In [1]:

```
import math
#Given:
n = 4. #Number of cylinders
d_o = 7.5 #Diameter of orifice in cm
Cd = 0.6 #Coefficient of discharge for orifice
d = 11.
l = 13. #Bore and stroke in cm
N = 2250. #Engine speed in rpm
bp = 36. #Brake power in kW
m_f = 10.5 #Fuel consumption in kg/hr
CV = 42000. #Calorific value in kJ/kg
deltaP_o = 4.1 #Pressure drop across orifice in cm of water
P = 1.013 #Atmospheric pressure in bar
T = 15+273 #Atmospheric temperature in K
g = 9.81 #Accelaration due to gravity in m/s**2
#Solution:
#(a)
eta_bt = bp*3600/(m_f*CV) #Brake thermal efficiency
#(b)
A = math.pi/4*d**2*10**-4 #Area of cylinder in m**2
bmep = bp*1000/(n*l/100*A*N/(2*60)) #Brake mean effective pressure in Pascal
#(c)
rho_w = 1000 #Mass density of water in kg/m**3
deltaP_o = rho_w*g*deltaP_o/100 #Pressure drop across orifice in N/m**2
R = 0.287 #Specific gas consmath.tant in kJ/kgK
rho_a = P*10**5/(R*10**3*T) #Mass density of air in kg/m**3
A_o = math.pi/4*d_o**2*10**-4 #Area of orifice in m**2
m_a = Cd*A_o*math.sqrt(2*deltaP_o*rho_a) #Air inhaled in kg/s
V_s = (math.pi/4)*d**2*l*n*N/(2*60)*10**-6 #Swept volume in m**3/s
eta_vol = m_a/V_s #Volumetric efficiency
#Results:
print " a)Brake thermal efficiency, eta_bt = %.3f"%(eta_bt)
print " b)Brake mean effective pressure, bmep = %.3f bar"%(bmep*10**-5)
print " c)Volumetric efficiency, eta_vol = %.3f"%(eta_vol)
```

In [2]:

```
import math
#Given:
d = 24.
l = 48. #Bore and stroke in cm
D_b = 1.25 #Effective diameter of the brake wheel in m
P = 1236. #Net load on brake in N
N = 77. #Average engine explosions in min
N1 = 226.7 #Average speed at output shaft in rpm
imep = 7.5 #Indicated mean effective pressure in bar
Vg1 = 13. #Gas used in m**3/hr
T1 = 15.+273
P1 = 771. #Temperature and pressure of the gas in K and mm of mercury
T2 = 0+273
P2 = 760. #Normal temperature and pressure (N.T.P.) in K and mm of mercury
CV = 22000. #Calorific value of the gas in kJ/m**3
m_w = 625. #Mass of cooling water used in kg/hr
T1_w = 25+273.
T2_w = 60.+273 #Inlet and outlet temperature of cooling water in K
#Solution:
#(a)
T = P*D_b/2 #Brake torque delivered in Nm
bp = 2*math.pi*N1/60*T #Brake power in W
ip = imep*10**5*l*math.pi/4*d**2*N/60*10**-6 #Indicated power in W
eta_m = bp/ip #Mechanical efficiency
#(b)
Vg2 = (P1/P2)*(T2/T1)*Vg1 #Gas consumption at N.T.P. in m**3/hr
#(c)
eta_it = ip/1000*3600/(Vg1*CV) #Indicated thermal efficiency
#Heat balance sheet
Q1 = Vg2/60*CV #Heat supplied in kJ/min
Q_bp = bp/1000*60 #Heat equivalent to brake power in kJ/min
cp = 4.1868 #Specfic heat of water in kJ/kgK
Q_w = m_w/60*cp*(T2_w-T1_w) #Heat in cooling water in kJ/min
Q_r = Q1-Q_bp-Q_w #Heat to exhaust, radiation in kJ/min
#Results:
print " a)The mechanical efficiency of the engine, eta_m = %.1f percent"%(eta_m*100)
print " b)The gas consumption at N.T.P. = %.1f m**3/hr"%(Vg2)
print " c)The indicated thermal efficiency, eta_it = %.1f percent"%(eta_it*100)
print " Heat balance sheet\t Heat supplied by the gas = %.1f kJ/min, %d percent"%(Q1,Q1/Q1*100)
print "\t Heat equivalent to b.p. = %.1f kJ/min, %.1f percent"%(Q_bp,Q_bp/Q1*100)
print "\t Heat in cooling water = %.1f kJ/min, %.1f percent"%(Q_w,Q_w/Q1*100)
print "\t Heat to exhaust, radiation = %.1f kJ/min, %.1f percent"%(Q_r,Q_r/Q1*100)
```

