1.

A piston and cylinder machine containing a fluid system has a stirring device in the cylinder the piston is frictionless, and is held down against the fluid due to the atmospheric pressure of 101.325kPa the stirring device is turned 10,000 revolutions with an average torque against the fluid of 1.275MN. Mean while the piston of 0.6m diameter moves out 0.8m. Find the net work transfer for the systems.

Answer»

Given that

Patm = 101.325 × 103 N/m2

Revolution = 10000

Torque = 1.275 × 106

Dia = 0.6m

Distance moved = 0.8m

Work transfer = ?

W.D by stirring device W1 = 2Π × 10000 × 1.275 J = 80.11 KJ ...(i)

This work is done on the system hence it is –ive.

Work done by the system upon surrounding

W2 = F.dx = P.A.d×

= 101.32 × Π/4 × (0.6)2 × 0.8

= 22.92 KJ ...(ii)

Net work done = W1 + W2

= –80.11 + 22.92 = –57.21KJ (–ive sign indicates that work is done on the system)

Wnet = 57.21KJ



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