CONSERVATION OF
MASS
Classification of a System
A system is a group of interacting objects and influences,
such as forces.
Two distinct ways of
interaction:
- Mass interaction
- Energy interaction
Classification of a System
In an open system, energy and
matter can pass through the
imaginary system boundary and
leave the system.
Examples:
Combustion
chamber, Heat exchangers,
Boilers, Pump transfers
Classification of a System
In a closed system, no energy
and matter can pass through
the system boundary.
The total energy in a closed
system remains unchanged wrt.
Time.
Examples: Refrigerant working
fluid, cyclical flow of water in
tube, Pressure vessel, Tires
Law of Conservation of mass
➢The law of conservation of mass states that mass is
indestructible.
➢“The mass in an isolated system can neither be created
nor destroyed but can be transformed from one form to
another.”
The quantity of fluid passing through a
given section is given by the formula:
𝑉ሶ = 𝐴𝑣
Mass flowrate
𝑉ሶ = 𝐴𝑣
𝑉ሶ 𝐴𝑣
𝑚ሶ = =
= 𝐴𝑣𝜌
v
v
𝑉ሶ = volume flow rate
A = cross section area of the stream
𝑣 = average velocity
𝑚ሶ = mass flow rate
Mass flowrate
Applying the law of conservation of mass:
𝐴1 𝑣1 𝐴2 𝑣2
𝑚ሶ =
=
v1
v2
𝐴1 𝑣1 𝜌1 = 𝐴2 𝑣2 𝜌2
Sample problem 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 in2, 𝒗𝟏 = 500 fps, v1 = 10 ft3/lb
• For the other gas: A2 = 50 in2, 𝒎ሶ 𝟐 = 16.67 lb/s, 𝝆𝟏 = 0.12
lb/ft3
• At exit: 𝒗𝟑 = 350 fps, v3 = 7 ft3/lb
a) The speed at section 2
b) The flow and area at the exit section
Sample problem 2
A 10-ft diameter by 15-ft height vertical tank is receiving
water
( 𝜌1 = 62.1 lb/ft3) 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 mins
later.
Given:
𝜌𝑤𝑎𝑡𝑒𝑟 = 62.1 𝑙𝑏/𝑓𝑡 3
ሶ
𝑉(𝑠𝑢𝑝𝑝𝑙𝑦)
= 300𝑔𝑝𝑚
𝐷𝑑𝑖𝑠𝑐ℎ𝑎𝑟𝑔𝑒 = 6"
𝓋𝑑𝑖𝑠𝑐ℎ𝑎𝑟𝑔𝑒 = 5𝑓𝑝𝑠
Required:
Water level after 15mins ,𝐻15𝑚𝑖𝑛
Solution:
𝑚ሶ 𝑛𝑒𝑡 + 𝑚ሶ 𝑠𝑢𝑝𝑝𝑙𝑦 − 𝑚ሶ 𝑑𝑖𝑠𝑐ℎ𝑎𝑟𝑔𝑒 = 0
𝑚ሶ 𝑠𝑢𝑝𝑝𝑙𝑦 = 𝑉ሶ𝑠𝑢𝑝𝑝𝑙𝑦 (𝜌𝑤𝑎𝑡𝑒𝑟 )
𝑚ሶ 𝑑𝑖𝑠𝑐ℎ𝑎𝑟𝑔𝑒 = 𝐴2 𝓋2 𝜌𝑤𝑎𝑡𝑒𝑟
𝑚ሶ 𝑛𝑒𝑡 =
𝜋𝐷2
𝓋2 𝜌𝑤𝑎𝑡𝑒𝑟
4
− 𝑉ሶ𝑠𝑢𝑝𝑝𝑙𝑦 (𝜌𝑤𝑎𝑡𝑒𝑟 )
Solution:
𝑉ሶ𝑛𝑒𝑡
𝑚ሶ 𝑛𝑒𝑡
=
𝜌𝑤𝑎𝑡𝑒𝑟
𝑉15𝑚𝑖𝑛 = 𝑉ሶ𝑛𝑒𝑡 (15min)
𝐻15𝑚𝑖𝑛
𝑉15𝑚𝑖𝑛
=
𝐴
Thermodynamics 1
Potential Energy, Kinetic Energy, Internal Energy, Heat,
Work, Flow Work, Enthalpy, General Energy Equation
Gravitational Potential Energy (P) of a body is
its energy due to its position or elevation
P = Fg z =
𝐦𝐠𝐳
𝐤
∆P = P2 – P1 =
z
𝐦𝐠
(z2
𝐤
– z1)
∆P = change in potential
energy
Fg
Datum Plane
Kinetic Energy (K) is the energy or stored
capacity for performing work possessed by a moving
body in the virtue of its momentum. Dependent on
the mass and the speed achieved.
m
v
K=
𝐦𝐯𝟐
𝟐𝐤
∆K = K2 – K1 =
𝐦
(v22
𝟐𝐤
– v12)
∆K = change in kinetic energy
Internal Energy (U, u) 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
∆𝑼 = 𝒏𝑪𝒗 ∆𝑻
Work (W)
• A product of the displacement of the body and
the component of the force in the direction of the
displacement.
