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Exam2s Michigan State University ME 201
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  • Title: Exam2s
  • Type: Notes
  • School: Michigan State University
  • Course: ME 201
  • Term: Spring

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2006 Spring Thermodynamics Exam #2 Solution Problem 1 Steam at 300 kPa with quality 0.96316 passes through a valve to convert it to saturated vapor. Determine the exit pressure required. Solution: We begin with our template. Substance Type: Compressible System Type: Control Volume Device: Valve Process: Isenthalpic Inlet State: Fixed Outlet State: UNKNOWN q=0 wsh = 0 1st Law: h2 - h1 = 0 State 1 State 2 T1 = 133.52 C T2 = 81.31 C P1 = 300 kPa P2 = 50 kPa h1 = 2645.2 kJ/kg h2 = 2645.2 kJ/kg Phase: 2 phase with Phase: sat.vap x1 = 0.96316 Italicized values are from steam tables. Bold values are calculated. Approach: Go to the steam tables and get hf and hg at 300 kPa. Use the quality to calculate h1. Set h2 = h1. Go to the saturation pressure table and find the pressure that gives an hg = h2. At 300 kPa, we have hf = 561.43 kJ/kg and hg = 2724.9 kJ/kg Then we can calculate h1 = (1-x1)hf +x1hg = (1-0.96316)(561.43) + (0.96316)(2724.9) = 2645.2 kJ/kg Since our valve is isenthalpic h2= h1= 2645.2 kJ/kg Now going to the saturation pressure table we find P2 = 50 kPa 1 ME 201 ME 201 Thermodynamics Spring 2006 Problem 2 A piston-cylinder device contains 0.001 kg of CO2 at 750 kPa and 1340 K. The device undergoes a polytropic process with polytropic exponent -0.3 or PV-0.3 = constant which decreases the pressure to 550.2 kPa. Determine the boundary work and heat transfer in kJ for this process. Solution: We begin with our template. Substance Type: Ideal Gas System Type: Closed System Process: Polytropic with n=-0.3 Initial State: Fixed Final State: UNKNOWN Q:UNKNOWN Wsh = 0 P V -P V Wbnd = 2 2 1 1 1- n Conserv. of mass: m1 = m2 = 0.001 kg 1st Law: m(u2 - u1) = Q - Wbnd State 2 State 1 T1 = 1340 K T2 = 350 K P1 = 750 kPa P2 = 550.2 kPa u1 = 1151.4 kJ/kg u2 = 191.75 kJ/kg 3 v1 = 0.3375 m /kg v2 = 0.1202 m3/kg V1 =3.375 x 10-4 m3 V2 = 1.202x 10-4 m3 m1 = 0.001 kg m2 = 0.001 kg Italicized values are from air tables or ideal gas relation. Bold values are calculated. Approach: At state 1, go to the air tables and get u1. Use the ideal gas relation to calculate v1. Calculate V1 from v1 and m1. Use polytropic relation to calculate V2. Calculate v2 from V2 and m2. Calculate T2 from ideal gas relation. Read u2 from air tables. Calculate the boundary work from PdV. Calculate the heat transfer from the 1st law. From the air tables for state 1 u1 = 50,672 kJ/kgmol = (50,672)/(44.01) = 1151.4 kJ/kg Using the ideal gas law RT (0.1889)(1340) v1 = 1 = = 0.3375 m 3 /kg P1 (750) 2 ME 201 Thermodynamics Spring 2006 Then V1 = m1v1 = (0.001)(0.3375) = 3.375 x 10-4 m 3 At state 2 we know the pressure and have a polytropic process so that n P2 V2 = P1V1n or solving P1 1/n V2 = V1 or substituting P2 750 -1/0.3 V2 = (3.375 x 10 ) = 1.202 10- x 4 m 3 550.2 -4 Then V2 1.202 x 10-4 v2 = = = 0.1202 m 3 m2 0.001 We can then determine the temperature at state 2 from Pv (550.2)(0.1202) T2 = 2 2 = = 350 K R (0.1889) From the air tables for state 2 u2 = 8,439 kJ/kgmol = (8,439)/(44.01) = 191.75 kJ/kg We can now use our equation for the boundary work P2 V2 - P1V1 (550.2)(1.202 x 10-4 ) - (750)(3.375 x 10-4 ) Wbnd = = = - 0.1439 kJ 1- n 1 - (-0.30) From the 1st law we calculate the heat transfer Q = m(u2 - u1) + Wbnd = (0.001)(191.75-1151.4) + (-0.1439) = -1.1035 kJ Problem 3 