Chapter 4
4.3 Applications of Energy Balance
We will discuss examples illustrating the analysis of serveral devices of interest in
engineering, including nozzles and diffusers, turbines, compressor and pumps, heat
exchangers, and throttling devices.
In a nozzle, the fluid velocity increases in the direction of flow due to the reduction in the
flow area. In a diffuser, the fluid velocity decreases in the direction of flow due to the
increase in the flow area.
Example 4.3-13. ---------------------------------------------------------------------------------Steam enters a converging-diverging nozzle operating at steady state with p1 = 40 bar, T1 =
400oC, and a velocity of 10 m/s. The steam flows through the nozzle with negligible heat
transfer and no significant change in potential energy. At the exit, p2 = 15 bar, and the
velocity is 665 m/s. The mass flow rate is 2 kg/s. Determine the exit area of the nozzle, in m2.
Solution ------------------------------------------------------------------------------------------
Apply the steady state energy balance between (1) and (2) gives
h2 − h1 + g(z2 − z1) +
h2 = h1 +
W
1 2
Q
V2 − V12 ) =
− s =0
(
m
m
2
1 2
V1 − V22 )
(
2
At p1 = 40 bar (4 MPa), T1 = 400oC, h1 = 3214 kJ/kg (Table E4.3-1)
h2 = 3214 kJ/kg + 0.5(102 − 6652)
3
m2
s2
1 kJ
1N
= 2993 kJ/kg
3
2
1 kg ⋅ m/s 10 N ⋅ m
Moran, M. J. and Shapiro H. N., Fundamentals of Engineering Thermodynamics, Wiley, 2008, pg. 162
4-21
Table E4.3-1 Steam properties
1
2
Temp Pressure
C
MPa
400
4
280.1
1.5
Specific
Volume
m3/kg
0.07341
0.1628
Internal Specific
Energy Enthalpy
kJ/kg
kJ/kg
2920
3214
2749
2993
Specific
Entropy
kJ/kg/K
6.769
6.839
Quality
Phase
Dense Fluid (T>TC)
Superheated Vapor
At p2 = 15 bar (1.5 MPa), h2 = 2993 kJ/kg, v2 = 0.1628 m3/kg
The exit area is then
mv2
(2 kg)(0.1628 m3 /kg)
A2 =
=
= 4.9×
×10-4 m2
V2
665 m/s
Example 4.3-24. ---------------------------------------------------------------------------------Steam enters a turbine operating at steady state with a mass flow rate of 4600 kg/h. The
turbine develops a power output of 1000 kW. At the inlet, the pressure is 60 bar, the
temperature is 400oC, and the velocity is 10 m/s. At the exit, the pressure is 0.1 bar, the
quality is 0.9, and the velocity is 30 m/s. Calculate the rate of heat transfer between the
turbine and surroundings, in kW.
Solution ------------------------------------------------------------------------------------------
Apply the steady state energy balance between (1) and (2) gives
h2 − h1 + g(z2 − z1) +
Solving for
W
1 2
Q
V2 − V12 ) =
− s
(
m
m
2
Q
with g(z2 − z1) = 0,we obtain
m
W
1
Q
= s + h2 − h1 + (V22 − V12 )
m
m
2
Properties of steam for states (1) and (2) are listed in Table E4.3-2
4
Moran, M. J. and Shapiro H. N., Fundamentals of Engineering Thermodynamics, Wiley, 2008, pg. 165
4-22
Table E4.3-2 Steam properties
1
2
Temp Pressure
C
MPa
400
6
45.81
0.01
Specific
Volume
m3/kg
0.04739
13.21
Internal Specific
Energy Enthalpy
kJ/kg
kJ/kg
2893
3177
2213
2345
Specific
Entropy
kJ/kg/K
6.541
7.4
Quality
Phase
Dense Fluid (T>TC)
0.9 Liquid Vapor Mixture
The change in kinetic energy is evaluated:
1 2
m2
2
2
2
V
−
V
=
0.5(30
−
10
)
(2 1)
2
s2
1 kJ
1N
= 0.4 kJ/kg
3
2
1 kg ⋅ m/s 10 N ⋅ m
The change in enthalpy is
h2 − h1 = 2345 kJ/kg − 3177 kJ/kg = − 832 kJ/kg
W
W
1
Q
= s + h2 − h1 + (V22 − V12 ) = s − 832 kJ/kg + 0.4 kJ/kg
m
m
m
2
W
Q
= s − 831.6 kJ/kg
m
m
Q = 1000 kW − 4600
kg
1 h 1 kW
(831.6 kJ/kg)
= − 62.6 kW
h
3600 s 1 kJ/s
Heat transfer from the turbine to the surroundings.