In [3]:

```
import math
#Given:
d = 18.
l = 36. #Bore and stroke in cm
N = 285. #Average engine speed in rpm
T = 393. #Brake torque delivered in Nm
imep = 7.2 #Indicated mean effective pressure in bar
m_f = 3.5 #Fuel consumption in kg/hr
m_w = 4.5 #Mass of cooling water used in kg/min
deltaT_w = 36. #Cooling water temperature rise in degreeC
A_F = 25. #Air-fuel ratio
T2 = 415.+273 #Exhaust gas temperature in K
P = 1.013 #Atmospheric pressure in bar
T1 = 21.+273 #Room temperature in K
CV = 45200. #Calorific value in kJ/kg
p = 15. #Perentage of hydrogen contained by the fuel
R = 0.287 #Specific gas consmath.tant in kJ/kgK
cv = 1.005
cp = 2.05 #Specific heat for dry exhaust gases and superheated steam in kJ/kgK
#Solution:
#(a)
ip = imep*10**2*l*math.pi/4*d**2*N/(2*60)*10**-6 #Indicated power in kW
ip = round(10*ip)/10
eta_it = ip*3600/(m_f*CV) #Indicated thermal efficiency
#(b)
m_a = m_f*A_F/60 #Mass of air inhaled in kg/min
m_a = round(100*m_a)/100
V_a = m_a*R*T1/(P*100) #Volume of air inhaled in m**3/min
V_s = (math.pi/4)*d**2*l*10**-6*N/2 #Swept volume in m**3/min
eta_vol = V_a/V_s #Volumetric efficiency
#Heat balance sheet
Q1 = m_f/60*CV #Heat input in kJ/min
bp = 2*math.pi*N/60*T*10**-3 #Brake power in W
Q_bp = bp*60 #Heat equivalent to brake power in kJ/min
cp_w = 4.1868 #Specific heat of water in kJ/kgK
Q_w = m_w*cp_w*deltaT_w #Heat in cooling water in kJ/min
m_e = m_a+m_f/60 #Mass of exhaust gases in kg/min
#Since, 2 mole of hydrogen gives 1 mole of water on combine with 1 mole of oxygen
#Thus, 1 mole of hydrogen gives 1/2 mole or 9 unit mass of water
m_h = m_f/60*p/100 #Mass of hydrogen in kg/min
m_s = 9*m_h #Mass of steam in exhaust gases in kg/min
m_d = m_e-m_s #Mass of dry exhaust gases in kg/min
Q_d = m_d*cv*(T2-T1) #Heat in dry exhaust gases in kJ/min
lv = 2256.9 #Latent heat of vapourisation of water in kJ/kg
Q_s = m_s*((373-T1)+lv+cp*(T2-373)) #Heat in steam in exhaust gases in kJ/min
Q_r = Q1-Q_bp-Q_w-Q_d-Q_s #Heat in radiation in kJ/min
#Results:
print " a)The indicated thermal efficiency, eta_it = %.1f percent"%(eta_it*100)
print " b)The volumetric efficiency, eta_vol = %.1f percent"%(eta_vol*100)
print " Heat balance sheet\t Heat input = %.1f kJ/min, %d percent"%(Q1,Q1/Q1*100)
print "\t Heat equivalent to b.p. = %.1f kJ/min, %.1f percent"%(Q_bp,Q_bp/Q1*100)
print "\t Heat in cooling water = %.1f kJ/min, %.1f percent"%(Q_w,Q_w/Q1*100)
print "\t Heat in dry exhaust gases = %.1f kJ/min, %.1f percent"%(Q_d,Q_d/Q1*100)
print "\t Heat in steam in exhaust gases = %.1f kJ/min, %.1f percent"%(Q_s,Q_s/Q1*100)
print "\t Heat in radiation = %.1f kJ/min, %.1f percent"%(Q_r,Q_r/Q1*100)
```

