Problems completed by Levi Lentz
Problem 4-4
The diameters of the ports in the accompanying figures are 10
cm, 5 cm, and 1 cm at port 1, 2, and 3 respectively. At a given
instant, water enters the tank at port 1 with a velocity of 2 m/s and
leaves through port 2 with a velocity of 1 m/s. Assuming air in
insoluble in water, determine(a) the mass flow rate of water in
kg/s at the inlet (b) at the exit and (c) the rate at which water
accumulates in the tank in kg/s. (d) Determine the mass flow rate
of air in kg/s coming out of port 3. (e) Determine the velocity of air
in m/s at port 3.
3
Water
2
1
Solution Balance the mass-flow equations for an unsteady system.
Assumptions Density of water and air to remain constant at 1000 kg/3 and 1 kg/m3 respectively,
the water and air do no mix.
Analysis Since the water and air do not mix, we can find the mass flow for the water. This can
then be converted into volume flow. Since the volume of the tank is constant, the volume flow of
the water must equal the volume flow of the air.
𝑑𝑚
= 𝑚̇1 − 𝑚̇2
𝑑𝑡
𝑚̇1 = 𝜌1 𝐴1 𝑉1 = (1000) (
𝜋0.12
) (2) = 15.71 kg/s
4
𝑚̇2 = 𝜌2 𝐴2 𝑉2 = (1000) (
𝜋0.052
) (1) = 1.96 kg/s
4
𝑑𝑚
= 𝑚̇1 − 𝑚̇2 = 15.71 − 1.96 = 13.75 kg/s
𝑑𝑡
𝑑𝑚
𝑑
𝑑𝑉
= (𝜌𝑉) = 𝜌
𝑑𝑡
𝑑𝑡
𝑑𝑡
⇒
𝑑𝑉 13.75 kg m3 m3
=
[
=
]
𝑑𝑡 1000 s kg
s
Since the volume of the tank is constant, this must also be the rate that air is expelled:
𝑉̇3 = 0.0137
m3
s
Given
𝜌𝑎𝑖𝑟 = 1
kg
kg
⇒
𝑚
̇
=
0.0137
3
m3
s
𝑚̇3 = 𝜌3 𝐴3 𝑉3 ⇒ 0.0137 = (1) (
⇒ 𝑉3 = 174
𝜋0.012
) 𝑉3
4
m
s
Problem 36
A rigid chamber contains 100 kg of water at 500 kPa, 100oC. A paddle
wheel stirs the water at 1000 rpm with a torque of 100 N.m. while an
internal electrical resistance heater heats the water, consuming 10
amps of current at 110 Volts. Because of thin insulation, the chamber
loses heat to the surroundings at 27oC at a rate of 1.2 kW. Determine
(a) the rate of shaft work transfer in kW, (b) the electrical work
transfer in kJ in 10 s, (c) what must be the rate of heat transfer in kW
so that the system does not receive or lose any net energy? (You must
include sign).
Solution Find work transfer across the boundary
Assumptions The volume of the chamber remains constant, the work transfer remains constant as
stated.
Analysis Since the energy transfer due to the shaft and electrical work are independent of the heat
transfer as well as the heat loss to the environment, we can find these to solve for (c).
