Department of Civil Engineering & Applied Mechanics
McGill University, Montreal, Quebec
Canada
CIVE 290 THERMODYNAMICS & HEAT TRANSFER
Assignment #4 SOLUTIONS
1. A 68-kg man whose average body temperature is 39oC drinks 1 L of cold water at 3oC in an
effort to cool down. Taking the average specific heat of the human body to be 3.6 kJ/kg·oC,
determine the drop in the average body temperature of this person under the influence of this
cold water.
A man drinks one liter of cold water at 3°C in an effort to cool down. The drop in the average body temperature of
the person under the influence of this cold water is to be determined.
Assumptions 1 Thermal properties of the body and water are constant. 2 The effect of metabolic heat generation and
the heat loss from the body during that time period are negligible.
Properties The density of water is very nearly 1 kg/L, and the specific heat of water at room temperature is c = 4.18
kJ/kg·°C (Table A-3). The average specific heat of human body is given to be 3.6 kJ/kg.°C.
Analysis. The mass of the water is
m w = ρV = (1 kg/L )(1 L ) = 1 kg
We take the man and the water as our system, and disregard any heat and mass transfer and chemical reactions. Of
course these assumptions may be acceptable only for very short time periods, such as the time it takes to drink the
water. Then the energy balance can be written as
E − Eout
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3
=
Net energy transfer
by heat, work, and mass
ΔEsystem
1
424
3
Change in internal, kinetic,
potential, etc. energies
0 = ΔU = ΔU body + ΔU water
or
[mc(T2 − T1 )]body + [mc(T2 − T1 )]water
Substituting
(68 kg)( 3.6 kJ/kg⋅o C)(T f − 39) o C + (1 kg)( 4.18 kJ/kg⋅o C)(T f − 3) o C = 0
=0
It gives
Tf = 38.4°C
Then
∆T=39 − 38.4 = 0.6°C
Therefore, the average body temperature of this person should drop about half a degree celsius.
1
2. The diffuser in a jet engine is designed to decrease the kinetic energy of the air entering the
engine compressor without any work or heat interactions. Calculate the velocity at the exit of
a diffuser when air at 100 kPa and 20oC enters it with a velocity of 500 m/s and the exit state
is 200 kPa and 90oC. The kinetic energy of a fluid decreases as it is decelerated in an
adiabatic diffuser. What happens to this lost kinetic energy?
Air is decelerated in an adiabatic diffuser. The velocity at the exit is to be determined.
Assumptions 1 This is a steady-flow process since there is no change with time. 2 Air is an ideal gas with constant
specific heats. 3 Potential energy changes are negligible. 4 There are no work interactions. 5 The diffuser is
adiabatic.
Properties The specific heat of air at the average temperature of (20+90)/2=55°C =328 K is cp = 1.007 kJ/kg⋅K
(Table A-2b).
Analysis There is only one inlet and one exit, and thus m& 1 = m& 2 = m& . We take diffuser as the system, which is a
control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in
the rate form as
E& − E&
=
ΔE& system Ê0 (steady)
=0
1in424out
3
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3
Rate of net energy transfer
by heat, work, and mass
Rate of change in internal, kinetic,
potential, etc. energies
100 kPa
20°C
500 m/s
E& in = E& out
m& (h1 + V12 / 2) = m& (h2 + V 22 /2)
AIR
200 kPa
90°C
h1 + V12 / 2 = h2 + V 22 /2
Solving for exit velocity,
[
V 2 = V12 + 2(h1 − h2 )
]
0.5
[
= V12 + 2c p (T1 − T2 )
]
0.5
⎡
⎛ 1000 m 2 /s 2
= ⎢(500 m/s) 2 + 2(1.007 kJ/kg ⋅ K)(20 − 90)K⎜
⎜ 1 kJ/kg
⎢⎣
⎝
= 330.2 m/s
⎞⎤
⎟⎥
⎟⎥
⎠⎦
0.5
The “lost” kinetic energy is mostly converted to internal energy as shown by a rise in the fluid temperature.
3. Refrigerant-134a at 700 kPa and 120oC enters an adiabatic nozzle steadily with a velocity of
20 m/s and leaves at 400 kPa and 300C. Determine a) the exit velocity, and b) the ratio of the
inlet to exit area A1/A2. In the case of a non-adiabatic nozzle, how would heat transfer affect
the fluid velocity at the nozzle exit?
