Respuesta :
Answer:
break power = 6.031 kW
indicated power = 6.859 kW
total frictional power = 0.827 kW
total friction imep = 111.26 kpa
Explanation:
given data
Vd = 0.496 L
revolving = 1800 rpm
brake torque T = 32 Nm
gross imep = 933 kPa
net imep = 922 kP
solution
first we get here break power that is
break power = [tex]\frac{2\pi N*T}{60}[/tex] ............1
put here value we get
break power = [tex]\frac{2\pi 1800*32}{60}[/tex]
break power = 6.031 kW
and
indicated power will be
indicated power = [tex]\frac{net\ imep* vd*N/2}{60}[/tex] ..........2
put here value
indicated power = [tex]\frac{922*10^{-3}0.496*900}{60}[/tex]
indicated power = 6.859 kW
and
total frictional power will be
total frictional power = indicated power - break power .......3
total frictional power = 6.859 - 6.031
total frictional power = 0.827 kW
and
total friction imep will be
total friction imep = [tex]\frac{fp* 60}{vd*N/2}[/tex] ..........4
put here value
total friction imep = [tex]\frac{0.827*60}{0.496*10^{-3}*900}[/tex]
total friction imep = 111.26 kpa
