THERMODYNAMICS 1
(MEEN 30024)
DR. DANTE V. GEDARIA
Chapter 1
BASIC PRINCIPLES, CONCEPTS AND
DEFINITIONS
Thermodynamics is that branch of the physical sciences that treats of various
phenomena of energy and the related properties of matter, especially of the laws of
transformation of heat into other forms of energy and vice versa.
Engineering Thermodynamics – is a branch of Thermodynamics where the emphasis is
focused on the engineering analysis and design of process devices and systems which
involve the beneficial utilization of energy and material.
Systems of Units
Newton’s law states that “the acceleration of a particular body is directly
proportional to the resultant force acting on it and inversely proportional to its mass.”
kF
ma
F=
k
a=
m
ma
k=
F
k is a proportionality constant
Systems of units where k is unity but not dimensionless:
cgs system: 1 dyne force accelerates 1 g mass at 1 cm/s ²
mks system: 1 newton force accelerates 1 kg mass at 1 m/s²
fps system: 1 lb force accelerates 1 slug mass at 1 ft/s²
1 𝑔𝑚
1 dyne
1 cm/s²
K=
𝑔𝑚 .cm
dyne.s²
1 𝑘𝑔𝑚
1 slug
1 newton
1 m/s²
1𝑘𝑔𝑚 .m
K=
newton.s²
Systems of units where k is not unity:
1
1ft/s²
K=
1 slug.ft
𝑙𝑏𝑓 .s²
If the same word is used for both mass and force in a given system, k is neither unity nor
dimensionless.
1 lb force accelerates a 1 lb mass at 32.174 ft/s²
1 g force accelerates a 1 g mass at 980.66 cm/s²
1 kg force accelerates a 1 kg mass at 9.8066 m/s²
1 𝑙𝑏𝑚
1𝑔𝑚
1𝑙𝑏𝑓
32.174 ft/s²
1g
980.66 cm/s²
𝑙𝑏𝑚 .ft
9.8066 m/s²
k = 980.66
𝑙𝑏𝑓 .s²
𝑔𝑓 .s²
kgmm
Therefore,
𝑘𝑔𝑚 .m
k=
N.s²
=
𝑘𝑔𝑓 .s ²
kgm.m
kgf.s²
1𝑘𝑔𝑓 = 9.8066 N
Relation between pound mass (𝑙𝑏𝑚 ) and slug
k=
slug.ft
𝑙𝑏𝑚 .ft
k=
𝑙𝑏𝑓 .s²
𝑙𝑏𝑓 .s²
Therefore,
1 slug = 32.174 𝑙𝑏𝑚
2
𝑘𝑔𝑚 m
k= 9.8066
Relation between kilogram force (𝑘𝑔𝑚 ) and Newton (N)
kgm.m
N.s²
1 1𝑘𝑔𝑓
𝑔𝑚 .cm
k = 32.174
k =
1 𝑘𝑔𝑚
𝑘𝑔𝑓 .s²
Relation between gm force (𝑔𝑚𝑓 ) and dyne
gm-cm
k= 1
gm-cm
k= 980.66
dyne-s²
Therefore,
1
gm-cm
dynes-s²
= 980.66
gmf-s²
gm-cm
gmf-s²
1 𝑔𝑚𝑓 = 980.66 dynes
Acceleration
A unit of force is one that produces unit acceleration in a body of unit mass.
1 𝑙𝑏𝑚
F=
m
1 poundal
1 ft/s²
a
k
1 poundal = (1𝑙𝑏𝑚 ) (1 ft/s²)
F is force in pondals
m/k is mass in pounds
a is acceleration in ft/s²
1 lbf
1 slug
ma
F=
k
1𝑙𝑏𝑓 . s²
1 pound = (1 slug) (1 ft/s²); 1 slug =
ft
3
1 ft/s²
F is force in pounds
m/k is mass in slugs
a is acceleration in ft/s²
Mass (m) and Weight (𝑭𝒈 )
The mass of a body is the absolute quantity of matter in it.
The weight of a body means the force of gravity 𝑭𝒈 on the body.
m
k
where:
F
=
𝑭𝒈
a
=
g
g = acceleration produced by force 𝑭𝒈
a = acceleration produced by another force F
k = standard acceleration
At or near the surface of the earth, k and g are numerically equal, so are m and 𝑭𝒈
Problems:
1. What is the weight of a 66-𝑘𝑔𝑚 man at standard condition? Express your
answer in 𝑘𝑔𝑓 and in Newton.
Solution
m = 66 𝑘𝑔𝑚
g = 9.8066 m/s²
m
66 𝑘𝑔𝑚
9.8066
mg
Fg =
s²
9.8066N
= 66 𝑘𝑔𝑓
=
𝑘𝑔𝑚 .m
k
x
𝑘𝑔𝑓
9.8066
𝑘𝑔𝑓 -s²
Fg = 647.23 N
4
2. The weight of an object is 50 lb. What is its mass at standard condition?
Express your answer in lbm and slugs.
Solution
g = 32.174 ft/s²
𝑭𝒈 = 50 𝑙𝑏𝑓
50𝑙𝑏𝑓
𝑙𝑏𝑚 -ft
32.174
lbf-s²
Fg k
m=
g
=
= 50lbm x
ft
32.174
slug
32.174 𝑙𝑏𝑚
s²
3. Five masses in a region where the acceleration due to gravity is 30.5 ft/s² are as
follows: m1 is 500 g of mass; m2 weighs 800 gf; m3 weighs 15 poundals; m4 weighs 3 lbf; m5
is 0.10 slug of mass. What is the total mass expressed (a) in grams, (b) in pounds, and (c) in
slugs.
Solution
g = (30 ft/s²) (12 in/ft) (2.54 cm/in) = 929.64 cm/s²
𝑔𝑚 .cm
800 𝑔𝑓
980.66
𝑔𝑓 .s²
𝐹𝑔2 k
(a) 𝑚2 =
= 843.91 𝑔𝑚
=
G
cm
929.64
s²
𝑙𝑏𝑚 .ft
15
𝑚𝑎
k
=
𝐹𝑔3
s²
g
ft
= 0.49𝑙𝑏𝑚 453.6
𝑔𝑚
= 222.26 𝑔𝑚
𝑙𝑏𝑚
30.5
s²
5
𝑙𝑏𝑚 .ft
3 𝑙𝑏𝑓 32.174
𝑚4 =
𝐹𝑔4 k
g
𝑙𝑏𝑓 s²
=
30.5
453.6
ft
𝑔𝑚
= 1435.49 𝑔𝑚
𝑙𝑏𝑚
s²
𝑚5
𝑙𝑏𝑚
= (0.10 slug) 32.174
k
Total mass
slug
453.6
𝑔𝑚
= 1459.41𝑔𝑚
𝑙𝑏𝑚
= 𝑚1 + 𝑚2 + 𝑚3 +𝑚4 +𝑚5
= 500+843.91+ 222.26+1435.49+1459.41
= 4461.07 𝑔𝑚
(b) Total mass =
4461.07 𝑔𝑚
453.6
= 9.83 𝑙𝑏𝑚
𝑔𝑚
𝑙𝑏𝑚
9.83 𝑙𝑏𝑚
(c) Total mass =
32.174
= 0.306 slug
𝑙𝑏𝑚
slug
4. Note that the gravity acceleration at equatorial sea level is g = 32.088 fps² and
that its variation is – 0.003 fps² per 1000 ft ascent. Find the height in miles above this point
for which (a) the gravity acceleration becomes 30.504 fps², (b) the weight of a given man is
decreased by 5%. (c) What is the weight of a 180 lbm man atop the 29, 131-ft Mt. Everest in
Tibet, relative to this point?
Solution
(a) Change in acceleration = 30.504 – 32.088 = -1.584 fps²
-1.584 fps²
Height, h =
= 528,000 ft or 100 miles
-0.003fps²
1000 ft
6
(b) F = 0.95 Fg
.a
Let Fg = weight of the man at sea level
𝐹𝑔
F
=
h
a
g
0.95 𝐹𝑔 𝐹𝑔
=
a
g
a = 0.95g = (0.95) (32.088) = 30.484 fps²
. Fg
g = 32.088 fps²
h=
(30.484 – 32.088) fps²
= 534,670 ft or 101.3 miles
0.003 fps²
1000 ft
(c)
F
a
29.131 ft
𝐹𝑔
g = 32.088 fps²
m = 180 𝑙𝑏𝑚
29.131 ft
a = 32.088 fps² -
1000 ft
180 𝑙𝑏𝑚 32.001
ma
F=
0.003 fps² = 32.001 fps²
ft
s²
=
k
32.174
𝑙𝑏𝑚 .ft
= 179.03𝑙𝑏𝑓
𝑙𝑏𝑓 . s²
7
Specific Volume (√), Density (ρ) and Specific Weight (µ)
The density ,ρ, of any substance is its mass (not weight) per unit volume.
m
ρ=
v
The specific volume ,v, is the volume of a unit mass.
V
v=
1
=
m
ρ
The specific weight ,γ, of any substance is the force of gravity on unit volume.
𝐹𝑔
µ=
V
Since the specific weight is to the local acceleration of gravity as the density is to the
standard acceleration, γ/g = ρ/k, conversion is easily made;
k
ρg
ρ=γ
or γ
g
k
At or near the surface of earth, k and g are numerically equal, so are ρ and γ.
Problems
1. What is the specific weight of water at standard condition?
Solution
g = 9.8066 m/s²
(1000 𝑘𝑔𝑚 /m³) (9.8066m/s²)
ρg
γ=
ρ = 1000 𝑘𝑔𝑚 /m³
= 1000 𝑘𝑔𝑓 /m³
=
k
9.8066𝑘𝑔𝑚 .m
𝑘𝑔𝑓 .s²
8
2. Two liquids of different densities (ρ1 = 1500 kg/m², ρ2 = 500 kg/m³) are poured
together into a 100-L tank, filling it. If the resulting density of the mixture is 800 kg/m³,
find the respective quantities of liquids used. Also, find the weight of the mixture; local g =
9.675 mps².
Solution
mass of mixture, 𝑚𝑚 = ρ𝑚 𝑣𝑚 = (800 kg/m³) (0.100 m³) = 80 kg
𝑚1 + 𝑚2 = 𝑚𝑚
ρ1 𝑣1 + ρ2 𝑣2 = 𝑚𝑚
1500 𝑣1 + 500 𝑣2 = 80
𝑣1 + 𝑣2 = 0.100
(1)
(2)
Solving equations (1) and (2) simultaneously
𝑣1 = 0.03𝑚3
𝑣2 = 0.07𝑚3
𝑚1 = ρ1 𝑣1 = (1500 kg/m³) (0.03𝑚3 ) = 45 kg
𝑚2 = ρ2 𝑣2 = (500 kg/m³) (0.07𝑚3 ) = 35 kg
Weight of mixture,
m
9.675
𝑚𝑚 g
𝐹𝑔𝑚 =
80 𝑘𝑔𝑚
s²
= 78.93 𝑘𝑔𝑓
=
𝑘𝑔𝑚 .m
k
9.8066
𝑘𝑔𝑓 .s²
Pressure (p)
Pressure (p) is defined as the normal force exerted by the system per unit area. The
standard reference atmospheric or barometric pressure (Po) is 760 mm Hg 1abs or 29.92 in.
Hg 1abs at 32°F, or 14.7 psia, or 1 atm.
9
Measuring Pressure
A. By using manometers
1. OPEN–TYPE MANUMETER – used to measure pressure in flow lines or
vessels.
(a) Absolute pressure is (p) greater than atmospheric pressure (Po).
p0
ℎ𝑔
pg
p
p = absolute pressure
𝑝𝑜 = atmospheric pressure
𝑝𝑔 = gage pressure, the pressure due to the liquid column ℎ𝑔
p = 𝑝𝑜 + 𝑝𝑔
(b) Absolute pressure is less than atmospheric pressure
Atmospheric Pressure (Po) is greater than Absolute Pressure (p)
p = 𝑝𝑜 - 𝑝𝑔
The gage reading is called vacuum pressure or the vacuum.
2. By using pressure gages
A pressure gage is a device for measuring gage pressure.
This picture shows the movement in one type of pressure gage, known as the
single tube gage. The fluid enters the tube through the threaded connection.
As the pressure increases, the tube with an elliptical section tends to
straighten, the end that is nearest the linkage toward the right. The linkage
causes the sector to rotate. The sector engages a small pinion gear. The index
hand moves with the pinion gear. The whole mechanism is of course
enclosed in a case, and a graduated dial, from which the pressure is read, and
is placed under the index hand.
10
Some Definitions
1.
Gage Pressure (pg) – is the defect or excess of absolute pressure (p)over the
barometric or atmospheric pressure (po).
2. Vacuum Pressure (Pvac = pg) – is the defect or excess of atmospheric pressure
(po)over the absolute pressure (p).
Air pressure
+ 𝑝𝑔
Atmospheric pressure
P
- 𝑝𝑔 vacuum
Absolute pressure
𝑝𝑜
(𝑝𝑔 = O, p = 𝑝𝑜 )
(p =𝑝𝑜 - 𝑝𝑔 )
P
Zero absolute or
total vacuum
(p = O, 𝑝𝑔 =𝑝𝑜 )
Fluid Pressure (Gage Pressure)
Open to atmosphere
𝐹𝑔
ℎ𝑔
p = 𝑝𝑜 + 𝑝𝑔
𝐹𝑔 γV γAℎ𝑔
𝑝𝑔 =
=
=
A
A
A
𝑝𝑜
𝑝𝑔
p
𝑝𝑔ℎ𝑔 𝑔ℎ𝑔
𝑝𝑔 = γℎ−𝑔 =
=
k
kv
11
Problem
A 25m vertical column of fluid (density 1878 kg/m³) is located where g= 9.65 mps².
Find the pressure at the base of the column.
Solution
𝑝𝑔 =
gz
g.65
=
Kv
1
m
s²
𝑘𝑔𝑚 -m
N - s²
25 m
KN
1000 N
𝑚32
5.324 x 10−4
𝑘𝑔𝑚
𝑝𝑔 = 424.9 KPa (gage)
Atmospheric Pressure
A barometer is used to measure atmospheric pressure, sometimes called barometric
pressure.
𝑝𝑜
Barometer
𝑝𝑜 = γ ℎ𝑜
where: ℎ𝑜 = the height of column of liquid supported by atmospheric pressure Po
Problems
1. A vertical column of water will be supported to what height by standard
atmospheric pressure.
12
Solution
At standard condition
γ 𝑤 = 62.4 lb/𝑓𝑡 3
14.7
ℎ𝑜 =
𝑝𝑜
lb
144
𝑖𝑛2
=
γ𝑤
62.4
𝑝𝑜 = 14.7 psi
𝑖𝑛.2
𝑓𝑡 2
lb
= 33.9 ft
𝑓𝑡 3
The specific gravity (sp gr) of a substance is the ratio of the specific weight of the
substance to that of water.
γ substance
sp gr =
γ𝑤
or s.g. =
ρ substance
ρ𝑤𝐻20
2. The pressure of a boiler is 9.5 kg/cm². The barometric pressure of the
atmospheric is 768 mm of Hg. Find the absolute pressure in the boiler. (ME
Board Problem – Oct.1987)
Solution
𝑝𝑔 9.5 𝑘𝑔𝑓 /𝑚3
𝑝𝑜 = 768 mm Hg
At standard condition
γ𝑤 = 1000 𝑘𝑔𝑓 /𝑚3
𝑝𝑜 = (γ𝐻𝑔 ) (Z0 ) = (sp gr) 𝐻𝑔 (γ𝑤 ) (Z0 )
(13.6) 1000
𝑘𝑔𝑓
𝑚
(0.768 m)
3
=
= 1.04
10,000
𝑐𝑚2
𝑘𝑔𝑓
𝑐𝑚2
𝑚3
13
p = 𝑝𝑜 + 𝑝𝑔 = 1.04 + 9.5 = 10.54
𝑘𝑔𝑓
𝑐𝑚
abs
2
Absolute Pressure
p=µ.z
where z = 𝑧𝑜 + 𝑧𝑔 , the height of column of liquid supported by absolute pressure p.
If the liquid used in the barometer is mercury, the atmospheric pressure becomes,
𝑝𝑜 = γ 𝐻𝑔 𝑧𝑔 = (sp gr)𝐻𝑔 (γ𝑤 )( Z0 )
(13.6)
lb
62.4
𝑓𝑡
=
1728
3
(ℎ𝑜 in)
𝑖𝑛3
𝑓𝑡 3
𝑝𝑜 = 0.491 ℎ𝑜
where ℎ𝑜
then,
𝑝𝑔
lb
𝑖𝑛2
= column of mercury in inches
lb
= 0.491 𝑧𝑔
𝑖𝑛2
and,
p
= 0.491 h
lb
𝑖𝑛2
14
Problems
1. A pressure gage registers 40 psig in a region where the barometer is 14.5 psia.
Find the absolute pressure in psia, and in kPa.
Solution
p = 14.5 + 40 = 54.5 psia
1𝑘𝑔𝑚
1 slug
1 newton
a = 1 m/ s²
2.205
1 𝑘𝑔𝑚
32.174
m
= 1
s²
𝑙𝑏𝑚
𝑘𝑔𝑚
1 𝑘𝑔𝑚 =
1
a = 1 ft/ s²
m
s²
3.28
= 0.06853 slug
𝑙𝑏𝑚
slug
ft
= 3.28
m
ft
s²
0.06853 slug
F, 𝑙𝑏𝑓
a = 3.28 ft/s²
F=
ma
k
= (0.06863 slug) 3.28
ft
s²
1 newton = 0.2248 𝑙𝑏𝑓
= 0.2248 𝑙𝑏𝑓
1 𝑙𝑏𝑓 = 4.4484 newtons
15
1 𝑙𝑏𝑓
(1 lb)
lb
1
4.4484
N
lb
39.37
in
m
=
𝑖𝑛2
𝑖𝑛2
lb
1
n
= 6895
𝑖𝑛
2
𝑚2
N
lb
𝑚2
p = 54.5
6895
𝑖𝑛
= 375,780 Pa or 375.78 kPa
2
lb
𝑖𝑛2
2. Given the barometric pressure of 14.7 psia (29.92 in. Hg abs), make these
conversations:
(a) 80 psig to psia and to atmosphere,
(b) 20 in. Hg vacuum to in. Hg abs and to psia,
(c) 10 psia to psi vacuum and to Pa,
(d) 15 in. Hg gage to psia, to torrs, and to Pa.