• An energy in transition; that is, it exists only when a
force is “moving through a distance”
Work of a Nonflow System
The work done as the piston moves
from e to f is
dW = Fxdx =
(pA)dL = pdV
which is the area under the curve e-f on the
pV plane. Therefore, the total work done as
𝟐
the piston moves from 1 to 2 is W = ‫𝑽𝒅𝒑 𝟏׬‬
which is the area under the curve 1-e-f-2.
The area under the curve of the
process on the pV plane represents the work
done during a non-flow reversible process.
•
Work done by the system is positive
(outflow of energy)
•
Work done on the system is negative
(inflow of energy)
Work of Expansion
Flow Work (Wf)
Flow Work or flow energy is work done in pushing a fluid across a
boundary, usually into or out of a system.
Wf = FL = pAL
Wf = pV
∆Wf = Wf2 – Wf1 = p2V2 – p1V1
∆Wf = change in flow work
Flow Work
Heat (Q)
• An 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 the heat is rejected by the body or system.
Classification of Systems
•
A closed system is one in which mass does not cross its boundaries.
•
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.
Characteristics of steady flow system:
1.
2.
3.
There is neither accumulation nor diminution of mass within the system.
There is neither accumulation nor diminution of energy within the system.
The state of the working substance at any point in the system remain
constant.
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)
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 + K 1 + H 1 + Q = P2 + K 2 + H 2 + W
Q = ∆P + ∆K + ∆Wf + ∆U + W
Sample Problem 1
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.
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)
Given:
p1 = 200 psia
p2 = 20 psia
ʋ1 = 200 fps
ʋ2 = 1000 fps
v1 = 1 ft3/lb
v2 = 8 ft3/lb
∆u = -25 Btu/lb
Q=0
Sample Problem 1
Solution
P1 + K1 + Wf1 + U1 + Q = P2 + K2 + Wf2 + U2 + W
Basis 1 lbm
K1 =
ʋ𝟏𝟐
K2 =
ʋ𝟐𝟐
𝟐𝐤
𝟐𝐤
=
𝐟𝐭
𝐬
𝐥𝐛𝐦 𝐟𝐭
𝟑𝟐.𝟏𝟕𝟒
𝐥𝐛𝐟 𝐬𝟐
(𝟐𝟎𝟎 )𝟐
𝟐
=
(𝟏𝟎𝟎𝟎)𝟐
𝟐 𝟑𝟐.𝟏𝟕𝟒 (𝟕𝟕𝟖)
𝐥𝐛
Wf1 = p1v1 =
Wf2 = p2v2 =
𝟐𝟎𝟎 𝟐
𝐢𝐧
𝐁𝐭𝐮
𝐟𝐭 𝐥𝐛𝐟
(𝟕𝟕𝟖
)
𝐁𝐭𝐮
= 0.80 𝐥𝐛
𝐦
𝐁𝐭𝐮
= 19.91 𝐥𝐛
𝐦
𝐢𝐧𝟐
𝐟𝐭 𝟐
𝟏𝟒𝟒 𝟐 (𝟏𝐥𝐛 )
𝐟𝐭
𝐦
𝐟𝐭 𝐥𝐛𝐟
𝟕𝟕𝟖 𝐁𝐭𝐮
𝟐𝟎 (𝟏𝟒𝟒)(𝟖)
𝟕𝟕𝟖
= 37.02
𝐁𝐭𝐮
𝐥𝐛𝐦
𝐁𝐭𝐮
= 29.61 𝐥𝐛
𝐦
K1 + Wf1 = K2 + Wf2 + ∆U + W
0.8 + 37.02 = 19.97 + 29.61 – 25 + W
W = 𝟏𝟑. 𝟐𝟒
𝐁𝐭𝐮
𝐥𝐛𝐦
𝐁𝐭𝐮
𝐛𝐲 =
𝟏𝟑.𝟐𝟒 𝐥𝐛
𝐦
𝐥𝐛
(𝟏𝟎𝐦𝐢𝐧)
𝐁𝐭𝐮
𝟒𝟐.𝟒 𝐦𝐢𝐧)(𝐡𝐩
= 3.12hp
Sample Problem 2
2. Steam is supplied to a fully loaded 100-hp turbine at 200 psia with
u1 = 1163.3 Btu/lb, v1 = 2.65 ft3/lb and ʋ1 = 400 fps. Exhaust is at
1 psia with u2 = 925 BTU/lb, v2 = 294 ft3/lb and ʋ2 = 1100 fps.
