Question

1. Find efficiency of a reversible power cycle consisting of two isochoric and two isobaric lines, if temperature is increased by a factor of n during the isochoric heating process and by the same factor n during the isobaric expansion process. The working fluid involved in the cycle is an ideal gas with isentropic exponent k. The answer must be expressed in terms of n and k only. Show the process on the P- V diagram.

Answer 1

efficiency eta = ? p.v diagram: - At point(1) pressure = p_1 Temperature = T_1 volume = v_1 At point (2) rightarrow T_2, p_2, v_2 At point (3) rightarrow T_3 p_3, v_3 At point (4) rightarrow T_4, p_4, v_4 process (1) to (2) Isobaric expansion (heating) and given T_2 = n T_1 and p_1 = p_2(Isobaric) process (4) to (1): - Isobaric heating T_1 = n T_4 and v_4 = v_1 \r\n process (1) to (2): - Reversible isobaric heat supplied Q_12 = c_p(T_2 - T_1) process (2) to (3): - Reversible isobaric heat rejection Q_23 = c_v(T_2 - T_3) process (3) to (4): - Reversible isobaric heat rejection Q_34 = c_p(T_3 - T_4) process (4) to (1) Reversible isochoric heat supplied Q_41 = c_v(T_1 - T_4) Total Heat supplied: H_s = Q_12 + Q_41 = c_p(T_2 - T_1) + c_v(T_1 - T_4) Total Heat Rejected: - H. R = Q_23 + Q_34 = c_v(T_2 - T_3) + c_p(T_3 - T_4) Hence Net work done = Heat supplied - Heat rejected = [c_p(T_2 - T_1) + c_v(T_1 - T_4)] - [c_v(T_2 - T_3) + c_p(T_3 - T_4)] \r\n = c_p T_2 - c_p T_1 + c_v T_1 - c_v