Question

A fluid contained in a horizontal cylinder with a frictionless piston continuously agitated by a stirrer. Cylinder diameter is 0.4 m and after 10 minutes the piston moves 0.485m. The net work done by the fluid is 2kJ. The speed of the electric motor is 840 rpm. Determine the Torque developed in the shaft and Power output of the motor

Answer 1

Since the piston compresses the fluid by 0.485m, there is a change in fluid volume. Hence, work is done by the piston.

Volume of cylinder = $\pi r^2l$

area of cylinder =0.1256 m²

Change in volume = 0.1256 x 0.485

= 0.061 m³

Since total work done by fluid = 2kJ

Work done by electric motor = work done by piston - work done by fluid.

Work done by piston = PΔV

= 1.013 x 10⁵ x 0.061

= 6.1793 kJ

∴ Work done by motor = 6.1793 - 2 = 4.174kJ

Power on the motor = Work / time

= 4.174 / 600 = 6.96 ≈ 7 W

Power = Torque x angular speed

T = P/ω

T = 7/( 2π × 840 ÷ 60)

T = 0.08 Nm