Modern Engineering Thermodynamics

350 CHAPTER 10: Availability Analysis
EXAMPLE 10.16
A student is using a bathroom sink with separate hot and cold water faucets. The student turns on both faucets and adjusts
them to create a pool of warm water in the sink. The sink’s drain is open, so the faucets must be kept running to maintain the
pool of water. The hot water faucet provides 130.°F water at 0.180 lbm/s, and the cold water faucet provides 60.0°F water at
0.270 lbm/s. The local environment (ground state) is at p 0 = 14.7 psia at T 0 = 55.0°F. Neglecting all flow stream kinetic and
potential energy and assuming the sink itself is insulated, determine
a. The temperature of the mixed water in the sink.
b. The second law availability efficiency of the sink as a mixing-type heat exchanger.
Solution
First, draw a sketch of the system (Figure 10.22).
The unknowns are the temperature of the mixed water in the
sink and the second law availability efficiency of the sink as
a mixing-type heat exchanger.
m hot = 0.180 1bm/s
T hot = 130.°F
m cold = 0.270 1bm/s
T cold = 60.0°F
a. Since the sink is insulated, the mixing process can be taken
as adiabatic, with all the heat transfer occurring inside the
system. The energy rate balance for an adiabatic, aergonic,
steady state, steady flow, double-inlet, single-outlet system
Ground state:
with negligible kinetic and potential energies is
T 0 =55.0°F
_m H h H + _m C h C = _m M h M = ð _m H + _m C Þh M
p 0 = 14.7 psia
a f = cðT − T 0 Þ + vðp − p 0 Þ − cT 0 ln T T 0
+ V2 + gZ
2g c g c
or
T mixed =?
_m H ðh H − h M Þ = _m C ðh M − h C Þ
where we use the conservation of mass, _m M = _m H + _m C :
FIGURE 10.22
Example 10.16.
The thermodynamic state of the water here is a slightly
compressed liquid, but since the compressibility of liquid water is so small, it can be treated as an incompressible
liquid with a constant specific heat c. Assuming p H =p C =p m , then Eq. (3.34) can be used in the previous equation to
produce
_m H cðT H − T M Þ = _m C cðT M − T C Þ
or
T M = _m HT H + _m C T C ð0:180 lbm/sÞð130: + 459:67 RÞ + ð0:270 lbm/sÞð60:0 + 459:67 RÞ
=
_m H + _m C
0:180 lbm/s + 0:270 lbm/s
= 548 R = 88:0°F
b. The specific flow availability is defined by Eq. (10.20) as
a f = h − h 0 − T 0 ðs − s 0 Þ + V 2
+ gZ
2g c g c
The liquid water exiting the faucet fits our definition of an incompressible fluid so that we can use Eq. (3.34)
h − h 0 = cðT − T 0 Þ + vðp − p 0 Þ
and Eq. (7.33)
s − s 0 = c ln T T 0
Combining these equations for the flow of an incompressible fluid gives