# Chapter 2 : First Law of Thermodynamics¶

## Example 2.1 Page No : 6¶

In [1]:

# Variables :
mc = 10.;			#Kg
Cpc = 0.4;			#KJ/KgK
Cpw = 4.187;			#KJ/KgK(Specific heat of water)
Vw = 0.35;			#m**3
density_water = 1000;			#Kg/m**3
mw = Vw*density_water;			#Kg

# Calculations
#mc*Cpc*(tc-t) = mw*Cpw*(t-tw)

# Results
print "Equillibrium temperature in degree_centigrade : %.4f"%t

Equillibrium temperature in degree_centigrade : 30.1633


## Example 2.2 Page No : 7¶

In [3]:

# Variables
Q1 = 2500.;			#KJ/Kg
Q2 = 1800.;			#KJ/Kg
Pdev = 210.;			#MW

# Calculations
#Power developed  =  Heat transfered: Pdev = m*(Q1-Q2)
m = Pdev*1000/(Q1-Q2);			#mass flow rate of steam in Kg/s

# Results
print "Mass flow rate of steam in Kg/s : ",m

Mass flow rate of steam in Kg/s :  300.0


## Example 2.3 Page No : 7¶

In [1]:

# Variables
WA = 20;			#KJ
QA = 15;			#KJ
QB = 10;			#KJ

# Calculations
U2subU1 = QA-WA;			#change in internal energy in KJ
WB = QB - U2subU1
QB = -10
WB = -15
QA = 15
WA = 20

dQ = QA + QB
dW = WA + WB

# Results
print "Change in internal energy in KJ : ",U2subU1
print "Workdone in process : %d KJ"%WB
print "d'Q = Qa + Qb = %d kJ"%dQ
print "d'W = Wa + Wb = %d kJ"%dW

Change in internal energy in KJ :  -5
Workdone in process : -15 KJ
d'Q = Qa + Qb = 5 kJ
d'W = Wa + Wb = 5 kJ


## Example 2.4 Page No : 8¶

In [5]:

# Variables
Q1 = 120.;			#KJ
Q2 = -16;			#KJ
Q3 = -48;			#KJ
Q4 = 12;			#KJ
W1 = 60000;			#N-m
W2 = 68000;			#N-m
W3 = 120000;			#N-m
W4 = 44000;			#N-m

# Calculations
Net_work = Q1+Q2+Q3+Q4;			#KJ

# Results
print "Net Work in N-m :  ",Net_work*1000
print ("Option (ii) is true.")

Net Work in N-m :   68000.0
Option (ii) is true.


## Example 2.5 Page No : 8¶

In [6]:
import scipy

# Variables
T1 = T1+273;			#kelvin
T2 = T2+273;			#kelvin

delQbydelT = 1.005;			#KJ/k

# Calculations
#delWbydelT = (4-0.12*T);			#KJ/k
def f12(T):
return 1.005

def f13(T):
return 4-0.12*T

U2subU1 = Q-W;			#change in internal energy in KJ

# Results
print "Change in internal energy in KJ : ",U2subU1

Change in internal energy in KJ :  4776.5


## Example 2.6 Page No : 9¶

In [7]:

# Variables
m = 20.;			#Kg
mw = 200.;			#Kg
Z1 = 15.;			#m
Z2 = 0.;			#m
g = 9.81;			#gravity consmath.tant

print ("(i) Stone is about to enter the water");
deltaPE = m*g*(Z2-Z1)/1000;			#KJ
Q = 0;			#Heat Transfer
W = 0;			#Work Transfer
deltaE = Q-W;			#Energy Transfer
#deltaE = deltaU+deltaKE+deltaPE
deltaU = 0;			#no change in temperature
deltaKE =  deltaE-deltaU-deltaPE;			#KJ
print "deltaU in KJ : ",deltaU
print "deltaPE in KJ : ",deltaPE
print "deltaKE in KJ : ",deltaKE
print "Q in KJ : ",Q
print "W in KJ : ",W

print ("(ii) Stone has come to rest near the math.tank.");
Q = 0;			#Heat Transfer
W = 0;			#Work Transfer
deltaE = Q-W;			#Energy Transfer
deltaKE = 0;			#rest condition

#deltaE = deltaU+deltaKE+deltaPE
deltaU =  deltaE-deltaKE-deltaPE;			#KJ
print "deltaU in KJ : ",deltaU
print "deltaPE in KJ : ",deltaPE
print "deltaKE in KJ : ",deltaKE
print "Q in KJ : ",Q
print "W in KJ : ",W

print ("(iii) Heat is transfered to surroundings.");
deltaKE = 0;			#Energy Transfered to water
deltaPE = 0;
W = 0;
deltaE = deltaU+deltaKE+deltaPE
Q = deltaE+W;			#KJ
print "deltaU in KJ : ",deltaU
print "deltaPE in KJ : ",deltaPE
print "deltaKE in KJ : ",deltaKE
print "Q in KJ : ",Q
print "W in KJ : ",W

(i) Stone is about to enter the water
deltaU in KJ :  0
deltaPE in KJ :  -2.943
deltaKE in KJ :  2.943
Q in KJ :  0
W in KJ :  0
(ii) Stone has come to rest near the math.tank.
deltaU in KJ :  2.943
deltaPE in KJ :  -2.943
deltaKE in KJ :  0
Q in KJ :  0
W in KJ :  0
(iii) Heat is transfered to surroundings.
deltaU in KJ :  2.943
deltaPE in KJ :  0
deltaKE in KJ :  0
Q in KJ :  2.943
W in KJ :  0


## Example 2.7 Page No : 10¶

In [8]:

# Variables
SigmaW = 30;			#KJ
n = 10;			#cycles/min
Q1_2 = 50;			#KJ
#Q2_3 = 0;			#KJ
#Q3_1 = 0;			#KJ
#W1_2 = 0;			#KJ
W2_3 = 30;			#KJ
#W3_1 = 0;			#KJ
deltaU1_2 = 20;			#KJ
deltaU2_3 = -10;			#KJ

