Question

Refrigerant-22 absorbs heat from a cooled space at \(50^{\circ} \mathrm{F}\) as it flows through an evaporator of a refrigeration system. \(R-22\) enters the evaporator at \(10^{\circ} \mathrm{F}\) at a rate of \(0.08 \mathrm{lbm} / \mathrm{s}\) with a quality of 0.3 and leaves as a saturated vapor at the same pressure. Determine \((a)\) the rate of cooling provided, in \(\mathrm{Btu} / \mathrm{h},(b)\) the rate of exergy destruction in the evaporator, and \((c)\) the second-law efficiency of the evaporator. Take \(T_{0}\) \(=77^{\circ} \mathrm{F} .\) The properties of \(\mathrm{R}-22\) at the inlet and exit of the evaporator are: \(h_{1}=107.5 \mathrm{Btu} / \mathrm{lbm}, s_{1}=0.2851 \mathrm{Btu} / \mathrm{lbm} \cdot \mathrm{R}\) \(h_{2}=172.1 \mathrm{Btu} / \mathrm{lbm}, s_{2}=0.4225 \mathrm{Btu} / \mathrm{lbm} \cdot \mathrm{R}\)

Short answer

Answer: To find these values, follow the 4-step solution provided above: 1. Calculate the heat transfer rate in the evaporator (Q) using the formula: \(Q = \dot{m} \times (h_{2} - h_{1})\) 2. Convert the heat transfer rate to Btu/h using the conversion factor 3600. 3. Calculate the rate of exergy destruction in the evaporator using the formula: \(\dot{E}_{\text{destruction}} = T_{0} \times \dot{m} \times (s_{2} - s_{1})\) 4. Calculate the second-law efficiency of the evaporator using the formula: \(\eta_{\text{II}} = 1 - \frac{\dot{E}_{\text{destruction}}}{\dot{E}_{\text{ideal}}}\) With the required calculations, you will be able to determine the rate of cooling provided (in Btu/h), the rate of exergy destruction in the evaporator, and the second-law efficiency of the evaporator.

Step-by-step solution

Calculate the heat transfer rate in the evaporator

Calculate the heat transfer rate in the evaporator, which can be determined by multiplying the mass flow rate of the refrigerant by the change in enthalpy between the inlet and outlet of the evaporator: $$Q = \dot{m} \times (h_{2} - h_{1})$$ Where: \(\dot{m}\) = mass flow rate of the refrigerant = 0.08 lbm/s \(h_{1}\) = enthalpy at the inlet = 107.5 Btu/lbm \(h_{2}\) = enthalpy at the outlet = 172.1 Btu/lbm

Convert the heat transfer rate to Btu/h

To find the rate of cooling provided in Btu/h, convert the Q value from Btu/s to Btu/h by multiplying it by the conversion factor 3600 (seconds in an hour): \((Q \:\text{in Btu/s}) \times 3600 = Q \:\text{in Btu/h}\)

Calculate the rate of exergy destruction in the evaporator

The rate of exergy destruction, can be calculated using the following formula for an open system: $$\dot{E}_{\text{destruction}} = T_{0} \times \dot{m} \times (s_{2} - s_{1})$$ Where: \(T_{0}\) = temperature of the environment = 77°F = 537.67 °R (use Rankine for consistency) \(s_{1}\) = entropy at the inlet = 0.2851 Btu/lbm·R \(s_{2}\) = entropy at the outlet = 0.4225 Btu/lbm·R

Calculate the second-law efficiency of the evaporator

The second-law efficiency of the evaporator is given by: $$\eta_{\text{II}} = \frac{\dot{E}_{\text{ideal}} - \dot{E}_{\text{actual}}}{\dot{E}_{\text{ideal}} - \dot{E}_{\text{destruction}}}$$ However, since the evaporator is a cooling device, the ideal rate of exergy transfer \(\dot{E}_{\text{ideal}}\) is equal to the actual rate of exergy transfer \(\dot{E}_{\text{actual}}\). Therefore, the second-law efficiency simplifies to: $$\eta_{\text{II}} = 1 - \frac{\dot{E}_{\text{destruction}}}{\dot{E}_{\text{ideal}}}$$ Now, we can calculate the second-law efficiency using the rate of exergy destruction calculated in step 3 and the actual and ideal rates of exergy transfer. After solving these steps, we will have the answers for the rate of cooling provided (in Btu/h), the rate of exergy destruction in the evaporator, and the second-law efficiency of the evaporator.