Chapter 15 : Engine Cooling

Example 15.1 Page No : 323

In [1]:
#Given:
bp  =  100.      #Brake power in kW
deltaT  =  30.      #Temperature raised of water in degreeC
p_p  =  30
p_d  =  26.      #Percentage of energy going to coolent in petrol and diesel
eta_p  =  26.
eta_d  =  31.      #Efficiency of petrol and diesel engine in percent
s  =  4.1868      #Specific heat capacity of water in J/kgK

#Solution:
#Petrol engine
CW_p  =  bp*(p_p/100)/((eta_p/100)*deltaT*s)      #Amount of cooling water required in petrol engine in kg/s
#Diesel engine
CW_d  =  bp*(p_d/100)/((eta_d/100)*deltaT*s)      #Amount of cooling water required in diesel engine in kg/s

#Results:
print " Amount of cooling water required in petrol engine   =   %d kg/hr"%(CW_p*3600)
print " Amount of cooling water required in diesel engine   =   %.1f kg/hr"%(CW_d*3600)
 Amount of cooling water required in petrol engine   =   0 kg/hr
 Amount of cooling water required in diesel engine   =   2403.9 kg/hr