R-134a is accelerated in a nozzle from a velocity of 20 m/s. The exit velocity of the refrigerant and the ratio of the
inlet-to-exit area of the nozzle are to be determined.
Assumptions 1 This is a steady-flow process since there is no change with time. 2 Potential energy changes are
negligible. 3 There are no work interactions. 4 The device is adiabatic and thus heat transfer is negligible.
Properties From the refrigerant tables (Table A-13)
P1 = 700 kPa ⎫ v 1 = 0.043358 m 3 /kg
⎬
T1 = 120°C ⎭ h1 = 358.90 kJ/kg
and
1
R-134a
2
2
P2 = 400 kPa ⎫ v 2 = 0.056796 m 3 /kg
⎬
T2 = 30°C
⎭ h2 = 275.07 kJ/kg
&1 = m
&2 = m
& . We take nozzle as the system, which is a
Analysis (a) There is only one inlet and one exit, and thus m
control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in
the rate form as
E& − E& out
=
ΔE& systemÊ0 (steady)
=0
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1442443
Rate of net energy transfer
by heat, work, and mass
Rate of change in internal, kinetic,
potential, etc. energies
E& in = E& out
& ≅ Δpe ≅ 0)
m& (h1 + V12 / 2) = m& (h2 + V22 /2) (since Q& ≅ W
0 = h2 − h1 +
V22 − V12
2
Substituting,
0 = (275.07 − 358.90 )kJ/kg +
It yields
V 22 − (20 m/s )2
2
⎛ 1 kJ/kg
⎜
⎜ 1000 m 2 /s 2
⎝
⎞
⎟
⎟
⎠
V2 = 409.9 m/s
(b) The ratio of the inlet to exit area is determined from the conservation of mass relation,
(
)
v V
A
0.043358 m 3 /kg (409.9 m/s )
1
1
⎯→ 1 = 1 2 =
= 15.65
A2V 2 =
A1V1 ⎯
v2
v1
A2 v 2 V1
0.056796 m 3 /kg (20 m/s )
(
)
Heat transfer to the fluid as it flows through a nozzle is desirable since it will probably increase the kinetic energy of
the fluid. Heat transfer from the fluid will decrease the exit velocity.
4. A steam turbine operates with 1.6 MPa and 350oC steam at its inlet and saturated vapour at
30oC at its exit. The mass flow rate of the steam is 16 kg/s, and the turbine produces 9000
kW of power. Determine the rate at which heat is lost through the casing of this turbine.
Steam expands in a turbine whose power production is 9000 kW. The rate of heat lost from the turbine is to be
determined.
Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy
changes are negligible.
Properties From the steam tables (Tables A-6 and A-4)
P1 = 1.6 MPa ⎫
⎬ h1 = 3146.0 kJ/kg
T1 = 350°C ⎭
T2 = 30°C ⎫
⎬ h2 = 2555.6 kJ / kg
x2 = 1
⎭
1.6 MPa
350°C
16 kg/s
Analysis We take the turbine as the system, which is a control volume since
mass crosses the boundary. Noting that there is one inlet and one exiti the
energy balance for this steady-flow system can be expressed in the rate form as
Heat
Turbine
30°C
sat. vap.
3
E& − E&
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3
ΔE& system ©0 (steady)
144
42444
3
=
Rate of net energy transfer
by heat, work, and mass
=0
Rate of change in internal, kinetic,
potential, etc. energies
E& in = E& out
m& 1 h1 = m& 2 h2 + W& out + Q& out
Q& out = m& (h1 − h2 ) − W& out
Substituting,
Q& out = (16 kg/s)(3146.0 − 2555.6) kJ/kg − 9000 kW = 446.4 kW
5. Air is compressed by an adiabatic compressor from 100 kPa and 20oC to 1.8 MPa and 400oC.
Air enters the compressor through a 0.15-m2 opening with a velocity of 30 m/s. It exits
through a 0.08-m2 opening. Calculate the mass flow rate of air and the required power input.
Why does the air exit the compressor at a higher temperature?
1.8 MPa
400°C
Compressor
100 kPa
20°C
30 m/s
Air is compressed in an adiabatic compressor. The mass flow rate of the air and the power input are to be
determined.
