Coursehero >>
Michigan >>
Michigan State University >>
ME 201 Course Hero has millions of student submitted documents similar to the one below including study guides, homework solutions, papers, and exam answer keys.
201 ME Thermodynamics Solutions to Transient System Practice Problems 1. A balloon initially contains 5 m3 CO2 at 100 kPa and 22 C. It is connected to a CO2 gas line that provides CO2 at 170 kPa and 30 C. The balloon is then filled to a pressure of 170 kPa. During this process the balloon is insulated and it is observed that the pressure and total volume are related by PV-1 = constant Determine the final temperature and volume of the balloon. Solution: We begin by setting up our table. System Type: Transeint System Substance Type: Ideal gas Process: Polytropic with n=-1 State 1: Fixed State 2: Unknown State In: Fixed Wsh = 0 Q=0 P V -P V Wbnd = 2 2 1 1 1- n Conservation of mass: m2 - m1 = min 1st Law: m2u2 - m1u1 = minhin -Wbnd State 1 P1 = 100 kPa T1 = 295 K V1 = 5 m3 m1 = 8.973 kg u1 = 154.49 kJ/kg State In Pin = 170 kPa Tin = 303 K min = 14.71 kg hin = 216.87 kJ/kg State 2 P2 = 170 kPa T2 = 324 K V2 = 8.5 m3 m2 = 23.6 kg u2 = 173.89 kJ/kg Italicized values are obtained from N2 tables or ideal gas equation. Bold values are calculated. At state 1 we know both the temperature and pressure, so the state is fixed. We can calculate the mass from PV (100)(5) m1 = 1 1 = = 8.973 kg RT1 (0.1889)(295) 1 ME 201 Thermodynamics The internal energy can be read from the CO2 tables, but will have to be converted to kJ/kg or 6799 u1 = 154.49 kJ/kg = u1 = MWN 2 44.01 At the inlet state we can find the enthalpy from the CO2 tables or hin = 216.87 kJ/kg To begin to fix our final state, we use the polytropic relation or -1 P2 V2 = P1V1-1 or 170 P = 8.5 m 3 V2 = V1 2 = (5) 100 P1 The boundary work can now be calculated P V - P V (170)(8.5) - (100)(5) Wbnd = 2 2 1 1 = 1 - (-1) 1- n = 472.5 kJ For the mass at state 2, we have PV m2 = 2 2 RT2 But since we do not have the temperature we cannot solve for this. The temperature could come from u2 as calculated from the 1st law, but the 1st law contains m2, so that we have three equations and three unknowns. We are going to have to solve this by trial error as follows: Guess a T2 Calculate m2, m 2 = P2 V2 RT2 Calculate min, min = m2 - m1 1 Calculate u2, u 2 = [m1u1 + min h in - Wbnd ] m2 Look T2 up in the table corresponding to u2 Repeat loop until T2 does not change 2 ME 201 Thermodynamics Now starting Guess T2 = 300 K (170)(8.5) = 25.5 kg (0.1889)(300) Calculate min, min = 25.5 - 8.973 = 16.526 kg Calculate u2, 1 [(8.973)(154.49) + (16.526)(216.87) - 472.5] = 176.38 kJ/kg u2 = 25.5 Look T2 = 327 K Repeat loop with T2 = 328 K Calculate m2, m 2 = Guess T2 = 328 K (170)(8.5) = 23.3 kg (0.1889)(328) Calculate min, min = 23.3 - 8.973 = 14.35 kg Calculate u2, 1 [(8.973)(154.49) + (14.35)(216.87) - 472.5] = 172.77 kJ/kg u2 = 23.3 Lookup T2 = 323 K Repeat loop with T2 = 323 K Guess T2 = 323 K (170)(8.5) Calculate m2, m 2 = = 23.6 kg (0.1889)(323) Calculate min, min = 23.6 - 8.973 = 14.71 kg Calculate u2, 1 [(8.973)(154.49) + (14.71)(216.87) - 472.5] = 173.89 kJ/kg u2 = 23.6 Lookup T2 = 324 K Stop loop since this should be close enough Calculate m2, m 2 = 3 ME 201 Thermodynamics 2. A pressure cooker used in cooking has a relief valve that normally opens at about 180 kPa and a volume of 3 liters. Initially, 