A proposed new ice maker is composed of two components. The first component is a cooling coil, while the second component is a freezing chamber. Water at 15 C and 100 kPa enters the cooling coil at 0.05 kg/s and exits at 0 C as a 50% - 50% mixture by mass of liquid and solid. Determine the heat transfer rate in kW for the cooling coil. Hint: The enthalpy for subcooled liquid or solid can be calculated from the tables with sub.liq.: h1= hf(at T1) + vf(at T1) x [P1-Psat(at T1)] sub.sol.: h1= hi(at T1) + vi(at T1) x [P1-Psat(at T1)] Liquid Water Cooling Coil Ice/Liquid Mixture 3 ME 201 Thermodynamics Spring 2006 Solution: We begin with the cooling coil. Setting up our template, Substance Type: Compressible System Type: Control Volume Device: Heat Device Process: Constant Pressure Inlet State: Fixed Outlet State: UNKNOWN q = ??? wsh = 0 1st Law: h2 - h1 = q State 1 State 2 T1 = 15 C T2 = 0 C P1 = 100 kPa P2 = 100 kPa h1 = 63.08 kJ/kg h2 = 2645.2 kJ/kg Phase: sub. liq. Phase: 2 phase with x2 = 0.50 Italicized values are from steam tables. Bold values are calculated. Approach: With state 1 fixed, we can go to the steam tables and get hf, vf, and Psat at 15 C and then use the equation in the hint to calculate h1. Next we use the constant pressure condition to fix state 2. We calculate the liquid and solid enthalpies using the equation from our hint and then use the quality to calculate h2. Use the 1st law to calculate the heat transfer rate. At 15 C we find hf = 62.982 kJ/kg, vf = 0.001001 m3/kg and Psat = 1.7057 kPa Then using our hint h1 = hf(at T1) + vf(at T1) x [P1-Psat(at T1)] = 62.982 + (0.001001)(100-1.7057) = 63.08 kJ/kg At 0 C we have hf = 0.001 kJ/kg, vf = 0.001000 m3/kg and Psat = 0.6117 kPa so that h2, = hf(at T2) + vf(at T2) x [P2-Psat(at T2)] = 0.001 + (0.001000)(100-0.6117) = 0.1004 kJ/kg and hi = -333.40 kJ/kg, vi =0.001091 m3/kg and Psat = 0.61169 kPa so that h2,s = hi(at T2) + vi(at T2) x [P2-Psat(at T2)] = -333.40 + (0.001091)(100-0.61169) = -333.292 kJ/kg Then h2 = (0.5)h2, + (0.5)h2,s = (0.5)(0.1004) + (0.5)(-333.292) = -166.596 kJ/kg 4 ME 201 Thermodynamics Spring 2006 Our 1st law then gives Q = m(h 2 - h1 ) = (0.05)(-166.596 - 63.08) = - 11.48 kW 5

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HW20s
Path: Michigan State University >> ME >> 201 Spring, 2006

Description: Spring 2006 Thermodynamics Homework #20 Solution 1. Consider an internal combustion engine operating on the ideal Dual cycle with the following conditions: Two cylinder, four stroke engine with displacement of 1.6 liters Compression ratio of 7.5 Cut...
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Description: Spring 2006 Thermodynamics Homework #10 Solutions 1. Three of the process that occur in the piston cylinder device of an internal combustion engine are: Process 1: Constant pressure heat addition during which the volume doubles Process 2: Isentropic...
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Path: Michigan State University >> ME >> 201 Spring, 2006
Description: Spring 2006 Thermodynamics Homework 18 Solutions 1. Determine the work per mass output of an adiabatic turbine with isentropic efficiency 0.83 that has a steam input of 15 MPa and 650C and an outlet pressure of 50 kPa. Solution: The actual work will...
HW8s
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: Spring 2006 Thermodynamics Homework #8 Solution 1. Ten grams of water at 15C and 100 kPa completely fills a balloon. The balloon is then heated on the stove top at constant pressure until the temperature reaches 125C. Determine the boundary work in ...
