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Unformatted text preview: PHYS 333 Winter 2009 Assignment 6 Due: 5:00 pm Monday March 3rd Most of these problems are from Schroeder’s book. Problems 2 and 3 use Tables 4.1, 4.2, 4.3 and 4.4 from the book: they are provided as a separate PDF file for those who need them. 1. Schroeder 4.6. Power of a Carnot engine . Consider a Carnot cycle in which the working substance is at temperature T hw as it absorbs heat from the hot reservoir, and at temperature T cw as it expels heat to the cold reservoir. Assume that the rates of heat transfer are directly proportional to the temperature differences: ˙ Q h = K ( T h T hw ) and ˙ Q c = K ( T cw T c ) Assume also that the time taken for each transfer is the same, so that Q h = ˙ Q h Δ t and Q c = ˙ Q C Δ t (a) If no new entropy is created during the cycle except during the two heat transfer pro cesses, derive an equation that relates T h , T c , T hw and T cw . In a complete cycle, the net entropy change of the working substance must be zero. Therefore Q h /T hw and Q c /T cw must be equal. But, from the information given, Q h Q c = T h T hw T cw T c so that T h T hw T cw T c = T hw T cw (b) If the time for the two adiabatic processes is negligible, write down an expresion 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 . Show that, using the result of part (a), the power output can be written as P = K 2 1 T C 2 T hw T h ( T h T hw ) The total time for one cycle is 2Δ t , where Δ t is the time for heat Q h to flow in. From the information given, it is Δ t = Q h K ( T h T hw ) Thus, the power is P = W 2Δ t = Q h Q c 2 K ( T h T hw ) Q h which, using the results from part (a), is the required result. 1 (c) Show that for fixed values of T h and T c , the power ouptut has a maximum when T hw = 1 2 ( T h + p T h T c ) Setting dP/dT hw equal to zero, we obtain 0 = K 2 2 T c (2 T hw T h ) 2 ( T h T hw ) 1 T c (2 T hw T h ) Simplifying, (setting everything over a common denominator, and throwing it away), we obtain: 0 = 4 T 2 hw 4 T h T hw T c T h + T 2 h a quadratic with two roots, one of which would be unphysical. (Imagine T c ∼ T h , and see that this solution would give T hw ∼ < T c , which is not allowed.) The physical root is as required. For the record, the corresponding expression for T cw is very similar: T cw = 1 2 ( T c + p T h T c ) (d) Show that the corresponding efficiency of the engine is then = 1 q T c /T h (The value of this expression is a much better approximation to the performance of a steam powerplant than is the “ideal” efficiency = 1 T c /T h .) The general expression for the efficiency of a heat engine is = 1 Q c Q h In the present case of maximum power, this gives = 1 T cw T hw = 1 ( T c + √ T h T c ) ( T h + √ T h T c ) = 1 s T c T h 2. Schroeder 4.23 and following, somewhat modified....
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 Winter '09
 Harris
 Physics, Thermodynamics, Entropy, Th, Carnot cycle, Schroeder

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