The heat loss from the steam in the turbine 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
Given:
p1 = 200 psia
u1 = 1163.3 Btu/lb
v1 = 2.65 ft3/lb
ʋ1 = 400 fps
p2 = 1 psia
u2 = 925 Btu/lb
v2 = 294 ft3/lb
ʋ2 = 1100 fps
Q = -10 Btu/lb
Sample Problem 2
Solution
P1 + K1 + Wf1 + U1 + Q = P2 + K2 + Wf2 + U2 + W
a) Basis 1 lbm2
K1 = ʋ1
2/2k
K2 = ʋ22 =
=
𝟒𝟎𝟎 𝟐
𝟐 𝟑𝟐.𝟏𝟕𝟒 𝟕𝟕𝟖
(𝟏𝟏𝟎𝟎)𝟐
𝟐 𝟑𝟐.𝟏𝟕𝟒 (𝟕𝟕𝟖)
𝐁𝐭𝐮
= 3.20 𝐥𝐛𝐦
𝐁𝐭𝐮
= 24.17𝐥𝐛𝐦
Wf1 = p1v1 =
𝟐𝟎𝟎 𝟏𝟒𝟒 𝟐.𝟔𝟓
𝟕𝟕𝟖
= 98.1
𝐁𝐭𝐮
𝐥𝐛𝐦
Wf2 = p2v2 =
𝟏 𝟏𝟒𝟒 (𝟐𝟗𝟒)
=
𝟕𝟕𝟖
54.24 𝐥𝐛𝐦
𝐁𝐭𝐮
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
𝐁𝐓𝐔
b) Steam flow =
𝟏𝟎𝟎 𝐡𝐩 (𝟐𝟓𝟒𝟒 𝐡 (𝐡𝐩)
𝐁𝐓𝐔
𝟐𝟓𝟏 𝐥𝐛
𝐦
= 1014 lbm/h
Sample Problem 3
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 kPa and
0.0051 m3/kg. The initial energy of the air is 1594 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. Compute the work.
Given:
p1 = 99.29 kPa
v1 = 0.026 m3/kg
u1 = 1594 J/kg
p2 = 689.5 kPa
v2 = 0.0051 m3/kg
u2 = 6241 J/kg
∆K = 896 J/kg
Q = -4383 J/kg
ṁ = 272 kg/min
Sample Problem 3
Solution
P1 + K1 + Wf1 + U1 + Q = P2 + K2 + Wf2 + U2 + W
Basis 1 kgm
Wf1 = p1 v1 = (99.29 KN/m2) (0.026 m3/kg) = 2.583 kJ/kg
Wf2 = p2 v2 = (689.5 kN/m2) (0.0051 m3/kg) = 3.516 kJ/kg
Wf1 + u1 + Q = ∆K + Wf2 + u2 + W
2.582 + 1.594 – 4.383 = 0.896 + 3.516 + 6.241 + W
W = -10.86 kJ/kg
W = (-10.86 kJ/kg) (272 kg/min)
W = -2954 kJ/min
Sample Problem 4
4. A centrifugal pump operating under steady flow condition delivers
2,270 kg/min of water from an initial pressure of 82,740 kPa to a final
pressure of 275,800 Pa. The diameter of the inlet pipe to the pump is
15.24cm and the diameter of the discharge pipe is 10.16 cm.
What is the work?
Given:
m = 2270 kg/min
ρ = 1000 kg/m3
p1 = 82,740 Pa
p2 = 275,800 Pa
d1 = 0.15 24 m
d2 = 0.1016m
Sample Problem 4
Basis 1 kgm
Solution
Area at entrance:
A1 =
𝛑
(. 𝟏𝟓𝟐𝟒)𝟐
𝟒
K1 = ʋ12 =
= 0.01824 m2
Area at exit:
K2 = ʋ2 / 2k =
𝛑
𝟒
= 2.074
= 4.667
𝐍. 𝐦
𝐤𝐠𝐦
= 10.89
𝐍. 𝐦
𝐤𝐠𝐦
𝐦
𝐬
=
𝟖𝟐,𝟕𝟒𝟎 𝟐
𝐦
𝐤𝐠
𝟏𝟎𝟎𝟎 𝟑
𝐦
= 82.74
𝐍. 𝐦
𝐤𝐠𝐦
Wf2 = p2 v2 = p2 / ƿ2
Speed at exit:
𝟏𝟎𝟎𝟎 (𝟎.𝟎𝟎𝟖𝟏𝟎𝟕)
𝟒.𝟔𝟔𝟕 𝟐
𝟐 (𝟏)
𝐍
𝟐𝟐𝟕𝟎 𝐤𝐠𝐦
𝟔𝟎
𝐬
𝐤𝐠𝐦
𝟏𝟎𝟎𝟎 𝟑 (𝟎.𝟎𝟏𝟖𝟐𝟒 𝐦𝟐
𝐦
𝟐𝟐𝟕𝟎
𝟔𝟎
= 2.151
Wf1 = p1 v1 = p1 / ƿ1
Speed at entrance:
m
ʋ2 =
=
p2 A2
𝟐
2
A2 = (𝟎. 𝟏𝟎𝟏𝟔)𝟐 = 0.008107 m2
m
ʋ1 =
=
p1 A1
𝐦 𝟐
𝐬
𝐤𝐠𝐦.𝐦
(𝟏
)
𝐍.𝐬 𝟐
𝟐.𝟎𝟕𝟒
𝐦
𝐬
= 275,800/1000 = 275.8
𝐍. 𝐦
𝐤𝐠𝐦
K1 + Wf1 = K2 + Wf2 + W
2.151 + 82.74 = 10.89 + 275.8 + W
W = -201.8 N-m/kgm
= (-201.8 N-m/kgm) (2270 kg/min)
= -458.81 kJ/min
Sample Problem 5
5. A turbine operates under steady flow conditions, receiving steam at
the following state: pressure 1200 kPa, temperature 1800 °C,
enthalpy 2785 kJ/kg, speed 33.3 m/s and elevation 3m. The steam
leaves the turbine at the following state; pressure 20kPa, enthalpy
2512 kJ/kg, speed 100 m/s and elevation 0m. 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?
