# OldFinalS - ME 201 Thermodynamics ME 201 Thermodynamics Old...

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ME 201 Thermodynamics 1 ME 201 Thermodynamics Old Final Exam Solutions Directions: Open book, open notes. Work all four problems. Problems are equally weighted. Problem 1 Consider applying our Carnot heat engine approach to a biological system, specifically, a hunting cheetah. The work output may be considered to be the kinetic energy change as the 45 kg cheetah accelerates from rest to 27 m/s. The heat input, Q H , may be considered to come from the digestion of meat. The outside temperature is 30 ° C and the cheetah’s internal temperature from which the Q H comes from is 42 ° C. Determine a. the heat that must be supplied b. the heat that must be rejected c. the kilograms of meat that must be digested if 200 kJ of heat is released for every kilogram of meat digested Solution: We recognize this as a heat engine problem, so we begin by drawing our interaction diagram. High Temperature Reservoir (digestion) at T H = 42 ° C Low Temperature Reservoir (ambient) at T L = 30 ° C Heat Engine (cheetah) Q H Q L W net =1/2 mV 2

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ME 201 Thermodynamics 2 Our net work output is said to be the kinetic energy change of the cheetah, or kJ 16.403 = J 16,403 = 27) (0.5)(45)( = v m 2 1 = W 2 2 net r The thermal efficiency is then given by 0.0472 = 273 + 45 273 + 30 - 1 = T T - 1 = H L η The heat transfer from the high temperature heat reservoir is kJ 347.74 = 0.0472 16.403 = W = Q net H η The heat transfer from the low temperature heat reservoir is kJ 331.34 = 16.403 - 347.74 = W - Q = Q net H L Finally the mass of meat that need to be consumed (and digested) is kg 17.4 = 200 347.74 = e Q = m meat H meat Problem 2 Steam at 8 MPa, 400 ° C, and 50 kg/s enters a turbine with isentropic efficiency of 78% and exhausts to a pressure of 40 kPa. Determine a. actual power output of the turbine b. second law efficiency for the turbine c. difference in steam exit quality between ideal and actual operation Solution: The actual power output for this turbine will be calculated from ideal s act W W η = so that we will need to do an ideal 1 st law turbine calculation. The second law efficiency will be given by W W = rev act II η
ME 201 Thermodynamics 3 So we proceed with our 1 st law calculation System Type: Control Volume Substance Type: Compressible (steam) Device: Non-Ideal Turbine(adiabatic with η s = 0.78) State 1 (steam inlet): Fixed State 2s (ideal steam outlet): unknown State 2a (actual steam outlet): unknown W sh : unknown Q = 0 W bnd = 0 conservation of mass: m

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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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OldFinalS - ME 201 Thermodynamics ME 201 Thermodynamics Old...

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