Assumptions 1 This is a steady-flow process since there is no change with time. 2 The compressor is adiabatic. 3
Air is an ideal gas with constant specific heats.
Properties The constant pressure specific heat of air at the average temperature of (20+400)/2=210°C=483 K is cp =
1.026 kJ/kg·K (Table A-2b). The gas constant of air is R = 0.287 kPa⋅m3/kg⋅K (Table A-1).
Analysis (a) There is only one inlet and one exit, and thus m& 1 = m& 2 = m& . We take the compressor as the system,
which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be
expressed in the rate form as
=
ΔE& system Ê0 (steady)
=0
E& − E&
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3
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3
Rate of net energy transfer
by heat, work, and mass
Rate of change in internal, kinetic,
potential, etc. energies
E& in = E& out
& ≅ Δpe ≅ 0
Since Q
⎛
V 2 − V12
W& in = m& ⎜⎜ h2 − h1 + 2
2
⎝
⎛
⎞
V 2 − V12
⎟⎟ = m& ⎜⎜ c p (T2 − T1 ) + 2
2
⎝
⎠
⎞
⎟⎟
⎠
The specific volume of air at the inlet and the mass flow rate are
v1 =
RT1 (0.287 kPa ⋅ m 3 /kg ⋅ K)(20 + 273 K)
=
= 0.8409 m 3 /kg
P1
100 kPa
m& =
A1V1 (0.15 m 2 )(30 m/s)
=
= 5.351 kg/s
v1
0.8409 m 3 /kg
4
Similarly at the outlet,
v2 =
RT2 (0.287 kPa ⋅ m 3 /kg ⋅ K)(400 + 273 K)
=
= 0.1073 m 3 /kg
P2
1800 kPa
V2 =
m& v 2 (5.351 kg/s)(0.1073 m 3 /kg)
=
= 7.177 m/s
A2
0.08 m 2
(b) Substituting into the energy balance equation gives
⎛
V 2 − V12 ⎞⎟
W& in = m& ⎜ c p (T2 − T1 ) + 2
⎟
⎜
2
⎠
⎝
⎡
(7.177 m/s) 2 − (30 m/s) 2 ⎛ 1 kJ/kg
= (5.351 kg/s) ⎢(1.026 kJ/kg ⋅ K)(400 − 20)K +
⎜
2
⎝ 1000 m 2 /s 2
⎢⎣
= 2084 kW
⎞⎤
⎟⎥
⎠⎥⎦
The air exits the compressor at a higher temperature because energy (in the form of shaft work) is being added to the
air.
6. An adiabatic capillary tube is used in some refrigeration systems to drop the pressure of the
refrigerant from the condenser level to the evaporator level. The R-134a enters the capillary
tube as a saturated liquid at 500C, and leaves at -12oC. Determine the quality of the
refrigerant at the inlet of the evaporator.
Refrigerant-134a is throttled by a capillary tube. The quality of the refrigerant at the exit is to be determined.
Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy
changes are negligible. 3 Heat transfer to or from the fluid is negligible. 4 There are no work interactions involved.
Analysis There is only one inlet and one exit, and thus m& 1 = m& 2 = m& . We take the throttling valve as the system,
which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be
50°C
expressed in the rate form as
Sat.
liquid
E& in − E& out = ΔE& system Ê0 (steady) = 0
E& in = E& out
m& h1 = m& h2
R-134a
h1 = h2
since Q& ≅ W& = Δke ≅ Δpe ≅ 0 .
-12°C
The inlet enthalpy of R-134a is, from the refrigerant tables (Table A-11),
T1 = 50°C ⎫
⎬ h1 = h f = 123.49 kJ/kg
sat. liquid ⎭
The exit quality is
h2 − h f 123.49 − 35.92
T2 = −12°C ⎫
=
= 0.422
⎬ x2 =
h2 = h1
h fg
207.38
⎭
5
7. A hot-water steam at 80oC enters a mixing chamber with a mass flow rate of 0.5 kg/s where
it is mixed with a stream of cold water at 20oC. If it is desired that the mixture leave the
chamber at 42oC, determine the mass flow rate of the cold-water stream. Assume all the
streams are at a pressure of 250 kPa. When two streams are mixed in a mixing chamber, can
the temperature of the exit stream ever be lower than the temperature of the colder stream?