Given:
z1 = 3 m
z2 = 0 m
h1 = 2785 kJ/kg
h2 = 2512 kJ/kg
ʋ1 = 33.3 m/s
ʋ2 = 100 m/s
Q = -0.29 kJ/s
m = 0.42 kg/s
Sample Problem 5
Solution
Basis 1 kgm
(9.8066 kJ )(3m)
kg
(g)(z1)
P1 =
=
k
1kg∙m
N ∙s2
= 0.0294
kJ
kg
(33.3 m
)2
ʋ12
kJ
s
K1 =
=
= 0.5544
2k (2)(1kg∙m )
kg
2
N ∙s
K2 =
ʋ22 (100)2
kJ
=
= 5.000
2k (2)(1)
kg
0.29kJ
s = 0.6905 kJ
Q=kg
0.42kg
s
P1 + K1 + h1 + Q = K2 + h2 + W
0.0294 + 0.5544 + 2785 + (-0.6905) = 5.00
+ 2512 + W
W = 267.9 kJ/kg
= (267.9 kJ/kg) (0.42 kg/s)
= 112.52 kW
Thermodynamics 1
Reference: Sta. Maria, H. B. (1990). Thermodynamics 1. Mandaluyong City, Philippines: National Book Store.
Presentation made by David Anthony C. Manalo & Gino Carlo O. Cadao
Thermodynamics 1
Boyle’s Law, Charles’ Law, Equation of State,
Gas Constant, Specific Heats of an Ideal Gas
An ideal gas is ideal only in the sense that it
conforms to the simple perfect gas laws.
Boyle’s Law
If the temperature of a given quantity of gas is held
constant, the volume of the gas varies inversely with the
absolute pressure during a change of state.
V ∝ 1/P or V = C/P
pV = C or p1V1 = p2V2
Charles’ Law
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/T = C
or
V1/T1 = V2/T2
If the volume of a particular quantity of gas is held
constant, then, with any change of state, the pressure will
vary directly as the absolute temperature.
P=T
or
p = CT
P/T = C
or
p1/T1 = p2/T2
Equation of State or Characteristic Equation of a Perfect Gas
Combining Boyle’s and Charles’ Laws
𝐏𝟏𝐕𝟏
𝐓𝟏
=
𝐩𝟐𝐕𝟐
𝐓𝟐
𝐩𝐕
=
𝐓
mR
= C, as constant
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
English units
lbf/ft2
ft3
lbm
°R
ftlbf / lbm °R
SI units
N/m2
m3
kg
K
Nm / kgK
Sample Problem 1
1. A drum 6 inch in diameter and 40 inch long contained acetylene at
250 psia and 90°F. After some of the acetylene was used, the
pressure was 200 psia and the temperature of 85°F.
(a) What proportion of the acetylene was used?
(b) What volume would the used acetylene occupy at 14.7 psia &
80°F? R for acetylene is 59.35 ft. lb/lb.°R.
Solution:
a) Let:
m1 = m of acetylene initially in the drum
m2 = m of acetylene left in the drum
m3 = m 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
𝟒 𝟏𝟕𝟐𝟖
= 0.6545 cu ft
m1 =
𝐩𝟏 𝐕𝟏
𝐑𝐓𝟏
=
𝟐𝟓𝟎 𝟏𝟒𝟒 𝟎.𝟔𝟓𝟒𝟓
𝟓𝟗.𝟑𝟓 𝟓𝟓𝟎
= 0.7218 lb
m2 =
𝐩𝟐 𝐕𝟐
𝐑𝐓𝟐
=
𝟐𝟎𝟎 𝟏𝟒𝟒 𝟎.𝟔𝟓𝟒𝟓
𝟓𝟗.𝟑𝟓 𝟓𝟒𝟓
= 0.5828 lb
m3 = m1 – m2 = 0.7218 – 0.5828 = 0.1390 lb
Acetylene used
=
𝐦𝟑
𝐦𝟏
=
𝟎.𝟏𝟑𝟗𝟎
𝐱
𝟎.𝟕𝟐𝟏𝟖
𝟏𝟎𝟎 = 19.26%
b) P3 = 14.7 psia
T3 = 80°F + 460 = 540°R
V3 =
𝐦𝟑 𝐑𝐓𝟑
𝐩𝟑
=
𝟎.𝟏𝟑𝟗 𝟓𝟗.𝟑𝟓 𝟓𝟒𝟎
𝟏𝟒.𝟕 𝟏𝟒𝟒
= 2.105 ft3
Sample Problem 2
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.
