, smoked
or dried. These processes in themselves extend shelf life beyond that of a
raw product. In addition, packaging processes such as canning, vacuumsealing, MAP, retort packaging, and sealing in blister packaging further
extend the shelf life of these products. If the plant has dated the product, it
promises the consumer it will be safe to use until at least that time,
assuming the consumer is storing the product at 40° F or below.
Storage Times
Since product dates aren't a guide for safe use of a product, the consumer
should follow these tips to store the food and use it at top quality.
Purchase the product before the date expires.
If perishable, take the food home immediately after purchase and refrigerate
it promptly. Freeze it if you can't use it within times recommended on chart.
Once a perishable product is frozen, it doesn't matter if the date expires
because foods kept frozen continuously are safe indefinitely.
Rotate frozen foods; use the oldest first for best flavor and quality.
Follow handling recommendations on products.
REFRIGERATOR HOME STORAGE (at 40° F or below) OF FRESH OR UNCOOKED PRODUCTS
If product has a "Use-By Date," follow that date.
If product has a "Sell-By Date" or no date, cook or freeze the product by the times on the following chart.
PRODUCT
STORAGE TIMES AFTER PURCHASE
Poultry
1 or 2 days
Beef, Veal, Pork and Lamb
3 to 5 days
Ground meat and ground poultry
1 or 2 days
Fresh Variety Meats (Liver, Tongue, Brain, Kidneys, Heart,
Chitterlings)
1 or 2 days
Cured Ham, cook-before-eating
5 to 7 days
Sausage from pork, beef or turkey, uncooked
1 or 2 days
Eggs
3 to 5 weeks
REFRIGERATOR HOME STORAGE (40° F or below) OF PROCESSED PRODUCTS SEALED AT
PLANT
If product has a "Use-By Date," follow that date.
If product has a "Sell-By Date" or no date, cook or freeze the product by the times on the following chart.
PROCESSED PRODUCT
UNOPENED, AFTER
PURCHASE
AFTER OPENING
Cooked Poultry
3 to 4 days
3 to 4 days
Smoked Turkey, whole frozen (after defrosting)
3 to 4 days
3 to 4 days
Cooked Sausage
3 to 4 days
3 to 4 days
Sausage, Hard/Dry, shelf stable
6 weeks/pantry
3 weeks
Corned Beef, uncooked, in pouch with pickling juices
5 to 7 days
3 to 4 days
Vacuum-packed Dinners, Commercial brand with USDA
seal
2 weeks
3 to 4 days
Bacon
2 weeks
7 days
Hotdogs
2 weeks (but no longer
than 1 week after a
"sell-by" date)
7 days
Lunch Meats
2 weeks (but no longer
than 1 week after a
"sell-by" date)
3 to 5 days
Ham, fully cooked
7 days
slices 3 days
whole 7 days
Ham, canned, labeled "keep refrigerated"
9 months
3 to 4 days
Ham, canned, shelf stable
2 years/pantry
3 to 5 days
Canned Meat and Poultry, shelf stable
2 to 5 years/pantry
3 to 4 days
For additional food safety information about meat, poultry or eggs, call the toll-free USDA Meat and Poultry
Hotline at 1 (800) 535-4555. It is staffed by home economists, dietitians and food technologists from 10 a.m.
to 4 p.m. Eastern time year round. An extensive selection of food safety recordings can be heard 24 hours a
day using a touch-tone phone.
PRODUCT LOAD
Q = Q 1 + Q2 + Q3
Q - product load
Q1 - heat to cool product from t2 to tf
Q2 - heat to freeze
Q3 - heat to cool product from tf to final storage temperature t2
Q1 = m Cpa (t1 - tf) KJ/min
Q2 = mhL KJ/min
Q3 = m Cpb (tf - t2)
where: m - mass rate in kg/min
t1 - entering temperature in C
tf - freezing temperature in C
t2 - storage temperature in C
Cpa - specific heat above freezing, KJ/kg-C or KJ/kg-K
Cpb - specific heat below freezing, KJ/kg-C or KJ/kg-K
hL - latent heat of freezing of product, KJ/kg
Q = m [Cpa (t1 - tf) + hL + Cpb (tf - t2)] KJ/min
Q
m [C pa (t1 - t f )  h L  C pb (t f - t 2 )]
211
Tons
AIR CONDITIONING
Instructor: Engr. Yuri G. Melliza
PROPERTIES OF MOIST AIR
Psychrometry - is the study of the properties of air and water vapor mixture.
FUNDAMENTAL PARAMETERS
1. Total Pressure
P = Pa + Pv KPa
2. Vapor Pressure
Pv = Pw - PA(td - tw)
where
A
A
KPa
6.66 x 10 4

5.94 x 10  4

 (For tw  than 0C)
C
 (For tw  than 0C)
C
3. Specific Humidity or Humidity Ratio
4. Relative Humidity
W
0.622 Pv
P  Pv
φ
Pv
x 100 %
Pd
kgm
kgda
5. Enthalpy
h  1.0045td  W2501.3  1.86td
KJ
kgda
h = 1.0045td + W[2501.3 + 1.86td] KJ/kgda
6. Specific Volume
υ
17. Degree of Saturation
0.287( t d  273) m 3
(P  Pv)
kgda
 P  Pd 
μ  φ

 P  Pv 
where:
P - total pressure of air and vapor mixture, KPa
Pa - Partial pressure of dry air, KPa
Pv - vapor pressure, KPa
Pw - saturation pressure corresponding tw, KPa
Pd - saturation pressure corresponding td, Kpa
td - dry bulb temperature, C
tw - wet bulb temperature, C
 - relative humidity, %
W - specific humidity or humidity ratio, kgm/kgda
h - enthalpy, KJ/kgda
hg - enthalpy at saturated vapor at td, KJ/kg
 = specific volume, m3/kgda
R = 0.287 KJ/kg-K
 -degree of saturation
Psychrometric Chart
P = 760 mm Hg
h

Dew Point
Temperature
(DPT)
tw
W
Saturation Curve

tw
td
DPT - dew point temperature (saturation temperature corresponding Pv)
Basic Psychrometric or Air Conditioning processes
1) SENSIBLE HEATING: Addition of heat to moist air. Sensible heating follows a constant specific humidity line on the
psychrometric chart.
m
h1
m
2
1
h2
Q
Sensible
Heating
h2
h1
Saturation Curve
1
W1 = W2
2
t1
t2
By Energy Balance
mh1  Q  mh 2  1
Q  m(h 2 - h1 )  2
Q  mC p (t 2 - t1 )  3
C p  1.0045  1.86W  4
W  W1  W2
2) SENSIBLE COOLING: It is the removal of heat from moist air without the removal of moisture. It also follows a
constant specific humidity line.
Q
m
h1
1
2
m
h2
Sensible
Cooling
h1
h2
Saturation Curve
1
2
t2
W1 = W2
t1
By Energy Balance
mh1  mh 2  Q  1
Q  m(h1 - h 2 )  2
Q  mC p (t1 - t 2 )  3
C p  1.0045  1.86W  4
W  W1  W2
3) COOLING AND DEHUMIDIFYING: It is the removal of heat and moisture from moist air. It involves
sensible and latent heat transfer.
Q
t1
m
h1
W1
2
1
m
h2
W2
mw
hw
t2
h1
Cooling and
Dehumidifying
DPT
ha
1
W1
h2
QL
Saturation Curve
2
QS
t2
By Energy Balance
mh1  mh 2  Q  m w h w  1
Q  m(h1 - h 2 ) - m w h w  2
By moisture balance
mW1  m w  mW2
m w  m(W1  W2 )  4
Q  Qs  Q L  5
Q s  mC p (t1 - t 2 )  5
C p  1.0045  1.86W1  6
Q L  m(W1 - W2 )h fg @t  7
2
where:
QS - sensible heat
QL - latent heat
hfg - latent heat at t2
For air conditioning design:
hfg = 2466 KJ/kg
Cp = 1.02 KJ/kg-C
a
t1
W2
4) HEATING AND HUMIDIFYING: It is the addition of heat and moisture to moist air.
Q
t1
m
h1
2
1
m
t2
h2
W2
W1
mw
hw
h2
Heating and
Dehumidifying
ha
2
W2
h1
QL
Saturation Curve
W1 = Wa
1
a
QS
W1
t1
t2
By Energy Balance
mh 1  Q  m w h w  mh 2  1
Q  m(h 2 - h1 ) - m w h w  2
By moisture balance
mW1  m w  mW2
m w  m(W2  W1 )  4
Q  Qs  Q L  5
Qs  mC p (t 2 - t1 )  5
C p  1.0045  1.86W1  6
Q L  m(W2 - W1 )h fg@t  7
2
W1  Wa
Q  m(h 2 - h1 ) - m(W2  W1 )h w
Q  m(h 2 - h1 ) - (W2  W1 )h w   8
m
mw
(W2  W1 )
Q
mw
(h 2 - h1 ) - (W2  W1 )h w 
(W2  W1 )
(h 2 - h1 )
Q

 hw  9
(W2  W1 ) m w
Δh
Q

 h w  10
ΔW m w
where:
hfg = hfg at t2 or hfg = 2466 KJ/kg
Cp = 1.0045 + 1.86W1
5) HUMIDIFYING: It is the addition of moisture to moist air without the addition of heat.
t1
m
h1
W1
2
1
m
t2
h2
W2
mw
hw
Humidifying
2
2
hw < hg at t1
hw = hf @ tw1 0r
tw2
2
hw = hg at t1
hw > hg at t1
2
Saturation Curve
1
By energy balance
m w h w  m ( h 2  h1 )  1
By moisture balance
m w  m( W2  W1 )  2
m
mw
3
( W2  W1 )
Equation 3 to equation 1:
Δh
 hw
ΔW
Conditions:
a. If hw = hg at t1, the process follows a constant dry bulb temperature line.
b. If hw  hg at t1, the air is heated as well as humidified.
c. If hw  hg at t1 , the air is cooled and humidified.
d. If hw is equal to hf at the wet bulb temperature, the process follows the constant wet bulb temperature line.
6) ADIABATIC MIXING OF TWO STREAMS OF MOIST AIR: It is the mixing of two streams of moist
without the addition of heat.
m2
W2
h2
W1
m1
h1
W3
m3
h3
air
h1
Adiabatic
Mixing
h3
1
W1
h2
W2
3
Saturation Curve
W3
2
t2
By energy balance:
m1h1 + m2h2 = m3h3  1
By moisture balance:
m1W1 + m2W2 = m3W3
By mass balance:
m1 + m2 = m3  3
t3
t1
2
m1 h 3  h 2 W3  W2 t 3  t 2



