Chapter9_B

# Chapter9_B - Chapter 9 - Section B - Non-Numerical...

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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

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## This note was uploaded on 04/18/2011 for the course CEM 497 taught by Professor Ku during the Spring '11 term at Punjab Engineering College.

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Chapter9_B - Chapter 9 - Section B - Non-Numerical...

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