# Chapter 16 : Reactive Systems¶

## Example 16.2 Page No : 657¶

In [2]:
import math

# Variables
eps_e = 0.27;
P = 1. ;

# Calculation
K = (4*eps_e**2*P)/(1-eps_e**2);
P1 = 100./760; 			# in Pa
eps_e_1 = math.sqrt((K/P1)/(4+(K/P1)));
T1 = 318.; T2 = 298.;
R = 8.3143; K1 = 0.664; K2 = 0.141;
dH = 2.30*R*((T1*T2)/(T1-T2))*(math.log(K1/K2));

# Results
print "K is",round(K,4),"atm"
print "epislon is ",round(eps_e_1,3)
print "The heat of reaction is",round(dH,0),"kJ/kg mol"


K is 0.3145 atm
epislon is  0.612
The heat of reaction is 140399.0 kJ/kg mol


## Example 16.3 Page No : 659¶

In [3]:
# Variables
v1 = 1.
v2 = v1;
v3 = v2;
v4 = v2;
e = 0.56; 			# Degree of reaction
P = 1.; 			# Dummy
T = 1200.
R = 8.3143;

# Calculation
x1 = (1-e)/2; x2 = (1-e)/2;
x3 = e/2; x4 = e/2;
K = (((x3**v3)*(x4**v4))/((x1**v1)*(x2**v2)))*P**(v3+v4-v1-v2); 			# Equillibrium consmath.tant
dG = -R*T*math.log(K);

# Results
print "Equillibrium constant is",round(K,2)
print "Gibbs function change is",round(dG,1),"J/gmol"

# note : rounding off error is there.

Equillibrium constant is 1.62
Gibbs function change is -4812.2 J/gmol


## Example 16.5 Page No : 662¶

In [6]:
# Variables
Veo = 1.777; 			# Ve/Vo
e = 1-Veo; 			# Degree of dissociation
P = 0.124; 			# in atm

# Calculation
K = (4*e**2*P)/(1-e**2);

# Results
print "The value of equillibrium constant is",round(K,3),"atm"

The value of equillibrium constant is 0.756 atm


## Example 16.6 Page No : 663¶

In [4]:
# Variables
v1 = 1.;
v2 = 0.;
v3 = 1.; v4 = 1./2;
dH = 250560.; e = 3.2e-03;
R = 8.3143; T = 1900.;

# Calculation
Cp = ((dH**2)*(1+e/2)*e*(1+e))/(R*T**2*(v1+v2)*(v3+v4));

# Results
print "Cp is",round(Cp,3),"j/gmol K"

# note : rounding off error.

Cp is 4.484 j/gmol K


## Example 16.7 Page No : 663¶

In [5]:
# Variables
a = 21.89;
y = 18.5;
x = 8.9;

# Calculation
PC = 100.*(x*12.)/((8.9*12)+(18.5*1));
PH = 100-PC;
AFR = ((32*a)+(3.76*a*28))/((12*x)+y);
EAU = (8.8*32)/((21.89*32)-(8.8*32));

# Results
print "carbon = ",round(PC,2),"% and Hydrogen = ",round(PH,2),"%"
print "Air fuel ratio is",round(AFR,0)
print "Percentage of excess air used is",round((EAU*100),2),"%"

carbon =  85.24 % and Hydrogen =  14.76 %
Air fuel ratio is 24.0
Percentage of excess air used is 67.23 %


## Example 16.8 Page No : 664¶

In [8]:
# Variables
hf_co2 = -393522;
hf_h20 = -285838;
hf_ch4 = -74874;

# Calculation
D = hf_co2 + (2*hf_h20);
QCV = D - (hf_ch4+1);

# Results
print "Heat transfer per kg mol of fuel is",D,"kJ"
print "Qcv = %.f kJ"%QCV

# note : rounding off error.

Heat transfer per kg mol of fuel is -965198 kJ
Qcv = -890325 kJ


## Example 16.9 Page No : 664¶

In [9]:
# Below values are taken fron table 16.4

# Variables
Hr = -249952.+(18.7*560)+(70*540);
Hp = 8.*(-393522.+20288)+9*(-241827+16087)+6.25*14171+70*13491;
Wcv = 150.; 			# Energy out put from engine in kW
Qcv = -205.; 			# Heat transfer from engine in kW

# Calculation
n = (Wcv-Qcv)*3600/(Hr-Hp);

# Results
print "Fuel consumption rate is",round((n*114),1),"kg/h"

# rounding error is there.

Fuel consumption rate is 38.5 kg/h


## Example 16.10 Page No : 665¶

In [10]:
from numpy import interp

# Variables
Hr1 = -249952.; 			# For ocmath.tane
Hp1 = Hr1;
# Below values are calculated umath.sing value fron table 16.4
T2 = 1000.;
Hp2 = -1226577.
T3 = 1200.;
Hp3 = 46537.;
T4 = 1100.;
Hp4 = -595964.;

# Calculation
Hp = [Hp2 ,Hp3, Hp4]
T = [T2 ,T3, T4]
T1 = interp(Hp1,Hp , T); 			# Interpolation to find temperature at Hp1