In [4]:

```
import math
#Given:
n = 4. #Number of cylinders
d_o = 5. #Diameter of orifice in cm
Cd = 0.6 #Coefficient of discharge for orifice
d = 10.5
l = 12.5 #Bore and stroke in cm
N = 1200. #Engine speed in rpm
T = 147. #Brake torque delivered in Nm
m_f = 5.5 #Fuel consumption in kg/hr
CV = 43100. #Calorific value in kJ/kg
deltaP_o = 5.7 #Head across orifice in cm of water
P1 = 1.013 #Atmospheric pressure in bar
T1 = 20.+273 #Atmospheric temperature in K
g = 9.81 #Accelaration due to gravity in m/s**2
#Solution:
#(a)
bp = 2*math.pi*N/60*T*10**-3 #Brake power in kW
eta_bt = bp*3600/(m_f*CV) #Brake thermal efficiency
#(b)
A = math.pi/4*d**2*10**-4 #Area of cylinder in m**2
bmep = bp*1000/(n*l/100*A*N/(2*60)) #Brake mean effective pressure in N/m**2
#(c)
rho_w = 1000 #Mass density of water in kg/m**3
deltaP_o = rho_w*g*deltaP_o/100 #Pressure drop across orifice in N/m**2
R = 0.287 #Specific gas consmath.tant in kJ/kgK
rho_a = P1*10**5/(R*10**3*T1) #Mass density of air in kg/m**3
rho_a = round(10*rho_a)/10
A_o = math.pi/4*d_o**2*10**-4 #Area of orifice in m**2
V_a = Cd*A_o*math.sqrt(2*deltaP_o/rho_a) #Air inhaled in m**3/s
V_s = (math.pi/4)*d**2*l*n*N/(2*60)*10**-6 #Swept volume in m**3/s
eta_vol = V_a/V_s #Volumetric efficiency
#Results:
print " a)Brake thermal efficiency, eta_bt = %.1f percent"%(eta_bt*100)
print " b)Brake mean effective pressure, bmep = %.2f bar"%(bmep*10**-5)
print " c)Volumetric efficiency, eta_vol = %.1f percent"%(eta_vol*100)
```

In [5]:

```
import math
#Given:
n = 6. #Number of cylinders
d = 7.5
l = 9. #Bore and stroke in cm
R_b = 38. #Torque arm radius of the brake wheel in cm
P1 = 324. #Net load when all cylinders operating on brake in N
N = 3300. #Engine speed in rpm
P2 = 245. #Net load when each cylinder is inoperative in N
m_f = .3 #Fuel consumption in kg/min
CV = 42000. #Calorific value in kJ/kg
m_w = 65. #Mass of cooling water used in kg/min
deltaT_w = 12. #Cooling water temperature rise in degreeC
m_a = 14. #Mass of air blown in kg/min
T1_a = 10.+273
T2_a = 55.+273 #Inlet and outlet temperature of air blown in K
#Solution:
bp = 2*math.pi*N/60*(P1*R_b/100)*10**-3 #Brake power when all the cylinders operating in kW
bp1 = 2*math.pi*N/60*(P2*R_b/100)*10**-3 #Brake power when each cylinder is inoperative in kW
ip = n*(bp-bp1) #Total ip of the engine in kW
A = math.pi/4*d**2*10**-4 #Area of cylinder in m**2
bmep = ip*1000/(n*l/100*A*N/(2*60)) #Brake mean effective pressure in N/m**2
#Heat balance sheet
Q1 = m_f*CV #Heat input in kJ/min
Q_bp = bp*60 #Heat equivalent to brake power in kJ/min
cp_w = 4.1868 #Specfic heat of water in kJ/kgK
Q_w = m_w*cp_w*deltaT_w #Heat in cooling water in kJ/min
cp_a = 1.005 #Specific heat of air in kJ/kgK
Q_a = m_a*cp_a*(T2_a-T1_a) #Heat to ventilating air in kJ/min (Wrong in book)
Q_e = Q1-Q_bp-Q_w-Q_a #Heat to exhaust and other losses in kJ/min
#Results:
print " a)The indicated mean effective pressure, bmep = %.1f bar"%(bmep*10**-5)
print " Heat balance sheet\t Heat input = %d kJ/min, %d percent"%(Q1,Q1/Q1*100)
print "\t Heat equivalent to b.p. = %d kJ/min, %.1f percent"%(Q_bp,Q_bp/Q1*100)
print "\t Heat in cooling water = %d kJ/min, %.1f percent"%(Q_w,Q_w/Q1*100)
print "\t Heat to ventilating air = %d kJ/min, %.1f percent"%(Q_a,Q_a/Q1*100)
print "\t Heat to exhaust and other losses = %d kJ/min, %.2f percent"%(Q_e,Q_e/Q1*100)
#Heat to ventilating air is wrong in book
```

