HW8_ME235

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University of Michigan, Dearborn *

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ME 230

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Mechanical Engineering

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Apr 3, 2024

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THE UNIVERSITY OF MICHIGAN Department of Mechanical Engineering ME 235: Thermodynamics I Section 002 Winter 2022 © Wooldridge 2022 Homework Set #8 - Solutions 24 total points, each section of each problem is worth 2 points HW#8 topics covered: Application of the second law 1. A heat engine receives 1200 kJ of heat from a source at 1200 K and rejects 300 kJ of heat to a sink at 300 K. The power cycle produces 200 kJ of work. Is this cycle reversible, irreversible or impossible? Justify your answer. Depends on the order of analysis, there are two correct solutions Solutions Option 1 First we evaluate if the cycle is violates the first law, COEnergy. W cyc must equal Q cycle . Here 200 kJ does not equal (Q in -Q out ) = 900 kJ So the cycle is impossible, based on the first law. Solution Option 2 How does the actual cycle efficiency compare with the Carnot cycle efficiency? Actual efficiency = Wcyc/Qin = 200 kJ/1200 K = 16% Carnot efficiency = 1-TL/TH = 1 - 300K/1200K=75% Actual efficiency < Carnot efficiency So the cycle is irreversible. 2. A refrigerator removes 0.75 Btu/s of heat from the interior of the refrigerator, maintaining the interior temperature at 3 o F. The heat from the refrigerator is rejected to the room at 67 o F. a) What is the maximum coefficient of performance (COP max ) of the refrigerator? b) What is the reversible power required by the refrigerator [Btu/s]? Solutions The reversible work is determined from the Carnot cycle for a refrigerator. a) COP max = COP carnot,ref = T L /(T H -T L ) = (3+459.67 R)/ (67 - 3 R) = 463/64 = 7.2

b) Ẇ = Q dot,L /COP = (0.75 BTU/s) / 7.2 = 0.10 Btu/s 3. A one-ton air conditioner can remove 12000 Btu/hr of heat from a building. a) If the operating temperatures of the air-conditioner are 70 o F (interior temperature) and 90 o F (outdoor temperature), what is the maximum coefficient of performance of the air conditioner? b) What is the minimum power required by the air conditioner [Btu/hr] for one ton of cooling? c) What is the flow rate of the working fluid [lbm/min] required for a specific work of 20 Btu/lbm (based on the minimum work required by the air conditioner)? d) The work required by the compressor of the air conditioner is 20% higher than the minimum work required by the cycle. R410a is the working fluid, and enters the compressor as a saturated vapor at 20 psia. If the compressor increases the pressure to 100 psia, what is the exit temperature of the compressor [ o F]? Use the CATT3 program for R410a properties. Solutions The reversible work is determined from the Carnot cycle for a refrigerator. a) COP max = COP carnot,actual = T L /(T H -T L ) = (70+459.67 R) / (90 - 70 R) = 530/20 = 26.5 b) Ẇ min = Q dot,L /COP = (12000 Btu/hr) / 26.5 = 453 Btu/hr c) ṁ = Ẇ min /w min = (453 Btu/hr) / (20 Btu/lbm) = (22.7 lbm/hr)(1 hr/60 min) = 0.38 lbm/min d) ẇ comp = ẇ min *1.2 = 1.2*(20 Btu/lbm) = 24 Btu/lbm = h 2 -h 1 From CATT3 using P 1 = 20 psia and x 1 = 1.0, h 1 = 111.9 Btu/lbm Solving for h 2 h 2 = h 1 +ẇ comp = 111.9 Btu/lbm + 24 Btu/lbm h 2 = 135.9 Btu/lbm Use h 2 and P 2 where P 2 = 100 psia and CATT3 to find T 2 T 2 = 94.1 o F 4. A large nuclear power plant has a thermal efficiency of 28% and has a rated capacity of 2 GW of electricity. The fission of nuclear fuel releases 8x10 7 MJ / kg(!) of uranium. a) How much uranium is consumed per year assuming the plant operates 80% of the time during the year (i.e., the power plant is online 80% of the time at the rated capacity)? b) How much heat is wasted from the plant by heat rejection [MW]? Solutions a) eff = Ẇ cycle /Q dot,in Solving for Q dot,in = Ẇ cycle /eff = 2 GW / 0.28 = 7.14 GW ṁ fuel = (capacity*Q dot,in )/mass uranium = (7.14x10 6 kW) /(8x10 10 kJ/kg)

= (8.925x10 -5 kg/s)*(0.8)*(3.15e7 s/year) = 2250 kg b) solve for Q dot,out Q dot,out = Q dot,in - Ẇ cycle = 7.14 GW - 2 GW = 5.14 GW = 5140 MW 5. Ford Lake is used as a heat sink for a power plant with a thermal efficiency of 35%. The lake has an average depth of 15 ft and covers an area of 975 acres. a) If the mass flow rate of water through the condenser of the power plant is 0.4x10 4 kg/s, what fraction of the lake water is used per day [%]? b) If the temperature of the lake water can only be heated by 1 o C. What is the maximum power out of the plant [W]? Solutions a) mass lake water = Vol * density assume density of lake water is 1000 kg/m 3 Vol = A*d, where A = 975 acre = 3.946x10 6 m 2 Vol = (3.946x10 6 m 2 )(4.57 m) = 18x10 6 m 3 m lake = (18x10 6 m 3 )(1000 kg/m 3 ) m lake = 18 x 10 9 kg The mass of water used by the power plant in one day is m pp = ṁDt = (0.4x10 4 kg/s)(3600 s/hr)(24 hr/day) m pp = 3.456x10 8 kg m pp /m lake = 1.9% b) Q dot,out = ṁc lake (ΔT) = (0.4x10 4 kg/s) (4.18 kJ/kgK)(1 K) = 16,720 kW eff = W cyc /Q in W cyc = Q in -Q out W cyc + Q out = Q in eff = W cyc /(W cyc +Q out ) (W cyc +Q out )eff = W cyc W cyc =(eff)(Q out ) / (1-eff) = (0.35)(16,720 kW)/(1-0.35) = 9003 kW

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6. We want to evaluate the effectiveness of a new heat pump for heating our home during our cold Michigan winters. The heat pump uses a 5 kW electric motor and has a coefficient of performance (COP) 60% of the Carnot COP. We want to maintain our home at 20 o C (67 o F) during the winter lows of -9.4 o C (15 o F). If our house loses heat at a rate proportional to 0.1 kW times the temperature difference between the indoor and outdoor temperatures, will the heat pump work for our needs in Michigan? Solutions We want to evaluate if Q H from the heat pump is greater than or equal to the heat losses from our home. Q dot,loss = (0.1 kW/K)(T inside - T outside ) = (0.1 kW/K)(29.4K) Q dot,loss = 2.94 kW COP max = COP carnot = T H /(T H -T L ) = (294 K)/(29.4 K) = 10 COP actual = 0.6*COP max = 6 COP actual = Q H /W in Q H,dot = W in,dot *COP actual = 5 kW*6 = 30 kW Q H > Q loss so yes! The heat pump will work for our Michigan weather.