2/3 of the volume of the pressure cooker is filled with liquid. What is the highest heat transfer rate allowed for the pressure cooker not to run out of water in one hour of cooking? Solution: We begin by setting up our table. System: Transient System Working Fluid: water (compressible substance) Process: Isobaric with P = 180 kPa State 1: Fixed State 2: Fixed State Out: Fixed Wsh = 0 Q: Unknown Wbnd = 0 Conservation of mass: m2 - m1 = -mout m u - m1u1 & & 1st Law: 2 2 = - m out h out + t Q State 1 P1 = 180 kPa T1 = 116.9 C V1 = 0.003 m3 m1 = 1.8914 kg v1 = 0.001586 m3/kg u1 = 491.589 kJ/kg Phase: 2 phase with x1 = 5.287 x 10-4 State Out Pout = 180 kPa Tout = 116.9 C State 2 P2 = 180 kPa T2 = 116.9 C V2 = 0.003 m3 m2 = 0.0031 kg v2 = 0.9775 m3/kg u2 = 2525.9 kJ/kg Phase: Sat. Vap. & m out =5.245 x 10-4 kg/s hout = 2701.8 kJ/kg Phase: Sat. Vap. Italicized values are obtained from steam tables. Bold values are calculated. We will assume that our mixture starts at 180 kPa. Then we need to determine the quality at state 1. The mass of vapor must be given by (% vapor by volume)(total volume) (1 / 3)(0.003) m v,1 = = specfic volume of vapor v g (@180 kPa) (1 / 3)(0.003) = 0.0010 kg 0.9775 Similarly for the mass of liquid we can write = 4 ME 201 Thermodynamics (2 / 3)(0.003) (% liquid by volume)(total volume) = v f (@180 kPa) specfic volume of liquid (2 / 3)(0.003) = = 1.8904 kg 0.001058 Then the total mass is m1 = m v,1 + m l,1 = 0.0010 + 1.8904 = 1.8914 kg m l,1 = and the quality and specific volume at state 1 are m v,1 0.0010 x1 = = = 5.287 x 10 -4 m1 1.8914 V 0.003 v1 = 1 = = 0.001586 m 3 / kg m1 1.8914 The internal energy is then calculated from u1 = (1- x1 )u f + x1u g = (1- 5.287 x 10 -4 )(490.5) + (5.287 x 10 -4 )(2525.9) = 491.58 kJ / kg At state 2 we have saturated vapor, so the properties can be read directly form the steam tables. The mass at state 2 is determined from V 0.003 m2 = 2 = = 0.0031 kg v2 0.9775 The mass flow rate is calculated from the conservation of mass or m1 - m 2 1.8914 - 0.0031 & = = 5.245 x 10 -4 kg m out = t 3600 The heat transfer rate is then determined from the 1st law or m 2 u 2 - m1u1 & & Q = m out h out + t (0.0031)(2525.9) - (1.8914)(491.589) = (5.245 x 10 -4 )(2701.8) + 3600 = 1.16 kW 5 ME 201 Thermodynamics 3. A 5 m3 tank contains saturated liquid R-134a at 1.0 MPa. A small leak occurs and the tank pressure falls to 0.32 MPa before it is sealed. It is determined that the Freon remaining in the tank has a quality of 0.3. Determine a. the mass that is discharged b. the heat transfer that occurs Solution: We begin by setting up our table System Type: Transient System Substance Type: Compressible Process: Isotropic with V = 5 m3 State 1: Fixed State 2: Fixed State Out: Unknown Wsh = 0 Q: Unknown Wbnd = 0 Conservation of mass: m2 - m1 = -mout m u - m1u1 & & 1st Law: 2 2 = - m out h out + Q t State 1 P1 = 1 MPa T1 = 39.39 C V1 = 5 m3 m1 = 5750 kg v1 = 0.0008695 m3/kg u1 = 104.42 kJ/kg Phase: sat.liq. State Out Pout = 0.66 MPa Tout = 24.7 C mout =5493.6 kg hout = 110.4 kJ/kg Phase: two phase with xout = 0.15 State 2 P2 = 0.32 MPa T2 = 2.48 C V2 = 5 m3 m2 = 256.4 kg v2 = 0.0195 m3/kg u2 = 105.67 kJ/kg Phase: 2 phase with x=0.3 Italicized values are obtained from R-134a tables. Bold values are calculated. At state 1 we have saturated liquid at 1 MPa, so we can go directly to the R-134a tables and find the