(1 atmosphere = 760 torrs)
Solution
(a) p
= 𝑝𝑜 + 𝑝𝑔 = 14.7 + 80 = 94.7 psia
80 psig
𝑝𝑔
=
= 5.44 atmospheres (gage)
14.7
psia
atm
(b)
Z𝑔 = 20 in.
Z0 = 29.92 psia
z
z = 9.92 in. Hg abs
p = 0.491 z
p = (0.491) (9.92) = 4.87 psia
16
(c)
𝑝𝑔 = 4.7 psi vacuum
𝑝𝑔
𝑝𝑜 = 14.7 psia
𝑝𝑔 = (4.7 psi)
p = 10 psia
6895
Pa
psi
= 32,407 Pa(gage)
(d)
𝑧𝑔 = 15 in.
z
z = 29.92 + 15 = 44.92 in. Hg abs
p = 0.491 z = (0.491) (44.92) = 22.06 psia
𝑝𝑔 =
𝑧𝑜 = 29.92 in
(15) (760)
= 381 torrs
29.92
𝑝𝑔 = 0.491 ℎ𝑔
= 0.491
psi
in.
15 in. 6895
Pa
psi
= 50, 780 Pa (gage)
Temperature
1. Derive the relation between degrees Fahrenheit and degrees Centigrade. (EE
Board Question)
212°F
100°F
t°F
t°C
32°F
0°C
t°F – 32
212 – 32
=
17
t°C – 0
100-0
9
t°F =
t°C + 32
5
9
t°C =
( t°F - 32)
5
Absolute temperature is the temperature measured from absolute zero.
Absolute zero temperature is the temperature at which all molecular motion ceases.
Absolute temperature will be denoted by T, thus
T°R = t°F + 460, degrees Rankine
T°K = t°C + 273, degrees Kelvin
Degrees Fahrenheit (°F) and degrees Centigrade (°C) indicate temperature reading
(t). Fahrenheit degrees (F°) and Centigrade degrees (C°) indicate temperature change or
difference (⧍t).
180 F° = 100 C°
5
1F° =
C°
9
9
1C° =
F°
5
It follows that,
1 F° = 1R°
and
1 C° = 1 K°
18
2. Show that the specific heat of a substance in Btu/(𝑙𝑏𝑚 ) (F°) is numerically equal
to cal/(𝑔𝑚 )(C°).
Solution
Btu
Btu
1
(𝑙𝑏𝑚 ) (F°)
252
cal
Btu
=
𝑔𝑚
454
𝑙𝑏𝑚
F°
𝑙𝑏𝑚
Btu
1
(𝑙𝑏𝑚 ) (F°)
5 C°
9 F°
cal
=1
(𝑔𝑚 ) (C°)
Conservation of Mass
The law of conservation of mass states that mass is indestructible.
The quantity of fluid passing through a given section is given section is given by the
formula
V=A𝑣
Aµ
V
m=
=
v
= A𝑣p
v
Where: V = volume flow rate, V/t
A = cross sectional area of the stream
𝑣 = average speed
m = mass flow rate, m/t
t = time
V = Volume
Applying the law of conservation of mass,
m=
A1 𝑣1
=
A2 𝑣2
𝑣1
𝑣2
A1 𝑣1 ρ1 = A2 𝑣2 ρ2
19
Problems
1. Two gaseous streams enter a combining tube and leave as a single mixture.
These data apply at the entrance section:
For one gas , A1 = 75 in.², 1= 500 fps, v1 = 10 ft³/lb
For the other gas, A2 = 50 in.², m2 = 16.67 lb/s
ρ 2 = 0.12 lb/ft ³
At exit, 3 = 350 fps, v3 = 7 ft³/lb
Find (a) the speed 2 at section 2, and
(b) the flow and area at the exit section.
Solution
(a)
𝑙𝑏𝑚
16.67
𝑣2 =
s
𝑚2
=
A2 ρ2
= 400 fps
50
𝑓𝑡 2 0.12
144
(b)
𝑙𝑏𝑚
𝑓𝑡 3
75
ft
𝑓𝑡
144
A1 𝑣1
𝑚1 =
500
2
s
=
𝑙𝑏𝑚
= 26.04
𝑉1
𝑓𝑡
s
3
10
𝑙𝑏𝑚
𝑙𝑏𝑚
𝑚1 = 𝑚1 + 𝑚2 = 26.04 + 16.67 = 42.71
s
20
42.71
m3 𝑣3
A3 =
=
ᵿ3
𝑓𝑡 3
𝑙𝑏𝑚
7
s
350
𝑙𝑏𝑚
ft
= 0.8542𝑓𝑡 2
s
2. A 10-ft diameter by 15-ft height vertical tank is receiving water (ρ = 62.1 lb/cu ft)
at the rate of 300 gpm and is discharging through a 6-in ID line with a constant speed of 5
fps. At a given instant, the tank is half full. Find the water level and the mass change in the
tank 15 min. later.
Solution
300 gpm
5fps
15
7.5’
10’
Area =
π
(10)² = 78.54 ft²
4
300
gal
min
Mass flow rate entering =
lb
62.1
7.48
gal
= 2490.6
𝑓𝑡 3
𝑓𝑡 3
21
lb
min
Mass flow rate leaving = A𝑣ρ =
= 3658
π
6
4
12
²
ft 5x60
ft
min
62.1
lb
𝑓𝑡 3
lb
min
Mass change = (3658 – 2490.6) (15) = 17, 511 lb (decreased)
⧍m = 17, 511 lbm
17,511 lb
Volume change =
lb
= 282 𝑓𝑡 3
62.1
𝑓𝑡 3
Decreased in height =
282 𝑓𝑡 3
= 3.59 ft
78.54 𝑓𝑡 2
Water level after 15 min. 7.5 – 3.59 = 3.91 ft
3. A fluid moves in a steady flow manner between two sections in a vessel.
At entrance: A1 = 20 cm², ρ1 = 1878 kgm/m³
At exit:
A2 = 10 cm², √2 = 249.5 cm³/gm
2 = 3048 cm/min
Compute (a) the mass flow rate and (b) the speed at entrance.
Solution
22
(a) Mass Flow Rate
At exit
At entrance
m1 = lacKing data
Apply the law of Conservation of Mass
m1 = m2 = 122.2
4. A turbine receives steam at a gage pressure of 2.5 MPa. After expansion in the turbine the
steam flows into a condenser which is maintained at a vacuum of 710 mmHg. The
atmospheric pressure is 760 mmHg abs. Express the inlet and exhaust steam pressure in
MPa abs. Ans.
Solution
Take density of mercury as 13, 600
kgm
m³
Pressure of steam in the turbine = 2.5 MPa gage
= 2.5x10⁶ Pa gage
Vacuum in the condenser
= 710 mmHg voc.
Atmospheric pressure
= 760 mmHg abs
Now the atmospheric pressure, P0
= ρgz0
=
p0 = 1.01x10⁵ Paa
Inlet Steam Pressure
= Gage pressure + Atmospheric Pressure
= 2.5x10⁶ Pa gage + 1.01x10⁵ Pa abs
= 2.6 MPaa
23
Condenser Pressure
Review Problems
1. What is the mass in grams and the weight in dynes and in gram-force of 12 oz of
salt? Local g is 9.65 m/s² 1 lbm = 16 oz.
Ans. 340.2 gm; 328,300 dynes; 334.8 gf
2. A mass of 0.10 slug in space is subjected to an external vertical force of 4 lb. If the
local gravity acceleration is g = 30.5 fps² and if friction effected are neglected, determine
the acceleration of the mass if the external vertical force is acting (a) upward and (b)
downward.
Ans. (a) 9.5 fps²; (b) 70.5 fbs²
3. The mass of a given airplane at sea level (g=32.1 fps²) is 10 tons. Find its mass in
lb, slugs, and kg and its (gravitational) weight in lb when it is travelling at a 50,000-ft
elevation. The acceleration of gravity g decreases by 3.33 X 10^-⁶ fps² for each foot of
elevation
Ans. =20,000 lbm ; 621.62 slugs; 19,850 lbf
4. A lunar excursion module (LEM) weights 1500 kgi on earth where g= 9.75 mps².
What will be its weight on the surface of the moon where gm= 1.70 mps². On the surface of
the moon, what will be the force in kgf and in newtons required to accelerate the module at
10 mps²?
Ans. 261.5 kgf; 1538.5 kgf; 15,087 N
5. The mas of a fluid system is 0.311 slug, its density is 30 lb/ft³ and g is 31.90 fps².
Find (a) the specific volume, (b) the specific weight, and (c) the total volume.
Ans. (a) 0.0333 ft³/lb; (b) 29.75 lb/ft³; (c) 0.3335 ft³
6. A cylindrical drum (2-ft diameter, 3-ft height) is filled with a fluid whose density is 40
lb/ft³. Determine (a) the total volume of fluid, (b) its total mass in pounds and slugs, (c) its
specific volume, and (d) its specific weight where g= 31.90 fps².
Ans. (a) 9.43 ft³; (b) 377.2 lb; 11.72 slugs; (c) 0.025 ft³/lb ; (d) 39.66 lb ft³
7. A weatherman carried an aneroid barometer from the ground floor to his office atop
the Sears Tower in Chicago. On the ground level, the barometer read 30.150 in. Hg absolute;
topside it read 28.607 in. Hg absolute. Assume that the average atmospheric air density was
0.075 lb/ft³ and estimate the height of the building.
Ans. 1455 ft
8. A vacuum gauge mounted ion a condenser reads 0.66 m Hg. What is the absolute
pressure in the condenser in kPa when the atmospheric pressure is 101.3 kPa?
Ans. 13.28 kPa
24
9. Convert the following readings of pressure to kPa absolute, assuming that the
barometer reads 760 mm Hg: (a) 90 cm Hg gage; (b) 40 cm Hg vacuum; (c) 100 psig; (d) 8 in.Hg
vacuum, and (e) 76 in. Hg gage.
Ans. (a) 221.24 kPa; (b) 48 kPa; (c) 790.83 kPa; (d) 74.213 kPa; (e) 358.591 kPa
10. A fluid moves in a steady flow manner between two sections in a flow line. At
section 1:A1 = 10ft², ℧1 = 100 fpm, v1 = 4 ft³/lb. At section 2: A2 = 2 ft², ρ2 = 0.20 lb/ft³, ℧2 =
120 fpm. Calculate (a) the mass flow rate, and the (b) change of mass stored in system, Δ
m
11. If a pump discharges 75 gpm of water whose specific weight is 61.5 lb/ft³ (g=
31.95 fps²), find (a) the mass flow rate in lb/min, and (b) and total time required to fill a
vertical cylinder tank 10 ft in diameter and 12 ft high.
Ans. (a) 621.2 lb/min, (b) 93.97 min
12. In the condenser, a vacuum recorded for a steam plant is 720 mm. Hg. Find the
absolute pressure in the condenser in psia. The atmosphere pressure is 760 mm Hg abs.
Express your answer in Paabs.
Ans. 5,334.6 Paabs.
Solution
Vacuum recorded in the condenser = 720 mmHg vac.
Barometer Reading………………..= 760 mmHg abs
Actual Pressure in the condenser
= Barometer Reading – Vacuum in the condenser
= 760 – 720
14.7 psia
6894 Paa
= 40 mmHg ab. x
x
760 mmHgab
psia
= 5,334.6 ρa abs.
13. In a pipeline the pressure of gas is measured with a mercury manometer having
one end open to atmosphere. If the difference in the height of mercury in the two legs is
550 mm, calculate the gas pressure in bar.
14. A Kelvin and a RanKie thermometer are both immersed in a fluid and indicate
identical numerical readings. What is the temperature expressed or ° C and °F?
25
15. A water manometer for the measurement of the pressure of a pipeline
containing oil with a specific gravity of 0.90 is shown with reading indicated below. The
density of water is 1000 kg/m³. For the manometer, calculate the pressure at the centreline
of the pipe. Barometer pressure is 100KPaa and local acceleration of gravity is standard.
Express your answer in KPaa.
16. The temperature of a system is 55°C. Express the temperature in of, °K and °R.
17. Define the temperature where the thermometers using the Fahrenheit scale and
Celsius scale give the same reading.
18. The temperature of a system increase by 50°F during heating process. Express
this wise in °K, °C and °R.
19. The standard barometric pressure at sea level in 14.7 psia. (a) Determine the
pressure on the mountain top having an elevation of 30,000 ft. above sea level. The average
air density between sea level and the mountain top can be assured to be constant value of
0.5 lbm/ft³. (b) Determine the pressure at the bottom of sea having a depth of 1300 ft. Take
the water density to be 62.4 lb/ft. Neglect variations in gravitational acceleration.
26
Chapter 2
CONSERVATION OF ENERGY
Energy - is the capacity for action or for doing work.
Mass-Energy Relation
In Physics, Albert Einstein states in his theory of relativity that “mass can be converted into energy
and energy into mass” by his equation:
Energy = (mass) (light speed)²
E = m c²
Where:
m = mass of object, gm kgm or lbm
c = light speed
10
c = 2.998 X 10
cm/sec
8
c = 2.998 X 10 m /sec
c = 186,000 mi/sec
c = 982,080,000 ft/sec
Albert Einstein relates his theory on the speed of light thinking that it is impossible for mankind to
invent a moving vehicle which can travel faster than light speed. He did not relate his theory on the speed
of sound since during his time, he is optimistic that mankind can travel faster than the speed of sound,
but not the speed of light.
Problem:
1. In relation to Einstein’s theory of relativity, view a system whose mass is 5 slugs. How
much energy may be derived out of this system if all of the mass could be converted into
energy? Express your answer in BTU.
Solution
m = 5 slugs
E = mc²
8
= (5 slugs) (2.998 X 10
m/s) ²
16
= 5 slug ( lbm-s²/ft )(8.98 X 10 m²/s²)
slug
1
16
E = 5 (lbm – s²) (8.98 X 10 ) m² X
Ft
ft
s²
X Kgm
3.28 m
2.205 lbm
16
E = 6.2 X 10
Kg ƒ – m X
N
9.8066 Kg ƒ
16
E = 0.63 X 10 J
X
KJ
1000 J
-4
X
BTU
1.055 KJ
16
= 6 X 10 X 10
BTU
12
E = 6 x 10 BTU
2. The amount of electrical energy for a residential house that you can be consumed per month
18
is 8 X 10
Watt-hour. What is its mass in Kgm?
18
: E = 8 X 10 Watt-hr
Solution
: E = mc²
E
m=
c²
18
=
8 X 10 Watt-hr
8
(2.998 X 10
m /s )²
18
=
8 X 10 ( J ) X hr X 3600 sec
sec
hr
16
8.98 X 10 m²/s²
18
8 X 10 N-m
m=
16
8.98 X 10 m²/s²
2
=
18
8 X 10 Kg m – m (m)
s²
16
8 .98 X 10 ( m²)
s²
m = 0.89 X 10² Kgm
m = 89 Kgm
Mass- Speed Relation
Albert Einstein also states that there is an effect of speed to the mass of the moving object, from
his formula.
When a body starts to move there is gain in mass, m.
mrm
m=
1- (℧/c)²
where:
m rm = rest mass of the object, gmm, Kgm or lbm
℧ = speed of moving object, m/sec
c = light speed, m/sec
Problem:
1. If a 2,110 Kg- object starts to move at a speed of 10,100 mph, what will be its
mass while travelling at that speed?
Solution
mrm = 2,110 Kgm
℧ = 10,100 mph
m while travelling
3
2,110 Kgm
m=
²
1 – 10,100 mph
8
6.696 X 10 mph
2,100 Kgm
=
1–
(10,100 )²
16
6.696 X 10
2,100 Kgm
=
16
8
6.696 X 10 - 1.02 X 10
16
6.696 X 10
2,100 Kgm
=
8
8
10 (6.96 X 10 - 1.02)
16
8
6.696 X 10
2,100 Kgm
2,100 Kgm
=
=
8
6.696 X 10 - 1.02
5.676
8
6.696 X 10
6.696
2100
=
=
2,277 Kgm
0.922
4
FORMS OF ENERGY
A. Gravitational Potential Energy (PE)
The gravitational potential energy of a body is its energy due to its position or elevation.
mgz
PE = Fgz =
k
mg
⧍PE = PE2 – PE1 =
(z2 – z1)
k
⧍PE = change in potential energy
= elevation
Kinetic Energy (KE)
The energy or stored capacity for performing work possessed by a moving body, by virtue of its
momentum is called kinetic energy.
℧
m
KE = m℧²
2K
⧍KE = KE2 – KE1 m (℧²2-℧²1)
2k
⧍KE = change in kinetic energy
℧ = speed or velocity
Internal Energy (U, u)
Internal energy is energy stored within a body or substance by virtue of the activity and
configuration of its molecules and of the vibration of the atoms within the molecules.
u = specific internal energy (unit mass)
⧍u = u2-u1
U = mu = total internal energy (m mass)
⧍U = U2 – U1
5
Work (W)
Work is the product of the displacement of the body and the component of the force in the
direction of the displacement. Work is energy in transition; that is, it exists only when force is “moving
through a distance”.
Work of a Nonflow System
The work done as the piston moves from e to f is
dW=Fx dx = (pA)dL = pdV
which is the area under the curve e-f on the pV plane.
Therefore, the total work done as the piston moves
from 1to 2 is
2
W=
∫
pdV
1
which is the area under the curve 1-e-f-2.
Fig. 2 Work of Expansion
The area under the curve of the process on the pV plane represents the work done during as nonflow
reversible process.
Work done by the system is positive (outflow of energy)
Work done on the system is negative (inflow of energy)
Steady Flow Energy Equation
Characteristics of steady flow system
1. There is neither accumulation nor diminution of mass within the system.
2. There is neither accumulation nor diminution of energy within the system
3. The state of the working substance at any point in the system remains constant.
6
Fig.4 Energy Diagram of a Steady Flow System
Energy Entering System = Energy Leaving System
P1 + K1 + Wf1 + U1 + Q = P2 + K2 + Wf2 + U2 + W
Q = ⧍P + ⧍K + ⧍Wf + ⧍U + W
(Steady Flow Energy Equation)
Enthalpy (H, h)
Enthalpy is a composite property applicable to all fluids and is defined by
h = u + pv and H = mh = U + pV
The steady flow energy equation becomes
P 1 + K1 + H 1 + Q = P 2 + K 2 + H 2 + W
Q = ⧍P + ⧍K + ⧍H + W
Flow Work (Wf)
Flow work or flow energy is work done in pushing a fluid across a boundary, usally into or out of a
system.