Given: 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
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
Sample Problem 2
Solution:
For the tank
m1 =
𝐩𝟏 𝐕𝟏
𝐑𝐓𝟏
=
𝟐𝟏𝟒.𝟕 𝟏𝟒𝟒 𝟑𝟑𝟗.𝟑
𝟓𝟑.𝟑𝟒𝟐 𝟓𝟒𝟓
m2 =
𝐩𝟐 𝐕𝟐
𝐑𝐓𝟐
=
(𝟔𝟒.𝟕)(𝟏𝟒𝟒)(𝟑𝟑𝟗.𝟑)
(𝟓𝟑.𝟑𝟒𝟐)(𝟓𝟒𝟓)
= 360.9 lb
= 108.7 lb
Mass of air that can be used = 360.9 – 108.7 = 252.2 lb
For the drums
m3 =
𝐩𝟑 𝐕𝟑
𝐑𝐓𝟑
=
𝟏𝟒.𝟕 𝟏𝟒𝟒 𝟏
𝟓𝟑.𝟑𝟒𝟐 𝟓𝟒𝟎
= 0.0735 lb
m4 =
𝐩𝟒 𝐕𝟒
𝐑𝐓𝟒
=
𝟔𝟒.𝟕 𝟏𝟒𝟒 𝟏
𝟓𝟑.𝟑𝟒𝟐 𝟓𝟒𝟎
= 0.3235 lb
Mass of air put in each drum = 0.3235 – 0.0735 = 0.25 lb
Number of drums filled up =
𝟐𝟓𝟐.𝟐
𝟎.𝟐𝟓
= 1008.8~1009
Sample Problem 3
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 gross lifting force of 20 metric tons?
Solution:
Let ma = mass of air displaced
by the balloon
mHe = mass of Helium
V = volume of the balloon
For the air
𝐉
𝐤𝐠.𝐊
Ra = 287.08
Pa = 101,325 Pa
Ta = 21.1 + 273 = 294.1 K
ma =
𝐩𝐚 𝐕
𝐑 𝐚 𝐓𝐚
=
𝟏𝟎𝟏,𝟑𝟐𝟓𝐕
𝟐𝟖𝟕.𝟎𝟖 𝟐𝟗𝟒.𝟏
= 1.2001 V kg
Sample Problem 3
Solution:
For the helium
RHe = 2,077.67
𝐉
𝐤𝐠.𝐊
PHe = 101,325 Pa
THe = 21.1 + 273 = 294.1K
mHe =
𝐩𝐇𝐞𝐕 =
𝟏𝟎𝟏,𝟑𝟐𝟓 𝐕
=
𝐑𝐇𝐞𝐓𝐇𝐞 (𝟐𝟎𝟕𝟕.𝟔𝟕) (𝟐𝟗𝟒.𝟏)
0.1658 V kg
ma = mHe + 20,000
1.2001 V = 0.1658 V + 20,000
V = 19,337 m3
𝟒
𝛑𝐫 𝟑
𝟑
= 19,337
r = 16.65 m
d = 2(16.65) = 33.3 m
Sample Problem 4
4. Two vessels A and 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. Vessel B,
of unknown volume, contains air at 68.95 kPa, 4.44°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
Tm = 43.33 + 273 = 316.33 K
mm = ma + mb
𝐏 𝐦 𝐕𝒎 𝐩 𝐚 𝐕𝒂
=
𝐑𝐓𝐦
𝐑𝐓𝒂
+
𝐩 𝐛 𝐕𝒃
𝐑𝐓𝒃
𝟏𝟑𝟕𝟖. 𝟗𝟔 𝑽𝒎
𝟐𝟕𝟔𝟕. 𝟗𝟐 𝟏𝟒𝟐
𝟔𝟖. 𝟗𝟓𝑽𝒃
=
+
𝟑𝟏𝟔. 𝟑𝟑
𝟑𝟔𝟔. 𝟑𝟑
𝟐𝟕𝟕. 𝟒𝟒
4.36Vm = 1072.9 + 0.25Vb
(1)
Vm = 142 + Vb
(2)
Solving equations 1 and 2 simultaneously
Vb = 110.4 Liters
The specific heat of a substance is defined as the quantity of
heat required to change the temp. of unit mass through one degree.
In dimensional form,
C→
heat energy units
mass (change of temperature)
In different quantities,
dQ
C = mdT
or dQ = mcdT
And for a particular mass m,
𝟐
Q = m ‫ 𝟏׬‬cdT
(the specific heat equation)
If the mean or instantaneous value of specific heat is used,
𝟐
Q = mc ‫ 𝟐𝑻( 𝐜𝐦 = 𝑻𝒅 𝟏׬‬− 𝑻𝟏 )
(constant specific heat)
Constant Volume Specific Heat (cv)
Qv = ∆U
Qv = mcv (T2 – T1)
Constant Pressure Specific Heat (cp)
Qp = mcp (T2 – T1)
Ratio of Specific Heats
c
K = cp > 1
v
Qp = ∆U + p(V2 – V1)
= U2 – U1 + p2V2 – p1V1
Qp = H2 – H1 = ∆H
2
Qp = ∆U + W = ∆U + ‫׬‬1 pdV
Internal Energy of an Ideal Gas
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.
Enthalpy of an Ideal Gas
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
cpdT = cvdT + RdT
Cp = cv + R
R
k−1
kR
Cp =
k−1
Cv =
Sample Problem
For a certain ideal gas R = 25.8 ft.lb/lb°R & k =1.09 (a) What are the
values of cp and cv? (b) What mass of 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, what are the resulting temperature & pressure?
