Chapter 7 : Comparison of SI and CI Engines

Example 7.1 Page No : 183

In [1]:
#Given:
#For SI engine
F_A1 = 1/13.5      #Fuel air ratio
CV1 = 44000.      #Calorific value in kJ/kg
eta_bt1 = 25.      #Brake thermal efficiency in percent
m_f1 = 1.      #Fuel consumption in kg/hr
#For CI engine
A_F2 = 25./1      #Air fuel ratio
CV2 = 42000.      #Calorific value in kJ/kg
eta_bt2 = 38.      #Brake thermal efficiency in percent

#Solution:
#(a)SI engine
bp1 = m_f1*CV1*eta_bt1/(100*3600)      #Brake power in kW
bsfc1 = m_f1/bp1      #Brake specific fuel consumption in kg/kWh
m_a1 = bsfc1/F_A1      #Air consumption in kg/kWh
#(a)CI engine
m_f2 = 1.      #Fuel consumption in kg/hr
bp2 = m_f2*CV2*eta_bt2/(3600*100)      #Brake power in kW
bsfc2 = m_f2/bp2      #Brake specific fuel consumption in kg/kWh
m_a2 = bsfc2*A_F2      #Air consumption in kg/kWh
#Comparison
R_bp = bp1/bp2      #Ratio of brake power of SI to CI
R_bsfc = bsfc1/bsfc2      #Ratio of brake specific fuel consumption of SI to CI
R_m_a = m_a1/m_a2      #Ratio of fuel consumption of SI to CI

#Results:
print " For SI engine\tBrake output, bp  =  %.2f kW/kg of fuel\tBrake specific fuel consumption, bsfc  =  %.3f kg/kWh"%(bp1,bsfc1)
print " For CI engine\tBrake output, bp  =  %.1f kW/kg of fuel\tBrake specific fuel consumption, bsfc  =  %.3f kg/kWh"%(bp2,bsfc2)
print " The air consumption\tfor SI engine, m_a  =  %.2f kg/kWh\tfor CI engine, m_a  =  %.2f kg/kWh"%(m_a1,m_a2)
print " Comparison of SI to CI\tbp  =  %.3f\tbsfc  =  %.3f\tair consumption  =  %.3f"%(R_bp,R_bsfc,R_m_a)

 For SI engine	Brake output, bp  =  3.06 kW/kg of fuel	Brake specific fuel consumption, bsfc  =  0.327 kg/kWh
 For CI engine	Brake output, bp  =  4.4 kW/kg of fuel	Brake specific fuel consumption, bsfc  =  0.226 kg/kWh
 The air consumption	for SI engine, m_a  =  4.42 kg/kWh	for CI engine, m_a  =  5.64 kg/kWh
 Comparison of SI to CI	bp  =  0.689	bsfc  =  1.451	air consumption  =  0.783

Example 7.2 Page No : 188

In [2]:
#Given:
#For SI engine
s1 = 0.72      #Specific gravity of gasoline fuel
CV1 = 44800.      #Calorific value of gasoline fuel in kJ/kg
eta_bt1 = 20.      #Brake thermal efficiency in percent
A_F1 = 14.      #Air fuel ratio
#For CI engine
s2 = 0.87      #Specific gravity of diesel oil
CV2 = 43100.      #Calorific value of diesel oil in kJ/kg
eta_bt2 = 30.      #Brake thermal efficiency in percent
A_F2 = 21.      #Air fuel ratio

#Solution:
#SI engine
bsfc_SI = 3600*100/(eta_bt1*CV1)      #Brake specific fuel consumption in kg/kWh
m_a_SI = A_F1*bsfc_SI      #Air consumption in kg/kWh
#CI engine
bsfc_CI = 3600*100/(eta_bt2*CV2)      #Brake specific fuel consumption in kg/kWh
m_a_CI = A_F2*bsfc_CI      #Air consumption in kg/kWh

#Results:
print " For SI engine\tBrake specific fuel consumption, bsfc_SI  =  %.3f kg/kWh\tAir consumption  =  %.2f kg/kWh"%(bsfc_SI,m_a_SI)
print " For CI engine\tBrake specific fuel consumption, bsfc_CI  =  %.3f kg/kWh\tAir consumption  =  %.2f kg/kWh"%(bsfc_CI,m_a_CI)

 For SI engine	Brake specific fuel consumption, bsfc_SI  =  0.402 kg/kWh	Air consumption  =  5.62 kg/kWh
 For CI engine	Brake specific fuel consumption, bsfc_CI  =  0.278 kg/kWh	Air consumption  =  5.85 kg/kWh