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FirsrLawSolns Michigan State University ME 201

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

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201 ME 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 25 C and inside pressure of 100 kPa. The diameter of the balloon is measured and found to be 0.13 m. The balloon is then taken outside and allowed to come to equilibrium with the outside temperature of -5 C and outside pressure of 100 kPa. Determine the boundary work, heat transfer, and final balloon diameter. Solution: System Type: Closed System Substance Type: Ideal gas Process: Isobaric Initial State: Fixed Final State: Fixed Q = UNKNOWN Wsh = 0 Wbnd = UNKNOWN Conservation of Mass: m1 = m2 1st Law: m(u 2 - u1 ) = Q - Wbnd Boundary Work: P(v 2 - v1 ) State 2 State 1 T1 = 25 C = 298 K T2 = -5 C = 268 K P1 = 100 kPa P2 = 100 kPa D1 = 0.13 m D2 = 0.1255 m -3 3 V1 = 1.150 x 10 m V2 = 1.034 x 10-3 m3 m1 = 1.345 x 10-3 kg m2 = 1.345 x 10-3 kg u1 = 209.06 kJ/kg u2 = 191.17 kJ/kg 1 = 1.6783 kJ/(kg K) 2 = 1.588 kJ/(kg K) v1 = 0.855 m3/kg v2 = 0.769 m3/kg Italicized values are from ideal gas relations. Bold values are calculated. Approach: We begin by evaluating the properties at state 1 and state 2 by using the air tables and the ideal gas law. We can then use our boundary work equation to calculate the boundary work. Finally, we use the conservation of energy to determine the heat transfer. 1 ME 201 Thermodynamics We start by determining our total volume at state 1. The volume of a sphere is given by 4 V1 = R 3 = 1.150 x 10 -3 m 3 3 Next we determine our mass. Using the ideal gas law we have P1V1 (100)(1.150 x 10 -3 ) m1 = = = 1.345 x 10 -3 kg RT1 (0.287)(298) V1 (1.150 x 10 -3 ) = = 0.855 m 3 /kg v1 = -3 m1 (1.345 x 10 ) We can go to the air tables and find u1 = 209.06 kJ/kg 1 = 1.6783 kJ/(kgK) u2 = 191.17 kJ/kg 2 = 1.5888 kJ/(kgK) The specific volume at state 2 must be given by the ideal gas law, so that RT2 (0.287)(268) = = 0.769 m 3 /kg v2 = P2 (100) The total final volume is then V2 = m2v2 = (1.345 x 10-3)(0.769) = 1.034 x 10-3 m3 The final diameter is 3V D 2 = 2 2 = 0.1255 m 4 The boundary work is calculated Wbnd = P(V2 - V1 ) = (100)(1.034 x 10 -3 - 1.150 x 10 -3 ) = - 0.0116 kJ and the heat transfer is Q = m(u 2 - u1 ) + Wbnd = (1.345 x 10 -3 )(191.17 - 209.06) + (-0.0116) 1/3 = - 0.0357 kJ 2 ME 201 Thermodynamics 2. Air at 1800 K and 800 kPa enters an ideal turbine at 2.3 kg/s. The power output required of this turbine is 700 kW. Determine the exhaust temperature and pressure. Solution: System Type: Control Volume System Substance Type: Ideal gas Process: Isentropic Initial State: Fixed Final State: Unknown Q=0 Wsh = 700 kW & & Conservation of Mass: m 2 = m1 = 2.3 kg/s & & 1st Law: m(h 2 - h1 ) = - Wsh State 1 State 2 T1 = 1800 K T2 = 1552 K P1 = 800 kPa P2 = 424 kPa h1 = 2003.3 kJ/kg h2 = 1698.95 kJ/kg 1 = 3.6684 kJ/(kg K) 2 = 3.4864 kJ/(kg K) Italicized values are from ideal gas relations. Bold values are calculated. Approach: We begin by evaluating the properties at state 1 by using the air tables and the ideal gas law. We can then use the conservation of energy to determine the enthalpy at state 2. The air tables will give T2. Finally, the isentropic relation will give the exit pressure At state 1 we can go to the air tables and find h1 = 2003.3 kJ/kg 1 = 3.6684 kJ/(kgK) From the first law we solve for h2 & W 700 h 2 = h1 - sh = 2003.3 = 1698.95 kJ/kg & m 2.3 The air tables can then give T2 = 1552 K 2 = 