m 2 h1  h 3 W1  W3 t1  t 3
m1 h 3  h 2 W3  W2 t 3  t 2



m 3 h1  h 2 W1  W2 t1  t 2
m 2 h1  h 3 W1  W3 t1  t 3



m 3 h1  h 2 W1  W2 t1  t 2
SENSIBLE HEAT FACTOR
Sensible Heat factor is the ratio of the sensible heat to the total heat transfer for a process.
QS
Q
QS
SHF 
QS  Q l
SHF 
APPARATUS DEW POINT
Cooling and
Dehumidifying
Process
ADP
(Apparatus Dew
Point)
LOAD RATIO LINE
Room Condition
Supply Air
LOAD RATIO
LINE
For Air Conditioning Design: Supply air temperature
is within 3°C to 11°C lower than room temperature
Sample Problems
1. Find the heat transfer rate and mass flow rate of a heating and adiabatic humidification process where 0.94 m 3/sec of air
enters at 5°C and 40% relative humidity and leaves at 43°C and a relative humidity of 30% and moisture is supplied at
43C. (Assume P = 101.325 KPa)
By energy balance
Q  m w h w  mh 2
Q  m(h 2  h1 )  m w h w
By water balance
mW1  m w  mW2
m w  m(W2  W1 )
Q  m(h 2  h1 )  (W2  W1 )h w 
From Psycrometric chart or by empirical formulas
At 5C and 40% RH
kgm
kgda
KJ
h1  10.43
kgda
W1  0.0021
m3
kgda
At 43C and 30% RH
υ1  0.7907
kgm
kgda
KJ
h 2  85.41
kgda
W2  0.0163
m3
kgda
h w  hf at 43C
υ 2  0.9192
h w  179.94
KJ
kg
2.
Find the heat transfer rate required to warm 42.5 m3/min of air at 16C and 90% relative humidity to 49C without the
addition of moisture.
From Psycrometr ic chart or by empirical formulas
At 16C and 90% RH
kgm
kgda
KJ
h1  41.93
kgda
W1  0.0102
υ1  0.8326
m3
kgda
At 49C and W1  W2  0.0102
h 2  75.71
kgm
kgda
KJ
kgda
m3
kgda
RH 2  13.91%
υ 2  0.9276
3.
Moist air at 27C db and 20C wb is cooled to 15C and 80% relative humidity. The volume flow rate is 0.95 m3/sec and
the condensate leave at 16C. Find the heat transfer rate in tons.
4.
Air at 14C and 90% RH is to be heated until the relative humidity is 50%. Determine the heat added per kilogram of dry
air. (10.2 KJ/kgda)
5.
Moist air at 32C and 60% relative humidity enter the refrigeration coils of a dehumidifier with a flow rate of 1.5 kg/sec.
The air leaves saturated at 15C. Calculate the condensate removed and the tons of refrigeration required.
From Psychromet ric chart or using emperical formulas, assuming P  101.325 KPa
At t d1  32C and 60% RH
W1 0.0180
h1  78.34
at t d2  15C and 100% RH (Saturated )
W2  0.0106
h 2  42.01
h w  hf at 15C
h w  62.867
m moist ai r  m  m Vapor
1.5  m  mW1
kg
of dry air
sec
by energy balance
m  1.47
mh 1  Q  mh 2  m w h w
Q  m ( h1  h 2 )  m w h w  1
by water balance
mW1  m w  mW2
m w  m(W1  W2 )  2
Q  m(h1  h 2 )  (W1  W2 )h w 
m w  0.0109
Q  52.72
6.
kg
sec
KJ
 15 TR
sec
Moist air enters a dehumidifier-heater unit at 26C and 80% RH. It is to leave at 26C and 50% RH. For an air flow rate
of 0.47 m3/sec, find the refrigeration in tons and the heating required in KW.
80%
1
50%
2
3
26C
At 26C and 80% RH
W1 = 0.017002 kgm/kgda
h1 = 69.474 KJ/kgda
v1 = 0.8711 m3/kgda
26C and 50% RH
W3 = 0.010518 kgm/kgda
h3 = 52.943 KJ/kgda
v3 = 0.8623 m3/kgda
At 2 to 3 W2 = W3 and RH2 = 100%, from chart or table,
h2 = 41.463 KJ/kgda
v2 = 0.83 m3/kgda
t2 = 14.794 C
hw = hf at 14.794C = 62.12 KJ/kg
6. Air enters a humidifier-heater at 10C and 10% RH. The air mixture leaves the heater at a temperature of
24C and a relative humidity of 50%. Determine the heat and water added if the airflow rate is 0.75
kg/sec. The water temperature entering is 15C.
7. Ten kilograms of air having a dry bulb temperature of 27C and a humidity ratio of 0.010 kgm/kgda are
mixed with 5 kilograms of air having a dry bulb temperature of 35C and a humidity ratio of 0.020
kgm/kgda. Determine the dry bulb temperature and humidity ratio of the mixture.
8. An air-water vapor mixture enters an air-conditioning unit at 30C and 50% RH. The mass flow rate of
dry air entering the unit is 1 kg/sec. The air vapor mixture leaves the air-conditioning unit at 10C and
100% RH and liquid condensate also leaves the unit at 10C. The total pressure of the air vapor mixture is
110 KPa. Determine the rate of heat transfer for this process.
At 30C and 50% RH and P = 110 KPa
W1 = 0.012266
h1 = 61.509
v 1= 0.8071
At 10C and 100% RH and P = 110 KPa
W2 = 0.007023
h2 = 27.745
v2 = 0.7447
hw = hf at 10C = 42.01 KJ/kg
9. A stream of outdoor air is mixed with a stream of return air in an air conditioning system that operates at
101 KPa pressure. The flow rate of outdoor air is 2 kg/sec, and its condition is 35C BD and 25C WB.
The flow rate of return air is 3 kg/sec, and its condition is 24C and 50% RH. Determine
a) the enthalpy of the mixture
b) the humidity ratio of the mixture
c) the dry bulb temperature of the mixture
SAMPLE PROBLEMS
1. An air conditioning system admits 24 m 3/sec of air at 32C db, 60% RH, and discharges it from the
cooling coil at 15C db and saturated. the refrigerant used is R-12, this leaves the receiver at 689 KPa
subcooled-12C and leaves the evaporator at 179 KPa superheated -12C. Compression efficiency is 80%
and mechanical efficiency is 90%. Find
a) the power required to drive the compressor in KW (240 KW)
b) the tons of refrigeration capacity (270 TR)
2. A space to be conditioned as shown has a total load of 29 KW and a sensible heat ratio of 0.80. It is to be maintain at a condition
not to exceed 27C and 50% RH. For proper ventilation 55% of the supply air is fresh air and the rest are re-circulated air.
Outside
air is 32C DB and 22C WB. The air is supplied at 21C, Find:
a) the kg/min of fresh air supplied
b) the tons of refrigeration required by the a-c unit
OA  Outside air
MA - Mixed air
CA - conditione d air
SA - Supply air
RA - Returned air(Recirc ulated air)
EA - Exhaust air
Room Air
Outside Air
tdo  32C
two  22C
Pwo  2.642 KPa
Pv  Pw - P(6.66 x 10 -4 )(td - tw)
Pvo  1.9952 KPa
0.622Pv
W
P - Pv
Wo  0.0125
h  1.0045td  W 2501.3  1.86td 
ho  64.16
td3  27C
RH 3  50%
Pd3  3.563 KPa
Pv
Pd
Pv3  1.7836 KPa
0.622Pv
W
P - Pv
W3  0.0111
RH 
h  1.0045td  W 2501.3  1.86td 
h 3  55.58
SHF 
Qs
Qs

 0.80
Q Qs  Q L
Qs  0.80(29)  23.2 KW
Q L  29 - 23.2  5.8 KW
For air conditioni ng design
KJ
kg - C
KJ
hfg  2466
kg
Qs  m(1.02)( 27  21)
Cp  1.02
m o  0.55m
23.3
kg
m
 3.81
(1.02)( 27  21)
sec
Q L  m(W3 - W2 )hfg  5.8 KW
m o  0.55(3.81)  2.096
5.8
W2  W3 3.81(2466)
W2  0.0105
mo
t1  t 3 h1  h 3 W1  W3
 0.55 


m
to  t 3 ho  h 3 Wo  W3
t1  29.75
at 21C and W2  0.0105
h1  60.3
h 2  47.48
W1  0.0119
m  mo  mR
m R  3.81  2.096  1.714 kg/sec
Qs = 0.8(29) = 23.2 KW
Ql = 29 - 23.2 = 5.8 KW
Wo = 0.0124 kgm/kg da
ho = 64.5 KJ/kg
o = 0.88 m3/kg
Qs = mCp(t4-t3)
W4 = 0.116 kgm/kg da
h4 = 56.6 KJ/kg
4 = 0.86 m3/kg
m = 3.79 kg/sec
Ql = m(W4-W3)hfg
hfg = 2437.6 KJ/kg at 27C
W3 = 0.009 kgm/kg da
Q = m(h4 - h3)
h3 = 48.95 KJ/kg
3 = 0.845 m3/kg
mo = 0.55(3.79) = 2.0845 kg/sec
m5 = 1.7055 kg/sec
a) Vo = 2.0485(0.88) = 1.835 m3/sec
= 110 m3/min
b) W2 = W3 = 0.009 kg m/kg da
Pv2 = 1.45 KPa
tdp2 = 12.5C
Note: the temperature of the spray water is equal to the temp. of the air leaving the air conditioner.
h2 = 35.3 KJ/kg
mo/m = h1 - h4/ho - h4
h1 = 60.945 KJ/kg
Qc = m(h1 - h2) =97.2 KW
= 5832 Kj/min
= 27.6 tons
3. A small store has a sensible heat load of 16 KW and a latent heat load of 4 KW. Indoor conditions of 24C
DB and 50% RH are to be maintained when outside air is 33C DB and 24C WB. All outside air supplied
with re-heater to satisfy the space conditions. The conditioned air leaves the fan at 16C Db. the fan
energy may be neglected. Find:
a) the tons of refrigeration required by the air conditioner
b) the supply fan capacity in m3/sec
c) the heat supplied in re-heater in KW
Wo = 0.0152 kgm/kgda
ho = 72.48 KJ/kg
W3 = 0.0094 kgm/kgda
h3 = 48 KJ/kg
Qs = mCp(t3-t2)
16 = m(1,02)(24-16)
m = 1.96 kg/sec
Ql = m(W3-W2)hfg
hfg at 24C = 2444.7 KJ/kg
W2 = 0.0086 kgm/kgda
Q = Qs + Ql = m(h 3-h2)
h2 = 37.8 KJ/kg
W1 = W2 =
0.622Pv2
kgm/kgda
101.325-Pv2
Pv2 = Pv1 = 1.38 KPa
t1 = 11.756C
h1 = 33.5 KJ/kgda
Qc = m(ho - h1) = 22 tons
Q1 = m2
2 = 0.82 m3/kg
Q1 = m2 = 1.61 m3/sec
Qh = m(h2 - h1) = 8.453 KW
4. A space to be conditioned has a sensible heat load of 633 000 KJ/hr and a moisture load of 1,584,000
grains per hour; 2300 lbs of air are to be introduced each minute to this space for its conditioning to 27C
DB and 50% RH. Determine the amount of air to be bypassed and the amount and temperature of air
leaving the dehumidifier.
re c ir c ula ted
a ir
o uts id e
at 27C and
a ir50% RH
Pd4 = 3.567 KPa
Pv4 = 1.7835 KPa
W4 = 0.0111 kgm/kg da
h4 = 55.6 KJ/kg da
4 = 0.865 m3/kg da
tdp4 = 15.69C
e xha us t a ir
b yp ass ed
a ir
co nd itio ne d
a ir
AC - unit
S pa
ce
Qs = 633,000 KJ/hr
Ql = m(W4-W3)hfg
m = 2300/2.205 = 1043.084 kg/min
m(W4-W3) = 1,584,000 Grains/min
1 lb = 7000 Grains
1 kg = 15,435 grain
5. A small store has a sensible heat load of 16 KW and a latent heat load of 4 KW, indoor conditions of 24C
DB and 50% RH are to be maintained when ventilating air amounting to 34 m 3/min at 33C DB and 24C
WB temperature is used. Duct work and apparatus are arranged with the by-pass system. The air
conditioner has an apparatus dew point of 10.3C and the conditioned air leaves the fan at 16C. The fan
energy may be neglected. Find
a) the tons of refrigeration required
b) the volume of by-pass air in m3/min
td3=16C
Vo = 34 m3/min
tdo = 33C
two = 24C
Wo = 0.0152 kgm/kgda
ho = 72.5 KJ/kg
o = 0.89 m3min
at 24C and 50% RH
W4 = 0.0094 kgm/kg da
h4 = 48 KJ/kg da
Q = Qs + Ql
Q = 20 KW
Q = mCp(t4 - t3)
m = 1.96 kg/sec
Q = m(h4-h3)
h3 = 37.8 KJ/kg
Ql = m(W4-W3)hfg
x+c=1
c=1-x
24x + ct2 = 16
24x + (1-x)t2 = 16
xtdp4 + ctdp2 = 1tdp3
12.8x + 10.3c = 11.756
4 = 0.854 m3/kg da
tdp4 = 12.8C
hfg @ 24C = 2444.7 KJ/kg
W3 = 0.0086 kgm/kg da
Pv3 = 1.38 KPa
tdp3 = 11.756C
24x + 10.3c = 16
x = 0.379
x = 37.9%
c = 62.1%
xh4 + ch2 = 1h3
h2 = 31.57 KJ/kg
Qc = m2 (h1 - h2)
Qc = 10.12 tons
Vx = xm4
Vx = 0.379(1,96)60(0.854)
=38.06m3/min
mo = Vo/o = 38.2 kg/min
m2 = 0.621(1.96)60 = 73.03 kg/min
mo/m2 = h1 - h4
ho - h4
h1 = 60.81 KJ/kg
6. An air conditioned theater is to be maintained at 26.7C DB and 50% RH. The calculated total sensible
heat load in the theater is 126,240 KCal/hr and latent heat load is 82,920 KCal/hr. the air mixture at
28.9C DB and 22.2C WB temperature is cooled to 17.22C DB and15C WB temperature by chilled
water cooling coils and delivered as supply air to the theater. Calculate the tons of refrigeration required.
recirculated air
exhaust air
4
3
AC unit
outside air
0
1
at 28.9C and 22.2C
WB
h1 = 65 KJ/kg
at 17.22C and 15C WB
h2 = 42 KJ/kg
at 26.7C and 50% RH
h3 = 54.5 KJ/kg
Q = Qs + Ql
Q = 875,753 KJ/hr
Q = m(h3 - h2)
m = 70,060 kg/hr
2
conditioned air
Qc = m(h1 - h2)
60(211)
Qc=127.3tons
7. An auditorium is to be maintained at a temperature of 26C DB and 50% RH. Air is to be supplied at a
temperature not lower than 15C DB. The sensible heat gain is 110 KW and the latent heat gain is
37.5 KW . Take ventilating air as 25% by weight of the air from the room and is at 35C DB and 60% RH.
Determine the refrigeration capacity in tons.
recirculated air
exhaust air
4
3
outside air
2
0
conditioned air
1
AC unit
8. Atmospheric air at 10C and 40% RH is heated to 25C in the heating section of an air-conditioning
device and then steam is introduced to increase the relative humidity to 50% while the temperature
increases to 26C. Determine the mass flow rate of water vapor added and the rate of heat transfer needed
in the heating section if the volume flow rate of inlet air is 50 m 3/min.(0.458 kg/sec; 19.33 KW)
Humidifier
m
0
1
2
m2
Heater
mw
9. Outside air in a dry climate enters an air conditioner at 40C and 10% RH and is cooled to 22C. (a)
Calculate the heat removed (b) Calculate the total energy required to conditioned outside (humid) air at
30C and 90% RH to 22C and 10% RH. ( 19 KJ/kgda; 98 KJ/kgda)
10. One hundred m3/min of outside air at 36C and 80% RH is conditioned for an office building by cooling
and heating. Estimate both the rate of cooling and the rate of heating required if the final state of the air is
25C and 40% RH. (152 KW ;26.8 KW)
11. Room air at 29C and 70% RH is cooled by passing it over coils through which chilled water at 5C
flows. The mass flux of the chilled water is 0.5 kg/sec and it experiences a 10C temperature rise. If the
room air exits the conditioner at 18C and 100% RH, Estimate (a) the mass flux of the room air (b) the
heat transfer rate. (0.91 kg/sec ; 20.9 KW)
12. Outside air at 40C and 20% RH is to be cooled using an evaporative cooler. If the flow rate of the air is
40 m3/min, estimate (a) the minimum possible temperature of the exit stream and (b) the minimum mass
flux needed for the water supply. (21.7C ; 0.329 kg/min)
13. Thirty m3/min of outside air at 0C and 40% RH is first heated and then passed through and evaporative
cooler so that the final state is 25C and 50% RH. Determine the temperature of the air when it enters the
cooler, the heat transfer rate needed during the heating process, and the mass flux of water required by
the cooler. (45C; 30 KW ; 0.314 kg/min)
14. Water is used to remove the heat from the condenser of a power plant. 10,000 kg/min of 40C water
enters a cooling tower. Water leaves at 25C. Air enters at 20C and 50% RH and leaves at 32C and
98% RH. Estimate (a) the volume flow rate of air into the cooling tower (b) the mass flow rate of water
that leaves the cooling tower from the bottom. (7450 m 3/min ; 9800 kg/min)
15. Cooling water leaves the condenser of a power plant at 38C with a mass flow rate of 40 kg/sec. It is
cooled to 24C in a cooling tower that receives atmospheric air at 25C and 60% RH. Saturated air exits
the tower at 32C. Estimate (a) the required volume flow rate of entering air (b) the mass flow rate of the
make up water. ( 37 m3/sec ; 0.8 kg/sec)
AIR OR GAS COMPRESSORS
By: Engr. Yuri G. Melliza
Compressor is a machine used to compressed air or gas to a final pressure exceeding 241.25 KPa gage (342.6 KPaa).
TYPES OF COMPRESSORS
 Centrifugal Compressors - For low pressure and high capacity application
 Rotary Compressors - For medium pressure and low capacity application
 Reciprocating Compressors - For high pressure and low capacity application
USES OF COMPRESSED AIR
 Operation of small engines
 Pneumatic tools
 Air hoists
 Industrial cleaning by air blast
 Tire inflation
 Paint Spraying
 Air lifting of liquids
 Manufacture of plastics and other industrial products
 To supply air in mine tunnels
 Other specialized industrial applications
ANALYSIS OF CENTRIFUGAL AND ROTARY TYPE
W
2
m
m
1
Q
Q = h + KE + PE + W (Steady state - steady flow equation)
W  Q  Δh  ΔKE  ΔPE

