Question

A refrigeration cycle is executed with \(\mathrm{R}-134 \mathrm{a}\) under the saturation dome between the pressure limits of 1.6 and \(0.2 \mathrm{MPa}\). If the power consumption of the refrigerator is \(3 \mathrm{kW},\) the maximum rate of heat removal from the cooled space of this refrigerator is \((a) 0.45 \mathrm{kJ} / \mathrm{s}\) (b) \(0.78 \mathrm{kJ} / \mathrm{s}\) \((c) 3.0 \mathrm{kJ} / \mathrm{s}\) \((d) 11.6 \mathrm{kJ} / \mathrm{s}\) \((e) 14.6 \mathrm{kJ} / \mathrm{s}\)

Short answer

Based on the given data and step-by-step calculations, the maximum rate of heat removal from the refrigeration cycle using R-134a is approximately (e) 14.6 kJ/s.

Step-by-step solution

Consider the given data

We are given: - High pressure, \(P_H = 1.6 \,\text{MPa}\) - Low pressure, \(P_L = 0.2 \,\text{MPa}\) - Power consumption, \(W_{in} = 3 \,\text{kW}\)

Determine the isentropic efficiency estimates

It is useful to determine the isentropic efficiency estimates for the process. We will use data available in R-134a tables (enthalpies and entropies in saturation conditions) to do this. At high pressure: - Entropy at state 1, \(s_1 = s_g(P_H) = -0.5890 \,\text{kJ/kg·K}\) - Enthalpy at state 1, \(h_1 = h_g(P_H) = 317.65 \,\text{kJ/kg}\) And for the low pressure: - Entropy at state 2, \(s_2' = s_g(P_L) = -0.5890 \,\text{kJ/kg·K}\) - Enthalpy at state 2, \(h_2' = h_f+(s_2'-s_f) \cdot \frac{h_{fg}}{s_{fg}} = 235.62 \,\text{kJ/kg}\) where \(h_f\) is the enthalpy of fluid, \(s_f\) is the entropy of fluid, and \(h_{fg}\), \(s_{fg}\) are the differences in enthalpies and entropies between vapor and liquid. Now we can calculate the isentropic efficiency (COP) estimates: - Coefficient of performance (COP)_ideal = \(\frac{h_1 - h_2'}{h_2' - h_f} = 4.788\)

Calculate actual COP

Now that we have the ideal COP, we'll calculate the actual COP using the relation: - COP_actual = \(\frac{Q_L}{W_{in}}\) And remembering that in a refrigeration cycle, \(Q_L = mh_2 - mh_f\), where: - \(m\) is the mass flow rate of the refrigerant The actual COP can now be expressed as: - COP_actual = \(\frac{m(h_2 - h_f)}{3000 \,\text{W}}\)

Calculate maximal heat removal rate

Since COP_actual cannot be higher than COP_ideal, we can find the maximum heat removal rate from the cooled space using the equations above: - \(m(h_2 - h_f) = \text{COP}_\text{actual} W_{in} \leq \text{COP}_\text{ideal} W_{in}\) Therefore, the maximum heat removal rate is: - \(Q_L = \text{COP}_\text{ideal} W_{in} = 4.788 \times 3000 \,\text{W} = 14364 \,\text{W}\) Converting this value to kilojoules per second: - \(Q_{L} = 14.364 \,\text{kJ/s}\) Now examine the given options and the calculated result shows that the maximal heat removal rate is approximately equal to the option (e). Answer: (e) \(14.6 \,\text{kJ/s}\)