Question

A system containing m = 5 kg of a liquid is stirred with a torque of T = 0.3 kgf-m
at a speed of X = 1000 RPM for t = 24 hr. The system meanwhile expands from
V1= 1m³ to V2 = 2m³ against a constant pressure of p = 4 kgf/cm². Determine
the net work done (W) in kJ.

Answer 1

Pressure is given, $P = 4kgf/cm^2 = 4000 \times 9.8 \times 10000 N/m^2 = 39.2 \times 10^7 N/m^2$

Initial Volume, $V_i = 1m^3$

Final Volume, $V_f = 2 m^3$

Work done by against pressure, $W_p = P(V_f - V_i) = 39.2\times 10^7 (2-1) J =39.2\times 10^7 J$

Torque acting on the water, $\tau = 0.3 kgfm = 0.3 \times 1000 \times9.8 = 2940 Nm$

Angular speed , $\omega = 1000RPM = \frac{2 \pi \times1000}{60 } rad/s = 104.72$ $rad/s$

time. $t = 24 hrs = 24 \times 3600 s = 86400s$

Work done by torque, $W_{\tau} = \tau \times \theta =\tau \times \omega t= (2940 \times 104.72 \times 86400) = 2660.1 \times 10^7 J$

Total work done, $W = W_p + W_{\tau} = (39.2 + 2660.1)\times 10^7 J = 2699.3\times 10^7 J$