Question

Consider a \(300 \mathrm{kJ} / \mathrm{min}\) refrigeration system that operates on an ideal vapor-compression refrigeration cycle with refrigerant- 134 a as the working fluid. The refrigerant enters the compressor as saturated vapor at \(140 \mathrm{kPa}\) and is compressed to \(800 \mathrm{kPa}\). Show the cycle on a \(T\) -s diagram with respect to saturation lines, and determine ( \(a\) ) the quality of the refrigerant at the end of the throttling process, ( \(b\) ) the coefficient of performance, and ( \(c\) ) the power input to the compressor.

Short answer

b) What is the coefficient of performance (COP) of the refrigeration cycle? c) What is the power input to the compressor in kW?

Step-by-step solution

Show the cycle on a T-s diagram

To do this, first plot the saturation lines for refrigerant-134a on a T-s diagram. Then locate the four points of the cycle: 1. Saturated vapor entering the compressor at 140 kPa 2. Compressed, superheated vapor leaving the compressor at 800 kPa 3. Saturated liquid leaving the condenser 4. The refrigerant after the throttling process and before entering the evaporator. Its pressure is equal to state 1 (since it's an ideal cycle).

Determine the state of refrigerant-134a in each phase

Using the properties of refrigerant-134a at the given pressures, find the temperatures, specific enthalpies, and specific entropies at each state: State 1: - Pressure: \(P_1 = 140 \mathrm{kPa}\) - Since it's a saturated vapor, use tables to find \(T_1, h_1,\text{ and } s_1\). State 2: - Pressure: \(P_2 = 800 \mathrm{kPa}\) - Since it's an ideal cycle, \(s_2 = s_1\). Use tables to find \(T_2\) and \(h_2\). State 3: - Pressure: \(P_3 = 800 \mathrm{kPa}\) - Since it's a saturated liquid, use tables to find \(T_3, h_3,\text{ and } s_3\).

The quality of the refrigerant at the end of the throttling process (state 4)

We know that in the throttling process the enthalpy remains constant, so \(h_4 = h_3\). Since we know the pressure and enthalpy of state 4 (\(P_4 = 140 \mathrm{kPa}\) and \(h_4 = h_3\)), we can use the refrigerant property tables to find the quality of the refrigerant at state 4, \(x_4\): $$ x_4 = \frac{h_4 - h_{f,4}}{h_{g,4} - h_{f,4}} $$

Calculate the coefficient of performance (COP)

The COP of a refrigeration cycle is defined as the ratio of the heat removed from the substance being cooled (Q_in) to the work input to the compression cycle (W_comp): $$ \text{COP} = \frac{Q_\text{in}}{W_\text{comp}} $$ We can find \(Q_\text{in}\) by using the mass flow rate, \(m_\text{dot}\) and the difference in enthalpies between state 1 and state 4: $$ Q_\text{in} = m_\text{dot}(h_1 - h_4) $$ We can find \(W_\text{comp}\) by using the mass flow rate, \(m_\text{dot}\) and the difference in enthalpies between state 1 and state 2: $$ W_\text{comp} = m_\text{dot}(h_2 - h_1) $$ To find the mass flow rate, we can use the cooling capacity, which is equal to \(Q_\text{in}\): $$ m_\text{dot} = \frac{300 \,\mathrm{kJ/min}}{(h_1 - h_4) \cdot 60 \,\mathrm{s/min}} $$ Finally, substitute the derived expressions for \(Q_\text{in}\) and \(W_\text{comp}\) in the COP equation.

Calculate the power input to the compressor

Using the derived expression for \(W_\text{comp}\): $$ W_\text{comp} = m_\text{dot}(h_2 - h_1) $$ Calculate the power input to the compressor by substituting the calculated mass flow rate, \(m_\text{dot}\) and the specific enthalpies, \(h_1\) and \(h_2\). Note that the unit of the power input will be kW.