Solution Manual For Fundamentals of Engineering Thermodynamics, 9th Edition
Preview Extract
Problem 2.1
A baseball has a mass of 0.3 lb. What is the kinetic energy relative to home plate of a 94 mile
per hour fastball, in Btu?
KE = ยฝ m V2
relative to home plate
= ยฝ (0.3 lb){(44
= 0.114 Btu
mi
5280 ft
h
1 mi
)

1h
2
} = 2851.1 ftโlb/s2 
60 s
1 lbf
32.2 ftโlb/s2

1 Btu
778 ftโlbf

Problem 2.2
Determine the gravitational potential energy, in kJ, of 2 m3 of liquid water at an elevation of
30 m above the surface of Earth. The acceleration of gravity is constant at 9.7 m/s2 and the
density of the water is uniform at 1000 kg/m3. Determine the change in gravitational potential
energy as the elevation decreased by 15 m.
KNOWN: The elevation of a known quantity of water is decreased from a given initial value by
a given amount.
FIND: Determine the initial gravitational potential energy and the change in gravitational
potential energy.
V = 2 m3
g = 9.7 m/s2
ฯ = 1000 kg/m3
SCHEMATIC AND GIVEN DATA:
ENGINERING MODEL:
(1) The water is a closed system. (2) The acceleration of
gravity is constant. (3) The density of water is uniform.
ฮz = – 15 m
z1 = 30 m
ANALYSIS: The initial gravitational potential energy is
??1 = ???1 = (?V)??1
= (1000
kg
m3
m
1N
?
1 kgโ 2
) (2 m3 ) (9.7 2 ) (30 m) 
m
s

1 kJ
= 582 kJ
The change in potential energy is
โ?? = ??(?2 โ ?1 ) = ??โ?
= (1000
kg
m3
m
1N
?
1 kgโ 2
) (9.7 2 ) (โ15 m) 
= โ145.5 kJ
m
s


103 Nโm
1 kJ
103 Nโm

Problem 2.3
An object whose weight is 100 lbf experiences a decrease in kinetic energy of 500 ftโlbf and an
increase in potential energy of 1500 ftโlbf. The initial velocity and elevation of the object, each
relative to the surface of the earth, are 40 ft/s and 30 ft, respectively. If g = 32.2 ft/s2, determine
(a) the final velocity, in ft/s.
(b) the final elevation, in ft.
KNOWN: An object experiences specified changes in kinetic and potential energy. The initial
velocity and elevation are known.
FIND: Determine the final velocity and the final elevation.
SCHEMATIC AND GIVEN DATA:
ฮKE = โ 500 ftโlbf
ฮPE = + 1500 ftโlbf
V1 = 40 ft/s
z1 = 30 ft
z
ENGINEERING MODEL: (1) The object is a
closed system. (2) The acceleration of gravity is
constant; g = 32.2 ft/s2. (3) Velocity and elevation
are measured relative to the surface of the earth.
Fgrav = mg = 100 lbf
ANALYSIS: (a) The change in kinetic energy is: ฮKE = 1/2 m(V22 โ V12). Solving for V2
1/2
2โKE
V2 = [(
?
) + V12 ]
The mass is
m=
Fgrav
?
=
(1)
100 lbf 32.2 ftโlb/s2

 = 100 lb
32.2 ftโs2
1 lbf
So, inserting values into (1) and converting units
2(โ500 ftโlbf) 32.2 ftโlb/s2
V2 = [
(100 lb)

? lbf
ft
1/2
 + (40 )2 ]
s
= 35.75 ft/s
(b) The change in potential energy is: ฮPE = mg(z2 โ z1). With mg = Fgrav,
z2 = ฮPE/Fgrav + z1 = (1500 ftโlbf)/(100 lbf) + (30 ft) = 45 ft
Problem 2.4
A construction crane weighing 12,000 lbf fell from a height of 400 ft to the street below during a
severe storm. For g = 32.05 ft/s2, determine mass, in lb, and the change in gravitational potential
energy of the crane, in ftโlbf.
KNOWN: A crane of known weight falls from a known elevation to the street below.
FIND: Determine the change in gravitational potential energy of the crane.
SCHMATIC AND GIVEN DATA:
Fgrav = 12,000 lbf
g = 32.05 ft/s2
z2 = 0
z1 = 400 ft
ENGINEERING MODEL: (1) The crane is the closed system. (2) The acceleration of gravity
is constant.
ANALYSIS:
To get the mass, note that Fgrrav = mg. Thus
?=
Fgrav
?
=
12000 lbf
32.174 lbโft/s2
32.05 ft/s
1 lbf

2
 = 12,046 lb
The change in gravitational potential energy is
โ?? = ??(?2 โ ?1 ) = Fgrav โ? = (12000 lbf)(โ400 ft) = 4.8 x 106 ft โ lbf
Problem 2.5
An automobile weighing 2500lbf increases its gravitational potential energy by 2.25 x 104 Btu
in going from an elevation of 5,183 ft in Denver to the highest elevation on Trail Ridge road in
the Rocky Mountains. What is the elevation at the high point of the road, in ft?
KNOWN: An automobile of known weight increases its gravitational potential energy by a
given amount. The initial elevation is known.
FIND: Determine the final elevation.
Fgrav = 2500 lbf
ฮPE = 2.25 x 104 Btu
z2 = ?
z1 = 5183 ft
ENGINEERING MODEL: (1) The automobile is the closed system. (2) The acceleration of
gravity is constant. (2) Velocity and elevation are each measured relative to a stationary observer
on the surface of the earth.
ANALYSIS: The change in gravitational potential energy is: ฮPE = mg(z2 โ z1). With
Fgrav = mg, we get
ฮPE = Fgrav(z2 โ z1)
Solving for z2
?2 =
โPE
F????
+ ?1 =
(2.25 ? 104 Btu) 778 ftโlbf
(2500 lbf)

1 Btu
 + 5183 ft = 12,185 ft
Problem 2.6
An object of mass 15 kg is at an elevation of 100 m relative to the surface of the Earth. What is
the potential energy of the object, in kJ? If the object were initially at rest, to what velocity, in
m/s, would you have to accelerate it for the kinetic energy to have the same value as the potential
energy you calculated above? The acceleration of gravity is 9.8 m/s2.
KNOWN: An object has known mass and elevation and is initially at rest. The acceleration of
gravity is given.
FIND: Determine the initial potential energy and the final velocity if it were accelerated to have
kinetic energy equal to the initial potential energy.
SCHEMATIC AND GIVEN DATA:
m = 15 kg
V2 = ?
V1 = 0
g = 9.8 m/s2
ENGINEERING MODEL: (1) The object is a
closed system. (2) The acceleration of gravity is
constant: g = 9.8 m/s2. (3) The final kinetic energy
is equal to the initial gravitational potential energy.
(4) Velocity and elevation are each measured relative
to a stationary observer on the surface of the earth.
m = 15 kg
z1 = 100 m
ANALYSIS: First, evaluate the initial potential energy.
PE1 = mgz1 = (15 kg)(9.8 m/s2)(100 m) 
For KE2 = PE1 โ
1N
1 kgโm/s2

1 kJ
103 Nโm
 = 14.7 kJ
ยฝ mV22 = PE1
So
V2 = โ
2PE1
?
=โ
2(14.7 kJ) 103 Nโm
(15 kg)

1 kJ

1 kgโm/s2
1N
 = 44.27 m/s
2.7 An automobile having a mass of 900 kg initially moves along a level highway at 100 km/h
relative to the highway. It then climbs a hill whose crest is 50 m above the level highway and
parks at a rest area located there. For the automobile, determine its changes in kinetic and
potential energy, each in kJ. For each quantity, kinetic and potential energy, specify your
choice of datum and reference value at that datum. Let g = 9.81 m/s2.
KNOWN: Data are provided for an automobile on the open road.
FIND: Determine the changes in kinetic and potential energy for the automobile and specify an
appropriate datum for each.
SCHEMATIC AND GIVEN DATA:
V2 = 0
stationary observer on
level road
V=0
z=0
z2 = 50m
m = 900 kg
V1= 100 km/h
ENGINEERING MODEL: (1) The automobile is the closed system. (2) The acceleration of
gravity is constant; g = 9.81 m/s2. (3) Velocities and elevations are measured relative to the
stationary observer on the level road.
ANALYSIS: The change in kinetic energy is
ฮKE = ยฝ m (V22 โ V12)
= ยฝ (900 kg)[0 โ (100 km/h)2
1h
2 103 m 2
1N
1 km
1 kgโm/s2
 
3600 s
 

1 kJ
103 Nโm

= โ 347.2 kJ (decrease)
The change in potential energy is
ฮPE = mg(z2 โ z1)
= (900 kg)(9.81 m/s2)(50 m โ 0) 
1N
1 kgโm/s2

1 kJ
103 Nโm
 = +441.5 kJ (increase)
2.8 Vehicle crumple zones are designed to absorb energy during an impact by deforming to
reduce transfer of energy to occupants. How much kinetic energy, in Btu, must a crumple
zone absorb to fully protect occupants in a 3000 lb vehicle that suddenly decelerates from
10 mph to 0 mph?
2.9 In a recent airline disaster, an airliner flying at 30,000 ft, 550 mi/h, lost power and fell to
Earth. The mass of the aircraft was 255,000 lb. If the magnitude of the work done by drag
force on the plane during the fall was 2.96 x 106 Btu, estimate the velocity of the aircraft at the
time of impact, in mi/h. Let g = 32.08 ft/s2.
KNOWN: An airliner falls from a known elevation and velocity to the surface of the Earth. The
magnitude of the work done against drag force is specified, and the average acceleration of
gravity is given.
FIND: Estimate the velocity at the time of impact with the Earth.
SCHEMATIC AND GIVEN DATA:
0
m = 255,000 lb
g = 32.08 ft/s2
V1 = 550 mi/h
5280 ft
1h
1 mi
3600 s
z2 = 0
z1 = 100 m
?drag = 2.96 x 106 Btu
z
ANALYSIS: The work done against drag is
?
Wdrag = โซ? 2 ?drag โ ??
= 806.7 ft/s
ENGINEERING MODEL: (1) The airliner is a
closed system. (2) The acceleration of gravity is
constant: g = 32.08 ft/s2. (3) The only forces acting
on the airliner as it falls are the drag force opposing
motion and the force of gravity.
Fdrag โ opposite to
direction of motion
1
Fdrag โds is negative.
So
ds – direction of
motion
Wdrag = โ 2.96 x 106 Btu
The work done against drag equals the change in kinetic energy plus the change in potential
energy
?
Wdrag = โซ? 2 ?drag โ ?? = ยฝ m(V22 โ V12) + mg(z2 โ z1)
1
Problem 2.9 (Continued)
Solving for V2
2?drag
V2 = โ
?
+ 2??1 + v12
lb
=
2
2(โ2.96 x 106 Btu) 778.17 ftโlbf 32.174 ftโs2
ft
2 ) ft
โ
(30,000