# Results
print "the adeabatic flame temperature is",T1,"K"


the adeabatic flame temperature is 1100.0 K


## Example 16.11 Page No : 666¶

In [11]:
# Variables
T0 = 298.;
Wrev = -23316-3*(-394374)-4*(-228583);
Wrev_ = Wrev/44.; 			# in kJ/kg
Hr = -103847.;
T = 980.; 			# Through trial and error

# Calculation
Sr = 270.019+20*205.142+75.2*191.611;
Sp = 3*268.194 + 4*231.849 + 15*242.855 + 75.2*227.485;
IE = Sp-Sr; 			# Increase in entropy
I = T0*3699.67/44;
Si = Wrev_ - I;

# Results
print "Reversible work is",Wrev_,"kJ/kg"
print "Increase in entropy during combustion is",round(Sp-Sr,2),"kj/kg mol K"
print "Irreversibility of the process",round(I,0),"kJ/kg"
print "Availability of products of combustion is",round(Si,1),"kJ/kg"

# note : there are rounding off errors. please check.

Reversible work is 47139.5 kJ/kg
Increase in entropy during combustion is 3699.67 kj/kg mol K
Irreversibility of the process 25057.0 kJ/kg
Availability of products of combustion is 22082.6 kJ/kg


## Example 16.12 Page No : 667¶

In [23]:
import math

# Variables
T0 = 298.15; P0 = 1; R = 8.3143;
xn2 = 0.7567
xo2 = 0.2035
xh2o = 0.0312
xco2 = 0.0003;

# Calculation and Results
# Part (a)
g_o2 = 0; g_c = 0; g_co2 = -394380;
A = -g_co2 + R*T0*math.log(xo2/xco2);
print "The chemical energy of carbon is",round(A,0),"kJ/k mol"

# Part (b)
g_h2 = 0; g_h2o_g = -228590;
B = g_h2 + g_o2/2 - g_h2o_g + R*T0*math.log(xo2**0.5/xh2o);
print "The chemical energy of hydrogen is",round(B,0),"kJ/k mol"

# Part (c)
g_ch4 = -50790;
C = g_ch4 + 2*g_o2 - g_co2 - 2*g_h2o_g + R*T0*math.log((xo2**2)/(xco2*xh2o));
print "The chemical energy of methane is",round(C,0),"kJ/k mol"

# Part (d)
g_co = -137150;
D =  g_co + g_o2/2 - g_co2 + R*T0*math.log((xo2**0.5)/xco2);
print "The chemical energy of Carbonmonoxide is",round(D,0),"kJ/k mol"

# Part (e)
g_ch3oh = -166240;
E = g_ch3oh + 1.5*g_o2 - g_co2 - 2*g_h2o_g + R*T0*math.log((xo2**1.5)/(xco2*(xh2o**2)))
print "The chemical energy of methanol is",round(E,0),"kJ/k mol"

# Part (f)
F = R*T0*math.log(1./xn2);
print "The chemical energy of nitrogen is",round(F,1),"kJ/k mol"

# Part (g)
G = R*T0*math.log(1./xo2);
print "The chemical energy of Oxygen is",round(G,0),"kJ/k mol"

# Part (h)
H = R*T0*math.log(1./xco2);
print "The chemical energy of carbondioxide is",round(H,0),"kJ/k mol"

# Part (i)
g_h2o_l = -237180;
I = g_h2o_l - g_h2o_g + R*T0*math.log(1./xh2o);
print "The chemical energy of water is",round(I,1),"kJ/k mol"

# note : rounding off error is there. please check.

The chemical energy of carbon is 410542.0 kJ/k mol
The chemical energy of hydrogen is 235212.0 kJ/k mol
The chemical energy of methane is 821580.0 kJ/k mol
The chemical energy of Carbonmonoxide is 275365.0 kJ/k mol
The chemical energy of methanol is 716699.0 kJ/k mol
The chemical energy of nitrogen is 691.1 kJ/k mol
The chemical energy of Oxygen is 3947.0 kJ/k mol
The chemical energy of carbondioxide is 20108.0 kJ/k mol
The chemical energy of water is 5.2 kJ/k mol


## Example 16.13 Page No : 669¶

In [25]:
# Variables
b = 8./(0.114+0.029); 			# By carbon balance
C = 18./2            			# By hydrogen balance
a = b*0.114 + (b/2)*0.029 + b*0.016 + C/2 ; 			# By oxygen balance
Wcv = 1. 			# Power developed by engine in kW

# Calculation and Results
n_fuel = (0.57*1)/(3600*114.22);
Qcv = Wcv-n_fuel*3845872; 			# 5.33
print "The rate of heat transfer from the engine is",round(Qcv,2),"kW"
# Part (b)
ach = 5407843.; 			# chemical energy of liquid ocmath.tane
n2 = Wcv/(n_fuel*ach);
print "The second law efficiency is",round((n2*100),1),"%"

The rate of heat transfer from the engine is -4.33 kW
The second law efficiency is 13.3 %