properties. The mass at state 1 can then be calculated V 5 m1 = 1 = = 5750 kg v1 0.0008695 At state 2 we know the pressure and the quality, so the properties can be determined. The mass at state 2 can then be calculated V 5 = 256.4 kg m2 = 2 = v 2 0.0195 6 ME 201 Thermodynamics The mass that is discharged now is calculated by the conservation of mass or mout = m1 - m2 = 5750 - 256.4 = 5493.6 kg To fix state out, we use our average approach. We will assume that the outlet state is at the average pressure and the average quality then Pout = 0.66 MPa and xout = 0.15 So with the outlet properties determined we can use the 1st law to calculate the heat transfer or Q = m out h out + m 2 u 2 - m1u1 = (5493.6)(110.4) + (256.4)(105.67) - (5750)(104.42) = 33,166 kJ 7
Find millions of documents here - Study Guides, Homework Solutions, Papers, Exam Answer Keys and more.
Course Hero has millions of course related materials that will enable you to learn better, faster and get an A in all your courses.
Below is a small sample set of documents:
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
Description: Spring 2006 ME 201 Thermodynamics Exam #4 Solution Problem 1 Often in an ideal jet propulsion cycle a second burner is used after the turbine, as shown in the figure. Consider the following operating conditions: Inlet conditions for the turbine: 900...
CLO
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: Spring 2006 Thermodynamics Course Learning Objectives 1. Basic Concepts a. Students can identify control volumes, closed systems, and transient systems b. Students can apply the state principle c. Students can work in different unit sets d. Students...
HW14
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: Spring 2006 Thermodynamics Homework 14, Due Wednesday, 3/22/2006 1. Consider the Carnot cycle occurring in a piston-cylinder device containing refrigerant-134a with operating conditions given below: Process A: Isothermal heat addition at TH = 30C to...
HW15
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: Spring 2006 Thermodynamics Homework 15 Due Friday, 3/24/2006 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 2...
Plagiarism
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: Spring 2006 Thermodynamics Plagiarism Policy Department of Mechanical Engineering Plagiarism is not tolerated in the Department of Mechanical Engineering. It shall be punished according to the student conduct code of the University. Integrity and ho...
HW2
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: Spring 2006 Thermodynamics Homework #2, Due Friday, January 20, 2006 Explain whether the following situations and should be modeled as closed systems, control volume systems, or transient systems. 1. Hot Water Heater 2. Refrigerator 3. Washing Machi...
HW9
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: Spring 2006 Thermodynamics Homework #9 Due Monday 2/20/06 1. A rigid wall container is divided into two regions by a removable wall. One region contains 1 lbm of kerosene at 100F, while the other region contains 2 lbm of kerosene at 150F. A stirrer ...
Exam1r
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: Spring 2006 Thermodynamics Exam #1 Results High Low Average Median 75 (100%) 15 (20%) 53 (70.7%) 54 (72%) ME 201 Distribution 5 4 Number of Students 3 2 1 0 10 15 20 25 30 35 40 45 50 55 60 65 70 75 Exam #1 Score 1 ...