+
2
(32.08
)
ft)
+
(806.7
(255,000 lb)
1 Btu
1 lbf
s2
s2
= 1412 ft/s
= 1412 ft/s
1 mi
3600 s
5280 ft
1h
= 962.7 mi/h
COMMENTS: The work done against drag is reduces the final kinetic energy, so it is negative
in this case. Also, be careful in applying the English system units in this problem.
Problem 2.10
Two objects having different masses are propelled vertically from the surface of Earth, each with
the same initial velocities. Assuming the objects are acted upon only by the force of gravity,
show that they reach zero velocity at the same height.
KNOWN: Two objects are propelled upward from the surface of Earth with the same initial
velocities and are acted upon only by the force of gravity.
FIND: Show that they reach zero velocity at the same height.
V2 = 0
SCHEMATIC AND GIVEN DATA:
ENGINEERING MODEL: (1) Each object is a closed system.
(2) The acceleration of gravity is constant. (3) The only force
acting is the force of gravity.
ANALYSIS: For an object moving vertically under the
influence of gravity only, Eq. 2.11 applies
1
2
z1 = 0
z2
V1
V1
m1
m2
??(V22 โ V12 ) + ??(?2 โ ?1 ) = 0
For V2 = 0 and z1 = 0
1
โ ?V12 + ???2 = 0
2
Thus
?2 = V12 /2?
Since the final height doesnโt depend on mass, both objects will reach zero velocity at the same
final height.
Problem 2.11
An object whose mass is 100 lb falls freely under the influence of gravity from an initial
elevation of 600 ft above the surface of Earth. The initial velocity is downward with a
magnitude of 50 ft/s. The effect of air resistance is negligible. Determine the velocity, in ft/s, of
the object just before it strikes Earth. Assume g = 31.5 ft/s2.
KNOWN: An object of known mass falls freely from a known elevation and with a given initial
velocity. The only force acting is the force of gravity.
m = 100 lb
g = 31.5 ft/s2
FIND: Determine the velocity of the object just before it strikes Earth.
SCHEMATIC AND GIVEN DATA:
ENGINEERING MODEL: (1) The object is a closed system.
(2) The acceleration of gravity is constant: g = 51.5 ft/s2.
(3) The only force acting on the object is the force of gravity.
ANALYSIS: Since the only force acting on the object is the force of
gravity, Eq. 2.11 applies. Thus
0
1
2
2)
1
?(V2 โ V1 + ??(?2 โ ?1 ) = 0
V1 = 50 ft/s
z1 = 600 ft
z2 = 0
V2
2
Solving for V2
V2 = โV12 + 2??1
Inserting values
ft2
ft
s
s
V2 = โ502 2 + 2 (31.5 2 ) (600 ft) = 200.7 ft/s
1. Note that the mass cancels out. Any object falling freely under the influence of gravity, with
no effects of air resistance, would reach the same final velocity.
2.12 During the packaging process, a can of soda of mass 0.4 kg moves down a surface inclined
at 20o relative to the horizontal, as shown in Fig. P2.12. The can is acted upon by a constant
force R parallel to the incline and by the force of gravity. The magnitude of R is 0.05 N.
Ignoring friction between the can and the inclined surface, determine the canโs change in
kinetic energy, in J, and whither it is increasing or decreasing. If friction between the can
and the inclined surface were significant, what would effect would that have on the value of
the change in kinetic energy? Let g = 9.8 m/s2.
Problem 2.12 (Continued)
Problem 2.13
Jack, who weighs 150 lbf, runs 5 miles in 43 minutes on a treadmill set at a onedegree incline
(Fig. P2.13). The treadmill display shows he has burned 620 kcal. For jack to break even
caloriewise, how much vanilla ice cream, in cups, may he have after his workout?
๏ท
๏ท
Exercise value = 620 kcal
Calorific value; 1 cup of vanilla ice cream = 264 kcal
To break even caloriewise, Jack may have
620 kcal
= 2.35 cups
264 kcal/cup
Problem 2.14
An object initially at an elevation of 5 m relative to Earthโs surface and with a velocity of 50 m/s
is acted on by an applied force R and moves along a path. Its final elevation is 20 m and its
velocity is 100 m/s. The acceleration of gravity is 9.81 m/s2. Determine the work done on the
object by the applied force, in kJ.
KNOWN: An object moves along a path due to the action of an applied force. The elevation
and velocities are known initially and finally.
V2 = 100 m/s
FIND: Determine the work of the applied force.
R
SCHEMATIC AND GIVEN DATA:
m = 50 kg
V1 = 50 m/s
ENGINEERING MODEL: (1) The object is a closed
system. (2) R is the only force acting on the object other
z
than the force of gravity. (3) g = 9.81 m/s2 and is constant.
z2 = 20 m
z1 = 5 m
ANALYSIS: To find the work of force R we use
2
1
Work = โซ1 ? โ ?? = ?(V22 โ V12 ) + ??(?2 โ ?1 )
2
Inserting values and converting units
1
m2
m
1N
1 kJ
Work = { (50 kg)(1002 โ 502 ) 2 + (50 kg) (9.81 2 ) (20 โ 5)m} 
 3

2
2
s
s
1 kg โ m/s 10 N โ m
= 187.5 + 7.36 = 194.9 kJ
Problem 2.15
An object of mass 10 kg, initially at rest, experiences a constant horizontal acceleration of 4 m/s2
due to the action of a resultant force applied for 20 s. Determine the total amount of energy
transfer by work, in kJ.
KNOWN:
A system of known mass experiences a constant horizontal acceleration due to an applied force
for a specified length of time.
m = 10 kg
ax = 4 m/s2
FIND: Determine the amount of energy transfer by work.
Fx
ฮt = 20 s
V1 = 0
SCHEMATIC AND GIVEN DATA:
ENGINEERING MODEL: (1) The object is a closed system.
(2) The horizontal acceleration is constant.
x
ANALYSIS: The work of the resultant force is determined using Eq. 2.6
0
?2
1
2
2
โซ? F? ?? = ?(V2 โ V1 )
2
1
To find V2, use the fact that the acceleration is constant
ax =
?V
โ dV = axdt
??
โ
V
?
1
1
2
2
โซV ?V = โซ? a? ??
0
or
(V2 โ V1) = ax(t2 โ t1) = axฮt
Thus
V2 = (4 m/s2) (20 s) = 80 m/s
Finally, the work of the resultant force is
?
1
2
โซ? F? ?? = 2 ?V22
1
1
m2
1N
2
s
1 kgโm/s2
= (10 kg)(802 ) 2 

1 kJ
 = 32 kJ
103 Nโm
Problem 2.16
An object with an initial horizontal velocity of 20 ft/s experiences a constant horizontal
acceleration due to the action of a resultant force applied for 10 s. The work of the resultant
force is 10 Btu. The mass of the object is 55 lb. Determine the constant horizontal acceleration,
in ft/s2.
KNOWN: Data are known for an object accelerating horizontally under the action of a constant
resultant force for a specified amount of time. The work of the force is given.
FIND: Determine the constant horizontal acceleration.
SCHEMATIC AND GIVEN DATA:
V1 = 20 ft/s
F
V2
m = 55 lb
Finally
ฮt = 10 s
Work of force F is 10 Btu
Initially
ENGINEERING MODEL: (1) The object is a closed system. (2) The resultant force is
constant over the time interval. (3) All forces and motions are horizontal.
ANALYSIS: The constant horizontal acceleration is a = dV/dt = (V2 โ V1)/ฮt. To find V2 = use
Eq. 2.6
?2
2
2
โซ? ? โ ?? = ยฝ m(V2 โ V1 )
1
Work, W, of the
resultant force F
So
2?
V2 = โ
?
2(10 Btu) 778 ftโlbf
+ V12 = โ
(55 lb)

1 Btu
32.2 ftโlb/s2

1 lbf
ft
 + (20 )2 = 97.52 ft/s
The acceleration is
a = (V2 โ V1)/ฮt = [(97.52 โ 20)ft/s](10s) = 7.752 ft/s2
s
Problem 2.17
A gas in a pistoncylinder assembly undergoes a process for which the relationship between
pressure and volume is pV2 = constant. The initial pressure is 1 bar, the initial volume is 0.1 m3,
and the final pressure is 9 bar. Determine (a) the final volume, in m3, and (b) the work for the
process, in kJ.
KNOWN: A gas in a pistoncylinder assembly undergoes a process during which pV2 =
constant. State data are provided.
FIND: Determine the final volume occupied by the gas and the work for the process.
SCHEMATIC AND GIVEN DATA:
p (bar)
9
Gas
.2
ENGINEERING MODEL: (1) The
gas is a closed system. (2) Volume
change is the only work mode. (3) the
process of the obeys pV2 = constant.
pV2 = constant
p1 = 1 bar
V1 = 0.1 m3
p2 = 9 bar
.1
1
0.1
V (m3)
ANALYSIS:
(a) We have pV2 = constant. Thus p1V12 = p2V22. Solving for V2
?1 1/2
V2 = [ ]
?2
1 1/2
V1 = [ ]
9
(0.1 m3) = 0.0333 m3
(b) Using Eq. 2.17 to determine the work
2
W = โซ1 ??? =
?2 ?2 โ?1 ?1
(1โ2)
(See Example 2.1(a) for the integration)
Inserting values and converting units
W=
(9 bar)(0.033 m3 )โ(1 bar)(0.1 m3 ) 105 N/m2
(โ1)

