Question 5 ( \( \mathbf{2 5} \) point) Refrigerant 134 a enters a compressor and a negligible velocity. The refrigerant enters the compressor as a saturated vapor at \( 12^{\circ} \mathrm{C} \) and leaves the compressor at 1400 kPa with an enthalpy of \( 281.4 \mathrm{~kJ} / \mathrm{kg} \) and a velocity of \( 50 \mathrm{~m} / \mathrm{s} \). The rate of work done on the refrigerant is measured to be \( 27.3 \mathrm{~kJ} / \mathrm{kg} \). If the diameter at the exit is equal to 4.6 cm , determine the rate of heat transfer associated with this process, in kW . Note: Don't do interpolation, take the closest value

To solve this problem, we will apply the first law of thermodynamics for a control volume, which can be expressed as: \[ \dot{Q} - \dot{W} = \dot{m} \left( h_2 - h_1 + \frac{V_2^2 - V_1^2}{2} \right) \] Where: - \(\dot{Q}\) = rate of heat transfer (kW) - \(\dot{W}\) = rate of work done on the refrigerant (kW) - \(\dot{m}\) = mass flow rate (kg/s) - \(h_1\) = specific enthalpy at the inlet (kJ/kg) - \(h_2\) = specific enthalpy at the outlet (kJ/kg) - \(V_1\) = velocity at the inlet (m/s) - \(V_2\) = velocity at the outlet (m/s) ### Step 1: Determine the specific enthalpy at the inlet (\(h_1\)) Given that the refrigerant enters as a saturated vapor at \(12^{\circ}C\), we can look up the specific enthalpy of saturated vapor at this temperature. The closest value for \(h_1\) (saturated vapor enthalpy at \(12^{\circ}C\)) is approximately \(h_1 = 49.0 \, \text{kJ/kg}\). ### Step 2: Identify the known values - \(h_2 = 281.4 \, \text{kJ/kg}\) - \(\dot{W} = 27.3 \, \text{kJ/kg}\) - \(V_1 = 0 \, \text{m/s}\) (negligible velocity) - \(V_2 = 50 \, \text{m/s}\) ### Step 3: Calculate the mass flow rate (\(\dot{m}\)) To find the mass flow rate, we need the area of the exit and the density of the refrigerant at the exit conditions. The diameter of the exit is given as \(4.6 \, \text{cm}\), which we convert to meters: \[ D = 0.046 \, \text{m} \] The area \(A\) at the exit can be calculated using the formula for the area of a circle: \[ A = \frac{\pi D^2}{4} \] Now, we will calculate \(A\): \[ A = \frac{\pi (0.046)^2}{4} \] Next, we need the density of refrigerant 134a at the exit conditions. At \(1400 \, \text{kPa}\) and \(281.4 \, \text{kJ/kg}\), we can look up the density. The closest value for density (\(\rho\)) is approximately \( \rho \approx 1000 \, \text{kg/m}^3\). Now we can calculate the mass flow rate: \[ \dot{m} = \rho \cdot A \cdot V_2 \] ### Step 4: Calculate the rate of heat transfer (\(\dot{Q}\)) Now we can rearrange the first law equation to solve for \(\dot{Q}\): \[ \dot{Q} = \dot{W} + \dot{m} \left( h_2 - h_1 + \frac{V_2^2 - V_1^2}{2} \right) \] ### Step 5: Perform the calculations Let's calculate \(A\), \(\dot{m}\), and then \(\dot{Q}\). 1. Calculate \(A\): \[ A = \frac{\pi (0.046)^2}{4} \] 2. Calculate \(\dot{m}\): \[ \dot{m} = 1000 \cdot A \cdot 50 \] 3. Calculate \(\dot{Q}\): \[ \dot{Q} = 27.3 + \dot{m} \left( 281.4 - 49.0 + \frac{50^2 - 0^2}{2} \right) \] Now, I will perform these calculations. Calculate the value by following steps: - step0: Calculate: \(\frac{1000\left(\pi \times 0.046^{2}\right)}{4}\times 50\) - step1: Remove the parentheses: \(\frac{1000\pi \times 0.046^{2}}{4}\times 50\) - step2: Convert the expressions: \(\frac{1000\pi \left(\frac{23}{500}\right)^{2}}{4}\times 50\) - step3: