Question

Refrigerant-134a enters the condenser of a residential heat pump at \(800 \mathrm{kPa}\) and \(35^{\circ} \mathrm{C}\) at a rate of \(0.018 \mathrm{kg} / \mathrm{s}\) and leaves at \(800 \mathrm{kPa}\) as a saturated liquid. If the compressor consumes \(1.2 \mathrm{kW}\) of power, determine \((a)\) the COP of the heat pump and \((b)\) the rate of heat absorption from the outside air.

Short answer

Based on the given information, determine the Coefficient of Performance (COP) of the heat pump and the rate of heat absorption from the outside air. Given: Refrigerant conditions at the condenser inlet: 800 kPa and 35°C Refrigerant state at the condenser outlet: Saturated liquid at 800 kPa Mass flow rate of the refrigerant: 0.018 kg/s Power consumed by the compressor: 1.2 kW Formula to find the COP of the heat pump: COP = Qc / Win Solution: 1. Using the Refrigerant-134a table, find specific enthalpies hin and hout for the inlet and outlet of condenser. 2. Calculate the cooling capacity: Qc = 0.018 kg/s (hin - hout) 3. Use the given work input for the heat pump: Win = 1.2 kW 4. Calculate the Coefficient of Performance (COP) of the heat pump: COP = (0.018 kg/s (hin - hout)) / 1.2 kW 5. Calculate the rate of heat absorption from the outside air: Qabs = 0.018 kg/s (hin - hout) + 1.2 kW Once the calculations are complete, you will have the COP of the heat pump and the rate of heat absorption from the outside air.

Step-by-step solution

Determine the properties of the refrigerant at the given states

Using a Refrigerant-134a table (or software), we can find the specific enthalpy (\(h_{in}\)) at the inlet of the condenser (\(800\,\text{kPa}\) and \(35^\circ\,\text{C}\)) and the specific enthalpy (\(h_{out}\)) at the outlet of the condenser where the refrigerant is a saturated liquid at \(800\,\text{kPa}\).

Calculate the cooling capacity

The cooling capacity (\(Q_c\)) can be found using the mass flow rate of the refrigerant (\(\dot{m}\)) and the difference in specific enthalpy between the inlet and outlet of the condenser: \(Q_c = \dot{m}(h_{in}-h_{out})\) Substitute the given values and the specific enthalpies found in Step 1: \(Q_c = 0.018\,\text{kg/s} (h_{in}-h_{out})\)

Calculate the work input for the heat pump

The work input (\(W_{in}\)) is given as \(1.2\,\text{kW}\). We will use this value when calculating the COP of the heat pump.

Calculate the Coefficient of Performance (COP) for the heat pump

Now, we can find the COP using the formula: \(\text{COP} = \dfrac{Q_c}{W_{in}}\) Substitute the values of \(Q_c\) and \(W_{in}\) found in Step 2 and Step 3: \(\text{COP} = \dfrac{0.018\,\text{kg/s} (h_{in}-h_{out})}{1.2\,\text{kW}}\)

Determine the rate of heat absorption from the outside air

Since we know the cooling capacity (\(Q_c\)) and the work input (\(W_{in}\)), we can now calculate the rate of heat absorption from the outside air (\(Q_{abs}\)) using the energy balance equation for the heat pump: \(Q_{abs} = Q_c + W_{in}\) Substitute the values of \(Q_c\) and \(W_{in}\): \(Q_{abs} = 0.018\,\text{kg/s} (h_{in}-h_{out}) + 1.2\,\text{kW}\) Now we have found both the COP of the heat pump and the rate of heat absorption from the outside air.