1 bar

1 kJ
103 Nโm

= โ 20 kJ
Negative sign denotes energy transfer to the
gas by work during the compression process.
Problem 2.18
Carbon dioxide (CO2) gas within a pistoncylinder assembly undergoes a process from a state
where p1 = 5 lbf/in.2, V1 = 2.5 ft3 to a state where p2 = 20 lbf/in.2, V2 = 0.5 ft3. The relationship
between pressure and volume during the process is given by p = 23.75 โ 7.5V, where V is in ft3
and p is in lbf/in.2 Determine the work for the process, in Btu.
KNOWN: CO2 gas within a pistoncylinder assembly undergoes a process where the pV
relation is given. The initial and final states are specified.
FIND: Determine the work for the process.
SCHEMATIC AND GIVEN DATA:
25
p1 = 5 lbf/in.2
V1 = 2.5 ft3
p (lbf/in^2)
p2 = 20 lbf/in.
V2 = 0.5 ft3
CO2
2
p = 23/75 โ 7.5 V
.
2
20
15
10
.
1
W
5
ENGINEERING MODEL: (1) The CO2 is the
closed system. (2) The pV relation during the
process is linear. (3) Volume change is the only
work mode.
0
0
0.5
1
1.5
2
2.5
3
V (ft^3)
ANALYSIS: The given pV relation can be used with Eq. 2.17 as follows:
?2
?
?2
7.5? 2 2
? = โซ ??? = โซ [23.75 โ 7.5?]?? = [23.75? โ
]
2 ?
?1
?1
1
7.5 2
[? โ ?12 ]
= 23.75[?2 โ ?1 ] โ
2 2
lbf 144 in.2
7.5 lbfโin.2 144 in.2
3
[0.5
? = (23.75 2 ) 

โ
2.5]ft
โ
(
)
 [0.52 โ 2.52 ](ft 3 )2
in.
1 ft 2
2
ft 3
1 ft 2
= 3600 ftโlbf
= (โ3600 ft โ lbf) 
1 Btu
 = โ4.63 Btu (negative sign denotes energy transfer in.)
778 ftโlbf
Alternative Solution
Since the pV relation is linear, W can also be evaluated geometrically as the area under the
process line:
? = ?ave (?2 โ ?1 ) = (
= 4.63 Btu
?1 + ?2
20 + 5 lbf 144 in2
1 Btu
 (0.5 โ 2.5)ft 3 

) (?2 โ ?1 ) = (
) 2
2
2
2
in
1 ft
778 ft โ lbf
PROBLEM 2.19
Problem 2.20
Nitrogen (N2) gas within a pistoncylinder assembly undergoes a process from p1 = 20 bar, V1 =
0.5 m3 to a state where V2 = 2.75 m3. The relationship between pressure and volume during the
process is pV1.35 = constant. For the N2, determine (a) the pressure at state 2, in bar, and (b) the
work, in kJ.
KNOWN: N2 gas within a pistoncylinder assembly undergoes a process where the pV relation
is pV1.35 = constant. Data are given at the initial and final states.
FIND: Determine the pressure at the final state and the work.
ENGINEERING MODEL: (1) The
N2 is the closed system. (2) The pV
relation is specified for the process.
(3) Volume change is the only work
mode.
SCHEMATIC AND GIVEN DATA:
pV 1.35= constant
p1 = 20 bar, V1 = 0.5 m3
V2 = 2.75 m3
N2
ANALYSIS: (a) ?1 ?1? = ?2 ?2?
0.5 m3
?2 = (20 bar) (
2.75 m
1.35
3)
โ
?
?
?2 = ?1 ( 1 ) ; n = 1.35. Thus
?2
= 2 bar
(b) Since volume change is the only work mode, Eq. 2.17 applies. Following the procedure of
part (a) of Example 2.1, we have
W=
?2 ?2 โ?1 ?1
1โ?
=
= 1285.7 kJ
(2 bar)(2.75 m3) โ(20)(0.5) 105 N/m2
1โ1.35

1 bar

1 kJ

103 Nโm
Problem 2.21
Air is compressed slowly in a pistoncylinder assembly from an initial state where p1 = 1.4 bar,
V1 = 4.25 m3, to a final state where p2 = 6.8 bar. During the process, the relation between
pressure and volume follows pV = constant. For the air as the closed system, determine the
work, in Btu.
KNOWN: Air is compressed in a pistoncylinder assembly from a known initial state to a
known final pressure. The pressurevolume relation during the process is specified.
FIND: Determine the work for the air as the closed system.
SCHEMATIC AND GIVEN DATA:
p1 = 1.4
V1 = 4.25 m3
Air
W=?
p2 = 6.8 bar
p
6.8 bar
.2
pV = constant
ENGINEERING MODEL: (1) The air is
a closed system. (2) The process is
polytropic, with pV = constant.
.1
1.4 bar
V
3
4.25 m
V2 = ?
ANALYSIS: The work for the polytropic process can be determined by integrating
?
W = โซ? 2 ??? . The pressurevolume relation is pV = constant.
1
?
?2 ????????
1
1
W = โซ? 2 ??? = โซ?
?
dV = (constant) ln (V2/V1)
The constant can be evaluated from the given data as constant = p1V1, and V2 can be determined
using p1V1 = p2V2. Thus
V2 = (p1/p2)V1 = (1.4/6.8)(4.25 m3) = 0.875 m3
And
W = (p1V1) ln (V2/V1)
W = (1.4 bar)(4.25 m3) ln (0.875/4.25)
105 N/m2
1 bar

1 kJ
103 Nโm
 = โ 940.4 kJ
Negative โ energy transfer by
work is in, as expected
Problem 2.22
Air contained within a pistoncylinder assembly is slowly compressed. As shown in Fig P2.32,
during this first process the pressure first varies linearly with volume and then remains constant.
Determine the total work, in kJ.
KNOWN: Air within a pistoncylinder assembly undergoes two processes in series.
FIND: Determine the total work.
SCEMATIC AND GIVEN DATA:
ENGINEERING MODEL: (1) The air within the
pistoncylinder assembly is the closed system. (2) The
twostep pV relation is specified graphically. (3)
Volume change is the only work mode.
3
150
2
+
+
1
p
(kPa)
+
100
1
2
50
0.015
0.055
0.07
V (m3)
ANALYSIS: Since volume change is the work mode, Eq. 2.17 applies. Furthermore, the
integral can be evaluated geometrically in terms of the total area under process lines:
1
2
?2
?1 + ?2
? = โซ ??? = ?ave (?2 โ ?1 ) + ?2 (?3 โ ?2 ) = (
) (?2 โ ?1 ) + ?2 (?3 โ ?2 )
2
?1
= [(
100+150
103 Nโm2
2
1 kPa
) kPa(0.055 โ 0.07)m3 + (150)(0.015 โ 0.055)] 
= (1.875 kJ) + (6 kJ) = 7.875 kJ (in)

1 kJ

103 Nโm
Problem 2.23
A gas contained within a pistoncylinder assembly undergoes three processes in series:
Process 12: Constant volume from p1 = 1 bar, V1 = 4 m3 to state 2, where p2 = 2 bar.
Process 23: Compression to V3 = 2 m3, during which the pressurevolume relationship is pV = constant.
Process 34: Constant pressure to state 4, where V4 = 1 m3.
Sketch the processes in series on pV coordinates and evaluate the work for each process, in kJ.
KNOWN: A gas contained within a pistoncylinder assembly undergoes three processes in
series. State data are provided.
FIND: Sketch the processes on pโV coordinates and evaluate the work for each process.
SCHEMATIC AND GIVEN DATA:
5
4
p (bar)
Gas
.
.
4
3
pV = constant
Note: p2V2 = p3V3
So p3 = (V2/V3)p2
= (4/3)(2 bar)
= 4 bar
3
.2
.1
2
1
V (m3)
1
2
3
4
ENGINEERING MODEL: (1) The gas in the pistoncylinder assembly is a closed system. (2)
The gas undergoes three processes in series, as illustrated. (3) Volume change is the only work
mode.
ANALYSIS: The work is evaluated using Eq. 2.17: W = โซ pdV
Process 12: The volume is constant, so W12 = 0.
? ????????
Process 23: W23 = โซ? 3
2
?
?? = constant ln (V3/V2) = (p2V2) ln (V3/V2)
= (2 bar)(4 m3) ln(2/4) 
105 N/m2
1 bar

1 kJ
 = โ 554.5 kJ (in)
103 Nโm
Process 34: For the constantpressure process
W34 = p(V4 โ V3) = (4 bar)(1 โ 2)m3 
105 N/m2
1 bar

1 kJ
103 Nโm
 = โ 400 k J (in)
Problem 2.24
Carbon dioxide (CO2) gas in a pistoncylinder assembly undergoes three processes in
series that begin and end at the same state (a cycle).
Process 12: Expansion from State 1 where p1 = 10 bar, V1 = 1 m3, to State 2 where V2 =
4 m3. During the process, pressure and volume are related by pV1.5 =
constant.
Process 23: Constant volume heating to State 3 where p3 = 10 bar.
Process 31: Constant pressure compression to State 1.
Sketch the processes on pV coordinates and evaluate the work for each process, in kJ.
What is net work for the cycle, in kJ?
KNOWN: Carbon dioxide gas undergoes a cycle in a pistoncylinder assembly. Data are
provided for the key states and the different process that make up the cycle.
FIND: Sketch the processes on pV coordinates, and determine the work for each process and
the net work for the cycle overall.
SCHEMATIC AND GIVEN DATA:
CO2
ENGINEERING MODEL: (1) The CO2 is a
closed system. (2) Processes 12 and 31 are
quasiequilibrium processes. (3) Process 23 is
at constant volume and Process 31 is at
constant pressure.
Process 12: Expansion from State 1 where
p1 = 10 bar, V1 = 1 m3, to State
2 where V2 = 4 m3. During the
process, pressure and volume
are related by pV1.5 = constant.
Process 23: Constant volume heating to
State 3 where p3 = 10 bar.
Process 31: Constant pressure compression
to State 1.
ANALYSIS: First, for Process 12, pV1.5 = constant. Thus
p1V11.5 = p2V21.5
โ
p2 = (V1/V2)1.5 p1 = (1/4)1.5 (10 bar) = 1.25 bar
With p3 = p2 = 10 bar, and V3 = V2 = 4 m3, all states are known and the pV diagram is
.
.
p (bar)
.
1
2
3
4
V (m3)
Problem 2.24 (Continued)
?
Process 12 The work is W = โซ? 2 ??? . For a polytropic process with pV1.5 = constant, the
1
integral becomes
W12 =
?2 ?2 โ?1 ?1
1โ1.5
(1.25 bar)(4 m3 )โ(10 bar)(1 m3 )
105 N/m2
(1โ1.5)
1 bar
=[
]

