Question 13.10: A new Ericsson cycle engine is being designed with a pressur...
A new Ericsson cycle engine is being designed with a pressure ratio of \text{PR} = p_4/p_1 = 2.85, a power piston outlet pressure of p_1 = 0.500 \text{MPa}, a maximum volume of \sout{V}_1 = 0.0110 \text{m}^3 , and a minimum volume of \sout{V}_3 = 3.00 × 10^{–3} \text{m}^3 . The engine will contain 0.0500 \text{kg} of air. Use the Ericsson cold ASC analysis shown in Figure 13.40 to complete this design, determining
a. The compressor inlet pressure and volume ( p_2 and \sout{V}_2).
b. The power piston outlet pressure and inlet volume ( p_4 and \sout{V}_4).
c. The compressor outlet pressure ( p_3).
d. The temperatures at the inlet and outlet of the power and displacer pistons ( T_1, T_2, T_3, and T_4).
e. The Ericsson cold ASC thermal efficiency of this engine.
Using the Ericsson cycle diagram shown in Figure 13.40 we can carry out the following analysis.
a. For this cycle, the compressor inlet pressure is the same as the power piston outlet pressure (see Figure 13.40), or p_2 = p_1 = 0.500 \text{MPa}. The compressor inlet volume is \sout{V}_2 = \sout{V}_3 × \text{(CR)}, where the isentropic compression ratio \text{CR} = \sout{V}_1/\sout{V}_4 and \sout{V}_4 = mRT_4/p_4. Now, from Figure 13.40,
T_4 = T_1 = \frac{p_1\sout{V}_1}{mR} = \frac{(0.500 \text{MPa})(1000 \text{kPa/MPa})(0.0110 \text{m}^3)}{(0.0500 \text{kg})(0.286 \text{kJ/kg.K})} = 385 \text{K}
and
p_4 = p_3 = p_2\text{(PR)} = p_1\text{(PR)} = (0.500 \text{MPa})(2.85) = 1.43 \text{MPa}
so
\sout{V}_4 = \frac{mRT_4}{p_4} = \frac{(0.0500 \text{kg})(0.286 \text{kJ/kg.K})(385 \text{K})}{(1.43 \text{MPa})(1000 \text{kPa/MPa})} = 0.00385 \text{m}^3
and the isentropic compression ratio is
\text{CR} = \frac{\sout{V}_1}{\sout{V}_4} = \frac{0.0110 \text{m}^3}{0.00385 \text{m}^3} = 2.86
Then, the compressor inlet volume is \sout{V}_2 = \sout{V}_3 × \text{(CR)} = (3.00 × 10^{–3})(2.86) = 0.00858 \text{m}^3
b. For this cycle, the power piston outlet pressure is the same as the compressor inlet pressure (see Figure 13.40):
p_4 = p_3 = 1.43 \text{MPa} , and from part a, \sout{V}_4 = 0.00385 \text{m}^3.
c. Since p_4/p_1 = p_3/p_2 = \text{PR}, the compressor outlet pressure is p_3 = p_2 \text{(PR)}, where the isentropic pressure ratio is given as \text{PR} = 2.86. Then, p_3 = 0.500(2.86) = 1.43 \text{MPa}.
d. Using the ideal gas equation of state, we have
T_3 = T_2 = \frac{p_2\sout{V}_2}{mR} = \frac{(0.500 \text{MPa})(1000 \text{kPa/MPa})(0.00858 \text{m}^3)}{(0.0500 \text{kg})(0.286 \text{kJ/kg.K})} = 300. \text{K}
e. The Ericsson cold ASC thermal efficiency for this engine is given by Eq. (13.20) as
(η_T)_{\substack{\text{Ericsson}\\\text{cold ASC}\\}} = 1- \frac{T_L}{T_H} = 1- \frac{T_2}{T_1} = 1- \frac{300. \text{K}}{385 \text{K}} = 0.221 = 22.1\%