# Calculations and Results
#deltaU3_1 = 0;			#KJ
#Q-W = deltaU
#For Proess 1-2 :
W1_2 = Q1_2-deltaU1_2;			#KJ
print "W1-2 in KJ : ",W1_2
#For Proess 2-3
Q2_3 = W2_3+deltaU2_3;			#KJ
print "Q2-3 in KJ :  ",Q2_3
#For Proess 3-1
W3_1 = SigmaW-W1_2-W2_3;			#KJ
print "-1 in KJ : ",W3_1

SigmaQ = SigmaW;			#KJ
Q3_1 = SigmaQ-Q1_2-Q2_3;			#KJ
print "Q3-1 in KJ : ",Q3_1

deltaU3_1 = Q3_1-W3_1;			#KJ
print "U1-U3 or deltaU3-1 in KJ : ",deltaU3_1

RateOfWork = SigmaW*n;			#KJ/min
RateOfWork = RateOfWork/60;			#KJ/sec or KW
print "Rate of work in KW : ",RateOfWork

W1-2 in KJ :  30
Q2-3 in KJ :   20
-1 in KJ :  -30
Q3-1 in KJ :  -40
U1-U3 or deltaU3-1 in KJ :  -10
Rate of work in KW :  5


## Example 2.8 Page No : 11¶

In [9]:

# Variables :
m = 50.;			#Kg
C1 = 10.;			#m/s
C2 = 30.;			#m/s
Z2subZ1 = 40.;			#m
Q = 30000.;			#J
W1 = -4500.;			#J
W2 = 0.002;			#KWh
g = 9.81;			#gravity constant
W2 = W2*3600.*1000;			#J

# Calculations
#sigmaQ-sigmaW = E2-E1 = (U2-U1)+(C2**2-C1**2)/2+g*(Z2-Z1)
U2subU1 = Q-(W1+W2)-(C2**2-C1**2)/2-g*(Z2subZ1);			#J

# Results
print "Change in Internal energy in J : ",U2subU1

Change in Internal energy in J :  26507.6


## Example 2.9 Page No : 11¶

In [10]:

# Variables
deltaU = -4000.;			#KJ
W = -1.2;	    		#KWh

# Calculations
W = W*3600.;			#KJ
Q = W+deltaU;			#KJ/hr

# Results
print "Net heat transfer in KJ/hr : ",Q

Net heat transfer in KJ/hr :  -8320.0


## Example 2.10 Page No : 11¶

In [3]:

# Variables
mw = 100;			#Kg
T = 30;			#min
T = T*60;			#sec
P = 1;			#KW
Q = -50;			#KJ
Sw = 4.19;			#KJ/KgK(Specific heat of water)
W = -P*T;			#KJ

# Calculations and Results
#Q = W+deltaU
deltaU = Q-W;			#KJ
print "Chnge in internal energy in  kJ : ",deltaU
delta_t = deltaU/mw/Sw;			#sec
print "Rise in temperature in degree C : %.3f"%delta_t

Chnge in internal energy in  kJ :  1750
Rise in temperature in degree C : 4.057


## Example 2.11 Page No : 12¶

In [12]:

# Variables
V = 12.;			#Volt
I = 6.;			#Ampere
t = 1.5;			#hr
t = t*3600.;			#sec
deltaU = -750.;			#KJ

# Calculations
W = V*I*t/1000;			#KJ
Q = W+deltaU;			#KJ

# Results
print "Heat transfer in KJ  : ",Q

Heat transfer in KJ  :  -361.2


## Example 2.12 Page No : 13¶

In [2]:

# Variables
Q = 82.;			#KJ
p1 = 4.;			#bar
m = 1.; 			#Kg
V1 = 0.21;			#m**3
R = 300.;			#Nm/KgK
W = 0.; 			#because V is consmath.tant.

# Calculations and Results
print "Work done in KJ : ",W

#Q-W = deltaU
deltaU = Q-W;			#KJ
print "Change in internal energy in KJ  : ",deltaU

#p1*V1 = m*R*T1
T1 = p1*10**5*V1/m/R;			#kelvin
Cv = deltaU/delta_t;			#KJ/KgK
print "Specific Heat in KJ/KgK  : %.3f"%Cv

Work done in KJ :  0.0
Change in internal energy in KJ  :  82.0
Specific Heat in KJ/KgK  : 0.683


## Example 2.13 Page No : 14¶

In [3]:

# Variables :
V1 = 250.;			#litres
V2 = 250.;			#litres
p1 = 3.;			#Mpa
p2 = 1.8;			#Mpa
Gamma = 1.4;			#
rho = 1.43;			#Kg/m**3
p = 0.1013;			#Mpa

# Calculations and Results
V1 = V1/1000;			#m**3
V2 = V2/1000;			#m**3
T1 = t1+273;			#Kelvin
T2 = t2+273;			#Kelvin

#p = rho*R*T
T = 0+273;			#Kelvin
R = p*10**6/rho/T;			#Nm/KgK

#p*V = m*R*T
m1 = p1*10**6*V1/R/T1;			#Kg
m2 = p2*10**6*V2/R/T2;			#Kg
Mass_oxygen = m1-m2;			#Kg
print "Mass of oxygen used in Kg : %.3f"%Mass_oxygen

#Cv*(Gamma-1) = R
Cv = R/(Gamma-1);			#Nm/KgK
Q = m2*Cv*(t1-t2);			#J
print "Heat transfered in J : %.1f"%Q

Mass of oxygen used in Kg : 3.864
Heat transfered in J : 15570.9


## Example 2.14 Page No : 16¶

In [4]:

# Variables :
m = 50./1000;			#Kg
t_heating = 300.;			#sec
Pheater = 10.04;			#Watts
Gamma = 1.4;

Q = Pheater*t_heating;			#J
#Q = m*Cp*(t2-t1)
Cp = Q/m/(t2-t1);			#J/KgK
print "Specific heat of air in J/KgK : ",Cp

