A test rig holds a fixed mass of air

PHYSICS · Multi-part · Fermi University Physics · Thermal, Statistical & Complex Systems · The laws of thermodynamics · The laws of thermodynamics

Thermal Physics, Ch Internal-combustion cycles and indicator diagrams

Problem

A test rig holds a fixed mass of air, treated throughout as an ideal gas with constant γ=1.4\gamma = 1.4 and NkB=0.20 J K−1N k_B = 0.20\,\mathrm{J\,K^{-1}}. Every leg is quasistatic; friction and heat leaks are ignored. Starting at state 1 with T1=300 KT_1 = 300\,\mathrm{K} and volume V1V_1, the gas is taken round a three-leg loop built from the legs of Chapter 3: 1→21 \to 2: adiabatic compression from V1V_1 to V2V_2, with V1/V2=8V_1/V_2 = 8; 2→32 \to 3: isothermal expansion, at the temperature reached at state 2, back to volume V1V_1; 3→13 \to 1: cooling at constant volume V1V_1 back to T1T_1. For a quasistatic isothermal change of an ideal gas the heat delivered to the gas is Q=NkBTln⁡(Vf/Vi)Q = N k_B T \ln (V_f/V_i) (Chapter 3), and on a quasistatic adiabat TVγ−1TV^{\gamma-1} is constant. Use e=1−Qc/Qhe = 1 - Q_c/Q_h (Chapter 4), with QhQ_h the heat taken in and QcQ_c the heat rejected. ((a)) Find the temperature T2T_2 at the end of the compression. ((b)) Find the heat taken in on 2→32 \to 3, the heat rejected on 3→13 \to 1, the net work delivered by the gas per cycle, and the efficiency of the loop. ((c)) An Otto cycle uses the same gas between the same two volumes, so with the same r=8r = 8. Compute its efficiency and verify that it equals 1−T1/T21 - T_1/T_2. Comment on what this says about the three-leg loop compared with the Carnot bound between the loop's own extreme temperatures.

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