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Unformatted text preview: 7B Review Questions on Thermodynamics Kevin Schaeffer February 22, 2010 1 Engines in a Chain (a) Consider two engines, each with an efficiency given by e connected so heat Q h is added to the first, and the second is driven by the output heat Q c released by the first. Thus, the second heat engine operates off of the waste heat of the first engine, but both engines do work. Calculate the overall effeciency of the combined twoengine system. (b) Now suppose we have a chain of N engines set up as shown in Figure 1, and assume that all engines have the same efficiency, e. Find the overall efficiency. Figure 1: (c) What would the efficiency be if we had an infinite set of engines? Before you calculate, try to think of what the answer should be on physical grounds. Now show that your intuition is correct by calculating the efficiency. (d) For simplicity, we have assumed that the efficiencies of all the engines in a chain are the same. Physically, why is this a poor assumption (Hint: think about the Carnot Cycle)? 1 1.1 Solution (a) Remember that efficiency is defined as “what I get out” divided by “what I put in.” The efficiency of an engine is given by, e = W Q h (1) I could rewrite this as, W = eQ h (2) All the heat that I put into an engine through Q h goes into either Work (W) or the heat released, Q c . In this solution I’ll take Q to always be positive so that, Q h = Q c + W (3) Therefore, using the above formula, Q c = (1 e ) Q h (4) This means that in the two engine system I can calculate the heats and works for either engine 1 or engine 2, which I indicate with a (1) or a (2), Q (1) c = Q (2) h = Q (1) h (1 e ) (5) We can find the total work by adding up the work done by each engine, W 1 = eQ (1) h (6) W 2 = eQ (2) h = e (1 e ) Q (1) h W tot = W 1 + W 2 = e + e (1 e ) Q (1) h Note that this result makes sense physically. Since e and 1 e are less than one, these formulas imply that the work done by the second engine is less than the work done by the first engine. This makes sense, because the second engine has less input heat to work with in the first place. I’ve rewritten everything in terms of Q (1) h because this is the only work that I actually put into the system. Q (2) h is really just energy being internally transferred between engines. So the efficiency of this engine system should be, e tot = W tot Q (1) h = e + e (1 e ) (7) (b) We can generalize the above result straightforwardly. Each engine has a certain amount of heat coming in. A fraction, e , of this heat goes into doing work, and the rest, a fraction 1 e goes into powering the next engine in the chain. (For you computer science majors, this is an example of recursion). So the total work is given by, W tot = Q (1) h e + (1 e ) e + (1 e ) 2 e + ··· + (1 e ) N 1 e (8) 2 This formula is perfectly fine, but we can use what we know about infinite series to simply this result. Recall that, 1 1 x = 1 + x + x 2 + ··· (9) where this series continues on forever. However, our series doesn’t continue onwhere this series continues on forever....
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 Spring '08
 ZETTLE
 Thermodynamics, Black Holes, Heat, Black hole, Heat engine

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