SOL7-2011

# SOL7-2011 - ECH 152A Fall 2011 Roland Faller Solutions for...

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Unformatted text preview: ECH 152A, Fall 2011, Roland Faller Solutions for Homework 7, November 9, 2011 Problem 1 A Stirling engine consists of two isothermal and two isochoric legs, i.e. we have a isochoric heating, an isothermal expansion, an isochoric cooling and an isothermal compression. (a) Plot it in a pV diagram. (b) Calculate its efficiency for an ideal gas in terms of the temperatures and the volumes. Solution (a) (b) Obviously there is no work performed in the isochoric steps (which we call 1 and 3). As we have an ideal gas the heat equals the change in internal energy which is Q = U = c V T . W 1 = W 3 = 0 Q 1 = Nc V ( T H- T C ) =- Q 3 In the isothermal steps the internal energy does not change so Q =- W = R pdV = R NRT V dV . W 2 =- Q 2 = NRT H ln V 3 V 2 &amp;gt; W 4 =- Q 4 = NRT C ln V 1 V 4 &amp;lt; So the effective work is | W 2 | - | W 4 | = NR ( T H ln V 3 V 2- T C ln V 4 V 1 ) The heat entering the system is Q 1 + Q 2 ....
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SOL7-2011 - ECH 152A Fall 2011 Roland Faller Solutions for...

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