W   VdP  ΔKE  ΔPE

Q  Δh   VdP
For a compressor work is done on the system, therefore
the work is (-W)
 W  Δh - Q  ΔKE  ΔPE

 W  VdP  ΔKE  ΔPE
let - W  Wc
Wc  Δh - Q  ΔKE  ΔPE

Wc  VdP  ΔKE  ΔPE
A) Isentropic Compression (PVk = C)
P
2
W    VdP
PV k  C
1
V
For Isentropic Compressio n
k 1
T2  P2  k
 
T1  P1 
Q0
Wc  Δh  ΔKE  ΔPE

Wc  VdP  ΔKE  ΔPE
Δh  mC p (T2  T1 )
k 1


k(P2 V2 - P1V1 ) kmR(T2 - T1 ) kmRT1  P2  k

 
VdP 


 1
k 1
k 1
k  1  P1 




if ΔKE  0 and ΔPE  0

Wc  Δh  mC p (T2  T1 )
k 1


k(P2 V2 - P1V1 ) kmR(T2 - T1 ) kmRT1  P2  k

 
Wc 


 1
k 1
k 1
k  1  P1 




where
m - mass flow rate of the gas in kg/sec
W - work in KW
P - pressure in KPa
T - temperature in K
R - gas constant in KJ/kg-K
V - volume flow rate in m3/sec
B) Polytropic Compression (PVn = C)
P
2
W    VdP
PV n  C
1
V
For Polytropic compression, Q  0
n 1
n 1
V 
T2  P2  n
  
  1 
T1  P1 
 V2 
Wc  Δh  Q  ΔKE  ΔPE

Wc  VdP  ΔKE  ΔPE
Δh  mC p (T2  T1 )
n 1


n (P2 V2  P1V1 ) nmR (T2  T1 ) nmRT1  P2  n



VdP 



1

n 1
n 1
n  1  P1 




Q  mC n (T2  T1 )

kn 
Cn  Cv 

 1 n 
If ΔKE  0 and ΔPE  0
Wc  Δh  Q
n 1


n (P2 V2  P1V1 ) nmR (T2  T1 ) nmRT1  P2  n



Wc  VdP 



1

n 1
n 1
n  1  P1 





n 1


nmRT1  P2  n



W
 1



n  1  P1 




C) Isothermal Compression (PV = C)
P
2
W    VdP
PV  C
1
V
P1V1  P2 V2
T1  T2
Wc  Δh  Q  ΔKE  ΔPE
Δh  0

Wc  VdP  ΔKE  ΔPE
 VdP  P V ln P
P2
1 1
1
if ΔKE  0 and ΔPE  0
Wc  Q
Wc  P1V1 ln
P1V1  mRT1
P2
P1
ANALYSIS OF RECIPROCATING TYPE COMPRESSOR
Cylinder
Valve
Piston
HE
Piston
Rod
CE
d
D
L
P
P2
2
2'
P1
1'
1
V1'
cVD
VD
V
HE - Head end
CE - Crank end
L - length of stroke, m
VD - Displacement volume, m3/sec
D - diameter of bore, m
d - diameter of piston rod, m
V1’ – Volume flow rate measured at intake in m3/sec
PERCENT CLEARANCE
Clearance Volume
C
Displaceme nt Volume
V
C  2' x 100%
VD
For compressor design, values of percent clearance C ranges from 3 to 10 %.
V2'  CVD
where:
V2’ - clearance volume
VOLUMETRIC EFFICIENCY
Volume flow rate at intake
ηv 
x 100%
Displaceme nt Volume
V
ηv  1' x 100%
VD
A) For Isentropic Compression (PVk= C)
1

 P2  k 

ηv  1  C  C   x 100%
 P1  



B) For Polytropic Compression (PVn = C)
1

 P2  n 

ηv  1  C  C   x 100%
 P1  



C) For Isothermal Compression (PV = C)

 P 
ηv  1  C  C 2  x 100 %
 P1 

ACTUAL VOLUMETRIC EFFICIENCY
ηva 
Va
x 100%
VD
Va - actual volume of air or gas drawn in
DISPLACEMENT VOLUME
A) For single acting
VD 
πLD 2 Nn' m3
4(60) sec
B) For Double acting without considering the volume of piston rod
VD 
2πLD 2 Nn' m3
4(60) sec
C) For Double acting considering volume of piston rod


πLNn'
m3
2D 2  d 2
4(60)
sec
where:
L - length of stroke, m
D - diameter of bore, m
d - diameter of piston rod, m
n' - no. of cylinders
N = no. of RPM
VD 
A) For Isentropic Compression and RE-expansion process (PVK = C),no heat is removed from the gas.
k 1
T2  P2  k
 
T1  P1 
Q0
Wc  Δh  ΔKE  ΔPE

Wc  VdP  ΔKE  ΔPE
Δh  mC p (T2  T1 )
k 1
k 1




kmRT1  P2  k
kP1V1'  P2  k


 
 
VdP 
 1 
 1
k  1  P1 
k  1  P1 








P1V1'  mRT1

m - mass flow rate measured at intake, kg/sec
V1'  volume flow rate at intake, m3/sec
if ΔKE  0 and ΔPE  0
Wc  Δh  mC p (T2  T1 )
k 1
k 1




kmRT1  P2  k
kP1V1'  P2  k






Wc 
 1 
 1






k  1  P1 
k  1  P1 








B) For Polytropic Compression and Re-expansion process (PVn = C), some amount of heat is removed from the gas.
n 1
n 1
V 
T2  P2  n
  
  1 
T1  P1 
 V2 
Wc  Δh  Q  ΔKE  ΔPE

Wc  VdP  ΔKE  ΔPE
Δh  mC p (T2  T1 )
n 1
n 1




nmRT1  P2  n
nP1V1'  P2  n


 
 
VdP 
 1 
 1
n  1  P1 
n  1  P1 








Q  mC n (T2  T1 )

 kn 
Cn  C v 

 1 n 
If ΔKE  0 and ΔPE  0
Wc  Δh  Q
n 1
n 1




nmRT1  P2  n
nP1V1'  P2  n


 
 
Wc  VdP 
 1 
 1
n  1  P1 
n  1  P1 









n 1
n 1




nmRT1  P2  n
nP1V1'  P2  n






Wc 

1


1


n  1  P1 
n  1  P1 








C) For Isothermal compression and re-expansion process (PV = C), an amount of heat equivalent to the
compression work is removed from the gas.
P1V1  P2 V2
T1  T2
Wc  Δh  Q  ΔKE  ΔPE
Δh  0

Wc  VdP  ΔKE  ΔPE
 VdP  P V ln P
P2
1 1'
1
if ΔKE  0 and ΔPE  0
Wc  Q
Wc  P1V1' ln
P2
P1
P1V1'  mRT1
MEAN EFFECTIVE PRESSURE
W
Pm 
KPa
VD
W in KJ, KJ/kg, KW
VD in m3, m3/kg, m3/sec
PISTON SPEED
PS 
2LN m
60 sec
EFFICIENCY
a.
COMPRESSION EFFICIENCY
Ideal Work
ecn 
x100%
Indicated Work
b.
MECHANICAL EFFICIENCY
em 
c.
Indicated Work
x100%
Brake or Shaft Work
COMPRESSOR EFFICIENCY
Ideal Work
x100%
Brake or Shaft Work
MOTOR EFFICIENCY
ec  ecnem 
BP
x100%
MP
BP  Brake or Shaft Power
emotor 
MP - Motor Power
MOTOR POWER
A. For Single Phase Motor
MP 
EI (Power Factor)
KW
1000
B. For 3 – Phase Motor
MP 
3 EI (Power Factor)
KW
1000
Where:
E – Volts
I – Current in Amperes
PF – Power Factor
MULTISTAGE COMPRESSION
Multi staging is simply the compression of air or gas in two or more cylinders in place of a single cylinder compressor. It is used in
reciprocating compressors when pressure of 300 KPa and above are desired, in order to:
 Save power
 Limit the gas discharge temperature
 Limit the pressure differential per cylinder
 Prevent vaporization of lubricating oil and to prevent its ignition if the temperature becomes too high.
It is common practice for multi-staging to cool the air or gas between stages of compression in an intercooler, and it is this cooling
that affects considerable saving in power.
A) 2 - Stage Compression without pressure drop in the intercooler
For an ideal multistage compression, with perfect inter-cooling and minimum work, the cylinder were properly designed so that:
 the work at each stage are equal
 the air in the intercooler is cooled back to the initial temperature
 no pressure drop occurs in the intercooler
 the pressure ratio at each stage are equal
W1 = W2 ;
T1 = T3 ; P2 = P3 = Px
where:
W1 - work of the LP cylinder (1st stage)
W2 - work of the HP cylinder (2nd stage)
Px = ideal intercooler pressure, optimum pressure
Assuming polytropic compression and expansion processes:
W1  W2
Work for LP cylinder (W1 )
T1  T3
n 1


nP1V1'  P2  n



W1 
 1



n  1  P1 




P1V1'  mRT1
n 1
 n
T2  P2
 
T1  P1 
n 1
T4  P4  n
 
T3  P3 
Work for HP cylinder (W2 )
n 1


nP3 V3'  P4  n

 
W2 
 1



n  1  P3 




P3 V3'  mRT3
T2  T4
P2 P4

P1 P3
P1V1'  P3V3'
Total Work (W) for Ideal 2 - stage
W  W1  W2
Pressure Ratio
W  2 W1
P2 P4

