Question 13.11: The small model airplane jet engine shown in Figure 13.43 op...
The small model airplane jet engine shown in Figure 13.43 operates on the Lenoir cycle. It has a maximum temperature of T_1 = 800. \text{R} and an intake temperature of T_3 = T_4 = 530. \text{R}. The expansion, exhaust, and intake pressures are all p_{2s} = p_3 = p_4 = 14.7 \text{psia}, and the engine contains 1.00 × 10^{–3} \text{lbm} of air. For this engine, determine
a. The combustion pressure p_1.
b. The isentropic compression ratio \text{CR} = v_{2s}/v_3.
c. The Lenoir cold ASC thermal efficiency.
Using the Lenoir cycle diagram shown in Figure 13.42, we can carry out the following analysis.
a. From the ideal gas equation of state, we can calculate p_1 = mRT_1/\sout{V}_1. For this cycle, we have
\sout{V}_1 = \sout{V}_4 = \frac{mRT_4}{p_4} = \frac{(1.00 × 10^{−3} \text{lbm})(53.34 \frac{\text{ft.lbf}}{\text{lbm.R}})(530 \text{R})}{(14.7 \frac{\text{lbf}}{\text{in}^2})(144 \frac{\text{in}^2}{\text{ft}^2} )} = 0.0134 \text{ft}^3
Then,
p_1 = \frac{mRT_1}{\sout{V}_1} = \frac{mRT_1}{\sout{V}_1} = \frac{(1.00 × 10^{−3} \text{lbm})(53.34 \frac{\text{ft.lbf}}{\text{lbm.R}})(800. \text{R})}{(0.0134 \text{ft}^3)(144 \frac{\text{in}^2}{\text{ft}^2} )} = 22.2 \text{psia}
b. For the Lenoir cycle, the isentropic compression ratio is \text{CR} = v_{2s}/v_3 = T_{2s}/T_3. From Eq. (7.38), we have
T_{2s} = T_1 (\frac{p_{2s}}{p_1} )^{\frac{k−1}{k} } = (800. \text{R}) (\frac{14.7 \text{psia}}{22.2 \text{psia}} )^{\frac{0.4}{1.4} } = 711 \text{R}
Then, the isentropic compression ratio for this engine is
\text{CR} = \frac{v_{2s}}{v_3} = \frac{T_{2s}}{T_3} = \frac{711 \text{R}}{530. \text{R}} = 1.34
c. Equation (13.21) gives the Lenoir cold ASC thermal efficiency as
(η_T)_{\substack{\text{Lenoir}\\\text{cold ASC}\\}} = 1- \frac{kT_3(\text{CR} – 1)}{T_1 − T_4} = 1- \frac{1.40 (530. \text{R}) (1.34 − 1)}{800. − 530. \text{R}} = 0.0656 = 6.56\%
