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2006 Spring 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, 100 F. Initially, the air in the tanks is at 20 psia and 70 F. Assuming that the tank is well insulated, determine the temperature and mass in the tanks when it is filled to 120 psia. Solution: System Type: Transient Working Fluid: Air (ideal gas) Process: Constant total volume, adiabatic State 2 (final) State 3 (inlet) State 1 (initial) T2 = 724 R T1 = 70 F = 530 R T3 = 100 F = 560 R P1 = 20 psia P2 = 120 psia P3 = 120 psia u1 = 90.33 Btu/lbm u2 = 123.95 Btu/lbm h3 = 133.86 Btu/lbm m1 = 0.20377 lbm m2 = 0.895 lbm m3 = 0.689 lbm Italicized values are from air table. Bold values are calculated. Initial State: Fixed Inlet State: Fixed Final State: UNKNOWN Q =0 Wsh = 0 Wbnd = 0 KE and PE are negligible 1st Law: m2u2 - m1u1 = m3h3 mass: m2 - m1 = m3 Approach: To fix the final state we will have to simultaneously solve the 1st law, conservation mass and ideal gas law. First we go to the air tables and find u1 = 90.33 Btu/lbm and h3 = 133.86 Btu/lbm Next we use the ideal gas law to calculate the initial mass or PV (20)(2) m1 = 1 1 = = 0.2038 lb m RT1 (10.73 / 28.97)(530) ME 201 1 ME 201 Thermodynamics Spring 2006 We begin by taking the 1st law and eliminating m3 with the use of the conservation of mass equation or m2 u 2 - m1u1 = (m 2 - m1 )h 3 Rearranging m u + (m 2 - m1 )h 3 u2 = 1 1 m2 Next we use our ideal gas law to eliminate m2 P V m2 = 2 2 RT2 Through the air tables we have a relationship between T2 and u2, but we don't have an equation to substitute with. Hence, we will have to solve this problem by iteration (also known as trial and error). The way an iterative solution works is that an initial guess is made for one of the unknowns. The other unknown is determined based on this guess and then the original unknown is calculated. This calculated value is compared with the guess and if the comparison is poor the process is repeated. For a well posed problem the unknowns will eventually not change. Specifically for our problem we will follow the following steps 1. Guess T2 P2 V2 RT2 m u + (m 2 - m1 )h 3 3. Calculate u2 with u 2 = 1 1 m2 4. Obtain T2 using the air tables and u2 5. Compare the just calculated T2 with the one used to evaluate m2. If the difference is greater than 1 R, re-guess T2 return to step 2 and repeat steps 2-4. 2. Calculate m2 with m2 = Now carrying out our calculations: Loop 1 Guess T2 = 100 F = 560 R PV (120)(2) Calculating m 2 = 2 2 = = 1.157 lb m RT2 (10.73/28.97)(560) m u + (m 2 - m1 )h 3 (0.2038)(90.33) + (1.157 - 0.2038)(133.86) u2 = 1 1 = m2 1.157 Calculating = 126.19 Btu/lbm Going to the air tables, T2 = 738 R Since the difference between the two T2's is greater than 1 R we need to re-guess T2. 2 ME 201 Thermodynamics Spring 2006 Loop 2 Guess T2 = 738 R Calculating m2 = 0.878 lbm Calculating u2 = 123.76 Btu/lbm Going to the air tables, T2 = 724 R Since the difference between the two T2's is greater than 1 R we need to re-guess T2. Loop 3 Guess T2 = 724 R Calculating m2 = 0.895 lbm Calculating u2 = 123.95 Btu/lbm Going to the air tables, = T2 725 R Since the difference between the two T2's is less than 1 R we are finished 2. Consider a rigid tank of volume 2 m3 containing saturated liquid