In [1]:

```
import math
# Variables
p = 1.0; # in MPa
p = p * 10**6; # in N per m**2
del_v = 1.5; # in m**3 per min
# Calculations
del_v = del_v * 60; # in m**3 per h
W = p * del_v; # in J
W = W * 10**-6; # in MJ
# Results
print "Work done by the pump upon the water in MJ is : %.0f"%W
```

In [1]:

```
# Variables
# w = 2*g*h
g = 9.81;
m =(0.2+10./1000)*10**3 ; # in gm
s = 1; # in cal/gm°C
del_T = 2; # in °C
# Calculations
H = m * s * del_T; # in cal
H = H * 10**-3; # kcal
J = 4.1868 * 1000;
# W= 2*g*h= J*H
h = J*H/(2 * g); # in m
# Results
print "Height from which the mass should fall in meter is : %.2f"%h
```

In [3]:

```
# Variables
W1 = -25; # in kJ
W2 = 45; # in kJ
Q1 = 65; # in kJ
Q2 = -40; # in kJ
# Calculations
# del_U = Q - W and but for a cycle del_U = 0, So
# Q = W
# Q1 + Q2 +Q3 = W1 +W2
Q3 = W1 + W2 - Q1 - Q2; # in kJ
# Results
print "Third Heat transfer in kJ is ",Q3
print "That is Third Heat transfer is of ",abs(Q3)," kJ from the fluid"
```

In [3]:

```
# Variables
m = 1.5; # in kg
T1 = 90; # in °C
T1 = T1 + 273; #in K
T2 = 225; # in °C
T2 = T2 + 273; # in K
C_p = 0.24;
C_v = 0.17;
# Calculations
Q = (m * C_p * (T2-T1)); # in kcal
del_U = (m * C_v * (T2-T1)); # in kcal
W = Q - del_U; # in kcal
# Results
print "The external work done in kcal is : %.1f"%W
```

In [1]:

```
import math
# Variables
v1 = 0.5; # in m**3
v2 = 0.125; # in m**3
p1 = 1.5; # in bar
p1 = p1 * 10**5; # in N per m**2
p2 = 9.; #in bar
p2 = p2 * 10**5; # in N per m**2
T1 = 100.; # in °C
T1 = T1 + 273; # in K
R = 8.31;
# Calculations and Results
# Formula p1*v1= n*R*T1
n= p1*v1/(R*T1); # in mole
print "Mass of gas in mole is : %.2f"%n
# Part (b)
# Formula p1*v1/T1 = p2*v2/T2
T2 = (p2 * v2 * T1)/(p1 * v1); # in K
print "Temperature at the end of compression in °C is : %.1f"%(T2-273)
# Part (c)
# Formula p1*v1**n = p2*v2**n
n1= math.log(p2/p1)/math.log(v1/v2)
print "Value of index n of compression is :",round(n1,1)
# Part (d)
F = 3;
C_v =1./2*R*F;
del_U = (n * C_v * (T2-T1)); # in J
print "Increase in internal energy of gas in kJ is : %.1f"%(del_U*10**-3)
# Part(e)
Gamma = 1.67;
Q_12 = n*(Gamma-n1)/(1-n1)*R*(T2-T1)/(Gamma-1); # in J
Q_12 = Q_12 * 10**-3; # in kJ
print "Heat interaction in kJ is : %.2f"%Q_12
if Q_12<0:
print "The -ve sign indicates heat rejection during the process"
# Note: There is some difference between the answer of book and coding . Both the answer is right but accurate answer is of coding.
# Because in the book, the intermediate values taken are appox. and in the coding the values taken are accurate
```

In [10]:

```
import math
# Variables
p1 = 0.01; # in N/mm**2
p1 = p1 * 10**3; # in kN/m**2
p2 = 50.; # in kN/m**2
v1 = 5.; # in m**3
v2 = 1.5; # in m**3
Gamma = 1.4;
# Calculations and Results
# Formula p1*v1**n = p2*v2**n
n= round(math.log(p2/p1)/math.log(v1/v2),2)
print "Part (a) The value of n is : %.2f"%n
print ("The process followed during air compression is POLYTROPIC");
# Part (b)
print "Part (b) The law of the process is %.2f p*v**"%n," = constant"
# Part (c)
W= (p1*v1-p2*v2)/(n-1); # in kNm or (kJ)
print "Part (c) Work done during the process in kJ is : %.1f"%W
print ("The -ve sign indicates that the work has been done on the system")
# Part (d)
Q = ((Gamma - n)/(Gamma - 1) * W); # in kJ
print "Part (d) Heat transfer during the process in kJ is : %.3f"%Q
print ("The -ve sign indicates that the heat is rejected from the system")
```

