# Chapter 9: Gases¶

## Example 9.1 pgno:193¶

In :
print"\t example 9.1 \t"
print"\t approximate values are mentioned in the book \t"
#given
T1=245; # inlet hot fluid,F
T2=95; # outlet hot fluid,F
t1=85; # inlet cold fluid,F
t2=95; # outlet cold fluid,F
W=9872; # lb/hr
w=78500; # lb/hr
#solution
from math import log
print"\t 1.for heat balance \t"
print"\t for ammonia gas \t"
c=0.53; # Btu/(lb)*(F)
Q=((W)*(c)*(T1-T2)); # Btu/hr
print"\t total heat required for ammonia gas is :  Btu/hr \t",Q
print"\t for water \t"
c=1; # Btu/(lb)*(F)
Q=((w)*(c)*(t2-t1)); # Btu/hr
print"\t total heat required for water is :  Btu/hr \t",Q
delt1=T2-t1; #F
delt2=T1-t2; # F
print"\t delt1 is :  F \t",delt1
print"\t delt2 is :  F \t",delt2
LMTD=((delt2-delt1)/((2.3)*(log(delt2/delt1))));
print"\t LMTD is : F \t",LMTD
R=((T1-T2)/(t2-t1));
print"\t R is :  \t",R
S=((t2-t1)/(T1-t1));
print"\t S is :  \t",S
print"\t FT is 0.837 \t" # from fig 18
delt=(0.837*LMTD); # F
print"\t delt is :  F \t",delt
Tc=((T2)+(T1))/2; # caloric temperature of hot fluid,F
print"\t caloric temperature of hot fluid is :  F \t",Tc
tc=((t1)+(t2))/2; # caloric temperature of cold fluid,F
print"\t caloric temperature of cold fluid is :  F \t",tc
print"\t hot fluid:shell side,ammonia at 83psia \t"
ID=23.25; # in
C=0.1875; # clearance
B=12; # baffle spacing,in
PT=0.937;
As=((ID*C*B)/(144*PT)); # flow area,ft**2,from eq 7.1
print"\t flow area is :  ft**2 \t",As
Gs=(W/As); # mass velocity,lb/(hr)*(ft**2),from eq 7.2
print"\t mass velocity is :  lb/(hr)*(ft**2) \t",Gs
mu1=0.012*2.42; # at 170F,lb/(ft)*(hr), from fig.15
De=0.55/12; # from fig.28,ft
Res=((De)*(Gs)/mu1); # reynolds number
print"\t reynolds number is :  \t",Res
jH=118; # from fig.28
k=0.017; # Btu/(hr)*(ft**2)*(F/ft),from table 5
Z=0.97; # Z=(Pr*(1/3)) prandelt number
ho=((jH)*(k/De)*(Z)*1); # using eq.6.15,Btu/(hr)*(ft**2)*(F)
print"\t individual heat transfer coefficient is :  Btu/(hr)*(ft**2)*(F) \t",ho
print"\t cold fluid:inner tube side,water \t"
Nt=364;
n=8; # number of passes
L=8; #ft
at1=0.302; # flow area, in**2,from table 10
at=((Nt*at1)/(144*n)); # total area,ft**2,from eq.7.48
print"\t flow area is :  ft**2 \t",at
Gt=(w/(at)); # mass velocity,lb/(hr)*(ft**2)
print"\t mass velocity is :  lb/(hr)*(ft**2) \t",Gt
V=(Gt/(3600*62.5)); # fps
print"\t V is :  fps \t",V
mu2=0.82*2.42; # at 90F,lb/(ft)*(hr),from fig 14
D=(0.62/12); # ft,from table 10
Ret=((D)*(Gt)/mu2); # reynolds number
print"\t reynolds number is :  \t",Ret
hi=900; # using fig 25,Btu/(hr)*(ft**2)*(F)
print"\t hi is :  Btu/(hr)*(ft**2)*(F) \t",hi
ID=0.62; # ft
OD=0.75; #ft
hio=((hi)*(ID/OD)); # using eq.6.5
print"\t Correct hi0 to the surface at the OD is :  Btu/(hr)*(ft**2)*(F) \t",hio
Uc=((hio)*(ho)/(hio+ho)); # clean overall coefficient,Btu/(hr)*(ft**2)*(F)
print"\t clean overall coefficient is :  Btu/(hr)*(ft**2)*(F) \t",Uc
A2=0.1963; # actual surface supplied for each tube,ft**2,from table 10
A=(Nt*L*A2); # ft**2
print"\t total surface area is :  ft**2 \t",A
UD=((Q)/((A)*(delt)));
print"\t actual design overall coefficient is :  Btu/(hr)*(ft**2)*(F) \t",UD
Rd=((Uc-UD)/((UD)*(Uc))); # (hr)*(ft**2)*(F)/Btu
print"\t actual Rd is :  (hr)*(ft**2)*(F)/Btu \t",Rd
print"\t pressure drop  for annulus \t"
f=0.00162; # friction factor for reynolds number 40200, using fig.29
Ds=23.25/12; # ft
phys=1;
N=(12*L/B); # number of crosses,using eq.7.43
print"\t number of crosses are :  \t",N
rowgas=0.209;
print"\t rowgas is  lb/ft**3 \t",rowgas
s=rowgas/62.5;
print"\t s is  \t",s
delPs=((f*(Gs**2)*(Ds)*(N))/(5.22*(10**10)*(De)*(s)*(phys))); # using eq.7.44,psi
print"\t delPs is :  psi \t",delPs
print"\t allowable delPs is 2 psi \t"
print"\t pressure drop  for inner pipe \t"
f=0.000225; # friction factor for reynolds number 21400, using fig.26
s=1;
D=0.0517; #ft
phyt=1;
delPt=((f*(Gt**2)*(L)*(n))/(5.22*(10**10)*(D)*(s)*(phyt))); # using eq.7.45,psi
print"\t delPt is :  psi \t",round(delPt,1)
X1=0.090; # X1=((V**2)/(2*g)), for Gt 1060000,using fig.27
delPr=((4*n*X1)/(s)); # using eq.7.46,psi
print"\t delPr is :  psi \t",round(delPr,1)
delPT=delPt+delPr; # using eq.7.47,psi
print"\t delPT is :  psi \t",round(delPT,1)
print"\t allowable delPT is 10 psi \t"
#end

