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201 ME Thermodynamics ME 201 Thermodynamics Solutions to Second Law Practice Problems 1. Ideally, which fluid can do more work: air at 600 psia and 600 F or steam at 600 psia and 600 F Solution: The maximum work a substance can do is given by its availablity. We will assume that we have a closed system so that = u - u o - To (s - s o ) We take the dead state to be at STP or 25 C and 100 kPa or 76.4 F and 14.7 psia. Then using the appropriate table we have 600 air = 183.30 - 91.53 - (537) 0.7649 - 0.5995 - 0.06855ln 14.7 = 139.48 Btu/lb m and steam = 1184.5 - 44.09 - (537) 1.5320 - 0.08215 = 361.84 Btu/lb m So the steam can do more work 2. A heat pump provides 30,000 Btu/hr to maintain a dwelling at 68 F on a day when the outside temperature is 35 F. The power input to the pump is 1 hp. If electricity costs 8 cents per kilowatt-hour, compare the actual operating cost per day with the minimum theoretical operating cost per day. 1 ME 201 Thermodynamics Solution: We sketch our device interactions Dwelling High Temperature Heat Reservoir at TH QH Heat Pump Wnet QL Outside Low Temperature Heat Reservoir at TL The cost is given by Cost = (0.08)Wnet For the actual cost we have (Cost)act = (0.08)(1)(0.7457 kW/hp)(24 hr/day) = $1.43 To calculate the minimum cost we will allow the heat pump to operate as a Carnot cycle, so that 1 1 COPCarnot = = = 16 TL 495 1 1 TH 528 Then the minimum possible power input is QH 30,000 Wnet min = = = 1875 Btu/hr COPCarnot 16 = 0.5495 kW 2 ME 201 Thermodynamics and the minimum cost is (Cost) min = (0.08)(0.5495)(24 hr/day) = $1.06 3. A cylinder/piston system contains water at 200 kPa, 200 C with a volume of 20 liters. The piston is moved slowly, compressing the water to a pressure of 800 kPa. The process is polytropic with a polytropic exponent of 1. Assuming that the room temperature is 20 C, show that this process does not violate the second law. Solution: To determine if this violate the 2nd law we will want to calculate the entropy change of the universe and compare it to zero. We have - Qsys S universe = m(s 2 - s1 ) + Tsurr We now work this as a first law problem Working Fluid: Water(compressible) System: Closed System Process: Polytropic with n=1.0 State 1 State 2 T1 = 200 C T2 = 214.7 C P1 = 200 kPa P2 = 800 kPa u1 = 2654.4 kJ/kg u2 = 2655.5kJ/kg 3 V1 = 0.020 m V2 = 0.005 m3 v1 = 1.0803 m3/kg v2 = 0.2703 m3/kg s1 = 7.5066 kJ/(kg K) s2 = 6.8811 kJ/(kg K) phase: sup.vap. phase: sup.vap. italicized values from tables, bold values are calculated Initial State: Fixed Final State: Unknown Wsh = 0 Q = ???? Wbnd = ???? We begin by calculating the mass V 0.020 m= 1 = = 0.0185 kg v1 1.0803 3 ME 201 Thermodynamics To fix the final state we use the polytropic relationship P1V1n V2 = P2 1/n (200)(0.020)1 = 800 1/1 = 0.005 m3 The specific volume at state 2 is then V 0.005 v2 = 2 = = 0.2703 kg/m 3 m 0.0185 which gives us superheated vapor. The boundary work can be shown to be V 0.005 Wbnd = P1V1 ln 2 = (200)(0.020)ln V1 0.020 = - 5.55 kJ We use the first law to determine the heat transfer Qsys = m(u 2 - u1 ) + W = (0.0185)(2655.5 - 2654.4) + (-5.55) = - 5.53 kJ Then S surrounds = and for the system - Qsys Tsurr = - (-5.53) = 0.0189 kJ/K 293 S system = m(s 2 - s1 ) = (0.0185)(6.8811 - 7.5066) = - 0.0129 kJ/K So that S universe = - 0.0129 + 0.0189 = 0.006 kJ/K Since this is greater than zero the second law is not violated. 4. When a man returns to his well-sealed house on a summer day, he finds that the house is at 32 C. He turns on the air conditioner which cools the entire house to 20 C in 15 minutes. If the COP of the heat pump system is 2.5, determine the power drawn by the heat pump. Assume the entire mass within the house is equivalent to 800 kg of air. 4 ME 201 Thermodynamics Solution: We begin by sketching our device interactions House High Temperature Heat Reservoir at TH QH Heat Pump Wnet QL AC System Low Temperature Heat Reservoir at TL By definition we have QH COP = W net So if the required heat transfer can be determined the power can be determined. From a first law analysis on the house, we can write u -u 209.06 - 217.67 Q = m 2 1 = (800) t (15)(60) = - 7.65 kW and QH = - Q = 7.65 kW Then the power required is Q 7.65 Wnet = H = = 3.06 kW COP 2.5 5 ME 201 Thermodynamics 5. An innovative