Wf = FL = pAL
Wf = pV
⧍Wf = Wf2 – Wf1 = p2V2 – p1V1
⧍Wf = change in flow work
Fig. 3 Flow Work
7
Heat (Q)
Heat is energy in transit (on the move) from one body or system to another solely because of a
temperature difference between the bodies or systems.
Q is positive when heat is added to the body or system.
Q is negative when heat is rejected by the body or system.
Classification of Systems
(1) A closed system is one in which mass does not cross its boundaries.
(2) An open system is one in which mass crosses its boundaries.
Conservation of Energy
The law of conservation of energy states that energy is neither created nor destroyed.
The first law of thermodynamics states that one form of energy may be converted into another.
Problems
1. During a steady flow process, the pressure of the working substance drops from 200 to 20
psia, the speed increases from 200 to 1000 fps, the internal energy of the open system
decreases 25 Btu/lb, and the specific volume increases from 1 to 8 ft³/lb. No heat is
transferred. Sketch an energy diagram. Determine the work per lb. Is it done on or by the
substance? Determine the work in hp for 10 lb per min. ( 1 hp = 42.4 Btu/min).
Solution
W?
𝐾1
𝑊𝑓1
System
𝑈1
𝑝1 = 200 psia
𝑝2 = 20 psia
𝐾2
𝑣1 = 200 fps
𝑣2 = 1000 fps
𝑊𝑓2
v1 = 1 𝑓𝑡 3 /𝑙𝑏
v2 = 8 𝑓𝑡 3 /𝑙𝑏
𝑈2
⧍u = -25Btu/lb
Energy Diagram
𝑃1 + 𝐾1 + 𝑊𝑓1 + 𝑈1 + Q = 𝑃2 + 𝐾2 + 𝑊𝑓2 + 𝑈2 + W
Basis 1 𝑙𝑏𝑚
8
Q =0
2
ft
200
𝐾1 =
𝑣2 ²
=
2k
s
= 0.80
𝑙𝑏𝑚 .ft
(2) 32.174
778
𝑙𝑏𝑓 . s²
ft. 𝑙𝑏𝑓
𝑙𝑏𝑚
Btu
2
𝐾1 =
𝑣2 ²
(1000)
Btu
=
= 19.97
2k
(2) (32.174) (778)
𝑙𝑏𝑚
lb
𝑖𝑛.2
200
144
𝑖𝑛.
𝑊𝑓1 = 𝑝1 v1 =
2
𝑊𝑓2 = 𝑝2 v2 =
1
𝑓𝑡
778
𝑓𝑡 3
2
Btu
𝑙𝑏𝑚
= 37.02
ft. 𝑙𝑏𝑓
𝑙𝑏𝑚
Btu
(20) (144) (8)
= 29.61
778
Btu
𝑙𝑏𝑚
K1 + 𝑊𝑓1 = 𝐾2 + 𝑊𝑓2 + ⧍u + W
0.8 + 37.02 = 19.97 + 29.61 – 25 + W
W = 13.24
Btu
(by)
lbm
13.24
W=
Btu
10
lbm
lb
min
Btu
42.4
= 3.12 hp
(min) (hp)
9
Btu
2. Steam is supplied to a fully loaded 100-hp turbine at 200 psia with u1, = 1163.3 Btu/lb, v1 =
2.65 ft 3/lb and 01= 400 fps. Exhaust is at 1 psia with u2 = 925 Btu/lb, v2 = 249 ft3/lb and v2
= 1100 fps. The heat loss from the steam in the turbin is 10 Btu/lb. Neglect potential energy
change and determine (a) the work per lb steam and (b) the steam flow rate in lb/h.
Solution
P1 = 200 psia
u1 = 1163.3 Btu/lb
v1 = 2.65 ft3/lb
P2 = 1 psia
u2 = 925 Btu/lb
v2 = 294 ft3/lb
V1 = 400 fps
v2 = 1100 fps
Q = - 10 Btu/lb
𝑃1 + 𝐾1 + 𝑊𝑓1 + 𝑈1 + Q = 𝑃2 + 𝐾2 + 𝑊𝑓2 + 𝑈2 + W
(a) Basis 1 𝑙𝑏²𝑚
𝐾1 =
𝐾2 =
𝑣²1
2k
=
𝑣²2
2k
𝑊𝑓1 = 𝑝1 v1 =
𝑊𝑓2 = 𝑝2 v2 =
=
(400)²
(2) (32.174) (778)
(1100)²
= 3.20
Btu
𝑙𝑏𝑚
= 24.17
(2) (32.174) (778)
(200) (144) (2.65)
Btu
𝑙𝑏𝑚
= 98.10
778
Btu
𝑙𝑏𝑚
(1)(144) (294)
= 54.24
778
Btu
𝑙𝑏𝑚
K1 + 𝑊𝑓1 + u1 + Q = 𝐾2 + 𝑊𝑓2 + u2 + W
3.20 + 98.10 + 1163.3 + (-10) = 24.17 + 54.42 + 925 + W
w = 251
Btu
𝑙𝑏𝑚
10
(100 hp)
2544
Btu
(h) (hp)
(b) Steam flow =
= 1014
Btu
251
𝑙𝑏𝑚
h
𝑙𝑏𝑚
3. An air compressor (an open system) receives 272 kg per min of air at 99.29 kPa and a
specific volume of 0.026 m3/kg. The air flows steady through the compressor and is
discharged at 689.5 J/kg; at discharge, the internal energy is 6241 J/kg. The cooling water
circulated around the cylinder carries away 4383 J/kg of air. The change in kinetic energy is
896 J/kg increase. Sketch an energy diagram. Compute the work.
Solution
P1 = 99.29 kPa
v1 = 0.026 𝑚3 /kg
u1 = 1594 J/kg
Q = -4383 J/kg
𝑚̇ = 272 kg/min
p2 = 689.5 kPa
v2 = 0.0051 𝑚3 /kg
u2 = 6241 J/kg
⧍K = 896 J/kg
𝑃1 + 𝐾1 + 𝑊𝑓1 + 𝑈1 + Q = 𝑃2 + 𝐾2 + 𝑊𝑓2 + 𝑈2 + W
Basis 1 𝑘𝑔𝑚
𝑚3
kN
𝑊𝑓1 = 𝑝1 v1 = 99.29
0.026
𝑚
2
= 2.583 kJ/kg
kg
11
kN
𝑊𝑓2 = 𝑝2 v2 = 689.5
𝑚
𝑚3
0.0051
= 3.516 kJ/kg
kg
2
𝑊𝑓1 + u1 + Q = ⧍K + 𝑊𝑓2 + u2 + W
2.582 + 1.594 – 4.383 = 0.896 + 3.516 + 6.241 + W
kJ
W = - 10.86
W = -10.86
W = -2954
kg
kJ
kg
272
kg
min
kJ
min
4. A centrifugal pump operating under steady flow conditions delivers 2,270 kg/min of water
from an initial pressure of 82, 740 Pa to a final pressure of 275,800 Pa. The diameter of the
inlet pipe to the pump is 15.24 cm and the diameter of the discharge pipe is 10.16 cm. What
is the work?
Solution
𝑚̇ = 2270 kg/min
d1 = 0.1524 m
p1 = 82,740 Pa
ρ = 1000 kg/𝑚3
d2 = 0.1016 m
p2 = 275,800 Pa
Area at entrance, A1 =
Area at entrance, A2 =
Speed at entrance, 𝑣1 =
π
(0.1524)² = 0.01824 m²
4
π
(0.1016)² = 0.008107 m²
4
𝑚̇
ρ1 A1
2270 𝑘𝑔𝑚
60
s
=
1000
𝑘𝑔𝑚
𝑚
3
12
0.01824 m²
= 2.074 m/s
Speed at exit, 𝑣2 =
2270/60
𝑚̇
=
= 2.074 m/s
ρ2 A2 (1000) (0.008107)
Basis 1 𝑘𝑔𝑚
m ²
2.074
K1 =
s
𝑣²1
=
2k
𝑘𝑔𝑚 .m
(2) 1
𝑣²2
K2 =
N.m
𝑘𝑔𝑚
N.s²
(4.667)²
=
2k
= 2.151
N.m
= 10.89
𝑘𝑔𝑚
(2) (1)
N
82,740
p1
𝑊𝑓1 = 𝑝1 v1 =
m2
=
N.m
= 82.74
kg
ρ1
𝑘𝑔𝑚
1000
𝑚3
𝑊𝑓2 = 𝑝2 v2 =
p2
275,800
=
N.m
= 275.8
1000
ρ2
𝑘𝑔𝑚
𝑃1 + 𝐾1 + 𝑢1 + 𝑊𝑓1 + q = 𝑃2 + 𝐾2 + 𝑢2 + 𝑊𝑓2 + W
K1 + 𝑊𝑓1 = 𝐾2 + 𝑊𝑓2+ W
2.151 + 82.74 = 10.89 + 275.8 + W
W = -201.8
N.m
𝑘𝑔𝑚
W = -201.8
N.m
𝑘𝑔𝑚
2270
kg
min
13
W = -458.1
kJ
min
5. A turbine operates under steady flow conditions, receiving steam at the following state:
pressure 1200 k Pa, temperature 188 C, enthalpy 2785 kJ/kg, speed 33.3 m/s and elevation 3
m. The steam leaves the turbine at the following state; pressure 20 kPa, enthalpy 2512 kJ/kg,
speed 100 m/s and elevation 0 m. Heat is lost to the surroundings at the rate of 0.29 kJ/s. If
the rate of steam flow through the turbine is 0.42 kg/s, what is the power output of the
turbine in kW?
Solution
z1 = 3m
z2 = 0 m
kJ
h1 = 2785
h2 = 2512
kg
m
𝑣1 = 33.3
kg
m
𝑣2 =
s
s
kJ
Q = -0.29
kJ
kg
𝑚̇ = 0.42
s
s
Basis 1 𝑘𝑔𝑚
9.8066
P1 =
𝑞𝑧1
k
=
m
(3 m)
s²
= 0.0294
kg.m
1
kJ
kg
N.s²
2
m
33.3
𝐾1 =
𝑣²1
2k
s
=
kg.m
(2) 1
𝐾2 =
𝑣²2
2k
=
kJ
= 0.5544
𝑘𝑔
N.s²
(100)²
(2) (1)
= 5.000
kJ
kg
14
-0.29
Q=
0.42
kJ
s
kg
kJ
= -0.6905
kg
s
𝑃1 + 𝐾1 + ℎ1 + 𝑄 = 𝑃2 + 𝐾2 + ℎ2 + W
𝑃1 + 𝐾1 + ℎ1 + 𝑄 = 𝐾2 + ℎ2 + W
0.0294 + 0.5544 + 2785 + (-0.6905) = 5.000 + 2512 + W
kJ
W = 267.9
W = 267.9
kg
kJ
kg
0.42
kJ
s
W = 112.52 kW
Review Problems
1. Assuming that there are no heat effects and no frictional effects, find the kinetic energy and
speed of a 3220-lb body after it falls 778 ft from rest. Start with the steady flow equation,
deleting energy terms which are irrelevant.
Ans. 224 fps
2. A reciprocating compressor draws in 500 cubes feet per minute of air whose density is 0.079
lb/cu ft and discharges it with a density of 0.304 lb/cu ft. At the suction, p1 = 15 psia; at
discharge, p2 = 80 psia. The increase in the specific internal energy is 33.8 Btu/lb. Determine
the work on the air in Btu/min and in hp. Neglect change in kinetic energy.
Ans. 56.25 hp
3. Steam enters a turbine with an enthalpy of 1292 Btu/lb and leaves with an enthalpy of 1098
Btu/lb. The transferred heat is 13 Btu/lb. What is the work in Btu/min and in hp for a flow of 2
lb/sec?
Ans. 512.3 hp
15
4. A thermodynamic steady flow system receives 4.56 kg per min of a fluid where p1 = 137.90
kPa, v1 = 0.0388 m3/kg, v1 = 122 m/s, and u1 = 17.16 kJ/kg. The fluid leaves the system at a
boundary where p2 = 551.6 kPa, v2 = 0.193m ³/kg,u2 =183 m/s and u2 = 52.80 kJ/kg. During
passage through the system the fluid receives 3,000 J/s of heat. Determine the work.
Ans. – 486kJ/min
5. Air flows steadily at the rate of 0.5 kg/s through an air compressor, entering at 7 m/s speed,
100 kPa pressure and , 0.95 m³/ kg specific volume, and leaving at 5m/s , 700 kPa, and 0.19
m³/kg. The internal energy of the air leaving is 90 kJ/kg greater than that of the air entering.
Cooling water in the compressor jackets absorbs heat from the air at the rate of 58 kW.
Compute the works in kW.
Ans. – 122kW
6. In steady flow apparatus, 135 kJ of work is done by each kg of fluid. The specific volume of the
fluid, pressure, and speed at the inlet are 0.37m³/kg, 600 kPa, and 16 m/s. The inlet is 32 m
above the floor, and the discharge [pipe is at floor level. The discharge conditions are0.62
m³/kg, 00 kPa, and 270 m/s. The total heat loss between the inlet and discharge is 9 kJ/kg of
fluid. In flowing this apparatus, does the specific internal energy increase or decrease, and by
how much?
Ans. – 20.01 kJ/kg
7. Steam enters a turbine stage with an enthalpy of 3628 kJ/kg at 70 m/s and leaves the same
stage with an enthalpy of 2846 kJ/kg and velocity of 124 m/s. Calculate the work done by the
steam.
Ans. 776.8 kJ/kg (ME Board Problem- Oct. 1986)
8. The properties of a closed system change following the relation between pressure and volume
as pV = 3 where p is in bar and V is in m³. Compute the work done when the pressure increases
from 1.5 bar to 7.5 bar. Express your answer in BTV.
9. A closed system has 150 KV of work supplied when the initial volume is 0.6³ and pressure of
the system changes in the relation p = 8-4V, where V is in m³ and p is in bar. Compute the final
pressure and volume of the system. Express your answer in Pa & m³.
5
Ans. 5.172 X 10 Pa, 0.707 m³
10. A fluid at a pressure of 3 bar and sp w/of 0.18 m³/kg contained in a piston cylinder
arrangement expands reversibly to a pressure of 0.6 bar according to relation p = c/v² where c
is constant. Compute the work done in cal/Kg fluid on the position.
Ans. 7127.66 cal/Kg
16
11. Determine the work done by a 1-Kgm fluid system as it expand slowly in a closed system from
p1 = Mpa, V1 = 0.08 m³ to V2 = 0.22m³ in accordance to the
1.4
defining relation: (a) p = c, (b) pV = c , (c) pV³ = c ,(d) pV (InV) = c (e) pV
.Express your answer in calories.
12. The flow of 0.124 m³/min of a fluid crossing a boundary to a system is 28 KW. Find the
pressure at this point.
13. A 10-lbm object is moving at a speed of 20 fps² at an elevation of 50 ft above the ground.
Calculate the weight of the object and its potential and kinetic energies.
14. A vehicle having a mass of 6,500 lb travels at a speed of 100miles per hour. Compute (a) the
mass of the vehicle in Kg, (b) its speed in km/hr, and (c) its kinetic energy in kcal.
15. A 60-Kgm body has a potential energy of 135 Kcal with respect to a datum plane X and a
potential energy of -99 Kcal with respect to datum place Y. If the local acceleration of gravity is
standard, compute the relative position of plane X relative top plane Y.
16. A gas of mass 1.5 Kg undergoes an expansion process which follows a relationship p= a + bV,
where a and b are constants. If p1 = 1 MPa, V1 = 0.20 m³ and P2 = 0.2 MPa, V2 = 1.20m³. The
interval energy of the gas is u= 1.5 pv- 85 kJ/kg where p is in KPa and v is in m³/Kg. Compute
the heat transfer and the maximum interval energy of the gas during expansion process.
Express your answer in BTV.
Ans. 625.6 BTV, 477.1 BTV
17. A 1500-Kg vehicle in Baguio climbs a 120-m long road uphill in Baguio with a 45° slope in 40
seconds. Determine the power required given:
(a) the velocity is constant
(b) from rest to a final velocity of 42 m/s, and
(c) from 40 m/s to a final velocity of 10 m/s
NOTE: Neglect rolling resistance, air drop and force of friction
18. An airplane weighing 20,000 lb at ground is flying at an altitude of 35,000 ft at 600 miles per
hour. Compute the potential and kinetic energies of the airplane. Express your answer in Keal.
19. A fluid undergoes a frictionless process in a closed system from V1 = 5ft³ to V2 = 2 ft³ in
accordance with the relation
V =
35
(p-8)
where p is in psia and V in cu.ft. During this process, the fluid rejects 30 BTV of heat.
Compute the change of enthalpy
17
20. A fluid is compressed in a closed system in such a way that pv= 14,400, where p is in psfa and v
in ft³/lbm , from an initial pressure of 110 psia to a final pressure of 210 psia. During the
compression process, the internal energy of the third changes from 70 to 75 BTV/lbm. Compute
the heat transfer.
21. Air is contained in a piston – cylinder arrangement. The following data were gathered: P1 =
700 KPa V1 = 0.10 m³/Kg, and u1 = 160 KJ/Kg. The air expands slowly with the pressure
remaining constant until the volume becomes, V2 = 0.20 m³/Kg and the internal energy
becomes, U2 = 310 Ks/Kg. Calculate the work done by the gas and the heat transferred.
22. A fluid in a closed system expands slowly according to the relation
U = 32+ v0.004 pV
where U is in BTV, p in lb/ft² abs, and V in ft³. When the fluid changes from an initial state
of 30 psia, 1 ft³,k to a final state of 65 psia, 3 ft³, compute the work and heat transfer in
calories.
23. A certain fluid in a closed system initially at a pressure of 200 psia and having a volume of 3 ft³
1.3
undergoes a frictionless process during which the product pV
the work done by or on the system.
= C. If p2 = 40 psia, compute
24. A piston-cylinder arrangement contains a gas at P1 = 600 KPa, V1 =0.10 m³ and expand to V2 =
1.4
0.5 m³.Compute the work done by the gas if the process follows the relation (a) pV
(b) p= -300V + 630, where p is in Kpa and V in m³.
= c, and
25. Work is done by a substance in reversible nonflow manner during which 32 KJ are added, the
pressure remains constant at 200 KPa while the volume changes from 0.5m³ to 0.25 m³.
Compute the change of internal energy is KJ .
18
Chapter 3
THE IDEAL GAS
An ideal, gas is ideal only in the sense that it conforms to the simple perfect
gas laws.
Examples:
1. Argon
2. Air
3. Methane
4. Carbon Dioxide
5. Oxygen
6. Nitrogen
Gas Laws:
A.