	 example 9.1
approximate values are mentioned in the book
1.for heat balance
for ammonia gas
total heat required for ammonia gas is :  Btu/hr 	784824.0
for water
total heat required for water is :  Btu/hr 	785000
delt1 is :  F 	10
delt2 is :  F 	150
LMTD is : F 	22.4772661868
R is :  	15
S is :  	0
FT is 0.837
delt is :  F 	18.8134717984
caloric temperature of hot fluid is :  F 	170
caloric temperature of cold fluid is :  F 	90
hot fluid:shell side,ammonia at 83psia
flow area is :  ft**2 	0.387706776948
mass velocity is :  lb/(hr)*(ft**2) 	25462.5417634
reynolds number is :  	40187.0924297
individual heat transfer coefficient is :  Btu/(hr)*(ft**2)*(F) 	42.4542545455
cold fluid:inner tube side,water
flow area is :  ft**2 	0.0954236111111
mass velocity is :  lb/(hr)*(ft**2) 	822647.551124
V is :  fps 	3.65621133833
reynolds number is :  	21418.7950051
hi is :  Btu/(hr)*(ft**2)*(F) 	900
Correct hi0 to the surface at the OD is :  Btu/(hr)*(ft**2)*(F) 	744.0
clean overall coefficient is :  Btu/(hr)*(ft**2)*(F) 	40.1624954017
total surface area is :  ft**2 	571.6256
actual design overall coefficient is :  Btu/(hr)*(ft**2)*(F) 	72.9943154025
actual Rd is :  (hr)*(ft**2)*(F)/Btu 	-0.0111991543302
pressure drop  for annulus
number of crosses are :  	8
rowgas is  lb/ft**3 	0.209
s is  	0.003344
delPs is :  psi 	2.03484815237
allowable delPs is 2 psi
pressure drop  for inner pipe
delPt is :  psi 	3.6
delPr is :  psi 	2.9
delPT is :  psi 	6.5
allowable delPT is 10 psi