1 kJ

103 Nโm
= 1000 kJ (out)
Process 23 Since there is no volume change, W23 = 0
?
Process 31 The work is W31 = โซ? 1 ??? . Since the pressure is constant, the integral becomes
3
W31 = p3(V1 โ V3)
= (10 bar) (1 โ 4) m3 
105 N/m2
1 bar

1 kJ
103 Nโm
 = โ 3000 kJ (in)
The net work for the cycle is:
Wcycle = W12 + W23 + W31 = (1000) + (0) + (โ 3000) = โ 2000 kJ (in)
Problem 2.25
Air contained within a pistoncylinder assembly undergoes three processes in series:
Process 12: Compression during which the pressurevolume relationship is pV = constant from
p1 = 10 lbf/in.2, V1 = 4 ft3 to p2 = 50 lbf/in.2
Process 23: Constant volume from state 2 to state 3 where p = 10 lbf/in.2
Process 31: Constant pressure expansion to the initial state.
Sketch the processes in series on a pV diagram. Evaluate (a) the volume at state 2, in ft3, and (b)
the work for each process, in Btu.
KNOWN: Air within a pistoncylinder assembly undergoes three processes in series.
FIND: Sketch the processes in series on a pV diagram. Evaluate (a) the volume at state 2, and
(b) the work for each process.
SCHEMATIC AND GIVEN DATA:
p
2
Air
pV = constant
Process 12: Compression during which the
pressurevolume relationship is pV = constant
from p1 = 10 lbf/in.2, V1 = 4 ft3 to p2 = 50 lbf/in.2
Process 23: Constant volume from state 2 to state
3 where p = 10 lbf/in.2
Process 31: Constant pressure expansion to the
initial state.
3
V
ENGINEERING MODEL: (1) The gas is the closed system. (2) Volume change is the only
work mode. (3) Each of the three processes is specified.
ANALYSIS: (a) For process 12; pV = constant. Thus p1V1 = p2V2, and
10lbfโin.2
) (4 ft 3 ) = 0.8 ft3
50lbfโin.2
?
?2 = ( 1 ) ?1 = (
?2
1
(b) Since volume change is the only work mode, Eq. 2.17 applies.
Process 12: For process 12, pV = constant = p1V1. Thus
?
? ?
?
?
1
1 ?
?1
?1
?12 = โซ? 2 ??? = โซ? 2 ?? = ?ln ( 2) = (?1 ?1 )ln ( 2 )
Problem 2.25 (Continued)
Inserting values and converting units
?12 = (10
lbf
in.2
0.8 ft3
) (4 ft 3 ) ln (
4 ft3
144 in.2
)
1 ft 2

1 Btu
778 ftโlbf
 = 11.92 Btu (in)
Process 23: Constant volume (piston does not move). Thus W23 = 0
?
Process 31: Constant pressure processes (p3 = p1): ?31 = โซ? 1 ??? = ?1 (?1 โ ?3 )
3
Noting that V3 = V2
?31 = (10
lbf
144 in.2
) (4 โ 0.8)ft 3 
in2 .
1 ft 2

1 Btu
778 ftโlbf
 = 5.92 Btu (out)
1. The net work for the three process is
Wnet = W12 + W23 + W31 = (11.92) + 0 + (5.92)) = – 6 kJ (net work is negative – in)
PROBLEM 2.26
A 0.15mdiameter pulley turns a belt rotating the driveshaft of a power plant pump. The torque
applied by the belt on the pulley is 200 N โ m, and the power transmitted is 7 kW. Determine the
net force applied by the belt on the pulley, in kN, and the rotational speed of the driveshaft, in
RPM.
Problem 2.27
A 10V battery supplies a constant current of 0.5 amp to a resistance for 30 min. (a) Determine the
resistance, in ohms. (b) For the battery, determine the amount of energy transfer by work, in kJ.
KNOWN: Operating data are given for a 10V battery providing current to a resistance.
FIND: Determine the resistance and the amount of energy transfer by work.
SCHEMATIC AND GIVEN DATA:
โ
+
ENGINEERING MODEL: (1) The resistor is
the closed system. (2) The current is constant
with time.
Welec
10V battery
resistance
i = 0.5 amp
ฮt = 30 min
ANALYSIS: For current flow through a resistor: Voltage = Resistance * Current (Ohmโs Law)
Thus
Resistance =
Voltage
Current
=
10 amps
1 ohm
 = 20 ohm
0.5 volts 1 voltโamp

With Eq. 2.21applied to the battery which is discharging
โ
1 watt amp
?ฬelec  = (voltage)(current) = (10 volt)(0.5 amp)
 = 5 watt
1 volt
So, for 30 min of continuous operation, the energy transfer by work to the resistor is
?
Welec = โซ? 2?ฬelec ?? = ?ฬelec ฮt
1
= (5 watt)(30 min)
60 s
1 min

1 J/s

1 watt
1 kJ
 = 9 kJ
103 J
Problem 2.28
An electric heater draws a constant current of 6 amp, with an applied voltage of 220 V, for 24 h.
Determine the instantaneous electric power provided to the heater, in kW, and the total amount
of energy supplied to the heater by electrical work, in kWโh. If electric power is valued at
$0.08/kWโh, determine the cost of operation for one day.
KNOWN: An electric heater draws a constant current at a specified voltage for a given length
of time. The cost of electricity is specified.
FIND: Determine the instantaneous power provided to the heater and the total amount of energy
supplied by electrical work. Determine the cost of operation for one day.
SHEMATIC AND GIVEN DATA:
i = 6 amp
E = 220 V
ฮt = 10 h
ENGINEERING MODEL: (1) The heater is a closed
system. (2) The current and voltage are constant.
ANALYSIS: The constant power input to the heater is
given by Eq. 2.21
1 Wโamp 1 kW
?ฬin = E I = (220 V)(6 amp) 
  3  = 1.320 kW
1V
10 W
Thus, the total energy input is
?
?in = โซ? 2 ?ฬin ?? = ?ฬin โ? = (1.320 kW)(24 h) = 31.68 kW โ h
1
Using the specified cost of electricity
Cost per day = (31.68 kWโh) ($0.08/kWโh) = $2.53
?ฬin
Problem 2.29
Problem 2.30
Problem 2.31
Figure P2.31 shows an object whose mass is 5 lb attached to a rope wound around a pulley. The
radius of the pulley is 3 in. If the mass falls at a constant velocity of 5 ft/s, determine the power
transmitted to the pulley, in hp, and the rotational speed of the shaft, in revolutions per minute
(RPM). The acceleration of gravity is 32.2 ft/ s2.
KNOWN: An object attached to a rope wound around a pulley falls at a constant velocity.
FIND: Find the power transmitted to the pulley and the rotational speed.
SCHEMATIC AND GIVEN DATA:
ENGINEERING MODEL: (1) The object falls at a constant
speed. (2) The acceleration of gravity is constant.
ANALYSIS: The power is obtained using Eq. 2.13
?ฬ = ? โ ? = (??)V
ft
= (5 lb) (32.2
s2
ft
1 lbf
s
32.2 lbโftโ? 2
) (5 ) 

= 25 ftโlb/s
Converting to horsepower
lbf
?ฬ = (25 ft โ ) 
s
1 hp
 = 0.0455 hp
550 ftโlbfโs
The rotational speed of the pulley is related to the velocity of the object and the radius by
V = Rฯ. Thus
?=
V
R
5 ftโs
=( โ
)
3 12 ft
1 rev
60 s
2?
1 min

 = 191 rev/min
Problem 2.32
Problem 2.32 (Continued)
Problem 2.33
Problem 2.34
A fan forces air over a computer circuit board with a surface area of 70 cm2 to avoid
overheating. The air temperature is 300 K while the circuit board surface temperature is 340
K. Using data from Table 2.1, determine the largest and smallest heat transfer rates, in W,
that might be encountered for this forced convection.
Problem 2.35
A 6in. insulated frame wall of a house has an average thermal conductivity of 0.04 Btu/hโftโoR.
The inner surface of the wall is at 68oF and the outer surface is at 40oF. Determine at steady
state the rate of heat transfer through the wall, in Btu/h. If the wall is 20 ft x 10 ft, determine the
total amount of energy transfer in 10 hours, in Btu.
KNOWN: The inner and outer temperatures, thermal conductivity, and thickness of a frame
wall are specified. The wall area is also known.
FIND: Determine the rate of energy transfer by heat through the wall and the total amount of
energy transferred in 10 hours.
SCHEMATIC AND GIVEN DATA:
20 ft
.
Tout = 40oF
10 ft
Tin = 68oF
ENGINEERING MODEL: (1) The wall is a
closed system at steady state. (2) Conduction
follows Fourierโs law. (3) The temperature
profile through the wall is linear.
ฮบ = 0.04 Btu/hโftโoR
.
Assume a linear
temperature profile.
L = 6 in.
ANALYSIS: For conduction through a plane wall with constant thermal conductivity and
assuming a linear temperature profile
Note: ฮT (oR) = ฮT (0F)
?ฬ =
?A(?in โ?out )
?
(0.04
=
Btu
)(20 ? 10 ft2 )(68โ40)o R
hโftโo R
(6โ12 ft)
= 448 Btu/h
The total amount of energy transfer by heat is
ฬ = ?ฬโ? = (448 Btu/h)(10 h) = 4480 Btu
Q = โซ ???
COMMENTS: Note that the temperature difference has the same value in oF and oR. Also,
since the heat transfer rate is constant for steady state conduction, the integral reduces to ?ฬโ?.
Problem 2.36
As shown in Fig. P2.36, an oven wall consists of a 0.635cmthick layer of steel (ฮบs = 15.1
W/mโK) and a layer of brick (ฮบb = 0.72 W/mโK). At steady state, a temperature decrease of 0.7oC
occurs over the steel layer. The inner temperature of the steel layer is 300 K. If the temperature
of the outer surface of the brick must be no greater than 40oC, determine the thickness of brick,
in cm, that ensures this limit is met. What is the rate of conduction, in kW per m2 of wall surface
area?
KNOWN: Steadystate data are provided for a composite wall formed from a steel layer and a
brick layer.
FIND: Determine the minimum thickness of the brick layer to keep the outer surface
temperature of the brick at or below a specified value.
SCHEMATIC AND GIVEN DATA:
Ti = 300 K
Tm
o
ENGINEERING MODEL: (1) The wall is
ฮT = 0.7 C
the system at steady state. (2) The temperature
varies linearly through each layer.
?ฬ
? โ?i
= โ?s [ m
A steel
?s
] and
ฮบs = 15.1 W/mโK
ฮบb = 0.72 W/mโK
Ls = 0.635 cm
ANALYSIS: Using Eq. 2.31 together with
model assumption 2
( )
To โค 40oC
?ฬ
( )
A brick
? โ?m
= โ?b [ o
?b
]
where Tm denotes the temperature at the steelbrick interface.
At steady state, the rate of conduction to the interface through the steel must equal the rate of
conduction from the interface through the brick: (?ฬโA)steel = (?ฬโA)brick . Thus
? โ?i
โ?s [ m
?s
? โ?m
] = โ?b [ o
?b
]
(300 โ 0.7) = 299.3 oC
And solving for Lb we get
?b =
?b ?o โ?m
[
?s ?m โ??
] ?s
= โ 0.7 oC
Problem 2.36 (Continued)
?b = (
0.72 WโmโK 299.3โ ?o
)[
] (0.635 cm)
15.1 WโmโK
0.7
Since To โค 40 oC
?b โฅ (
0.72
15.1
299.3โ 40
)[
0.7
] (0.635 cm)
Lb โฅ 11.22 cm
The rate of conduction is
?ฬ
( )
A steel
? โ?i
= โ?s [ m
?s
299.3โ300
] = โ(15.1 Wโm โ K) [
0.635 cm
]
100 cm
1m