Boyle’s Law
If the temperature of a given quantity of gas is held constant, the volume of
gas varies inversely with the absolute pressure during a chance of state.
V∞
1
or V =
p
C
p
pV = C or p1 V1 = p2 V2
B.
Charles’ Law
(1) If the pressure on a particular quantity of gas is held constant, then, with
any change of state, the volume will vary directly as the absolute temperature.
V ∞ T or
V = CT
V
𝑉1
T
= C or
T1
=
V2
T2
(2) If the volume of a particular quantity of gas is held constant, the, change
of state, the pressure will vary directly as the absolute temperature.
P ∞ T or
p = CT
P
p1 p2
=
T1 T2
=C
T
or
1
Equation of State or Characteristic Equation of a Perfect Gas
Combining Boyle’s and Charles’ laws,
p1 V1 p2V2
=
= C, a constant
T1
T2
pV
= mR
T
pV = mRT
pv = RT
(unit mass)
where p = absolute pressure
V = volume
v = specific volume
m = mass
T = absolute temperature
R = specific gas constant or simply gas constant
p
V
m
T
R
lbf
ft³
lbm
°R
ft.lbf
English units
ft²
SI units
N
lbm°R
m³
kg
°K
m²
J
Kgm - °K
Problems
1. A drum 6 in. in diameter and 40 in. long contained acetylene at 250 psia
and 90°F. After some of the acetylene was used, the pressure was 200 psia and the
temperature was 85°F, volume would the used acetylene occupy at 14.7 psia and
80°F? R for acetylene is 59.35 ft.lb/lb.°R.
2
Solution
(a) Let
m1 = mass of acetylene initially in the drum
m2 = mass of acetylene left in the drum
m3 = mass of acetylene used
p1 = 250 psia
T1 = 90°F + 460 = 550°R
p2 = 200 psia
T2 = 85°F + 460 = 545°R
volume of drum =
π (6)² (40)
= 0.6545 cu ft
(4) (1728)
m1 = p1 V1 = (250)(144)(0.6545) = 0.7218 lb
(59.35)(550)
RT1
(200) (144) (0.6545)
p2 V2
m2 =
=
(59.35) (545)
RT2
= 0.5828 lb
m3 = m1 –m2 = 0.7218 – 0.5828 = 0.1390 lb
Acetylene used =
m3
m1
=
0.1390
0.7218
= 0.1296 or 19.26%
(b) p3 =14.7 psia
T3 = 80°F = 460 = 540°R
(0.139) (59.35) (540)
m3 RT3
V3 =
=
p3
= 2.105 ft³
(14.7) (144)
3
2. The volume of a 6 x 12-ft tank is 339.3 cu ft. It contains air at 200 psig and
85°F. How many 1-cu ft drums can be filled to 50 psig and 80°F if it is assumed that the air
temperature in the tank remains at 85°F? The drums have been sitting around in the
atmosphere which is at 14.7 psia and 80°F.
Solution
Let
𝑚1 = mass of air initially in the tank
𝑚2 = mass of air left in the tank
𝑚3 = mass of air initially in the drum
𝑚4 = mass of air in the drum after filling
𝑝1 = 200 + 14.7 = 214.7 psia
𝑇1 = 85 + 460 = 545°R
𝑝2 = 50 + 14.7 = 64.7 psia
𝑇2 = 85 + 460 = 545°R
𝑝3 = 14.7 psia
𝑇3 = 80 + 460 = 540°R
𝑝4 = 50 + 14.7 = 64.7 psia
𝑇4 = 80 + 460 = 540°R
For the tank
(214.7) (144) (339.3)
p1 V1
𝑚1 =
=
𝑅𝑇1
(64.7) (144) (339.3)
p2 V2
𝑚2 =
= 360.9 lb.
(53.34) (545)
=
𝑅𝑇2
(53.34) (545)
= 108.7 lb
mass of air that can be used = 360.9 – 108.7 = 252.2 lb.
For the drums
(14.7) (144) (1)
p3 V3
𝑚3 =
=
= 0.0735 lb
𝑅𝑇3
(53.34) (540)
p4 V4
(64.7) (144) (1)
𝑚4 =
=
𝑅𝑇4
= 0.3235 lb
(53.34) (540)
mass of air put in each drum = 0.3235 – 0.0735 = 0.25 lb
4
Number of drums filled up =
252.2
0.25
= 1009
3. It is planned to lift and move logs from almost inaccessible forest areas by
means of balloons. Helium at atmospheric pressure (101.325 kPa) and temperature 21.1°C
is to be used in the balloons. What minimum balloon diameter (assume spherical shape)
will be required for a gross lifting force of 20 metric tons?
Solution
20,000 kg
Let
𝑚𝑎 = mass of air displaced by the balloon
𝑚𝐻𝑒 = mass of Helium
V = volume of the balloon
m
He
𝑚𝑎
For the air
𝑅𝑎 = 287.08
J
kg . K
𝑝𝑎 = 101,325 Pa
𝑇𝑎 = 21.1 + 273 = 294.1 K
𝑚𝑎 =
𝑝𝑎 V
𝑅𝑎 𝑇𝑎
101,325V
=
= 1.2001 V kg
(287.08) (294.1)
For the helium
𝑅𝐻𝑒 = 2,077.67
J
kg . K
5
𝑝𝐻𝑒 = 101,325 Pa
𝑇𝐻𝑒 = 21.1 + 273 = 294.1 K
𝑚𝐻𝑒 =
𝑝𝐻𝑒 V
101,325V
=
= 0.1658 V kg
𝑅𝐻𝑒 𝑇𝐻𝑒 (2077.67) (294.1)
𝑚𝑎 = 𝑚𝐻𝑒 + 20,000
1.2001 V = 0.1658 V + 20,000
V = 19,337 m³
4
πr³ = 19,337
3
r = 16.65 m
d = 2 (16.65) = 33.3 m
4. Two vessels A and V=B of different sizes are connected by a pipe with a
valve. Vessel A contains 142 L of air at 2,767.92 kPa, 93.33°C. The valve is opened
and, when the properties have been determined, it is found that pm = 1378.96 kPa,
tm = 43.33°C. What is the volume of vessel B?
Solution
For vessel A
𝑃𝐴 = 2,767.92 kPa
𝑉𝐴 = 142 liters
𝑇𝐴 = 93.33 + 273 = 366.33 K
For vessel B
𝑃𝐵 = 68.95 kPa
𝑇𝐵 = 4.44 + 273 = 277.44 K
For the mixture
𝑝𝑚 = 1378.96 kPa
𝑇𝐵 = 43.33 + 273 = 316.33 K
𝑚𝑚 =𝑚𝐴 + 𝑚𝐵
𝑝𝑚 𝑉𝑚
𝑅𝑇𝑚
=
𝑝𝐴 𝑉𝐴
𝑅𝑇𝐴
+
𝑝𝐵 𝑉𝐵
𝑅𝑇𝐵
6
(1378.96) 𝑉𝑚
=
(2767.92) (142)
316.33
366.33
4.36 𝑉𝑚 = 1072.9 + 0.25 VB
𝑉𝑚 = 142 + 𝑉𝐵
+
68.95 VB
277.44
(1)
(2)
solving equations 1 and 2 simultaneously
𝑉𝐵 = 110.4 liters
C.
Specific Heat
The specific heat of a substance is defined as the quantity of heat required to
change the temperature of unit mass through one degree.
In dimension form,
heat (energy units)
c
(mass) (change of temperature)
In differential quantities,
dQ
c=
or dQ =mcdT
mdT
and for a particular mass m,
2
Q=m
∫
cdT
1
(The specific heat equation)
If the mean or instantaneous value of specific heat is used,
2
Q = mc
∫
dT = mc (𝑇2 -𝑇1 )
1
(constant specific heat)
7
Constant Volume Specific Heat (𝑪𝒗 )
⧍U
Volume
Constant
𝑄𝑣 = ⧍U
𝑉𝑣 = 𝑚𝑐𝑣 (𝑇2 -𝑇1 )
𝑄𝑣
Constant Pressure Specific Heat (𝒄𝒑 )
𝑄𝑝 = mcp (𝑇2 -𝑇1 )
2
𝑄𝑝 = ⧍U + W = ⧍U +
∫
pdV
-1
𝑄𝑝 = ⧍U + p (𝑉2-𝑉1)
= 𝑈2 – 𝑈1 + p2 p2 – p1 p1
𝑄𝑝 = 𝐻2 – 𝐻1 = ⧍H
Ratio of Specific Heats (K)
𝑐𝑝
k=
>1
𝐶𝑣
Internal Energy of an Ideal Gas (U, u)
Joule’s law states that “the change of internal energy of an ideal gas is a function
of only the temperature change.” Therefore, ⧍U is given by the formula,
⧍U = 𝑚𝑐𝑣 (𝑇2 -𝑇1 )
whether the volume remains constant or not.
8
Enthalpy of an Ideal Gas (H, h)
The change of enthalpy of an ideal gas is given by the formula,
⧍H = 𝑚𝑐𝑝 (𝑇2 -𝑇1 )
whether the pressure remains constant or not.
Relation Between cp and cv
From h = u + pv and pv = RT
dh = du + Rdt
𝑐𝑝 dT = 𝑐𝑣 dT + RdT
𝑐𝑝 = 𝑐𝑣 + R
R
𝑐𝑣 =
k–1
kR
𝑐𝑝 =
k–1
Problems
1. For a certain ideal gas R = 25.8 ft.lb/lb.°R and k = 1.09 (a) What are the values of cp
and cv? (b) What mass of this gas would occupy a volume of 15 cu ft at 75 psia and 80°F?
(c) If 30 Btu are transferred to this gas at constant volume in (b), what are the resulting
temperature and pressure?
Solution
(a) 𝑐𝑝 =
𝑐𝑣 =
kR
(1.09) (25.8)
=
= 312.47
k–1
𝑐𝑝
k
f t.lb
1.09 – 1
0.4016
=
1.09
(b) V = 15 cu ft
m=
pV
RT
or 0.4016
lb.R°
Btu
= 0.3685
p = 75 psia
=
Btu
lb.R°
T = 80 + 460 = 540°R
(75) (144) (15)
= 11.63 lb
(25.8) (540)
(c) Q = 𝑚𝑐𝑣 (𝑇2 -𝑇1 )
9
lb.R°
30 = 11.36 (0.3685) (𝑇2 - 540)
𝑇2 = 547°R
𝑝2 = 𝑝1 (𝑇2 /𝑇1 ) = 75 (547/540) = 76 PSIA
2. For a certain gas R = 320 J/kg. K and c = 0.84 kJ/kg. K° (a) Find cp and k. (b) If 5 kg of
this gas undergo a reversible non flow constant pressure process from V1 = 1.133 m³ and p1
= 690 kPa to a state where t2 = 555°C, ⧍U and ⧍H.
Solution
kJ
(a) 𝑐𝑝 = 𝑐𝑣 + R = 0.84 + 0.32 = 1.16
kg . k°
R
0.32
k=
+1=
+ 1 = 1.381
𝑐𝑣
0.84
𝑝1 𝑣1
(690,000) (1.133)
(b) 𝑇1 =
=
= 488.6 K
mR
(5) (320)
⧍U = 𝑚𝑐𝑣 (𝑇2 -𝑇1 ) = 5 (0.84) (828 – 488.6)
= 1425.5 kJ
⧍H = 𝑚𝑐𝑝 (𝑇2 -𝑇1 ) = 5(1.16) (828 – 488.6)
= 1968.5 kJ
Entropy (S, s)
Entropy is that property of a substance which remains constant if no heat enters
or leaves the substance, while it does work or alters its volume, but which increases or
diminishes should a small amount of heat enter or leave.
The change of entropy of a substance receiving (or delivering) heat is defined by
dQ
dS =
2
or
⧍S =
T
∫
1
10
dQ
T
where: dQ = heat transferred at the temperature T
⧍S = total change of entropy
2 mcdT
⧍S =
∫
T
1
2
⧍S = mc
∫
1
dT
𝑇2
= mc ln
T
𝑇1
(constant specific heat)
Temperature-Entropy Coordinates
dQ = TdS
2
Q=
∫ TdS
1
“The area under the curve of the process on the TS “
plane represents the quantity of heat transferred during
the process.”
11
Other Energy Relations
2
-∫
Vdp = 𝑊𝑠 + ⧍K
1
(Reversible steady flow, ⧍P = 0)
“The area behind the curve of the process on the pV
planes represents the work of a steady flow process when
⧍K = 0, or it represents ⧍K when 𝑊𝑠 = 0.”
Any process that can be made to go in the reverse direction by an infinitesimal
change in the conditions is called a reversible process.
Any process that is not reversible is irreversible.
Review Problems
1. An automobile tire is inflated to 32 psig pressure at 50°F. After being driven the
temperature assuming the volume remains constant.
Ans. 34.29 psig (EE Board Problem)
2. If 100 ft³ of atmospheric at zero Fahrenheit temperature are compressed to a volume of
1 ft³ at temperature of 200°F, what will be the pressure of the air in psi?
Ans. 2109 psia (EE Board Problem)
3. A 10-ft³ tank contains gas at a pressure of 500 psia, temperature of 85° F and a weight of
25 pounds. A part of the gas was discharged and the temperature and pressure changed to
70° F and 300 psia , respectively. Heat was applied and the temperature was back to 85° F>
find the final weight, volume, and pressure of the gas.
Ans. 15.43 lb; 10 ft³; 308.5 psia (EE Board Problem)
4. Four hundred cubic centimeters of a gas at 740 mm Hg absolute and 18°C undergoes a
process until the pressure becomes 760 mm Hg absolute and the temperature 0°C. What is
the final volume of the gas?
Ans. 365 cc (EE Board Problem)
5. A motorist equips his automobile tires with a relief-type valve so that the pressure inside
the tire never will exceed 240 kPa (gage). He starts a trip with a pressure of 200 kPa (gage)
and a temperature of 23°C in the tires. During the long drive, the temperature of the air in
the tires reaches 83°C. Each tire contains 0.11 kg of air. Determine
(a) the mass of air escaping each tire, (b) the pressure of the tire when the temperature
returns to 23°C.
Ans. (a) 0.0064 kg; (b) 182.48 kPa (gage)
12
6. A 6-m³ tank contains helium at 400 K and is evacuated from atmospheric pressure to a
pressure of 740 m Hg vacuum. Determine (a) mass of helium remaining in the tank, (b)
mass of helium pumped out, (c) the temperature of the remaining helium falls to 10°C.
What is the pressure in kPa?
Ans. (a) 0.01925 kg; (b) 0.7123 kg; (c) 1.886 kPa
7. An automobile tire contains 3730 cu in. of air at 32 psig and 80°F. (a) What mass of air is
in the tire? (b) In operation, the air temperature increases to 145°C. If the tire is inflexible,
what is the resulting percentage increase in gage pressure? (c) What mass of the 145°F air
must be bled off to reduce the pressure back to its original value?
Ans. (a) 0.5041 lb; (b) 17.53%; (c) 0.0542 lb
8. A spherical balloon is 40 ft in diameter and surrounded by air at 60°F and 29.92 in Hg
abs. (a) If the balloon is filled with hydrogen at a temperature of 70°F and atmospheric
pressure, what total load can it lift? (b) If it contains helium instead of hydrogen, other
conditions remaining the same, what load can it lift? (c) Helium is nearly twice as heavy as
hydrogen. Does it have half the lifting force? R for hydrogen is 766.54 and for helium is
386.04 ft.lb/lb.°R.
Ans. (a) 2381 lb; (b) 2209 lb
9. A reservoir contains 2.83 cu m of carbon monoxide at 6895 kPa and 23.6°C. An
evacuated tank is filled from the reservoir to a pressure of 3497 kPa and a temperature of
12.4°C, while the pressure in the reservoir decreases to 6205 kPa and the temperature to
18.3°C. What is the volume of the tank? R for CO is 296.92 J/kg. K°.
Ans. 0.451 m³
10. A gas initially at 15 psia and 2 cu ft undergoes a process to 90 psia and 0.60 cu ft, during
which the enthalpy increases by 15.5 Btu; cv = 2.44 Btu/lb. R°. Determine (a) ⧍U, (b) cp, and
(c) R.
Ans. (a) 11.06 Btu; (b) 3.42 Btu/lb.R°; (c) 762.4 ft.lb/lb°R
11. For a certain gas, R = 0.277 kJ/kg.K and k = 1.384. (a) What are the value of cp and cv?
(b) What mass of this gas would occupy a volume of 0.425 cu m at 517.11 kPa and 26.7°C?
(c) If 31.65 kJ are transferred to this gas at constant volume in (b), what are the resulting
temperature and pressure?
Ans. (a) 0.7214 and 0.994 kJ/kg.R°; (b) 2.647 kg; (c) 43.27°C, 545.75 kPa
12. An unknown gas at p1 = 0.65 KPa, V1 = 0.4m³ undergoes a process to p2 = 0.103 KPa, V2
= 1.54 m³, during which the enthalpy decreases 87.6 KJ and cv = 0.659 KJ/ Kgm-°K.
13
Chapter 4
PROCESSES OF IDEAL GASES
There are five processes involved in any system undergoing a change in a state for
an ideal gas where it can be nonflow for a closed system or steady flow for an open system,
namely:
1. Isometric or Isochoric Process (v = c)
2. Isobaric or Isopiestic Process (p = c)
3. Isothermal Process (T = c)
4. Isentropic or Adiabatic Process (s = c)
m
5. Polytropic Process (pV = c)
A. Constant Volume Process (v = c)
An isometric or Isochoric process is a reversible constant volume process. A
constant volume process may be reversible or irreversible. In this process, the working
substance is contained in a rigid vessel.
(a) Relation between p and T.
T2
T1
=
Fig. 5 Isometric Process
p2
p1
(b) Nonflow work.
2
Wn =
∫pdV = 0
1
1
(c) The change of internal energy.
⧍U = mcv (T2 – T1 )
(d) The heat transferred.
Q = mcv (T2 – T1 )
(e) The change of enthalpy.
⧍H = mcp (T2 – T1 )
(f) The change of entropy.
T2
⧍S = mcv 1n
T1
(g) Reversible steady flow constant volume.
(a) Q = ⧍U + ⧍K + ⧍wf + Ws + ⧍P
Ws = - (⧍wf + ⧍K + ⧍P)
Ws = - ⧍wf = V(p1 –p2 )
(⧍P = 0, ⧍K = 0)
(b) - Vdp = Ws + ⧍K
- V(p2 – p1 ) = Ws + ⧍K
V(p1 – p2 ) = Ws + ⧍K
V(p1 – p2 ) = Ws
(⧍K = 0)
(h) Irreversible nonflow constant volume process.
