Question

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.

Answer 1

Answer:

a) W=85.225 kW

b) $0.02\frac{kW}{K}$

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):

$W+m*h_1=Q+m*h_2\\W=Q+m*(h_2-h_1)$

Consider the enthalpy data from van Wylen 6th edition, Table B.2.2. According to that, $h_1=h(200kPa,-10C)=1440.6\frac{kJ}{kgK}$,  $h_2=h(1200kPa,140C)=1757.5\frac{kJ}{kgK}$

So, the power input to the compressor is:

$W=6kW+15\frac{kg}{min}*\frac{min}{60s}*(1757.5-1440.6)\frac{kJ}{kg}\\W=85.225kW$

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

$dS=\frac{dQ}{T}$

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

$S_2-S_1=\frac{Q}{T}=\frac{6kW}{300K}=0.02\frac{kW}{K}$