ControlVolume
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: ME 201 Thermodynamics First Law for Control Volume Systems Guide Recall that for a control volume system there is no accumulation or depletion of mass so that the mass inflow must equal the mass outflow or inflows m out outflows Also ...
SecondLaw
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: ME 201 Thermodynamics Second Law Guide The second law of thermodynamics really consists of a number of statements that one might consider rules of reality that help explain physical observations that are not explained by the conservation of mass or c...
GasTurbineCyles
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: ME 201 Thermodynamics Gas Turbine Cycles 1. Gas Turbine Power Cycles All gas turbine power plants are based upon the ideal Brayton cycle shown below. Compressor Burner Turbine The three devices are Isentropic Compressor Constant Pressure Burner (...
HW1
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: Spring 2006 Thermodynamics Homework #1: Conservation of Mass Due Wednesday, January 18, 2006 1. Describe mass conservation for a real world system such as the human body or a jet aircraft engine. 2. During an attack, the asthma sufferer actually acc...
AirWaterVapor
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: ME 201 Thermodynamics Air Processing Cycles (Air -Water Vapor Cycles) Basic Definitions Dry Bulb Temperature (TDB): This is the temperature of the air/water vapor mixture that would be measured with a standard thermometer. It is the temperature that ...
RevWork
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: ME 201 Thermodynamics Reversible Work, Irreversibility, and Availability Guide The concepts of reversible work, irreversibility and availability allow us to apply the second law of thermodynamics in a useful way. In particular, these concepts will he...
MassSolutions
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: ME 201 Thermodynamics Conservation of Mass Practice Problems 1. A human being can blow air out of their mouth at a rate of 10-4 kg/s. How long will it take for this human to blow up a balloon to a volume of 5 x 10-4 m3? The air may be taken to be at ...
IdealGas
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: ME 201 Thermodynamics ME 201 Thermodynamics Ideal Gas Property Evaluation Guide (For ME 201see summary at end) Most normal gases at normal pressures and temperature can be treated as ideal gases provided that there are not phase changes occurring. T...
PreFinalGrades
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: Spring 2006 ME 201 Thermodynamics Pre-Final Exam Grades PID A32213067 A32705194 A33771282 A33904427 A34191051 A34237404 A34273614 A34433300 A34438470 A34458339 A34947034 A35165679 A35306249 A35323701 A35532202 A35536130 A35642829 A35654142 A35804172...
HW22s
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: Spring 2006 ME 201 Thermodynamics Homework 22 Solution 1. An ideal vapor compression refrigeration cycle with refrigerant 134a as the working fluid operates with an evaporator temperature of 20C and a condenser pressure of 1.2 MPa. For a refrigerant...
mass
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: ME 201 Thermodynamics Handout: Conservation of Mass The general form of our conservation of mass equation is: dm sys dt = m out inflows outflows where dm sys : change in mass within the system per time dt & m in : sum of all the mass i...
2005exam2
Path: Michigan State University >> PHY >> 321 Spring, 2006
Description: PHYSICS 321 EXAM 2 Mar 21, 2005 NAME 1. [6 pts] A particle of mass M = 1 moves in one dimension in the potential U (x), where U (x) = 3 3 + 2x if x > 0 . if x < 0 (Units have been chosen to keep things simple, so don\'t worry about the dimensi...
week4
Path: Michigan State University >> CSE >> 860 Spring, 2004
Description: Week 4 Reduction via the history of Computation: Linear Bounded Turing Machine (Automata) Definition: A linear bounded automaton is a restricted type of Turing machine where in the tape head isn\'t permitted to move off the portion of the tape contai...
Final
Path: Michigan State University >> CSE >> 860 Spring, 2004
Description: CSE860 Final Due: Saturday 12 noon, May 1. PART I. Solve problem 1 and 2. 1. For each of the following assertions, state whether they are True, False, or Open according to our current state of knowledge of computability and complexity theory, as desc...
hw3
Path: Michigan State University >> CSE >> 860 Spring, 2004
Description: CSE860 HW 3. Problem set (No grading) 1. Solve 6.3 2. Solve 6.9 3. Solve 7.1 4. Solve 7.8 5. Solve 7.12 6. Solve 7.16 7. Solve 7.23 8. Solve 7.29 ...
syllabus
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...
