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# HW5-336 - ME 336 HOMEWORK SET 5 10.16 Find the availability...

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ME 336 HOMEWORK SET 5 10.16 Find the availability of 100 kW delivered at 500 K when the ambient is 300 K. Solution: The availability of an amount of heat transfer equals the possible work that can be extracted. This is the work out of a Carnot heat engine with heat transfer to the ambient as the other reservoir. The result is from Chapter 7 as also shown in Eq. 10.1 and Eq. 10.36 Φ . = W . rev HE = (1 – T o T )Q . = (1 – 300 500 ) 100 kW = 40 kW 10.20 A household refrigerator has a freezer at T F and a cold space at T C from which energy is removed and rejected to the ambient at T A as shown in Fig. P10.27. Assume that the rate of heat transfer from the cold space, Q . C , is the same as from the freezer, Q . F , find an expression for the minimum power into the heat pump. Evaluate this power when T A = 20 ° C, T C = 5 ° C, T F = 10 ° C, and Q . F = 3 kW. Solution: C.V. Refrigerator (heat pump), Steady, no external flows except heat transfer. Energy Eq.: Q . F + Q . c + W . = Q . A (amount rejected to ambient) Q W A Q C REF Q F Reversible gives minimum work in as from Eq. 10.1 or 10.9 on rate form.

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