In [1]:

```
from scipy.integrate import quad
# Variables
# Relation of specific internal energy of the gas
# U= 1.5*p*v-85 kJ/kg
p1 = 1000; # in kpa
p2 = 200; # in pa
v1 = 0.20; # in m**3
v2 = 1.20; # in m**3
m = 1.5; # in kg
# Calculations and Results
U1= 1.5*p1*v1-85; # kJ/kg
U2= 1.5*p2*v2-85; # kJ/kg
delU= U2-U1; # in kJ
print "Change in internal energy in kJ is",delU
# p1= a+b*v1 (i)
# p2= a+b*v2 (ii)
# From eq(i) and (ii)
b= (p1-p2)/(v1-v2); # in kN/m**2
a= p1-b*v1; # in kN/m**2
print "The value of a in kN/m**2 is ",a
print "The value of b in kN/m**2 is ",b
# Part (c)
# Work done = integration of p w.r.t. v and p = a+b*v1
def f5(v):
return a+b*v
W= quad(f5,v1,v2)[0]
print "Work done in kJ is :",W
# Part (d)
Q= delU+W; # in kJ
print "The net heat transfer in kJ is : ",Q
```

In [2]:

```
# Variables
a= 1160.; # in kN/m**2
b= -800.; # in kN/m**2
# Calculations
v= -a/(2*b)
Umax= 1.5*(a*v+b*v**2)-85; # in kJ/kg
# For 1.5 kg mass of gas it is
Umax= Umax*1.5; # in kJ/kg
# Results
print "The maximum internal energy of the gas in kJ/kg is : ",Umax
```

In [8]:

```
# Exa 4.11
import math
# Variables
T1 = 127.; # in °C
T1 = T1 + 273; # in K
R = 287.;
V1 = 300.; # in m/s
p1 = 2.; # in MPa
p2 = 0.5; # in MPa
p1 = p1 * 10**6; # in Pa
p2 = p2 * 10**6; # in Pa
C_P = 1.005*10**3; # in J/ kg-K
Gamma = 1.4;
# Calculations and Results
V2 = math.sqrt(2 * C_P *T1 *(1-(p2/p1)**((Gamma-1)/Gamma)) + V1**2); # in m/s
print "The exit velocity of air in m/s is : %.3f"%V2
m = 600.; # in kg/hr
m = m / 3600; # in kg/sec
v1 = (R * T1)/p1; # in m**3 per kg
# m = (A1*V1)/v1 = (A2* V2)/v2
A1 = (m * v1)/V1; # in m**2
A1 = A1 * 10**6; # in mm**2
print "Inlet area of the nozzle in square milimeter is : %.2f"%A1
T2 = T1*(p2/p1)**((Gamma-1)/Gamma); # in K
v2 = (R * T2)/(p2); # in m**3/kg
A2 = (m * v2)/V2; # in m**2
A2 = A2 * 10**6; # in mm**2
print "Exit area of nozzel in square milimeter is : %.2f"%A2
```

In [13]:

```
# Variables
W = -1; # in kWh
W = W * 10**3 * 3600; # in J
del_U = -5000; # in kj
# Calculations
del_U = del_U * 10**3; # in J
Q = del_U + W; # in J
Q = Q * 10**-6; # in MJ
# Results
print "Net heat transfer for the system in MJ is : ",Q
```

In [2]:

```
# Variables
Q_acb = 84; #in kJ
W_acb = 32; # in kJ
# Calculations and Results
#Formula Q_acb = del_U+W_acb where del_U = U_b - U_a;
del_U = Q_acb - W_acb; # in kJ
# Part (a) Path a b d
W_abd = 10.5; # in kJ
Q_abd = del_U + W_abd; # in kJ
print "Heat flows into the system along the path a b d in kJ is : ",Q_abd
# Part (b) curved path b a
W_ba = -(21); # in kJ
Q_ba = -(del_U) + W_ba; # in kJ
print "Heat liberated by the system in kJ is : ",(Q_ba)
# Part (c) process a b and d b
W_ad = 10.5; # in kJ
del_U1 = 42; # in kJ
Q_ad = del_U1 + W_ad; # in kJ
print "Heat absorbed in processes ad in kJ is : ",Q_ad
W_db = -(42); # in kJ
del_U2 = 52; # in kJ
Q_bd = del_U2 + W_db; # in kJ
print "Heat absorbed in processes bd in kJ is : ",Q_bd
W_db = 0;
W_abd = W_ad + W_db; # in kJ
print "Heat absorbed in processes in ad and db in kJ is : ",W_abd
```

In [11]:

```
# Variables
v1 = 5.; # in m**3
p1 = 2.; # in bar
p2 = 6.; # in bar
p3 = 2.; # in bar
# Calculations
p1 = p1 * 10**5; # in N/m**2
p2 = p2 * 10**5; # in N/m**2
p3 = p3 * 10**5; # in N/m**2
n = 1.3;
v2 = v1 * ((p1/p2)**(1/1.3)); # in m**3
W1_2 = ((p2 * v2)-(p1 * v1))/(1-n); # in J
Gamma = 1.4;
v3 = v2 * ((p2/p3)**(1/Gamma)); # in m**3
W2_3 = ((p3 * v3) - (p2 * v2))/(1-Gamma); # in J
W_net = W1_2 + W2_3; # in J
W_net = W_net * 10**-3; # in kJ
# Results
print "net work done in kJ is : %.2f"%W_net
```