HW17s
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: Spring 2006 Thermodynamics Homework 17 Solution 1. Two kilograms of Refrigerant-134a is contained in a piston-cylinder system. It is initially at 160 kPa and 0C and is compressed to saturated vapor at 0C. The heat transfer from the cylinder is repor...
HW7s
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: Spring 2006 Thermodynamics Homework #7 Solutions 1. Refrigerant -134a as saturated vapor at 0.5 MPa is isentropically compressed by a compressor in a refrigeration plant to 1.2 MPa. Determine the enthalpy change for the process and the final fluid p...
Exam3s
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: Spring 2006 Thermodynamics Exam #3 Solution Problem 1 Steam at 0.5 MPa and 350C is used to fill a 0.1 m3 tank, which is initially empty. After filling, the tank is cooled to 50C and the contents become saturated liquid. Determine (a) the heat transf...
syllabus
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: Spring 2006 Thermodynamics Section 1 MWF 8:00-8:50 2400 Engineering Building Instructor: Professor Craig W. Somerton Office: 2439 Engineering Building Telephone: 353-6733 email: somerton@egr.msu.edu Hours: Mon. 1:30-2:30, Tues. 1:30-2:30,Wed. 9-10, ...
HW11s
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: Spring 2006 Thermodynamics Homework 11 Solution 1. One component in a household refrigerator is the compressor where refrigerant 134-a enters as saturated vapor at -24F and is isentropically compressed to 30 psia. Determine the work required in Btu/...
HW4
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: Spring 2006 Thermodynamics Homework #4 Due Monday, January 30, 2006 1. Assuming an ideal gas calculate the specific volume in the appropriate units for: a. N2 at 500 kPa and 900 K b. Neon at 1 psia and 500 R c. Air at 14.7 psia and 72F 2. Calculate ...
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Path: Michigan State University >> ME >> 201 Spring, 2006
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Path: Michigan State University >> ME >> 201 Spring, 2006
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HW13s
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: Spring 2006 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...
Exam4s
Path: Michigan State University >> ME >> 201 Spring, 2006
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Path: Michigan State University >> ME >> 201 Spring, 2006
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HW14
Path: Michigan State University >> ME >> 201 Spring, 2006
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HW15
Path: Michigan State University >> ME >> 201 Spring, 2006
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Path: Michigan State University >> ME >> 201 Spring, 2006
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HW2
Path: Michigan State University >> ME >> 201 Spring, 2006
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HW9
Path: Michigan State University >> ME >> 201 Spring, 2006
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Path: Michigan State University >> ME >> 201 Spring, 2006
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Path: Michigan State University >> ME >> 201 Spring, 2006
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SecondLaw
Path: Michigan State University >> ME >> 201 Spring, 2006
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Path: Michigan State University >> ME >> 201 Spring, 2006
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Path: Michigan State University >> ME >> 201 Spring, 2006
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Path: Michigan State University >> ME >> 201 Spring, 2006
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Path: Michigan State University >> ME >> 201 Spring, 2006
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Path: Michigan State University >> ME >> 201 Spring, 2006
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Path: Michigan State University >> ME >> 201 Spring, 2006
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Path: Michigan State University >> ME >> 201 Spring, 2006
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Path: Michigan State University >> ME >> 201 Spring, 2006
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Path: Michigan State University >> PHY >> 321 Spring, 2006
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Path: Michigan State University >> CSE >> 860 Spring, 2004
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Path: Michigan State University >> CSE >> 860 Spring, 2004
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Path: Michigan State University >> CSE >> 860 Spring, 2004
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Path: Michigan State University >> CSE >> 860 Spring, 2004
Description: Computer Science 860 Foundations of Computing Spring, 2004 Instructor: Moon Jung Chung chung@cse.msu.edu Office Hours: Tu, Th 1-2pm&by appointment Text: Introduction to the Theory of Computation by Michael Sipser Reference: Computers and Intractabil...
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Path: Michigan State University >> PHY >> 410 Spring, 2007
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Path: Michigan State University >> PHY >> 410 Spring, 2007
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Path: Michigan State University >> PHY >> 102 Spring, 2006
Description: Worksheet #7 - PHY102 (Spr. 2006) Collisions Due Thursday 9pm March 2th, 2006 In this worksheet, we will return to solving equations and solving differential equations. Often there are multiple ways of accomplishing something in M athematica. Usually...