refrigerant-134a at 20 C. The tank is connected to an ideal turbine and as the refrigerant leaves the tanks it flashes to saturated vapor. If the tank is maintained at 20 C and the turbine exit pressure is 100 kPa, how much work will be produced when two thirds of the initial mass has been used? Solution: We first recognize that we will have four states to worry about in this problem: State 1: Initial State of Tank State 2: Outlet State of Tank and Inlet to Turbine State 3: Outlet State of Turbine State 4: Final State of Tank System Type: Transient Working Fluid: R-134a (compressible) Process: Constant total volume, turbine is isentropic State 2 P2 = 572.07 kPa T2 = 20 C s2 = 0.9223 kJ/(kgK) h2 = 261.49 kJ/kg m2 = 1634 kg phase: sat.vap. State 3 P3 = 100 kPa T3 = -26.37 C s3 = 0.9223 kJ/(kgK) h3 = 227.16 kJ/kg m3 = 1634 kg phase: two phase with x3 = 0.966 State 1 P1 = 572.07 kPa T1 = 20 C V1 = 2 m3 u1 = 76.86 kJ/kg m1 = 2451 kg phase: sat.liq. Italicized values are obtained from R-134a tables. Bold values are calculated. State 4 P4 = 572.07 kPa T4 = 20 C V4 = 2 m3 u4 = 86.39 kJ/kg m4 = 817 kg phase: two phase with x4 = 0.0464 3 ME 201 Thermodynamics Spring 2006 State 1: Fixed State 2: Fixed State 3: UNKNOWN State 4: UNKNOWN Q = UNKNOWN Wbnd = 0 KE and PE are negligible 1st Law (tank): m4u4 - m1u1 = -m2h2 + Q mass (tank): m4 - m1 = m2 1st Law (turbine): Wsh = m2(h2-h3) At state 1 the state is fixed. Going to the R-134a tables we have u1 = 78.86 kJ/kg and v1 = 0.0008161 m3/kg and P1 = 572.07 kPa We can then calculate the mass with V 2 m1 = 1 = = 2451 kg v1 0.0008161 At state 2 we have a fixed state so that going to the R-134a tables we have h2 = 261.59 kJ/kg and s2 = 0.92234 kJ/(kgK) and P2 = 572.07 kPa and we are told that the process continues until 2/3 of the mass has been removed, then m2 = (2/3)m1 = (2/3)(2451) = 1634 kg At state 3 we know the pressure and since the turbine is isentropic s3 = s2 = 0.92234 kJ/(kgK) Going to the saturation pressure table at 100 kPa we have sf = 0.07188 kJ/(kgK) and sg = 0.95183 kJ/(kgK) Since s3 lies between these two values, we have a two phase mixture with quality s -s 0.92234 - 0.07188 x3 = 3 f = = 0.966 s g - s f 0.95183 - 0.07188 The enthalpy is then h3 = hf + x3hfg = 17.28 + (0.966)(217.16) = 227.16 kJ/kg with a temperature of -26.37 C. By conservation of mass we have m3 = m2 = 1635 kg Then the work output of the turbine is given by Wsh = m2(h2-h3) = (1635)(261.59-227.16) = 56,293 kJ To fix state 4, we note that using the conservation of mass m4 = m1-m2 = 2451-1634 = 817 kg Then V2 2 v2 = = = 0.002448 m 3 / kg m2 817 4 ME 201 Thermodynamics Spring 2006 which fixes our state 4. We see that we have a two phase mixture with quality v -v 0.002448 - 0.0008161 x4 = 4 f = = 0.0464 v g - v f 0.035969 - 0.0008161 The internal energy is then u4 = uf + x4ufg = 78.86 + (0.0464)(162.16) = 86.39 kJ/kg The heat transfer to the tank is then Q = m4u4 - m1u1 + m2h2 = (817)(86.39)-(2451)(76.86) + (1634)(261.59) = 309,635 kJ 5
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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...
HW21
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: Spring 2006 Thermodynamics Homework #21, Due Friday, April 21, 2006 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 ...
