# Chapter 4 : Heat Effects¶

## Example 4.2 page no : 61¶

In [2]:
import math

def ICPH(T0,T,A,B,C,D):
t = T/T0;
return  (A+((B/2)*T0*(t+1))+((C/3)*T0*T0*((t**2)+t+1))+(D/(t*T0*T0)))*(T-T0)

# Variables for methane
R = 8.314;
T0 = 533.15;
T = 873.15;
A = 1.702;
B = 9.081*(10**-3);
C = -2.164*(10**-6);
D = 0;

# Calculations
Q = round(R*ICPH(T0,T,A,B,C,D),0);

# Results
print 'Heat Required',Q,'J'

Heat Required 19778.0 J


## Example 4.3 page no : 62¶

In [3]:
def MCPH(T0,T,A,B,C,D):
t = T/T0;
return (A+((B/2)*T0*(t+1))+((C/3)*T0*T0*((t**2)+t+1))+(D/(t*T0*T0)))

# Variables for Ammonia
R = 8.314;
T0 = 533.15;
A = 3.578;
B = 3.020*(10**-3);
C = 0;
D = -0.186*(10**5);
Q = 422*(10**3);
n = 11.3;
del_H = Q/n;

# Calculations
i = -1;
a = (T0);			#Initial
while (i == -1):
b = R*MCPH(T0,a,A,B,C,D);
c = b*(a-T0);
flag = del_H-c;
if(flag <= 100):
T = a-1;
i = 1;
else:
a = a+1;
i = -1;

# Results
print 'Temperature Required(Approx)',T,'K'

Temperature Required(Approx) 1250.15 K


## Example 4.4 page no : 63¶

In [4]:
# Variables
del_H1 = 2257.;			#latent Heat of Vapourisation of water at 373.15K(100C)[KJ/Kg]
Tr1 = 373.15/647.1;
Tr2 = 573.15/647.1;

# Calculations
del_H2 = round(del_H1*((1-Tr2)/(1-Tr1))**0.38,0);			#KJ/Kg

# Results
print 'latent Heat at 573.15K',del_H2,'KJ/Kg'
print ('Note: The Value as given in steam tables at 573.15K is 1406 KJ/Kg')

latent Heat at 573.15K 1372.0 KJ/Kg
Note: The Value as given in steam tables at 573.15K is 1406 KJ/Kg


## Example 4.5 page no : 64¶

In [5]:
# Variables
#4HCL + O2 --> 2H2O + 2Cl2
del_H_HCL = -92.307;			#KJ Heat Of Formation
del_H_H2O = -241.818;			#KJ

# Calculations
#4HCL --> 2H2 + 2Cl2
del_H_298_HCL = 4*(-1)*del_H_HCL;
#2H2 + O2 --> 2H2O
del_H_298_H2O = 2*del_H_H2O;
#Final
del_H_298 = del_H_298_HCL+del_H_298_H2O;

# Results
print 'Standard Heat',del_H_298,'KJ'

Standard Heat -114.408 KJ


## Example 4.6 page no : 65¶

In [6]:
import math
from numpy import array

def IDCPH(T0,T,dA,dB,dC,dD):
t = T/T0;
return (dA+((dB/2)*T0*(t+1))+((dC/3)*T0*T0*((t**2)+t+1))+(dD/(t*T0*T0)))*(T-T0)

# Variables
#Methanol Synthesis @ 1073.15K(800C)
#CO + 2H2 --> CH3OH
del_H_CO = -110.525			#@298K from Table C.4
del_H_CH3OH_g = -200.660;			#@298K from Table C.4
del_H_298 = ((1)*del_H_CH3OH_g)-((1)*del_H_CO);			#KJ/mol
T0 = 298.15;
T = 1073.15;
R = 8.314;
#Moles (CH3OH,CO,H2)
n = array([1,-1,-2]);
#A..from Table C.1
A = array([2.211,3.376,3.249]);
#B..from Table C.1
B = (10**-3)*array([12.216,0.557,0.422]);
#C..from Table C.1
C = (10**-6)*array([-3.450,0,0]);
#D..From table C.1
D = (10**5)*array([0,-0.031,0.083]);

# Calculations
del_A = 0;
del_B = 0;
del_C = 0;
del_D = 0;
for i in range(3):
del_A = del_A+n[i]*A[i];
del_B = del_B+n[i]*B[i];
del_C = del_C+n[i]*C[i];
del_D = del_D+n[i]*D[i];