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Path: Michigan State University >> PHY >> 102 Spring, 2006
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Path: Michigan State University >> CSE >> 422 Spring, 2008
Description: CSE 422 Lab 2: Creating a Multi-Threaded Chat Room Server In this lab you will be making a server application for a chat room. You will need to use threading to listen for a client\'s message, as well as wait for any number of clients to connect. We w...
HW12s
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: Spring 2006 Thermodynamics Homework #12 Solution 1. A 2 ft3 scuba diver\'s air tank is to be filled with air from a compressed air line at 120 psia, 100F. Initially, the air in the tanks is at 20 psia and 70F. Assuming that the tank is well insulated...
HW5
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: Spring 2006 Thermodynamics Homework #5 Due Wednesday, February 1, 2006 1. Calculate the entropy change for N2 as it goes from 250 K and 1000 kPa to 1300 K and 60 kPa. 2. For the two processes given below, determine the final temperature, pressure, s...
Exam1
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: Spring 2006 Thermodynamics Exam #1 Open Book, Open Notes Problem 1 As shown in the drawing below, two pipes merge into one. Determine the velocity (in m/s) of water in the merged pipe under the following conditions: Pipe #1: diameter: 0.03 m, water ...
HW1s
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: Spring 2006 Thermodynamics Homework #1: Conservation of Mass Solution 1. Describe mass conservation for a real world system such as the human body or a jet aircraft engine. (5 pts) Solution: Various answers possible 2. During an attack, the asthma s...
HW19
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: Spring 2006 Thermodynamics Homework 19, Due Monday, April 17, 2006 1. Consider a steam power plant operating on a Rankine cycle with reheat as shown below. Steam leaves the boiler at 20 MPa and 700C. The first turbine exhausts to 0.4 MPa and the ste...
Exam3
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: Spring 2006 Thermodynamics Exam #3 Open Book, Open Notes Problem 1 Steam at 0.5 MPa and 350C is used to fill a 0.1 m3 tank, which is initially empty. After filling, the tank is cooled to 50C and the contents become saturated liquid. Determine (a) th...
HW3
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: Spring 2006 Thermodynamics Homework #3, Due Monday, January 23, 2006 1. Convert the following temperatures to F, C, K, R a. 98.6 F b. 298 K c. 5715 F d. 460 R e. 100 C 2. Convert the following pressures to psia and kPa. a. 760 mm of Hg b. 101 bar c....
Exam2
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: Spring 2006 Thermodynamics Exam #2 Open Book, Open Notes Problem 1 Steam at 300 kPa with quality 0.96316 passes through a valve to convert it to saturated vapor. Determine the exit pressure required. Problem 2 A piston-cylinder device contains 0.001...
HW13
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: Spring 2006 Thermodynamics Homework 13 Due Friday, March 17, 2006 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 restric...
HW12
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: Spring 2006 Thermodynamics Homework #12 Due Wednesday, 3/15/06 1. A 2 ft3 scuba diver\'s air tank is to be filled with air from a compressed air line at 120 psia, 100F. Initially, the air in the tanks is at 20 psia and 70F. Assuming that the tank is ...
Exam2r
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: Spring 2006 Thermodynamics Exam #2 Results High Low Average Median 74 (97%) 25 (33%) 53.9 (71.8%) 54 (72%) ME 201 Distribution 5 4 Number of Students 3 2 1 0 25 30 35 40 45 50 55 60 65 70 75 Exam #2 Score 1 ...
Exam3r
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: Spring 2006 Thermodynamics Exam #3 Results High Low Average Median 75 (100%) 28 (37%) 57.3 (76.3%) 60 (80%) ME 201 Distribution 5 4 Number of Students 3 2 1 0 25 30 35 40 45 50 55 60 65 70 75 Exam #3 Score 1 ...
HW18
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: Spring 2006 Thermodynamics Homework 18, Due Monday, 4/10/2006 1. Determine the work per mass output of an adiabatic turbine with isentropic efficiency 0.83 that has a steam input of 15 MPa and 650C and an outlet pressure of 50 kPa. 2. Refrigerant-13...
