Please be clear with work. An automobile drive shaft rotates at 3000 rpm and del
ID: 2994500 • Letter: P
Question
Please be clear with work.
An automobile drive shaft rotates at 3000 rpm and delivers 75 kW of power from the engine to the wheels. Determine the torque (N m and ft lbf) in the drive shaft. [Be sure to use the angular velocity with units of radians/second.) A gas is contained in a cylinder-piston. The initial pressure and volume are 14 kPa and 0.028 m3. Determine the work by the gas when the volume is increased to 3.08 m3 in a polytropic process with n=1.4. One cubic meter of an ideal gas expands in an isothermal process from 760 to 350 kPa. Determine the work done by this gas in kJ and BtuExplanation / Answer
1.
Angular velocity w = 2*pi*N/60
= 2*3.14*3000/60
= 314 rad/s
Power = Torque*w
75*10^3 = Torque*314
Torque = 238.85 Nm = 176.17 ft-lbf
2.
P2 / P1 = (V1 / V2)^n
P2 / 14 = (0.028 / 0.08)^1.4
P2 = 3.22 kPa
W = (P1*V1 - P2*V2) / (n - 1)
W = (14*0.028 - 3.22*0.08) / (1.4 - 1)
W = 0.336 kJ
3.
For isothermal process,
W = P1*V1 ln (P1 / P2)
W = 760*1 ln (760 / 350)
W = 589.29 kJ = 558.5 Btu