I = IDCPH(T0,T,del_A,del_B,del_C,del_D);
del_H = round(del_H_298+(R*I/10**3),3);

# Results
print 'Standard Heat Of Enthalpy',del_H,'KJ'

Standard Heat Of Enthalpy -103.566 KJ


## Example 4.7 page no : 66¶

In [7]:
import math
from numpy import array

def MCPH(T0,T,A,B,C,D):
t = T/T0;
return  (A+((B/2)*T0*(t+1))+((C/3)*T0*T0*((t**2)+t+1))+(D/(t*T0*T0)))

# Variables
#Combustion Of methane
#CH4 + 2O2 --> CO2 + 2H2O
R = 8.314;
del_H_CO2 = -393509;			#from table C.4
del_H_O2 = -241818;			#from table C.4
del_H_CH4 = -74520;			#from table C.4
del_H_298 = del_H_CO2+(2*del_H_O2)-del_H_CH4;
del_Hp = -del_H_298;

# Calculations
#moles of reacmath.tants
n_CH4 = 1;
n_O2 = 2+(0.2*2);			#20% Excess
n_N2 = n_O2*(79./21);
#Moles Of Products..(CO2,H2O,O2,N2)
np = array([1,2,0.4,9.03]);
#A..from Table C.1
A = array([5.457,3.470,3.639,3.280]);
#B..from Table C.1
B = (10**-3)*array([1.045,1.450,0.506,0.593]);
#C..from Table C.1
C = (10**-6)*array([0,0,0,0]);
#D..From table C.1
D = (10**5)*array([-1.157,0.121,-0.227,0.040]);

E_A = 0;
E_B = 0;
E_C = 0;
E_D = 0;
for i in range(4):
E_A = E_A+np[i]*A[i];
E_B = E_B+np[i]*B[i];
E_C = E_C+np[i]*C[i];
E_D = E_D+np[i]*D[i];

T0 = 298.15;
a = (T0);			#Initial
i = -1
while (i == -1):
b = R*MCPH(T0,a,E_A,E_B,E_C,E_D);
c = b*(a-T0);
flag = del_Hp-c;
if(flag <= 100):
T = a-1;
i = 1;
else:
a = a+1;
i = -1;

# Results
print 'Temperature Required(Approx)',T,'K'

Temperature Required(Approx) 2065.15 K


## Example 4.8 page no : 68¶

In [8]:
import math

def MCPH(T0,T,A,B,C,D):
t = T/T0;
return (A+((B/2)*T0*(t+1))+((C/3)*T0*T0*((t**2)+t+1))+(D/(t*T0*T0)))

#CH4 + H2O --> CO + 3H2 (A)
#CH4 + 2H2O --> CO2 + 4H2  (B)
del_H_A = 205813.			#J
del_H_B = 164647.			#J
#0.87 mol of CH4 for (A) (1-0.87)mol of CH4 for (B)
del_H_298 = (0.87*del_H_A)+(0.13*del_H_B);
R = 8.314;
T0 = 298.15;
T_A = 600.; 			#Cooled
T_B = 1300.;			#Heated

# Calculations
#Moles of reactants (CH4,H2O)
nr = [1,2];
#Moles of Products (CO,H2,CO2,H2O)
np = [0.87,3.13,0.13,0.87];
#For Reacmath.tants
#for CH4
I1 = MCPH(T0,T_A,1.702,9.081*(10**-3),-2.164*(10**-6),0);
#For H2O
I2 = MCPH(T0,T_A,3.470,1.450*(10**-3),0,0.121*(10**5));
del_Hr = R*((nr[0]*I1)+(nr[1]*I2))*(T0-T_A);			#J
#For Products
#for CO
I1 = MCPH(T0,T_B,3.376,0.557*(10**-3),0,-0.031*(10**5));
#For H2
I2 = MCPH(T0,T_B,3.249,0.422*(10**-3),0,0.083*(10**5));
#for CO2
I3 = MCPH(T0,T_B,5.457,1.045*(10**-3),0,-1.157*(10**5));
#For H2O
I4 = MCPH(T0,T_B,3.470,1.450*(10**-3),0,0.121*(10**5));
del_Hp = R*((np[0]*I1)+(np[1]*I2)+(np[2]*I3)+(np[3]*I4))*(T_B-T0);			#J
#del_H
del_H = del_H_298+del_Hr+del_Hp;
Q = round(del_H,-1);

# Results
print 'Heat Required',Q,'J'

Heat Required 328010.0 J