Chapter 15 Vapor cycle and processes

Ex:15.1 Pg:697

In [1]:
#Initialization of variables
p1=600 #psia
p2=0.2563 #psia
t1=486.21 #F
t2=60 #F
fur=0.75
#calculations
print "from steam tables,"
h1=1203.2
hf1=471.6
hfg1=731.6
h2=1088
hf2=28.06
hfg2=1059.9
s1=1.4454
sf1=0.6720
sfg1=0.7734
s2=2.0948
sf2=0.0555
sfg2=2.0393
xd=(s1-sf2)/sfg2
hd=hf2+xd*hfg2
xa=0.3023
ha=hf2+xa*hfg2
wbc=0
wda=0
wcd=h1-hd
wab=ha-hf1
W=wab+wcd+wbc+wda
Wrev=hfg1- (t2+459.7)*sfg1
etat=(t1-t2)/(t1+459.7)
eta=fur*etat
#results
print "Thermal efficiency = %d percent"%(etat*100)
print "\n Furnace efficiency = %.1f percent"%(eta*100)

from steam tables,
Thermal efficiency = 45 percent

 Furnace efficiency = 33.8 percent

Ex:15.2 Pg:698

In [2]:
#Initialization of variables
dhab=-123.1
etac=0.5
ha=348.5
etaf=0.75
eta=0.85
hf=471.6
hfg=731.6
hc=1203.2
dhcd=452.7
#calculations
dwabs=dhab/etac
hbd=ha-dwabs
dwcds=dhcd*eta
dqa=hc-hbd
etat=(dwcds+dwabs)/dqa
eta=etat*etaf
#results
print "Thermal efficiency = %.1f percent"%(etat*100)
print "\n Overall efficiency = %.1f percent"%(eta*100)

Thermal efficiency = 22.8 percent

 Overall efficiency = 17.1 percent

Ex:15.3 Pg:699

In [3]:
#Initialization of variables
t=60 #F
J=778.16
p1=600 #psia
p2=0.2563 #psia
etaf=0.85 
#calculations
print "From steam tables,"
vf=0.01604 #ft**3/lbm
dw=-vf*(p1-p2)*144/J
ha=28.06 #Btu/lbm
hb=29.84 #Btu/lbm
hd=1203.2 #Btu/lbm
he=750.5 #Btu/lbm
dqa=hd-hb
dqr=ha-he
dw=dqa+dqr
dwturb=hd-he
dwpump=ha-hb
etat=dw/dqa
eta=etat*etaf
#results
print "Thermal efficiency = %.1f percent"%(etat*100)
print "\n Overall efficiency = %.1f percent"%(eta*100)
    

From steam tables,
Thermal efficiency = 38.4 percent

 Overall efficiency = 32.7 percent

Ex:15.4 Pg:699

In [4]:
#Initialization of variables
dhab=-1.78
etac=0.5
ha=28.06
eta=0.85
hf=471.6
hfg=731.6
hd=1203.2
dhcd=452.7
#calculations
dwabs=dhab/etac
hbd=ha-dwabs
dwcds=dhcd*eta
dqa=hd-hbd
etat=(dwcds+dwabs)/dqa
eta=etat*eta
#results
print "Thermal efficiency = %.1f percent"%(etat*100)
print "\n Overall efficiency = %.1f percent"%(eta*100)

Thermal efficiency = 32.5 percent

 Overall efficiency = 27.7 percent

Ex:15.5 Pg:700

In [5]:
#Initialization of variables
sh=1.6070
ph=94.8 #psia
th=324 #F
tr=60 #F
hh=1186.2 
pi=94.8 #psia
hi=1399.5
si=1.8265
#calculations
Q=hi-hh
Hr=-(tr+459.7)*(si-sh)
work= Q+Hr
eff=work/Q
Qa1=1557.5
W1=637.1 
etat=W1/Qa1
he=1374
hj=948
Whp=he-hh
Wlp=hi-hj
Wnet=Whp+Wlp
#results
print "Thermal efficiency in case 1= %.1f percent"%(eff*100)
print "\n Thermal efficiency in case 1= %.1f percent"%(etat*100)
print "\n High pressure work = %.1f Btu/lbm"%(Whp)
print "\n Low pressure work = %.1f Btu/lbm"%(Wlp)
print "\n Net work = %.1f Btu/lbm"%(Wnet)

Thermal efficiency in case 1= 46.5 percent

 Thermal efficiency in case 1= 40.9 percent

 High pressure work = 187.8 Btu/lbm

 Low pressure work = 451.5 Btu/lbm

 Net work = 639.3 Btu/lbm

Ex:15.6 Pg:701

In [6]:
#Initialization of variables
p2=600 #psia
p1=44 #psia
te=486.21 #F
tb=273.1 #F
J=778.16
p3=0.25 #psia
#calculations
hc=241.9
hj=834.6
y=1-0.805
v1=0.0172
v2=0.016
ha=28.06
hd=hc+v1*(p2-p1)*144/J
hb=ha+v2*(p1-p3)*144/J
hh=1374
Qa=hh-hd
Qr=(ha-hj)*(1-y)
etat=(Qa+Qr)/Qa
#results
print "thermal efficiency = %.1f percent"%(etat*100)

thermal efficiency = 42.6 percent