(a)
|𝑊̇𝑠ℎ | = 2𝜋𝑛𝑇
= 2𝜋
1000 100 rev min
kN
Nm
= kW]
[
60 1000 min 60s
1000N
= 10.47kW
With sign
𝑊̇𝑠ℎ = −10.47kJ
(b)
|𝑊̇𝑒𝑙 | =
𝑉𝐼
110 ∗ 10
=
= 1.1kW
1000
1000
With sign
𝑊̇𝑒𝑙 = −1.1kW
∴ 𝑊𝑒𝑙 = 𝑊̇𝑒𝑙 ∆𝑡 = −11kJ
(c)
Energy coming in by work:
= 10.47+1.1 = 11.57kW
∴ 𝑄̇ = −11.57kW
Problem 37
An insulated piston-cylinder device contains steam at 300 kPa,
200 deg-C, occupying a volume of 1 m3, and having a specific
100kPa
volume of 0.716 m3/kg. It is heated by an internal electrical
heater until the volume of steam doubles due to an increase in
temperature. (a) Determine the final specific volume of steam in
m3/kg. (b) If the diameter of the piston is 20 cm and the outside
pressure is 100 kPa, determine the mass of the weight placed on
the piston to maintain a 300 kPa internal pressure. (c) Calculate
Steam
the boundary work (magnitude only) done by the steam in kJ. (d)
Calculate the amount of work (magnitude only) transferred into
the weight in kJ. (e) If you are asked to choose one of the three
values - 1300 kJ, 300 kJ, 200 kJ - as your guess for the magnitude
of the electrical work transfer, which one will you pick? (enter magnitude only).
Solution Use the free body diagram and accompanying information to deduce the formulas required for
a-e
Assumptions No heat is lost to the surroundings due to the insulation and the mass of the system is
constant
Analysis The solutions can be determined by examining the free body diagram of the piston
a)
𝜈1 =
𝑉1
m3
= 0.712
𝑚1
kg
Final Volume = 2𝑉1and 𝑚1 = 𝑚2
⇒ Final Specific Volume = 2
𝑉1
m3
= 2𝜈1 = 1.432
𝑚1
kg
b)
mg/1000
𝑃0 𝐴
From the free body diagram of the piston:
𝑝𝐴 = 𝑝0 𝐴 +
⇒𝑚=
𝑚𝑔
[kN]
1000
(𝑝 − 𝑝0 )𝐴(1000) (300 − 100)𝜋(0.2)2 (1000)
=
= 640.5kg
𝑔
4(9.81)
𝑃𝐴
c)
𝑓
𝑊𝐵 = ∫ 𝑝𝑑𝑉 = Area of the rectangle, 𝑃 = Const.
P
b
f
𝑏
= 300 ∗ (2 − 1) = 300kJ
V
d)
The weight is displaced upward by a distance of
∆𝑉
1
=
= 31.8m
𝐴
𝜋(22 /4)
The work done:
𝑊 = ∫ 𝐹𝑑𝑥 =
𝑚𝑔
640.5 ∗ 9.81
∗ ∆𝑥 =
∗ 31.8 = 200kJ
1000
1000
e)
Since the steam temperature increases, it must be receiving more energy from the heater than it is
delivering through work (300kJ). So my pick is 1300kJ
Problem 33
A chamber contains a mixture of 2 kg of oxygen and 2 kmol of hydrogen. (a)
Determine the average molar mass of the mixture in kg/kmol. (b) If the specific
volume of the mixture is 2 m3/kg, determine the volume of the chamber in m3.
O2 : 2kg
H2 : 2 kmol
Solution Use the relationship between molar mass to convert the given
measurements to a single mass
Assumptions The mass and volume of the system remain unchanged
Analysis From the chemical definition of oxygen gas and hydrogen gas, we can determine the mass of
the system.
Mass of O2 = 2 kg
𝑚 𝑂2
Mole of O2 = 𝑀̅
𝑂2
2
= 32 = 0.0625 kmol
Mole of H2 = 2 kmol
Mass of H2:
̅𝐻 = 2 ∗ 2 = 4 kg [kmol ∗
𝑛𝐻2 ∗ 𝑀
2
kg
= kg]
kmol
For the mixture:
𝑚 = 2 + 4 = 6 kg
𝑛 = 2 + 0.0625 = 2.0625kmol
a)
̅𝑀𝑖𝑥𝑡𝑢𝑟𝑒 =
⇒𝑀
𝑚
6
kg
=
= 2.91
𝑛 2.0625
kmol
b)
Since:
𝜈=
𝑉
m3
=2
𝑚
kg
⇒ 𝑉 = 𝑚𝜈 = 6 ∗ 2 = 6 m3
Problem 34
The nutrition label on a granola bar, which costs $1.00, reads - Serving size 42 g; Calories Per Serving
180. Determine (a) the heating value in MJ/kg, and (b) price in cents per MJ of heat release. (c) If
gasoline with a heating value of 44 MJ/kg and a density of 750 kg/m3 costs $2.50 a gallon, what is the
gasoline price in cents/MJ?