Solution:
𝐤𝐑
𝟏.𝟎𝟗 𝟐𝟓.𝟖
𝟏.𝟎𝟗−𝟏
𝐟𝐭 𝐥𝐛
𝐁𝐭𝐮
312.47 𝐥𝐛𝐑° or 0.4016 𝐥𝐛 𝐑⁰
𝐜𝐩
𝟎.𝟒𝟎𝟏𝟔
𝐁𝐭𝐮
=
=
0.3685
𝐤
𝟏.𝟎𝟗
𝐥𝐛 𝐑⁰
a) Cp =𝐤−𝟏 =
=
Cv =
b) V = 15 cu ft
p = 75 psia
T = 80+460 = 540⁰R
𝐩𝐯
m = 𝐑𝐓 =
𝟕𝟓 𝟏𝟒𝟒 𝟏𝟓
𝟐𝟓.𝟖 𝟓𝟒𝟎
= 11.63 lb
c) Q = mcv (T2-T1)
30 = 11.63(0.3685)(T2 – 540)
T2 = 547⁰R
P2 = p1 (T2/T1) = 75(547/540)
= 76 psia
Sample Problem
For a certain gas R = 320 J/kg°K and cv = 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 & p1 = 690 kPa to a state where t2 = 555°C,
find ∆U and ∆H.
Solution:
a)Cp = cv + R = 0.84 + 0.32 = 1.16 kJ/kg K
𝐑
𝟎.𝟑𝟐
K = 𝐜 + 𝟏 = 𝟎.𝟖𝟒 + 𝟏 = 1.381
𝐯
b) T1 =
𝐩𝟏 𝐕𝟏
𝐦𝐑
=
𝟔𝟗𝟎,𝟎𝟎𝟎 𝟏.𝟏𝟑𝟑
𝟓 𝟑𝟐𝟎
= 488.6 K
∆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) 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:
𝟐 𝐝𝐐
dS = dQ/T or ∆S = ‫𝟏׬‬
𝐓
where dQ = heat transferred at the temperature T
∆S = total change of entropy
𝟐 𝐦𝐜𝐝𝐓
𝐓
∆S = ‫𝟏׬‬
𝟐 𝐝𝐓
𝐓
= mc‫𝟏׬‬
𝐓
= 𝐦𝐜 𝐥𝐧 𝐓𝟐
(constant specific heat)
𝟏
Internal Energy, Enthalpy and Entropy of an Ideal Gas
∆𝑈 = 𝑚𝐶𝑣 ∆𝑇
∆𝐻 = 𝑚𝐶𝑝 ∆𝑇
Where:
𝑇1
∆𝑆 = 𝑚 𝐶 ln( )
𝑇2
m = mass of the ideal gas
Cv = specific heat at constant volume
Cp = specific heat at constant pressure
C = specific heat depending on the process
∆T = change in temperature
U = internal energy
H = enthalpy
S = entropy
Universal Gas Constant
A value that determines the gas constant of
any specific ideal gas in relation to their molecular
composition.
𝑅ത = 8.3145
𝑓𝑡 − 𝑙𝑏𝑓
𝐽
𝐿 − 𝑎𝑡𝑚
= 1545
= 0.0821
𝑚𝑜𝑙. 𝐾
𝑚𝑜𝑙. °𝑅
𝑚𝑜𝑙. 𝐾
Finding a specific gas constant using the universal
gas constant
𝑅ത
𝑅𝑔𝑎𝑠 =
𝑀𝑊𝑔𝑎𝑠
Universal Gas Constant
The equation of state for an ideal gas for a
specific number of molecules can be denoted by:
ത
𝑝𝑉 = 𝑛𝑅𝑇
Where:
p = absolute pressure of ideal gas
V = volume of ideal gas
n = number of mols
𝑅ത = universal gas constant
T = absolute temperature
Some typical elements and their atomic weight
Atomic No.
Symbol
Name
Atomic Weight
(AMU)
1
H
Hydrogen
1.00797
2
He
Helium
4.00260
6
C
Carbon
12.011
7
N
Nitrogen
14.0067
8
O
Oxygen
15.9994
10
Ne
Neon
20.179
16
S
Sulfur
32.06
Where: Atomic Mass Unit = 1AMU = 1g/mol = 1kg/mol = 1lbm/mol
Composition of a mixture of gases
If a mixture of a gas is composed of fractions of
several gases, the molecular weight of that mixture is
equal to:
𝑀𝑊𝑚 = ෍ 𝑋𝑖 𝑀𝑊𝑖
𝑛𝑚 = ෍ 𝑛𝑖
Where:
MWm = molecular weight of the mixture
Xi = fraction/composition of a specific gas
MWi = molecular weight of a specific gas
nm = total number of mols in a mixture
ni = number of mols of a specific gas
Sample Problem
Find the value of the specific gas
constant of Oxygen. From the periodic
table, the atomic weight of oxygen is
15.9994.