#Cp*(1-1/Gamma) = R
R = Cp*(1-1/Gamma);			#Gas Constant in Nm/KgK
print "Gas constant of air in Nm/KgK : %.2f"%R

#p = rho*R*T
p = 0.1;			#Mpa
T = 0+273;			#kelvin
rho = p*10**6/R/T;			#Kg/m**3
print "Density of air in Kg/m**3 : %.4f"%rho

Specific heat of air in J/KgK :  1004.0
Gas constant of air in Nm/KgK : 286.86
Density of air in Kg/m**3 : 1.2769


## Example 2.15 Page No : 17¶

In [5]:

# Variables :
m = 1.;			#Kg
V1 = 0.3;			#m**3
p = 3.2*100;			#Kpa
p1 = 3.2*100;			#Kpa
p2 = 3.2*100;			#Kpa
V2 = 2.*V1;			#m**3
Cp = 1.003;			#KJ/KgK
R = 0.2927;			#KJ/kgK

# Calculations and Results
#p*V = m*R*T
T1 = p1*V1/m/R;			#kelvin
T2 = p2*V2/m/R;			#kelvin
Q = m*Cp*(T2-T1);			#KJ
print "Heat Added in KJ : %.2f"%Q
W = p*(V2-V1);			#KJ

print "Work done in KJ : ",W
print "Initial temperature of air in kelvin : ",round(T1)
print "Final temperature of air in kelvin : ",round(T2)

Heat Added in KJ : 328.96
Work done in KJ :  96.0
Initial temperature of air in kelvin :  328.0
Final temperature of air in kelvin :  656.0


## Example 2.16 Page No : 18¶

In [5]:
import scipy

# Variables :
p = 105;			#Kpa
p1 = 105;			#Kpa
p2 = 105;			#Kpa
V1 = 0.25;			#m**3
V2 = 0.45;			#m**3
T1 = 10+273;			#kelvin
T2 = 240+273;			#kelvin

# Calculations and Results
def f5(T):
return 0.4+18/(T+40)

print "Heat Transfer in KJ : %.2f"%Q
W = p*(V2-V1);			#KJ
print "Work Transfer in KJ : ",W
deltaU = Q-W;			#KJ
print  "in internal energy in KJ L : %.2f"%deltaU
deltaH = Q;			#KJ
print "Change in enthalpy in KJ : %.2f"%deltaH

Heat Transfer in KJ : 101.68
Work Transfer in KJ :  21.0
in internal energy in KJ L : 80.68
Change in enthalpy in KJ : 101.68


## Example 2.17 Page No : 18¶

In [8]:
import math

# Variables :
N = 250.;			#rpm
tau = 10.;			#min
Q1 = -5.;			#KJ
deltaU = 2.;			#KJ
p = 1.2;			#bar
p = p*100.;			#KJ
E = 24.;			#volt
I = 0.45;			#Ampere
A = 0.1;			#m**2
T = 0.5;			#Nm
Q2 = E*I*tau*60./1000;			#KJ
Q = Q1+Q2;			#KJ

# Calculations
#Consider piston moves through a distance y
#Q-(W1+W2) = deltaU where W1 = p*A*y
W2 = -T*2*math.pi*N*tau;			#Nm
W2 = W2/1000;			#KJ
y = (Q-W2-deltaU)/A/p;			#meter

# Results
print "Distance in cm : %.2f"%(y*100)

#Ans is wrong in the book.

Distance in cm : 61.12


## Example 2.18 Page No : 20¶

In [10]:
import math

# Variables :
m = 0.8;			#Kg
p1 = 1.;			#bar
p2 = 5.;			#bar
T1 = 25.+273;			#kelvin
R = 287.;			#KJ/kgK

# Calculations and Results
W = m*R*T1*math.log(p1/p2);			#J
print "Work done in KJ : %.3f"%(W/1000)

U2subU1 = 0;			#change in internal energy
Q = W+U2subU1;			#J
print "Heat Transfer in KJ : %.3f"%(Q/1000)

Work done in KJ : -110.119
Heat Transfer in KJ : -110.119


## Example 2.19 Page No : 21¶

In [11]:
import math

# Variables :
m = 1.;			#Kg
p1 = 100.;			#Kpa
T1 = 300.;			#kelvin
V_ratio = 1./2;			#V2/V1
T = 1.;			#Nm
tau = 1.;			#hr
tau = tau*60;			#min
N = 400.;			#rpm
R = 0.287;			#KJ/kgK

# Calculations
W1 = m*R*T1*math.log(V_ratio);			#KJ
W2 = -T*2*math.pi*N*tau/1000;			#KJ
W = W1+W2;			#KJ

# Results
print "Net work transfer in KJ : %.2f"%W

Net work transfer in KJ : -210.48


## Example 2.20 Page No : 26¶

In [6]:

# Variables :
m = 2.;			#Kg
T1 = 125.+273;			#kelvin
T2 = 30.+273;			#kelvin
W = 152.;			#KJ
deltaH = -212.8;			#KJ
Q = 0;			#KJ(For adiabatic process)

# Calculations and Results
#Q = W+m*Cv*(T2-T!)
Cv = (Q-W)/m/(T2-T1);			#KJ/KgK
print "Specific heat at constant volume in KJ/KgK : %.1f"%Cv

#deltaH = m*Cp*(T2-T1);
Cp = deltaH/m/(T2-T1);			#KJ/KgK
print "Specific heat at cinstant pressure in KJ/KgK : ",Cp
R = Cp-Cv;			#KJ/KgK
print "Characteristic gas constant in KJ/KgK : ",R

Specific heat at constant volume in KJ/KgK : 0.8
Specific heat at cinstant pressure in KJ/KgK :  1.12
Characteristic gas constant in KJ/KgK :  0.32


## Example 2.21 Page No : 26¶

In [7]:
import math
# Variables :
V1 = 0.5;			#m**3
p1 = 1.5;			#bar
T1 = 100+273;			#kelvin
V2 = 0.125;			#m**3
p2 = 9;			#bar
R = 287;			#KJ/KgK