P1 P3
n 1


2nP1V1'  P2  n

 
W1 
 1
n  1  P1 




1
P2 Px


P1
P1
P1P4
P1
1
Px  P2  P3
1
1
n 1


2nP1V1'  P4  2 n



W1 

1

n  1  P1 




By Energy Balance in the Intercoole r
Q x  m(h 2 - h1 )  mC p (T2 - T1 )
Heat produced during compressio n (PV n  C)
Q1  mC n (T2 - T1 )
Q 2  mC n (T4 - T3 )
 k-n 
Cn  Cv 

 1- n 
1
P 2P 2
P 2
P 2  P 2
 1 4  4 1  4 1   4 
P1
P
P1P1 2
P1 2  1 
PX P4

P1 PX
Px  P1P4  Ideal intercoole r pressure
B) 2 stage compressor with pressure drop in the intercooler For 2 stage compression with
pressure drop in the intercooler,
P2  P3
The air in the intercooler may or may not be cooled to the initial temperature, and the work at each stage may or may not be equal,
thus the work W = W1 + W2
For 2 - stage ompression with pressure drop
P2  P3
W1  W2 or W1  W2
T1  T3
or T1  T3
P2 P4
P
P

or 2  4
P1 P3
P1 P3
T2 T4
T
T

or 2  4
T1 T3
T1 T3
W  W1  W2
where
Work for LP cylinder (W1 )
n 1


nP1V1'  P2  n

 
W1 
 1

n  1  P1 




P1V1'  mRT1
Work for HP cylinder (W2 )
n 1


nP3 V3'  P4  n



W2 
 1



n  1  P3 




P3 V3'  mRT3
By Energy Balance in the Intercoole r
Q x  m(h 2 - h1 )  mC p (T2 - T1 )
Heat produced during compressio n (PV n  C)
Q1  mC n (T2 - T1 )
Q 2  mC n (T4 - T3 )
 k-n 
Cn  Cv 

 1- n 
C. Three-Stage compressor without pressure drop in the intercooler
Considering Polytropic compression and expansion processes and with perfect inter-cooling and minimum work
For 3 - stage ompression without pressure drop
P2  P3  Px
P4  P5  Py
W1  W2  W3
T1  T3  T5
P
P2 P4

 6
P1 P3
P5
P
Px Py

 6
P1 Px
Py
Px  3 P12 P6  LP Intercoole r pressure
Py  3 P1P6 2  HP Intercoole r pressure
T
T2 T4

 6
T1 T3
T5
W  W1  W2  W3
where
Work for 1st Stage cylinder (W1 )
n 1


nP1V1'  P2  n



W1 

1

n  1  P1 




P1V1'  mRT1
Work for 2nd Stage cylinder (W2 )
n 1


nP3 V3'  P4  n

 
W2 
 1
n  1  P3 




P3 V3'  mRT3
Work for 3rd Stage cylinder (W3 )
n 1


nP5 V5'  P6  n

 
W3 
 1
n  1  P5 




P5 V5'  mRT5
n 1


3nP1V1'  P2  n

 
W
 1
n  1  P1 




Total W
W  W1  W2  W3
but
W1  W2  W3
W  3W1
1
 P2   P6  3
    
 P1   P1 
n 1


3nP1V1'  P6  3n

 
W
 1
n  1  P1 




For multistage compression with minimum work and perfect inter-cooling and no pressure drop in the inter-coolers between stages,
the following conditions apply:
1. the work at each stage are equal
2. the pressure ratio between stages are equal
3. the air temperature in the inter-coolers are cooled to the
original temperature T1
4. the total work W is equal to
n 1