Q = ⧍U + wn
For reversible nonflow, wn = 0.
For irreversible nonflow, wn = 0.
wn = nonflow work
Ws = steady flow work
2
Problems
1. Ten cu ft of air at 300 psia and 400°F is cooled to 140°F at constant volume. What
are (a) the final pressure, (b) the work, (c) the change of internal energy, (d) the
transferred heat, (e) the change of enthalpy, and (f) the change of entropy?
Solution
1
p
T
2
1
2
V
(a) p2 =
s
p1 T2
T1
=
(300) (600)
860
V = 10 cu ft
p1 = 300 psia
T1 = 400 + 460 = 860°R
T2 = 140 + 460 = 600°R
= 209 psia
(b) W = 0
p1 v1 (300) (144) (10)
(c) m =
=
= 9.147 lb
RT1
(53.34) (860)
⧍U = mcv (T2 – T1 )
= (9.417) (0.1714) (600 - 860)
= -420 Btu
(d) Q = mcv (T2 – T1 ) = -420 Btu
(e) ⧍H = mcp (T2 – T1 )
= (9.417) (0.24) (600 - 860)
= -588 Btu
T2
(f) ⧍S = mcv 1n
T1
= (9.417) (0.1714) 1n 600
860
= -0.581 Btu
°R
2. There are 1.36 kg of gas, for which R = 377 J/kg.k and k = 1.25, that undergo a
nonflow constant volume process from p1 = 551.6 kPa and t1 = 60°C to p2 = 1655 kPa.
During the process the gas is internally stirred and there are also added 105.5 kJ of heat.
Determine (a) t2, (b) the work input and (c) the change of entropy.
3
Solution
2
p
T
1
2
1
V
s
k = 1.25
R = 377 J/kg.k
m = 1.36 kg
Q = 105.5 kJ
p1 = 551.6 kPa
p2 = 1655 kPa
T1 = 60 + 273 = 333K
(a) T2 =
T1 p2
=
(333) (1655)
(b) cv =
R
= 999 K
551.6
p1
377
J
=
= 1508
k-1 1.25-1
kg.K°
⧍U = mcv (T2 – T1 )
= (1.36) (1.508) (999 - 333)
= 1366 kJ
Wn = Q - ⧍U = 105.5 – 1366
= -1260.5 kJ
T2
999
(c) ⧍S = 𝑚cv 1n =
(1.36) (1.508)1n
T1
333
kJ
= 2.253
K
3. A group of 50 persons attended a secret meeting in a room which is 12 meters
wide by 10 meters long and a ceiling of 3 meters. The room is completely sealed off and
insulated. Each person gives off 150 kcal per hour of heat and occupies a volume of 0.2
cubic meter. The room has an initial pressure of 101.3 kPa and temperature of 16°C.
Calculate the room temperature after 10 minutes. (ME Board Problem – April 1948)
4
Solution
2
p
T
1
2
p1 = 101.3 kPa
T1 = 16 + 273 = 289 K
1
V
s
cv = 0.1714
Btu
lb.F°
= 0.1714
cal
g.C°
= 0.1714
kcal
kg.K°
Q = (50 persons) (150 kcal/ person.hour) = 7500 kcal/h
volume of room = (12) (10) (3) = 360 m³
volume of air, V = 360 – (0.2) (50) = 350 m³
mass of air, m =
p1 V
RT1
Q = 7500
kcal 10
h
=
(101.3) (350)
= 427.34 kg
(0.28708) (289)
h = 1250 kcal
60
Q = mcv (T2 –T1 )
1250 = (427.34) (0.1714) (T2 - 289)
T2 = 306.1 K
t 2 = 33.1°C
4. A 1-hp stirring motor is applied to a tank containing 22.7 kg of water. The stirring
action is applied for 1 hour and the tank loses 850 kJ/h of heat. Calculate the rise in
temperature of the tank after 1 hour, assuming that the process occurs at constant volume
and that cv for water is 4.187 kJ/(kg) (C°).
5
Solution
2
p
T
1
2
1
V
s
Irreversible Constant Volume Process
Q
= (-850 kJ/h) (1 h) = -850 kJ
W
= (-1 hp) (h) = (-1hp) (0.746 kW/hp) (h) (3600 s/h)
= -2685.6 kJ
Q
= ⧍U + W
⧍U
= Q – W = -850 – (-2685.6) = 1835.6 kJ
⧍U
= mcv (⧍T)
⧍U
1835.6 kJ
=
=
=19.3 C°
mcv (22.7 kg) (4.187 kJ/kg.C°)
⧍T
5. A closed constant-volume system receives 10.5 kJ of paddle work. The system
contains oxygen at 344 kPa, 278 K, and occupies 0.06 cu m. Find the heat (gain or loss) if
the final temperature is 400 K. (EE Board Problem – April 19, 1988)
Solution
p
2
1
p1 V
RT1
cv = 0.6595 kJ/(kg) (K)
R = 259.90 J/(kg) (K)
p1 = 344 kPa
T1 = 278 K
V = 0.06 m³
T2 = 400 K
1
V
m =
2
T
s
=
(344)(0.06)
= 0.2857 kg
(0.2599)(278)
6
⧍U = mcv (T2 –T1 )
= (0.2857)(0.6595)(400-278)
= 22.99 kJ
Q = ⧍U + W
= 22.99 + (-10.5)
= 12.49 kJ
B. Constant Pressure (p=c)
An isobaric or isopiestic process is an internally reversible process of a substance
during which the pressure remains constant. In this process, the boundary of the system is
inflexible as in a v=c process.
Fig. 6 Isobaric Process
(a) Relation between V and T.
T2
=
T1
V2
T1
(b) Nonflow work.
2
Wn =
∫pdV = p(V –V )
2
1
1
(c) The change of internal energy.
⧍U = mcv (T2 –T1 )
7
(d) The heat transferred.
Q = mcp (T2 –T1 )
(e) The change of enthalpy.
⧍H = mcp (T2 –T1 )
(f) The change of entropy.
T2
⧍S = mcp 1n
T1
(g) Reversible steady flow constant volume.
(a) Q = ⧍P + ⧍K + ⧍H + Ws
Ws = - (⧍K + ⧍P)
Ws = - ⧍K
(⧍P = 0)
2
(b) -
∫
Vdp = Ws + ⧍K
1
0 = Ws + ⧍K
Ws = - ⧍K
Problems
1. A certain gas, with cp = 0.529 Btu/lb. R° and R = 96.2 ft. lb/lb. °R, expands from 5 cu
ft and 80°F to 15 cu ft while the pressure remains constant at 15.5 psia. Compute (a)
T2 , (b) ⧍H, (c) ⧍U and (d) ⧍S. (e) For an internally reversible nonflow process, what
is the work?
8
Solution
p
1
2
2
T
1
V
(a) T2 =
T1 V2
s
=
(540) (15)
= 1620°R
5
V2
(b) m =
p = 15.5 psia
𝑉1 = 5 cu ft
V2 = 15 cu ft
T1 = 80 + 460 = 540 °R
p1 V1 (15.5)(144)(5)
=
= 0.2148 lb
RT1
(96.2)(540)
⧍H = mc𝑝 (T2 –T1 )
= (0.2148)(0.529) (1620-540)
= 122.7 Btu
96.2
(c) cv = cp – R = 0.529 -
Btu
= 0.4059
778
lb. °R
⧍U = mc𝑣 (T2 –T1 )
= (0.2148)(0.4053)(1620-540)
= 94 Btu
(d) ⧍S = mc𝑝 1n
T2
T1
= (0.2148) (0.529) 1n
Btu
= 0.1249
1620
540
°R
9
(e) W𝑛 = p(V2 –V1 )
= (15.5) (144) (15 - 5)
778
= 28.7 Btu
2. A perfect gas has a value of R = 319.2 J/kg.K and k = 1.26. If 120 kJ are added to 2.27 kg
of this gas at constant pressure when the initial temperature is 32.2°C, find (a) T2, (b) ⧍H,
(c) ⧍U, and (d) work for a nonflow process.
Solution
p
1
2
2
T
1
V
(a) c𝑝 =
kR
k = 1.26
m = 2.27 kg
R = 319.2 J/kg.K
Q = 120 kW
T1 = 32.2 + 273 = 305.2 K
s
=
(1.26) (0.3192)
k-1
= 1.5469
1.26 – 1
kJ
kg.K°
Q = mc𝑝 (T2 –T1 )
120 = (2.27) (1.5469) (T2 – 305.2)
T2 = 339.4 K
(b) ⧍H = mc𝑝 (T2 –T1 ) = 120 kJ
(c) CV =
R
0.3192
kJ
=
= 1.2277
k-1 1.26 – 1
kg.K°
⧍U = mc𝑉 (T2 –T1 )
= (2.27) (1.2277) (339.4 – 305.2)
= 95.3 kJ
10
(d) W𝑛 = p(V2 –V1 ) = p
mRT2
-
p2
mRT1
p1
= mR(T2 –T1 )
= (2.27) (0.3192) (339.4 – 305.2)
= 24.78 kJ
C.
Isothermal Process (T = c)
An isothermal process is an internally reversible constant temperature process of a
substance.
Fig. 7. Isothermal Process
(a) Relation between p and V.
p1 V1 = p2 p2
(b) Nonflow work.
2
Wn =
2
∫ pdV =
1
CdV
V2
= C1n
V
1
p1
= p1V11n
V1
(c) The change of internal energy.
⧍U = 0
(d) The heat transferred.
Q = ⧍U + Wn = p1 V1 1n
V2
V1
= mRT1n
p1
p2
11
p2
(e) The change of enthalpy.
⧍H = 0
(f) The change of entropy.
Q
⧍S =
T
= mR1n
p1
p2
(g) Steady flow isothermal.
(a) Q = ⧍P + ⧍K + ⧍H + Ws
Ws = Q - ⧍P - ⧍K
Ws = Q
(⧍P = 0, ⧍K = 0 )
(b)
2
-
∫ Vdp = W
s + ⧍K
1
From pV = C, pdV + Vdp = 0, dp –pdV/V
2
2
2
∫ Vdp =-∫V (-pdV/V) =∫ pdV
-
1
p1 V1 1n
V2
1
1
= Ws + ⧍K
V1
Wn = Ws
(⧍K=0)
12
Problems
1. During an isothermal process at 88°F, the pressure on 8 lb of air drops from 80 psia to 5
psig. For an internally reversible process, determine (a) the ∫ pdV and the work of a
nonflow process, (b) the -∫Vdp and the work of a steady flow process during which ⧍K=0,
(c) Q, (d) ⧍U and ⧍H, and (e) ⧍S
T = 88 + 460 = 548
m = 8lb
p1 = 80 psia
p2 = 5 + 14.7 = 19.7 psia
(a)
∫pdv = p
V2
1
V1 1n
p1
= mRT 1n
V1
=
p2
(8) (53.34) (548)
778
1n
80
19.7
= 421.2 Btu
∫ pdv = 421.2 Btu
Wn =
∫ Vdp = p
(b) -
1
V2
V1 1n
V1
= 421.2 Btu
(c) Q = ⧍U + Wn = 421.2 Btu
(d) ⧍U = 0
⧍H = 0
Q
(e) ⧍S =
421.2
=
T
Btu
= 0.7686
548
°R
13
2. During a reversible process there are abstracted 317 kJ/s from 1.134 kg/s of a certain
gas while the temperature remains constant at 26.7°C . For this gas, cp = 2.232and cv = 1.713
kJ/kg.K . The initial pressure is 586 kPa. For both nonflow and steady flow (⧍P = 0, ⧍K = 0)
process, determine (a) V1, V2 and p2, (b) the work and Q, (c) ⧍S and ⧍H.
Solution
Q = -317 kJ/s
𝑚̇ = 1.134 kg/s
p1 = 586 kPa
T = 26.7 + 273
(a) R = c𝑝 - c𝑣 = 2.232 – 1.713 = 0.519 kJ/kg.K
𝑚̇ RT1
V1 =
(1.134) (0.519) (299.7)
=
586
p1
Q = p1 V1 1n
= 0.301𝑚3 /s
V2
V1
1n
V2
=
V1
V2
Q
-317
=
p1 V1 (586) (0.307)
-1.80
e
=
= -1.80
= 0.1653
V1
V2 = (0.1653) (0.301) = 0.0498 𝑚3 /s
p2 =
p1 V1
=
(586) (0.301)
= 3542 kPa
0.0498
V2
(b) Since ⧍P = 0 and ⧍K = 0, Wn = Ws = Q = -317 kJ/s
Q
-317
(c) ⧍S = =
= -1.058 kJ/K.s
T 299.7
⧍H = 0
14
3. Air flows steadily through an engine at constant temperature, 400 K. Find the
work per kilogram if the exit pressure is one-third the inlet pressure ant the inlet pressure
is 207 kPa. Assume that the kinetic and potential energy variation is neglible. (EE Board
Problem – April 1988)
Solution
p 1
T
pV = C 1
T = 400
R = 287.08 kJ/ (kg) (K)
p1 = 207 kPa
2
2
V
p1
s
=3
p2
RT1 (0.28708) (400)
V1 =
=
= 0.5547 m³/kg
p1
207
W = p1 V1 1n
V2
=
V1
p1
p1 V1 1n
p2
= (207) (0.5547) 1n 3
= 126.1 kJ
D. Isentropic Process (S=c)
An isentropic process is a reversible adiabatic process. Adiabatic simply means no
heat. A reversible adiabatic is one of constant entropy.
Fig. 8. Isenteropic Process
1. Relation among p, V, and T.
(a) Relation between p and V.
k
k
p1V1 = p2V2 = C
(b) Relation between T and V.
15
k
k
From p1 V1 = p2 V2 and
p1 V1
=
p2 V2
T1
, we have
T2
k-1
T2 V1
T1 V2
(c) Relation between T and p.
k-1
k
T2 p2
T1 p1
2. Nonflow work.
From 𝑝𝑉 𝑘 = C, p = 𝐶𝑉 +𝑘
2
W𝑛 =
2
∫ pdV = ∫ 𝐶𝑉
1
2
−𝑘
dV = C1
1
∫𝐶𝑉
−𝑘
1
Integrating and simplifying,
W𝑛 =
p2 V2 – p1 V
1–k
=
mR (T2 –T1 )
1- k
3. The change of internal energy.
⧍U = mc𝑣 (T2 –T1 )
4. The heat transferred.
Q=0
5. The change of enthalpy.
⧍H = mc𝑝 (T2 –T1 )
6. The change of entropy.
⧍S = 0
16
dV
7. Steady flow isentropic.
(a) Q = ⧍P + ⧍K + ⧍H + W𝑠
W𝑠 = -⧍P - ⧍K - ⧍H
W𝑠 = -⧍H
(⧍P = 0, ⧍K = 0)
W𝑠 = K. W𝑛
2
(b) –
∫ Vdp = W + ⧍K
𝑠
1
Let C = 𝑝1/𝑘 V or V = C 𝑝1/𝑘
2
-
2
∫ Vdp = ∫C 𝑝
1
1/𝑘
dp
1
Integrating and simplifying,
2
-
∫ Vdp =
k (p2 V2 – p1 V1 )
1
2
∫ pdV
=k
1- k
1
Problems
1. From a state defined by 300 psia, 100 cu ft and 240°F, helium undergoes and
isentropic process to 0.3 psig. Find (a) V2 and t2, (b) ⧍U and ⧍H, (c) ∫ pdV, (d) - ∫ Vdp, (e)
Q and ⧍S. What is the work (f) if the process is nonflow, (g) if the process is steady flow
with ⧍K = 10 Btu?
Solution
p1 = 300 psia
p2 = 0.3 + 14.7 = 15 psia
V1 = 100 cu ft.
T1 = 240 + 460 = 700°R
17
1
(a) V2 = V1
𝑝1
𝑝2
300
= 100
1.666
= 603.4 𝑓𝑡 3
15
𝑘−1
T2 = T1
𝑝2
1.666−1
𝑘
= 700
𝑝1
1.666
15
300
= 211.3 °R
t 2 = - 248.7°F
(b) m =
𝑝1 𝑉1
𝑅𝑇1
=
(300)(144)(100)
(386.04)(700)
= 15.99 lb
⧍H = 𝑚𝑐𝑝 (T2 –T1 ) = (15.99) (1.241) (211.3 - 700) = - 9698 Btu
⧍U = 𝑚𝑐𝑣 (T2 –T1 ) = (15.99) (0.745) (211.3 - 700) = - 5822 Btu
(c)
∫pdV =
(d) -
𝑝2 𝑉2 − 𝑝1 𝑉
2
1−𝑘
=
(144)(15 𝑋 603.4−300 𝑥 100)
(778)(1−1.666)
= 5822 Btu
∫Vdp = k ∫pdV = (1.666) (5822) = 9698 Btu
(e) Q = 0
⧍S = 0
(f) Q = ⧍U + 𝑊𝑛
𝑊𝑛 = - ⧍U = - (- 5822) = 5822 Btu
(g) -
∫ Vdp = 𝑊 + ⧍K
𝑠
9698 = 𝑊𝑠 + 10
𝑊𝑠 = 9688 Btu
2. An adiabatic expansion of air occurs through a nozzle from 828 kPa and 71°C to
138 kPa. The initial kinetic energy is negligible. For an isentropic expansion, compute the
specific volume, temperature and speed at the exit section.
18
Solution
p1 = 828 kPa
T1 = 71 + 273 = 344 K
p2 = 138 kPa
p2
T2 = T1
𝑘−1
𝑘
= 344
p1
138
1.4−1
1.4
= 206 K
828
t 2 = - 67°C
v1 =
𝑅𝑇1
𝑝1
=
𝑝
v2 = v1 𝑝1
2
(0.28708)(344)
= 0.1193 𝑚3 /kg
828
1
𝑘
= 0.1193
1
1.4
828
= 0.429 𝑚3 /kg
138
⧍h = cp (T2 –T1 ) = 1.0062 (206 - 344) = - 138.9 kJ/kg
Q = ⧍P + ⧍K + ⧍h + Ws
⧍K = - ⧍h = 138,900 J/kg
⧍K = K 2 - K1 =
𝑣²2
2𝑘
𝑣²2
2𝑘
= (2k) (⧍K) = 2(1
𝑘𝑔.𝑚
𝑁.𝑠2
) (138,900
𝑁.𝑚
𝑘𝑔
) = 277,800 m²/s²
𝑣2 = 527.1 m/s
E. Polytropic Process
A polytropic process is an internally reversible process during which
pV n = C and p1 V1n = p2 V2n = pi Vin
where n is any constant.
n=1, n=0 & n=K
Missing Figure
19
1. Relation among p, V, and T
(a) Relation between p and V.
p1 V1n = p2 V2n
(b) Relation between T and V.
n-1
T2
T1
V1
=
V2
(c) Relation between T and p.