In [8]:

```
import math
#Given:
N = 450. #Engine speed in rpm
P = 450. #Net load on brake in N
imep = 2.9 #Indicated mean effective pressure in bar
m_f = 5.4 #Fuel consumption in kg/h
deltaT_w = 36.1 #Cooling water temperature rise in degreeC
m_w = 440. #Mass of cooling water used in kg/h
A_F = 31. #Air-fuel ratio
T1_g = 20.+273
T2_g = 355.+273 #Inlet and outlet temperature of exhaust gases blown in K
P1 = 76. #Atmospheric pressure in cm of Hg
d = 22.
l = 27. #Bore and stroke in cm
D_b = 1.5 #Effective diameter of the brake wheel in m
CV = 44000. #Calorific value in kJ/kg
p = 15. #Percentage of hydrogen by mass contained by the fuel
R = 0.287 #Specific gas consmath.tant in kJ/kgK
cp_g = 1.005
cp_s = 2.05 #Specific heat for dry exhaust gases and superheated steam in kJ/kgK
#Solution:
ip = imep*10**2*l*math.pi/4*d**2*N/(60)*10**-6 #Indicated power in kW
eta_it = ip*3600/(m_f*CV) #Indicated thermal efficiency
bp = 2*math.pi*N/60*(P*D_b/2)*10**-3 #Brake power in kW
bp = round(10*bp)/10
bsfc = m_f/bp*1000 #Brake specific fuel consumption in gm/kWh
V_s = (math.pi/4)*d**2*l*10**-6*N #Swept volume in m**3/min
m_a = m_f*A_F/60 #Mass of air inhaled in kg/min
P1 = 1.0132 #Atmospheric pressure equivalent to 76 cm of Hg in bar
T1 = 293 #Atmospheric temperature in K
V_a = m_a*R*T1/(P1*100) #Volume of air inhaled in m**3/min
V_a = round(100*V_a)/100
eta_vol = V_a/V_s #Volumetric efficiency
#Heat balance sheet
Q1 = m_f/60*CV #Heat input in kJ/min
Q_bp = bp*60 #Heat equivalent to brake power in kJ/min
cp_w = 4.1868 #Specfic heat of water in kJ/kgK
Q_w = m_w/60*cp_w*deltaT_w #Heat in cooling water in kJ/min
m_e = m_a+m_f/60 #Mass of exhaust gases in kg/min
#Since, 2 mole of hydrogen gives 1 mole of water on combine with 1 mole of oxygen
#Thus, 1 mole of hydrogen gives 1/2 mole or 9 unit mass of water
m_h = m_f/60*p/100 #Mass of hydrogen in kg/min
m_s = 9*m_h #Mass of steam in exhaust gases in kg/min
m_d = m_e-m_s #Mass of dry exhaust gases in kg/min
Q_d = m_d*cp_g*(T2_g-T1_g) #Heat in dry exhaust gases kJ/min
lv = 2256.9 #Latent heat of vapourisation of water in kJ/kg
Q_s = m_s*((373-T1_g)+lv+cp_s*(T2_g-373)) #Heat in steam in exhaust gases in kJ/min
Q_r = Q1-Q_bp-Q_w-Q_d-Q_s #Heat in radiation in kJ/min
#Results:
print " a)The indicated thermal efficiency, eta_it = %.1f percent"%(eta_it*100)
print " b)Brake specific fuel consumption = %.1f gm/kWh"%(bsfc)
print " c)The volumetric efficiency, eta_vol = %.1f percent"%(eta_vol*100)
print " Heat balance sheet\t Heat input = %.1f kJ/min, %d percent"%(Q1,Q1/Q1*100)
print "\t Heat equivalent to b.p. = %.1f kJ/min, %.1f percent"%(Q_bp,Q_bp/Q1*100)
print "\t Heat in cooling water = %.1f kJ/min, %.1f percent"%(Q_w,Q_w/Q1*100)
print "\t Heat in dry exhaust gases = %.1f kJ/min, %.1f percent"%(Q_d,Q_d/Q1*100)
print "\t Heat in steam in exhaust gases = %.1f kJ/min, %.1f percent"%(Q_s,Q_s/Q1*100)
print "\t Heat in radiation = %.1f kJ/min, %.1f percent"%(Q_r,Q_r/Q1*100)
```

In [9]:

```
import math
from numpy import array,sum
#Given:
n = 12. #Number of cylinders
def f(W):
return W*N/180 #Power law of engine
d = 38.
l = 50. #Bore and stroke in cm
N = 200. #Engine speed in rpm
Wall1 = 2000.
Wall2 = 2020. #Brake loads when all cylinders are firing in N
Wn = array([1795 ,1814, 1814, 1795, 1804, 1819, 1800, 1824, 1785, 1804, 1814, 1795]) #Brake load when cylinder number 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 are out in N
#Solution:
W = (Wall1+Wall2)/2 #Average of brake loads when all cylinders are firing in N
bp = f(W) #Total brake power in kW
ipn = bp-f(Wn) #Indicated power of cylinders number 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 in kW
ip = sum(ipn) #Total indicated power equal to sum of individual in kW
eta_m = bp/ip #Mechanical efficiency
A = math.pi/4*d**2*10**-4 #Area of cylinder in m**2
bmep = bp*1000/(n*l/100*A*N/(60)) #Brake mean effective pressure in Pascal
#Results:
print " The brake mean effective pressure, bmep = %.2f bar"%(bmep*10**-5)
print " The mechanical efficiency, eta_m = %.1f percent"%(eta_m*100)
```

