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## Q. 2.8

A piston and cylinder machine containing a fluid system has a stirring device as shown in Fig. 2.24. The piston is frictionless and is held down against the fluid due to atmospheric pressure of 1 bar. The stirring device is turned 9600 revolutions with an average torque against the fluid of 1.5 N.m. In the mean time the piston of 0.6 m diameter moves out 0.8 m. find the net work done for the system.

## Verified Solution

Work done on the system by stirring device,

$W_1=-2 \pi N T=-2 \pi \times 9600 \times 1.5=-90478 \text { N.m }$

Work done by the system, $W_2=p A L=\frac{1 \times 10^5 \times \pi \times(0.6)^2 \times 0.8}{4}=22619 \text { N.m }$

Net work done = $W_1+W_2=-90478+22619=-67859 \text { N.m }$