Reduce the fraction: \(\frac{529\pi }{1000}\times 50\) - step4: Reduce the numbers: \(\frac{529\pi }{20}\) Calculate or simplify the expression \( (\pi*(0.046)^2)/4 \). Calculate the value by following steps: - step0: Calculate: \(\frac{\left(\pi \times 0.046^{2}\right)}{4}\) - step1: Remove the parentheses: \(\frac{\pi \times 0.046^{2}}{4}\) - step2: Convert the expressions: \(\frac{\pi \left(\frac{23}{500}\right)^{2}}{4}\) - step3: Multiply the numbers: \(\frac{\frac{23^{2}}{500^{2}}\times \pi }{4}\) - step4: Evaluate the power: \(\frac{\frac{529\pi }{500^{2}}}{4}\) - step5: Multiply by the reciprocal: \(\frac{529\pi }{500^{2}}\times \frac{1}{4}\) - step6: Multiply the fractions: \(\frac{529\pi }{500^{2}\times 4}\) Calculate or simplify the expression \( 27.3 + (1000*(\pi*(0.046)^2)/4*50) * (281.4 - 49 + (50^2)/2) \). Calculate the value by following steps: - step0: Calculate: \(27.3+\left(\frac{1000\left(\pi \times 0.046^{2}\right)}{4}\times 50\right)\left(281.4-49+\frac{50^{2}}{2}\right)\) - step1: Remove the parentheses: \(27.3+\frac{1000\pi \times 0.046^{2}}{4}\times 50\left(281.4-49+\frac{50^{2}}{2}\right)\) - step2: Divide the terms: \(27.3+\frac{1000\pi \times 0.046^{2}}{4}\times 50\left(281.4-49+1250\right)\) - step3: Calculate: \(27.3+\frac{1000\pi \times 0.046^{2}}{4}\times 50\times 1482.4\) - step4: Convert the expressions: \(27.3+\frac{1000\pi \left(\frac{23}{500}\right)^{2}}{4}\times 50\times 1482.4\) - step5: Reduce the fraction: \(27.3+\frac{529\pi }{1000}\times 50\times 1482.4\) - step6: Multiply the terms: \(27.3+\frac{980237\pi }{25}\) - step7: Convert the expressions: \(\frac{273}{10}+\frac{980237\pi }{25}\) - step8: Reduce fractions to a common denominator: \(\frac{273\times 5}{10\times 5}+\frac{980237\pi \times 2}{25\times 2}\) - step9: Multiply the numbers: \(\frac{273\times 5}{50}+\frac{980237\pi \times 2}{25\times 2}\) - step10: Multiply the numbers: \(\frac{273\times 5}{50}+\frac{980237\pi \times 2}{50}\) - step11: Transform the expression: \(\frac{273\times 5+980237\pi \times 2}{50}\) - step12: Multiply the numbers: \(\frac{1365+980237\pi \times 2}{50}\) - step13: Multiply the terms: \(\frac{1365+1960474\pi }{50}\) ### Step 1: Calculate the Area \(A\) The area at the exit is calculated as: \[ A \approx 0.001662 \, \text{m}^2 \] ### Step 2: Calculate the Mass Flow Rate \(\dot{m}\) The mass flow rate is calculated as: \[ \dot{m} \approx 83.095 \, \text{kg/s} \] ### Step 3: Calculate the Rate of Heat Transfer \(\dot{Q}\) Now, substituting the values into the equation for \(\dot{Q}\): \[ \dot{Q} \approx 27.3 + 83.095 \left( 281.4 - 49 + \frac{50^2}{2} \right) \] Calculating the term inside the parentheses: \[ 281.4 - 49 + \frac{2500}{2} = 281.4 - 49 + 1250 = 1482.4 \] Now substituting this back into the equation for \(\dot{Q}\): \[ \dot{Q} \approx 27.3 + 83.095 \times 1482.4 \] Calculating the final value: \[ \dot{Q} \approx 27.3 + 123207.514319 \] Now, let's calculate this final value. Calculate the value by following steps: - step0: Calculate: \(27.3+83.095\times 1482.4\) - step1: Multiply the numbers: \(27.3+123180.028\) - step2: Add the numbers: \(123207.328\) The final calculation for the rate of heat transfer \(\dot{Q}\) is: \[ \dot{Q} \approx 123207.328 \, \text{kW} \] ### Conclusion The rate of heat transfer associated with this process is approximately **123.2 kW**.