Question

A fluid, contained in a horizontal cylinder fitted with a frictionless leak proof piston, is continuously agitated by means of a stirrer passing through the cylinder cover. The cylinder diameter is 0.40 m. During the stirring process lasting 10 minutes, the piston slowly moves out a distance of 0.485 m against the atmosphere. The net work done by the fluid during the process is 2 kJ. The speed of the electric motor driving the stirrer is 840 rpm. Determine the torque in the shaft and the power output of the motor.

Answer 1

Change of volume = A L

\(= \frac{\pi d^2}4 \times L\)

\(= \frac{\pi \times 0.4^2}4 \times 0.485\) m³

\(= 0.061\) m³

As piston moves against constant atmospheric pressure then work done p \(\Delta\)V

= 101.325 x 0.061 kJ 

= 6.1754 kJ 

Net work done by the fluid = 2 kJ

∴ Net work done by the Motor = 4.1754 kJ

There for power of the motor 

= \(\frac{4.1754 \times 10^3}{10 \times 60}W\)

\(= 6.96\) W

Torque on the shaft = \(\frac {P}W\)

\(= \frac{6.96 \times 60}{2\pi \times 840}\)

\(= 0.0791\) mN