Solution Use the definition of heating value as well as fluids definitions to obtain the proper units.
Assumptions The values are not changing with time
Analysis We must use our intuition to determine the proper values, using mainly unit analysis
Heat released from 42g = 180 ∗ 4.2 = 720kJ
a)
Heating Value =
𝐸𝑛𝑔𝑒𝑟𝑔𝑦
𝑀𝑎𝑠𝑠
720
kJ
kJ
= 0.042 [kg] = 18,000 kg = 180
b)
Heat released from a bar
MJ
kg
=
180 ∗ 4.2
MJ = 0.756MJ
1000
Price per MJ:
=
100 ¢
¢
= 132
0.756 MJ
MJ
c)
1 Gallon = 3.785 ∗ 10−3 m3
Mass of a gallon:
𝑚 = 𝜌𝑉 = (750)(3.785 ∗ 10−3 ) = 2.84 kg
∴ Heat released from 1 Gallon = (44 ∗ 2.84) MJ = 124.9MJ
250
¢
∴ Price per MJ = 123.9 = 2 MJ
Problem 35
𝑄̇𝑖𝑛
(a) What is a system with no mass transfer called?
(b) What is a system with no heat transfer called?
(c) What is a system with no mass or energy transfer
called?
Solution Use the generic system diagram with the
definition of thermodynamics terms to solve the
problem
Assumptions The system is at steady state
a)
No mass transfer:
𝑚̇𝑖 = 𝑚̇𝑒 = 0 ⇒ 𝐶𝑙𝑜𝑠𝑒𝑑
b)
No heat transfer:
𝑄̇𝑖𝑛 = 𝑄̇𝑜𝑢𝑡 = 0 ⇒ 𝐴𝑑𝑖𝑎𝑏𝑎𝑡𝑖𝑐
c)
𝑚̇𝑖
𝑊̇𝑖𝑛
𝑊̇𝑜𝑢𝑡
𝑚̇𝑒
𝑄̇𝑜𝑢𝑡
No mass or heat transfer:
𝑚̇𝑖 = 𝑚̇𝑒 = 𝑄̇𝑖𝑛 = 𝑄̇𝑜𝑢𝑡 = 0 ⇒ 𝐼𝑠𝑜𝑙𝑎𝑡𝑒𝑑
Problem 38
Betz law states that only 53% of the kinetic energy transported by wind can
be converted to shaft work by a perfect wind turbine. For a 50 m diameter
turbine in a 20 mph wind, determine (a) the mass flow rate of air (in kg/s)
intercepted by the turbine? (b) the specific kinetic energy of the flow (in
kJ/kg), (c) the rate of transport of kinetic energy (kW), and (d) the maximum
possible shaft power? Assume density of air as 1.1 kg/m3. (1 mph = 0.447
m/s).
Solution Use thermodynamic principles along with the given information to solve the problem
Assumptions The air velocity is not changing with time and the system is at steady state.
Analysis We must us the energy equations to convert the winds energy to shaft work using Benz’ law.
a)
𝑚̇ = 𝜌𝐴𝑉 = 1.1 ∗
𝜋(50)2
kg
∗ (20 ∗ 0.447) = 19300
4
s
b)
𝑘𝑒 =
𝑉2
(20 ∗ 0.447)2 kJ
kJ
=
= 0.04
2000
2000
kg
kg
c)
̇ = 𝑚̇(𝑘𝑒) = 771.6 kW
𝐾𝐸
d)
̇ , the shaft work can be solved as follows:
Since there only energy available to the shaft is the 𝐾𝐸
̇ = 0.53 ∗ 771.6 = 408.95 kW
𝑊̇𝑠ℎ = 𝜂𝐾𝐸
Problem 39
A closed system interacts with its surroundings and the
following data are supplied: Wdot_sh= -10 kW, Wdot_el =
𝑄̇ = 5
𝑊̇𝑒𝑙 = 5
𝑊̇𝑠ℎ = 10
5 kW, Qdot = -5 kW. If there are no other interactions, determine dE/dt.