way of power generation involves the utilization of geothermal energy, the energy of hot water that exists naturally underground (hot springs), as the heat source. If a supply of hot water at 140 C is discovered at a location where the environmental temperature is 20 C, determine the maximum thermal efficiency a geothermal plant built at that location can have. If the power output of the plant is to be 5 MW, what is the minimum mass flow rate of hot water needed? Solution: We begin by sketching our device interactions Geothermal Source High Temperature Heat Reservoir at TH QH Heat Engine Wnet QL Environment Low Temperature Heat Reservoir at TL The maximum thermal efficiency will occur when the heat engine operates as a Carnot cycle, TL (20 + 273) =1= 0.291 th = Carnot = 1 TH (140 + 273) The minimum mass flow rate of hot water corresponds to the maximum thermal efficiency or 6 ME 201 Thermodynamics QH min = Wnet Carnot = 5000 = 17,208 kW 0.291 7 ME 201 Thermodynamics Performing a first law analysis on the hot water stream we have Q = m h out - h in For minimum the flow rate we will assume that the hot water is cooled down to the environment temperature, then Q - 17,208 m= = c P Tout Tin (4.1978) 20 140 = 34.2 kg/s 6. Air enters an adiabatic non-ideal nozzle at 9 m/s, 300 K, and 120 kPa and exits at 100 m/s and 100 kPa. Determine the irreversibility and the reversible work on a per mass basis. Solution: We first solve this as a first law problem Working Fluid: Air(ideal gas) System: Control Volume System Process: Nozzle State 1 State 2 T1 = 300K T2 = 295 C P1 = 120 kPa P2 = 100 kPa h1 = 300.19 kJ/kg h2 = 295.04 kJ/kg = 1.70203 kJ/(kg K) = 1.68515 kJ/(kg K) 2 1 v 2 = 100 m/s v1 = 9 m/s italicized values from tables, bold values are calculated Initial State: Fixed Final State: ??? Wsh = 0 Q=0 We use the first law to fix the final state v1 v2 h1 + = h 2 + 2 2 Then solving for h2 2 v1 v 2 (9)2 (100) 2 2 h 2 = h1 + = 300.19 + (10-3 ) 2 2 = 295.04 kJ/kg which allows us to determine the temperature and 2. Then the reversible work is 8 ME 201 Thermodynamics w rev = h1 - h 2 - THR 1 - 2 - R ln P1 P2 120 100 = 300.19 - 295.04 - (298) 1.70203 - 1.68515 - (0.287)ln = 15.71 kJ/kg Since the actual work is zero the irreversibility is i = w rev = 15.71 kJ/kg 7. Determine if a tray of ice cubes could remain frozen when placed in a food freezer having a COP of 9, operating in a room where the temperature is 32 C. Solution: We begin by sketching our device interactions Surroundings High Temperature Heat Reservoir at TH QH Refrigerator Wnet QL Freezer Compartment Low Temperature Heat Reservoir at TL 9 ME 201 Thermodynamics Assuming that the refrigerator operates on the Carnot cycle, we have 1 COP = COPCarnot = TH 1 TL Solving for TL TH 305 TL = = = 274.5 K 1 1 1 1 COP 9 and since this is greater than 0 C the ice cubes will not remain frozen. 8. Air is compressed in a closed system from a state where the pressure is 100 kPa and the temperature is 27 C to a final state at 500 kPa and 177 C. Can this process occur adiabatically? If yes, determine the work per mass. If no, determine the direction of the heat transfer. Solution: To determine if the process can occur, we must calculate - Qsys S universe = m(s 2 - s1 ) + Tsurr and compare it to zero. Since the process is adiabatic P s universe = s 2 - s1 = 2 - 1 - R ln 2 P1 Going to the air tables we find 500 s universe = s 2 - s1 = 2.11161 - 1.70203 - (0.287)ln 100 = - 0.523 kJ/kg Since this is less than zero, the process cannot be adiabatic. To make suniverse greater than zero will require Qsys to be negative, so that the direction of heat transfer is out of the system. 9. The pressure of water is increased by the use of a pump from 14 to 40 psia. A rise in the water temperature from 60 F to 60.2 F is observed. Determine the irreversibility, the second law efficiency, and the isentropic efficiency of the pump. 10 ME 201 Thermodynamics Solution: We first solve this as a first law problem Working Fluid: Water (incompressible) System: Control Volume System Process: Pump State 1 State 2s (ideal) State 2a (actual T2s = T1 = 60 F T2a = 60.2 F P1 = 14 psia P2 = 40 psia P2 = 40 psia bold values are calculated Initial State: Fixed Final State: fixed Wsh = ???? Q=0 To calculate the irreversibility, we use i = THR (s 2 - s1 ) - q = (537)cp,avg ln