Ammonia enters the compressor of an industrial refrigeration plant at 2 bar, −10°c with a mass flow rate of 15 kg/min and is compressed to 12 bar, 140°c. heat transfer occurs from the compressor to its surroundings at a rate of 6 kw. for steady-state operation with negligible kinetic and potential energy effects, determine (a) the power input to the compressor, in kw, and (b) the rate of entropy production, in kw/k, for a control volume enclosing the compressor and its immediate surroundings such that the heat transfer occurs at 300 k.

a) W=85.225 kW

b)

Explanation:

First, consider the energy balance for the compressor: The energy that enters to the system (W and enthalpy of the feed flow) is equal to the energy that goes out from it (Heat Q and enthalpy of the exit flow):

Consider the enthalpy data from van Wylen 6th edition, Table B.2.2. According to that, ,  

So, the power input to the compressor is:

b) The differential entropy change dS for a reversible heat transfer dQ at a temperature T is:

This equation can be integrated if the heat transfer surface temperature remains constant, which is the case, giving as a result:

96.77 amp

explanation:

rl = 250/100 = 2.5 ohms.

egl = 250 + (100 × 0.1) + 2 = 262 v.

at 700 r.p.m., eg2 = 262 × 700/1000 = 183.4 v.

if ia is the new current, then eg2 - 2 - (ia × 0.1) = 2.5 ia

which means this will give ia = 96.77 amp.