T1 = 80°C
·
m1 = 0.5 kg/s
H2O
(P = 250 kPa)
T3 = 42°C
T2 = 20°C
·
m2
A hot water stream is mixed with a cold water stream. For a specified mixture temperature, the mass flow rate of
cold water is to be determined.
Assumptions 1 Steady operating conditions exist. 2 The mixing chamber is well-insulated so that heat loss to the
surroundings is negligible. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Fluid
properties are constant. 5 There are no work interactions.
Properties Noting that T < Tsat @ 250 kPa = 127.41°C, the water in all
three streams exists as a compressed liquid, which can be
approximated as a saturated liquid at the given temperature. Thus,
h1 ≅ hf @ 80°C = 335.02 kJ/kg
h2 ≅ hf @ 20°C =
83.915 kJ/kg
h3 ≅ hf @ 42°C = 175.90 kJ/kg
Analysis We take the mixing chamber as the system, which is a control volume. The mass and energy balances for
this steady-flow system can be expressed in the rate form as
Mass balance:
m& in − m& out = Δm& systemÊ0 (steady) = 0 ⎯
⎯→ m& 1 + m& 2 = m& 3
Energy balance:
E& − E& out
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424
3
=
Rate of net energy transfer
by heat, work, and mass
ΔE& system©0 (steady)
1442443
=0
Rate of change in internal, kinetic,
potential, etc. energies
E& in = E& out
m& 1h1 + m& 2 h2 = m& 3h3 (since Q& = W& = Δke ≅ Δpe ≅ 0)
& 2 gives
Combining the two relations and solving for m
m& 1h1 + m& 2h2 = (m& 1 + m& 2 )h3
&2 =
m
h1 − h3
&1
m
h3 − h2
Substituting, the mass flow rate of cold water stream is determined to be
m& 2 =
(335.02 − 175.90) kJ/kg (0.5 kg/s ) = 0.865
(175.90 − 83.915) kJ/kg
kg/s
When two streams are mixed in a mixing chamber, the temperature of the exit stream can be lower than the
temperature of the colder stream if the mixing chamber is losing heat to the surroundings.
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8. Steam is to be condensed on the shell side of a heat exchanger at 85oF. Cooling water enters
the tubes at 60oF at a rate of 138 lbm/s and leaves at 73oF. Assuming the heat exchanger to be
well-insulated, determine the rate of heat transfer in the heat exchanger and the rate of
condensation of the stream.
Steam is condensed by cooling water in a condenser. The rate of heat transfer in the heat exchanger and the rate of
condensation of steam are to be determined.
Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the
surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3
Changes in the kinetic and potential energies of fluid streams are negligible. 4 Fluid properties are constant.
Properties The specific heat of water is 1.0 Btu/lbm.°F (Table A3E). The enthalpy of vaporization of water at 85°F is 1045.2
Btu/lbm (Table A-4E).
Analysis We take the tube-side of the heat exchanger where cold
water is flowing as the system, which is a control volume. The
energy balance for this steady-flow system can be expressed in the
rate form as
E& − E& out
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424
3
=
Rate of net energy transfer
by heat, work, and mass
ΔE& systemÊ0 (steady)
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Steam
85°F
73°F
=0
Rate of change in internal, kinetic,
potential, etc. energies
60°F
E& in = E& out
Q& in + m& h1 = m& h2 (since Δke ≅ Δpe ≅ 0)
Q& in = m& c p (T2 − T1 )
Water
85°F
Then the rate of heat transfer to the cold water in this heat exchanger becomes
Q& = [m& c (T − T )]
= (138 lbm/s)(1.0 Btu/lbm.°F)(73°F − 60°F) = 1794 Btu/s
p
out
in
water
Or one could approximate the compressed liquid as a saturated liquid at the same temperature.
Q& in = m& (h2 - h1 )
Table A-4E: h1 ≈ hf @ 60oF = 28.08 Btu/lbm
h2 ≈ hf@73oF = 41.07 Btu/lbm
Q& in = 138 lbm/s (41.07 - 28.08)Btu/lbm = 1793 Btu/s
Noting that heat gain by the water is equal to the heat loss by the condensing steam, the rate of condensation of the
steam in the heat exchanger is determined from
1794 Btu/s
Q&
⎯→ m& steam =
=
= 1.72 lbm/s
Q& = (m& h fg )steam = ⎯
h fg 1045.2 Btu/lbm
7