Physics 410 Homework 14
Path: Michigan State University >> PHY >> 410 Spring, 2007
Description: Physics 410 Homework 14: 1. (7 pts) Reflective heat shield and Kirehhoff\'s law. Consider a plane sheet of material of absorptivity a, emissivity e, and reflectivity r = 1-a. Let the sheet be suspended between and parallel with two black sheets mainta...
Physics 410 Homework 8
Path: Michigan State University >> PHY >> 410 Spring, 2007
Description: Physics 410 Homework 8: 1. (5 pts) 2. (8 pts) 3. (8 pts) 4. (4 pts) 5. (12 pts) 6. (4 pts) 7 . (4 pts) Problem 5.1 Problem 5.5 Problem 5.12 Problem 5.13 Problem 5.14 Problem 5.20 Problem 5.21 from Schroeder. from Schroeder. from Schroeder. from Schro...
worksheet07
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...
worksheet12
Path: Michigan State University >> PHY >> 102 Spring, 2006
Description: Worksheet #12 - PHY102 (Spr. 2006) DC and AC circuits Due Thursday April 13th 9pm In earlier worksheets we have studied the behavior of damped, massspring systems. We also took a brief look at the linear and non-linear pendulum problems. The equation...
lab2
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 ...
HW2s
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: Spring 2006 Thermodynamics Homework #2 Solutions Explain whether the following situations and should be modeled as closed systems, control volume systems, or transient systems. 1. Hot Water Heater 2. Refrigerator 3. Washing Machine 4. Catalytic Conv...
HW23
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: Spring 2006 Thermodynamics Homework 23, Due Friday, May 5, 2006 Extra Credit worth 10 points 1. Complete the following table for the properties of an air/water vapor mixture. Tdry bulb (C) 15 35 10 20 . Twet bulb (C) . 25 10 . . Rel.Hum. (%) 20 . . ...
HW8
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: Spring 2006 Thermodynamics Homework #8 Due Wednesday, 2/15/06 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 bou...
HW16
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: Spring 2006 Thermodynamics Homework 16 Due Monday, 3/27/2006 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...
HWScores
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: Spring 2006 ME 201 Thermodynamics Homework Scores PID A32213067 A32705194 A33771282 A33904427 A34191051 A34237404 A34273614 A34433300 A34438470 A34458339 A34804667 A34947034 A35165679 A35306249 A35323701 A35532202 A35536130 A35642829 A35654142 A3580...
HW17
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: Spring 2006 Thermodynamics Homework 17, Due Wednesday, 3/31/2006 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 ...
HW6
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: Spring 2006 Thermodynamics Homework #6, Due Friday, 2/3/06 1. What are the enthalpy, internal energy, specific volume, and entropy for steam at 1107C and 27 MPa? 2. Determine the internal energy change as saturated liquid refrigerant-134a at -6F goe...
HW7
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: Spring 2006 Thermodynamics Homework #7 Due Monday 2/6/06 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...
HW11
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: Spring 2006 Thermodynamics Homework 11 Due Friday 2/24/06 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 requir...
HW22
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: Spring 2006 Thermodynamics Homework 22, Due Friday, May 5, 2006 Extra Credit worth 15 points 1. An ideal vapor compression refrigeration cycle with refrigerant 134a as the working fluid operates with an evaporator temperature of 20C and a condenser ...
ICEngineCycles
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: ME 201 Thermodynamics Piston-Cylinder Cycles (Internal Combustion Engine Cycles) Most internal combustion engines can be modeled as one of three piston/cylinder cycles. At the start of these cycles, the piston is out as far as possible, so that we ha...
HW10
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: Spring 2006 Thermodynamics Homework #10 Due Wednesday 2/22/06 1. Three of the processes 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...