# Calculations and Results
m = p1*10**5*V1/R/T1;			#Kg
print "Mass of air in Kg : %.4f"%m

#p1*V1**n = p2*V2**n
n = math.log(p2/p1)/math.log(V1/V2);			#
print "Value of index : %.4f"%n

W = (p1*V1-p2*V2)*10**5/(n-1);	    		#Nm
print "Work done in KJ : %.4f"%(W/1000)


Mass of air in Kg : 0.7006
Value of index : 1.2925
Work done in KJ : -128.2133


## Example 2.22 Page No : 27¶

In [8]:
import scipy

# Variables :
p1 = 1.;			#bar
V1 = 0.14;			#m**3
V2 = 0.07;			#m**3
R = 287.;			#KJ/KgK

# Calculations and Results
#p*V = R*k1*V**(-2/5) or p*V**(7/5) = K
K = p1*10**5*V1**(7./5);			#Nm/Kg

def f17(V):
return K*V**(-7./5)

print "Work done in Nm : %.1f"%W
p2 = K*V2**(-7./5);			#N/m**2
p2 = p2/10**5;			#bar
print "Final pressure in bar : %.2f"%p2

#Ans in the book is wrong.

Work done in Nm : -11182.8
Final pressure in bar : 2.64


## Example 2.23 Page No : 28¶

In [16]:

# Variables :
m = 2.;			#Kg
Q = 0.;			#KJ(because of adiabatic process)
p1 = 1.;			#Mpa
p1 = p1*10.**6/1000;			#Kpa
T1 = t1+273;			#kelvin
p2 = 100.;			#Kpa
n = 1.2;
R = 0.196;			#KJ/KgK

# Calculations and Results
T2 = T1*(p2/p1)**((n-1)/n);			#kelvin
u1 = 196+0.718*t1;			#KJ
u2 = 196+0.718*t2;			#KJ
deltau = u2-u1;			#KJ
deltaU = m*deltau;			#KJ
print "Change in internal energy in KJ : %.1f"%deltaU
W = Q-deltaU;			#KJ
print "Work transfer in KJ : %.1f"%W

W1 = m*R*(T1-T2)/(n-1);			#KJ
print "Displacement work in KJ : %.2f"%W1

Change in internal energy in KJ : -216.5
Work transfer in KJ : 216.5
Displacement work in KJ : 295.47


## Example 2.24 Page No : 29¶

In [8]:
import scipy

# Variables :
m = 1.5;			#Kg
V1 = 0.06;			#m**3
p1 = 5.6*10;			#Kpa
T2 = t2+273;			#kelvin
a = 0.946;
b = 0.662;
K = 10.**-4;

# Calculations
#p*V = m*R*T = m*(a-b)*T
T1 = p1*10**5*V1/m/(a-b)/1000;			#Kelvin

def f28(T):
return m*(b+K*T)

Q = 0;			#isentropic process
W = Q-U2subU1;			#KJ

# Results
print "Work done in KJ : %.1f"%W

#Answer in the book is wrong.

Work done in KJ : 300.7


## Example 2.25 Page No : 30¶

In [21]:
import math
from numpy import *

# Variables :
m = 1.5;			#Kg
p1 = 1000;			#Kpa
p2 = 200;			#Kpa
V1 = 0.2;			#m**3
V2 = 1.2;			#m**3
#p = a+b*v
#solving for a and b by matrix
A = array([[1, V1],[1, V2]]);
B = array([p1,p2]);
X = linalg.solve(A,B)
a = X[0]
b = X[1]

def f16(V):
return a+b*V

print "Work transfer in KJ/Kg : ",W

u2SUBu1 = (1.5*p2*V2+35)-(1.5*p1*V1+35);			#KJ/Kg
print "Change in internal energy in KJ/Kg : ",u2SUBu1

q = W+u2SUBu1;			#KJ/Kg
print "Heat transfer in KJ/Kg : ",q

#u = 1.5*(a+b*V)*V+35;
#1.5*a+2*V*1.5*b = 0;			#for max value putting du/dV = 0
V = -1.5*a/2/1.5/b;			#m**3/Kg
p = a+b*V;			        #KPa
u_max = 1.5*p*V+35;			#KJ/Kg
print "Maximum internal energy in KJ/Kg : ",u_max

#Answer in the book is wrong because a is 1160 instead of 1260.

Work transfer in KJ/Kg :  600.0
Change in internal energy in KJ/Kg :  60.0
Heat transfer in KJ/Kg :  660.0
Maximum internal energy in KJ/Kg :  665.75


## Example 2.26 Page No : 31¶

In [19]:
import math

# Variables :
V1 = 5.;			#m**3
p1 = 2.;			#bar
T1 = t1+273;			#kelvin
p2 = 6.;			#bar
p3 = p1;			#bar
R = 287.;			#KJ/KgK
n = 1.3;

# Calculations
#p*V**(1/3) = C
V2 = V1*(p1/p2)**(1/1.3);			#m**3
#p*V = m*R*T1
m = p1*10**5*V1/R/T1;			#Kg
W1_2 = 10**5*(p1*V1-p2*V2)/(n-1);			#Nm
W1_2 = W1_2/1000;			#KJ
Gamma = 1.4;			#for air
#p*V**Gamma = C
V3 = (p2/p3)**(1/Gamma)*V2;			#m**3
W2_3 = 10**5*(p2*V2-p3*V3)/(Gamma-1);			#Nm
W2_3 = W2_3/1000;			#KJ
W = W1_2+W2_3;			#KJ

# Results
print "Net work done in KJ : %.2f"%W

Net work done in KJ : -94.02


## Example 2.27 Page No : 32¶

In [30]:

# Variables :
Q1_2 = 85;			#KJ
Q2_3 = -90;			#KJ
W2_3 = -20;			#KJ
W1_2 = 0;			#consmath.tant volume process

# Calculations
W3_1 = Q1_2+Q2_3+Q3_1-W1_2-W2_3;			#KJ

# Results
print "Direction is 3-1 and work in KJ : ",W3_1

Direction is 3-1 and work in KJ :  15


## Example 2.28 Page No : 33¶

In [22]:

