Question

Refrigerant-134a enters the compressor of a refrigerator as superheated vapor at \(0.20 \mathrm{MPa}\) and \(-5^{\circ} \mathrm{C}\) at a rate of \(0.07 \mathrm{kg} / \mathrm{s},\) and it leaves at \(1.2 \mathrm{MPa}\) and \(70^{\circ} \mathrm{C}\). The refrigerant is cooled in the condenser to \(44^{\circ} \mathrm{C}\) and \(1.15 \mathrm{MPa}\), and it is throttled to 0.21 MPa. Disregarding any heat transfer and pressure drops in the connecting lines between the components, show the cycle on a \(T\) -s diagram with respect to saturation lines, and determine ( \(a\) ) the rate of heat removal from the refrigerated space and the power input to the compressor, \((b)\) the isentropic efficiency of the compressor, and \((c)\) the \(C O P\) of the refrigerator.

Short answer

Question: Determine (a) the rate of heat removal from the refrigerated space and the power input to the compressor, (b) the isentropic efficiency of the compressor, and (c) the coefficient of performance (COP) of the refrigerator for a vapor-compression refrigeration cycle with given states at various components.

Step-by-step solution

Identifying the states and drawing the T-s diagram

First, we need to identify the states of the refrigerant in each component of the cycle. Assign the state numbers as follows: 1. Refrigerant entering the compressor (superheated vapor) 2. Refrigerant leaving the compressor 3. Refrigerant leaving the condenser 4. Refrigerant leaving the throttling valve On a T-s diagram, plot these states with respect to the saturation lines. Since we know the temperatures and pressures at each state, this should be straightforward. Take note of the isentropic process (constant entropy) from state 1 to state 2s (theoretical state where refrigerant would leave the compressor if the process was isentropic).

Determining specific enthalpies and entropies

Using the given pressure and temperature values at the states and property tables for Refrigerant-134a, determine the specific enthalpies (h) and specific entropies (s) at each state: 1. \(h_{1}, s_{1}\) 2. \(h_{2}, s_{2}\) 3. \(h_{3}, s_{3}\) 4. \(h_{4}, s_{4}\) For state 2s (refrigerant leaving the compressor if the process was isentropic), we know that the entropy is the same as state 1: \(s_{2s} = s_{1}\). Use this information to find the specific enthalpy at state 2s: \(h_{2s}\).

Rate of heat removal from the refrigerated space and power input to the compressor

Calculate the rate of heat removal from the refrigerated space (Q) and the power input to the compressor (W) using the mass flow rate and the specific enthalpies. Rate of heat removal from refrigerated space: \(Q = \dot{m} (h_{1} - h_{4})\) Power input to the compressor: \(W = \dot{m} (h_{2} - h_{1})\) Where \(\dot{m}\) is the mass flow rate.

Isentropic efficiency of the compressor

Calculate the isentropic efficiency of the compressor using the actual and isentropic specific enthalpies: Isentropic efficiency = \(\frac{h_{1} - h_{4}}{h_{1} - h_{2s}}\)

Coefficient of Performance (COP) of the refrigerator

Calculate the Coefficient of Performance (COP) of the refrigerator using the rate of heat removal from the refrigerated space and the power input to the compressor: COP = \(\frac{Q}{W}\) Now we have found (a) the rate of heat removal from the refrigerated space and the power input to the compressor, (b) the isentropic efficiency of the compressor, and (c) the COP of the refrigerator.