# CHAPTER 5: AIR COMPRESSORS AND MOTORS REFRIGERATIONS¶

## Example 5.1 Page 66¶

In [2]:
#initialisation of variable
from math import pi,sqrt,acos,asin,atan,cos,sin,tan,log
a=7.0#in
b=10.0#in
c=12.0#in
r=96.0#in
p1=15.0#lb/in^2
p2=100.0#lb/in^2
T=16.0#Degree C
gama=1.4#in
h=120.0#r.p.m
T1=T+273#C absolute

#CALCULATIONS
v1=(pi/4)*(a/c)**2*(b/c)#ft^3
w=(p1*144*v1)/(r*T1)#lb
w1=h*w#lb
W=1680*(1.72-1)#ft lb
I=144*p1*v1*log(p2/p1)#ft lb
E=I/W*100#percent

#RESULTS
print"The ideal efficiency is defined as the ratio of tthis work is",round(E,4),"%"

The ideal efficiency is defined as the ratio of tthis work is 75.4482 %


## Example 5.2 Page 68¶

In [3]:
#initialisation of variable
from math import pi,sqrt,acos,asin,atan,cos,sin,tan,log
h1=16#i.h.p
p1=100#lb/in^2
p2=15#lb/in^2
R=275#R.p.m
h=550#ft/min
q=33000#in^2
v1=4.85#lb
B=8.53#in

#CALCULATIONS
M=(p1/v1)-p2+(p1/v1-p2)*1/0.2
S=h/(2*R)#ft
I=(q*h1)/(M*S*R)#in^2

#RESULTS
print"The effect of the clearance volume is",round(I,3),"in^2"

The effect of the clearance volume is 56.954 in^2


## Example 5.3 Page 69¶

In [6]:
#initialisation of variable
from math import pi,sqrt,acos,asin,atan,cos,sin,tan,log
h=100#ft^3
t=15#degree C
p=120#lb/in^2
gama=1.3#in
t1=15#Degree C
M=((144*t*h*2.6)/(0.3)*(1.271-1))#ft lb

#CALCULATIONS
V=sqrt(p/t)#ft lb

#RESULTS
print"Compare the values of the two cylinders is",round(V,3),"ft lb"

Compare the values of the two cylinders is 2.828 ft lb


## Example 5.5 Page 71¶

In [7]:
#initialisation of variable
from math import pi,sqrt,acos,asin,atan,cos,sin,tan,log
h=0.2#ft^3
v=10#percent
T=15#degree c
p=30#lb/in62
t1=15#Degree C
p1=60#lb/in^2
v1=2.2#ft^3
v3=0.328#ft^3
A=(v1-v3)#ft^3
v2=1.341#ft^3
V=v2-h#ft^
t2=288#Degree C

#CALCULATIONS
T2=(t2*p*v2)/(t1*v1)#Degree C absolute
v5=(t2/T2)*V#ft^3
v7=0.164#ft^3
v8=v5-(v7/11)*v5
v6=v8/(1-v7/11)#ft^3

#RESULTS
print"The required volume of the H.P cylinder including clearance is",round(v6,3),"ft^3"

The required volume of the H.P cylinder including clearance is 0.936 ft^3


## Example 5.6 Page 73¶

In [8]:
#initialisation of variable
from math import pi,sqrt,acos,asin,atan,cos,sin,tan,log
p1=80#lb/in^2
p2=20#lb/in^2

#CALCULATIONS
P=sqrt(p1*p2)#lb/in62
V=P/p1#stroke
W=p2/P#stroke

#RESULTS
print"the ratio of cut off to length of stroke is",round(W,3),"stroke"

the ratio of cut off to length of stroke is 0.5 stroke


## Example 5.9 Page 75¶

In [10]:
#initialisation of variable
from math import pi,sqrt,acos,asin,atan,cos,sin,tan,log
p1=25.0#lb/in^2
p2=50.0#lb/in^2
p3=75.0#lb/in^2
p4=100.0#lb/in^2
v1=29.2#ft^3
v2=28.8#ft^3
v3=28.1#ft^3
v4=27.2#ft^3
h=14.7#lb/in^2
v=3.0#percent
s=5.0#stroke

#CALCULATIONS
V=(pi*p1)/(4)*4#in^3
V1=v/p4*V#in^3

#RESULTS
print"The volume of efficiency of pressure is",round(V1,3),"in^3"