CycleGuide
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: ME 201 Thermodynamics Cycle Analysis Guide Basic Principles Three basic types of thermal cycles Control Volume Power or Propulsion Cycles Closed System Power or Propulsion Cycles Control Volume Refrigeration Cycle All three systems have Heat Added He...
OldFinalExam
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: ME 201 Thermodynamics ME 201 Thermodynamics Old Final Exam 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, specifically, a ...
IncompSubstanceTable
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: ME 201 Thermodynamics ME 201 Thermodynamics Use of Compressible Substance Tables for the Evaluation of Incompressible Substance Properties Guide When a substance that has been identified as an incompressible substance, say subcooled liquid water, ha...
CompSubstance
Path: Michigan State University >> ME >> 201 Spring, 2006
Description: ME 201 Thermodynamics ME 201 Thermodynamics Compressible Substance Property Evaluation Guide Substances that are undergoing a phase change or have the potential to undergo a phase change must be considered compressible substances. Occasionally, we c...
week3
Path: Michigan State University >> CSE >> 860 Spring, 2004
Description: Week 3: Some undecidable 1. 2. ATM = {<M,w> | M is a TM and M accepts w} is undecidable. ATM is Turing recognizable (recursively enumerable). Diagonalization Method Countable, uncountable Z is countable, Z Z is countable. Theorem: A is a subset of B...
week2
Path: Michigan State University >> CSE >> 860 Spring, 2004
Description: Week 2: Lecture 3 TM: tape: two way, can write. Formally, Turing machine M is a 7-tuple, (Q, , , , q0, qaccept, qreject,), where 1. 2. 3. 4. 5. 6. 7. Q is the set of states is the input alphabet not containing the special blank symbol is the tape alp...
hw2
Path: Michigan State University >> CSE >> 860 Spring, 2004
Description: CSE860 HW 2. Due Feb. 17 1. Construct \"and-or\" graph for the following CFG G and Using this graph, decide (i) if L(G) is empty or not, and (ii) if L(G) is finite or not. G: S -> AB| CA B -> BC | AB A -> a C -> aB | b 2. Solve 3.6 3. Solve 3.7 4. Solv...
week1
Path: Michigan State University >> CSE >> 860 Spring, 2004
Description: Week 1: Lecture 1 1. 2. Languages, Machines, Functions Languages Regular Languages, Context Free Language Context Sensitive Languages Recursively Enumerable Languages How to define? Grammars: Regular Grammars, CFG, CSG etc. Grammar G = (V, , P, S), w...
course_rules
Path: Michigan State University >> PHY >> 410 Spring, 2007
Description: Course rules and grading procedure: Homework: Reading and homework assignments will be given at the beginning of each week. All assigned problems will also be posted on the Physics 410 home page. Homework will be due at the beginning of class on Mond...
Physics 410 Homework 10
Path: Michigan State University >> PHY >> 410 Spring, 2007
Description: Physics 410 Homework 10: 1. 2. 3. 4. 5. 6. 7. (6 pts) (5 pts) (5 pts) (4 pts) (10 pts) (5 pts) (10 pts) Problem 5.54 Problem 6.10 Problem 6.14 Problem 6.18 Problem 6.22 Problem 6.28 Problem 6.30 from Schroeder. from Schroeder. from Schroeder. from Sc...
worksheet13
Path: Michigan State University >> PHY >> 102 Spring, 2006
Description: # Worksheet 13 - 2006 Due Thursday 20th April 9pm This worksheet gives you some more practice at problem solving using mathematica. PROBLEM 1. Make a 3-d plot of the function Sin(xy) for x [0, 3] and y [0, 3]. PROBLEM 2. Numerically find the roots to...
worksheet04
Path: Michigan State University >> PHY >> 102 Spring, 2006
Description: Worksheet #4 - PHY102 (Spr. 2006) Solving equations Solving equations in Mathematica Look up how to solve algebraic equations exactly(Solve) and numerically(NSolve). If you have a transcendental equation (e.b. x = sin(x) you need to use \"FindRoot\". I...