3.4864 kJ/(kgK) The isentropic relation is P s = 0 = 2 - 1 - R ln 2 P 1 Solving for P2 gives - 3.4864 - 3.6684 P2 = P1 exp 2 1 = (800)exp = 424 kPa 0.287 R 3 ME 201 Thermodynamics 3. Refrigerant-134a as saturated liquid at 32 C enters a valve and exits at 0.16 MPa. What is the fluid phase at the exit? Solution System Type: Control Volume System Substance Type: Compressible Process: Isenthalpic Initial State: Fixed Final State: Unknown Q=0 Wsh = 0 1st Law: h2 - h1 = 0 State 2 State 1 T1 = 32 C T2 = -15.62 C P1 = 0.81528 MPa P2 = 0.16 MPa h1 = 94.39 kJ/kg h2 = 94.39 kJ/kg Phase: sat.liq. Phase: 2 phase, x2 = 0.31 Italicized values are fromR-134a tables. Bold values are calculated. Approach: We begin by evaluating the properties at state 1 by using the refrigerant tables. We can then use the conservation of energy to determine the enthalpy at state 2. The refrigerant tables will give the fluid phase. At state 1 we can go to the refrigerant tables and find h1 = 94.39 kJ/kg P1 = 0.81528 MPa From the first law we solve for h2 h2 = h1 = 94.39 kJ/kg At 0.16 MPa, we find from the tables hf = 29.78 kJ/kg hg = 237.97 kJ/kg Since h2 is between these two values we have a two phase mixture at -15.62 C and with quality h -h 94.39 - 29.78 x2 = 2 f = = 0.31 h fg 208.18 4 ME 201 Thermodynamics 4. A tank contains Refrigerant-134a as saturated vapor at 100 kPa. There is heat transfer to the tank of 66 kJ/kg. Determine the final temperature and pressure. Solution: System Type: Closed System Substance Type: Compressible Process: Isotropic Initial State: Fixed Final State: Unknown q = 66 kJ/kg Wsh = 0 Wbnd = 0 Conservation of Mass: m1 = m2 1st Law: u 2 - u1 = q State 2 State 1 T1 = -26.43 C T2 = 60 C P1 = 100 kPa P2 = 140 kPa u1 = 212.18 kJ/kg u2 = 278.18 kJ/kg 3 v1 = 0.1917 m /kg v2 = 0.1917 m3/kg phase: sat.vap. phase: sup.vap. Italicized values are from tables. Bold values are calculated. Approach: We begin by evaluating the properties at state 1 by using the refrigerant tables. We can then use isotropic process to determine the specific volume at state 2. Next, we use the conservation of energy to determine the final internal energy. Finally, we go to the refrigerant tables and find the T and P. We can go to the refrigerant tables and find u1 = 212.18 kJ/kg T1 = -26.43 C v1 = 0.1917 m3/kg Since our process is isotropic, we have v2 = v1 = 0.1917 m3/kg The final internal energy is given by the 1st law or u2 = q + u1 = 66 + 212.18 = 278.18 kJ/kg So now we must go to the refrigerant tables and find the T and P that will correspond to these values of u and v. Since we had saturated vapor and we added heat, we will assume that we have gone to superheated vapor. Then scanning the tables we find that at 0.14 MPa and 60 C, we have u = 278.74 kJ/kg v = 0.1902 m3/kg which is close enough. 5 ME 201 Thermodynamics 5. Consider a 300 gallon hot water heater which is to provide water a 180 F and 20 psia. If a typical shower consumes 3 gallons/minute , last 15 minutes, and requires that the hot water should stay above 160 F, determine the heat transfer rate required. Water is supplied at 63 F and 22 psia and 2 gallons/minute. Solution: System Type: Transient System Substance Type: Incompressible Process: Isotropic Initial State: Fixed Final State: Fixed Inlet State: Fixed Exit State: Fixed Q = UNKNOWN Wsh = 0 Wbnd = 0 m - m1 & & Conserv. of Mass: 2 = m in - m out t m u - m1u1 & & & 1st Law: 2 2 = m in h in - m out h out + Q t State 1 State In State Out State 2 T1 = 180 F Tin = 63 F Tout = 170 F T2 = 160 F P1 = 20 