T2 -0 T1 520.2 520 = (537)(1.0014)ln = 0.2068 Btu/lb m To determine the second law efficiency we need both the actual work and the ideal work. Starting with the actual work we have w act = h1 - h 2 + q = c p,avg (T1 - T2 ) + v avg (P1 - P2 ) - 0 = (1.0014)(60 - 60.2) + (0.016035)(14 - 40)/(5.40395 psia ft 3 /Btu) = - 0.2774 Btu/lb m The reversible work is given by w rev = i + w act = (0.2068) + (-0.2774) = - 0.0706 Btu/lb m which allows us to determine the second law efficiency as w rev (-0.0706) = = 0.255 II = w act (-0.2774) 11 ME 201 Thermodynamics To determine the isentropic efficiency, we must first calculate the ideal work. Recognizing that in a isentropic process, the water will not change temperature, we can write w ideal = v avg (P1 - P2 ) = (0.016035)(14 - 40)/(5.40395 psia ft 3 /Btu) = - 0.0771 Btu/lb m Then our isentropic efficiency is w ideal (-0.0771) = = 0.2781 s= w act (-0.2774) 10. Carbon dioxide undergoes an isothermal reversible process from 250 kPa and 300 C to 500 kPa. Determine the heat transfer per mass by using the first law and evaluating the boundary work from Pdv . Compare this to the heat transfer per mass calculated from the entropy change and the second law. Solution: We first solve this as a first law problem Working Fluid: CO2 (ideal gas) System: Closed System Process: Isothermal, Reversible State 1 State 2 T1 = 300C = 573K T2 = T1 = 573K P1 = 250 kPa P2 = 500 kPa u1 = 369.23 kJ/kg u2 = 369.23 kJ/kg 1 = 5.478 kJ/(kg K) 2 = 5.478 kJ/(kg K) 3 v1 = 0.433 m /kg v2 = 0.2165 m3/kg italicized values are from ideal gas relationships Initial State: Fixed Final State: fixed Wsh = 0 Q = ??? Wbnd = ???? 12 ME 201 Thermodynamics We first go to the CO2 tables and get our properties. Our boundary work for an ideal gas undergoing an isothermal process is v w bnd = RT ln 2 v1 = (0.1889)(573)ln 0.2165 0.433 = - 75.03 Btu/lb m Using the first law our heat transfer is q = u 2 - u1 + w bnd = 369.23 - 369.23 + (-75.036) = - 75.036 Btu/lb m From the second law we have q = T1 (s 2 - s1 ) = T1 2- 1 - R ln P2 P1 500 250 = (573) 5.478 - 5.478 - (0.1889)ln = - 75.03 Btu/lb m So the two calculations for heat transfer agree. 13
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Exam4
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: Spring 2006 ME 201 Thermodynamics Exam #4 Open Book, Open Notes 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...
HW14s
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: Spring 2006 Thermodynamics Homework 14 Solution 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 convert saturate...
FirsrLawSolns
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: ME 201 Thermodynamics Solutions 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 ...
HW4s
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: Spring 2006 Thermodynamics Homework #4 Solutions 1. (3 pts)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 R uT Solution: We will us...
HW5s
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: Spring 2006 Thermodynamics Homework #5 Solutions 1. (5 pts) Calculate the entropy change for N2 as it goes from 250 K and 1000 kPa to 1300 K and 60 kPa. Solution: Substance Type: Ideal Gas (N2) Problem Type: Process State2 State 1 T1 = 250 K T2 = 13...
Exam2s
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: Spring 2006 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: Compress...
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...
HW10s
Path: Michigan State University >> ME >> 201 Spring, 2006
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...
HW18s
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 ...
FinalExamS
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: Spring 2006 ME 201 Thermodynamics Final Exam Solutions Directions: Work all three problems. The exam is open notes and open text book. All problems have equal weight. Note that you may round where appropriate to avoid interpolation. Problem 1 A more...
TransientSolns
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: ME 201 Thermodynamics Solutions to Transient System Practice Problems 1. A balloon initially contains 5 m3 CO2 at 100 kPa and 22C. It is connected to a CO2 gas line that provides CO2 at 170 kPa and 30C. The balloon is then filled to a pressure of 170...
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...
