In the Cannot cycle, a system undergoes the following reversible, quasistatic pr
ID: 1718683 • Letter: I
Question
In the Cannot cycle, a system undergoes the following reversible, quasistatic process that extracts energy from a high temperature reservoir at T_h and transfers a fraction of this energy, epsilon, into a reversible work source (RWS). The system Expands isothermally at T_h from a volume V_a to a volume V_b; Expands adiabatically from (T_h,V_b) to (Tc,V_c); Is compressed isothermally at T_c from a volume V_c to a volume V_d; Is compressed adiabatically from (T_c,V_d) to (T_h,V_d). The efficiency of the Cannot cycle is defined epsilon = delta__TU_T_RWS/q_1, where delta_TU_rws is the total energy transferred into the RWS and q_1 is the heat extracted from the thermal reservoir at T_h during step In class, we showed that, if the system is a simple ideal gas, then q_1 = nR(T_hln(V_b/V_a) delta_TU_RWS=nR(T_h-T_c)lnV_b/V_a) epsilon=1-T_c/T_h Recall the key step: for an adiabatic reversible process for a simple ideal gas, TV^gamma-1=const. We stated without proof, that the efficiency of the Cannot cycle is independent of substance. Now we shall demonstrate this to be true for a van der Waals (vdW) fluid. The equations of state for a vdW fluid are: P = nRT/V-nb-an^2/V^2, where a, b, and k are all positive constants. U=knRT-an^2/V For a vdW fluid, the relevant equation for an adiabatic reversible process (V-nb)T^k=const. Consider the Carnot cycle for a vdW fluid. Demonstrate the following: q_1=nRT_hln[V-b-nb/V-a-nb] delta_TU_RWS=nRT_hln[V_b-nb/V_a-nb]+nRT_cln[V_d-nb/V_c-nb] epsilon=1-T_c/T_h, as promised.Explanation / Answer
In the Cannot cycle, a system undergoes the following reversible, quasistatic pr