Solution:
Getting the Molecular Weight of Oxygen (O2):
Molecular Weight = No. of Atoms(Atomic Weight)
Atomic Mass Unit = 1AMU = 1g/mol = 1kg/mol = 1lbm/mol
MWoxygen = 2 (15.9994 AMU) = 31.9988 AMU = 31.9988 kg/mol = 31.9988 lbm/mol
= 31.9988 g/mol
Roxygen =
Roxygen =
ഥ
𝑹
𝑴𝑾𝒐𝒙𝒚𝒈𝒆𝒏
ഥ
𝑹
𝑴𝑾𝒐𝒙𝒚𝒈𝒆𝒏
=
=
𝟖.𝟑𝟏𝟒𝟓
𝑱
𝒎𝒐𝒍−𝑲
𝟑𝟏.𝟗𝟗𝟖𝟖 𝒈/𝒎𝒐𝒍
𝟏𝟓𝟒𝟓
𝒇𝒕−𝒍𝒃
𝒎𝒐𝒍−𝑹
𝟑𝟏.𝟗𝟗𝟖𝟖 𝒍𝒃/𝒎𝒐𝒍
= 𝟎. 𝟐𝟓𝟗𝟖𝟒
= 𝟒𝟖. 𝟐𝟖
𝑱
𝒈𝑲
𝒇𝒕−𝒍𝒃𝒇
𝒍𝒃𝒎 °𝑹
Sample Problem
What is the value for the specific heat at constant pressure in SI units of
Carbon Dioxide if its specific heat ratio is 1.288.
Solution:
Getting the Molecular Weight of Carbon Dioxide (CO2):
Molecular Weight = 1(AWCarbon) + 2(AWoxygen)
𝑴𝑾𝑪𝑶𝟐 = 𝟏 𝟏𝟐. 𝟎𝟏𝟏𝒂𝒎𝒖 + 𝟐 𝟏𝟓. 𝟗𝟗𝟗𝟒𝒂𝒎𝒖 = 𝟒𝟒 𝑨𝑴𝑼 = 𝟒𝟒
𝑹𝑪𝑶𝟐
𝑱
𝟖. 𝟑𝟏𝟒
ഥ
𝑹
𝒎𝒐𝒍 − 𝑲 = 𝟎. 𝟏𝟖𝟖𝟗𝟔 𝑱
=
=
𝑴𝑾𝑪𝑶𝟐
𝟒𝟒 𝒈/𝒎𝒐𝒍
𝒈𝑲
𝒌=
𝑪𝒑
𝒔𝒐, 𝑪𝒑 = 𝒌(𝑪𝒗)
𝑪𝒗
𝑱
𝑹𝑪𝑶𝟐 𝟎. 𝟏𝟖𝟖𝟗𝟔 𝒈𝑲
𝑱
𝑪𝒗 =
=
= 𝟎. 𝟔𝟓𝟔𝟏
𝒌−𝟏
𝟏. 𝟐𝟖𝟖 − 𝟏
𝒈𝑲
𝑪𝒑 = 𝟏. 𝟐𝟖𝟖 𝟎. 𝟔𝟓𝟔𝟏
𝑱
𝑱
= 𝟎. 𝟖𝟒𝟓𝟏
𝒈𝑲
𝒈𝑲
𝒈
𝒎𝒐𝒍
Other Fundamental Laws of Ideal Gases
Dalton’s Law of Partial Pressure
The pressure exerted in a vessel by a mixture of gases is equal
to the sum of the pressures that each separate gas would exert if it
alone occupied the whole volume of the vessel.
𝑃 = 𝑃1 + 𝑃2 + 𝑃3 + ⋯ + 𝑃𝑛
Similarly,
𝑃 = 𝑋1 𝑃 + 𝑋2 𝑃 + 𝑋3 𝑃 + ⋯ + 𝑋𝑛 𝑃
P = total pressure of the mixture
P1, P2, P3, Pn = partial pressure of individual gases
Partial Pressure = pressure exerted by each gas
X1, X2, X3, Xn = Percentage composition of a specific gas
Other Fundamental Laws of Ideal Gases
Avogadro’s Law
At equal volume, at the same temperature
and pressure conditions, the gases contain the same
number of molecules.
𝑚1 𝑀𝑊1
=
𝑚2 𝑀𝑊2
Sample Problem
A tank contains 80ft3 of air at a pressure of 350 psig. If the air is cooled
until its pressure and temperature decreases to 200 psig and 70oF
respectively, what is the decrease in internal energy in BTU?
Solution:
Since no specified:
∆𝑈 = 𝑚𝐶𝑣 ∆𝑇
Rair = 53.342 ft-lb/lbmR
Where:
k = 1.4
m=?
𝒄𝒗 =
Cv = ?
T2 = 70+460 = 530oR
Solving for Cv:
𝒄𝒗 =
𝑹𝒂𝒊𝒓
𝒌−𝟏
53.342 ft−lb/lbmR
𝟏. 𝟒 − 𝟏
𝒇𝒕 − 𝒍𝒃
𝒄𝒗 = 𝟏𝟑𝟑. 𝟑𝟓𝟓
𝒍𝒃°𝑹
Applying Charles’ Law:
𝑉1 = 𝑉2
𝑃1 𝑇1
=
𝑃2 𝑇2
350 + 14.7
𝑇1
=
200 + 14.7 530oR
𝑇1 = 900.28oR
Applying equation of state at state 1:
𝑝𝑉 = 𝑚𝑅𝑇
𝑝𝑉
350 + 14.7 (144)(80)
𝑚=
=
𝑓𝑡 − 𝑙𝑏𝑓
𝑅𝑇
53.342
(900.28)
𝑙𝑏 − °𝑅
𝑚 = 87.49 𝑙𝑏𝑚
Solving for ∆𝑈:
∆𝑈 = 𝑚𝐶𝑣 ∆𝑇
∆𝑈 = 87.49 𝑙𝑏(133.355
𝑓𝑡−𝑙𝑏𝑓
)(530 −
𝑙𝑏°𝑅
900.28)°𝑅
∆𝑈 = −4,320,141.5356 𝑓𝑡 − 𝑙𝑏𝑓
∆𝑼 = −𝟓𝟓𝟓𝟏. 𝟔𝟕 𝑩𝑻𝑼
Sample Problem
A large mining company was provided with a 3000cm3 of
compressed air tank. Air pressure in the tank drops from 700kPa to
180kPa while temperature remains unchanged at 28oC. What
percentage has the mass of air in the tank been reduced?