SnP1V1'  P2S  Sn



WS 

1

n  1  P1 




where S - number of stages
INTERCOOLER PRESSURES
1
Pa Pb Pc Pd Pe P2S  P2S  S







P1 Pa Pb Pc Pd
Pe  P1 
1
Pa  P2S  S


P1  P1 
1
 P S
Pa  P1  2S 
 P1 
1
1
1
1
1
 P S
 P S  P S
Pb  Pa  2S   P1  2S   2S 
 P1 
 P1   P1 
2
 P S
Pb  P1  2S 
 P1 
1
 P S
 P S  P S
Pc  Pb  2S   P1  2S   2S 
 P1 
 P1   P1 
3
 P S
Pc  P1  2S 
 P1 
4
 P S
Pd  P1  2S 
 P1 
5
 P S
Pe  P1  2S 
 P1 
Sample Problems
1. A 2-stage, double acting air compressor 41cm and 25cm bore x 18 cm running 600 RPM has a free air un-loader at
each end for capacity control. It is driven by a 150 HP motor, 460V,3 phase, 60 hertz, 1175 RPM thru super-high
capacity V-belts at sea level installation.P1 = 101 KPa and P4= 984 KPa, calculate the power assuming the LP cylinder
discharges the air to the ideal pressure and intercooler cools the air to the initial temperature of 21C.
2. A single acting compressor has a volumetric efficiency of 87% operates at 500 RPM. It takes in air at 100 KPa and
30C and discharges it to 600 KPa. The air handled is 6 m3/min measured at discharge condition. If compression and
expansion processes are isentropic, find the piston displacement per stroke.
3. A two-stage reciprocating air compressor is required to deliver 0.70 kg/sec of air from 98.6 Kpa and 305K to 1276
KPa. The compressor operates at 205 RPM, compression and re-expansion processes are PV1.25 = C, and both
cylinders have 3.5% clearance. There is a 20 KPa pressure drop in the intercooler. The LP cylinder discharges at the
optimum pressure into the intercooler. The air enters the HP cylinder at 310K. the intercooler is water cooled with the
water entering at 295K and leaving at 305K. Determine the motor power required if m= 85%.( 260 KW)
4. A 36 cm x 36 cm , horizontal double acting air compressor w/ 5% clearance operates at 120 RPM and draws air at 100
KPa and 31C and discharges it to 397 KPa. Compression and expansion processes follows PV1.3 = C. Determine
compressor power required in KW.(22 KW)
5. An ideal 3-stage air compressor with intercoolers handles air at the rate of 2 kg/min. The suction pressure is 101 Kpa,
suction temperature is 21C, delivery pressure is 5000 KPa. Assuming perfect inter-cooling and minimum work,
calculate total power required if compressor efficiency is 60% and both compression and expansion processes are
PVn=C, where n = 1.2. (21 KW)
6. A two stage compressor with 5% clearance delivers 40 kg/min of air to 965 Kpa. At intake the pressure is 98.6 KPa and
the temperature is 16C. the compression is polytropic with n = 1.31 and the intercooler cools the air to 16C. Determine
the mean effective pressure in KPa.
7. A double acting, 2 stage air compressor running at 150 RPM has a suction pressure and temperature of 100 KPa and
27C, respectively. The low pressure cylinder is 36 cm x 38 cm and connected in tandem to the high pressure cylinder.
The LP cylinder discharges the air at 386 KPa. The intercooler cools back the air to its initial temperature and the air
enters the HP cylinder at 370 KPa and leaves at 1 480 KPa. Both LP and HP cylinders have 4% clearance and
compression follows PV1.3 = C. Neglect effect of piston rods. Standard air is at 101 KPa and 16C. Determine:
a) the volume of free air based on apparent volumetric efficiency
b) the heat removed by cooling water in the intercooler ( 24 KW)
c) the diameter of the HP cylinder in cm (19 cm)
d) the power required to drive the compressor if the aggregate losses is 20% of the power of the conventional diagram. (72
KW)
8. A turbine driven compressor handles 10 kg/sec of air from 100 KPa to 600 KPa with an inlet temperature of 300K and
a discharge temperature of 530K. The inlet tubing has a 0.5 m inside diameter and the discharge piping a 0.20 m
diameter. The compression is adiabatic. Determine:
a) the air inlet and exit velocities in m/sec
b) the isentropic compression efficiency
c) the power required in KW
9. A large centrifugal compressor handles 9 kg/sec of air, compresses it from 100 KPa and 15C and with an initial
velocity of 110 m/sec to discharge pressure. The velocity of the high pressure air stream is 90 m/sec. The pressure
ratio is 4:1 the compressor has an isentropic compression efficiency of 80%. Determine
a) the exit pressure in KPa
b) the exit temperature in K
c) the power required
10.A centrifugal compressor handling air draws 6 m 3/sec of air at a pressure of 96.5 KPa and a temperature of 15.6C.
The air delivered from the compressor at a pressure 482.5 KPa and a temperature of 73C. The area of the suction
pipe is 0.2 m2, the area of the of the discharge pipe is 0.4 m2, the discharge pipe is located 6 m above the suction
pipe. The weight of the jacket water that enters at 16C and leaves at 43C is 3.45 kg/sec. Find the power required
to drive this compressor assuming no loss from radiation.
11.A low pressure water jacketed air compressor, compresses 6.8 kg/min of air from 101.325 KPa and 21C to 136 Kpa
and 43C.
a) let the process to be polytropic, neglecting KE and PE find W and the mass flow rate of cooling water circulating
if tw = 4C.
b) considering the process to be irreversible adiabatic with a final temperature of 54C, find the value of m in
PVm = C and W.
ANSWER: a) 3 KW; 1.7 kg/sec; b) W = 4 KW; m = 1.6
12.There are required 1903 KW of compressor power to handle air adiabatically from 1001 KPa, 27C to 305 KPa. The
initial air velocity is 21 m/sec and the final velocity is 85 m/sec.
a) if the process is isentropic, find the volume of air handled in m3/min measured at inlet conditions.
b) If the compression is an irreversible adiabatic to a final temperature of 158C,with the capacity found in (a)
find the power input.
13.A 36 cm x 30 cm, single cylinder, double acting air compressor with 5.5% clearance operates at 125 RPM. The
suction pressure and temperature are 96 KPa and 38C, respectively. The discharge pressure is 290 KPa.
Compression and re-expansion processes are PVn=C with n = 1.3. /neglecting piston rod effects, determine:
a) the volumetric efficiency ( 93%)
b) the volume flow rate at suction conditions (7.07 m3/min)
c) the work in KW (15 KW)
d) the heat rejected during compression (3 KW)
e) the indicated power in KW in the compression efficiency is 75%. (20 KW)
14.An industrial plant operates an air compressor system for cooling. The air entering the compressor has a temperature
of 16 C at 345 KPa and is compressed to 1206 KPa. The compressed air is then cooled by circulating water to a
temperature of 38C and then allowed to expand to the original pressure. Assume adiabatic compression and dry air.
Find:
a) temperature of the compressed air
b) final temperature of air after expansion
c) diameter of the pipe necessary to conduct 0.02 m3/sec if the piping system has a 12- 90 elbow long radius
elbow , 2-globe valves, 3 tees on run, and 61 m of straight pipe. Friction drop in the pipeline is limited to 14
KPa. Assume specific weight and velocity remains constant along the pipe
15.Air is compressed from 103.4 KPa and 21 C to 4132 KPa by a three-stage compressor with a value of n =
1.32.Determine the theoretical work of the compressor and the heat absorbed by the intercoolers.
16.A single acting compressor with a clearance of 6% takes in air at atmospheric pressure and at a temperature of 29C
and discharges it at a pressure of 586 KPa. The air handled is 0.0071 m3/cycle measured at the discharge pressure. If
the compression is isentropic,find:
a) Piston displacement per cycle (0.03 m3/cycle)
b) Compressor power if n = 750 RPM ( 72 KW)
17.A single cylinder, double acting air compressor running at 200 RPM has a piston speed of 183 m/min. It compresses
27 kg/min of air from 97 KPa and 16C to 654 KPa. Clearance is 5.5%. For isentropic compression,
Find:
a) the volumetric efficiency (84%)
b) the piston displacement in m3/min (28 cu.m./min)
c) the work in KW (96 KW)
d) the mean effective pressure in KPa (207 Kpa)
e) the bore and stroke of the compressor in cm. (D=44 cm;L=46 cm)
18.An air compressor with a clearance of of 6% is to compress 0.43 m3/sec of free air where the atmospheric pressure is
101 KPa and the atmospheric temperature is 24C. At the end of the suction stroke, the pressure and temperature of
the air in the cylinder are 97 KPa and 32C. For a discharge pressure of 520 KPa, Determine the conventional
volumetric efficiency and displacement of the compressor when the compression process is:
a) Isothermal (74%; 0.6 cu.m./sec)
b) Isentropic (86%; 0.5 cu.m./sec)
c) polytropic with n = 1.33 (84.8%; 0.51 cu.m./sec)
19.A single stage reciprocating air compressor has two double acting cylinders each 46 cm diameter and 36 cm stroke
and the piston rods are 50 mm in diameter. The intake and discharge pressure are 101 KPa and 445 KPa, respectively
and the speed is 257 RPM. If the volumetric efficiency is 78%, mechanical efficiency is 94% and the isothermal
compression efficiency is 66%, Find:
a) the metered flow per second (0.795 cu.m./sec)
b) the indicated power in KW (181 KW)
c) the power required to drive this compressor in KW ( 193 KW)
d) the adiabatic compression efficiency ( 82%)
20. An air pollution control device requires 0.23 cu.m./sec of free air. To provide this demand a double acting, single stage
reciprocating air compressor having a clearance of 3% was bought. Suction conditions are 101 KPa and 21C and a
discharge pressure of 724 KPa is desired. When tested under these conditions, performance was within the terms of
guarantee.
a) what is the piston displacement in cu.m./sec of the compressor if n =1.34
b) if the compression ratio is held constant and the unit is operated at an altitude of 1800 m, what will be its
capacity and discharge pressure
d) what would be the displacement of a similar compressor to handle the same mass of air as in (a) when operated at
1800 m elevation.
21. A 2-stage, double acting air compressor, 410mm x 250 mm bore x 180 mm
stroke, running at 600 RPM, has a free air unloader at each end for
capacity control. It is driven by a 150 HP electric motor, 460 volts,
3 phase, 60 Hz, 1175 RPM thru superhigh capacity V-belts at sea level
installation. Suction pressure is 1.034 kg/ cm2 absolute and discharge
pressure is 9 kg/cm2 gage. Calculate:
a) piston displacement in m3/hr ( 1711; 636)
b) piston speed in m/sec ( 3.6)
c) ideal pressure gage reading in kg.cm2 ( 2.2)
22. A reciprocating 2-stage air compressor takes in air at atmospheric
pressure and 27C.The flash point of the oil used in the cylinder is
260C. Safety precautions limit the temperature of the air in the HP
cylinder to be 28 C below the flash point of the oil. Assuming perfect
intercoooling, what would be the allowable working pressure of this
compressor if the compression curve follows PV1.34 = C.
(P4 = 6144.3 KPa)
23. The following data were obtained from a performance test of a 10 cm x
4 cm x 9 cm single acting, water jacketed two stage air compressor;
Pressures:
Atmospheric – 102 KPa
Leaving 1st stage – 634 KPa
Entering 2nd stage – 634 KPa
Leaving 2nd stage – 4200 KPa
Temperature:
Atmospheric – 24C
Leaving 1st stage – 108C
Entering 2nd stage –25C
Leaving 2nd stage – 127C
Compressor speed – 648 RPM
Power input to motor – 5.65 KW
Motor and belt efficiency – 81.5%
The following are calculated from test data:
Air flow rate – 0.387 kg/min
Compression exponents: 1st stage, 1.15: 2nd stage, 1.2
Calculate:
a) approximate indicated power (2.4 KW)
b) power for isothermal compression ( 2.04 KW)
c) Shaft power ( 4.6 KW)
d) 1st stage piston displacement ( 0.458 m3/min)
e) actual volumetric efficiency ( 70.52%)
f) isothermal compression efficiency ( 86.61%)
g) mechanical efficiency based on the approximate indicated
power ( 51.3%)
h) overall efficiency of the compressor ( 44.49%)
24. An industrial plant requires 127.5 m3 of free air per hour for its
processes. Two steam engine driven three stage air compressor has to
compress the under the following conditions:
Intercooler pressure between 1st and 2nd stage – 353 KPa
Barometer reading – 792.5 mm Hg
Exponent n = 1.3
Mechanical efficiency of compressor – 80%
Thermal efficiency of the steam engine – 14%
Boiler, Furnace, An piping efficiency – 70%
Heating value of Malangas coal – 32 103 KJ/kg
Assuming the intermediate pressure between stages is that required
for minimum work , find the amount of Malangas coal to produced
enough steam to operate two compressors ( 3 stage) of the same
diameter to compresses the required capacity. ( 45 kg/hr)
25. Determine the percent clearance of a single cylinder, double acting
air compressor with bore and stroke of 45 cm x 45 cm x 180 RPM. The
compressor is tested and it delivers air from 101.3 KPa and 300 K to
675 KPa at a rate of 0.3 m3/sec. The exponent n = 1.33 for compression
and expansion processes. There is a 5% pressure drop through the inlet
and discharge valves. The air is warmed to 38 C during the intake
stroke. (c = 6.78%)
26. A two stage, reciprocating air compressor is required to deliver 0.70
kg/sec of air from 986 KPa and 32 C to 1276 KPa. The compressor
operates at 205 RPM , compression and expansion processes follow PV1.25
= C, and both cylinders have 3.5% clearance. There is a 20 KPa
pressure drop in the intercooler. The LP cylinder discharges at the
optimum pressure into the intercooler. The air enters the HP cylinder
at 37 C. The intercooler is water cooled, with the water entering at
22 C and leaving at 32 C. Determine:
a) the free air capacity in m3/sec 0.583)
b) the LP and HP discharge temperature ( 121 ; 405)
c) the optimum interstage pressure (354.7 KPa)
d) the cooling water required for the intercooler in
kg/sec ( 1.41)
e) the theoretical power required in KW (185)
f) the LP cylinder dimensions if L/D = 0.70 (706.5 mm x
494.6 mm)
g) the motor requirements in KW, if the adiabatic
compression efficiency is 83% and mechanical efficiency
is 85%. (278.14 KW)
27. A single acting, 2-stage air compressor with 3% clearance is required
to compressor has a mean piston speed of 140 m/min, and that each
piston has the same stroke. Take free air conditions to be the same as
suction conditions and for perfect intercooling, Calculate:
a) piston areas for minimum power (0.0435 m2; 0.0251 m2)
b) total power required in KW (6 KW)
c) heat loss in the intercooler in KW ( 3 KW)
28. Find the indicated power of a compressor delivering NH3 at 2.3 kg/min.
The gas enters the compressor at 207 KPa and –7C and is compressed
to 1103 KPa at 121C. Five kilograms of cooling water enters the
compressor jacket at 27C and leaves at 30C.(12 KW)
29. A 2 cylinder, single acting air compressor is directly coupled to an
electric motor running at 1000 RPM. Other data are as follows:
Size of each cylinder – 15 cm x 20 cm
Clearance volume - 10% of displacement
Exponent n for PVn = C - 1.6
Calculate the following:
a) the volume rate of air delivery in terms of standard air
for a delivery pressure of 8 times ambient pressure under ambient air
conditions of 100 KPa and 300 K.
b) Shaft power required if mechanical efficiency is 81%.
30. A single acting air compressor operates at 300 RPM with an initial
conditions of air at 98 KPa and 27 C, and discharges the air at 380
KPa to a cylindrical tank. The bore and stroke are 355 mm and 381 mm,
respectively, with a percent clearance of 5%. If the surrounding air
is at 100 KPa and 20 C while the compression and expansion processes
are PV1.3= C. Determine:
a) the free air capacity in m3/sec (0.164)
b) power of compressor in KW ( 27 KW)
31. A closed metal tank is designed to be safe when subjected to an
internal pressure of 689 KPa. It is used to hold compressed air and is
filled with this material at a temperature of 16C and a pressure of
551 KPa. The tank stands in the sun and its contents may obtained a
temperature of 52 C.
a) Assuming the tank does not expand with temperature and
pressure changes will the design pressure be exceeded
b) what temperature will have to be attained to raise the
pressure of the air to the value for which the tank was designed.
32. An air compressor receives air from the surroundings at 100 KPa and
21C. There is a 2 KPa drop through the intake valves, and the
temperature at the end intake stroke is 38 C. The discharge pressure
is 480 KPa and there is a 20 KPa pressure drop through the discharge
valves. Determine
a) the ideal volumetric efficiency (80%)
b) the actual volumetric efficiency (74.1%)
c) the power if the piston displacement is 14 Liters and
the ideal intake volume is 11.2 Liters. The compressor is operating
at 200 RPM and n = 1.35 ( 7 KW)
33. Required an air compressor to supply 36 kg of air per minute at a
pressure of 1034 KPa; Temperature of air entering the compressor is 16
C, atmospheric. Not counting frictional losses , Find:
a) the power require to operate the compressor if;
1. Single stage with n = 1.35 (160 KW)
2. 2-stage with air entering 2nd stage cylinder at 339
KPa and 18 C. ( 137 KW)
b) the temperature of the air going to the receiver in each
case.( 255 C ; 119 C)
34. An air compressor is direct connected to a heat engine. The compressor
handles 28.5 m3/min of free air and delivers it to 1203 KPa.
Atmospheric pressure and temperature being 97 KPa and 16 C,
respectively and the overall isothermal efficiency of the compressor
is 70%. The brake thermal efficiency of the heat engine is 25% and all
the engine comes from an oil, the combustion of each liter of which
liberates 41 000 KJFind:
a) The liters of oil per day required by the engine if run
continuously at the given load (1386.5 L/day)
b) The combined thermal efficiency based on the indicated
power of the compressor if the compression efficiency is 77%. ( 23%)
35. Determine the percent clearance of a single cylinder, double acting
air compressor with bore and stroke of 45 cm x 45 cm by 3 RPS. The
compressor is tested and it delivers air from 101.3 KPa and 300 K
to 675 KPa at a rate of 0.3 m3/sec. The exponent n = 1.33 for
compression and expansion processes. There is a 5% pressure drop
through the inlet and
discharge valve. The air is warmed to 311 K during the intake stroke.
(7%)
36. A reciprocating compressor with 3 % clearance receives air at 100 KPa
and 27 C and discharges it at 1000 KPa. The compression and expansion
are polytropic with n = 1.25. There is a 5% pressure drop through the
inlet and discharge valves. The air is warmed to 38 C by the cylinder
walls by the end of the intake stroke. Determine:
a) the theoretical and actual volumetric efficiency (82.5%; 75.6%)
b) the percentage of the work needed to overcome the throttling
process.
37. A single acting compressor has a volumetric efficiency of 87% and
operates at 500 RPM. It takes in air at 100 KPa and 30 C and
discharges it at 600 KPa. the air handled is0.10 m3/sec measured at
discharge condition. If the compression is PVk = C, find the mean
effective pressure in KPa. (204 KPa)
38. A reciprocating, 2-stage, single acting air compressor has a rated
capacity of 0.02 m3/sec of free air at 101 KPa and 27 C when running
at 600RPM. The discharge pressure is 30 kg/cm2 and the air is
discharge into an air receiver of 1250 Liters capacity. The compressor
has two LP cylinders each 13 cm diameter and one HP cylinder 7 cm in
diameter; piston stroke is 10 cm. the intercooler is of the finned
tube type with the cooling fan mounted on the compressor shaft. The
compressor is driven by a 1750 RPM,3-phase,60 Hz, 460 V motor thru Vbelts with transmission efficiency of 95%. Calculate:
a) Optimum intercooler pressure for equal work in the LP and HP
cylinders
b) Volumetric efficiency of compressor
c) Brake power at compressor shaft if compressor efficiency is 85%
d) KW of the driving motor in KW
e) Time in minutes to fill air receiver from empty condition.
39. A two stage air compressor receives 0.20 m3/sec of air at 103 KPa and
27 C and discharges it at 828.5 KPa. All clearances are 3.5%. The
compression and expansion processes on both cylinders is polytropic
with n = 1.35. Determine:
a)the ideal intercooler pressure in KPa
b) the minimum power required
in KW
c) the displacement volume of each cylinder in m3/min
d) the LP and HP cylinder dimensions if the compressor operates at 205 RPM and L/D = 1.2 for both cylinder
e) the cooling water required in the intercooler for perfect inter- cooling assuming 10 C temperature rise in cooling water.
f) the percentage of the work saved for single stage compression to
the same pressure.
40. A natural pipeline distributes gas through the country, pumped by a
large gas-turbine driven centrifugal compressors. Assume that the
natural gas is methane, the pipe diameter is0.2m and the gas enters
the compressor at 300 K and 105 KPa. the velocity of the methane
entering the compressor is 4 m/sec. The compression process is
isentropic and the discharge pressure is 700 KPa. Determine:
a) the discharge temperature (475.7 K)
b) the mass flow rate in kg/sec ( .085 kg/sec)
c) the power required in KW (32 KW)
For Methane: R = 0.5183 KJ/kg-K ; k = 1.321
41. A 2 stage, double acting air compressor, operates at 150 RPM. The
conditions of the air at the beginning of compression are P1 = 97.9
KPa and t1 = 27 C. The LP cylinder with a bore and stroke of 35.5 cm x
38.1 cm discharges the air at 379 KPa into the intercooler. The air in
the intercooler suffers a 17.2 KPa pressure drop and enters the HP
cylinder at 29 C. the discharge pressure from the compressor is
2000 KPa. Compression and expansion process in both cyclinders are
PV1.3 = C. The surrounding are at
100 KPa and 20 C. The percent clearance of each cylinder is 5%.
Determine:
a) the free air capacity in m3/sec ( 0.164)
b) the heat loss in the intercooler in KJ/min (1270)
c) the total power required in KW (62 KW)
d) the optimum interstage pressure in KPa. ( 442.5)
e) the diameter of the HP piston if the stroke is the
same as the LP piston (19 cm)
f) the heat loss in the LP and the high pressure
compression processes.(Q1 = 308 KW; Q2 408 KW)
42. There are compress 0.2 m3/sec 0f air from 103 KPa, 27 C to 828 KPa.
All clearances are 8%.
a) Find the isentropic power and piston displacement required for a
single stage compression
b) Using the same data, find the minimum ideal power for 2-stage
compression when the intercooler cools the air to the initial
temperature.
c) Find the displacement of each cylinder for the conditions of
part (b).
d) How mush heat is exchanged in the intercooler.
e) For a compressor efficiency of 78%, what driving motor output is
required for 2-stage compression.
43. A single cylinder, double acting air compressor running at 200 RPM
has a piston speed of 183 m/min. It compresses 27 kg/min of air from
97 KPa and 16 C to 654 KPa. Clearance is 5.5%. For isentropic
compression, Find:
a) the volumetric efficiency (84%)
b) the piston displacement in m3/min ( 28 cu.m./min)
c) the work in KW ( 96 KW)
d) the mean effective pressure in KPa (207 Kpa)
e) the bore and stroke of the compressor in cm. (D=44
cm;L=46 cm)
FANS
Fan is a machine used to add energy to the gaseous fluid to increase its pressure. Fans are used where low pressures (from a few mm
of water to 50 mm Hg or 6.70KPa) and comparatively large volume are required. They run at relatively low speed, the casing and
impeller usually built of sheet iron.
FAN TYPES
1) AXIAL FLOW FANS - the flow of the gases is parallel to the fan shaft.
a. tube axial
b. vane axial
c. Propeller
2) RADIAL OR CENTRIFUGAL FLOW FANS- the flow of gases depends upon the
centrifugal action of the impeller or rotor.
a. Straight blades
b. Forward curved blades
c. Backward curved blades
d. Double curved blades
COMMON USES OF FANS
1. Ventilation and air conditioning
2. Forced and induced draft service for boilers
3. Dust collection
4. Drying and cooling of materials
5. Cooling towers
6. Mine and tunnel ventilation
7. Pneumatic conveying and other industrial process work
AIR FLOW MEASUREMENT
Static pressure
Static pressure is the potential energy put into the system by the fan. It is given up to friction in the ducts and at the duct inlet as it is
converted to velocity pressure. At the inlet to the duct, the static pressure produces an area of low pressure (see Figure 5.15).
Velocity pressure
Velocity pressure is the pressure along the line of the flow that results from the air flowing through the duct. The velocity pressure is
used to calculate air velocity.
Total pressure
Total pressure is the sum of the static and velocity pressure. Velocity pressure and static pressure can change as the air flows though
different size ducts accelerating and de-accelerating the velocity. The total pressure stays constant, changing only with friction losses.
The illustration that follows shows how the total pressure changes in a system. The fan flow is measured using pitot tube manometer
combination, or a flow sensor (differential pressure instrument) or an accurate anemometer. Care needs to be taken regarding number
of traverse points, straight length section (to avoid turbulent flow regimes of measurement) up stream and downstream of measurement
location. The measurements can be on the suction or discharge side of the fan and preferably both where feasible.
MEASUREMENT BY PITOT TUBE
The Figure 5.16 shows how velocity pressure is measured using a pitot tube and a manometer. Total pressure is measured using the
inner tube of pitot tube and static pressure is measured using the outer tube of pitot tube. When the inner and outer tube ends are
connected to a manometer, we get the velocity pressure. For measuring low velocities, it is preferable to use an inclined tube
manometer instead of U tube manometer.
HEAD CALCULATIONS
Note: Usually Fans are designed to overcome fluid friction (No cooling system required). (Q = 0)
W (Fan Work or
Power)
P2
P1
m
m
v2
v1
P P v  v
h   m of gas
 2g
2 2
2 1 2 1
a) For fans installed with both suction and discharge duct
P E