In [10]:

```
import math
#Given:
n = 6. #Number of cylinders
def f(W):
return W*N/20000 #Power law of engine
d = 95.
l = 120. #Bore and stroke in mm
N = 2400. #Engine speed in rpm
C_H = 83./17 #Carbon Hydrogen ratio by mass in fuel
d_o = 30. #Diameter of orifice in mm
Cd = 0.6 #Orifice coefficient of discharge
P = 550. #Net load on brake in N
P1 = 750. #Ambient pressure in mm of Hg
T1 = 25.+273 #Ambient temperature in K
deltaP_o = 14.5 #Head over orifice in cm of Hg
s = 0.831 #Specific gravity of fuel
t = 19.3 #Time to use 100 cc fuel in s
V_f = 100. #Volume of fuel used in t seconds in cc
#Solution:
#(a)
bp = f(P) #Brake power at brake load in kW
A = math.pi/4*d**2*10**-6 #Area of cylinder in m**2
bmep = bp*1000/(n*l/1000*A*N/(2*60)) #Brake mean effective pressure in Pascal
#(b)
T = bp*1000/(2*math.pi*(N/60)) #Brake torque in Nm
#(c)
rho_f = s*1000 #Fuel density in kg/m**3
m_f = V_f*10**-6/t*3600*rho_f #Fuel flow rate in kg/hr
bsfc = m_f/bp #Brake specific fuel consumption in kg/kWh
#(e)
R = 0.287 #Specific gas consmath.tant in kJ/kgK
P1 = P1/760*1.01325 #Ambient pressure in bar
rho_a = P1*10**5/(R*10**3*T1) #Mass density of air in kg/m**3
deltaP_o = 13.6*1000*9.81*deltaP_o/100 #Pressure drop across orifice in N/m**2
A_o = math.pi/4*d_o**2*10**-6 #Area of orifice in m**2
V_a = Cd*A_o*math.sqrt(2*deltaP_o/rho_a) #Air inhaled in m**3/s
V_s = (math.pi/4)*d**2*l*n*N/(2*60)*10**-9 #Swept volume in m**3/s
eta_vol = V_a/V_s #Volumetric efficiency
#(d)
pH = 17.
pC = pH*C_H #Percentage of Hydrogen and Carbon in fuel
pO = 23.3 #Percentage of Oxygen in air
H = 1.
C = 12.
O = 16. #Atomic masses of Hydrogen, Carbon, Oxygen in gm
mO2 = pC/100*(2*O/C)+pH/100*(O/(2*H)) #Oxygen required in kg/kg of fuel
m_a = mO2/(pO/100) #Mass of air in kg/kg of fuel
A_F_t = m_a #Theoritical air fuel ratio
m_a_act = V_a*rho_a #Actual air mass flow rate in kg/s
A_F_act = m_a_act/m_f*3600 #Actual air fuel ratio
P_e = (A_F_act-A_F_t)/A_F_t*100 #Percentage excess air
#Results:
print " a)The brake mean effective pressure, bmep = %.3f bar"%(bmep*10**-5)
print " b)The brake torque, T = %.1f Nm"%(T)
print " c)The brake specific fuel consumption, bsfc = %.3f kg/kWh"%(bsfc)
print " d)The percentage excess air = %.1f percent"%(P_e)
print " e)The volumetric efficiency, eta_vol = %.1f percent"%(eta_vol*100)
```

