Question

The evaporator of a refrigeration cycle is basically a heat exchanger in which a refrigerant is evaporated by absorbing heat from a fluid. Refrigerant-22 enters an evaporator at \(200 \mathrm{kPa}\) with a quality of 22 percent and a flow rate of 2.65 L/h. \(R-22\) leaves the evaporator at the same pressure superheated by \(5^{\circ} \mathrm{C}\). The refrigerant is evaporated by absorbing heat from air whose flow rate is \(0.75 \mathrm{kg} / \mathrm{s}\). Determine (a) the rate of heat absorbed from the air and ( \(b\) ) the temperature change of air. The properties of \(R-22\) at the inlet and exit of the condenser are \(h_{1}=220.2 \mathrm{kJ} / \mathrm{kg}, v_{1}=0.0253 \mathrm{m}^{3} / \mathrm{kg}\) and \(h_{2}=398.0 \mathrm{kJ} / \mathrm{kg}\).

Short answer

Answer: The rate of heat absorbed from the air is 463 W, and the temperature change of the air is a decrease of 0.000613°C.

Step-by-step solution

Convert given variables to SI units if necessary

The refrigerant flow rate is given as 2.65 L/h. We need to convert it to a standard unit like kg/s. The specific volume at the inlet (v1) is provided as 0.0253 m³/kg. Flow rate = 2.65 L/h = 2.65 * (10^(-3)) m³/h = 2.65 * (10^(-3)) / 3600 m³/s Now we can find the mass flow rate of the refrigerant as follows: Flow rate = Mass flow rate × specific volume => Mass flow rate = Flow rate / specific volume

Calculate the mass flow rate of the refrigerant

Using the converted flow rate and given specific volume v1, we calculate the mass flow rate. \(m_{\mathrm{R22}} = \dfrac{2.65 \times (10^{-3})}{3600 \times 0.0253} = 2.606 \times 10^{-6} \thinspace \mathrm{kg/s}\)

Apply the conservation of energy

We use the energy conservation equation, where the net rate of heat transfer equals the rate of heat absorbed from the air: \(Q_{\text{air}} = m_{\mathrm{R22}}(h_2 - h_1)\). We know the mass flow rate of the refrigerant and the specific enthalpies \(h_1 = 220.2 \thinspace \mathrm{kJ/kg}\) and \(h_2 = 398.0 \thinspace \mathrm{kJ/kg}\).

Calculate the rate of heat absorbed from the air

Using the known values: \(Q_{\text{air}} = (2.606 \times 10^{-6}) \times (398.0 - 220.2) \thinspace \mathrm{kJ/s}\) \(Q_{\text{air}} = 0.000463 \thinspace \mathrm{kJ/s}\) or \(463 \thinspace \mathrm{W}\) Thus, the rate of heat absorbed from the air is 463 W.

Calculate the specific heat of air

Air is classified as an ideal gas, and we can assume constant specific heat capacity \(c_{\mathrm{p}}\) at room temperature as \(c_{\mathrm{p,air}} = 1.005 \thinspace \mathrm{kJ/kg^{\circ}C}\).

Compute the temperature change of the air

Now, we apply the energy conservation equation for the air, considering that the mass flow rate of the air is 0.75 kg/s: \(Q_{\text{air}} = m_{\mathrm{air}} \text{ }c_{\mathrm{p,air}} \text{ }\Delta T_{\text{air}}\) Solving for \(\Delta T_{\text{air}}\): \(\Delta T_{\text{air}} = \dfrac{Q_{\text{air}}}{m_{\mathrm{air}} \text{ }c_{\mathrm{p,air}}} = \dfrac{0.000463 \thinspace \mathrm{kJ/s}}{0.75 \thinspace \mathrm{kg/s} \times 1.005 \thinspace \mathrm{kJ/kg^{\circ}C}}\) \(\Delta T_{\text{air}} = -0.000613 \thinspace\mathrm{^{\circ}C}\) The negative sign suggests that the air's temperature decreases by 0.000613°C due to the heat absorption. In conclusion, the rate of heat absorbed from the air is \(463 \thinspace \mathrm{W}\), and the temperature change of the air is a decrease of \(0.000613^{\circ} \mathrm{C}\).