1 kW
 = 1.665 kW/m2
103 W
or
?ฬ
( )
A brick
? โ?m
= โ?s [ o
?b
40โ299.3
100 cm
11.22 cm
1m
] = โ(0.72 Wโm โ K) [
The slight difference is due to roundoff.
]

1 kW
 = 1.664 kW/m2
103 W
Problem 2.37
A composite plane wall consists of a 3in.thick layer of insulation (ฮบi = 0.029 Btu/hโftโoR) and a
0.75in.thick layer of siding (ฮบs = 0.058 Btu/hโftโoR). The inner temperature of the insulation is
67oF. The outer temperature of the siding is 8oF. Determine at steady state (a) the temperature
at the interface of the two layers, in oF, and (b) the rate of heat transfer through the wall in Btu
per ft2 of surface area.
KNOWN: Energy transfer by conduction occurs through a composite wall consisting of two
layers.
FIND: Determine the temperature at the interface between the two layers and the rate of heat
transfer per unit area through the wall.
SCHEMATIC AND GIVEN DATA:
T1 = 67 oF
T3 = 8 oF
ENGINEERING MODEL: (1) The wall is
the system at steady state. (2) The temperature
varies linearly through each layer.
T2 = ?
ฮบi = 0.029 Btu/hโftโoR
ANALYSIS: With Eq. 2.17, and recognizing that
at steady state the rates of energy conduction must be
equal through each layer
insulation
?ฬ
Li = 3 in.
?
? โ?1
= โ?i [ 2
?i
? โ?2
] = โ?s [ 3
?s
]
(*)
siding
ฮบs = 0.058 Btu/hโftโoR
Ls = 0.75 in.
Solving for T2
?
?i
?
?s
( i ?1 + s ?3 )
?2 = (? โ )+(? โ
i
?i
?i
=
?i
s
?s )
0.029 BtuโhโftโR 12 in.

3 in.
Thus
?2 =
1 ft
?s
 = 0.116 Btu/hโoR
(0.116)(527)+(0.928)(452)
(0.116)+(0.928)
=
?s
0.058 BtuโhโftโR 12
.75 in.
= 460.3 oR = 0.33 oF
Thus, using Eq. (*)
?ฬ
?
?ฬ
?
? โ?1
= โ?i [ 2
?i
] = (โ0.029
? โ?2
= โ?s [ 3
?2
] = (โ0.058
Btu
(0.33โ67)R
)[
hโftโR
Btu
3
12
)[
hโftโR
ft
] = 7.73 Btu/ft2
(โ8โ0.33)R
0.75
12
ft
] = 7.73 Btu/ft2
  = 0.928 Btu/hโoR
1
Problem 2.38
Complete the following exercise using heat transfer relations:
(a) Referring to Fig. 2.12, determine the rate of conduction heat transfer, in W, for ฮบ = 0.07
W/mโK, A = 0.125 m2, T1 = 298 K, T2 = 273 K.
(b) Referring to Fig. 2.14, determine the rate of convection heat transfer from the surface to the
air, in W, for h = 10 W/m2, A = 0.125 m2, Tb = 305 K, Tf = 298 K.
(a) Referring to Fig. 2.12, determine the rate of conduction heat transfer, in W, for ฮบ = 0.07
W/mโK, A = 0.125 m2, T1 = 298 K, T2 = 273 K.
Using Eq. 2.31 and noting that the temperature varies linearly
through the wall
?2 โ ?1
?ฬ? = โ?A [
]
?
= โ (0.07
W
mโK
(273โ298)K
) (0.125 m2 ) [
(0.127 m)
] = 1.722 W
(b) Referring to Fig. 2.14, determine the rate of convection heat transfer from the surface to the
air, in W, for h = 10 W/m2, A = 0.125 m2, Tb = 305 K, Tf = 298 K.
Using Eq. 2.34
?ฬc = hA[Tb โ Tf]
= (10
W
m2
) (0.125 m2 )[305 โ 298]K
= 8.75 W
Problem 2.39
At steady state, a spherical interplanetary electronicsladen probe having a diameter of 0.5 m
transfers energy by radiation from its outer surface at a rate of 150 W. If the probe does not
receive radiation from the sun of deep space, what is the surface temperature, in K? Let ฮต = 0.8.
KNOWN: Steadystate operating data are provided for a spherical interplanetary probe.
FIND: Determine the surface temperature of the sphere.
SCHEMATIC AND GIVEN DATA:
?ฬe
= 150 W
Ts
ENGINEERING MODEL: (1) The probe is at
steady state. (2) The probe emits but does not
receive radiation.
D = 0.5 m
ฮต = 0.8
ANALYSIS: In this case, Eq. 2.32 applies:
?ฬe = ??A? 4 ,
where A = ฯ d 2 = 0.7854 m2
The StefanBoltzmann constant is ฯ = 5.67 x 108 W/m2โK4. Thus
Ts = [
1/4
?ฬe
??A
]
=[
150 W
1/4
]
10โ8 W
)(0.7854 m2 )
m2 K4
(0.8)(5.67 x
= 254.7 K
Problem 2.40
A body whose surface area is 0.5 m2, emissivity is 0.8, and temperature is 150oC is placed in a
large, evacuated chamber whose walls are at 25oC. What is the rate at which radiation is emitted
by the surface, in kW? What is the net rate at which radiation is exchanged between the body
and the walls, in kW?
KNOWN: Data are provided for a body placed in a large, evacuated chamber.
FIND: Determine the rate at which radiation is emitted from the surface and the net rate at which
radiation is exchanged between the body and the chamber.
SCHEMATIC AND GIVEN DATA:
?ฬe
Body at surface
temperature Tb
ENGINEERING MODEL: (1) The area of
body is much less than that of the chamber
walls. (2) The chamber is evacuated.
A = 0.5 m2
ฮต = 0.8
Tb = 150oC = 423 K
Ts = 25oC = 298 K
Chamber
surface at Ts
ANALYSIS:
The rate radiation is emitted from the surface is given by Eq. 2.32, where ฯ = 5.67 x 108
W/m2โK4 is the StefanBoltzmann constant. That is
?ฬe = ฮตฯATb4 = (0.8)(5.67 x 108 W/m2โK4 )(0.5 m2)(423 K)4 = 726 W
With the assumptions in the Engineering Model, the net rate at which energy is exchanged by
radiation between the body and the chamber walls is given by Eq. 2.33. Thus
(?ฬe )net = ฮตฯA(Tb4 โ Ts4)
= (0.8)(5.67 x 108 W/m2โK4 )(0.5 m2)[(423 K)4 โ (298 K)4]
= 547 W
Problem 2.41
The outer surface of the grill shown in Fig. P2.41 is at 47oC and the emissivity is 0.93. The heat
transfer coefficient for convection between the hood and the surroundings at 27oC is 10 W/m2 โ K.
Determine the rate of heat transfer between the grill hood and the surroundings by convection
and radiation, in kW per m2 of surface area.
Problem 2.42
Each line of the following table gives data for a process of a closed system. Each entry has the
same energy units. Determine the missing entries.
Process
a
b
c
d
e
Process
a
b
c
d
e
Q
+50
E1
20
W
+20
60
40
+50
Q
+50
+50
40
90
+50
E2
+50
+60
+40
+50
0
+150
W
20
+20
60
90
+150
E1
20
+20
+40
+50
+20
Process a:
W = Q – ฮE = +50 โ (+ 70) = 20
ฮE = E2 โ E1
E2 = ฮE + E1 = +70 + (20) = +50
Process b:
Q = ฮE + W = +30 + (+20) = +50
ฮE = E2 โ E1
E1 = E2 โ ฮE = +50 โ (+30) = +20
Process c:
ฮE = E2 โ E1 = +60 โ (+40) = +20
Q = ฮE + W = +20 + (60) = 40
Process d:
W = Q โ ฮE = (90) โ 0 = 90
ฮE = E2 โ E1
E2 = ฮE + E1 = 0 +50 = +50
Process e:
ฮE = Q โ W = +50 โ (+150) = 100
E1 = E2 โ ฮE = (80) โ (100) = +20
E2
+50
+50
+60
+50
80
80
ฮE
+70
+30
+20
0
100
ฮE
+70
+30
๏ฝ
Problem 2.43
Each line of the following table gives data for a process of a closed system. Each entry has the
same energy units. Determine the missing entries.
Process
a
b
c
d
e
Process
a
b
c
d
e
Q
E1
+15
+7
+6
W
โ 25
+27
โ4
+10
โ 10
E2
+30
+15
+10
โ8
+3
Q
โ 10
+27
โ4
+10
+3
W
โ 25
โ 12
+10
โ 10
+3
E1
+15
+7
+6
โ 10
โ8
Process a:
ฮE = E2 โ E1 = + 30 โ (+15) = +15
Q = ฮE + W = +15 + (โ 25) = 10
Process b:
W = ฮE โ Q = +15 โ (+27) = โ 12
ฮE = E2 โ E1
E2 = ฮE + E1 = +15 + (+7) = +22
Process c:
ฮE = Q โ W = (โ 4) โ (+10) = โ 14
ฮE = E2 โ E1
E2 = ฮE + E1 = (โ 14) + (+6) = โ 8
Process d:
Q = ฮE + W = (+20) + (โ 10) = +10
ฮE = E2 โ E1
E1 = E2 โ ฮE = (+10) โ (+20) = โ 10
Process e:
W = Q โ ฮE = (+3) โ (0) = +3
ฮE = E2 โ E1
E1 = E2 โ ฮE = (โ 8) โ (0) = โ 8
E2
+ 30
+22
โ8
+10
8
ฮE
+15
+15
โ 14
+20
0
ฮE
๏ฝ
+20
0
Problem 2.44
A closed system of mass of 10 kg undergoes a process during which there is energy transfer by
work from the system of 0.147 kJ per kg, an elevation decrease of 50 m, and an increase in
velocity from 15 m/s to 30 m/s. The specific internal energy decreases by 5 kJ/kg and the
acceleration of gravity is constant at 9.7 m/s2. Determine the heat transfer for the process, in kJ.
KNOWN: Data are provided for a closed system undergoing a process involving work, heat
transfer, change in elevation, and change in velocity.
FIND: Determine the heat transfer for the process.
V1 = 15 m/s
SCHEMATIC AND GIVEN DATA:
ฮu = 5 kJ/kg
W/m = + 0.147 kJ/kg
50 m
V2 = 30 m/s
ENGINEERING MODEL: (1) The system is a
closed system. (2) The acceleration of gravity is
constant.
z
m = 10 kg
g = 9.7 m/s2
ANALYSIS:
ฮU + ฮPE + ฮKE = Q – W
โ
Q = ฮU + ฮPE + ฮKE – W
W = m [W/m] = 10 kg [0.147 kJ/kg] = 1.47 kJ
ฮU = mฮu = 10 kg [ 5 kJ/kg] = 50 kJ
ฮKE =
?
2
(V22 โ V12 ) =
10 kg
2
[(30
m 2
m 2
s
s
) โ (15
ฮPE = mg(z2 โ z1) = (10 kg) (9.7 m/s2)(50 m)
1N
1 kJ