In [12]:

```
import math
#Given:
n = 6. #Number of cylinders
d = 125.;l = 190. #Bore and stroke in mm
pC = 82./100;pH2 = 18./100 #Composition of Carbon and Hydrogen in petrol
pCO2 = 11.19/100;pO2 = 3.61/100;pN2 = 85.2/100 #Composition of Carbon di oxide, Oxygen, Nitrogen in dry exhaust
P1 = 1. #Pressure of mixture entering the cylinder in bar
T1 = 17.+273 #Temperature of mixture entering the cylinder in K
m_f = 31.3 #Mass of the petrol used in kg/hr
N = 1500. #Engine speed in rpm
m = 1.;T = 0.+273;P = 1.013;V = 0.773 #Mass, temperature, pressure, volume, of air in kg, K, bar, m**3
p = 23./100 #Composition of Oxygen in air by mass
#Solution:
C = 12. #Atomic mass of Carbon(C)
H = 1. #Atomic mass of Hydrogen(H)
O = 16. #Atomic mass of Oxygen(O)
N2 = 14. #Atomic mass of Nitrogen(N)
A_F_s = (pC*2*O/C+pH2*O/(2*H))/(p) #Stoichiometric air fuel ratio
#Stoichiometric equation of combustion of fuel (petrol)
# 0.82/12[C] + 0.18/2[H2] + [0.21[O2] + 0.79[N2]]*x = a[CO2] + b[CO] + c[H2O] + d1[N2]
#Equating coefficients
a = pC/C;c = pH2/(2*H) #On balancing C and H
d1 = pN2/pCO2*a #On taking composition of CO2 and N2 in exhaust
x = d1/0.79 #On balancing N
m_a = (p*2*O)+((1-p)*2*N2) #Mass of air per mole air in kg/mole
A_F_act = x*m_a #Actual air fuel ratio
P_e = (A_F_act-A_F_s)/A_F_s*100 #Percentage excess air
R_a = P*100*V/(m*T) #Specific gas consmath.tant for air in kJ/kgK
V_a = A_F_act*R_a*T1/(P1*100) #Volume of air in m**3
#Given, rho_f = 3.35 * rho_a, V_f = 1/3.35 * V_a
V_f = V_a/A_F_act*1/3.35 #Volume of fuel in m**3/kg of fuel
V_m = V_a+V_f #Total volume of mixture in m**3/kg of fuel
V_m1 = V_m*m_f/60 #Mixture aspirated in m**3/min
V_s = (math.pi/4)*d**2*l*n*N/2*10**-9 #Swept volume in m**3/s
eta_v = V_m1/V_s*100 #Volumetric efficiency in percent
#Results:
print " The mass of air supplied per kg of petrol, m_a = %.2f kg/kg of fuel"%(A_F_act)
print " The percentage excess air = %.1f percent"%(P_e)
print " The volume of mixture per kg of petrol, V_m = %.2f m**3/kg fuel"%(V_m)
print " The volumetric efficiency of the engine, eta_v = %.0f percent"%(eta_v)
```

In [15]:

```
import math
#Given:
d = 27.
l = 45. #Bore and stroke in cm
D_b = 1.62 #Effective diameter of the brake wheel in m
t = 38.5 #Duration of test in min
N = 8080.
N1 = 3230. #Number of revolutions and explosions
P = 903. #Net load on brake in N
imep = 5.64 #Indicated mean effective pressure in bar
Vg1 = 7.7 #Gas used in m**3
T1 = 27.+273 #Temperature of the gas in K
deltaP1 = 135. #Pressure difference of gas above atmospheric pressure in mm of water
Patm = 750. #Atmospheric pressure in mm of Hg
CV = 18420. #Calorific value of the gas in kJ/m**3 at N.T.P.
m_w = 183. #Mass of cooling water used in kg
deltaT_w = 47. #Cooling water temperature rise in degreeC
#Solution:
P1 = Patm+deltaP1/13.6 #Gas pressure in mm of Hg
P1 = P1/750 #Gas pressure in bar
T2 = 0+273;P2 = 1.013 #Normal temperature and pressure (N.T.P.) in K and bar
Vg2 = (P1/P2)*(T2/T1)*Vg1 #Gas consumption at N.T.P. in m**3
Q1 = Vg2/t*CV #Heat supplied in kJ/min
T = P*D_b/2 #Brake torque delivered in Nm
bp = 2*math.pi*(N/t*1/60)*(T)*10**-3 #Brake power in kW
bp = round(10*bp)/10
Q_bp = bp*60 #Heat equivalent to brake power in kJ/min
A = math.pi/4*d**2*10**-4 #Area of cylinder in m**2
ip = imep*10**2*l/100*A*(N1/t*1/60) #Indicated power in kW
ip = round(10*ip)/10
Q_ip = ip*60 #Heat equivalent to indicated power in kJ/min
fp = ip-bp #Frictional power in kW
Q_fp = fp*60 #Heat equivalent to frictional power in kJ/min
cp = 4.1868 #Specfic heat of water in kJ/kgK
Q_w = m_w/t*cp*(deltaT_w) #Heat in cooling water in kJ/min
Q_e = Q1-Q_bp-Q_w #Heat to exhaust, radiation in kJ/min
eta_it = Q_ip/Q1 #Indicated thermal efficiency
eta_bt = Q_bp/Q1 #Brake thermal efficiency
#Results:
print " The indicated thermal efficiency, eta_it = %.1f percent"%(eta_it*100)
print " The brake thermal efficiency, eta_bt = %.1f percent"%(eta_bt*100)
print " Heat balance sheet\t Heat supplied by the gas = %d kJ/min, %d percent"%(Q1,Q1/Q1*100)
print "\t Heat equivalent to b.p. = %d kJ/min, %.1f percent"%(Q_bp,Q_bp/Q1*100)
print "\t Heat in cooling water = %d kJ/min, %.1f percent"%(Q_w,Q_w/Q1*100)
print "\t Heat to exhaust, radiation = %d kJ/min, %.1f percent"%(Q_e,Q_e/Q1*100)
```