# Variables :
V1 = 200./1000;			#m**3
p1 = 4.;	    		#bar
T1 = 400.;	    		#K
p2 = 1.;	    		#bar
H3subH2 = 72.;			#KJ
Cp = 1.		       	    #KJ/KgK
Cv = 0.714;		    	#KJ/KgK

# Calculations and Results
Gamma = Cp/Cv;
R = Cp-Cv;			    #KJ/KgK
#p*V = m*R*T
m = p1*10**5*V1/R/1000/T1       			#Kg
T2 = T1*(p2/p1)**((Gamma-1)/Gamma);			#K
V2 = p1*V1/T1*T2/p2;			            #m**3
W1_2 = m*R*(T1-T2)/(Gamma-1);			    #KJ
print "Work done W1-2 in KJ : %.1f"%W1_2

#H3subH2 = m*Cp(T3-T2);
T3 = (H3subH2+m*Cp*T2)/m/Cp;			#K
W2_3 = m*R*(T3-T2);		            	#KJ
W = W1_2+W2_3;			                #KJ
print "Workdone in KJ : %.0f"%W

#W = m*R*(T1-T3)/(n-1)
n = m*R*(T1-T3)/W+1;			#
print "Index of expansion : %.5f"%n

Work done W1-2 in KJ : 65.4
Workdone in KJ : 86
Index of expansion : 1.06507


## Example 2.29 Page No : 34¶

In [9]:
from scipy.integrate import quad
from numpy import *

# Variables :
p1 = 10.;			#bar
p2 = 2.;			#bar
V1 = 0.1;			#m**3
V2 = 0.9;			#m**3
R = 300.;			#Nm/Kg-K
m = 1.; 			#Kg

# Calculations
#p = a*v+b
#solving for a and b by matrix
A = array([[V1, 1],[V2, 1]]);
B = array([p1,p2]);
X = linalg.solve(A,B);
a = X[0]
b = X[1]

#p = a*v+b = a*R*T/p+b
#2*p-b = 0;			    #on differentiating
p = b/2;			    #bar
#p = a*v+b
v = (p-b)/a;			    #m**3/Kg
T = p*10**5*v/R;			#K
print "Maximum temperature in K : %.2f"%T

def f23(v):
return (a*v+b)*10**5

W = W/10**3;			    #KJ/KgK
print "Work done in KJ : %.2f"%W

T1 = p1*10**5*V1/R;			#K
T2 = p2*10**5*V2/R;			#K
Gamma = 1.4;
Cv = R/(Gamma-1);			#Nm/KgK
Cv = Cv/1000;			    #KJ/KgK
deltaU = m*Cv*(T2-T1);		#KJ/Kg
Q = W+deltaU;			    #KJ
print "Net Heat transfer in KJ : %.2f"%-Q

Maximum temperature in K : 1008.33
Work done in KJ : 480.00
Net Heat transfer in KJ : -680.00


## Example 2.29 Page No : 36¶

In [24]:

# Variables :
m = 5.;			#Kg
#u = 3.62*p*v
p1 = 550.;			#KPa
p2 = 125.;			#KPa
V1 = 0.25;			#m**3

# Calculations and Results
#p*V**(1/2) = C
n = 1.2;
V2 = (p1/p2)**(1/n)*V1;			#m**3/Kg
W = (p1*V1-p2*V2)*10**5/(n-1)/1000;			#KJ
delta_u = (3.62*p2*V2)-(3.62*p1*V1);			#KJ/Kg
deltaU = m*delta_u;		    	#KJ
print "Change in internal energy in KJ : %.4f"%deltaU

Q = W+deltaU;	        		#KJ
Q = Q/1000;			            #MJ
print "Heat transfer in MJ : %.3f"%Q

Change in internal energy in KJ : -544.5580
Heat transfer in MJ : 14.498


## Example 2.30 Page No : 44¶

In [26]:
import math

# Variables :
Vdot = 0.032;			#m**3/s
d = 1.5;			#m
L = 4.2;			#m
m = 3500.;			#Kg

# Calculations and Results
V = math.pi/4*d**2*L;			#m**3
rho = m/V;			#Kg/m**3
print "Density of liquid in Kg/m**3 : %.1f"%rho

m_dot = rho*Vdot;			#Kg/s
print "Mass flow rate in Kg/s : %.2f"%m_dot

Density of liquid in Kg/m**3 : 471.6
Mass flow rate in Kg/s : 15.09


## Example 2.31 Page No : 44¶

In [28]:
import math

# Variables :
p1 = 1.;    			#bar
T1 = 20.+273;			#K
p2 = 6.;		    	#bar
m = 1.;		    	    #Kg
R = 287.;	    		#Nm/Kg
Gamma = 1.4;
Cp = 1.005;		    	#KJ/KgK
Cv = 0.7175;			#KJ/KgK

#T2 = T1 : Isothermal compression
T2subT1 = 0;
deltaU = m*Cv*(T2subT1);    			#KJ
print ("Isothermal :");
print "Change in internal energy in KJ : ",deltaU

Wsf = m*R/1000*T1*math.log(p1/p2);			#KJ/Kg
print "Work done in KJ/Kg : %.2f"%Wsf

p2V2subp1V1 = 0;        			#isothermal process
Q = Wsf+deltaU+p2V2subp1V1;			#KJ/Kg
print "Heat transfer in KJ/Kg : %.2f"%Q
print ("Isentropic :");

T2 = T1*(p2/p1)**((Gamma-1)/Gamma);			#K
U2subU1 = m*Cv*(T2-T1);			#KJ/Kg
print "Change in internal energy in KJ/Kg : %.2f"%U2subU1

H2subH1 = m*Cp*(T2-T1);			#KJ/Kg
print "Change in heat in KJ/Kg : %.2f"%H2subH1

print "Heat transfer in KJ/Kg : %.2f"%Q

Wsf = Q-H2subH1;			#KJ/Kg
print "Work done in KJ/Kg : %.2f"%Wsf
print ("Polytropic : ");

n = 1.25;			                #index
T2 = T1*(p2/p1)**((n-1)/n);			#K
deltaU = m*Cv*(T2-T1);		    	#KJ/Kg
print "Change in internal energy in KJ/Kg : %.2f"%deltaU

H2subH1 = m*Cp*(T2-T1);			    #KJ/Kg
Wsf = (n/(n-1))*m*R/1000*(T1-T2);			#KJ/Kg
print "Work done in KJ/Kg : %.2f"%Wsf

Q = Wsf+H2subH1;			#KJ/Kg
print "Heat transfer in KJ/Kg : %.2f"%Q

#Answer of chane in internal energy for last part is wrong in the book.

