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ME235_Exam2_F10.pdf

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Department
Mechanical Engineering
Course
MECHENG 235
Professor
Donald Siegel
Semester
Winter

Description
ME235 EXAM II, NOVEMBER 30, 2010 C. BORGNAKKE and D. SIEGEL 1. [30%] In a remote location you run a heat engine to provide the power to run a refrigerator. The input to the heat engine is at 800 K and the low T is 400 K, it has an actual efficiency equal to ½ of the corresponding Carnot unit. The refrigerator hasLa T = -10C and T H T ambient 10, with a COP that is 1/3 of the corresponding carnot unit. Assume the ambient is at 25C and that a cooling capcity of 2 kW is needed. a. Find the rate of heat input to the heat engine b. Find the entropy generation rate in the heat engine c. Find the entropy generation rate in the refrigerator 2. [30%] Air at 100 kPa, 20C is compressed in a piston cylinder to a volume 9 times smaller in a polytropic process where n = 1.25. If heat transfer goes out it goes into the ambient at 20C and if the heat transfer comes in it comes from a 250C energy reservoir. a) Find the final T and P of the air b) Find the specific work and specific heat transfer c) Find the entropy generation per unit mass of air 3. [40%] When large concrete pipes for storm drains are made they are placed in a sealed perfectly insulated constant P flexible bag. Assume the bag contains 1000 kg concrete at 30C which releases 1.2 W/kg due to chemical reaction. This chemical heat release is like a work term in and not heat transfer. The bag also contains 25 kg of saturated water vapor at 100 kPa. The pipe cures for 6 hr and not all the steam condenses. a) Find the final T (assumed uniform) and quality x of the water. b) Find the total entropy generation. EXAM II, SOLUTION NOVEMBER 30, 2010 C. BORGNAKKE and D. SIEGEL 1. [30 points] T = 800 K T = -10 C H The coolin. Q H1 QL2 capacity isL2 . W HE REF. The only term Q Q that does the L1 H2 cooling. 400 K 35 C Definition of the heat engine efficiency TL1 400 HE = 0.5 CARNOT = 0.5 (1 T ) = 0.5 ( 1 800) = 0.25 H Definition of the refrigerator COP 1 1 T L 1 263.15  =  =  =  = 1.949 3 CARNOT 3 T H T L 3 35 - (-10) From the required cooling load . . W = Q /  = 2 kW = 1.026 kW L2 1.949 Heat engine output determined so find the input . . 1.026 Q = W /  = kW = 4.104 kW H1 HE 0.25 Entropy equation for the two control volumes . . . QL1 Q H 4.104 - 1.0264.104 Sgen totT T = 400  800 = 0.002565 kW/K L1 H . . . Q H2 Q L2 2 + 1.026 2 Sgen refT T = 308.15 263.15 = 0.00222 kW/K H2 L2 Notice there is entropy generated outside of the heat engine (from the 400 K to the 25C ambient redistribution of heat) and outside the refrigerator (from the 35C to the 25C redistribution of heat). 2. [30 points] C.V. Air of constant mass m 2 m = 1. Energy Eq.5.11: m(u2 u 1 = 1 2 W 1 2 Entropy Eq.8.3 (37): m(s2 s1) =  dQ/T + 1 2 gen 1.25 Process: Pv = constant, v2/v1= 1/9 State 1: P 1 100 kPa, T =1293.15 K State 2:
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