Example 4.3-35. ---------------------------------------------------------------------------------Air enters a compressor operating at steady state at a pressure of 1 bar, a temperature of 290
K, and a velocity of 6 m/s through an inlet area of 0.1 m2. At the exit, the pressure is 7 bar,
the temperature is 450 K, and the velocity is 2 m/s. Heat transfer from the compressor to its
surroundings occurs at a rate of 180 kJ/min. Employing the ideal gas model, calculate the
power input to the compressor, in kW.
Solution ------------------------------------------------------------------------------------------
5
Moran, M. J. and Shapiro H. N., Fundamentals of Engineering Thermodynamics, Wiley, 2008, pg. 168
4-23
Apply the steady state energy balance between (1) and (2) gives
h2 − h1 + g(z2 − z1) +
W
1 2
Q
V2 − V12 ) =
− s
(
m
m
2
Solving for Ws with g(z2 − z1) = 0,we obtain
V12 − V22
Ws = Q + m ( h1 − h2 ) +
2
The mass flow rate is given by
m =
AV
1 1
v1
The specific velocity can be determine from ideal gas law
v1 =
( R / M )T
1
p1
=
8314 N ⋅ m
(290 K)
28.97 kg ⋅ K
= 0.8324 kg/m3
5
2
10 N/m
The mass flow rate is then
0.1 m 2 ) ( 6 m/s )
(
AV
1 1
m =
=
= 0.7209 kg/s
v1
0.8324 kg/m3
The change in air enthalpy can be obtained from CATT2 program
h1 − h2 = 290.6 kJ/kg − 452.3 kJ/kg = −161.7 kJ/kg
Table E4.3-3 Air properties from CATT2
Temp
K
290
450
Pressure
MPa
0.1
0.7
Specific Enthalpy (Mass)
kJ/kg
290.6
452.3
The change in kinetic energy is evaluated:
1 2
m2
2
2
2
V
−
V
=
0.5(6
−
2
)
( 1 2)
2
s2
1 kJ
1N
= 0.02 kJ/kg
3
2
1 kg ⋅ m/s 10 N ⋅ m
The power input to the compressor is then
4-24
Ws = Q + m ( h1 − h2 ) +
Ws = −
V12 − V22
2
180 kJ
+ (0.7209 kg/s)(−161.7 + 0.02) kJ/kg = − 119.6 kW
60 s
4.4 Heat Exchanger
Heat exchangers are devices for transferring heat between two fluid streams. Heat
exchangers can be classified as indirect contact type and direct contact type. Indirect contact
type heat exchangers have no mixing between the hot and cold streams, only energy transfer
is allowed as shown in Figure 4.4-1.
Hot fluid
Cold fluid
Wall separates streams
Figure 4.4-1 Indirect contact type heat exchanger.
Direct contact type heat exchangers have no wall to separate the cold from the hot streams as
shown in Figure 4.4-2.
Noncondensible
bleed
Warm water
Steam
Cold water
Figure 4.4-2 Direct contact type heat exchanger.
We will briefly discuss two types of tubular heat exchangers: concentric tube and shell-andtube heat exchangers. A concentric tube or double pipe heat exchanger is the simplest heat
exchanger for which the hot and cold fluids move in the same or opposite directions as
shown in Figure 4.4-3.
4-25
Hot fluid
Hot fluid
Cold fluid
Cold fluid
Counter flow
Parallel flow
Figure 4.4-3 Concentric tube heat exchangers.
Shell-and-tube heat exchanger is the most common configuration. There are many different
forms of shell-and-tube heat exchangers according to the number of shell-and-tube passes. A
common form with one shell pass and two tube passes is shown in Figure 4.4-4. Baffles are
usually installed to increase the heat transfer coefficient of the fluid by introducing
turbulence and cross-flow in the shell side.