Isothermal :
Change in internal energy in KJ :  0.0
Work done in KJ/Kg : -150.67
Heat transfer in KJ/Kg : -150.67
Isentropic :
Change in internal energy in KJ/Kg : 140.54
Change in heat in KJ/Kg : 196.85
Heat transfer in KJ/Kg : 0.00
Work done in KJ/Kg : -196.85
Polytropic :
Change in internal energy in KJ/Kg : 90.60
Work done in KJ/Kg : -181.20
Heat transfer in KJ/Kg : -54.30


## Example 2.32 Page No : 46¶

In [38]:
import scipy

# Variables :
p1 = 5.;			#bar
p2 = 50.;			#bar
V = 0.001;			#m**3/Kg
m_dot = 10.;			#Kg/s

# Calculations
def f20(p):
return -V

wsf = wsf/1000      			#KJ/Kg
Wsf = abs(wsf)*m_dot;			#KW(leaving -ve sign as it is to indiacte heat is supplied)

# Results
print "Power required in KW : ",Wsf

Power required in KW :  45.0


## Example 2.33 Page No : 46¶

In [29]:
import math

# Variables :
p1 = 10.**5;			#Pa
p2 = 5.*10**5;			#Pa
T1 = 25.+273;			#K
V1 = 1.8;			#m**3/Kg

# Calculations and Results
V2 = p1/p2*V1;			#m**3/Kg
W = -p1*V1*math.log(p2/p1);			#J/kg
W = W/1000.;			#KJ/Kg
print "Workdone in KJ : %.1f"%W

deltaU = 0;			#As in a isothermal process T2-T1  = 0
print "Change in internal energy in KJ : ",deltaU

Q = -W;			#KJ/Kg(As in a isothermal process T2-T1  = 0 )
print "Heat Transfered in KJ/Kg : %.1f"%Q

Workdone in KJ : -289.7
Change in internal energy in KJ :  0
Heat Transfered in KJ/Kg : 289.7


## Example 2.34 Page No : 51¶

In [32]:

# Variables :
p = 6.;			#bar
m = 18.;			#Kg
v = 260.;			#m/s
rho = 4.;			#Kg/m**3
Q = 42.;			#KJ/Kg
W = 261.;			#KW
Cv = 0.715;			#KJ/KgK
pA = 1.;			#bar
vA = 60.;			#m/s
mdotA = 14.;			#Kg/s
CvA = 0.835;			#m**3/Kg
TA = 115.+273;			#K
pB = 5.5;			#bar
vB = 15.;			#m/s
mdotB = 4.;			#Kg/s
CvB = 0.46;			#m**3/Kg
TB = 600.+273;			#K
v1 = 1./rho;			#m**3/Kg

# Calculations
#m*(Cv*T+p*10**5*v1/1000+v**2/2000)+Q*rho-W = mdotA*(Cv*TA+pA*10**5*CvA/1000+vA**2/2000)+m_dotB*(Cv*TB+pB*10**5*CvB/1000+vB**2/2000);
T = (((mdotA*(Cv*TA+pA*10**5*CvA/1000+vA**2/2000)+mdotB*(Cv*TB+pB*10**5*CvB/1000+vB**2/2000))+W-Q*rho)/m-v**2/2000-p*10**5*v1/1000)/Cv;			#K

# Results
print "Temperature of air at inlet in K : %.2f"%T

#Answer in the book is wrong.

Temperature of air at inlet in K : 417.40


## Example 2.35 Page No : 52¶

In [33]:
import math

# Variables :
h1 = 3000.;			#KJ/Kg
C1 = 60.;			#m/s
h2 = 2762.;			#KJ/Kg
Q = 0.;			#KJ
m = 1.;			#Kg
W = 0.;			#in case of nozzle

# Calculations and Results
#Q-W = m*[(h2-h1)+(C2**2-C1**2)/2/1000+g*(Z2-Z1)/1000]
Z2subZ1 = 0;			#as Z1 = Z2 for horizontal nozzle
C2 = math.sqrt(-(h2-h1)*2*1000+C1**2);			#m/s
print "Velocity at exit of nozzle in m/s : %.2f"%C2

A1 = 0.1;			#m**3
v1 = 0.187;			#m**3/Kg
mdot = A1*C1/v1;			#Kg/s
print "Mass flow rate through the nozzle in Kg/s : %.4f"%mdot

v2 = 0.498;			#m**3/Kg
#mdot = A2*C2/v2 = math.pi/4*d**2*C2/v2
d2 = math.sqrt(mdot/math.pi*4*v2/C2);			#m
print "Diameter of nozzle at exit in meter : %.4f"%d2

Velocity at exit of nozzle in m/s : 692.53
Mass flow rate through the nozzle in Kg/s : 32.0856
Diameter of nozzle at exit in meter : 0.1714


## Example 2.36 Page No : 53¶

In [34]:

# Variables :
p1 = 4.;		    	#bar
p2 = 1.;			    #bar
T1 = 40.+273;			#K
T2 = 2.5+273;			#K
C1 = 40.;	    		#m/s
C2 = 200.;		    	#m/s
W = 52.;		    	#KJKg
m = 1.;			        #Kg
Cp = 1.005;		    	#KJ/KgK
Z2subZ1 = 0.;			#as Z1 = Z2

