Modern Engineering Thermodynamics

10.10 Modified Availability Rate Balance (MARB) Equation 337
The unknowns are the inlet specific flow availability, the exit specific flow availability, and the irreversibility rate inside the
nozzle. The system is open, and material is steam.
a. The station data at the inlet and outlet of the nozzle are
Station 1 Station 2s Ground State
p 1 = 100: psia p 2s = p 2 = 10:0 psia x 0 = 0
T 1 = 500:°F s 2s = s 1 = 1:7087 Btu/lbm.R T 0 = 70:0°F
h 1 = 1279:1 Btu/lbm x 2s = 0:94735 s 0 = 0:0746 Btu/lbm.R
s 1 = 1:7087 Btu/lbm.R h 2s = 161:2 + 0:94735ð982:1Þ
= 1091:6 Btu/lbm
Since this an aergonic, adiabatic, steady state, steady flow, single-inlet, single-outlet system, we can use the energy rate
balance with V 1 = 0andZ 1 = Z 2 = 0 to determine the isentropic exit velocity as
V 2s = ½2g c ðh 1 − h 2s ÞŠ 1/2
= ½2ð32:174 lbm.ft/lbf .s 2 Þð1279:1 − 1091:6 Btu/lbmÞð778:16 ft.lbf/BtuÞŠ 1/2
= 3064 ft/s
and the actual exit velocity is V 2 = 0.95 × V 2s = 0.95(3064) = 2911 ft/s. We can use the energy rate balance to calculate the
actual exit specific enthalpy as
h 2 = h 1 − V2 2
2g c
= 1279:1 Btu/lbm −
Then, we can determine the actual quality of the exit steam as
from which we can calculate the actual entropy at the exit as
The specific flow availability is defined by Eq. (10.20) as
ð2911 ft/sÞ 2
2 32:174 lbm .ft
lbf .s 2
778:16 ft .lbf
Btu
x 2 = ðh 2 − h 2f Þ/h 2fg = ð1110 − 161:2Þ/982:1 = 0:9660
s 2 = s 2f + x 2 s 2fg = 0:2836 + ð0:9660Þð1:5043Þ = 1:7368 Btu/lbm.R
a f = h − h 0 − T 0 ðs − s 0 Þ + V2
2g c
+ gZ
g c
from which we can now calculate the entrance specific flow availability as
= 1110 Btu/lbm
a f 1 = ð1279:1 − 38:1 Btu/lbmÞ − ð70:0 + 459:67 RÞð1:7087 − 0:0746 Btu/lbm.RÞ
+ 0 + 0 = 375 Btu/lbm
b. The specific flow availability at the exit can now be determined from Eq. (10.20) as
a f 2 = ð1109:9 − 38:1 Btu/lbmÞ − ð70:0 + 450:67 RÞð1:7368 − 0:0746 Btu/lbm.RÞ
ð2910:9ft a sÞ 2
+
2 32:174 lbm
.ft
lbf .s 2 778:16 ft + 0 = 361Btu/lbm
.lbf
Btu
c. The irreversibility rate for this system is given by Eq. (10.26) as
_I = 1 − T 0
T b
_Q − _W + _mða f 1 − a f 2 Þ
and, since _Q = _W = 0 here, this equation reduces to
or
_I = _m ða f 1 − a f 2 Þ
_I = ð2:80 lbm/sÞð375 − 361 Btu/lbmÞ = 39:2 Btu/s
(Continued )