Question

A refrigeration unit operates on the ideal vapor compression refrigeration cycle and uses refrigerant-22 as the working fluid. The operating conditions for this unit are evaporator saturation temperature of \(-5^{\circ} \mathrm{C}\) and the condenser saturation temperature of \(45^{\circ} \mathrm{C}\). Selected data for refrigerant- 22 are provided in the table below. $$\begin{array}{lcccc}\hline T,^{\circ} \mathrm{C} & P_{\text {sat }}, \mathrm{kPa} & h_{f}, \mathrm{kJ} / \mathrm{kg} & h_{g}, \mathrm{kJ} / \mathrm{kg} & s_{g}, \mathrm{kJ} / \mathrm{kg} \cdot \mathrm{K} \\ \hline-5 & 421.2 & 38.76 & 248.1 & 0.9344 \\\45 & 1728 & 101 & 261.9 & 0.8682 \\\\\hline\end{array}$$ For \(R-22\) at \(P=1728\) kPa and \(s=0.9344 \mathrm{kJ} / \mathrm{kg} \cdot \mathrm{K}\) \(T=68.15^{\circ} \mathrm{C}\) and \(h=283.7 \quad \mathrm{kJ} / \mathrm{kg} .\) Also, take \(c_{p, \text { air }}=1.005 \mathrm{kJ} / \mathrm{kg} \cdot \mathrm{K}\) (a) Sketch the hardware and the \(T\) -s diagram for this heat pump application. (b) Determine the COP for this refrigeration unit. (c) The evaporator of this unit is located inside the air handler of the building. The air flowing through the air handler enters the air handler at \(27^{\circ} \mathrm{C}\) and is limited to a \(20^{\circ} \mathrm{C}\) temperature drop. Determine the ratio of volume flow rate of air entering the air handler \(\left(m_{\text {ail }}^{3} / \min \right)\) to mass flow rate of \(\mathrm{R}-22\left(\mathrm{kg}_{\mathrm{R}-22} / \mathrm{s}\right)\) through the air handler, in \(\left(m_{\mathrm{air}}^{3} / \mathrm{min}\right) /\left(\mathrm{kg}_{\mathrm{R}-22} / \mathrm{s}\right) .\) Assume the air pressure is \(100 \mathrm{kPa}\).

Short answer

Question: Sketch the hardware and T-s diagram for the given heat pump application, calculate the COP, and determine the ratio of volume flow rate of air entering the air handler to mass flow rate of R-22 through the air handler. Answer: The hardware sketch for the vapor compression refrigeration cycle should include four main components: evaporator, compressor, condenser, and expansion valve. The T-s diagram should be plotted based on the provided data and saturation temperatures. The COP for this refrigeration unit is 5.88, and the ratio of volume flow rate of air entering the air handler to mass flow rate of R-22 through the air handler is 0.00824 \(\frac{m^3/min}{kg/s}\).

Step-by-step solution

T-s Diagram and Hardware

A vapor compression refrigeration cycle consists of four main components: evaporator, compressor, condenser, and expansion valve. The T-s diagram for this cycle can be plotted using the data from the provided table and the saturation temperatures. Plot the points for condenser saturation temperature (45°C) with entropy \(s_g=0.8682 \ kJ/kg \cdot K\) and evaporator saturation temperature (-5°C) with entropy \(s_g=0.9344 \ kJ/kg \cdot K\). The hardware sketch should include these 4 main components and their respective relations. (b): Determine the COP for this refrigeration unit.

Coefficient of Performance Calculation

To calculate the COP, we need to use the following formula: $$\text{COP}=\frac{Q_{in}}{W_{in}}$$ Where \(Q_{in}\) is the heat absorbed in the evaporator and \(W_{in}\) is the work input to the compressor. From the provided table, we find the enthalpy values for the evaporator and the condenser. Calculate \(Q_{in}\) and \(W_{in}\) as follows: $$Q_{in}=h_1-h_4=248.1-38.76=209.34 \ kJ/kg$$ $$W_{in}=h_2-h_1=283.7-248.1=35.6 \ kJ/kg$$ Now, compute the COP: $$\text{COP}=\frac{209.34}{35.6}=5.88$$ The COP of this refrigeration unit is 5.88. (c): Determine the ratio of volume flow rate of air entering the air handler to mass flow rate of R-22 through the air handler.

Volume Flow Rate to Mass Flow Rate Ratio Calculation

We know the temperature drop of the air across the evaporator is 20°C. From this information, we can calculate the heat transfer rate of air, assuming constant pressure and known specific heat capacity: $$Q_{air}=m_{air}c_{p, air}(T_{in}-T_{out})$$ We also have the heat transfer rate in the evaporator, which is equal to \(Q_{in}\) calculated before: $$Q_{in}=m_{R-22}(h_1-h_4)$$ Since \(Q_{air}=Q_{in}\), we can write: $$m_{air}c_{p, air}(T_{in}-T_{out})=m_{R-22}(h_1-h_4)$$ Now, we can find the ratio of volume flow rate of air to mass flow rate of R-22: $$\frac{\dot{V}_{air}}{m_{R-22}} = \frac{c_{p, air} (T_{in} - T_{out})}{(h_1 - h_4) \rho_{air}}$$ Assuming air pressure is 100 kPa, and using the ideal gas law, we can calculate the density of air: $$\rho_{air} = \frac{P_{air}}{RT_{air}} = \frac{100 \times 10^3}{287 \times 300} =1.16 \ kg/m^3$$ Now we can calculate the ratio: $$\frac{\dot{V}_{air}}{m_{R-22}} = \frac{1.005 (20)}{(209.34) (1.16)} = 0.00824 \ \frac{m^3/min}{kg/s}$$ The ratio of volume flow rate of air entering the air handler to mass flow rate of R-22 through the air handler is 0.00824 \(\frac{m^3/min}{kg/s}\).