 = +3.38 kJ
2
1 kgโmโs
103 Nโm
) ]
1N
1 kgโmโs2
Q = (50) + (4.85) + (3.38) โ (1.47) = 50 kJ (out)

1 kJ
103 Nโm
 = – 4.85 kJ
Problem 2.45
Problem 2.46
Problem 2.47
An electric motor operating at steady state draws a current of 20 amp at a voltage of 110 V. The
output shaft rotates at a constant speed of 2000 RPM and exerts a torque of 9.07 Nโm.
Determine
(a) the magnitude of the power input, in W.
(b) the output power, in W.
(c) the cost of 24 hours of operation if electricity is valued at $0.09 per kWโh.
KNOWN: An electric motor operates at steady state. The input current and voltage are known
and the rotational speed and torque of the exit shaft are specified. The cost of electricity is also
specified.
FIND: Determine (a) the magnitude of the power input, (b) the output power, and (c) the cost of
24 hours of operation.
SCHEMCATIC AND GIVEN DATA:
20 amps
T = 9.07 Nโm
110 V
2000 RPM
ENGINEERING MODEL: (1) The motor operates at steady state.
ANALYSIS: (a) The magnitude of the input power is
?ฬelec  = (20 amp) (110 V) 
1W

1 voltโamp
1 kW
103 W
 = 2.2 kW (input)
(b) The shaft power is
?ฬshaft  = T x ฯ
= (9.07 Nโm)(2000
rev
min
) (2ฯ
1
rev
1 min
1 kJ
60 s
103 Nโm
)


1 kW
 = 1.90 kW (out)
1 kJ/s
(c) For steady operation (constant with time) the total amount of energy transferred in by
electricity is
Welec = โซ?ฬelec ?? = ?ฬelec  ฮt = (2.2 Kw)(24 h) = 52.8 kWโh
Cost = (52.8 kWโh)($0.09/kWโh) = $4.75 (per day)
COMMENT: Note, for steady state operation the inputs and outputs are constant with time.
Problem 2.48
An electric motor draws a current of 10 amp with a voltage of 110 V, as shown in Fig. P2.48.
The output shaft develops a torque of 9.7 Nโm and a rotational speed of 1000 RPM. For
operation at steady state, determine for the motor
(a) the electric power required, in kW.
(b) the power developed by the output shaft, in kW.
(c) the average surface temperature, Ts, in oC, if heat transfer occurs by convection to the
surroundings at Tf = 21oC.
KNOWN: Operating data are provided for an electric motor at steady state.
FIND: Determine (a) the electric power required, (b) the power developed by the output shaft,
and (c) average the surface temperature.
Tf = 21oC
Ts
T = 9.7 Nโm
1000 RPM
10 amp
?ฬ = hA(?f โ ?s )
hA = 3.9 W/K
110 V
ENGINEERING MODEL: (1) The motor is the closed system. (2) The system is at steady
state.
ANALYSIS: (a) Using Eq. 2.21
1 Wโamp 1 kW
?ฬelectric = – (voltage) (current) = – (110 V)(10 amp)
  3  = 1.1 kW (in)
1V
10 W
(b) Using Eq. 2.20
?ฬshaft = (torque) (angular velocity)
= (9.7 N โ m) (1000
rev
2ฯ rad
1 min
rev
60 s
)
min


1 kW
 = 1.016 kW (out)
103 Nโmโs
(c) To determine the surface temperature, first find the rate of energy transfer by heat using the
energy balance
0
??
= ?ฬ โ ?ฬ = ?ฬ โ (?ฬelectric + ?ฬshaft )
??
?ฬ = (?ฬelectric + ?ฬshaft ) = (1.1 kW) + (1.016 kW) = 0.084 kW
The surface temperature of the motor is
103 W
?s = (?ฬโhA) + ?? = (0.084 kW)/(3.9 W/K)
 + 294 K
1 kW
= 315.5 K = 42.5 oC
Problem 2.49
A gas is contained is a vertical pistoncylinder assembly by a piston with a face area of 40 in.2
and a weight of 100 lbf. The atmosphere exerts a pressure of 14.7 lbf/in.2 on top of the piston. A
paddle wheel transfers 3 Btu of energy to the gas during a process in which the elevation of the
piston increases. The change in internal energy of the gas is 2.12 Btu. The piston and cylinder
are poor thermal conductors, and friction between them can be neglected. Determine the
elevation increase of the piston, in ft.
KNOWN: A rotating shaft transfers a specified amount of energy to a gas contained in a vertical
pistoncylinder assembly and the piton elevation increases. Data are provided for the piston and
the change in internal energy of the gas.
FIND: Determine the elevation increase of the piston.
SCHEMATIC AND GIVEN DATA:
State 1
patm = 14.7 lbf/in.2
State 2
ฮzpist
Apist = 40 in.2
Fgrav, pist = 100 lbf
Gas
Gas
ฮUgas = 2.12 Btu
Q=0
Wpw = โ 3 Btu
Process
ENGINEERING MODEL: (1) The gas and the piston are the closed system. (2) Energy
transfer by heat is negligible; Q = 0. (3) Kinetic energy effects are negligible. (4) The potential
energy change of the gas is negligible, but the potential energy change of the piston in
considered. (5) The change of internal energy of the piston is negligible. (6) Friction between
the piton and the cylinder wall is neglected.
ANALYSIS: The change in elevation of the piston is related to its change in potential energy by
ฮPEpist = (mpist g) ฮzpist = (Fgrav, piston) ฮzpist
(*)
To evaluate the change in internal energy of the piston, we apply the energy balance to the
system consisting of the gas and the piston:
ฮEgas + ฮEpist = Q โ W
Problem 2.49 (Continued)
For the piston: ฮEpist = ฮUpist + ฮKEpist + ฮPEpist
For the gas: ฮEgas = ฮUgas + ฮKEgas + ฮPEgas
There are two mechanisms for mechanical work for the system chosen: the paddle wheel work
and the work of the piston against the atmospheric pressure: W = Wpw + Watm
Since the atmospheric pressure is constant: Watm = patmApist ฮzpist
Combining these results
ฮUgas + (Fgrav, piston) ฮzpist = Q โ (Wpw + patmApist ฮzpist)
Solving for ฮzpist
ฮzpist =
โ โ?gas โ?pw
Fgrav,piston+patm Apist
Inserting values and converting units
ฮzpist =
โ (2.12 Btu)โ(โ3 Btu)
778 Btu

