Refrigerant 134a at p1 = 30 lbf/in2, T1 = 40oF enters a compressor operating at steady state with a mass flow rate of 200 lb/h and exits as saturated vapor at p2 = 160 lbf/in2. Heat transfer occurs from the compressor to its surroundings, which are at T0 = 40oF. Changes in kinetic and potential energy can be ignored. The power input to the compressor is 2 hp. Determine the heat transfer rate for the compressor, in Btu/hr, and the entropy production rate for the compressor, in Btu/hr·oR.

a)

b)

Explanation:

From the properties of Super-heated Refrigerant 134a Vapor at , ; we obtain the following properties for specific enthalpy and specific entropy.

So; specific enthalpy

specific entropy

Also; from the properties of saturated Refrigerant 134 a vapor (liquid - vapor). pressure table at ; we obtain the following properties:

Given that the power input to the compressor is 2 hp;

Then converting to  Btu/hr ;we known that since 1 hp = 2544.4342 Btu/hr

2 hp = 2 × 2544.4342 Btu/hr

2 hp = 5088.8684 Btu/hr

The steady state energy for a compressor can be expressed by the formula:

By neglecting kinetic and potential energy effects; we have:

b)  To determine the entropy generation; we employ the formula:

In a steady state condition

Hence;

it was a fall down when it happens

explanation: