#Given:
bp = 80. #Brake power in kW
eta_m = 80. #Mechanical efficiency in percent
bsfc = 258. #Brake specific fuel consumption in gm/kWh
Reduction = 3.7 #Reduction in friction power in kW
#Solution:
ip1 = bp*100/eta_m #Initial indicated power in kW
fp1 = ip1-bp #Initial friction power in kW
fp2 = fp1-Reduction #Final friction power in kW
ip2 = bp+fp2 #Final indicated power in kW
eta_m2 = bp/ip2 #Final mechanical efficiency
bsfc2 = bsfc*(eta_m/(100*eta_m2)) #Final brake specific fuel consumption in gm/kWh
Saving = bp*(bsfc-bsfc2)/1000 #Saving in fuel in kg/hr
#Results:
print " a)The new mechanical efficiency, eta_m = %.3f"%(eta_m2)
print " b)The new bsfc = %.1f gm/kWh"%(bsfc2)
print " c)The saving in fuel per hour = %.2f kg/hr"%(Saving)
#Answers in the book are wrong
#Given:
eta_it = 30. #Indicated thermal efficiency in percent
fp_1500 = 18. #Friction power at 1500 rpm in kW
fp_2500 = 45. #Friction power at 2500 rpm in kW
bp = 75. #Brake power in kW
CV = 44000. #Calorific value of fuel in kJ/kg
#Solution:
isfc = 3600/(CV*eta_it/100) #Indicated specific fuel consumption in kg/kWh
eta_m_1500 = bp/(bp+fp_1500) #Mechanical efficiency at 1500 rpm
bsfc_1500 = isfc/eta_m_1500 #Brake specific fuel consumption at 1500 rpm in kg/kWh
eta_m_2500 = bp/(bp+fp_2500) #Mechanical efficiency at 2500 rpm
bsfc_2500 = isfc/eta_m_2500 #Brake specific fuel consumption at 2500 rpm in kg/kWh
#Results:
print " The brake specific fuel consumption\tat 1500 rpm, bsfc_1500 = %.3f kg/kWh\tat 2500 rpm, bsfc_2500 = %.3f kg/kWh"%(bsfc_1500,bsfc_2500)