SM_9B - T C . For a real refrigeration system, increasing T...

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Chapter 9 - Section B - Non-Numerical Solutions 9.1 Since the object of doing work | W | on a heat pump is to transfer heat | Q H | to a heat sink, then: What you get =| Q H | What you pay for =| W | Whence φ | Q H | | W | For a Carnot heat pump, φ = | Q H | | Q H |−| Q C | = T H T H T C 9.3 Because the temperature of the finite cold reservoir (contents of the refrigerator) is a variable, use differential forms of Carnot’s equations, Eqs. (5.7) and (5.8): dQ H dQ C =− T H T C and dW = µ 1 T C T H dQ H In these equations Q C and Q H refer to the reservoirs . With dQ H = C t dT C , the first of Carnot’s equations becomes: dQ H =− C t T H dT C T C Combine this equation with the second of Carnot’s equations: dW =− C t T H dT C T C + C t dT C Integration from T C = T H to T C = T C yields: W =− C t T H ln T C T H + C t ( T C T H ) or W = C t T H µ ln T H T C + T C T H 1 9.5 Differentiation of Eq. (9.3) yields: µ ∂ω T C T H = 1 T H T C + T C ( T H T C ) 2 = T H ( T H T C ) 2 and µ ∂ω T H T C =− T C ( T H T C ) 2 Because T H > T C , the more effective procedure is to increase
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Unformatted text preview: T C . For a real refrigeration system, increasing T C is hardly an option if refrigeration is required at a partic-ular value of T C . Decreasing T H is no more realistic, because for all practical purposes, T H is fixed by environmental conditions, and not subject to control. 9.6 For a Carnot refrigerator, is given by Eq. (9.3). Write this equation for the two cases: = T C T H T C and = T C T H T C Because the directions of heat transfer require that T H > T H and T C < T C , a comparison shows that < and therefore that is the more conservative value. 581...
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This note was uploaded on 01/29/2011 for the course CHEM 101 taught by Professor Brown during the Spring '10 term at The University of Akron.

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