Modern Engineering Thermodynamics

186 CHAPTER 6: First Law Open System Applications
EXAMPLE 6.4 (Continued )
Solution
First, draw a sketch of the system (Figure 6.10).
Steam
3
From
river
1 2
1
To river
Liquid
4
Assumptions:
1. Q = 0
2. W = 0
3. Water vapor m = 12.0 kg/min
4. p 4 = p 3
5. Neglect all changes in
flow stream ke and pe terms
6. Steady state, steady flow
FIGURE 6.10
Example 6.4.
The system is open, and the unknown is the river water mass flow rate _m w : We treat this as a steady state, steady flow, double-inlet,
double-outlet open system problem. Since no information is given about heat loss or gain from the environmental
surface of the condenser we assume that it is insulated and therefore adiabatic. Also, since no pressure loss information is
given for the river cooling water flow stream, we assume that it is negligible.
This particular case is not covered by any of the equations developed in the previous discussion of heat exchangers. It is a
combination of a pure substance (water vapor) and an incompressible fluid (river water). However, its ERB can be quickly
arrived at by beginning with the general ERB for heat exchangers, Eq. (6.27), and adding the auxiliary formulae for the specific
enthalpy change of an incompressible fluid, given in Eq. (6.19). This gives
_Q ↘0
+ _m s ðh 1 − h 2 Þ = _m w ðh 4 − h 3 Þ
Then, solving for the river water mass flow rate _m w , we get
= _m w ½c w ðT 4 − T 3 Þ + v w ðp 4 − p 3 ⎵ ÞŠ
j
0
_m w = _m s fðh 1 − h 2 Þ/ ½c w ðT 4 − T 3 ÞŠg
This equation tells us what properties we need to calculate the unknown. The station data are in Table 6.4.
Table C.3b gives us h 1 = 3478.4 kJ/kg, and Table C.2b gives us h 2 = h f (1.0 MPa) = 762.8 kJ/kg. Both these values have been
added to the station data list in Table 6.4. Also, Table 3.5 in Chapter 3 gives the specific heat of liquid water as c = 4.2 kJ/(kg · K).
The condensate flow rate is given in the problem statement to be _m s = 12.0 kg/min, so that the required river water flow rate is
_m w = ð12:0kg/minÞ½ð3478:4 − 762:8Þ kJ/kgŠ/f½4:2 kJ/ðkg.KÞŠð20 − 15 KÞg
= 1552 kg/min
Table 6.4 Data for Example 6.4
Station 1 Station 2 Station 3 Station 4
p 1 = 1.0 MPa x 2 = 0.0 (sat. liq.) T 3 =15°C T 4 =20°C
T 1 = 500°C
p 2 = 1.0 MPa
h 1 = 3478.4 kJ/kg
h 2 = 762.8 kJ/kg
Exercises
9. What cooling water flow rate would be required from the river in Example 6.4 if the steam flow rate is 500. kg/s instead
of 12.0 kg/min and everything else remained the same? Answer: _m w = 3.88 × 10 6 kg/min.
10. Suppose air instead of river water is used to cool the steam in Example 6.4. Determine the mass flow rate of air required
if the air has an inlet temperature of 20.0°C and an exit temperature of 30.0°C. Answer: _m air = 3250 kg/min.
11. If the pumps that supply the river cooling water in Example 6.4 could deliver only 1000. kg/min, what would be the
new temperature of the cooling water as it exits the heat exchanger? Answer: (T w ) exit = 22.8°C.