Solution Use the energy balance equation with the proper sign convention to solve for dE/dt.
Assumptions As we are solving for dE/dt, we cannot assume it is steady or unsteady; because of this, the
only assumption we can make is that we are solving the problem at this instance.
Analysis We will simply solve the energy balance equation
𝑑𝐸
= ∑𝑚̇𝑖 𝑗𝑖 + ∑𝑚̇𝑒 𝑗𝑒 + 𝑄̇𝑒𝑥𝑡 − 𝑊̇𝑒𝑥𝑡
𝑑𝑡
Since there is no other interactions, this implies that it is a closed system:
𝑑𝐸
= 𝑄̇𝑒𝑥𝑡 − 𝑊̇𝑒𝑥𝑡
𝑑𝑡
𝑑𝐸
= [−5] − [−10 + 5]
𝑑𝑡
⟹
𝑑𝐸
= 0 kW
𝑑𝑡
Problem 40
A gas trapped in a piston-cylinder device is heated as shown in figure. The load on the massless
piston of area 0.2 m2 is such that the initial pressure of the gas is 200 kPa while the outside pressure
is 100 kPa. Also, the initial temperature is 300 K. When the temperature of the gas reaches 600 K,
the piston velocity is measured as 0.5 m/s and dE/dt is measured as 30 kW. At that instant,
determine (a) the rate of boundary work transfer, (b) the rate of external work transfer, (c)the rate
of heat transfer, and (d) the load (in kg) mass on the piston. Assume the ambient atmospheric
pressure to be 100 kPa. (e) What type of system is this?
Solution We will use the energy balance equation along with the thermodynamic definition of work to
solve the problem
Assumptions We are solving this problem for this instance only, every instance will have a different
solution. The pressure inside the piston will not chance due to the equilibrium condition.
Analysis The piston will have the heat going into the system while performing boundary work as well as
work to raise the mass.
We first need to determine the 𝑉̇:
𝑉̇ = 𝐴𝑉 = .2 ∗ 0.5 = 0.1
a)
m3
s
100kPa
200kPa
m3 kN kN ∗ m
𝑊̇𝐵 = 𝑉̇ ∗ 𝑝 = 0.1 ∗ 200 = 20 kW [
∗
=
= kW]
s m2
s
b)
Since 𝑊̇𝐵 is the only work the system is doing
⟹ 𝑊̇𝐵 = 𝑊̇𝑒𝑥𝑡 = 20 kW
c)
The rate of heat transfer can be determined from the energy balance equation:
𝑑𝐸
= 30 kW = ∑𝑚̇𝑖 𝑗𝑖 + ∑𝑚̇𝑒 𝑗𝑒 + 𝑄̇𝑒𝑥𝑡 − 𝑊̇𝑒𝑥𝑡 = 𝑄̇𝑒𝑥𝑡 − 𝑊̇𝑒𝑥𝑡
𝑑𝑡
⇒ 30 kW = 𝑄̇𝑒𝑥𝑡 − 20 kW
⇒ 𝑄̇𝑒𝑥𝑡 = 50 kW
d)
From the force-balance equation:
mg/1000
𝑃0 𝐴
𝑚𝑔
𝑝𝐴 = 𝑝0 𝐴 +
[kN]
1000
⇒𝑚=
(𝑝 − 𝑝0 )𝐴(1000) (200 − 100)𝜋(0.2)2 (1000)
=
= 320.2kg
4𝑔
4(9.81)
𝑃𝐴
e)
𝑑𝐸
≠ 0 ⇒ 𝑈𝑛𝑠𝑡𝑒𝑎𝑑𝑦 𝑆𝑦𝑠𝑡𝑒𝑚
𝑑𝑡
Problem 41
A rigid cylindrical tank stores 100 kg of a substance at 500 kPa and 500 K while the outside temperature
is 300 K. A paddle wheel stirs the system transferring shaft work at a
rate of 0.5 kW. At the same time an internal electrical resistance
m=100kg
heater transfers electricity at the rate of 1 kW. (a) Do an energy
T=500K
analysis to determine the rate of heat transfer Qdot in kW for the
𝑊̇𝑠ℎ =
P=500kP
tank. (b) Determine the absolute value of the rate at which entropy
0.5kW
a
leaves the internal system (see attached diagram). (c) Determine
the rate of entropy generation for the system's universe.