Physics 410 Homework 11
Path: Michigan State University >> PHY >> 410 Spring, 2007
Description: 1. (6 pts) Free energy of a two state system. (a) Find an expression for the free energy as a function of =kBT of a system with two states, one at energy 0 and one at energy . (b) From the free energy, find expressions for the energy and entropy of t...
Physics 410 Homework 9
Path: Michigan State University >> PHY >> 410 Spring, 2007
Description: Physics 410 Homework 9: 1. 2. 3. 4. 5. 6. (8 pts) (6 pts) (5 pts) (6pts) (4 pts) (5 pts) Problem 5.32 Problem 5.34 Problem 5.48 Problem 5.49 Problem 5.51 Problem 5.53 from Schroeder. from Schroeder. from Schroeder. from Schroeder. from Schroeder. fro...
Physics 410 Homework 13
Path: Michigan State University >> PHY >> 410 Spring, 2007
Description: Physics 410 Homework 13: 1. 2. 3. 4. 5. 5. (10 pts) (5 pts) (4 pts) (4 pts) (5 pts) (8 pts) Problem 7.28 Problem 7.31 Problem 7.37 Problem 7.39 Problem 7.45 Problem 7.46 from Schroeder. from Schroeder. from Schroeder. from Schroeder. from Schroeder. ...
Physics 410 Homework 6
Path: Michigan State University >> PHY >> 410 Spring, 2007
Description: Physics 410 Homework 6: 1. (10 pts) 2. (3 pts) 3. (12 pts) 4. (6 pts) 5. (4 pts) 6 . (7 pts) Problem 3.25 from Schroeder. Problem 3.33 from Schroeder. Problem 3.34 from Schroeder. Problem 3.37 from Schroeder. Problem 3.38 from Schroeder. Problem 3.39...
Physics 410 Homework 7
Path: Michigan State University >> PHY >> 410 Spring, 2007
Description: Physics 410 Homework 7: 1. (4 pts) 2. (8 pts) 3. (8 pts) 4. (4 pts) 5. (4 pts) 6 . (5 pts) 7 . (6 pts) 8 . (10 pts) Problem 4.4 Problem 4.6 Problem 4.14 Problem 4.17 Problem 4.19 Problem 4.20 Problem 4.30 Problem 4.33 from Schroeder. from Schroeder. ...
Physics 410 Homework 5
Path: Michigan State University >> PHY >> 410 Spring, 2007
Description: Physics 410 Homework 5: 1. (4 pts) Problem 3.1 from Schroeder. 2. (3 pts) Problem 3.4 from Schroeder. 3. (5 pts) Problem 3.8 from Schroeder. 4. (6 pts) Problem 3.10 from Schroeder. 5. (4 pts) Problem 3.14 from Schroeder. 6. (3 pts) Problem 3.19 from ...
LIR 832 Q_Exam_Answer_key_ Spring, 2004
Path: Michigan State University >> LIR >> 832 Spring, 2007
Description: LIR 832 Qualifying Examination: Spring, 2004 This examination consists of 10 equally weighted questions which require basic knowledge of statistics. The examination lasts two hours, additional time will not be provided. Z and T tables are attached at...
Homework7
Path: Michigan State University >> PHY >> 321 Spring, 2006
Description: Physics 321 Spring 2006 Homework #7, Due at beginning of class Wednesday Mar 15. 1. [8 pts] A \"triangle wave\" can be defined by F (t) = 1 - 2|t|/ for -/ < t < +/, with F (t) defined at all other values of the time t by the property of having period ...
2005exam3
Path: Michigan State University >> PHY >> 321 Spring, 2006
Description: PHYSICS 321 EXAM 3 April 18, 2005 NAME 1. [5 pts] Imagine a peculiar cloud of cosmic dust whose mass distribution is spherically symmetric with a density (mass per unit volume) given by (r) = k r 2 where r is the distance from the center and k is...