psia Pin = 22 psia Pout = 20 psia P2 =20 psia u1 = 147.97 Btu/lbm hin = 31.43 Btu/lbm hout =138.01 Btu/lbm u2 = 127.94 Btu/lbm v1 = 0.01651 ft3/lbm vin = 0.01606 ft3/lbm vout = 0.01645 ft3/lbm v2 = 0.01640 ft3/lbm m1 = 51.08 lbm min = 0.0055 lbm/s mout = 0.0083 lbm/s m2 = 48.56 lbm Note that we have used the average temperature at state 1 and 2 to fix our outlet temperature Italicized values are from steam tables. Bold values are calculated. Approach: We begin by evaluating the properties at all states by using the steam tables. We can then use the conservation of mass to determine the mass at state 2. Finally the 1st law is used to determine the heat transfer rate. Even though we will treat water as an incompressible substance in this case, we can use the steam tables to determine the specific volume and internal energy since for an incompressible substance they only depend on temperature, we can take the values for saturated liquid at the given temperature. For the enthalpy, we can use the enthalpy of saturated liquid and then bring it up to the appropriate pressure or hincompressible sub. = hf + vf(P-Psat) 6 ME 201 Thermodynamics Our values are then entered on the table. To obtain the masses and mass flows we convert gallons to ft3 and divide by the specific volume. These values are entered on the table. The final state mass is given by conservation of mass or & & m 2 = t (m in - m out ) + m1 = (15)(60)(0.055 - 0.0083) + 51.08 = 48.56 lb m Then the heat transfer rate becomes & m u - m1u1 - m h + m h & in in & out out Q= 2 2 t (48.56)(127.94) - (51.08)(147.97) = (15)(60) - (0.0055)(31.43) + (0.0083)(138.01) = - 0.522 Btu/s = - 1881 Btu/hr 6. A piston cylinder system contains gas at 2300 K, 2500 and kPa, 0.03 liters. The gas then undergoes a polytropic expansion with a polytropic exponent of 1.15 to 0.3 liters. Compare the work performed in kJ for air as the gas versus hydrogen as the gas. Solution: We start this problem by working in air and then in hydrogen System Type: Closed System Substance Type:: Ideal gas Process: Polytropic with n=1.15 Initial State: Fixed Final State: UNKNOWN Q = UNKNOWN (but not asked for) Wsh = 0 Wbnd = UNKNOWN Conservation of Mass: m1 = m2 1st Law: m(u 2 - u1 ) = Q - Wbnd P2 v 2 - P1v1 (1 - n) State 1 State 2 T1 = 2300 K T2 = NA P1 = 2500 kPa P2 = 162.8 kPa -5 3 V1 = 3 x 10 m V2 = 3 x 10-4 m3 Italicized values are from ideal gas relations. Bold values are calculated. Boundary Work: Wbnd = Approach: At state 2 we can determine the pressure from the polytropic relation. We can then use our boundary work equation to calculate the boundary work 7 ME 201 Thermodynamics From the polytropic relationship we have n P2 V2 = P1V1n or solving for P2 1.15 V 0.03 1 P2 = P1 = (2300) = 162.8 kPa V 0.3 2 The boundary work is then given by (162.8)(3 x 10 -4 ) - (2300)(3 x 10 -5 ) = 0.134 kJ Wbnd = (1 - 1.15) Since the calculation above never used the fact that air was our ideal gas, we will obtain the same boundary work for hydrogen as the gas. n 7. In an open feedwater, subcooled liquid water is heated to saturated liquid by mixing directly with steam. In a given situation water at 10 MPa and 200 C enters an open feedwater heater at 12 kg/s. Steam at 10 MPa and 350 C is available to be added to this water to produce saturated liquid at 10 MPa. How much steam in kg/s must be added? Solution: System Type: Control Volume System Substance Type: Water (compressible substance) Process: Isobaric Inlet State: Fixed Exit State: Fixed Q=0 Wsh = 0 & & & Conserv. of Mass: m