# Calculations
Q = W+m*(Cp*(T2-T1)+(C2**2-C1**2)/2/1000)			#KJ/Kg

# Results
print "Heat transfered per Kg of air in KJ/Kg : %.1f"%Q

Heat transfered per Kg of air in KJ/Kg : 33.5


## Example 2.37 Page No : 53¶

In [35]:

# Variables :
m1dot = 0.01;			#Kg/s
h1 = 2950.  			#KJ/Kg
C1 = 20.    			#m/s
m2dot = 0.1;			#Kg/s
h2 = 2565.  			#KJ/Kg
C2 = 120.   			#m/s
m3dot = 0.001;			#Kg/s
h3 = 421.   			#KJ/Kg
C3 = 0.     			#m/s
C4 = 0.     			#m/s
Wsf_dot = 25.   		#KW
Qdot = 0.   			#KJ

# Calculations
#m1dot+m2dot = m3dot+m4dot
m4dot = m1dot+m2dot-m3dot;			#Kg/s
#m1dot*(h1+C1**2/2/1000)+m2dot*(h2+C2**2/2/1000) = m3dot*(h3+C3**2/2/1000)+m4dot*(h4+C4**2/2/1000)+Wsf_dot
h4 = (m1dot*(h1+C1**2/2/1000)+m2dot*(h2+C2**2/2/1000)-m3dot*(h3+C3**2/2/1000)-Wsf_dot)/m4dot-C4**2/2/1000;			#KJ/Kg

# Results
print "Enthalpy of 2nd exit stream in KJ/Kg : %.2f"%h4

Enthalpy of 2nd exit stream in KJ/Kg : 2397.26


## Example 2.38 Page No : 54¶

In [36]:
# Variables :
mdot = 0.5;			#kg/s
p1 = 1.4;			#bar
rho1 = 2.5;			#kg/m**3
u1 = 920.;			#kJ/kg
C1 = 200.;			#m/s
p2 = 5.6;			#bar
rho2 = 5.;			#kg/m**3
u2 = 720.;			#kJ/kg
C2 = 180.;			#m/s
Qdot = -60.;			#kW
Z21 = 60.;			#m
g = 9.81;			#gravity consmath.tant

# Calculations and Results
h21 = u2-u1+(p2*10**5/(rho2*1000)-p1*10**5/(rho1*1000));			#kJ/kg(change in enthalpy)
H21 = mdot*h21;			#kW(total change in enthalpy)
print "Change in enthalpy, H2-H1 in kW :  ",H21

Wsf = Qdot-mdot*(h21+(C2**2-C1**2)/2/1000+g*(Z21)/1000);			#kW
print "Rate of workdone, Wsf in kW : %.3f"%Wsf

Change in enthalpy, H2-H1 in kW :   -72.0
Rate of workdone, Wsf in kW : 13.606


## Example 2.39 Page No : 54¶

In [45]:

# Variables :
mdot = 0.4;			    #Kg/s
C1 = 6.;			    #m/s
p1 = 1.;			    #bar
p1 = p1*100;			#KPa
V1 = 0.16;		    	#m**3/Kg
u2subu1 = 88.;			#KJ/Kg
Qdot = -59.;			#W
Qdot = Qdot/1000.;			#KJ/s
W = 0.059;			#KJ/
Gamma = 1.4;
Z2subZ1 = 0;

# Calculations
h2subh1 = Gamma*u2subu1;			#KJ
Wdot = Qdot-mdot*(h2subh1);			#As C1 = C2, C2**2-C1**2 = 0 & Z2-Zi = 0

# Results
print "Power in KW : ",Wdot

Power in KW :  -49.339


## Example 2.40 Page No : 55¶

In [37]:

# Variables :
mdot = 1;			#Kg/s
p1 = 40.;			#bar
T1 = 1047.+273;			#K
C1 = 200.;			#m/s
C2 = 100.;			#m/s
p2 = 1.;			#bar
Qdot = 0.;			#W
Cp = 1.05;			#KJ/KgK
R = 300.;			#Nm/KgK
Gamma = 1.4;

# Calculations
#p*v = m*R*T
v1dot = mdot*R*T1/p1/10**5;			#m**3/s
v2dot = (p1/p2)**(1/Gamma)*v1dot;			#m**3/s
T2 = p2*v2dot/p1/v1dot*T1;			#K
Wsf_dot = Qdot-mdot*(Cp*(T2-T1)+(C2**2-C1**2)/2/1000);			#KJ/s or KW

# Results
print "Output of turbine in KJ/s or KW : %.0f"%Wsf_dot

Output of turbine in KJ/s or KW : 918


## Example 2.41 Page No : 56¶

In [38]:
import math

# Variables :
A1C1 = 0.7;			#m**3/s
p1 = 85.;			#KPa
p2 = 650.;			#KPa
v1 = 0.35;			#m**3/Kg
v2 = 0.1;			#m**3/Kg
d1 = 10./100;			#m
d2 = 6.25/100;			#m

# Calculations and Results
mdot = A1C1/v1;			#Kg/s
p2v2SUBp1v1 = mdot*(p2*v2-p1*v1);			#KJ/s
print "Change in flow work in KJ/s : %.1f"%p2v2SUBp1v1
print "Mass flow rate in Kg/s : %.0f"%mdot

C1 = A1C1/(math.pi/4*d1**2);			#m/s
A2C2 = mdot*v2;	                		#m**3/s
C2 = A2C2/(math.pi/4*d2**2);			#m/s
C2subC1 = C2-C1;			            #m/s
print "Velocity change in m/s : %.3f"%C2subC1

Change in flow work in KJ/s : 70.5
Mass flow rate in Kg/s : 2
Velocity change in m/s : -23.937