Homework6
Path: Michigan State University >> PHY >> 321 Spring, 2006
Description: Physics 321 Spring 2006 Homework #6, Due at beginning of class Wednesday Mar 1. 1. [4 pts] A hook is at height y above the floor, where y is constant for all negative times: y = y0 for t < 0. For positive times, y oscillates: y = y0 + A sin t for t ...
2005final
Path: Michigan State University >> PHY >> 321 Spring, 2006
Description: PHYSICS 321 FINAL EXAM May 4, 2005 NAME 1. [5 pts] A particle of mass M moves in one dimension in the potential V (x) = a + bx where a and b are constants. Find all possible motions x(t). 2. [5 pts] The motion of a system with two degrees of freedom...
schedule_2007
Path: Michigan State University >> PHY >> 410 Spring, 2007
Description: Tentative Schedule: Week 1 2 3 4 5 6 7 8 Spring 9 10 11 12 13 14 15 Finals Month Jan Jan Jan Jan/Feb Feb Feb Feb Feb/Mar Break Mar Mar Mar/Apr Apr Apr Apr Apr April 22 29 5 12 19 26 5 12 19 26 2 9 16 23 30: M 8 W 10 17 24 31 7 14 21 28 7 14 21 28 4 1...
Physics 410 Homework 12
Path: Michigan State University >> PHY >> 410 Spring, 2007
Description: Physics 410 Homework 12: 1. 2. 3. 4. 5. (6 pts) (6 pts) (12 pts) (7 pts) (14 pts) Problem 7.2 Problem 7.6 Problem 7.8 Problem 7.16 Problem 7.23 from Schroeder. from Schroeder. from Schroeder. from Schroeder. from Schroeder. ...
Physics 410 Homework 2
Path: Michigan State University >> PHY >> 410 Spring, 2007
Description: Physics 410 Homework 2: 1. 2. 3. 4. 5. 6. 7. Problem 1.36 from Schroeder. Problem 1.40 from Schroeder. Problem 1.41 from Schroeder. Problem 1.44 from Schroeder. Problem 1.50from Schroeder. Problem 1.60 from Schroeder. Problem 1.63 from Schroeder. ...
Physics 410 Homework 4
Path: Michigan State University >> PHY >> 410 Spring, 2007
Description: Physics 410 Homework 4: 1. 2. 3. 4. 5. 6. (7 pts) Problem 2.30 from Schroeder. (4 pts) Problem 2.31 from Schroeder. (4 pts) Problem 2.32 from Schroeder. (4 pts) Problem 2.34 from Schroeder. (4 pts) Problem 2.37 from Schroeder. (6 pts) Problem 2.39 fr...
Physics 410 Homework 1
Path: Michigan State University >> PHY >> 410 Spring, 2007
Description: Physics 410 Homework 1: 1. 2. 3. 4. 5. 6. 7. Problem 1.8 from Schroeder. Problem 1.16 from Schroeder. Problem 1.21 from Schroeder. Problem 1.22 from Schroeder. Problem 1.25 from Schroeder. Problem 1.26 from Schroeder. Problem 1.33 from Schroeder. ...
Physics 410 Homework 3
Path: Michigan State University >> PHY >> 410 Spring, 2007
Description: Physics 410 Homework 3: 1. 2. 3. 4. 5. 6. 7. Problem 2.2 from Schroeder. Problem 2.6 from Schroeder. Problem 2.13 from Schroeder. Problem 2.17 from Schroeder. Problem 2.19 from Schroeder. Problem 2.23 from Schroeder. Problem 2.27 from Schroeder. ...
worksheet02
Path: Michigan State University >> PHY >> 102 Spring, 2006
Description: Worksheet #2 - PHY102 (Spr. 2006) Formats and List operations (Vectors) Formating From the toolbar do the following: \"format\" followed by \"style\", the dropdown menu offers you lots of options. \"input\" is the default, and is the \"format\" in which you ...