## Example 9.2 pgno:196¶

In :
print"\t example 9.2 \t"
print"\t approximate values are mentioned in the book \t"
#given
V1=4670; # inlet air volume,cfm
Pp=0.8153; # Saturation partial pressure of water at 95F,psi,from table 7
Ps=404.3;#  Saturation specific volume of water at 95F,ft**3/lb, from table 7
#solution
print"\t The air and water both occupy the same volume at their respective partial pressures \t"
Vw1=(V1*60/Ps); # water entering per hr,lb
print"\t volume of water entering is :  lb \t",Vw1
print"\t for first stage \t"
c=2.33; # compression ratio
P1=14.7; # psi
P2=(P1*c); # (c=(P2/P1)),psi
print"\t P2 is :  psi \t",P2
gama=1.4; # for air
T1abs=95; # F
T2absr=((T1abs+460)*(P2/P1)**((gama-1)/gama));
print"\t T2absr is :  R \t",T2absr
T2abs=(T2absr-459.67); # F
print"\t T2abs is :  F \t",T2abs
print"\t for intercooler \t"
V2=(V1*60*P1/P2); # ft**3/hr
print"\t final gas volume is :  ft**3/hr \t",V2
Vw2=(V2/Ps); # water remaining in air, lb/hr
print"\t water remaining in air is :  lb/hr \t",Vw2
C=(Vw1-Vw2); # condensation in inter cooler, lb/hr
print"\t condensation in inter cooler is :  lb/hr \t",C
Vs=14.8; # Specific volume of atmospheric air,ft**3/lb
print"\t Specific volume of atmospheric air is :  ft**3/lb \t",Vs
Va=(V1*60/Vs); # air in inlet gas, lb/hr
print"\t air in inlet gas is :  lb/hr\t",Va
print"\t heat load(245 to 95F) \t)"
print"\t sensible heat \t"
Qair=((Va)*(0.25)*(245-T1abs)); # Btu/hr
print"\t Qair is :  Btu/hr \t",Qair
Qwaters=(Vw1*0.45*(245-T1abs)); # Btu/hr
print"\t Qwaters is :  Btu/hr \t",Qwaters
print"\t latent heat \t"
l=1040.1; # latent heat
Qwaterl=(C*l); # Btu/hr
print"\t Qwater1 is :  Btu/hr \t",Qwaterl
Qt1=Qair+Qwaters+Qwaterl;
print"\t total heat is :  Btu/hr \t",Qt1
print"\t for second stage \t"
c=2.33; # compression ratio
P3=(P2*c); # (c=(P3/P1)),psi
print"\t P3 is :  psi \t",P3
V3=(V1*60*P1/P3); # ft**3/hr
print"\t final gas volume is :  ft**3/hr \t",V3
Vw3=(V3/Ps); # water remaining in air, lb/hr
print"\t water remaining in air is :  lb/hr \t",Vw3
C1=(297-Vw3); # condensation in inter cooler, lb/hr
print"\t condensation in inter cooler is :  lb/hr \t",C1
print"\t heat load(245 to 95F) \t)"
print"\t sensible heat \t"
Qair=(Va*0.25*(245-T1abs)); # Btu/hr
print"\t Qair is :  Btu/hr \t",round(Qair)
Qwaters=(Vw2*0.44*(245-T1abs)); # Btu/hr
print"\t Qwater is :  Btu/hr \t",round(Qwaters)
print"\t latent heat \t"
l=1040.1; # latent heat
Qwaterl=(C1*l); # Btu/hr, calculation mistake in book
print"\t Qwater is :  Btu/hr \t",round(Qwaterl)
Qt1=Qair+Qwaters+Qwaterl;
print"\t total heat is :  Btu/hr ",round(Qt1)
# end

	 example 9.2
approximate values are mentioned in the book
The air and water both occupy the same volume at their respective partial pressures
volume of water entering is :  lb 	693.049715558
for first stage
P2 is :  psi 	34.251
T2absr is :  R 	706.727218318
T2abs is :  F 	247.057218318
for intercooler
final gas volume is :  ft**3/hr 	120257.51073
water remaining in air is :  lb/hr 	297.446229853
condensation in inter cooler is :  lb/hr 	395.603485705
Specific volume of atmospheric air is :  ft**3/lb 	14.8
air in inlet gas is :  lb/hr	18932.4324324
sensible heat
Qair is :  Btu/hr 	709966.216216
Qwaters is :  Btu/hr 	46780.8558001
latent heat
Qwater1 is :  Btu/hr 	411467.185481
total heat is :  Btu/hr 	1168214.2575
for second stage
P3 is :  psi 	79.80483
final gas volume is :  ft**3/hr 	51612.6655492
water remaining in air is :  lb/hr 	127.659326117
condensation in inter cooler is :  lb/hr 	169.340673883
sensible heat
Qair is :  Btu/hr 	709966.0
Qwater is :  Btu/hr 	19631.0
latent heat
Qwater is :  Btu/hr 	176131.0
total heat is :  Btu/hr  905729.0


## Example 9.3 pgno:197¶

In :
print"\t example 9.3 \t"
print"\t approximate values are mentioned in the book \t"
#given
Va=18900.; # air in inlet gas
Vw1=692.; # water entering
#solution
Ma=(Va/29.); # moles
Mw=(Vw1/18.); # moles
M=(Ma+Mw); # moles
print"\t total number of moles re :  \t",M
print"\t Moles of air is :  \t",Ma
print"\t Moles of water is :  \t",Mw
print"\t after compression \t"
P=34.2; # pressure,psi
pw=(Mw/M)*(P); # partial pressure
print"\t partial pressure is : psi \t",round(pw,2)
Td=124; # F, table table 7
print"\t dew point is :  F \t",Td
# end

	 example 9.3
approximate values are mentioned in the book
total number of moles re :  	690.168582375
Moles of air is :  	651.724137931
Moles of water is :  	38.4444444444
after compression
partial pressure is : psi 	1.91
dew point is :  F 	124