HW20
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: Spring 2006 Thermodynamics Homework #20, Due Wednesday, April 19, 2006 1. Consider an internal combustion engine operating on the ideal Dual cycle with the following conditions: Two cylinder, four stroke engine with displacement of 1.6 liters Compre...
FirsrLawProbs
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: ME 201 Thermodynamics First Law Practice Problems 1. Consider a balloon that has been blown up inside a building and has been allowed to come to equilibrium with the inside temperature of 25C and inside pressure of 100 kPa. The diameter of the balloo...
SecondLawProbs
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: ME 201 Thermodynamics Second Law Practice Problems 1. Ideally, which fluid can do more work: air at 600 psia and 600F or steam at 600 psia and 600F 2. A heat pump provides 30,000 Btu/hr to maintain a dwelling at 68F on a day when the outside temperat...
Energy
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: ME 201 Thermodynamics Conservation of Energy Guide The most general equation for the conservation of energy is d & (m e) = (min ein ) - (m out eout ) + Q - Wsh - Wbnd dt inflows outflows The time derivative portion represents the change...
hw4
Path: Michigan State University >> CSE >> 860 Spring, 2004
Description: CSE860 HW 4. Due: April 23, 5pm 1. Solve 7.28 2. Solve 8.5 3. Solve 8.12 4. Solve 8.20 5. Solve 9.9 6. Solve 9.18 7. Show that if NP is a subset of BPP, then RP = NP. ...
exam1
Path: Michigan State University >> CSE >> 860 Spring, 2004
Description: CSE860 Exam Due: 5 pm March 19. PART I. Solve the following three problems. 1. Suppose that (i) A and B are problems in P, (ii) C and D are in NP, (iii) E is NP-complete. (iv) F is co-NP. For each of the following questions, answer either \"false\" (i...
HW21s
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: Spring 2006 Thermodynamics Homework #21 Solution 1. Consider a jet aircraft flying at 300 m/s at an altitude of 3,000 m (use Table A-16 in the text to determine the pressure and temperature). The jet operates with a simple, ideal turbojet engine. Th...
OldFinalS
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: ME 201 Thermodynamics 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, specif...
HW6s
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: Spring 2006 Thermodynamics Homework #6 Solution 1. (10 pts) What is the enthalpy, internal energy, specific volume, and entropy for steam at 1107C and 27 MPa? Solution: Substance Type: Compressible (steam) Problem Type: State We are given steam at 2...
HW19s
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: Spring 2006 Thermodynamics Homework 19 Solution 1. Consider a steam power plant operating on a Rankine cycle with reheat as shown below. Steam leaves the boiler at 20 MPa and 700C. The first turbine exhausts to 0.4 MPa and the steam is then reheated...
HW15s
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: Spring 2006 Thermodynamics Homework 15 Solution 1. A Carnot heat engine produces power of 2.5 kW. It rejects heat to a river that is flowing at 2 kg/s, resulting in a temperature increase of 2C. The average temperature of the river is 20C. Determine...
HW3s
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: Spring 2006 Thermodynamics Homework #3, Solutions 1. Convert the following temperatures to F, C, K, R (each temperature pt) a. 98.6 F T(F)=98.6 F, T(C)=(98.6-32)/1.8=37C, T(K)=(98.6+460)/1.8=310.3 K, T(R)=98.6+460=558.6 R b. 298 K T(F)=(298)1.8-460...
Readings
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: Spring 2006 Thermodynamics Readings All citations to sections and pages refer to Thermodynamics: An Engineering Approach, 5th edition, by Y.A. engel and M.A. Boles. Topic Introduction Basic Definitions Units Mass Conservation Properties Types of Sub...
HW16s
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: Spring 2006 Thermodynamics Homework 16 Solution 1. Consider a power plant that is producing 1 MW of electric power as it operates with a high temperature of 1800 K and a low temperature of 290 K. If we can sell the electric power for $0.04 per kWhr,...
SteamPowerCycles
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: ME 201 Thermodynamics Steam Power Plants All steam power plants are based upon the ideal Rankine cycle shown below. Boiler Turbine Pump Condenser The four devices are Constant Pressure Boiler Isentropic Turbine Constant Pressure Condenser (fluid ...

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