Modern Engineering Thermodynamics

338 CHAPTER 10: Availability Analysis
EXAMPLE 10.8 (Continued )
Exercises
22. Determine the actual exit enthalpy in Example 10.8 if the nozzle efficiency is 85.0% instead of 95.0%. Assume all the
other variables remain unchanged. Answer: h 2 = 1144 Btu/lbm.
23. In other thermodynamics problems, we often neglect the effect of flow stream velocity in our analysis. Recompute the
irreversibility rate in Example 10.8 neglecting the exit velocity and observe the effect flow stream velocity has on the
answer. Note that both h 2 and s 2 change, and that station 2 is now superheated. Answer: = 373 Btu/s (a very significant
error is produced by neglecting the flow stream velocity).
24. If the inlet pressure is increased from 100. psia to 400. psia, determine the new actual exit velocity. Assume all the other
variables remain unchanged. Answer: V 2 = 2640 ft/s (to three significant figures).
One further case needs to be introduced. To be able to deal with the inequality that appears in the entropy
production term of the entropy balance, we introduce the simplifying assumption of a reversible process. Even
though reversible processes do not occur in practical engineering problems, it is often a useful simplifying
assumption to invoke. However, when it is used, one should not expect calculated values to closely match the
results of experimental measurements.
■ Case 6 Reversible Processes
In any system undergoing reversible processes, _S P = _I = 0: Then, the resulting SS, SF, SI, SO, CV, IB availability
rate balance for a system undergoing an internally reversible processes is
The modified availability rate balance for a SS, SF, SI, SO, CV, IB, reversible system
1 − T
0 _Q − _W + _m ða f Þ
T in − ða f Þ out = 0 (10.27)
b
Note that the first term on the left-hand side of this equation represents external heat transfer irreversibilities,
even though we made the system internally reversible.
■
EXAMPLE 10.9
A large uninsulated steam turbine receives superheated steam
at 18.0 kg/s, 500.°C, and 3.00 MPa and exhausts it to 0.0100
MPa with a quality of 96.0%. If the turbine is assumed to be
internally reversible, determine the rate of heat loss from the
surface of the turbine if the power output is 20.0 × 10 3 kW.
The surface temperature of the turbine is uniform at 350.°C,
and the local environment (ground state) is taken to be saturatedliquidwateratT
0 = 20.0°C. Neglect all flow stream
kinetic and potential energies in this problem.
Solution
First, draw a sketch of the system (Figure 10.14).
m = 18.0 kg/s
T 1 = 500.°C
p 1 = 3.00 MPa
T s = 350.°C
Ground state:
Water at x 0 = 0.00 and
T 0 = 20.0°C
Q = ?
W = 20.0 × 10 3 kw
x 2 = 0.960
p 2 = 0.0100 MPa
The unknown is the rate of heat loss from the surface of the
turbine. This is an open system, and the material is steam.
The station data at the inlet and outlet of the turbine are
FIGURE 10.14
Example 10.9.
Station 1 Station 2 Ground State
p 1 = 3:00 MPa p 2 = 0:0100 MPa x 0 = 0:00
T 1 = 500:°C x 2 = 0:960 T 0 = 20:0°C
h 1 = 3456:5 kJ/kg h 2 = 191:8 + 0:96ð2392:8Þ h 0 = 83:9 kJ/kg
s 1 = 7:2346 kJ/kg .K = 2488:9 kJ/kg s 0 = 0:2965 kJ/kg .K
s 2 = 0:6491 + 0:96ð7:5019Þ
= 7:8509 kJ/kg .K