CHAPTER 1: HEATING AND EXPANSION OF GASES ENTROPY

Example 1.1 Page 1

In [2]:
#initialisation of variable
from math import pi,sqrt,acos,asin,atan,cos,sin,tan
p1=280#lb/in^2
v=2#ft^3
p2=20#lb/in^2
v2=18.03#ft^3

#calculation
W=144*(p1*v-p2*v2)/(1.2-1)#ft/lb

#result
print"The volume and work done during the expansion is",round(W,2),"ft/lb"

The volume and work done during the expansion is 143568.0 ft/lb

Example 1.2 Page 2

In [1]:
#initialisation of variable
from math import pi,sqrt,acos,asin,atan,cos,sin,tan
v=2#ft^3
v2=20#ft^3
p=100000#ft lb
v2=10.41#lb/in^2
v3=10#lb/in^2
p1=1.3#lb
p2=(v2*199.5)/9.95#lb/in^2

#calculation
P=(p2/v3-v2)#lb/in^2

#result
print"The initial andfinal pressure is",round(P,4),"lb/in^2"

The initial andfinal pressure is 10.4623 lb/in^2

Example 1.4 Page 3

In [5]:
#initialisation of variable
from math import pi,sqrt,acos,asin,atan,cos,sin,tan
Cp=0.24#lb/in^2
Cv=0.18#ft^3
p1=5#lb/in^2
T1=20#Degree C
T2=150#Degree C

#CALCULATIONS
W=p1*Cp*(T2-T1)#C.H.U
H=p1*Cv*(T2-T1)#C.H.U
Gamma=Cp/Cv#lb/in^2

#RESULTS
print"the constant pressure is",round(W,2),"C.H.U"
print"the constant volume the value of gas is",round(Gamma,2),"lb/in^2"

the constant pressure is 156.0 C.H.U
the constant volume the value of gas is 1.33 lb/in^2

Example 1.5 Page 4

In [6]:
#initialisation of variable
from math import pi,sqrt,acos,asin,atan,cos,sin,tan
Gama=1.33#ft/lb
p=100#lb/in^2
p1=20#lb/in^2
v2=10.05#ft^3
v=3#ft/lb

#CALCULATIONS
W=144*(p*v-p1*v2)/0.33#ft lb

#RESULTS
print"The work done is",round(W,2),"lb"

The work done is 43200.0 lb

Example 1.8 Page 5

In [39]:
#initialisation of variable
from math import pi,sqrt,acos,asin,atan,cos,sin,tan
p=3.74#ft/lb
p1=2.48#ft/lb
v=5.70#ft lb
Cv=0.21#ft/lb
P=440.00#lb/in^2
P1=160.00#lb/in^2
P2=14.00#lb/in^2
T=25.00#degree C
T1=100.00#F
vs=(pi*(p1)**2/4)*(p/1728.00)#ft^3
vc=5.70#ft^3
v1=4.70#ft^3
v2=vs/v1#ft^3
v3=0.01273#ft^3
T2=298.00#F

#CALCULATIONS
W=(P2*144.00*v3)/(T2*T1)#lb
T3=((P1*144.0*1.0)/(P2*144.0*7.0)*T2)#Degree C
T4=(P*T3)/P1#Degree C
H=W*Cv*(T4-T3)#C.H.U

#RESULTS
print"The heat supplied during explosion is",round(H,5),"C.H.U"

The heat supplied during explosion is 0.15398 C.H.U

Example 1.9 Page 6

In [37]:
#initialisation of variable
from math import pi,sqrt,acos,asin,atan,cos,sin,tan,log
v=10.0#ft^3
p=100.0#lb/in^2
p1=18.0#lb/in^2
v1=50.0#ft^3
n=log(p/p1)/log(5)
gama=1.40#air

#CALCULATIONS
W=(144.0*(p*v-p1*v1))/(n-1)#ft lb
H=(gama-n)/(gama-1)*W#ft lb
E=W-H#ft lb

#RESULTS
print"The heat supplied and the change of internal energy",round(E,2),"ft lb"

The heat supplied and the change of internal energy 36000.0 ft lb

Example 1.11 Page 7

In [42]:
#initialisation of variable
from math import pi,sqrt,acos,asin,atan,cos,sin,tan,log
v=2.0#ft^3
p=1100.0#lb/in^2
t1=44.0#Degree C
t2=15.0#Degree C
p1=300.0#lb/in^2
t3=3.0#Degree c
Cv=0.17#ft/lb
T=273.0#F
R=96.0#ft lb
p3=300.0#lb/in^2
n=1.12#lb
gama=1.404#lb
W=((144*p*v)/(T+t1))/R#lb

