Question

An ideal Ericsson engine using helium as the working fluid operates between temperature limits of 550 and \(3000 \mathrm{R}\) and pressure limits of 25 and 200 psia. Assuming a mass flow rate of \(14 \mathrm{lbm} / \mathrm{s}\), determine (a) the thermal efficiency of the cycle, \((b)\) the heat transfer rate in the regenerator, and \((c)\) the power delivered.

Short answer

2. How does the ideal gas law help in the calculation of specific heat capacities of helium? 3. How is the work done in processes 1-2 and 3-4 calculated in the Ericsson cycle? 4. What is the formula for the thermal efficiency of the Ericsson cycle and how is this calculated? 5. How can the heat transfer rate in the regenerator and the power delivered be calculated using the thermal efficiency and other parameters? 6. What are the final calculated values for the thermal efficiency, the heat transfer rate in the regenerator, and the power delivered in this example?

Step-by-step solution

Calculate the specific heat capacities and ratios of helium using given temperatures and pressures

We will be using the ideal gas law with the specific heat capacities \(c_p\) and \(c_v\) for helium. For Helium, \(c_p = 5.193 \frac{\mathrm{Btu}}{\mathrm{lbm} \cdot \mathrm{R}}\) and \(c_v = 3.116 \frac{\mathrm{Btu}}{\mathrm{lbm} \cdot \mathrm{R}}\). Using these specific heat capacities, we can find the ratio of specific heat capacities \(\gamma = \frac{c_p}{c_v}\). \(\gamma = \frac{5.193 \frac{\mathrm{Btu}}{\mathrm{lbm} \cdot \mathrm{R}}}{3.116 \frac{\mathrm{Btu}}{\mathrm{lbm} \cdot \mathrm{R}}} = 1.67\)

Calculate work done in processes 1-2 and 3-4

The work done in process 1-2 isothermal expansion can be given as: \(W_{12} = \frac{R \cdot T1 \cdot \ln{\frac{P2}{P1}}}{(\gamma - 1)}\) And, the work done in process 3-4 isothermal compression can be given as: \(W_{34} = \frac{R \cdot T2 \cdot \ln{\frac{P2}{P1}}}{(\gamma - 1)}\)

Calculate thermal efficiency of the Ericsson cycle

The thermal efficiency of the Ericsson cycle can be given as: \(\eta = 1 - \frac{W_{12}}{W_{34}}\) Calculating the thermal efficiency, $\eta = 1 - \frac{\frac{R \cdot T1 \cdot \ln{\frac{P2}{P1}}}{(\gamma - 1)}}{\frac{R \cdot T2 \cdot \ln{\frac{P2}{P1}}}{(\gamma - 1)}} = 1 - \frac{T1}{T2}$ Using the given values, T1 = 550 R and T2 = 3000 R: \(\eta = 1 - \frac{550}{3000} = 0.817 = 81.7 \%\)

Calculate the heat transfer rate in the regenerator

The regenerator heat transfer rate can be given as: \(q_{regenerator} = \dot{m} \cdot c_p \cdot (T2 - T1)\) Using the given mass flow rate = 14 lbm/s, specific heat capacity (cp) = 5.193 Btu/lbm.R, and temperatures T1 = 550 R and T2 = 3000 R: \(q_{regenerator} = 14 \cdot 5.193 \cdot (3000 - 550) = 14 \cdot 5.193 \cdot 2450 = 178402.99 \frac{\mathrm{Btu}}{\mathrm{s}}\)

Calculate the power delivered

The power delivered can be given as: \(P = W_{net} \cdot \dot{m} = \eta \cdot q_{regenerator}\) Using our calculated values for thermal efficiency η = 0.817 and heat transfer rate \(q_{regenerator} = 178402.99 \frac{\mathrm{Btu}}{\mathrm{s}}\): \(P = 0.817 \cdot 178402.99 = 145684.84 \frac{\mathrm{Btu}}{\mathrm{s}}\) In conclusion, the ideal Ericsson engine has a thermal efficiency of 81.7%, the heat transfer rate in the regenerator is 178402.99 BTU/s, and the power delivered is 145684.84 BTU/s.