PROBLEM 6P

To get more than an infinitesimal amount of work out of a Carnot engine, we would have to keep the temperature of its working substance below that of the hot reservoir and above that of the cold reservoir by non-infinitesimal amounts. Consider, then, a Carnot cycle in which the working substance is at temperature Thw as it absorbs heat from the hot reservoir, and at temperature Tcw as it expels heat to the cold reservoir. Under most circumstances the rates of heat transfer will be directly proportional to the temperature differences:

I’ve assumed here for simplicity that the constants of proportionality (K) are the same for both of these processes. Let us also assume that both processes take the same amount of time, so the Δt’s are the same in both of these equations.

(a) Assuming that no new entropy is created during the cycle except during the two heat transfer processes, derive an equation that relates the four temperatures Th, Tc, Thw, and Tcw.

(b) Assuming that the time required for the two adiabatic steps is negligible, write down an expression for the power (work per unit time) output of this engine. Use the first and second laws to write the power entirely in terms of the four temperatures (and the constant K), then eliminate Tcw using the result of part (a).

(c) When the cost of building an engine is much greater than the cost of fuel (as is often the case), it is desirable to optimize the engine for maximum power output, not maximum efficiency. Show that, for fixed Th, and Tc, the expression you found in part (b) has a maximum value at Thw = . (Hint: You’ll have to solve a quadratic equation.) Find the corresponding expression for Tcw.

(d) Show that the efficiency of this engine is . Evaluate this efficiency numerically for a typical coal-fired steam turbine with Th = 600° C and Tc = 25° C, and compare to the ideal Carnot efficiency for this temperature range. Which value is closer to the actual efficiency, about 40%, of a real coal-burning power plant?