z
0
h


Q

0
0
P2
 P
1

hs

P2
hv

v2
h

hs
v2

 P
1

 v1
2g
2
 hv
 v1
2g
2
2
 S tatic
m
of
g as
He ad
2
m

V e lo c ity
of
g as
He ad
b) For fans installed with only a suction duct, P2 = 0 gage
 P1 v 2  v1
h


2g
2
2
m of gas
c) For fans installed with only a discharge duct, P1 = 0 gage and v1 = 0.
2
P v
h 2  2

2g
m of gas
hs - static head at which a fan operates, m of gas
hv - velocity head at which a fan operates, m of gas
h - total head added to the fluid, m of gas
h = hS + hV m of gas
HEAD CONVERSION
hwater 
gashgas
water

 gashgas
 water
m of water
where
h refers to h, hs or hv
h tw  hsw  h vw
Where
h - stands for ,total head, static head or velocity head
w - refers to water; g - refers to gas
FAN POWER
FP  Qhtw KW
Where
Q – capacity of fan in m3/sec
 - specific weight of water (Gage Fluid) in KN/m3
htw – total head in m of water
STATIC POWER: Part of the total Fan Power that is used to produced the change in
static head.
SP  Qhsw KW
where
Q - capacity in m3/sec
w - specific weight of water (gage fluid) in KN/m3
htw - total head in m of WG
hsw - static head in m of WG
FP - total fan power in KW
SP - Static power in KW
FAN EFFICIENCY
Fan Power
x 100%
Brake or Shaft Power
FP
e fan  X %
BP
e fan 
STATIC EFICIENCY
Static Power
x 100%
Brake or Shaft Power
SP
e Static 
X%
BP
e Static 
Where
BP - Brake or shaft power in KW
MOTOR EFFICIENCY
Brake or Shaft Power
x 100%
Motor Power
BP
eMotor 
X%
MP
eMotor 
MOTOR POWER: (For motor driven Fan)
a. For Single Phase Motor
EI (Power Factor)
KW
1000
where
MP 
E - Volts
I - Current Amperes
PF - Power Factor
b.
For 3 – Phase Motor
MP 
SPECIFIC FAN POWER
3 EI (Power Factor)
KW
1000
AFFINITY LAWS FOR FANS
A. Variation in speed and impeller diameter
Q  ND3
H  N2D2
B. Variation in impeller Speed
Q  N ; H  N2 ; Power  N3
C. Variation in impeller size; Tip speed = C ;  = C and same proportions; H = C
Q  D2 ; Power  N2 ; N  1/D
D. Variation in impeller size; N = C;  = C ; Same proportions
Q  D3 ; Power  D5 ; H  D2 ; Tip Speed  D
E. Variation in density; Q = C; N =C; D = C; system = C
H   ; Power  
F. Variation in Density; D = C; H = C
1
Q
ρ
Power 
1
N
ρ
1
ρ
G. Variation in density; m = C;D = C; system = C
Q
1
1
1
; H ; N ;
ρ
ρ
ρ
Power 
COMMON LAWS FOR ACTUAL FAN APPLICATION
A. Variation in speed and constant impeller diameter
Q 2  N2 
 