T2
T1
p2
=
𝑛−1
𝑛
p1
2. Nonflow work
2
𝑊𝑛 =
∫pdV =
𝑝2 𝑉2 − 𝑝1 𝑉1
1−𝑛
=
𝑚𝑅 (𝑇2 − 𝑇1 )
1−𝑛
1
3. The change of internal energy
⧍U = 𝑚𝑐𝑣 (T2 –T1 )
4. The heat transferred
Q = ⧍U + 𝑊𝑛
= 𝑚𝑐𝑣 (T2 –T1 ) +
=m
𝐶𝑣 − 𝑛𝑐𝑣 +𝑅
=m
𝐶𝑝 − 𝑛𝑐𝑣
1−𝑛
1−𝑛
= 𝑚𝑐𝑣
𝑘−𝑛
1−𝑛
𝑚𝑅 (𝑇2 − 𝑇1 )
1−𝑛
(T2 –T1 )
(T2 –T1 )
(T2 –T1 )
Q = 𝑚𝑐𝑛 (T2 –T1 )
𝑐𝑛 = 𝑐𝑣
𝑘−𝑛
1−𝑛
, the polytropic specific heat
20
5. The change of enthalpy
⧍H = 𝑚𝑐𝑝 (T2 –T1 )
6. The change of entropy
⧍S = 𝑚𝑐𝑛 1n
T2
T1
7. Steady flow polytropic
(a) Q = ⧍P + ⧍K + ⧍H + 𝑊𝑠
𝑊𝑠 = Q - ⧍P - ⧍K - ⧍H
𝑊𝑠 = Q - ⧍H
(⧍P = 0, ⧍K = 0)
2
(b) - ∫ Vdp = 𝑊𝑠 + ⧍K
1
2
- ∫ -Vdp =
2
𝑛 (𝑝2 𝑉2 − 𝑝1 𝑉1 )
1
1−𝑛
= n ∫ pdV
1
Problems
1. During a polytropic process, 10 lb of an ideal gas, whose R = 40 ft.lb/lb. R and cp =
0.25 Btu/lb.R, changes state from 20 psia and 40°F to 120 psia and 340°F. Determine (a) n,
(b) ⧍U and ⧍H, (c) ⧍S, (d) Ω (e) ∫ pdV, (f) - ∫ Vdp. (g) If the process is steady flow during
which ⧍K = 0, what is Ws?
What is ⧍K if Ws = 0? (h) What is the work for a
nonflow process?
Solution
p1 = 20 psia
m = 10 lb
p2 = 120 psia
R = 40
𝑓𝑡.𝑙𝑏
𝑙𝑏.°𝑅
T1 = 40 + 460 = 500°R
21
𝐵𝑡𝑢
T2 = 340 + 460 = 800°R
𝑝
𝑛−1
𝑛
(a) 𝑝2
1
𝑛−1
𝑛
800
= 500
20
𝑛−1
𝑛
𝑇
= 𝑇2
1
120
𝑐𝑝 = 0.25 𝑙𝑏.𝑅°
1n 6 = 1n 1.9
𝑛−1
𝑛
0.4700
= 1.7918
n = 1.356
40
𝐵𝑡𝑢
(b) 𝑐𝑣 = 𝑐𝑝 – R = 0.25 - 778 = 0.1986 𝑙𝑏.𝑅°
⧍U = mcv (T2 –T1 )
= (10) (0.1986) (800 - 500)
= 595.8 Btu
⧍H = mcp (T2 –T1 )
= (10) (0.25) (800 - 500)
= 750 Btu
𝑐𝑝
0.25
(c) k = 𝑐 = 0.1986 = 1.259
𝑣
cn = cv
𝑘−𝑛
= 0.1986
1−𝑛
1.259−1.356
1−1.356
T
𝐵𝑡𝑢
= 0.0541 𝑙𝑏.𝑅°
800
⧍S = mcn 1n T2 = (10) (0.0541) 1n 500 = 0.2543
1
(d) Q = mcn (T2 –T1 )
= (10) (0.0541) (800 - 500)
= 162.3 Btu
22
𝐵𝑡𝑢
°𝑅
(e) ∫pdV =
𝑚𝑅 (T2 –T1 )
1−𝑛
(10)(40)(800−500)
=
(778)(1−1.356)
= - 433.3 Btu
(f) -∫Vdp = n∫pdV = (1.356) (- 433.3) = - 587.6 Btu
(g) 𝑊𝑠 = -∫Vdp = - 587.6 Btu
⧍K = -∫Vdp = - 587.6 Btu
(h) 𝑊𝑛 = ∫pdV = - 433.3 Btu
MISSING FIGURE
1.2
2. Compress 4 kg/s of CO2gas polytropically (pV = C) from p1 = 103.4 kPa, t1 =
60°C to t2 = 227°C. Assuming ideal gas action, find p2, W, Q, ⧍S (a) as nonflow, (b) as a
steady flow process where ⧍P = 0, ⧍K = 0.
Solution
𝑘𝑔
𝑝1 = 103.4 kPa
m=4
𝑇1 = 60 + 273 = 33 K
𝑇2 = 227 + 273 = 500 K
𝑠
(a) Nonflow
n
1.2
n−1
𝑇
𝑝2 = 𝑝1 𝑇2
= (103.4)
1
𝑊̇ 𝑛 = 𝑚̇R
(T2 –T1 )
1−𝑛
= - 631.13
𝑐𝑛 = 𝑐𝑣
𝑘−𝑛
1−𝑛
=
500
1.2−1
333
= 1184.9 kPa
(4)(0.18896)(500−333)
1−1.2
𝑘𝐽
𝑠
= (0.6561)
1.288−1.2
1−1.2
𝑘𝐽
= - 0.2887 𝑘𝑔.𝐾
𝑄̇ = mcn (T2 –T1 ) = (4) (- 0.2887) (500 - 333)
23
𝑘𝐽
= - 193.8 𝐾.𝑠
T
500
⧍𝑆̇ = mcn 1n T2 = (4) (- 0.2887)1n 333
1
kJ
= - 0.4694 K.s
(b) Steady Flow
p2 = 1184.9 kPa
𝑘𝐽
𝑄̇ = - 193.8 𝑠
𝑘𝐽
⧍𝑆̇ = - 0.4694 𝐾.𝑠
⧍𝐻̇ = 𝑚̇𝑐𝑝 (T2 –T1 )
= (4) (0.8452) (500 - 333)
= 563.6
𝑘𝐽
𝑠
𝑄̇ = ⧍𝑃̇ + ⧍𝐾̇ + ⧍𝐻̇ + 𝑊̇𝑠
𝑊̇𝑠 = 𝑄̇ - ⧍𝐻̇ = - 193.8 – 563.6
= - 757.4
𝑘𝐽
𝑠
Curves for Different Values of n
Polytropic processes are all inclusive in that many of the prior equations can be
obtained by choosing proper values of n.
Let n = 0; then pV° = C, or p = C, an isobaric process.
n
Let n = ∞; then, from pV = C, we have
1/n
p
1/∞
V=p
k
V = V = C, an isometric process.
Let n = k; then pV = C, an isentropic process.
Let n = 1; then pV = C, an isothermal process.
Missing Figure
24
The isentropic curve on the pV plane is steeper than the isothermal curve and on the
TS plane the constant volume curve is steeper than the constant pressure curve when both
are drawn between the same temperature limits.
Missing Formula
Review Problems
1. A perfect gas has a value of R = 58.8 ft.lb/lb.R and R = 1.26. If 20 Btu are added to
5 lb of this gas at constant volume when the initial temperature is 90°F, find (a) T2, (b) ⧍H,
(c) ⧍S, (d) ⧍U, and (e) work for a nonflow process.
Ans. (a) 563.8°R; (b) 25.27 Btu; (c) 0.036 Btu/°R; (d) 20.06 Btu
2. A reversible, nonflow, constant volume process decreases the internal energy by
316.5 kJ for 2.268 kg of a gas for which R = 430 J/kg.K and k = 1.35. For the process,
determine (a) the work, (b) Q, and (c) ⧍S. The initial temperature is 204.4°C.
Ans. (a) 0; (b) -316.5 kJ; (c) -0.7572 kJ/ K.
3. A 10-ft³ vessel of hydrogen at a pressure of 305 psia is vigorously stirred by
paddles until the pressure becomes 400 psia. Determine (a) ⧍U and (b) W. No heat is
transferred, cv = 2.434 Btu/lb.R
Ans. (a) 434 Btu; (b) -434 Btu
4. Three pounds of a perfect gas with R = 38 ft.lb/lb.R and k = 1.667 have 300 Btu of
heat added during a reversible nonflow constant pressure change of state. The initial
temperature is 100°F. Determine the (a) final temperature, (b) ⧍H, (c) W, (d) ⧍U, and (e)
⧍S.
Ans. (a) 919°F; (c) 120 Btu; (d) 180 Btu; (e) 0.3301 Btu/°R
25
5. While the pressure remains constant at 689.5 kPa the volume of a system of air
changes from 0.567 m³ to 0.283 m³. What are (a) ⧍U, (b) ⧍H, (c) Q, (d) ⧍S? (e) If the
process is nonflow and internally reversible, what is the work?
Ans. (a) – 490.2 kJ; (b) -686.3 kJ; (c) -686.3 kJ;
(d) -0.6974 kJ/kg.K; (e) -195.8 kJ
6. Four pounds of air gain 0.491 Btu/°R of entropy during a nonflow isothermal
process. If p1 = 120 psia and V2 = 42.5 ft³, find (a) V1 and T1, (b) W, (c) Q, and (d) ⧍U.
Ans. (a) 7.093 ft³, 574.5°R; (b) 282.1 Btu; (c) 282.1 Btu; (d) 0
7. If 10 kg/min of air are compressed isothermally from p1 = 96 kPa and V1 = 7.65
m³/min to p2 = 620 kPa, find the work, the change of entropy and the heat for (a) nonflow
process and (b) a steady flow process with υ1 = 15 m/s and υ2 = 60 m/s.
Ans. (a) -1370 kJ/min, -5.356 kJ/K.min; (b) -1386.9kJ/min
8. One pound of an ideal gas undergoes an isentropic process from 95.3 psig and a
volume of 0.6 ft³ to a final volume of 3.6 ft³. If cp = 0.124 and cv = 0.093 Btu/lb.R, what are
(a) t2, (b) p2, (c) ⧍H and (d) W.
Ans. (a) -243.1°F; (b) 10.09 psia; (c) -21.96 Btu; (d) 16.48 Btu
9. A certain ideal gas whose R = 278.6 J/kg.K and cp = 1.015 kJ/kg.K expands
isentropically from 1517 kPa, 288°C to 965 kPa. For 454 g/s of this gas determine, (a) Wn,
(b) V2, (c) ⧍U and (d) ⧍H.
Ans. (a) 21.9 kJ/s; (b) 0.06495 m³/s; (d) -30.18 kJ/s
in
10. A polytropic process of air from 150 psia, 300°F, and 1 ft³ occurs to p2 = 20 psia
1.3
accordance with pV = C. Determine (a) t2 and V2, (b) ⧍U, ⧍H and ⧍S, (c) ∫ pdV and – ∫
Vdp. (d) Compute the heat from the polytropic specific heat and check by the equation Q =
⧍U + pdV. (e) Find the nonflow work and (f) the steady flow work for ⧍K =0.
Ans. (a) 17.4°F, 4.711 ft³; (b) -25.81 Btu, -36.14 Btu, 0.0141 Btu/°R; (c) 34.41 Btu,
44.73 Btu; (d) 8.60 Btu; (e) 34.41 Btu; (f) 44.73 Btu
1.30
26
11. The work required to compress a gas reversibly according to pV = C is 67, 790
J, if there is no flow. Determine ⧍U and Q if the gas is (a) air, (b) methane. For methane, k =
1.321, R = 518.45 J/kg.K, cv = 1.6187, cp = 2.1377 kJ/kg.K.
Ans. (a) 50.91 KJ, -16.88 kJ; (b) 63.50 kJ, -4.29 kJ
12. Air at 22°C, 1.02 bar initially occupying a cylinder volume of 0.015m³, is
compressed reversibly and isentropically by a piston to a pressure of 6.8 bar. Compute t2,
V2 and Wn.
13. 0.44 Kg of air 180° expands adiabatically to three times its original volume and
during the process, there is a decrease in temperature to 15°C. The work done
during the process is
1.3
14. 1 Kg of ethane is compressed from 1.1 bar, 27°C according to a law p1V = C,
until the pressure is 6.6 bar. Calculate the heat flow to or from the cylinder walls.
15. A cylinder contains 0.45 m² of a gas at 1 X 10⁵ N/m² and 80°C. The gas is
compressed to a volume of 0.13m³, the fluid pressure being 5 X 10⁵ N/m².
Compute: (a) The mass of gas
(b) The value of index “n” for compression
(c) The increase interval energy of the gas
(d) The heat added or rejected bv
The gas during compression.
Take R = 294.z J/Kg°C
k = 1.4
16. A fluid undergoes a reversible adiabatic compression from 0.5 mPa, 0.2m³
1.3
according to the law, pV = C. Calculate the change in enthalpy, entropy and interval energy.
Also, compute the heat transfer and work nonflow during the process.
ft -lbf
17. Fine pounds of a certain ideal gas with K = 1.30 and R = 35
is compressed
lbm-°n
in polytropic process is a closed system from 15 psia, 80°F to 60psia, 315°F. Compute the
polytropic index n, the work nonflow and the heat interactions with the surroundings.
ft -lbf
27
18. One pound of air R = 53.3
is compressed isothermally in a closed
lbm - °R
system from 15 psia and 80°F to 90 psia. Compute the heat transfer.
19. 1/10 pound Nitrogen with g1 = 0.248 BTV/ lbm - °R is contained in a closed
system with V1 = 1.0 cu ft. Heat transfer is allowed to take place until V2 = 90% of its initial
value while the pressure is kept constant at 20 psia. Compute the final temperature and the
heat transfer.
20. Helium is compressed in an adiabatic cylinder from 70°F, 10 ft³, 15 psia to 1 ft³
according to relation pV = c. Find the work nonflow, the first temperature and the change of
enthalpy.
1.3
21. Air is compressed in a cylinder in a given relation pV = constant. The process
start with 14 psia, 80°F and the pressure increases to 80 psia. If the process is frictionless,
compute the heat transferred per lbm of air.
22. Ammonia rapor is compressed in a closed system from 15 psia, 3 cu ft, to 150
n
psia, 0.4 cu ft. according to the relation pV = constant. Assuming the rapor to be an
ideal gas
lbm – ft
BN
with the gas constant 90.72
and cv = 0.384
, compute the heat transferred
lbf - °R
lbm - °R
and the work.
23. Air with Cv = 0.723
kJ
is compressed in a piston-cylinder machine according
Kgm - °K
to pV = c from an initial temperature of 17°c and pressure of 1 bar, to a final pressure of 5
bars. Determine the final temperature, the heat transferred and the work.
1.30
KJ
24. There are 1.5 Kgm of a gas where K = 1.3 and R = 0.38
28
Kgm - °K
that undergo an
isochoric process from p1 = 0.552 MPa, t1 = 58.5°c to p2 = 1.66 MPa. During the process,
there added 100 Ks of heat. Compute the heat transferred, change if interval energy and the
change of entropy.
25. Helium at 10.14 MPa, -108°C expands at constant temperature to 1 MPa. For 1.5
Kgm, calculate the work nonflow, work steady flow.
26. There are added 300 BN of heat from 2 Kgm of a certain gas that undergoes an
KJ
isothermal process at 27°c. For this gas R = 0.519
and K = 1.3. The initial pressure is
Kg - °K
0.590 MPa. For both nonflow and steady flow, compute V1, V2, p2, Work, ⧍S, ⧍H and Q.
27. 1000CFM of air are compressed at constant temperature of 85°F and 198 psia to
580 psia. For both nonflow and steady flow, compute ∫p & v, ∫ Vdp, ⧍S and ⧍H.
Kg
28. During an isentropic of 2
of oxygen, the temperature increases from 20°c to
sec
120°c. Calculate ⧍H, ⧍W, Q, ⧍S, ∫ pdv and - ∫ vdp.
29. 10
lb
of COgas are compressed polytropically with index 1.3 from p1 = 14 psia,
sec
t1 = 138°F, to 338°F. Compute for p2, W, ⧍S and Q assuming ideal gas action for a nonflow
process.
lb
30. 0.98
ft³
of Nitrogen initially at 390 psia, 9
, 299°F are cooled isometrically
sec
min
to 118°F in an intermally reversible manner. Compute p2, ∫ pdv, - ∫ vdp, ⧍U, ⧍H, Q and
⧍S for a nonflow process.
29
Chapter 5
GAS CYCLES
In this section, gas serves as the working fluid which does not undergo any phase
charge.
Heat engine or thermal engine is a closed system (no mass crosses its boundaries)
that exchanges only heat and work with its surrounding and that operates in cycles.
Elements of a thermodynamic heat engine with a fluid as the working substance:
1. a working substance, matter that receives heat, rejects heat, and does work;
2. a source of heat (also called a hot body, a heat reservoir, or just source); from
which the working substance receives heat;
3. a heat sink (also called a receiver, a cold body, or just sink), to which the working
substance can reject heat; and
4. an engine, wherein the working substance may do work or have work done on it.
A thermodynamic cycle occurs when the working fluid of a system experiences a
number of processes that eventually return the fluid to its initial state.
W
QA
QA = heat added
QR = heat rejected
W = net work
ENGINE
QR
Available energy is that part of the heat that was converted into mechanical work.
Unavailable energy is the remainder of the heat that had to be rejected onto the
receiver (sink).
The Second Law of Thermodynamics
work.
All energy received as heat by a heat-engine cycle cannot be converted into mechanical
Work of a Cycle
(a) W = ΣQ
W = QA + (-QR )
W = QA – QR
(Algebraic sum)
(Arithmetic difference)
(b) The net work of a cycle is the algebraic sum of the works done by the individual
processes.
W = ΣW
W = W1−2 + W2−3 + W3−4 + …
1
A. The Carnot Cycle (1824)
The Carnot cycle is the most efficient cycle conceivable. There are other ideal cycles
as efficient as the Carnot cycle, but none more so, such a perfect cycle forms a standard of
comparison for actual engines and actual cycles and also for other less efficient ideal cycles,
permitting as to judge how much room there might be for improvement.