In [17]:

```
#Given:
Li = 100. #Length of indicator diagram in mm
Ai = 2045. #Area of indicator diagram in mm**2
Pi = 2./10 #Pressure increment in cylinder from indicator pointer in bar/mm
d = 100.;l = 100. #Bore and stroke in mm
N = 900. #Engine speed in rpm
eta_m = 75. #Mechanical efficiency in percent
#Solution:
Hi_av = Ai/Li #Mean height of indicator diagram in mm
imep = Hi_av*Pi #Mean effective pressure in bar
ip = imep*100*math.pi/4*d**2*l*N/(2*60)*10**-9 #Indicated power in kW
bp = ip*eta_m/100 #Brake power in kW
#Results:
print " The mean effective pressure, mep = %.2f bar"%(imep)
print " The indicated power, ip = %.2f kW"%(ip)
print " The brake power, bp = %.2f kW"%(bp)
```

In [20]:

```
from scipy.optimize import fsolve
import math
#Given:
n = 6. #Number of cylinders
bp = 110. #Brake power in kW
N = 1600. #Engine speed in rpm
CV = 43100. #Calorific value in kJ/kg
pC = 86.2/100;pH2 = 13.5/100;pNC = 0.3/100 #Composition of Carbon, Hydrogen and non combustibles in fuel
eta_v = 78. #Volumetric efficiency in percent
eta_it = 38. #Indicated thermal efficiency in percent
eta_m = 80. #Mechanical efficiency in percent
MS = 110. #Mixture strength in percent
l_d = 1.5 #Stroke bore ratio (l/d)
v_a = 0.772 #Specific volume of air in m**3/kg
p_m = 23.1/100;p_v = 20.8/100 #Composition of Oxygen in air by mass and volume
#Solution:
C = 12. #Atomic mass of Carbon(C)
H = 1. #Atomic mass of Hydrogen(H)
O = 16. #Atomic mass of Oxygen(O)
N2 = 14. #Atomic mass of Nitrogen(N)
A_F_s = (pC*2*O/C+pH2*O/(2*H))/p_m #Stoichiometric air fuel ratio
A_F_act = (1+MS/100)*A_F_s #Actual air fuel ratio
Ma = (p_m*2*O)+((1-p_m)*2*N2) #Molecular mass of air per mole air in kg/mole
#Stoichiometric equation of combustion of fuel (petrol)
# 0.862/12[C] + 0.135/2[H2] + [p_v[O2] + (1-p_v)[N2]]*x = a[CO2] + b[H2O] + c[O2] + d[N2]
#Equating coefficients
a = pC/C;b = pH2/(2*H) #On balancing C and H
x = A_F_act/Ma #Moles of air
c = p_v*x-a-b/2 #On balancing O
d = (1-p_v)*x #On balancing N
pCO2 = a/(a+c+d);pO2 = c/(a+c+d);pN2 = d/(a+c+d) #Composition of Carbon di oxide, Oxygen, Nitrogen in dry exhaust
ip = bp/eta_m*100 #Indicated power in kW
m_f = ip/(eta_it/100*CV)*60 #Mass of fuel in kg/min
m_a = m_f*A_F_act #Mass of air in kg/min
V_a = m_a*v_a #Volume of air in m**3/min
V_s = V_a/eta_v*100 #Swept volume in m**3/min
V_s = V_s/(n*N/2) #Swept volume in m**3
def f(d): #Defining a function, f of unknown bore, d
l = l_d*d #Stroke in terms of bore
return math.pi/4*d**2*l-V_s
d = fsolve(f,1)
l = l_d*d #Stroke in m
#Results:
print " The volumetric composition of dry exhaust gas, \tCO2 = %.2f percent \
\nO2 = %.2f percent\
\nN2 = %.2f percent"%(pCO2*100,pO2*100,pN2*100)
print " The bore of the engine, d = %.2f cm The stroke of the engine, l = %.2f cm"%(d*100,l*100)
```