Q1  N1 
h 2  N2 
 
h1  N1 
2
Power2  N2 
 
Power1  N1 
3
B. Variation in impeller Diameter and Constant Speed
Q2  D2 
 
Q1  D1 
h2  D2 
 
h1  D1 
2
Power2  D2 
 
Power1  D1 
3
1
ρ2
SAMPLE PROBLEMS
1.
Find the air power of an industrial fan that delivers 25 m3/sec of air through a duct 90 cm x 120 cm outlet. Static pressure is
127 mm WG. Air temperature is 21C and barometric pressure is 760 mm Hg.
A  0.90(1.20)  1.08 m2
v
Q
m
 23.15
A
sec
v2
 hva  27.31m of air
2g
kg
ρ w  1000 3
m
P
kg
ρ air 
 1 .2 3
RT
m
h (ρ )
h vw  va air  0.03 m of H2O
ρ water
h tw  hsw  h vw  0.16 m of WG
FP  Q( γ w ater )h tw  39.2 KW
2.
3.
A fan filled with only a discharge duct delivers 2400 m3/min of standard air against a total pressure of 10.8 cm of WG when
the static pressure is 10.2 cm WG. The temperature of the gage fluid is 40 C ( = 992.26 kg/m3). The fan receives a 60 KW
input. Find the fan efficiency. (70%)
A large centrifugal fan handles 7.5 m3/sec of air from 100 KPa and with an initial velocity of 110 m/sec to discharge
pressure. the velocity at exit is 90 m/sec and the pressure ratio is 3:1. The fan efficiency is 80%. For a gage fluid density of
996 kg/m3, determine the brake power required in KW. (Pair = 100 KPa ; tair = 21 C)
m3
sec
m
m
v 1  110
; v 2  90
sec
sec
Q  7 .5
v  v1
hva  2
 203.87 m of air
2g
P
kg
kg
ρ air 
 1.19 3 ; ρ w ater  992.96 3
RT
m
m
992.96(9.81)
KN
γ w ater 
 9.77 3
1000
m
h ρ
hvw  va air  -0.24 m WG
ρ w ater
2
2
P1  100 KPa
P2
3
P1
P2  300 KPa
hsw 
P2 - P1
 20.47 m WG
γ w ater
htw  20.23 m WG
Fan Power  Qγ w aterhtw  1,482.22
Fan Power
x100%
Brake Power
Brake Power  1,852.8
eFan 
4.
A fan operating with standard air and a system with fixed ducts gives the following performance data: Q = 8.5 m 3/sec; hsw
= 5 cm; N = 583 RPM; BP =6 KW. The fan is to be used with the system unchanged and is to handle the same rated weight
of air when it exists at a temperature of 82C and Pa = 759 mm Hg. Find the necessary performance data under the required
conditions of operation.
m3
sec
hs1  5 cm  0.05 m
Q1  8.5
Q
1
ρ
H
N1  583 RM
N
1
ρ
Power 
BP1  6 KW
For standard air
P  101.325 KPa
T  21 273  294K
kg
m3
at T  82  273  355K
ρ 21 C  1.2
P  759 mm Hg  101.2 KPa
ρ 82C  0.99
kg
m3
1
ρ
Q 2 = 10.3 m3/sec
N2 = 707 RPM
h sw 2 = 6.06 cm WG
BP2 = 8.82 KW
1
ρ2
A fan filled with only a discharge duct delivers 2400 m3/min of standard air against a total pressure of 10.8 cm of WG when
the static pressure is 10.2 cm WG. The temperature of the gage fluid is 40 C ( = 992.26 kg/m3). The fan receives a 60 KW
input. Find the fan efficiency. (70%)
6. An induced draft fan handles 1700 m3/min of flue gas at apparent molecular weight of 30 from a boiler at 260 C against a
static pressure of 10.2 cm WG. The discharge duct has an area of 2.8 m 2 and a total fan efficiency of 65%. Determine the
power required to drive the fan. (45KW)
7. Compute the required motor capacity needed for the forced draft fan under the following conditions:
Coal rate - 10 metric tons /hr
Coal analysis: C = 78%;H2 = 3%;O2 = 3% ;S = 1% ;M = 7%; A= 8%
Excess air = 30%
Plenum chamber pressure = 18 cm H2O
Atmospheric condition - P = 101.325 KPa ; t = 21 C
Mechanical efficiency = 60%
( BP = 87 KW)
8. Air enters a fan through a duct at a velocity of 6.3 m/sec and an inlet static pressure of 2.5 cm of water less than atmospheric
pressure. The air leaves the fan through a duct at a velocity of 11.25 m/sec and a discharge static pressure of 7.62 cm of water
above the atmospheric pressure. If the density of the air is 1.20 kg/m 3 and the fan delivers 9.45 m3/sec, what is the fan
efficiency when the power input to the fan is 13.75 KW at the coupling? (71.81%)
9. Find the motor size needed to provide the FD fan service to a boiler that burns coal at the rate of 10 Metric Tons per hour.
The air requirements are 28 m3/sec air is being provided under 15.24 cm WG by the fan which has a mechanical efficiency
of 60%. Assume fan to deliver at a total pressure of 15.24 cm WG.
10. A fan with the inlet open to the atmosphere delivers 1.5 m3/sec of air at a static pressure of 5.1 cm WG through a duct 28 cm
in diameter, the manometer being attached to the discharge duct at the fan. Air temperature is 21C and the barometer pressure
is 767 mm Hg. Calculate the brake power of the drive motor if the efficiency is 65%.
11. A certain fan delivers 340 m3/min of air at a static pressure of 25.4 mm WG when operating at a speed of 400 RPM and requires
an input of 3 KW. If in the same installation 425 m3/min of air are desired, what will be the new Q, hsw and Fan power
required? (40 mm WG;500 RPM;6 KW )
10. A certain fan delivers 6 m3/sec of air at 21 C and normal barometric pressure at a static pressure of 2.5 cm WG when
operating at 400 RPM and requires 3 KW input. If the air temperature is increased to 93 C and the speed of the fan remains
the same, what will be the new static pressure and power. ( 2.05 cm WG; 2.5 KW )
11. A certain fan delivers 6 cu.m./sec at 21 C and normal barometric pressure at a static pressure of 3 cm WG when operating
at 400 RPM and requires 3 KW. If the speed of the fan is increased so as to produce 3 cm WG static pressure at 93 C, what
will be the new capacity, speed and power needs? (450 RPM; 6.4 m3/sec; 3.4 KW)
12. If the speed of the fan of (8) is increased so as to deliver the same weight of air at 93 C as at 21 C, what will be the new
speed;new capacity; new static pressure and new power? (498 RPM; 7 m3/sec; 3.2 cm; 5 KW)
13. Two forced draft fans in parallel with a capacity of 73 m3/sec each supplying combustion air to a steam generator. Air inlet
is at 45C, a static pressure of 26 cm WG is developed and the fan speed is 1200 RPM. If the fan input is 3 KW, determine
the static efficiency; capacity; head and power requirements for a speed increase of 20%. (71%; 90 m 3/sec;370 RPM; 450
KW)
14. A fan discharges 5 m3/sec of air through a duct 90 cm in diameter against a static pressure of 22 mm WG. The gage fluid
density is 995 kg/m3, the air temperature is 30 C and the barometric pressure is 730 mm Hg. If the power input to the fan
is measured as 3 KW, what is the overall and static efficiency of the fan. (41%; 38%)
15. A fan manufacturer rates his fan at 12 cm of water static pressure for 3 m3/sec of 21C air, 1500 RPM, 760 mm Hg barometric
pressure, 71% static efficiency. What will be the flow, static pressure, and brake power at 95 C air and 660 mm Hg
barometric pressure. (4 m3/sec; 12 cm of water; 6 KW)
16. A fan manufacturer rates his fan at 15.2 cm WG static pressure for 10 m3/sec of 21C air at 1200 RPM and static efficiency
of 69%. At what speed should this fan operate to developed 13 cm WG static pressure when the temperature is 316C. Also
find the capacity and brake power at this temperature. ( 1571 RPM; 13.1 m3/sec; 24.2 KW)
17. A fan is driven by a belt and pulley so that its speed can be changed by changing the pulley diameter. The fan is rated at 4
m3/sec and 127 mm WG static pressure for 1700 RPM and 15.56C. The static efficiency is 73%. What is the maximum
speed at which the fan can operate without overloading a 7.46 KW motor if the temperature is 93 C. (1895 RPM)
18. A fan whose static efficiency is 40% has a capacity of0.47 m 3/sec at 15.56 C and barometer of 762 mm Hg and gives a
static pressure of 5.081 cm of WG on full delivery. What size of electric motor should be used to drive this fan? ( 0.784 HP)
19. A fan discharges 5 m3/sec through a duct 90 cm in diameter against a static pressure of 20 mm WG. The gage fluid density
is 995 kg/cu.m. The air temperature is 30 C and the barometric pressure is 730 mm Hg. If the power input to the fan is
measured to be 3 KW, what is the overall efficiency of the fan. (38.3)
20. A certain fan delivers 6 cu.m./sec at a static pressure of 25 mm WG when operating at a speed of 400 RPM and requires an
input of 3 KW. If in the same installation 7 cu.m./sec are desired, what will be the power input required in KW. (4.8)
21. What capacity would you recommend for a forced draft fan to be used with a boiler operating under the following conditions:
Maximum boiler rating = 137 000 kg/hr
Steam pressure = 6 MPa
Steam temperature = 480 C
Feedwater temperature = 150 C
Boiler efficiency = 90%
Ambient air temperature = 38 C
Flue gas exit temperature = 150C
Assume 15% excess air is required for combustion. Heating value of fuel is equal to 43 000 KJ/kg. Allow 20%
excess capacity to cover air leakages and slagging. The ultimate analysis of the fuel are: C = 83%; H = 12.7%; O =
1.2%; N = 1.6%; S = 0.4%; M = 1%; Ash= 1%.
5.
21.A fan operating with standard air and a system with fixed ducts gives the following
performance data: Q = 8.5 m3/sec; hsw = 5 cm; N = 583 RPM; BP =6 KW. The
fan is to be used with the system unchanged and is to handle the same rated weight of air when it exists at a temperature of 82
C and Pa = 759 mm Hg. Fin the necessary performance data under the required conditions of operation. (10.3 m3/sec; 707
RPM; 8.82 KW)
BLOWERS
Blower is a machine used to compressed air or gas by centrifugal force to a final pressure not exceeding 241 KPa gage. Usually blower
has no cooling system or it is not water cooled.
COMPRESSION OF GASES
The design of blower is usually based upon either an adiabatic or isothermal compression.
A. Isothermal Compression (PV = C) or (T1 = T2 = T)
W (Work/Power
of Blower
P
P2 T2
P1 T1
P2
m
m
P1 V1  P2 V2
T1  T2
P
W  P1 V1 ln 2
P1
V1  Q
P
Wc  P1 Q ln 2
P1
P1 Q  mRT
W  QH
H  IsothermalHead
g

1000
P

RT
Pg

(1000)RT
 P1 g 
P
H
P1Q ln 2  Q 
P1
 (1000)RT1 
H
(1000)RT1  P2 
 ln  meters
g
 P1 
2
W    VdP
PV  C
P1
1
V
B. Isentropic Compression (PVk = C; 0r S = C)
P1 V1k  P2 V2k
T2  P2 
 
T1  P1 
k 1
k
Q 0
W  Q  h  KE  PE
W    VdP KE  PE
For a compressor work is done
on the system;let - W  Wc
Wc  Ideal compressor work/Power
 W  WC  h  Q  KE  PE
 W  WC   VdP  KE  PE
k 1


kP1 V1  P2  k
 VdP  k  1  P1   1 KW


v 2 2  v 12
KE 
KW
2(1000)
mgz2  z1 
PE 
KW
1000
k 1


kP1 V1  P2  k
   1  KE  PE
Wc 

k  1  P1 


If KE & PE are negligible
k 1


kP1 V1  P2  k
   1
Wc 


k  1  P1 


P1 V1  mRT1
Q0
k 1




kP1 V1  P2 k
   1  QH
Wc 

k  1  P1 


V1  Q

P1 g
(1000)RT1
k 1


 P1 g 
kP1Q  P2  k
H 
   1
Q 


k  1  P1 
 (1000)RT1 


k 1




1000kRT1  P2 k
   1  Adiabatic Head
H


g(k  1)  P1 


EFFICIENCY
A. Adiabatic Efficiency
Isentropic Work
x 100%
Actual
Work
B. Isothermal efficiency
ek 
eI 
IsothermalWork
x 100%
Actual Work
RATIO OF THE ADIABATIC TEMPERATURE RISE TO THE
ACTUAL TEMPERATURE RISE
Y
 P  K 1K

2
T1  
 1
 P1 



T  T 
'
2
1
RELATIONSHIP FOR CORRECTING PERFORMANCE CURVES
1. Volume Flow
Q B NB

Q A NA
2. Weight Flow
m B  NB   P1B   T1A 




m A  N A   P1A   T1B 
3. Pressure Ratio
 P  K 1K 
 2 
 1
 P1 



4. Head
2
 NB   T1A 
  


N
T
 P  K 1K 
 A   1B 
 2 


1
 P1 


A
P2
 rp (pressureratio)
P1
B
HB NB2
 2
HA NA
5. Brake Power
3
BPB  NB   P1B  T1A 
 



BPA  N A   P1A  T1B 
 P  K 1K

 2 
 1
 P1 

BPB  P1B  Q B  
B


 
K 1
BPA  P1A  Q A   P  K

 2 
 1
 P1 



A
1 - suction
2 - discharge
A - 1st condition
B - 2nd condition
R - gas constant, KJ/kg-K
P - absolute pressure in KPa
 - density, kg/m3
T - absolute temperature, K
 - specific weight, KN/m3
Q - capacity, m3/sec
BP - brake power, KW
N - speed, RPM
W - work, KW
m - mass flow rate, kg/sec
H - head, m
PREPARED BY: YURI G. MELLIZA
SAMPLE PROBLEMS
1. A radial type Boeing 747 blower is designed to deliver 1.3 m3/sec of air at a pressure of 107 KPa when operating at an altitude of
4000 m where the temperature is – 10 C and thepressure is 60 KPa. It rotates at 18 000 RPM and is to have an adiabatic
efficiency of 72%. It is to be tested at sea level ( P= 101 KPa and t = 21 C) at a speed of 14 000 RPM.Assuming that the efficiency
at the design point does not change, determine for thedesign under test conditions the power required to drive it in KW. ( 61 KW)
2. A blower compresses 0.02 m3/sec of gaseous mixture that has the following volumetric analysis: CO2 = 11% ; O2 = 8% ; CO =
1% ; N2 = 80% ; from 101 Kpa and 27 C to 180 KPa. Assuming the compression to be isentropic, find the work in KW.
For CO2 :
For O2 :
For CO:
For N2 :
M =44
M = 32
M = 28
M = 28
K = 1.288
K = 1.295
K = 1.399
K = 1.399
ANSWER: W = 1.3 KW
3. It requires 41 KW to compress 28 m3/min of air at 15.56C and 101.325 KPa to a pressure of 69 KPag. The temperature of the air
leaving the blower is 85C.
a) what is the flow in m3/min from the blower discharge? (20.654 m3/min)
b) What is the head of the blower in meters on an adiabatic basis. (4727 m)
c) What is the head of the blower in meters on an isothermal basis. (4385 m)
d) What is the value of n for the compression. ( n = 1.709)
e) What power would be required if the compression were adiabatic. (26.5 KW)
f) What power would be required if the compression were isothermal. (24.56 KW)
g) What is the adiabatic efficiency. (64.6%)
h) What is the isothermal efficiency. (60%)
i) What is the adiabatic temperature rise ratio. (0.665)
4. (A) A blower operating at 15000 RPM compresses air from 20C and 101.325 KPa to 69 KPag. The design flow is 38.3 m3/min
and at this point the brake power is 60 KW. Determine the efficiency of the blower at the design point. (60.4%)
(B) If the discharge pressure of the blower is increased to 83 KPag, find the new speed and the corresponding values of Q and
Brake power. (16 379 RPM; 41.8 m3/min; 78.12 KW)
(C) If the suction of the blower is throttled to 83 KPa, determine the head and the discharge pressure in KPag, the flow (referred
to 20C and 101.325 KPa) and the brake power. (4799.4 m; 38.195 KPag; 31.4 m 3/min; 49.15 KW)
(D) If the temperature at the inlet of the blower is reduced to 5C, determine the head, discharge pressure in KPag, the flow
(referred to 20C and 101.325 KPa) and the brake power.
5. A booster blower takes 340 m3/min of natural gas having a volumetric chemical composition of 15% C 2H6 and 85% CH4 at a gage
pressure of 13.8 KPa and a temperature of 21C and compresses it to 63 KPagage. What power is required if the adiabatic
efficiency of the blower is 60%. (BP = 445.2 KW)
For C2H6: M = 30; Cp = 1.7549 KJ/kg-K; Cv = 1.4782 KJ/kg-K
For CH4: M = 16; Cp = 2.1377 KJ/kg-K; Cv = 1.6187 KJ/kg-K
6. A two stage radial type airplane supercharger is designed to deliver 80 kg of air per minute at a pressure of 107 KPa when
operating at an altitude of 6000 m where the temperature is -20C and the pressure is 46 KPa. It rotates at 20,000 RPM and is to
have an adiabatic efficiency of 70%. It is to be tested at sea level ( P = 101.325 KPa; t = 21C) at a speed of 12,000 RPM.
Assuming that the efficiency at the design point does not change, determine for the design point under test conditions:
a) the cu. m. of air taken in per minute
b) the discharge pressure in KPa
c) the brake power required for driving
7. A blower is design to draw in 0.5 m3/sec of CO2 at 32C and 122 KPa and compresses it to 138 KPa when operating at 4000
RPM. It is to be tested with air at 101 KPa and 21C and driven by a motor running at 3550 RPM. Determine the flow and
discharge pressure which the machine should deliver at the design point on test to be acceptable.
8. A large centrifugal fan handles 7.5 m3/sec of air from 100 KPa and with an initial velocity of 110 m/sec to discharge pressure.
the velocity at exit is 90 m/sec and the pressure ratio is 3:1. The fan efficiency is 80%. For a gage fluid density of 996 kg/m3,
determine the brake power required in KW.
9. A fan operating with standard air and a system with fixed ducts gives the following performance data: Q = 8.5 m3/sec; hsw = 5
cm; N = 583 RPM; BP =6 KW. The fan is to be used with the system unchanged and is to handle the same rated weight of air
when it exists at a temperature of 82 C and Pa = 759 mm Hg. Find the necessary performance data under the required
conditions of operation.
ME 514P
Quiz # 1
1. A large centrifugal fan handles 7.5 m3/sec of air from 100 KPa and with an initial velocity
of 110 m/sec to discharge pressure. the velocity at exit is 90 m/sec and the pressure ratio
is 3:1. The fan efficiency is 80%. For a gage fluid density of 996 kg/m3, determine the
brake power required in KW.
2. A fan operating with standard air and a system with fixed ducts gives the following
performance data: Q = 8.5 m3/sec; hsw = 5 cm; N = 583 RPM; BP =6 KW. The fan is to be
used with the system unchanged and is to handle the same rated weight of air when it
exists at a temperature of 82 C and Pa = 759 mm Hg. Find the necessary performance data
under the required conditions of operation.
---------------------------------------------------------------------------------------------------#1:
Q = 7.5 m3/sec
P1 = 100 KPa
v1 = 110 m/sec
v2 = 90 m/sec
rp = 3:1
F = 0.80
w = 996 kg/m3
P2 = 100(3) = 300 KPa
w = 996(9.81)/1000 = 9.77 KN/m3
a = 1.2 kg/m3
a = 11.772 x 10-3 KN/m3
#2:
Given:
Q1 = 8.5 m3/sec
hsw1 = 5 cm WG
BP1 = 6 KW
1
ρ
1
N
ρ
Q
H
1
ρ
Power 
1
ρ2
For standard air:
P = 101.325 KPa ; t = 21 C ;
21 = 1.2 kg/m3
P = 759 mm Hg = 101.2 KPa
82 = 0.99 kg/m3
Q2 = 10.3 m3/sec
N2 = 707 RPM
hsw2 =6.06 cm WG
BP2 = 8.82 KW
A radial type Boeing 747 blower is designed to deliver 1.3 m3/sec of air at a pressure of
107 KPa when operating at an altitude of 4000 m where the temperature is – 10 C and the
pressure is 60 KPa. It rotates at 18 000 RPM and is to have an adiabatic efficiency of
72%. It is to be tested at sea level ( P= 101 KPa and t = 21 C) at a speed of 14 000 RPM.
Assuming that the efficiency at the design point does not change, determine for the
design under test conditions the power required to drive it in KW. ( 61 KW)
Solution:
Condition A:
k = 0.72
K 1
Q = 1.3 m3/sec

