# HW13s - determine the required power input(in kW ME 201 Thermodynamics Spring 2006 2 Solution We begin by drawing the interaction diagram for this

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Spring 2006 1 ME 201 Thermodynamics Homework 13 Solution 1. A reversible process has been defined as a process, which having taken place, can be reversed and in so doing leaves no change in either the system or the surroundings. Six restrictions were imposed a. no friction b. heat transfer occurs only for infinitesimal temperature differences c. unrestrained expansion does not occur d. no mixing e. no turbulence f. no combustion Choose a process for which one of these restrictions is relaxed and discuss how this process is not reversible. (5 pts) Solution: This will vary from student to student, but one needs to describe a process that includes one of the restrictions listed above and then describe the reverse of the process and show that the reverse can only occur if the system or surroundings are changed. 2. Thirty five (35) kilograms of chicken is to be frozen in a household freezer. The chicken is at 15 ° C when it is placed in the freezer and reaches -20 ° C (well below the freezing point) in 3 hours. If the COP of the freezer is 4.3 during this process,

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Unformatted text preview: determine the required power input (in kW). ME 201 Thermodynamics Spring 2006 2 Solution: We begin by drawing the interaction diagram for this refrigerator High Temperature Heat Reservoir (kitchen) at T H Low Temperature Heat Reservoir (freezer) at T L Refrigerator Q H Q L W net By definition we have for our refrigerator COP = Q W L net If we can determine the required heat transfer for freezing the chickens, Q chick , then we can use Q L = -Q chick and solve for W net , In the freezing process for the chicken, we have three processes: Lowering the chicken temperature from 15 ° C to its freezing point Q 1 = mc(T 2-T 1 ) = (35)(3.32)[(-2.8)-15] = -2068.4 kJ Freezing the chicken Q 2 = m(-latent heat of fusion) = (35)(-247) = -8645 Lowering the chicken temperature from its freezing point to -20 ° C Q 3 = mc(T 3-T 2 ) = (35)(1.77)[-20-(-2.8)] = -1065.5 kJ Then Q L = -Q chick = -(Q 1 +Q 2 +Q 3 ) = 11,779 kJ ME 201 Thermodynamics Spring 2006 3 and W = Q COP = 11,779 4.3 = 2739 kJ net L The power required is & W = W t = 2739 (3)(3600) = 0.25 kW net net Δ...
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## This note was uploaded on 07/25/2008 for the course ME 201 taught by Professor Somerton during the Spring '06 term at Michigan State University.

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HW13s - determine the required power input(in kW ME 201 Thermodynamics Spring 2006 2 Solution We begin by drawing the interaction diagram for this

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