#CALCULATIONS
Wc=W*Cv*(t1-t2)#C.H.U
p2=p*(T+t2)/(T+t1)#lb /in^2
A=(144*p3*v)/(R*276)#lb
W1=(A/W)*v#ft^3
H=((gama-n)/(gama-1))*(144*(p*0.65-p1*v)/(n-1))#ft lb
H1=H/1400#C.H.U

#RESULTS
print"the heat was lost by all the air in the vessel before leakage began",round(Wc,4),"C.H.U"
print"the heat was lost or gainned leakage by the air",round(H1,4),"C.H.U"

the heat was lost by all the air in the vessel before leakage began 51.3218 C.H.U
the heat was lost or gainned leakage by the air 69.2928 C.H.U

Example 1.13 Page 9

In [43]:
#initialisation of variable
from math import pi,sqrt,acos,asin,atan,cos,sin,tan,log
h=0.218#ft^3
h1=0.156#ft^3
n=0.249#lb
h2=0.178#lb
c=0.208#lb
c1=0.162#lb
w1=1.0#ft^3
p=150.0#lb/in^2
T=100.0#Degree C
T1=373.0#F
Cp=(h*0.2312)+(n*0.3237)+(c*0.4451)#C.H.U/lb
Cv=(h1*0.2312)+(h2*0.3237)+(c1*0.4451)#C.H.U//lb
R=1400*(Cp-Cv)#ft lb units

#CALCULATIONS
W=(144*p*w1)/(R*T1)#lb

#RESULTS
print"The characteristic constant of the gas is",round(W,4),"lb"

The characteristic constant of the gas is 0.7157 lb

Example 1.20 Page 12

In [44]:
#initialisation of variable
from math import pi,sqrt,acos,asin,atan,cos,sin,tan,log
T=200.0#Degree C
p=150.0#lb/in^2
v=12.0#ft^3
R=96.0#Lb
T1=473.0#F
T2=273.0#F
j=1400.0#lb
Cv=0.169#lb/in^2
v1=(R*T1)/(p*144)#ft^3

#CALCULATIONS
Fhi=(R/j)*log(v/v1)+Cv*log(T2/T1)#rank

#RESULTS
print"The change of entropy is",round(Fhi,4),"rank"

The change of entropy is 0.0266 rank

Example 1.22 Page 13

In [46]:
#initialisation of variable
from math import pi,sqrt,acos,asin,atan,cos,sin,tan,log
v=10.0#ft^3
T=20.0#Degree C
p=15.0#lb in^2
p1=200.0#lb//in^2
gama=1.41 #lb
Cv=0.169#lb
v2=1.153#ft^3
j=1400.0#lb
T1=293.0#F
T2=451.0#F
T1=((p1*v2)/(p*v))*T1#Degree C

#CALCULATIONS
R=Cv*j*(gama-1)
W=0.816#lb
Fhi=Cv*((gama-1.2)/(1.2-1))*log(T1/T2)*W#rnak

#RESULTS
print"The change of entropy is",round(Fhi,5),"rank"

The change of entropy is -0.00018 rank

Example 1.23 Page 15

In [47]:
#initialisation of variable
from math import pi,sqrt,acos,asin,atan,cos,sin,tan,log
p=1.0#lb
T=200.0#Degree C
p1=15.0#lb/in^2
v1=4.0#ft^3
gama=1.41#lb
Cv=0.169#lb
J=1400.0#lb
n=1.2
T=473.0#F
v2=16.1#ft^3
T1=473.0#F

#CALCULATIONS
T2=(p1*v2)/(p*v1)*T1#Degree C
R=Cv*J*(gama-p)#lb/in^2
Fhi=0.1772*log(1.317)#rank

#RESULTS
print"the change of entropy from intial condition is",round(Fhi,5),"rank"

the change of entropy from intial condition is 0.04879 rank

Example 1.26 Page 16

In [48]:
#initialisation of variable
from math import pi,sqrt,acos,asin,atan,cos,sin,tan,log
w=0.066#ft^3
p=14.7#lb/in^2
w1=14.2#lb/in^2
w2=2780.0#lb/in^2
g=0.038#lb
a=28.9#lb
R=w2/w1#for gas
R1=93.0#for air
T=273.0#F
V=0.4245#ft^3

#CALCULATIONS
W=(p*144*w)/(T*R)#lb
m=(g-W)#lb gas
T2=(V+w)#ft^3

#RESULTS
print"The volume of mixture is",round(T2,4),"ft^3"

The volume of mixture is 0.4905 ft^3