Solution Use the thermodynamic definitions to analyze the system
𝑊̇𝑒𝑙 =
1kW
Assumptions Due to the way the system is setup, the tank can be assumed to be operating at steadystate. Since it is a ridged tank, there is also no mass transfer.
Analysis we will use the energy balance equation as well as the entropy balance equation to determine
the numerical values.
a)
𝑑𝐸
= 0 = ∑𝑚̇𝑖 𝑗𝑖 + ∑𝑚̇𝑒 𝑗𝑒 + 𝑄̇𝑒𝑥𝑡 − 𝑊̇𝑒𝑥𝑡
𝑑𝑡
Since there is no mass transfer:
0 = 𝑄̇𝑒𝑥𝑡 − 𝑊̇𝑒𝑥𝑡
⇒ 𝑄̇𝑒𝑥𝑡 = 𝑊̇𝑒𝑥𝑡 = [−1 − .5] = −1.5 kW
b)
𝑑𝑆
𝑄̇𝑒𝑥𝑡
= 0 = ∑𝑚̇𝑖 𝑠𝑖 + ∑𝑚̇𝑒 𝑠 +
+ 𝑆̇
𝑑𝑡
𝑇
⇒ 𝑆̇ = −
𝑄̇𝑒𝑥𝑡
1.5
kW
= − [−
] = .003
𝑇
500
k
c)
The rate of entropy generation in the systems universe will be when the heat reaches the ambient
boundary:
𝑑𝑆
𝑄̇𝑒𝑥𝑡
= 0 = ∑𝑚̇𝑖 𝑠𝑖 + ∑𝑚̇𝑒 𝑠 +
+ 𝑆̇
𝑑𝑡
𝑇
⇒ 𝑆̇ = −
𝑄̇𝑒𝑥𝑡
1.5
kW
= − [−
] = .005
𝑇
300
k
Problem 42
A horizontal piston-cylinder device contains air at 90 kPa while the outside
pressure is 100 kPa. This is made possible by pulling the piston with a hanging
weight through a string and pulley arrangement. If the piston has a diameter of
20 cm, (a) determine the mass of the hanging weight in kg. (b) The gas is now
heated using an electrical heater and the piston moves out by a distance of 20
cm. Determine the boundary work in kJ. (c) What fraction of the boundary work
performed by the gas goes into the hanging weight?
Solution Use force balance equation to determine the mass of the weight and
the definition of boundary work to determine the work done.
Assumptions The pressure in the piston remains constant and there is no mass transfer.
Analysis The problem has given us enough information to solve all required inputs in one step each.
