Thermodynamics Problem 1. An automobile drive shaft rotates at 3000 rpm and deli

Question

Thermodynamics

Problem 1. 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.) Problem 2. 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 0.08 m3 in a polytropic process with n-1.4 Problem 3. 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 ku and Btu. (Answer: 589 kJ)

Explanation / Answer

1) Given

N = 3000 rpm

Power(P) = 75 kW

Angular Velocity(w) = 2*pi*(N/60) = 314 rad/s

We know

Power = Torque*angular velocity

Torque = 0.2388 kN.m = 238.8 N.m = 176.13 ft.lbf

2) Given

P1 = 14 kPa

V1 = 0.028 m^3

V2 = 0.08

n = 1.4

In a polytropic process

P*V^n = constant

P1*V1^n = P2*V2^n

P2 = 3.22 kPa

Work done in a polytropic process

W = (P1*V1 - P2*V2)/(1-n)    = (14*0.028 - 3.22*0.08)/(1-1.4) = 0.336 kJ = 336 J

3) Given

V1 = 1 m^3

P1 = 760 kPa

P2 = 350 kPa

For an Ideal gas

P*V = n*R*T

Work done in an isothermal process

W = n*R*T1*Ln(P1/P2) = P1*V1*Ln(P1/P2) = 589.29 kJ = 558.54 Btu

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