The volume of efficiency of pressure is 2.356 in^3


## Example 5.12 Page 78¶

In [14]:
#initialisation of variable
from math import pi,sqrt,acos,asin,atan,cos,sin,tan,log
p1=15.0#lb/in^2
p2=60.0#lb/in^2
t=16.0#Degree C
Ta=273.0+t#Degree C absolute
T=1.486#lb/in^2
Td=Ta/T#Degree C absolute

#CALCULATIONS
P=Td/(Ta-Td)#Degree C absolute

#RESULTS
print"The lowest temperature and coefficient of per formance is",round(P,3),"Degree C absolute"

The lowest temperature and coefficient of per formance is 2.058 Degree C absolute


## Example 5.14 Page 79¶

In [15]:
#initialisation of variable
from math import pi,sqrt,acos,asin,atan,cos,sin,tan,log
T1=30#Degree c
T2=-10#degree C
t1=263#F
t2=303#F
h1=20#Units
h2=79#C.H.U/lb
h=24#hours
T3=1#Degree C
p=2.2046#C.H.U/sec

#CALCULATIONS
P=h1*p#C.H.U/sec
T=t1/(t2-t1)#F
H=P*60#C.H.U
W=(H*1400)/T#ft/lb
hp=W/33000#h.p
W1=(H*60*h)/(80*2240)#tons

#RESULTS
print"the cycle is a perfect one",round(W1,3),"tons"

the cycle is a perfect one 21.259 tons


## Example 5.15 Page 80¶

In [16]:
#initialisation of variable
from math import pi,sqrt,acos,asin,atan,cos,sin,tan,log
p1=930#lb/in^2
p2=440#lb/in^2
T=268#F
t1=25#F
t2=5#F
h1=19.4#C.H.U
h2=-1.8#C.H.U
h3=29#C.H.U
h4=58.6#C.H.U
d=0.6#C.H.U
d1=0.06#lb
d2=-0.01#lb
c=40#percent
h=24#hour
t3=10#C
d3=15#lb
h5=80#C.H.U

#CALCULATIONS
A=(h1-(h2))-(d1-(d2))*T#C.H.U
FD=A/T#units of entropy
W=4.28#C.H.U
P=0.4*T#C.H.U
H=P*W*d3#C.H.U
H1=P*W*d3*60*h#C.H.U
H2=t3+h5#C.H.U
W1=H1/(H2*2240)#tond

#RESULTS
print"The many tons of ice would a machine working between the same limit and having a relative coefficient is",round(W1,3),"tons"

The many tons of ice would a machine working between the same limit and having a relative coefficient is 0.598 tons


## Example 5.16 Page 82¶

In [20]:
#initialisation of variable
from math import pi,sqrt,acos,asin,atan,cos,sin,tan,log
t1=20.0#Degeree C
t2=-10.0#degree C
h=0.95#dry
t3=35.0#Degree C
h1=0.066#lb
h2=1.089#lb
v1=-0.033#lb
v2=1.193#lb
v3=0.508#lb
T1=263.0#F
T2=293.0#F

#CALCULATIONS
T=T1/(T2-T1)#F
E=h1-(v1)#lb
C=0.1079#lb
CP=E/C#lb
A=CP*(T2-T1)-E*T1#C.H.U
F=A/T1#units of entropy
H=254.212#C.H.U
H2=274.447#C.H.U
W=(CP*(T2-T1)+h*1.023*(T2-T1)-E*T1)#C.H.U
P=H/W#C.H.U
V=A+v3*15-T1*v3*0.0507#C.H.U
H1=T1*(v3*0.0507+0.05*1.023)#C.H.U
N=H2/(W+V)#C.H.U

#RESULTS
print"The upper and lower temperature limits respectively are",round(T,4),"F"
print"The vapour compression cycle work done is",round(H,3),"C.H.U"
print"The vapour is now additional work done is",round(N,3),"C.H.U"

The upper and lower temperature limits respectively are 8.7667 F
The vapour compression cycle work done is 254.212 C.H.U
The vapour is now additional work done is 8.322 C.H.U


## Example 5.18 Page 85¶

In [21]:
#initialisation of variable
from math import pi,sqrt,acos,asin,atan,cos,sin,tan,log
h=0.8#dry
p=120#lb/in^2
p1=1#lb/in^2
t=100#Degree C
A=99.6-38.6-0.178*311.8#C.H.U
G=311.8#units of entropy
AF=440.52#C.H.U
H=399.82#lb/in^2
p=307#lb

#CALCULATIONS
T=H/p#C.H.U

#RESULTS
print"theoretical coefficient pf performance as a refrigeratior is",round(T,3),"C.H.U"

theoretical coefficient pf performance as a refrigeratior is 1.302 C.H.U