in -1 + m in -2 - m out = 0 & & & 1st Law: m in -1h in -1 + m in -2 h in -2 - m out h out = 0 State In-1 State In-2 State Out Tin-1 = 200 C Tin-2 = 350 C Tout = 311 C Pin-1 = 10 MPa Pin-2 = 10 MPa Pout = 10 MPa hin-1 = 856 kJ/kg hin-2 = 2923.4 kJ/kg hout =1407.56 kJ/kg & & & m in -1 = 12 kg/s m in -2 = 4.366 kg/s m out = 16.366 kg/s phase: sub.liq. phase: sub.liq. phase: sat.liq. Italicized values are from steam tables. Bold values are calculated. Approach: We begin by evaluating the properties at all states by using the steam tables. We can then use the conservation of mass to algebraically solve for the exit mass flow rate. Substituting this into the 1st law steam flow rate can be determined. 8 ME 201 Thermodynamics The enthalpies are read from the steam tables and entered into our table above. Solving for mout from conservation of mass & & & m out = m in -1 + m in -2 Substituting into the energy equation & & & & m in -1h in -1 + m in -2 h in -2 - (m in -1 + m in -2 )h out = 0 Solving for min-2 & m (h - h ) 12(856 - 1407.56) & m in -2 = in -1 in -1 out = = 4.366 kg h out - h in - 2 1407.56 - 2923.4 8. Most of the time during the winter Dr. Somerton turns down the thermostat to 50 F when he leaves in the morning. When he is in the house he likes to have the temperature at 68 F. The house may be considered to be composed of air, occupying a volume of 10,000 ft3, and structural material (mostly wood) of 11,000 lbm. Determine the total heat transfer required to bring the house up to 68 F. What fraction of this total goes to heating up the air and what fraction goes to heating up the structural material? Solution System Type: Closed System Substance Type: Ideal gas (air) and Incompressible (wood) Process: Isobaric Initial State: Fixed Final State: Fixed Q = UNKNOWN Wsh = 0 Wbnd = 0 Conservation of Mass: m1 = m2 1st Law: U2 - U1 = Q State 2 State 1 T1 = 50 F T2 = 68 F 3 V1,a = 10,000 ft V2,a = 10,000 ft3 m1,a = 779 lbm m2,a = 779 lbm m1,w = 11,000 lbm m2,w = 11,000 lbm Italicized values are from ideal gas relations. Bold values are calculated. Approach: We begin by determining the mass of air in the house. Then we can use the 1st law to calculate the heat transfer required. We determine our mass of air assuming a pressure of 14.7 psia. Using the ideal gas law we have PV (14.7)(10,000) m1,a = 1 1,a = = 779 lb m RT1 (10.73 / 29)(510) 9 ME 201 Thermodynamics We can calculate the heat transfer from the energy equation Q = U 2 - U1 = m a c v,a (T2 - T1 ) + m w c P, w (T2 - T1 ) = (779)(0.171)(68 - 50) + (11,000)(0.42)(68 - 50) = 85,558 Btu of which 97% goes into the structural material and 3% into the air. 9. Oil enters an ideal pump at 100 F and 12 psia and leaves at 17 psia. The oil flow rate is 0.04 lbm/s. The pump inlet has a diameter of 5 inches and the pump outlet has a diameter of 2 inches. What pumping power is required? Solution: System Type: Control Volume System Substance Type: Incompressible Process: Isentropic, can't neglect KE Initial State: Fixed Final State: Unknown Q=0 Wsh = UNKNOWN & & Conservation of Mass: m 2 = m1 = 0.04 lb m /s r2 r2 & h 2 + v 2 - h1 - v1 = - Wsh & 1st Law: m 2 2 State 1 State 2 T1 = 100 F T2 = 100 F P1 = 12 psia P2 = 17 psia r r v1 = 0.005 ft/s v 2 = 0.032 ft/s Bold values are calculated. Approach: We begin by using our process description to fix state 2. Then from the diameter and mass flow information the velocities are obtained. Finally, the 1st law is used to calculate the power. Since the process is isentropic and we have an incompressible