import math
#Given:
n = 4. #Number of cylinders
d = 68./10 #Bore in cm
l = 75./10 #Stroke in cm
r = 8. #Compression ratio
#Solution:
V_s = (math.pi/4)*d**2*l #Swept volume of one cylinder in cm**3
cubic_capacity = n*V_s #Cubic capacity in cm**3
#Since, r = (V_c + V_s)/V_c
V_c = V_s/(r-1) #Clearance volume in cm**3
#Results:
print " The cubic capacity of the engine = %.1f cm**3"%(cubic_capacity)
import math
#Given:
ip = 10. #Indicated power in kW
eta_m = 80. #Mechanical efficiency in percent
#Solution:
#Since, eta_m = bp/ip
bp = (eta_m/100)*ip #Brake power in kW
fp = ip-bp #Friction power in kW
#Results:
print " The brake power delivered, bp = %d kW"%(bp)
print " The friction power, fp = %d kW"%(fp)
import math
#Given:
bp = 100. #Brake power at full load in kW
fp = 25. #Frictional power in kW (printing error)
#Solution:
eta_m = bp/(bp+fp) #Mechanical efficiency at full load
#(a)At half load
bp = bp/2 #Brake power at half load in kW
eta_m1 = bp/(bp+fp) #Mechanical efficiency at half load
#(b)At quarter load
bp = bp/2 #Brake power at quarter load in kW
eta_m2 = bp/(bp+fp) #Mechanical efficiency at quarter load
#Results:
print " The mechanical efficiency at full load, eta_m = %d percent"%(eta_m*100)
print " The mechanical efficiency, \
\na)At half load, eta_m = %.1f percent \
\nb)At quarter load, eta_m = %d percent"%(eta_m1*100,eta_m2*100)
#Data in the book is printed wrong
import math
#Calculations on four stroke petrol engine
#Given:
bp = 35. #Brake power in kW
eta_m = 80. #Mechanical efficiency in percent
bsfc = 0.4 #Brake specific fuel consumption in kg/kWh
A_F = 14./1 #Air-fuel ratio
CV = 43000. #Calorific value in kJ/kg
#Solution:
#(a)
ip = bp*100/eta_m #Indicated power in kW
#(b)
fp = ip-bp #Frictional power in kW
#(c)
#Since, 1 kWh = 3600 kJ
eta_bt = 1/(bsfc*CV/3600) #Brake thermal efficiency
#(d)
eta_it = eta_bt/eta_m*100 #Indicated thermal efficiency
#(e)
m_f = bsfc*bp #Fuel consumption in kg/hr
#(f)
m_a = A_F*m_f #Air consumption in kg/hr
#Results:
print " a)The indicated power, ip = %.2f kW \
\nb)The friction power, fp = %.2f kW"%(ip,fp)
print " c)The brake thermal efficiency, eta_bt = %.1f percent \
\nd)The indicated thermal efficiency, eta_it = %.1f percent"%(eta_bt*100,eta_it*100)
print " e)The fuel consumption per hour, m_f = %.1f kg/hr \
\nf)The air consumption per hour, m_a = %d kg/hr"%(m_f,m_a)
#Given:
F_A = 0.07/1 #Fuel-air ratio
bp = 75. #Brake power in kW
eta_bt = 20. #Brake thermal efficiency in percent
rho_a = 1.2 #Density of air in kg/m**3
rho_f = 4*rho_a #Density of fuel vapour in kg/m**3
CV = 43700. #Calorific value of fuel in kJ/kg
#Solution:
m_f = bp*3600/(eta_bt*CV/100) #Fuel consumption in kg/hr
m_a = m_f/F_A #Air consumption in kg/hr
V_a = m_a/rho_a #Volume of air in m**3/hr
V_f = m_f/rho_f #Volume of fuel in m**3/hr
V_mixture = V_f+V_a #Mixture volume in m**3/hr
#Results:
print " The air consumption, m_a = %.1f kg/hr"%(m_a)
print " The volume of air required, V_a = %.1f m**3/hr"%(V_a)
print " The volume of mixture required = %.1f m**3/hr"%(V_mixture) #printing error)
#Answer in the book is printed wrong
#Given:
bp = 5. #Brake power in kW
eta_it = 30. #Indicated thermal efficiency in percent
eta_m = 75. #Mechanical efficiency in percent (printing error)
#Solution:
ip = bp*100/eta_m #Indicated power in kW
CV = 42000. #Calorific value of diesel(fuel) in kJ/kg
m_f = ip*3600/(eta_it*CV/100) #Fuel consumption in kg/hr
#Density of diesel(fuel) = 0.87 kg/l
rho_f = 0.87 #Density of fuel in kg/l
V_f = m_f/rho_f #Fuel consumption in l/hr
isfc = m_f/ip #Indicated specific fuel consumption in kg/kWh
bsfc = m_f/bp #Brake specific fuel consumption in kg/kWh
#Results:
print " The fuel consumption of engine, m_f in, \
\na)kg/hr = %.3f kg/hr \
\nb)litres/hr = %.2f l/hr"%(m_f,V_f)
print " c)Indicated specific fuel consumption, isfc = %.3f kg/kWh"%(isfc)
print " d)Brake specific fuel consumption, bsfc = %.3f kg/kWh"%(bsfc)
#Data in the book is printed wrong
#Given:
bp = 5000. #Brake power in kW
fp = 1000. #Friction power in kW
m_f = 2300. #Fuel consumption in kg/hr
A_F = 20./1 #Air-fuel ratio
CV = 42000. #Calorific value of fuel in kJ/kg
#Solution:
#(a)
ip = bp+fp #Indicated power in kW
#(b)
eta_m = bp/ip #Mechanical efficiency
#(c)
m_a = A_F*m_f #Air consumption in kg/hr
#(d)
eta_it = ip*3600/(m_f*CV) #Indicated thermal efficiency
#(e)
eta_bt = eta_it*eta_m #Brake thermal efficiency
#Results:
print " a)The indicated power, ip = %d kW"%(ip)
print " b)The mechanical efficiency, eta_m = %d percent"%(eta_m*100)
print " c)The air consumption, m_a = %d kg/hr"%(m_a)
print " d)The indicated thermal efficiency, eta_it = %.1f percent \
\ne)The brake thermal efficiency, eta_bt = %.1f percent"%(eta_it*100,eta_bt*100)