 B:
Condition
kP1Q  P2  P1 =K 101 KPa
P1 = 60 KPa
 
WA 
1
h = 4000 m
 P1  T1 = 294 K
k

1
T1 = 263 K
N = 14000

R
P2 = 107 KPa
WA  49 KW
N = 18000 RPM
BPA 
49
 68 KW
0.72
INDUSTRIAL PLANT DESIGN
1. A fan discharges 5 m3/sec of air through a duct 90 cm in diameter against a static
pressure of 22 mm WG. The gage fluid density is 995 kg/m 3, the air temperature is 30 C
and the barometric pressure is 730 mm Hg. If the power input to the fan is measured as 3
KW, what is the overall and static efficiency of the fan. (41%; 38%)
2. A fan manufacturer rates his fan at 12 cm of water static pressure for 3 m 3/sec of 21C
air, 1500 RPM, 760 mm Hg barometric pressure, 71% static efficiency. What will be the
flow, static pressure, and brake power at 95 C air and 660 mm Hg barometric pressure.
(4 m3//sec; 12 cm of water; 6 KW)
3. A fan manufacturer rates his fan at 15.2 cm WG static pressure for 10 m 3/sec of 21C air
at 1200 RPM and static efficiency of 69%. At what speed should this fan operate to
developed 13 cm WG static pressure when the temperature is 316C. Also find the
capacity and brake power at this temperature. ( 1571 RPM; 13.1 m3/sec; 24.2 KW)
4. A fan is driven by a belt and pulley so that its speed can be changed by changing the
pulley diameter. The fan is rated at 4 m3/sec and 127 mm WG static pressure for 1700
RPM and 15.56 C. The static efficiency is 73%.What is the maximum speed at which the
fan can operate without overloading a 7.46 KW motor if the temperature is 93 C.
(1895 RPM)
5. A 2-stage, double acting air compressor 41cm and 25cm bore x 18 cm running 600 RPM has
a free air unloader at each end for capacity control. It is driven by a 150 HP motor, 460V,
3 phase, 60 hertz, 1175 RPM thru superhigh capacity V-belts at sea level installation. P1 =
101 KPa and P4= 984 KPa, calculate the power assuming the LP cylinder discharges the air to
the ideal pressure and intercooler cools the air to the initial temperature of 21 C.
6. A single acting compressor has a volumetric efficiency of 87% operates at 500 RPM. It
takes in air at 100 KPa and 30 C and discharges it to 600 KPa. The air handled is 6 m3/min
measured at discharge condition. If compression and expansion processes are isentropic ,
find the piston displacement per stroke.
7. A double acting, 2 stage air compressor running at 150 RPM has a suction pressure and
temperature of 100 KPa and 27 C, respectively. The low pressure cylinder is 36 cm x 38
cm and connected in tandem to the high pressure cylinder. The LP cylinder discharges the
air at 386 KPa. The intercooler cools back the air to its initial temperature and the air
enters the HP cylinder at 370 KPa and leaves at 1 480 KPa. Both LP and HP cylinders have
4% clearance and compression follows PV1.3= C. Neglect effect of piston rods. Standard air
is at 101 KPa and 16 C. Determine:
a) the volume of free air based on apparent volumetric efficiency
b) the heat removed by cooling water in the intercooler ( 24 KW)
c) the diameter of the HP cylinder in cm (19 cm)
d) the power required to drive the compressor if the aggregate losses is 20 % of the
power of the conventional diagram. (72 KW)
8. A turbine driven compressor handles 10 kg/sec of air from 100 KPa to 600 KPa with an
inlet temperature of 300 K and a discharge temperature of 530 K. The inlet tubing has a
0.5 m inside diameter and the discharge piping a 0.20 m diameter. The compression is
adiabatic. Determine:
a) the air inlet and exit velocities in m/sec
b) the isentropic compression efficiency
c) the power required in KW
9. A single cylinder, double acting air compressor running at 200 RPM has a piston speed of
183 m/min. It compresses 27 kg/min of air from 97 KPa and 16 C to 654 KPa. Clearance is
5.5%. For isentropic compression, Find:
a) the volumetric efficiency (84%)
b) the piston displacement in m3/min ( 28 cu.m./min)
c) the work in KW ( 96 KW)
d) the mean effective pressure in KPa (207 Kpa)
e) the bore and stroke of the compressor in cm. (D=44 cm;L=46 cm)
10. An atmospheric cooling tower has an efficiency of 50% when handling 4 L/s of water. It is
connected to the water cooling system of an engine developing 56 KW. If the wet bulb
temperature is 21 C, what are the temperature of the water entering and leaving the
cooling tower. Heating value of fuel is 43 000 KJ/kg, fuel consumption is 0.24 kg/KW-hr,
1
27 % of total heat value of fuel goes to jacket water. Assume no heat loss in connections.
( 27C; 24C)
11. A cooling tower receives 3.2 L/sec of water at 46 C. Atmospheric air at 16C DB and 55%
RH enters the tower at 3 m3/sec and leaves at 32 C saturated.
Determine:
a) the volume of water leaving the tower in L/sec ( 3.07 L/sec)
b) exit temperature of water in C. (25 C)
12. A cooling tower receives 6 kg/sec of water at 60 C. Air enters the tower at 32C
DB and 27 C WB temperature and leaves at 50 C and 90% RH.The coolingtower
efficiency is 61% and make up water is available at 30C. Determine:
a) the mass flow of air entering ( 3.41 kg/sec)
b) the quantity of make up water required ( 0.191 kg/sec)
13. It is desired to designed a drying plant to have a capacity of 680 kg/hr of product
3.5% moisture content from a wet feed containing 42% moisture. Fresh air at 27C
with 40% RH will be preheated to 93 C before entering the dryer and will leave the
dryer with the same temperature but with a 60% RH. Find:
a) the amount of air to dryer in m3/sec ( 0.25)
b) the heat supplied to the preheater in KW (16)
At 27 C DB and 40% RH
At 93 C and W = .0089 kgm/kgda
W = .0089 kgm/kgda
h = 117.22 KJ/kgda
h = 49.8 KJ/kgda
 = 1.05 m3/kgda
At 93 C and 60% RH
W = 0.54 kgm/kgda
h = 1538.94 KJ/kgda
14. A 400 kg/hr of ceramic powder is to be produced. The powder has a specific heat of
0.921 KJ/kg-C and a density of 1568 kg/m3. Initial moisture content is 19% and
final moisture content is 5%. A continuous belt dryer is used. the drying time is 45
minutes at 71C DB and 38C WB. Product temperature is 38 C. Compute:
a) the weight of moisture to be removed and weight of material entering the
dryer
b) the weight of air required in kg/hr if make up air is at 27C DB and
W = 0.0168 kgm/kgda.
c) heat required by the dryer
15. A rotary dryer is fired with bunker oil of 41 870 KJ/kg HHV is to produce 20 metric
tons per hour of dried sand with 0.5% moisture from a wet feed containing 7%
moisture, specific heat of sand is 0.879 KJ/kg-C, temperature of wet feed is 30 C
and temperature of dried product is 115 C.
a) Calculate the kg/hr of wet feed
b) Calculate the kg/hr of moisture removed from sand
c) Calculate the heat in KJ/hr required
d) Calculate the L/hr of bunker oil consumed if the specific gravity of bunker
oil is 0.90 and dryer efficiency of 60%
16. An industrial furnace needs an actual draft of 64 mm of water column. The frictional
losses in the stack or chimney are 15% of the theoretical draft. Assume that the
flue gas average temperature is 149C with average molecular weight of 30. Outside
air temperature is 10C and barometer reading of 759.5 mm Hg, calculate the height
of the stack or chimney in meters.
a) 196
b) 96
c) 106
d) 276
17. A blower is designed to draw in 0.5 m3/sec of CO2 at 32 C and 122 KPa and compress
it to 136 KPa when operating at 4000 RPM. It is to be tested with air at 101 KPa and
21 C and driven by a motor running at 3550 RPM. Determine the flow and discharge
pressure which the machine should deliver at the design point on test to be acceptable.
For CO2:
R = 44
k = 1.288
2
ME 514P
Semi Final Exam
1. A certain fan delivers 6 cu.m./sec at 21 C and normal barometric pressure at a static
pressure of 3 cm WG when operating at 400 RPM and requires 3 KW. If the speed of the
fan is increased so as to produce 3 cm WG static pressure at 93 C, what will be the new
capacity, speed and power needs?
2. If the speed of the fan of (1) is increased so as to deliver the same weight of air at 93 C
as at 21 C, what will be the new speed; new capacity; new static pressure and new power?
3. A single cylinder, double acting air compressor running at 200 RPM has a piston speed of
183 m/min. It compresses 27 kg/min of air from 97 KPa and 16 C to 654 KPa. Clearance is
5.5%. For isentropic compression, Find:
a) the volumetric efficiency
b) the piston displacement in m3/min
c) the work in KW
d) the mean effective pressure in KPa
e) the bore and stroke of the compressor in cm.
4. A two stage, reciprocating air compressor is required to deliver 0.70 kg/sec of air from
986 KPa and 32 C to 1276 KPa. The compressor operates at 205 RPM , compression and
expansion processes follow PV1.25 = C, and both cylinders have 3.5% clearance. There is a
20 KPa pressure drop in the intercooler. The LP cylinder discharges at the optimum
pressure into the intercooler. The air enters the HP cylinder at 37 C. The intercooler is
water cooled, with the water entering at 22 C and leaving at 32 C. Determine:
a) the free air capacity in m3/sec
b) the LP and HP discharge temperature
c) the optimum interstage pressure
d) the cooling water required for the intercooler in kg/sec
e) the theoretical power required in KW
f) the LP cylinder dimensions if L/D = 0.70
g) the motor requirements in KW, if the adiabatic compression efficiency is 83%
and mechanical efficiency is 85%.
PREPARED BY: ENGR. YURI G. MELLIZA
TO BE CONTINUED……………….
3
4