a)
From the force balance equation:
𝑝𝐴 +
𝑚𝑔
= 𝑝0 𝐴 [kN]
1000
⇒𝑚=
𝑃𝐴
(𝑝0 − 𝑝)𝐴 ∗ 1000 (100 − 90)𝜋(0.2)2 1000
=
= 32 kg
𝑔
4 ∗ 9.81
𝑃0 𝐴
mg/1000
b)
𝑉𝑓
𝑊𝐵 = ∫ 𝑝𝑑𝑉
𝑉𝑖
Since the internal pressure is not changing:
𝑊𝐵 = 𝑝∆𝑉 = 𝑝𝐴∆𝐿 = 90 ∗
𝜋
∗ (0.2)2 ∗ 0.2 = .565 kJ
4
Since it is being done by the system:
𝑊𝐵 = +.565 kJ
c)
Since gravity is supplying the force on the weight, the boundary work is not going into the weight
⇒ 0%
Problem 43
Steam flows into a steady adiabatic turbine at 10 MPa,
600°C and leaves at 58 kPa and 90% quality. The mass
flow rate is 9 kg/s. Additional properties at the exit that
are known are: A = 1.143 m2, v = 2.54 m3/kg, u = 2275.2
kJ/kg, e= 2275.4 kJ/kg, h=2422.4 kJ/kg. If the turbine
produces 1203 kW of shaft power, determine at the exit
(a) the velocity in m/s. (b) the rate of flow of kinetic
energy, (c) the rate of transfer of stored energy, and (d)
the rate of transfer of flow energy. Neglect potential
energy.
Simplified view of our turbine:
𝑚̇𝑖
Solution Apply thermodynamic equations to the given properties to determine the given values
𝑊̇𝑜𝑢𝑡
𝑚̇𝑒
Assumptions The turbine is operating at steady state and we can assume the given values to be constant
and not changing with time. The turbine is also not losing heat to the environment.
Analysis After drawing a free-body diagram, we can determine the numerical solutions for the problem.
a)
𝑚 = 𝜌𝐴𝑉
⇒𝑉=
𝑚 𝑚𝜈 9 ∗ 2.54
𝑘𝑔 𝑚3 1
m
m
=
=
= 20 [ ∗
∗ 2 = ] = 20
𝜌𝐴
𝐴
1.143
𝑠 𝑘𝑔 𝑚
s
s
b)
̇ = 𝑚̇(𝑘𝑒) = 𝑚̇ (
𝐾𝐸
̇ =9∗
⇒ 𝐾𝐸
𝑣2
)
2000
202 𝑘𝑔 𝑘𝐽
= 𝑘𝑊] = 1.8 kW
[ ∗
2000 𝑠 𝑘𝑔
c)
𝑈̇ = 𝑚̇(𝑢)
⇒ 𝑈̇ = 9 ∗ 2275.2 [
𝑘𝑔 𝑘𝐽
∗
= 𝑘𝑊] = 20480 𝑘𝑊 = 20.48 𝑀𝑊
𝑠 𝑘𝑔
d)
𝐽 ̇ = 𝑚̇(𝑗) = 𝑚̇(𝑘𝑒 + 𝑝𝑒 + ℎ)
⇒ 𝐽̇ = 9 ∗ (
202
𝑘𝑔 𝑘𝐽
+ 0 + 2422.4) [ ∗
= 𝑘𝑊] = 21800𝑘𝑊 = 21.8𝑀𝑊
2000
𝑠 𝑘𝑔
Test 1-3
An electric motor, operating at steady state, produces 100 kW of shaft
output while consuming electricity at the rate of 120 kW. (a)
Determine Qdot (include sign). (b) The rate at which entropy enters
the surroundings (in kW/K) if the motor surface temperature is 400 K
and the ambient temperature is 300 K. (c) Determine the rate at which
entropy is generated (in kW/K) in the system's universe. (d) What is
the energetic efficiency of the motor (in percent)?
Solution Balance thermodynamic equations for a steady system
T=300K
T=400K
Assumptions Since we would not design a motor to work unsteady, the system is assumed to be steady
Analysis We are given enough information to solve the energy and entropy balance equations since it is
operating at steady-state.
a)
𝑑𝐸
= 0 = ∑𝑚̇𝑖 𝑗𝑖 + ∑𝑚̇𝑒 𝑗𝑒 + 𝑄̇𝑒𝑥𝑡 − 𝑊̇𝑒𝑥𝑡
𝑑𝑡
⇒ 𝑄̇ 𝑒𝑥𝑡 = 𝑊̇ 𝑒𝑥𝑡 = [−𝑊̇ 𝑒𝑙 + 𝑊̇ 𝑠ℎ ] = [−120 + 100] = −20kW
b)
Since we are looking for the entropy entering the surrondings, we need the temp of the surroundings to
find the entropy.