 = 0.995 ft
(100 lbf)+(40 in.2 )(14.7 lbfโin.2 ) 1 ftโlbf
COMMENTS: Note that the gas and the piston can be considered individually as closed
systems. With this approach, the work done by the gas on the piston must be considered, which
involves evaluating the change in volume of the gas and the gas pressure. The approaches are
algebraically equivalent, but the approach chosen leads to a simpler analysis.
Problem 2.50
A gas undergoes a process in a pistoncylinder assembly during which the pressurespecific
volume relation is pv1.2 = constant. The mass of the gas is 0.4 lb and the following data are
known: p1 = 160 lbf/in.2, V1 = 1 ft3, and p2 = 390 lbf/in.2 During the process, heat transfer from
the gas is 2.1 Btu. Kinetic and potential energy effects are negligible. Determine the change in
specific internal energy of the gas, in Btu/lb.
KNOWN: A gas is compressed in a pistoncylinder assembly. The pressurespecific volume
relation is specified.
p
FIND: Determine the change in specific internal energy.
2
.
pv1.2 = constant
SCHEMATIC AND GIVEN DATA:
Q = 2.1 Btu
gas
m = 0.4 lb
.
p1 = 160 lbf/in.2
V1 = 1 ft3
p2 = 390 lbf/in.2
W
1
v
ENGINEERING MODEL: (1) The gas is a closed system.
(2) The process follows pv1.2 = constant. (3) Kinetic and
potential energy effects are negligible.
ANALYSIS: The change in specific internal energy will be found from an energy balance.
First, determine the work. Since volume change is the only work mode, Eq. 2.17 applies:
?
?2 ?????
1
? 1.2
W = โซ? 2 ??? = = โซ?
1
?? =
(?2 ?2 โ?1 ?1 )
1โ1.2
Evaluating V2
?1
1โ
1.2
V2 = ( )
?2
1โ
1.2
160 lbfโin.2
?1 = (
)
(1 ft 3 ) = 0.4759 ft3
390 lbfโin.2
Thus
(390 lbfโin.2 )(0.4759 ft3 )โ(160)(1)
144 in.2
1โ1.2
1 ft2
W=[
]

1 Btu
 = 23.69 Btu (in)
778 ftโlbf
Now, writing the energy balance: โ?? + โ?? + โ? = ? โ ?
With ฮU = mฮu
1
ฮu =
?โ?
?
=
(โ2.1 Btu)โ (โ23.69 Btu)
0.4 lb
= 54.0 Btu
1. The amount of energy transfer in by work exceeds the amount of energy transfer out by heat,
resulting in a net increase in internal energy.
Problem 2.51
Four kilograms of carbon monoxide (CO) is contained in a rigid tank with a volume of 1 m3. The tank
is fitted with a paddle wheel that transfers energy to the CO at a constant rate of 14 W for 1 h.
During the process, the specific internal energy of the carbon monoxide increases by 10 kJ/kg. If no
overall changes in kinetic or potential energy occur, determine (a) the specific volume at the final
state, in m3/kg. (b)the energy transfer by work, in kJ. (c) the energy transfer by heat, in kJ, and the
direction of the heat transfer.
Problem 2.52
Steam in a pistoncylinder assembly undergoes a polytropic process. Data for the initial and
final states are given in the accompanying table. Kinetic and potential energy effects are
negligible. For the process, determine the work and heat transfer, each in Btu per lb of steam.
p (lbf/in.2)
100
40
State
1
2
v (ft3/lb)
4.934
11.04
u (Btu/lb)
1136.2
1124.2
KNOWN: Steam undergoes a polytropic process in a pistoncylinder assembly. Data are
known at the initial and final states.
FIND: Determine the work and heat transfer, each per unit mass of steam.
p (lbf/in.2)
100
40
State
1
2
Steam
v (ft3/lb)
4.934
11.04
p
1
.
pvn = constant
u (Btu/lb)
1136.2
1124.2
.
2
v
ENGINEERING MODEL: (1) The steam is a closed system. (2) The process is polytropic,
and volume change is the only work mode. (3) Kinetic and potential energy effects are
negligible.
ANALYSIS: Since the process is polytropic, Eq 2.17 applies for the work:
?2 ?????
?
W/m = โซ? 2 ??? = โซ?
1
??
1
?? =
(?2 ?2 โ?1 ?1 )
1โ?
The pressures and specific volumes are known at each state, but n is unknown. To find n, pvn =
constant , as follows:
?1 ?1? = ?2 ?2?
โ
?1
?2
?
?
= ( 2)
?1
โ
?=
ln(?1 โ?2 )
ln(100โ40)
=
= 1.1377
ln(?2 โ?1 )
ln(11.04โ4.934)
Thus
W/m =
(40 lbfโin.2 )(11.04 ft3 โlb)โ(100)(4.934) 144 in.2
1โ1.1377

The heat transfer is obtained using the energy balance.
1 ft2

1 Btu
 = 69.63 Btu/lb (out)
778 ftโlbf
Problem 2.52 (Continued)
ฮU + ฮKE + ฮPE = Q โ W
โ
Q = ฮU + W
With ฮU = m ฮu = m(u2 โ u1)
Q/m = (u2 โ u1) + (W/m) = (1124.2 โ 1136.2) Btu/lb + (69.63 Btu/lb)
= 57.63 Btu/lb (in)
Problem 2.53
Air expands adiabatically in a pistoncylinder assembly from an initial state where p1 = 100
lbf/in.2, v1 = 3.704 ft3/lb, and T1 = 1000 oR, to a final state where p2 = 50 lbf/in.2 The process is
polytropic with n = 1.4. The change in specific internal energy, in Btu/lb, can be expressed in
terms of temperature change as ฮu = (0.171)(T2 โ T1). Determine the final temperature, in oR.
Kinetic and potential energy effects can be neglected.
KNOWN: Air undergoes a polytropic process with known n in a pistoncylinder assembly. Data
are known at the initial and final states, and the change in specific internal energy is expressed as
a function of temperature change.
FIND: Determine the final temperature.
p
1
SCHEMATIC AND GIVEN DATA:
p1 = 100 lbf/in.2
v1 = 3.704 ft3/lb
Air
T1 = 1000 oR
p2 = 50 lbf/in.2
ฮu = (0.171Btu/lbโoR)(T2 โ T1)
Q=0
.
pv1.4 = constant
.2
ENGINEERING MODEL: (1) The air is a closed system.
(2) The process is polytropic with n = 1.4 and volume change
is the only work mode. (3) The process is adiabatic: Q = 0.
(4) Kinetic and potential energy effects are negligible.
v
ANALYSIS: To find the final temperature, we will use the energy balance with the given
expression for change in specific internal energy as a function of temperature change. First,
determine the work using Eq. 2.17
?
? ?????
1
? 1.4
W/m = โซ? 2 ??? = โซ? 2
1
?? =
(?2 ?2 โ?1 ?1 )
1โ1.4
For the polytropic proess, ?1 ?11.4 = ?2 ?21.4 . Thus
1
? 1.4
?2 = ( 1 ) ?1 =
?2
1
100 lbfโin.2 1.4
(
) (3.704 ft 3 /lb) = 6.077 ft3/lb
50 lbfโin.2
So, the work is
W/m =
(50 lbfโin.2 )(6.077 ft3 โlb)โ(100 )(3.704)
1โ1.4

144 in.2
1 ft2

1 Btu
 = 30.794 Btu/lb
778 ftโlbf
The energy balance is: ฮU + ฮKE + ฮPE = Q โ W. With ฮU = m(u2 โ u1)
Problem 2.53 (Continued)
(u2 โ u1) = โ W/m
Inserting values
(0.171 Btu/lbโoR)(T2 โ 1000 oR) = โ (30.794 Btu/lb)
Solving; T2 = (30.794 Btu/lb)/(0.171 Btu/lbโoR) + 1000 = 819.9 oR
Problem 2.54
Problem 2.55
Process
12
23
34
41
ฮE
1200
400
Q
0
800
200
W
200
400
Problem 2.56
A gas within a pistoncylinder assembly undergoes a thermodynamic cycle consisting of three
processes:
Process 12: Constant volume V1 = 2 m3, p1 = 1 bar, to p2 = 3 bar, U2 โ U1 = 400 kJ.
Process 23: Constant pressure compression to V3 = 1 m3.
Process 31: Adiabatic expansion, with W31 = 150 kJ.
There are no significant changes in kinetic or potential energy. Determine the net work for the
cycle, in kJ, and the heat transfers for Processes 12 and 23, in kJ. Is this a power cycle or
refrigeration cycle? Explain.
KNOWN: Data are provided for a gas undergoing a thermodynamic cycle consisting of three
processes.
FIND: Determine the net work for the cycle and the heat transfers for processes 12 and 23.
Explain whether it is a power cycle or a refrigeration cycle.
SCHEMATIC AND GIVEN DATA:
p
.3
3
Gas
ENGINEERING MODEL: (1) The gas is a closed
system. (2) Kinetic and potential energy effects are
neglected. (3) Volume change is the only work mode.
(4) Process 31 is adiabatic, so Q31 = 0.
.2
.1
1
V
1
2
ANALYSIS:
Beginning with Process 12, the volume is constant, so W12 = 0. The energy balance reduces to ฮU12 =
Q12 โ W12. Thus
Q12 = ฮU12 = 400 kJ (in)
Process 23 is at constant pressure. So, with Eq. 2.17
105 N/m2
?
1 kJ
W23 = โซ? 3 ??? = p2(V3 โ V2) = (3 bar)(1 โ 2)m3
  3  = โ 300 kJ (in)
1 bar
10 Nโm
2
The energy balance becomes ฮU23 = Q23 โ W23. So
Q23 = ฮU23 + W23
To get ฮU23, letโs consider that for a cycle, ฮUcycle = 0. So
ฮU12 + ฮU23 + ฮU31 = 0
โ
ฮU23 = โ ฮU12 โ ฮU31
Problem 2.56 (Continued)
Now, for Process 31: Q31 = 0, so ฮU31 = โ W31 = โ 150 kJ
And
ฮU23 = โ ฮU12 โ ฮU31 = โ (400 kJ) โ (โ 150 kJ) = โ 250 kJ
Finally
Q23 = ฮU23 + W23 = (โ 250 k J) + ( 300 kJ) = โ 550 kJ
The net work for the cycle is
Wcycle = W12 + W23 + W31 = 0 + (โ 300 k J) + (150 kJ) = โ 150 kJ (in)
From the pV diagram, we see that the cycle is executed in a counter clockwise fashion. And, the
net work is in. So, the cycle is a refrigeration cycle.
Problem 2.57
A gas undergoes a cycle in a pistoncylinder assembly consisting of the following three
processes:
Process 12: Constant pressure, p = 1.4 bar, V1 = 0.028 m3, W12 = 10.5 kJ
Process 23: Compression with pV = constant, U3 = U2
Process 31: Constant volume, U1 โ U3 = 26.4 kJ
There are no significant changes in kinetic or potential energy.
(a) Sketch the cycle on a pV diagram.
(b) Calculate the net work for the cycle, in kJ.
(c) Calculate the heat transfer for process 12, in kJ
KNOWN: A gas undergoes a cycle consisting of three processes.
FIND: Sketch the cycle on a pV diagram and determine the net work for the cycle and the heat
transfer for process 12.
SCHEMATIC AND GIVEN DATA:
Process 12: Constant pressure, p = 1.4 bar, V1 = 0.028 m3,
W12 = 10.5 kJ
Process 23:
Compression with pV = constant, U3 = U2
Process 31:
Constant volume, U1 โ U3 = 26.4 kJ
Gas
ENGINEERING MODEL: (1) The gas is a closed system. (2) Kinetic and potential energy
effects are negligible. (3) The compression from state 2 to 3 is a polytropic process.
ANALYSIS: (a) Since W12 > 0, the process is an expansion. Thus
p
3
1
.
.
.
2
V
Problem 2.57 (Continued)
0
(b) The net work for the cycle is Wcycle = W12 +W23 + W31. W12 = 10.5 kJ, so we need W23.
?
? ?????
2
2
W23 = โซ? 3 ??? = โซ? 3
?
?
?
?2
?2
?? = (?2 ?2 )ln ( 3) = (?2 ?2 )ln ( 1)
(*)
where V3 = V1 has been incorporated. But, we still need to evaluate V2. For Process 12 at
constant pressure
?
W12 = โซ? 2 ??? = ?(?2 โ ?1 )
1
or
V2 =
?12
?
+ ?1 =
(10.5 kJ) 103 Nโm
(1.4 bar)