Fig 11. The Carnot Cycle
Operation of the Carnot Engine
A cylinder C contains m mass of a substance. The cylinder head, the only place
where heat may enter or leave the substance (system) is placed in contact with the source
of heat or hot body which has a constant temperature T1. Heat flows from the hot body into
the substance in the cylinder isothermally, process 1-2, and the piston moves from 1’ to 2’.
Next, the cylinder is removed from the hot body and the insulator I is placed over the head
of the cylinder, so that no heat may be transferred in or out. As a result, any further process
is adiabatic. The insentropic change 2-3 now occurs and the piston moves from 2’ to 3’.
When the piston reaches the end of the stroke 3’, the insulator I is removed and the
cylinder head is placed in contact with the receiver or sink, which remains at a constant
temperature T3. Heat then flows from the substance to the sink, and the isothermal
compression 4-1 returns the substance to its initial condition, as the piston moves from 4’
to 1’.
MISSING FIGURE 12
Analysis of the Carnot Cycle
𝑄𝐴 = 𝑇1 (𝑆2 – 𝑆1), area 1-2-n-m-1
2
𝑄𝑅 = 𝑇3 (𝑆4 – 𝑆3 ), area 3-4-m-n-3
= -𝑇3 (𝑆3 – 𝑆4 ) = -𝑇3 (𝑆2 – 𝑆1)
W = 𝑄𝐴 – 𝑄𝑅 = 𝑇1 (𝑆2 – 𝑆1) – 𝑇3 (𝑆2 – 𝑆2 )
= (𝑇1 – 𝑇3 ) (𝑆2 – 𝑆1), area 1-2-3-4-1
𝑊
(𝑇1− 𝑇3 )(𝑆2− 𝑆1 )
e=𝑄 =
𝑇1 (𝑆2− 𝑆1 )
𝐴
e=
𝑇1− 𝑇3
𝑇1
The thermal efficiency is defined as the fraction of the heat supplied ta a
thermodynamic cycle that is converted into eork.
Work from the TS plane
𝑉
𝑄𝐴 = mR𝑇1 1n𝑉2
1
𝑉
𝑉
𝑄𝑅 = mR𝑇3 1n 𝑉4 = -mR𝑇3 1n 𝑉3
3
4
From process 2-3,
k-1
𝑇3
𝑇2
=
𝑉2
𝑉3
From process 4-1,
k-1
𝑇4
𝑇1
𝑉2
=𝑉
4
but 𝑇4 = 𝑇3 and 𝑇1 = 𝑇2
k-1
therefore,
𝑉2
𝑉3
then,
=
𝑉3
𝑉4
k-1
𝑉1
𝑉4
𝑉
= 𝑉2
1
𝑉
𝑄𝑅 = -mR𝑇3 1n 𝑉2
1
W = 𝑄𝐴 – 𝑄𝑅 = mR𝑇1 1n
𝑉2
𝑉1
𝑉
-mR𝑇3 1n 𝑉2
1
3
𝑉
W = (𝑇1− 𝑇3 )m𝑅1 n 𝑉2
1
e=
e=
𝑊
𝑄𝐴
=
𝑉
(𝑇1− 𝑇3 )𝑚𝑅1𝑛 2
𝑚𝑅 𝑇1 1𝑛
𝑉2
𝑉1
𝑉1
𝑇1− 𝑇3
𝑇1
Work from the pV plane.
W = ΣW = 𝑊1−2+ 𝑊2−3 + 𝑊3−4 + 𝑊4−1
W = 𝑝1 𝑉1 1n
𝑉2
𝑉1
+
𝑝3 𝑉3− 𝑝2 𝑉2
1−𝑘
+ 𝑝3 𝑉3 1𝑛
𝑉4
𝑉3
+
𝑝1 𝑉1− 𝑝4 𝑉4
1−𝑘
Mean Effective Pressure (pm or mep)
𝑃𝑚
𝑊
=𝑉
𝐷
VD = displacement volume, the volume swept by the piston in one stroke.
Mean effective pressure is the average constant pressure that, acting through one
stroke, will do on the piston the network of a single cycle.
Ratio of Expansion, Ratio of Compression
𝑣𝑜𝑙𝑢𝑚𝑒 𝑎𝑡 𝑡ℎ𝑒 𝑒𝑛𝑑 𝑜𝑓 𝑒𝑥𝑝𝑎𝑛𝑠𝑖𝑜𝑛
Expansion ratio = 𝑣𝑜𝑙𝑢𝑚𝑒 𝑎𝑡 𝑡ℎ𝑒 𝑏𝑒𝑔𝑖𝑛𝑛𝑖𝑛𝑔 𝑜𝑓 𝑒𝑥𝑝𝑎𝑛𝑠𝑖𝑜𝑛
𝑉
Isothermal expansion ratio = 𝑉2
1
𝑉
Isentropic expansion ratio = 𝑉3
2
𝑉
Overall expansion ratio = 𝑉3
1
Compression ratio =
𝑣𝑜𝑙𝑢𝑚𝑒 𝑎𝑡 𝑡ℎ𝑒 𝑏𝑒𝑔𝑖𝑛𝑛𝑖𝑛𝑔 𝑜𝑓 𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛
𝑣𝑜𝑙𝑢𝑚𝑒 𝑎𝑡 𝑡ℎ𝑒 𝑒𝑛𝑑 𝑜𝑓 𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛
4
𝑉
Isothermal compression ratio = 𝑉2
4
𝑉
Isentropic compression ratio, 𝑟𝑘 = 𝑉4
1
𝑉
Overall compression ratio =𝑉3
1
The isentropic compression ratio 𝑟𝑘 is the compression ratio most commonly used.
B. Stirling Cycle (1827)
The stirling Cycle is composed of two reversible isothermos and two reversible
isomers. For 1 kgm of ideal gas
Missing Figure
Missing Formulas
C. The Ericsson cycle is composed of two reversible isotherms and two reversible isobars.
For 1 kgm of ideal gas.
Missing Figure
Missing Formulas
Problems
1. A Carnot power cycle operates on 2 lb of air between the limits of 70°F and 500°F.
The pressure at the beginning of isothermal expansion is 400 psia and at the end of
isothermal expansion is 185 psig. Determine (a) the volume at the end of isothermal
compression, (b) ⧍S during an isothermal process, (c) QA, (d) QR, (e) W, (f) e, (g) the ratio of
expansion during isotheral heating and the overall ratio of expansion, and (h) the mean
effective pressure.
Missing Figure &
m = 2 lb
𝑝1 = 400 psia
𝑇1 = 960°R
𝑝2 = 199.7 psia
𝑇3 = 530°R
5
Point 1:
𝑉1 =
𝑚𝑅𝑇1
𝑝1
=
(2)(53.34)(960)
=
(2)(53.34)(960)
(400)(144)
= 1.778 𝑓𝑡 3
Point 2:
𝑉2 =
𝑚𝑅𝑇2
𝑝2
(199.7)(144)
= 3.561 𝑓𝑡 3
Point 3:
𝑇
𝑘
𝑘−1
𝑝3 = 𝑝2 𝑇𝑎
= (199.7)
2
𝑉3 =
𝑚𝑅𝑇3
𝑝3
=
530
1.4
1.4−1
= 24.57 psia
960
(2)(53.34)(530)
(24.97)(144)
= 15.72 𝑓𝑡 3
Point 4:
𝑉
1.778
𝑉4 = 𝑉3 𝑉1 = (15.72) 3.561 = 7.849 𝑓𝑡 3
2
(a) 𝑉4 = 7.849 𝑓𝑡 3
𝑉
(b) ∆ 𝑆1−2 = Mr1n 𝑉1 =
2
(2)(53.34)
(778)
3.561
1n 1.778 = 0.0952
Btu
°R
(c) 𝑄𝐴 = - 𝑇1 (∆S) = (960) (0.0952) = 91.43 Btu
(d) 𝑄𝑅 = - 𝑇3 (∆S) = - (530) (0.0952) = - 50.46 Btu
(e) W = 𝑄𝐴 – 𝑄𝑅 = 91.43 – 50.46 = 40.97 Btu
𝑊
40.97
(f) e = 𝑄 = 91.43 = 0.4481 or 44.81%
𝐴
𝑉
(g) Isothermal expansion ratio = 𝑉1 = 1.778 = 2
2
𝑉
15.72
Overall expansion ratio = 𝑉3 = 1.778 = 8.84
1
6
𝐸
(h) 𝑝𝑚 = 𝑉 = 𝑉
(40.97)(778)
𝑊
3− 𝑉1
𝐷
= (15.72−1.778)(144) = 15.88 psi
2. A Carnot engine operating between 775 K and 305 K produces 54 kJ of work.
Determine (a) QA, (b) ⧍S during heat rejection, and (c) e.
Missing Figure
Solution
𝑇1 = 775 K
𝑇3 = 305 K
W = 54 kJ
(a) e =
𝑇1− 𝑇3
𝑄𝐴 =
𝑊
𝑒
775−305
=
𝑇1
775
= 0.6065 or 60.65%
54
= 0.6065 = 89.04 kJ
(b) 𝑄𝑅 = 𝑄𝐴 - W = 89.04 – 54 = -35.04 Kj
⧍𝑆3−4 =
𝑄𝑅
𝑇3
=
−35.04
305
= -0.115
𝑘𝐽
𝐾
(c) e = 60.65%
Three-Process Cycle
Problems
1. Ten cu ft of helium at 20 psia and 80°F are compressed isentropically 1-2 to 80
psia. The helium is then expanded polytropically 2-3 with n = 1.35 to the initial
temperature. An isothermal 3-1 returns the helium to the initial state. Find 𝑇2 , 𝑉2, 𝑃3 , 𝑄𝐴 ,
𝑄𝑅 , W, ⧍𝑆3−1, and 𝑃𝑚 .
Missing Figure
p1 = 20 psia
𝑇1 = 540°R
𝑉1= 10 cu ft
𝑝2 = 80 psia
𝑇3 = 540°R
7
m=
𝑝1𝑉1
𝑅𝑇1
(20)(144)(10)
= (386.04)(540) = 0.1382 lb
Point 2:
𝑘−1
1.666−1
𝑘
𝑝2
𝑇2 = 𝑇1 𝑝
= (540)
1
𝑝
80
1.666
= 393.9°R
20
1
1.35
𝑘
1.35−1
𝑉2= 𝑉1 𝑝1
20
= (10)
2
= 4.351 ft3
80
Point 3:
𝑇
𝑛
1.35
𝑛−1
1.35−1
𝑃3 = 𝑃2 𝑇3
540
= (80)
2
𝑃
= 9.435 psia
939.9
1
1
𝑛
1.35
𝑉3= 𝑉2 𝑃2
= (4.351)
3
𝑘−𝑛
𝑐𝑛 = 𝑐𝑣 1−𝑛 = (0.754)
80
= 21.2 ft3
4.435
1.666−1.35
1−1.35
𝐵𝑡𝑢
= -0.6808 𝑙𝑏.°R
𝑄𝐴 = (m) (𝑐𝑛 ) (T3 –T2 ) = (0.1982) (-0.6808) (540 – 939.9)
= 37.63 Btu
𝑉
𝑄𝑅 = 𝑚𝑅𝑇31n 𝑉1 =
(0.1382)(386.04)(540)
778
3
10
1n 21.2
= - 27.82 Btu
W = 𝑄𝐴 - 𝑄𝑅 = 37.63 – 27.82 = 9.81 Btu
∆𝑆3−1=
𝑄𝑅
𝑇1
=
− 27.82
540
= -0.0515
𝐵𝑡𝑢
°𝑅
(9.81)(778)
𝑊
𝑝𝑚 = 𝑉 − 𝑉 = (21.2−4.351)(144) = 3.15 psi
3
2
8
2. Two and a half kg of an ideal gas with R = 296.9 J(kg) (K) and cv = 0.7442
kJ/(kg)(K) at a pressure of 827.4 kPa and a temperature of 667°C reject 132.2 kJ of heat at
1.25
constant pressure. The gas is then expanded according to pV = C to a point where a
constant volume process will bring the gas back ti its orinal state. Determine p3, QA, and the
power in kW for 100 Hz.
Missing Figure
𝑝1 = 827.4 kPa
𝑇1 = 677 + 273 = 950K
𝑄𝑅 = - 132.2 kJ
𝑘𝐽
𝑐𝑝 = 𝑐𝑣 + R = 0.7442 + 0.2969 = 1.0411 𝑘𝑔.𝐾°
𝑐𝑝
1.0411
k = 𝑐 = 0.7442 = 1.399
𝑣
Point 1:
𝑉1 =
𝑚𝑅𝑇1
𝑝1
=
(2.5)(0.2969)(950)
827.4
= 0.8522 𝑚3
Point 2:
𝑄𝑅 = mcp (T2 –T1 )
-132.2 = (2.5) (1.0411) (T2 –950)
T2 = 899.2 K
T
V2 = V1 T2 = (0.8522)
899.2
950
1
= 0.8066 𝑚3
Point 3:
n–1
V2
T3 = T2 V
1.25 - 1
= (899.2)
1
n
V2
p3 = p2 V
3
0.8066
0.8522
= 886.9 K
1.25
= (827.4)
cn = cv
𝑘−𝑛
1−𝑛
0.8066
0.8522
= (0.7742)
= 772.4 kPa
1.399−1.25
1−1.25
𝑘𝐽
= - 0.4435 𝑘𝑔.𝐾°
9
𝑄𝐴 = mcn (T2 –T1 ) + mcv (T1 –T3 )
𝑄𝐴 = (2.5) (-0.4435)(886.9-899.2) + (2.5)(0.7442)(950 – 886.9)
𝑄𝐴 = 131 kJ
W = 𝑄𝐴 - 𝑄𝑅 = 131 – 132.2 = - kJ
𝑘𝐽
W = 1.2 𝑐𝑦𝑐𝑙𝑒 100
𝑐𝑦𝑐𝑙𝑒𝑠
𝑠
= - 120kW
Review Problems
1. The working substance for a Carnot cycle is 8 lb of air. The volume at the
beginning of isothermal expansion is 9 cu ft and the pressure is 300 psia. The ratio of
expansion during the addition of heat is 2 and the temperature of the cold body is 90°F.
Find (a) QA, (b) QR, (c) V3, (d) p3, (e) V4, (f) p4, (g) pm, (h) the ratio of expansion during the
isentropic process, and (i) the overall ratio of compression.
Ans. (a) 346.4 Btu; (b) -209.1 Btu; (c) 63.57 cu ft; (d) 25.64 psia; (e) 31.79 cu ft; (f)
51.28 psia; (g) 13.59 psia; (h) 3.53; (i) 7.06
2. Gaseous nitrogen actuates a Carnot power cycle in which the respective volumes
at the four corners of the cycle, starting at the beginning of the isothermal expansion, are V1
= 10.10 L, V2 =14.53 L, V3 = 226.54 L, and V4 = 157.73 L. The cycle receives 21.1 kJ of heat.
Determine (a) the work and (b) the mean effective pressure.
Ans. (a) 14.05 kJ; (b) 64.97 kPa
3. Show that the thermal efficiency of the Carnot cycle in terms of the isentropic
compression ratio rk is given by
1
e=1–
k-1
rk
4. Two and ne-half pounds of air actuate a cycle composed of the following
processes: polytropic compression 1-2, with n = 1.5; constant pressure 2-3; constant
volume 3-1. The known data are: p1 = 20 psia, t1 = 100°F, QR = -1682 Btu. Determine (a) T2
and T3, (b) the work of the cycle using the pV plane, in Btu; (c) QA, (d) the thermal
efficiency, and (e) pm.
Ans. (a) 1120°R, 4485°R; (b) 384.4 Btu; (c) 2067 Btu; (d) 18.60%; (e) 106.8 psia
5. A three-process cycle of an ideal gas, for which cp = 1.064 and cv = 0.804 kJ/kg.K°,
is initiated by an isentropic compression 1-2 from 103.4 kPa, 27°C to 608.1 kPa. A constant
10
volume process 2-3 and a polytropic 3-1 with n = 1.2 complete the cycle. Circulation is a
steady rate of 0.905 kg/s, compute (a)QA, (b) W, (c) e, and (d) pm.
Ans. (a) 41.4 kJ/s; (b) -10 kJ/s; (c) 24.15%; (d) 19.81 kPa
BTU
BTU
6. An ideal gas with R = 0.062 lbm - °R and Cv =0.158 lbm - °R undergoes a
three process cycle in closed system. From the initial state of 20 psia and 570 °R, the gas is
compressed at constant temperature to 1/5 of its initial volume, process 1-2; it is then
heated isochoric ally to state 3,process 2-3; and finally, it expands polytropically with index
n= 1.5 back to its initial state, process 3-1. Determine the pressures, specific volumes and
temperatures at cardinal points around the cycle and the cycle thermal efficiency.
7. A three process cycle involving 1 lbm of a gas is made up of the following process:
1.4
Process 1-2: The gas is compressed according to pv = c from p1 = 14.7 psia, p1= 0.007
Lbm to p2 = 29.4 psia,
ft³
Process 2-3: The gas in heated isobarically until ρ1 = ρ2.
Process 3-1: The gas is cooled isometrically
Draw the pV and Ts diagram and calculate the network of the cycle.
8. A thermodynamic cycle is composed of the following reversible process: 1-2:
Isothermal compression; 2-3: Isometric Heating; 3-1: Polytrophic process, with index n=
1.45. Sketch the pV and Ts diagram and write the expression for QA, QR, WNET and e.
9. Consider the following data for problem #8 if it operates on 5 lb of Nitrogen with
p1= 15 psia, t1= 100 °F. During the constant temperature process, there are 316 BTU of heat
transferred from N2. Compute the isothermal compression ratio, p2, p3,T3, QA, QR, WNET and
e.
10. In a Stirling cycle with an ideal regenerator, the volume varies between 0.03 and
0.006 cum, the maximum pressure is 2 atmosphere, and the temperature varies between
500°C and 250 °C. The working fluid is air. Compute the network and the cycle thermal
efficiency.
11. An Ericsson cycle operated on 0.75 lbm O2 from 60 psia and 1200 °F. If the
isothermal expansion ratio is 3, compute QA, QR, WNET, e and Pm.
12. Solve the same problem above, except that the working fluid 0.5 N2.
11
13. In a stirling cycle, the volume varies between 1 in ft and 2 in ft while the
temperature varies between 1000°F and 500 °F, and the maximum pressure is 30 psia. The
working fluid is air
(a) Find the thermal efficiency
(b) Find the work per cycle.