In [21]:

```
import math
#Given:
d = 150.;l = 250. #Bore and stroke in mm
Li = 50. #Length of indicator diagram in mm
Ai = 450. #Area of indicator diagram in mm**2
ISR = 1.2 #Indicator spring rating in mm
N = 420. #Engine speed in rpm
T = 217. #Brake torque delivered in Nm
m_f = 2.95 #Fuel consumption in kg/hr
CV = 44000. #Calorific value in kJ/kg
m_w = 0.068 #Mass of cooling water used in kg/s
deltaT_w = 45. #Cooling water temperature rise in K
cp = 4.1868 #Specfic heat capacity of water in kJ/kgK
#Solution:
Hi_av = Ai/Li #Mean height of indicator diagram in mm
imep = Hi_av/ISR #Mean effective pressure in bar
ip = imep*100*math.pi/4*d**2*l*N/(2*60)*10**-9 #Indicated power in kW (Error in book)
bp = 2*math.pi*(N/60)*(T)*10**-3 #Brake power in kW
eta_m = bp/ip #Mechanical efficiency (Error in book)
eta_bt = bp*3600/(m_f*CV) #Brake thermal efficiency
bsfc = m_f/bp #Brake specific fuel consumption in kg/kWh (Error in book)
#Energy balance
Power_f = m_f/3600*CV #Power in fuel in kW
Power_w = m_w*cp*deltaT_w #Power to cooling water in kW
Power_e = Power_f-bp-Power_w #Power to exhaust, radiation in kW
#Results:
print " The mechanical efficiency, eta_m = %d percent"%(eta_m*100)
print " The brake thermal efficiency, eta_bt = %.1f percent"%(eta_bt*100)
print " The specific fuel consumption, bsfc = %.3f kg/kWh"%(bsfc)
print " Energy balance\t Power in fuel = %.1f kW, %d percent"%(Power_f,Power_f/Power_f*100)
print "\t Brake power = %.2f kW, %.1f percent"%(bp,bp/Power_f*100)
print "\t Power to cooling water = %.1f kW, %.1f percent"%(Power_w,Power_w/Power_f*100)
print "\t Power to exhaust, radiation = %.1f kW, %.1f percent"%(Power_e,Power_e/Power_f*100)
#Answers in the book are wrong
```

In [22]:

```
import math
#Given:
n = 6. #Number of cylinders
d = 70.
l = 100. #Bore and stroke in mm
V_c = 67. #Clearance volume in cm**2
N = 3960. #Engine speed in rpm
m_f = 19.5 #Fuel consumption in kg/hr
T = 140. #Brake torque delivered in Nm
CV = 44000. #Calorific value in kJ/kg
g = 1.4 #Specific heat ratio for air (gamma)
#Solution:
bp = 2*math.pi*N/60*T*10**-3 #Brake power in kW
A = math.pi/4*d**2*10**-6 #Area of cylinder in m**2
bmep = bp*1000/(n*l/1000*A*N/(2*60)) #Brake mean effective pressure in Pascal
eta_bt = bp*3600/(m_f*CV) #Brake thermal efficiency
V_s = (math.pi/4)*d**2*l/1000 #Swept volume of one cylinder in cm**3
r = (V_s+V_c)/V_c #Compression ratio
eta = 1-1/r**(g-1) #Air standard efficiency
eta_r = eta_bt/eta #Relative efficiency
#Results:
print " a)The brake power, bp = %d kW"%(bp)
print " b)The brake mean effective pressure, bmep = %.2f bar"%(bmep*10**-5)
print " c)The brake thermal efficiency, eta_bt = %.1f percent"%(eta_bt*100)
print " d)The relative efficiency, eta_r = %.1f percent"%(eta_r*100)
```

In [23]:

```
import math
#Given:
d = 178.;l = 330. #Bore and stroke in mm
N = 400. #Engine speed at full load in rpm
wmep = 6.2 #Working loop mep in bar
pmep = 0.35 #Pumping loop mep in bar
mep_dc = 0.62 #Mean effective pressure from the dead cycles in bar
N_f = 47. #Number of firing strokes at no load in rpm
#Solution:
imep = wmep-pmep #Net indicated mean effective pressure per cycle in bar
N_d = N/2-N_f #Number of dead cycles at no load in rpm
ip1 = imep*100*l*math.pi/4*d**2*N_f/60*10**-9 #Indicated power at no load in kW
pp_dc = mep_dc*100*l*math.pi/4*d**2*N_d/60*10**-9 #Pumping power of dead cycles when no load in kW
fp = ip1-pp_dc #Friction power in kW
ip = imep*100*l*math.pi/4*d**2*N/(2*60)*10**-9 #Indicated power at full load in kW
bp = ip-fp #Brake power at full load in kW
eta_m = bp/ip #Mechanical efficiency at full load
#Results:
print " The brake power at full load, b.p. = %.2f kW"%(bp)
print " The mechanical efficiency at full load, eta_m = %.1f percent"%(eta_m*100)
```

In [24]:

```
import math
#Given:
d = 200.;l = 250. #Bore and stroke in mm
imep = 4.5*10**5 #Indicated mean effective pressure in N/m**2
m_f = 7. #Fuel consumption in kg/hr
CV = 43500. #Calorific value in kJ/kg
N = 180. #Engine speed in rpm
#Solution:
#(a)
ip = imep*l*math.pi/4*d**2*N/60*10**-9*10**-3 #Indicated power in kW
#(b)
eta_it = ip*3600/(m_f*CV) #Indicated thermal efficiency
#Results:
print " a)The indicated power, ip = %.1f kW"%(ip)
print " b)The indicated thermal efficiency, eta_it = %.1f percent"%(eta_it*100)
```