1 kJ

1 bar
 + 0.028 m3 = 0.103 m3
105 Nโm2
Thus, with Eq. (*)
0.028
W23 = (1.4 bar)(0.103 m3 )ln (
)
0.103
105 Nโm2
1 kJ
 3
 = 18.78 kJ
1 bar
10 Nโm
Thus
Wcycle = 10.5 kJ + (18.78 kJ) + 0 = 8.28 kJ
0
0
(c) To get Q12, we apply the energy balance to process 12: ฮKE + ฮPE + (U2 โ U1) = Q12 โ W12
With U2 = U3,
Q12 = (U3 โ U1) + W12 = (+26.4 kJ) + (10.5 kJ) = 36.9 kJ
Problem 2.58
The net work of a power cycle operating as in Fig. 2.17a is 10,000 kJ, and the thermal efficiency
is 0.4. Determine the heat transfers Qin and Qout, each in kJ.
?=
?cycle
?in
โ
Qin =
?cycle
?
Qin = (10,000 kJ) / (0.4) = 25,000 kJ
Wcycle = 10,000 kJ
ฮท = 0.4
Wcycle = Qcycle = Qin – Qout
Thus
Qout = Qin โ Wcycle = 25,000 โ 10,000 = 15,000 kJ
Problem 2.59
For a power cycle operating as shown in Fig. 2.17a, the energy transfer by heat into the cycle,
Qin, is 500 MJ. What is the net work developed, in MJ, if the cycle thermal efficiency is 30%?
What is the value of Qout, in MJ?
ฮท=
?cycle
?in
Wcycle = ฮทQin = (0.3)(500 MJ) = 150 MJ
Qin = 500 MJ
Wcycle = Qcycle = Qin – Qout
n
Thus
ฮท = 30%
Qout = Qin โ wcycle = 500 MJ โ 150 MJ = 350 MJ
Problem 2.60
For a power cycle operating as in fig. 2.17a, Qin = 17 x 106 Btu and Qout = 12 x 106 Btu.
Determine Wcycle, in Btu, and ฮท.
Wcycle = Qcycle = Qin โ Qout
= (17 x 106) โ (12 x 106) = 5 x 106 Btu
Qin = 17 x 106 Btu
ฮท=
?cycle
?in
=
5 x 106 Btu
17 x 106 Btu
= 0.294 (29.4%)
Alternatively
Qout = 12 x 106 Btu
ฮท=1โ
?out
?in
=1โ
12 x 106
17 x 106
= 0.294
Problem 2.61
A system undergoing a power cycle develops a steady power output of 0.3 kW while receiving
energy input by heat transfer of 2400 Btu/h. Determine the thermal efficiency and the total
amount of energy developed by work, in kW โ h, for one full year of operation.
The thermal efficiency, based on constant rates of energy
transfer, is ฮท = ?ฬcycle /?ฬin
?ฬin = 2400 Btu/h
ฮท = [(0.3 kW)/(2400 Btu/h)]
3.413 Btu/h
1W

103 W
1 kW
= 0.427 (42.7 %)
?ฬcycle = 0.3 kW
?ฬout
For one year of steady operation
W = โซ ?ฬcycle ?? = ?ฬcycle ฮt
= (0.3 k W) (365 days) (24 h/day) = 2628 kW โ h

Problem 2.62
)
Problem 2.63
A concentrating solar collector system, as shown in Fig. P2.63, provides energy by heat transfer
to a power cycle at a rate of 2 MW. The cycle thermal efficiency is 36%. Determine the power
developed by the cycle, in MW. What is the work output, in MWโh, for 4380 hours of steadystate operation? If the work is valued at $0.08/kWโh, what is the total dollar value of the work
output?
Power
Cycle
?ฬin = 2 MW
Atmosphere
?ฬout
?ฬcycle
The power developed is
?ฬcycle = ฮท?ฬin = (0.36) (2 MW) = 0.72 MW
For 4380 hours of steadystate operation
Wcycle = ?ฬcycle ฮt = (0.72 MW)(4380 h) = 3153.6 MWโh
The total dollar value is
$ Value = (3153.6 MWโh)($0.08/kWโh)
103 kW
1 MW
 = $252,300
Problem 2.64
Problem 2.65
Problem 2.66
Problem 2.67
Problem 2.68
For a refrigerator with automatic defrost and a topmounted freezer, the electric power required is
approximately 420 watts to operate. If the coefficient of performance is 2.9, determine the rate
that energy is removed from its refrigerated space, in watts. Evaluating electricity at $0.10/kW โ h,
and assuming the units runs 60% of the time, estimate the cost of one monthโs operation, in $.
KNOWN: Data are provided for steady operation of a refrigeratorfreezer.
FIND: Determine the rate that energy is removed from the refrigerated space, and estimate the cost
of one monthโs operation.
SCHEMATIC AND GIVEN DATA:
?ฬ out
Kitchen
(Hot body)
Refrigerator
components
?ฬ in = ?
ฮฒ = 2.9
ENGINEERING MODEL:
(1) The refrigeration components
execute a refrigeration cycle.
(2) Data are given for steady
operation.
= 420 W
Refrigerated space
(Cold body)
?ฬin
ANALYSIS: The coefficient of performance for steady operation is ฮฒ = ฬ
. Thus
?cycle
?ฬin = ฮฒ ?ฬcycle = (2.9)(420 W) = 1218 W
The total amount of electric energy input for one monthโs operation is determined for steady
operation as
Welec = ?ฬcycle ฮt
Where
ฮt = (0.6)(30 days/month)(24 h/day) = 432 h/month
And
Welec = (420 W)(432 h/month)
1 kW
 = 181.4 kW โ h/month
103 W
Cost = ($0.1/ kW โ h) (181.4 kW โ h/month) = $18.14 / month
Problem 2.69
A windowmounted room air conditioner removes energy by heat transfer from a room and
rejects energy by heat transfer to the outside air. For steady operation, the air conditioner cycle
requires a power input of 0.434 kW and has a coefficient of performance of 6.22. Determine the
rate that energy is removed from the room air, in kW. If electricity is valued at $0.1/kWโh,
determine the cost of operation for 24 hours of operation.
KNOWN: Steadystate operating data are provided for an air conditioner.
FIND: Determine the rate energy is removed from the room and air the cost of 24 hours of
operation.
SCHEMATIC AND GIVEN DATA:
Room air
?ฬin
Outside air
?ฬout
Air
Conditioner
?ฬin = 0.434 kW
ENGINEERING MODEL:
(1) The system shown in the schematic
undergoes a refrigeration cycle.
(2) Energy transfers are positive in the
directions of the arrows.
(3) The cycle operates steadily for 24
hours.
(4) Electricity is valued at $0.1/kWโh.
Refrigeration
Cycle, ฮฒ = 6.22
Electric cost:
$0.1/kWโh
ฮฒ = 6.22
ANALYSIS: Using Eq. 2.45 on a time rate basis
?ฬin
ฮฒ= ฬ
?cycle
โ
?ฬin = ??ฬcycle = (6.22)(0.434 kW) = 2.70 kW
The total amount of electric energy input by work for 24 h of operation is
Wcycle = ?ฬcycle โ? = (0.434 kW)(24 h) = 10.42 kWโh
Thus, the total cost is
Total cost = (10.42 kWโh)($0.1/kWโh) = $1.04 (for 24 hours)
Problem 2.70
Problem 2.71
A heat pump delivers energy by heat transfer to a dwelling at a rate of 11.7 kW. The coefficient
of performance is 2.8.
(a) Determine the power input to the cycle, in kW.
(b) Evaluating electricity at $0.1/kW โ h, determine the cost of electricity during the heating
season when the heat pump operates for 1800 hours.
KNOWN: Operating data are provided for a residential heat pump.
FIND: Determine the power input to the cycle, and the seasonal operating cost.
SCHEMATIC AND GIVEN DATA:
Seasonal hours of
operation = 1800 h
ENGINEERING MODEL: (1) The closed
system undergoes a heat pump cycle. (2) The
cycle operates steadily for 1800 h during the
heating season. (3) Electricity is valued at
$0.1/kW โ h.
?ฬ in
?ฬ cycle = ?
ฮฒ = 2.8
?ฬ out = 11.7 kW
ANALYSIS:
(a) The coefficient of performance for steady operation of the heat pump cycle is: ฮณ = ?ฬout /?ฬcycle .
Thus
ฬ
?out 11.7 kW
?ฬcycle =
=
= 4.179 kW
?
2.8
(b) Based upon modeling assumptions, the cost to operate the heat pump is estimated to be
Cost = (4.179 kW) (1800 h/season) ($0.1/kW โ h)
= $752.22/season
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