14. Air is made to pass through a stirling cycle. At the beginning of the constant
temperature expansion, p1= 105 psia, v1 = 2 in ft, t1= 600°F. The isothermal expansion ratio
is r= 1.5, and the minimum temperature in the cycle k is t2 = 80 °F. Compute △S during the
isothermal process, QA, QR, WNET and e. Also, compute the Mep.
15. Consider 2 lbm of O2 actuate a Stirling cycle between the temperature limits of
240°F and -160°F for cryogenic use. If the maximum pressure in the cycle is 200 psia and
the isothermal compression ratio is rk = 4, compute QA, QR, WNET , e and Mep.
16. An air standard Ericsson Cycle operates under steady- flow conditions. Air is at
120 KPa and 27°C at the beginning of the isothermal compression process, during which
150 KJ/Kg of heat is rejected. The air is at 927°C during the isothermal heat-addition
process. Compute the maximum pressure in the cycle, the network and thermal efficiency
of the cycle.
12
Chapter 6 INTERNAL COMBUSTION ENGINES
Any type of engine that derives energy from combustion of fuel and converts this
energy into mechanical work is called a heat engine.
Internal combustion engine is a heat engine deriving its power from the energy
liberated by the explosion of a mixture of some hydrocarbon, in gaseous or vaporized form,
with atmospheric air.
A. Spark-Ignition (SI) or Gasoline Engine
Fig. 13. Four-Stroke Cycle Gasoline Engine
A cycle begins with the intake stroke as the piston, moves down the cylinder and draws in a fuel-air mixture.
Next, the piston compresses the mixture while moving up the cylinder. At the top of the compression stroke,
the spark plug ignites the mixture. Burning gases push the piston down for the power stroke. The piston then
moves up the cylinder again, pushing the burned gases out during the exhaust stoke.
The four- stroke cycle is one wherein four strokes of the piston, two revolution, are
required to complete a cycle.
Otto Cycle
The Otto cycle is the ideal prototype of spark-ignition engines.
1
Fig. 14. Air-Standard Otto Cycle
Air-Standard Cycle means that air alone is the working medium.
PROCESSES:
1-2: isentropic compression (S1 = S2)
2-3: constant volume addition to heat (V1 = V2)
3-4: isentropic expansion (S3 = S4)
4-1: constant volume rejection of heat (V4 = V1)
Analysis of Otto Cycle
QA
QR
W
=
=
=
mcv(T3-T2)
mcv (T1-T4) = -mcv (T4-T1)
QA – QR = mcv (T3-T2) – mcv (T4-T1)
W
e
=
QA
mcv(T3-T2) -mcv (T4-T1)
=
mcv(T3-T2)
T4-T1
e
=
1–
T3-T2
(1)
1
2
e
=
1-
k-1
rk
V1
where rk = V2 , the isentropic compression ratio
Derivation of the formula for e
Process 1-2:
k-1
T2=[V1/V2]
T1
k-1
T2 = T1 rk
(2)
Process: 3-4:
k-1
T3=[V4/V3]
T4
k-1
= [V1/V2]
k-1
T3 = T4 rk
(3)
Substituting equations (2) and (3) in equation (1)
e
=
1–
T4-T1
k-1
T4 rk
1
e
=
1-
k-1
- T1 rk
k-1
rk
Work from the pV plane
W = ∑W =
p2 V2 –p1 V1
1-k
p4V4-p3V3
+
1-k
Clearance volume, per cent clearance
3
rk =
V1
=
V2
VD + V3
V3
=
V D + c VD
c Vd
Where c = per cent clearance
V3 = clearance volume
VD = displacement volume
Ideal standard of comparison
Cold-air standard, k = 1.4
Hot-air standard, k < 1.4
The thermal efficiency of the theoretical Otto cycle is
1. Increased by increased in rk
2. Increased by increased in k
3. Independent of the heat added
The average family car has a compression ratio of about 9.1
The economical life of the average car is 8 years or 80,000 miles of motoring.
Problems
1. An Otto cycle operates on 0.1 lb/s of air from 13 psia and 130 F at the beginning
of compression. The temperature at the end of combustion is 5000 R;
compression ratio is 5.5;hot-air standard, k = 1.3 (a) Find V1, P, t2,p3 V3, t4, and p4.
(b) Compute QA, QR, W, e and the corresponding hp.
2. The conditions at the beginning of compression in an Otto engine operating on
hot-air standard with k = 1.34, are 101.3 kPa, 0.038 m3 and 32 C. The clearance
is 10% and 12.6 kJ are added per cycle. Determine (a) V2, T2, p2, T3, p3, T4 and p4
(b) W, (c) e, and (d) pm.
4
A cycle begins with the intake stroke the piston moves down and draws air into the
cylinder. The piston rises and compresses the air. During the compression stroke, the
temperature of the air rises to about 900 F (480 C). When oil is injected into the cylinder, it
mixes with the hot air and burns explosively. Gases produced by this combustion action
push the piston down for the power stroke. During the exhaust stroke, the piston moves up
again and forces the burned gases out of the cylinder.
Diesel Cycle (1892)
5
Fig. 16. Air-Standard Diesel Cycle
PROCESSES:
1-2: isentropic compression
2-3: constant-pressure addition to heat
3-4: isentropic expansion
4-1: constant volume rejection of heat (V4 = V1)
Analysis of Otto Cycle
QA
QR
W
=
=
=
e
=
mcp(T3-T2)
mcv (T1-T4) = -mcv (T4-T1)
QA – QR = mcv (T3-T2) – mcv (T4-T1)
W
=
mcp(T3-T2) -mcv (T4-T1)
QA
e
=
mcp(T3-T2)
T4-T1
1–
(4)
k( T3-T2)
missing solution
V1
where rk = V2 , the compression ratio
V3
r c = V2 , the cutoff ratio
Points 3 is called the cutoff point.
Derivation of the formula for e
6
Process 1-2:
k-1
T2=[V1/V2]
T1
k-1
T2 = T1 rk
(5)
Process: 2-3:
missing solution
Process: 3-4:
missing solution
Substituting equations (5) ,(6) and (7) in equation (4)
missing solution
The efficiency of the Diesel cycle differs from that of the Otto cycle by the bracketed
factor rc k -1. This factor is always
k(rc - 1)
greater than 1, because rc is always greater than 1. Thus , for a particular compression ratio
rk , the Otto cycle is more efficient. However, since the diesel engine compressor air only,
the compression ratio is higher than in an Otto engine. An actual Diesel engine with a
compression ratio of 15 is more efficient than actual Otto engine with a compression ratio
of 9.
Relation among rk, rc and re (expansion ratio)
missing solution
7
Problems
1. A diesel cycle operates with a compression ratio of 13.5 and with a cut off occurring
at 6 % of the stroke. State 1 is defined by 14 psia and 140° F. For the hot-air
standard with k= 1.34 and for an initial 1 cu ft, compute (a) t2,p2,V2,t3,V 3,p4 and t4(b)
QR, (c) W, (d) e and pm . (e) For a rate of circulation of 1000 cfm, compute the
horsepower.
Solution
missing solution
2. There are supplied 317 kJ/cycle to an ideal Diesel engine operating on 227g air: =
p1 = 97.91 kPa, t1 = 48.9°. At the end of compression, p2 = 3930 kPa Determine (a) rk
, (b) c, (c) rc , (d) W, (e) e, and (f) pm .
missing solution
Problems
1. At the beginning of compression in an ideal dual combustion cycle, the working fluid is 1
lb of air at 14.1 psia and 80°F. The compression ratio is 9, the pressure at the
constant volume addition of heat is 470 psia, and there are added 100 Btu during
the constant pressure expansion. Find (a) rp (b) rc (c) the percentage clearance ,
(d) e, and (e) pm.
missing solution
2. An ideal dual combustion cycle operates on 454 g of air. At the beginning of
compression, the air is at 96.53 kPa, 43.3 C. Let rp = 1.5, rc = 1.60, and rk = 11.
Determine (a) the percentage clearance, (b) p, V, and T at each corner of the cycle,
(c) QA, (d) e, and (e) pm.
missing solution
Comparison of Otto, Diesel & Oval Cycles
8
The three Cycles can be compared below for the same compression ratio and heat
rejection
Comparison of Otto, Diesel & Dual Cycles for the same rk.
CYCLES
1-2-6-5
- OTTO CYCLE
1-2-7-5
- DIESEL CYCLE
1-2-3-4-5 - DUAL CYCLE
The Fs Diagram shows for area 2-6 , at represents QA for the otto cycle, The area
under 2-7 represents QA for the Diesel cycle, and the area 2-3-4 represents QA for RE Dual
cycle. For the same QR , The higher the QA , the higher is the cycle efficiency. Therefore, for
the same rk and QR.
e
> e
> e
Otto
Dual
Diesel
Figure below shows a comparison of the three air standard cycles for the same maximum
temperature and pressure (point 4), The heat rejection being the same.
missing solution
Comparison of Otto, Diesel and Dual cycle for the same maximum temperature and
pressure.
CYCLES
1-6-4-5
1-7-4-5
1-2-3-4-5
-
Otto Cycle
Diesel Cycle
Dual Cycle
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QA is represents by the area under 6-4 for the Otto cycle, area under 7-4 for Diesel cycle and
by the area by the area under 2-3-4 for the Dual or Mixed Cycle in the Ts diagram, QR being
the same.
e
>
Diesel
e
Dual
> e
Otto
This comparison is being important, suie The Diesel cycle would give higher rk than
the Otto cycle.
Review Problems
1. An ideal Otto engine, operating on the hot-air standard with k = 1.34, has a
compression ratio of 5. At the beginning of compression the volume is 6 cu ft, the
pressure is 13.75 psia and the temperature is 100 F. During the constant-volume
heating, 340 Btu are added per cycle. Find (a) c, (b) T3, (c) p3 , (d) e, and (e) pm .
Ans. (a) 25%; (b) 520 R; (c) 639.4 psia; (d) 42.14%
(e) 161.2 psi
2. An ideal Otto cycle engine with 15% clearance operates on 0.227 kg/s of air; intake
state is 100.58 kPa, 37.7 C. The energy released during combustion is 110 kJ/s. For
hot-air standard with k = 1.32, compute (a) p, V, and T at each corner, (b) W, (c) e,
and (d) pm.
Ans. (a) 0.2013 m3/s, 0.02626 m3/s, 596.2 K, 1479.85
kPa, 1136.4 K, 2820.7 kPa, 592.2K, 191.71 kPa;
( b) 52.7 kJ/s; (c) 47.91%; (d) 301.1 kPa
3. In an ideal Diesel engine compression is from 14.7 psia, 80 F, 1.43 cu ft to 500 psia,
80 F, 1.43 cu ft to 500 psia. Then 16 Btu/cycle are added as heat. Make
computations for cold-air standard and find (a) T2 ,V2, T3 , V3, T4, and p4 , (b) W, (c) e
and pm, and (d) the hp for 300 cycles/min.
Ans. (a) 1479 R, 01152ft3, 2113 R, 0.1646 ft3, 890 R, 24.2psia; (b) 9.7 Btu;
(e) 60.63%, 39.9 psi; (d) 68.6 hp
4. For an ideal Diesel cycle with the overall value of k = 1.33, rk = 15, rc = 2.1, p1 = 97.9
kPa, find p2 and pm.
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Ans. 3589 kPa, 608 kPa
5. State 1 for a dual combustion engine is p1 = 1 atm and T1 = 60.3 C; rk = 18; at the end
constant volume combustion process the pressure is 7695 kPa, rc = 1.5. Base on
1kg/cycle of a hot-air standard with k = 1.31, determine (a) the percentage
clearance, (b) p, V, and T at each corner point on the cycle, (c) W, (d) e, and (e) pm
Ans. (a) 5.88% (b) 0.9443 m,0.05246 m³, 4468 kPa, 816.5 K, 1406.2 K,
0.07869 m³, 2109 3 K, 296.8 kPa, 976.3 K; (c) 803.5 kJ; (d) 57.43%; (e) 900 kPa
6. At the beginning of the compression process is a hot-air-standard Otto cycle, the air
is at 100 kPa and 300° K. The heat added to the air is 0.750 J/Kg-K and rk = 5. If cv
=718 J/Kg-K K=1.38. Calculate the pressure, temperature and volume at the cardinal
points. Also compute heat transfer and work done per Kgm of air for each process
and the cycle thermal efficiency.
7. An air-standard Otto cycle operates at 100°F, 14 psia and an rk = 7. The maximum
temperature of the cycle is 3000° F. Determine the heat supplied per lb of air, the
work done per lb of air, the cycle thermal efficiency, the maximum pressure of the
cycle and the temperature at the end of Isentropic expression. Assume K = 1.4.
8. An Otto cycle using Nitrogen as the working fluid operates at 100° F and 14 psia
with Vk = 7. The maximum temperature of the cycle is 2,000° F. Compute
temperature and pressure at cardinal points of the cycle, the heat supplied per lb,
the work done per lb, the cycle terminal efficiency and ⋀S of the heat addition and
rejection processes.
9. An Otto cycle with air as the working fluid has a maximum temperature of 3,000° F
with 100 kPa and 20° C and rk=5 at the beginning of compression process. Compute
the work done per kg of air and the cycle thermal efficiency.
10. An Otto cycle using air start at the 14 psia and 80° F with rk = 9 at its compression
process. The heat added to the air is 800 BN/lb per cycle. Compute the maximum
pressure and temperature, the mean effective pressure and the cycle thermal
efficiency.
11. An ideal Otto cycle using air has a the following conditions. At point 1, p1 = 14.5 psia
and 80° F. The heat added to the air is 370 BN/lb per cycle. If the maximum
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temperature is 3,400 °R and K = 1.4, Compute the pressure, temperature and
volume at each corner of the cycle; the compression ratio, mean effective pressure
and the cycle thermal efficiency if cv = 0.172 BTU/lb-°R.
12. At the beginning of compression in an ideal air standard Diesel cycle, the
compression ratio is 16 with p1 = 100 kPa and t1 = 300° K. The maximum
temperature of the cycle is 2000° K. If cp = 1.005 KJ/Kg-°K and cv = 0.718 KJ/Kg-°K
for the air, compute (a) the pressure, temperature and volume at four corner of the
cycle. (b) the heat transfer (c) the work done per Kg. for each process, and (d) the
cycle thermal efficiency.
13. A Diesel cycle using Nitrogen as the working fluid has 100 kPa and 20° C at the
beginning of compression process with maximum temperature of 200° K. Compute
the rest work per Kg of the gas. For rk = 15, what is its cycle thermal efficiency? Use
K = 1.4 for the cycle.
14. An air-standard Diesel cycle operating at 15 psia, 140° F, rk = 13 has a maximum
temperature of 2540° F. Compute the heat supplied per lb of air, the work per lb air,
the cut-off ratio, the cycle thermal efficiency and the maximum pressure of the cycle.
Use K = 1.4 for air.
15. A Diesel cycle using Nitrogen as working fluid operates at 100 kPa, 20° C with rk =
15 at the beginning of the compression process. The maximum temperature of the
cycle is zero °K. Compute cycle thermal efficiency.
16. An ideal Diesel cycle was Argon as the working fluid. At the beginning of
compression process, p1 = 15 psia, t1 = 140° F and rk = 5. The maximum
temperature is 2540° F. Calculate the work heat supplied per lb of Argon, the cut-off
ratio, the maximum pressure of the cycle, the temperature at the end of isentropic
expression and the cycle thermal efficiency.
17. Nitrogen is used as a working fluid for Diesel cycle operating with rk = 18 and the
heat input is 600 KJ/Kg. At the beginning of compression, p1 = 100 kPa and T1 =
300° K. Compute the temperature pressure and specific volume at cardinal points of
the cycle, the cut-off ratio and the cycle thermal efficiency.
18. Compute the thermal efficiency of the Diesel cycle if the temperature and pressure
before compression is an air-standard cycle are 535° R and 14.2 psia, and the
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temperature before and after the heat-addition process are 1260° R and 3460° R,
respectively.
19. Compute the pressure, temperature, and specific volume at cardinal points of the
cycle, the cut-off ratio and the thermal efficiency for a Diesel cycle using air as the
working fluid with rk = 18 and the heat input is 800 KJ/Kg. The pressure and
temperature at the beginning of compression are 100 kPa and 45° R, respectively
20. Nitrogen is used as a working fluid for an ideal diesel or mixed cycle with rk = 14.
The gas is at 27° C and 100 kPa at the beginning of the compression process. At the
end of the heat addition process, the gas is at 1927° C. The heat added to the gas is
1550 KJ/Kg. Compute the pressure ratio, the cut-off ratio and thermal efficiency of
the cycle.
21. A dual cycle using air operates at 14.7 psia, 70°F and rk = 7. The cylinder diameter of
the engine is 10 inches and the stroke is 12 inches. The pressure at the end of
isometric heating process is 800 psia. If heat is added at constant pressure during
3% of the stroke, compute the net work and cycle thermal efficiency. The
atmosphere is at 14.7 psia and 70° F.
22. In an air-standard Otto cycle, the maximum temperature is 2727° C. At the
beginning of compression, the gas is at 20° C and 100 kPa. Compute the net work
output and the cycle thermal efficiency for a compression ratio of (a) 5, (b) 6, (c) 7,
(d) 8, (e) 9, (f) 10, (g) 11, (h) 12. Use K = 1.4 and cr = 0.718 KJ/Kg°K for air.
23. Nitrogen is used as a working fluid for Diesel cycle with 20°c and 100kPa at the
beginning of compression process and the maximum temperature of the cycle is 1727°c.
Compute the work output per Kg of the gas and the cycle thermal efficiency for a
compression ratio of (a) 10, (b) 11, (c) 12, (d) 13, (e) 14, (f) 15, (g) 16, (h) 17, (i) 18, (j)
19,
KJ
KJ
and (k) 20. Use Cp = 1.039
and cv = 0.743
KJ°K
kg - °K
24. A thermodynamic cycle is composed of the following reversible process:
1-2: Isentropic
2-3: Isothermal
3-1: Isobaric
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Sketch the pV and Ts diagrams for the cycle. If the cycle operates on 113 gm of
Nitrogen; the expansion ratio 1-2 is re = 5, t1 = 149 °c, p1 = 689.5 kPa, Compute the p, V and
T at cardinal points, QA, QR and the cycle thermal efficiency.
25. If there are four reversible processes given: Isochoric, Isopiestic, Isothermal and
Isotropic. Consider four different three-process cycles. Draw their pV and Ts diagrams and
write expressions for QA, QR, WNET and e
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