Question 13.9: A new Stirling cycle engine using air as the working fluid i...
A new Stirling cycle engine using air as the working fluid is being designed to power the small portable electric generator shown Figure 13.39. The piston displacement is to be 0.0110 \text{m}^3 and the minimum volume in the cylinder is to be \sout{V}_3 = \sout{V}_4 = 1.00 × 10^{–3} \text{m}^3 . The design calls for a maximum power piston pressure of p_1 = 0.300 \text{MPa}, a minimum displaced piston pressure of p_2 = 0.100 \text{MPa}, and a displacer inlet temperature of T_2 = 30.0°\text{C}. Use the Stirling cold ASC analysis shown in Figure 13.38 to complete the design, determining
a. The displacer piston maximum pressure (p_3).
b. The power piston maximum pressure (p_4).
c. The mass of air in the engine (m).
d. The heat addition temperature (T_1 = T_4).
e. The Stirling cold ASC thermal efficiency of the engine.
Using the Stirling cycle diagram shown in Figure 13.38, we can carry out the following analysis.
a. Since T_2 = T_3, the ideal gas equation of state gives p_3 = p_2(\sout{V}_2/\sout{V}_3), where \sout{V}_2 = \sout{V}_1 = \text{Piston displacement} – \sout{V}_3 = 0.011 – 0.001 = 0.010 \text{m}^3 . Then, p_3 = (0.100 \text{MPa})(0.0100/0.00100) = 1.00 \text{MPa}
b. Since T_1 = T_4, the ideal gas equation of state gives p_4 = p_1(\sout{V}_1/\sout{V}_4), where \sout{V}_4 = \sout{V}_3 = 0.001 \text{m}^3. Then, p_4 = (0.300 \text{MPa}) × (0.0100/0.00100) = 3.00 \text{MPa}.
c. Again using the ideal gas equation of state, we find that m = p_2\sout{V}_2/RT_2 and
m = \frac{(0.100 \text{MPa})(1000 \text{ kPa}/\text{MPa})(0.0100 \text{m}^3 )}{(0.286 \text{kJ/kgK})(30.0 + 273.15 \text{K}} = 0.0115 \text{kg}
d. The ideal gas equation of state gives
T_1 = \frac{p_1\sout{V}_1}{mR} = \frac{(0.300 \text{MPa})(1000 \text{ kPa}/\text{MPa})(0.0100 \text{m}^3 )}{(0.0115 \text{kg})(0.286 \text{kJ/kg.K})} = 912 \text{K}
e. Equation (13.19) gives the Stirling cold ASC thermal efficiency of this engine as
(η_T)_{\substack{\text{Stirling}\\\text{cold ASC}\\}} = 1- \frac{T_L}{T_H} = 1- \frac{T_2}{T_1} = 1- \frac{30.0 + 273.15 \text{K}}{912 \text{K}} 0.668 = 66.8\%