T
v 1
2 T1 = 400°C
p = 60 bar
p = 0.1 bar 1
2 m1 = 4600 kg/h
p1 = 60 bar T1 = 400°C V1 = 10 m/s
ã
Wãcv = 1000 kW
p2 = 0.1 bar x2 = 0.9 (90%) V2 = 30 m/s
Fig. E4.4
A A
Turbine A.19 – Tabs a, b, & c
c c c c EXAMPLE 4.4 c
162 Chapter 4 Control Volume Analysis Using Energy
Analysis: To calculate the heat transfer rate, begin with the one-inlet, one-exit form of the energy rate balance for a control volume at steady state, Eq. 4.20a. That is
05Q#
cv2W#
cv1m# c(h12h2)1 (V122V22)
2 1g(z12z2)d where m# is the mass flow rate. Solving for Q#
cv and dropping the potential energy change from inlet to exit
Q#
cv5W#
cv1m# c(h22h1)1aV222V12
2 bd (a)
To compare the magnitudes of the enthalpy and kinetic energy terms, and stress the unit conversions needed, each of these terms is evaluated separately.
First, the specific enthalpy difference h2 2 h1 is found. Using Table A-4, h1 5 3177.2 kJ/kg. State 2 is a two- phase liquid–vapor mixture, so with data from Table A-3 and the given quality
h25hf 21x2(hg22hf 2)
5191.831 (0.9)(2392.8)52345.4 kJ/kg Hence
h22h152345.423177.25 2831.8 kJ/kg
Consider next the specific kinetic energy difference. Using the given values for the velocities
➊ aV222V12
2 b5 c(30)22(10)2 2 d am2
s2b ` 1 N
1 kg?m/s2` ` 1 kJ 103 N?m` 50.4 kJ/kg
Calculating Q#
cv from Eq. (a)
➋ Q#
cv5(1000 kW)1a4600kg
h b(2831.810.4)akJ kgb ` 1 h
3600 s` `1 kW 1 kJ/s` 5 262.3 kW
➊ The magnitude of the change in specific kinetic energy from inlet to exit is much smaller than the specific enthalpy change. Note the use of unit conver- sion factors here and in the calculation of Q#
cv to follow.
➋ The negative value of Q#
cv means that there is heat transfer from the turbine to its surroundings, as would be expected. The magnitude of Q#
cv is small relative to the power developed.
Ability to…
❑ apply the steady-state energy rate balance to a control volume.
❑ develop an engineering model.
❑ retrieve property data for water.
✓Skills Developed
If the change in kinetic energy from inlet to exit were neglected, evaluate the heat transfer rate, in kW, keeping all other data unchanged. Comment. Ans. 262.9 kW
4.8 Compressors and Pumps 163
4.8 Compressors and Pumps
Compressors and pumps are devices in which work is done on the substance flowing through them in order to change the state of the substance, typically to increase the pressure and/or elevation. The term compressor is used when the substance is a gas (vapor) and the term pump is used when the substance is a liquid. Four compressor types are shown in Fig. 4.11. The reciprocating compressor of Fig. 4.11a features reciprocating motion while the others have rotating motion.
The axial-flow compressor of Fig. 4.11b is a key component of turbojet engines (Sec. 9.11). Compressors also are essential components of refrigeration and heat pump systems (Chap. 10). In the study of Chap. 8, we find that pumps are important in vapor power systems. Pumps also are commonly used to fill water towers, remove water from flooded basements, and for numerous other domestic and industrial applications.
4.8.1 Compressor and Pump Modeling Considerations
For a control volume enclosing a compressor, the mass and energy rate balances reduce at steady state as for the case of turbines considered in Sec. 4.7.1. Thus, Eq. 4.20a reduces to read
05Q#
cv2W#
cv1m#(h12h2) (a)
Heat transfer with the surroundings is frequently a secondary effect that can be neglected, giving as for turbines
W#
cv5m#(h12h2) (b)
For pumps, heat transfer is generally a secondary effect, but the kinetic and potential energy terms of Eq. 4.20a may be significant depending on the application. Be sure to note that for compressors and pumps, the value of W#
cv is negative because a power input is required.
4.8.2 Applications to an Air Compressor and a Pump System
In this section, modeling considerations for compressors and pumps are illustrated in Examples 4.5 and 4.6, respectively. Applications of compressors and pumps in energy storage systems are described in Sec. 4.8.3.
In Example 4.5 the objectives include assessing the significance of the heat trans- fer and kinetic energy terms of the energy balance and illustrating the appropriate use of unit conversion factors.
compressor, pump
Calculating Compressor Power
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 with an 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.
Fig. 4.11 Compressor types.
Inlet
Outlet
(a) Reciprocating
(b) Axial flow Rotor Inlet
Inlet
Outlet
Outlet Stator
Inlet
Outlet
Driveshaft Impeller
(c) Centrifugal Outlet
Inlet (d) Roots type
c c c c EXAMPLE 4.5 c
164 Chapter 4 Control Volume Analysis Using Energy
SOLUTION
Known: An air compressor operates at steady state with known inlet and exit states and a known heat transfer rate.
Find: Calculate the power required by the compressor.
Schematic and Given Data:
Engineering Model:
1. The control volume shown on the accompanying figure is at steady state.