Baffles
Shell inlet
Tube oulet
Tube inlet
Shell outlet
Figure 4.4-4a Shell-and-tube heat exchanger with one shell pass and two tube passes.
Figure 4.4-4b Details construction of a shell-and-tube heat exchanger.
4-26
Example 4.4-1. ---------------------------------------------------------------------------------Saturated steam at 99.63oC condenses on the outside of a 5-m long, 4-cm-diameter thin
horizontal copper tube by cooling liquid water that enters the tube at 25oC at an average
velocity of 3 m/s and leaves at 45oC. Liquid water density is 997 kg/m3, cp of liquid water is
4.18 kJ/kgoC. (a) Determine the rate of heat transfer to water. (b) If the rate of heat transfer
to water is 200 kW, determine the rate of condensation of steam
Solution -----------------------------------------------------------------------------------------Condensing steam Q
Te
Ti
(a) The rate of heat transfer to water is given by
Q = m cp(Te − Ti)
In this equation, the mass flow rate of water is given by
m = ρVvelAtube
m = (997 kg/m3)(3 m/s)(π×0.022 m2) = 3.7586 kg/s
The heat transfer rate is then
Q = (3.7586 kg/s)(4.18 kJ/kgoC)(45 − 25)oC = 314.2 kW
(b) If the rate of heat transfer to water is 200 kW, determine the rate of condensation of
steam.
We need the enthalpy for saturated liquid and saturated vapor
Temp
C
99.63
99.63
Specific
Pressure Enthalpy
MPa
kJ/kg
0.1
2675
0.1
417.5
Quality
Phase
1
0
Saturated Vapor
Saturated Liquid
The rate of heat transfer to water can also be determined from
Q = msteam (hg − hf)
msteam =
msteam =
Q
hg − h f
200 kJ/s
= 0.0886 kg/s
(2675 - 417.5) kJ/kg
4-27
Example 4.4-2. ---------------------------------------------------------------------------------A light oil is to be preheated before being fed to a distillation tower. The preheating is to be
accomplished in a heat exchanger in which the heating medium is hot oil. The following data
are available:
Light oil
Hot oil
Mass flow rate, kg/hr
160,000
120,000
1.7
1.8
Heat capacity, kJ/kg⋅K
300
500
Entry temperature, K
1) If the light oil leaves the heat exchanger at 360 K, determine the temperature of the
exit hot oil.
2) If the two fluids flow co-currently through the heat exchanger, determine the
maximum attainable temperature of the light oil.
3) If the two fluids flow counter-currently through the heat exchanger, determine the
maximum attainable temperature of the light oil.
Solution -----------------------------------------------------------------------------------------1) Determine the temperature of the exit hot oil.
Q = mc cpc(Tce − Tci) = mh cph(Thi − The)
The = 500 K −
The = Thi −
mc c pc
mh c ph
(Tce − Tci)
(160, 000)(1.7)
(360 − 300) K = 424.4 K
(120, 000)(1.8)
2) If the two fluids flow co-currently through the heat exchanger, determine the
maximum attainable temperature of the light oil.
For cocurrent flow, the maximum attainable temperatrue of the light oil is Tce = The
mc cpc(Tce − Tci) = mh cph(Thi − Tce)
Tce =
Tce =
mc c pcTci + mh c phThi
mc c pc + mh c ph
(16)(1.7)(300 K) + (12)(1.8)(500 K)
= 388.5 K
(16)(1.7) + (12)(1.8)
3) If the two fluids flow counter-currently through the heat exchanger, determine the
maximum attainable temperature of the light oil.
Since mc cpc = (160,000)(1.7) = 272,000 kJ/hr⋅s > mh cph = (120,000)(1.8) = 216,000 kJ/hr⋅s
The = Tci
mc cpc(Tce − Tci) = mh cph(Thi − Tci)
Tce = 300 K +
Tce = Tci +
(120, 000)(1.8)
(500 − 300) K = 458.8 K
(160, 000)(1.7)
4-28
mh c ph
mc c pc
( Thi − Tci)