substance, we write T s 2 - s1 = c P,avg ln 2 T 1 which yields T2 = T1 =100 F The velocities can be calculated from one of our continuity relations & & r m m v= = Ac D 2 4 10 ME 201 Thermodynamics Using a density for oil of 57 lbm/ft3, we find r v1 = 0.005 ft/s r v 2 = 0.032 ft/s The enthalpy change for an incompressible substance is given by h 2 - h1 = c P,avg (T2 - T1 ) + v avg (P2 - P1 ) Then the shaft work is calculated (using the density rather than the specific volume) r2 r2 & sh = - m P2 - P1 + v 2 - v1 & W 2 17 - 12 1 Btu 3 57 5.404 psia ft = - (0.04) (3600 s/hr) 2 2 (0.032) - (0.005) 1 Btu / lb m + 2 25.037 ft 2 / s 2 = - 2.338 Btu/hr 10.The exhaust process for an internal combustion engine may be modeled as transient system undergoing an isobaric process with boundary work. Just before the exhaust valve opens the cylinder of 0.5 liters contains air at 200 kPa and 500 K. At the end of exhaust the volume is 0.0833 liters. Assume that the process is adiabatic. Determine the final temperature and mass and the boundary work. Solution: System Type: Transient System Substance Type: Ideal gas Process: Isobaric Initial State: Fixed Outlet State: UNKNOWN Final State: UNKNOWN Q=0 Wsh = 0 Wbnd = P(V2-V1) Conserv. of Mass: m2-m1 = - mout 1st Law: m2u2-m1u1 = - mouthout - Wbnd State 1 State Out State 2 T1 = 500 K Tout = T2 = P1 = 200 kPa Pout = 200 kPa P2 =200 kPa -4 3 V1 = 5 x 10 m Vout = NA V2 =8.33 x 10-5 m3 Bold values are calculated. Approach: We begin by calculating the boundary work. We then use the ideal gas law and constant specific heat to determine the enthalpy and internal energies. 11 ME 201 Thermodynamics The boundary work is then Wbnd = P(V2-V1) = (200)[8.33 x 10-5 - 5 x 10-4] = -0.0833 kJ Now we want to solve for the final temperature, but we note that in the first law four different things depend on T2: u2, m2, hout, and mout. We can eliminate mout with the help of our conservation of mass so that m2u2-m1u1 = - (m1-m2)hout - Wbnd Since the exhaust air temperature is changing during the process, we will use our linear average approach to determine hout, or T + T2 h out = h air at 1 2 The mass at the final state will be given by PV m2 = 2 2 RT2 So substituting PV RT2 T + T2 m1u1 - m1 - 2 2 h air at 1 u2 = - Wbnd 2 RT2 P2 V2 When we consider that the air tables provide the relationships among u and T and h and T, we see that we do have a well posed problem, but we can't do the algebra, so we will have to solve the problem by iteration. Our process will be Guess T2 Evaluate hair and u2 from the air tables Calculate u2 from the above equation Compare the u2 at the guessed value of T2 to the calculated value of u2 Re-guess T2 and repeat the process until the difference between the values of u2 at the guessed T2 and the value of the calculated u2 is negligible I set this up in an Excel spreadsheet and the results are shown below T2,guess 500 525 550 600 650 625 u2,guess 147.31 165.99 184.82 222.95 261.71 242.25 Tout 500 512.5 525 550 575 562.5 hout 205.28 218.19 231.14 257.14 283.3 270.2 m2 0.000117 0.000111 0.000106 9.71E-05 8.96E-05 9.32E-05 u2 u2,guess-u2,calc 572.3208 -425.0108163 522.2173 -356.2273109 464.1406 -279.3206002 324.0476 -101.097563 151.6535 110.0564586 241.9062 0.343817527 So it would appear that our final temperature is about 625 K. 12

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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...

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...

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