Modern Engineering Thermodynamics

342 CHAPTER 10: Availability Analysis
EXAMPLE 10.11 (Continued )
Solution
First, draw a sketch of the system (Figure 10.16).
T
The unknowns are the first and second law efficiencies for heating a liquid
from temperature T to temperature T + ΔT in an open uninsulated heating
tank.
T +ΔT
An energy balance on this system with _W = 0 and Δpe = Δke = 0is
_Q in − _Q loss − 0 = _mðh 2 − h 1 + 0 + 0Þ
Ground state:
or
Temperature = T 0
Q loss
_Q in = _m ðh 2 − h 1 Þ+ _Q loss
We can treat the liquid as incompressible with a negligible pressure drop
and write
Q in
h 2 − h 1 = u 2 − u 1 + vðp 2 − p 1 Þ = cðT 2 − T 1 Þ + 0 = c½ðT + ΔTÞ − TŠ = cðΔTÞ
The “desirable result” in this problem is the increase in enthalpy of the FIGURE 10.16
liquid flowing through the system, _mðh 2 − h 1 Þ, so the first law (thermal) Example 10.11.
efficiency becomes
Desirable result
η T = = _mðh 2 − h 1 Þ _mc ΔT
= ð Þ
Cost
_Q in
_Q in
The second law efficiency is given by Eq. (10.29) as
_A
ε = desired result
_A initial or net input
where
_A desirable result = _mða f 2 − a f 1 Þ = _m h 2 − h 1 − T 0 ðs 2 − s 1 Þ + ðV2 2 − V2 1 Þ/2g
c + gðZ 2 − Z 1 Þ/g c
and, since V 1 = V 2 and Z 1 = Z 2 here, this equation reduces to
_A desirable result = _m ½h 2 − h 1 − T 0 ðs 2 − s 1 ÞŠ
For a constant pressure incompressible liquid, h 2 − h 1 = u 2 − u 1 + v ( p 2 − p 1 ) = c (T 2 − T 1 )+0 = c (ΔT )ands 2 − s 1 =
c ln(T 2 /T 1 ), then
_A desirable result = _mc T 2 − T 1 − T 0 ln T
h
2
= _mc ΔT − T 0 ln T + ΔT i
T 1
T
so
2
ΔT − T 0 ln T + ΔT 3
0
ΔT − T
6
ε = _mc4
T 7
0 ln 1 + ΔT 1
B
5 = _mc@
T C
A
_Q in
_Q in
But, from the first law efficiency defined previously, _Q in = _mcðΔTÞ/η T , so we have
h
ε = η T 1 − T i
0
ln 1 +
ΔT
ΔT T
as was found in Example 10.10. If we again set T 0 = 70.0°F = 530. R and T = 120.°F = 580. R, then this equation gives
h
ε = η T 1 − 530: i
ln 1 +
50:0
= 0:124 × η
50:0 580:
T