⇒
𝑄̇ 𝑒𝑥𝑡
20
kW
=
= 0.0667
𝑇𝐵
300
K
c)
Once again, since we are looking for the entropy generation in the universe, T=300K:
𝑑𝑆
𝑄̇𝑒𝑥𝑡
̇
= 0 = ∑𝑚̇𝑖 𝑠𝑖 + ∑𝑚̇𝑒 𝑠 +
+ 𝑆𝑔𝑒𝑛,𝑢𝑛𝑖𝑣
𝑑𝑡
𝑇
̇
⇒ 𝑆𝑔𝑒𝑛,𝑢𝑛𝑖𝑣
=−
𝑄̇𝑒𝑥𝑡
20
kW
= − [−
] = 0.0667
𝑇
300
K
d)
𝜂𝑇 =
𝐷𝑒𝑠𝑖𝑟𝑒𝑑 𝐸𝑛𝑒𝑟𝑔𝑦
100𝑘𝑊
∗ 100% =
∗ 100% = 83.33%
𝑅𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝐼𝑛𝑝𝑢𝑡
120𝑘𝑊
Test 1-4
A heat engine with a 40% thermal efficiency operate between a 2000
K reservoir and atmosphere at 300 K. The output of the engine is
used by a refrigeration system with a COP of 2, which operates
between the atmosphere and a 200 K cold storage. Determine the
cooling load (𝑄̇𝐶 in kW) of the refrigerator if 50 kW of heat is supplied
to the heat engine.
Solution Use the efficiency equations to solve for the heat
transferred from the cold side of the refrigeration cycle.
Assumptions Both engines are operating at steady state and
all losses are accounted for.
300K
300K
𝑊̇𝑁𝑒𝑡
2000K
200K
Analysis Since we know the efficiency for the heat engine and the COP of the refrigeration engine, we do
not need to employ the energy equations.
For the heat engine:
𝜂=
𝑂𝑢𝑡𝑝𝑢𝑡 𝑊̇𝑛𝑒𝑡
=
𝐼𝑛𝑝𝑢𝑡
𝑄̇𝑛𝑒𝑡
⇒ 𝑊̇𝑛𝑒𝑡 = 𝜂 ∗ 𝑄̇𝑛𝑒𝑡 = 0.40 ∗ 50kW = 20kW
For the refrigeration engine:
𝐶𝑂𝑃 =
𝑂𝑢𝑡𝑝𝑢𝑡
𝑄̇𝐶
=
𝐼𝑛𝑝𝑢𝑡
𝑊̇𝑖𝑛
⇒ 𝑄̇𝐶 = 𝐶𝑂𝑃 ∗ 𝑊̇𝑖𝑛 = 2 ∗ 20kW = 40kW
Test 1-8
A piston inside a rigid cylinder, closed at both ends, separates two
chambers: one contains a two-phase mixture of H2O and the other
contains a two-phase mixture of R-134a. If the temperature inside
the chamber containing H2O is 85.9 C, determine the temperature
inside the other chamber.
H2O
R-134a
Solution Use the equilibrium conditions to find the temperature of the R-134a chamber
Assumptions Each side of the piston is isolated from the other so no heat is transferred across the
boundary or to the surroundings. Since it is in equilibrium, the pressures on either side are equal.
Analysis Since both sides are said to be a two-phase mixture, this implies that both sides are between
their respective f- and g-points.
Since we know the temperature of the H2O side, we need to determine the pressure of the H2O using
the Saturation Table:
𝑝𝐻2 𝑂 = 𝑝𝑆𝑎𝑡@85.9°𝐶 = 70kPa
This will be the saturation pressure of R-134a, which will yield the saturation temperature from the
Saturation Table:
⇒ 𝑇𝑅−134𝑎 = 𝑇𝑆𝑎𝑡@70kPa = −34℃