Given:
V1 = V2 and T1 = T2
V1 = 3000 cm3
Let:
P1 = 700kPa
m1 = mass of air at state 1
P2 = 180kPa
m2 = mass of air at state 2
T1 = 28oC +273 = 301K Solving for masses:
𝑝1 𝑉1
700𝑘𝑃𝑎(3𝑚3 )
𝑚1 =
=
𝑅𝑇1 0.28708 𝑘𝐽 (301𝐾)
𝑘𝑔𝐾
𝑝2 𝑉2
180𝑘𝑃𝑎(3𝑚3 )
𝑚2 =
=
𝑘𝐽
𝑅𝑇2
0.28708
(301𝐾)
𝑘𝑔𝐾
Percent reduction:
%Air reduced =
%Air reduced =
𝑚1 −𝑚2
𝑥100%
𝑚1
24.31−6.25
24.31
𝑥100%
%Air reduced = 74.29%
Sample Problem
A mixture is formed by bringing together gases and results to a state
of 689.45 kPa, 37.8oC. The mixture is composed of 3mol CO2, 2mol N2
and 4.5mol O2. Find the partial pressure of CO2 after mixing.
Given:
Getting the percentage of CO2 in
Pm = 689.45 kPa
the mixture:
Tm = 37.8oC
%CO2 =
nco2 = 3
nn2 = 2
no2 = 4.5
𝑛𝐶𝑂2
𝑥
𝑛𝑡𝑜𝑡𝑎𝑙
𝑃𝐶𝑂2 = %𝐶𝑂2 ∗ 𝑃𝑚
𝑃𝐶𝑂2 = 0.3158 ∗ 689.45𝑘𝑃𝑎
100% =
3
9.5
= 31.58%
From Dalton’s Law:
𝑃𝑚 = 𝑃𝐶𝑂2 + 𝑃𝑁2 + 𝑃𝑂2
Getting the Pressure of CO2:
𝑃𝐶𝑂2 = %𝐶𝑂2 ∗ 𝑃𝑚
𝑷𝑪𝑶𝟐 = 𝟐𝟏𝟕. 𝟕𝟑 𝒌𝑷𝒂
Sample Problem
An air with a mass of 0.454kg and an unknown mass of CO2 occupy
an 85Liters tank at 2068.44 kPa. If the partial pressure of the CO2 is
344.74kPa, determine its mass.
Given:
mair = 0.454kg
Vtank = 85L
Since both air and CO2 are in
equilibrium, they have equal state:
𝑝𝑎𝑖𝑟 𝑉𝑎𝑖𝑟
𝑝𝐶𝑂2 𝑉𝐶𝑂2
=
𝑚𝑎𝑖𝑟 𝑅𝑎𝑖𝑟 𝑇𝑎𝑖𝑟 𝑚𝐶𝑂2 𝑅𝐶𝑂2 𝑇𝐶𝑂2
Ptank = 2068.44kPa
Where:
PCO2 = 344.74kPa
Vair = VCO2 and Tair = TCO2
Pair = PT - PCO2 = 2068.44-344.74
= 1723.7kPa
mCO2 = 𝑚𝑎𝑖𝑟
𝑃𝐶𝑂2
𝑃𝑎𝑖𝑟
𝑅𝑎𝑖𝑟
𝑅𝐶𝑂2
mCO2 = 𝑚𝑎𝑖𝑟
𝑃𝐶𝑂2
𝑅𝑎𝑖𝑟
𝑃𝑎𝑖𝑟
𝑅𝐶𝑂2
𝑅𝑎𝑖𝑟 = 287.08
𝐽
𝑘𝑔𝐾
𝑅𝐶𝑂2 = 188.96
𝐽
𝑘𝑔𝐾
𝑚𝐶𝑂2 = 0.454𝑘𝑔
344.74
1723.7
287.08
188.96
𝒎𝑪𝑶𝟐 = 𝟎. 𝟏𝟑𝟖 𝐤𝐠
“The area under the curve of the process on the TS plane represents the
quantity of heat transferred during the process.”
dQ = TdS
𝟐
Q = ‫)𝐒𝐝𝐓( 𝟏׬‬
“The area behind the curve of the process on the pV plane represents the
work of a steady flow process when ∆K = 0 or it represents ∆K when Ws = 0.”
2
− න Vdp = Ws + ∆K
1
(Reversible steady flow, ∆P = 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.
Thermodynamics 1
Reference: Sta. Maria, H. B. (1990). Thermodynamics 1. Mandaluyong City, Philippines: National Book Store.
Presentation made by David Anthony C. Manalo & Gino Carlo O. Cadao