## Example 2.42 Page No : 57¶

In [39]:
import math

# Variables :
m = 12./60;			#Kg/s
C1 = 12.;			#m/s
p1 = 1.*100;			#KPa
v1 = 0.5;			#m**3/Kg
C2 = 90.;			#m/s
p2 = 8.*100;			#KPa
v2 = 0.14;			#m**3/Kg
deltah = 150.;			#KJ/Kg
Qdot = -700./60;			#KJ/s

# Calculations and Results
#Assuming deltaPE = 0 = g*(Z2-Z1)
#Qdot-Wdot = mdot*(deltah+(C2**2-C1**2)/2/1000+g*(Z2-Z1)/1000)
Wdot = Qdot-m*(deltah+(C2**2-C1**2)/2/1000);			#KW
print "Power required to drive the compressor in KW : %.3f"%abs(Wdot)

#A1C1/v1 = A2C2/v2
d1BYd2 = math.sqrt(C2/v2*v1/C1);
print "Ratio of inlet to outlet pipe diameter : %.4f"%d1BYd2

Power required to drive the compressor in KW : 42.462
Ratio of inlet to outlet pipe diameter : 5.1755


## Example 2.43 Page No : 57¶

In [40]:

# Variables :
h1 = 160.;			#KJ/Kg
h2 = 2380.;			#KJ/Kg
m1dot = 10.;			#Kg/s
m2dot = 0.8;			#Kg/s
Qdot = 10.;			#KJ/s
Wdot = 0.;			#KJ
deltaKE = 0.;
deltaPE = 0.;
m3dot = m1dot+m2dot;			#Kg/s
print "Mass flow of heated water in Kg/s : ",m3dot
#m1dot*h1+m2dot*h2 = m3dot*h3+Qdot
h3 = (m1dot*h1+m2dot*h2-Qdot)/m3dot;			#KJ/Kg
print "Specific enthalpy of heated water in KJ/Kg : %.3f"%h3

Mass flow of heated water in Kg/s :  10.8
Specific enthalpy of heated water in KJ/Kg : 323.519


## Example 2.44 Page No : 58¶

In [41]:
import math

# Variables :
v = 0.001;			#m**3/Kg
DisRate = 10./60;			#m**3/s
p1 = 100.;			#KN/m**2
p2 = 300.;			#KN/m**2
Z1 = 3.;			#m
Z2 = 9.;			#m
d1 = 0.25;			#m
d2 = 0.17;			#m

# Calculations
#A1*C1 = A2*C2 = DisRate
C1 = DisRate/(math.pi/4*d1**2);			#m/s
C2 = DisRate/(math.pi/4*d2**2);			#m/s
mdot = DisRate/v;			#Kg/s
g = 9.81;			#gravity consmath.tant
delta_u = 0;
#Qdot-Wdot = mdot*(delta_u+p2*v2-p1*v1+C2**2-C1**2+g*(Z2-Z1))
Wdot = mdot*(delta_u+p2*10**3*v-p1*10**3*v+(C2**2-C1**2)/2+g*(Z2-Z1))-Qdot;			#J/s
Wdot = Wdot/1000;			#KJ/s or KW

# Results
print "Power required to drive the pump in KW : %.3f"%Wdot

Power required to drive the pump in KW : 46.676


## Example 2.45 Page No : 59¶

In [42]:

# Variables :
mdot = 5.;			#Kg/s
T1 = 27.+273;			#K

#Z1 = Z2
deltaPE = 0.;
Wdot = -100.;			#KW
C1 = 60.;			#m/s
C2 = 150.;			#m/s
q = -2.;			#KJ/Kg
Cp = 1.05;			#KJ/Kg
Qdot = mdot*q;			#KJ/s
delta_h = Cp;			#KJ/Kg

# Calculations
#Qdot-Wdot = mdot*(delta_h*(T2-T1)+(C2**2-C1**2)/2/1000+g*(Z2-Z1))/1000)
T2 = ((Qdot-Wdot)/mdot-(C2**2-C1**2)/2/1000)/delta_h+T1;			#K

# Results
print "Exit temperature in K : %.3f"%T2

Exit temperature in K : 308.143


## Example 2.46 Page No : 59¶

In [43]:
import math

# Variables :
t1 = 90.;			#degreeC
t2 = 30.;			#degreeC
modot = 3.;			#Kg/s

# Calculations
#h = 1.7*t+11*10**-4*t**2
h1 = 1.7*t1+11*10**-4*t1**2;			#KJ/Kg
h2 = 1.7*t2+11*10**-4*t2**2;			#KJ/Kg
tw1 = 27.;			#degreeC
tw2 = 67.;			#degreeC
Cp = 4.2;			#KJ/KgK
#h = Cp*tw;			#KJ/Kg
hw1 = Cp*tw1;			#KJ/Kg
hw2 = Cp*tw2;			#KJ/Kg

#modot*(h1-h2) = mwdot*(hw2-hw1)
mwdot = modot*(h1-h2)/(hw2-hw1);			#Kg/s

# Results
print "Rate of flow of water in Kg/s : %.4f"%mwdot

Rate of flow of water in Kg/s : 1.9629


## Example 2.47 Page No : 61¶

In [44]:
import math

# Variables :
V1 = 6.;			#m**3
p1 = 20.*100;			#Kpa
T1 = 37.+273;			#K
p2 = 10.*100;			#Kpa
V2 = V1;			#m**3
R = 0.287;			#KJ/KgK
m1 = p1*V1/R/T1;			#Kg

# Calculations
#T2 = T1*(p2/p1)**((Gamma-1)/Gamma)
Gamma = 1.4;
T2 = T1*(p2/p1)**((Gamma-1)/Gamma);			#K
m2 = p2*V2/R/T2;			#Kg
m = m1-m2;			#mass of air discharged in Kg

# Results
print "Mass of air discharged in Kg : %.2f"%m

Mass of air discharged in Kg : 52.67


## Example 2.48 Page No : 61¶

In [46]:
import scipy

# Variables :
V1 = 1.5;			#m**3
V2 = 0.;			#m**3
p = 1.02;			#bar

def f21(V):
return 1


Work done by the air in KJ : -153