Exam is Open Textbook, Open Class Notes, Computers can be used (Computer limited to class notes, lectures, homework, book material, calculator, conversion utilities, etc. No searching for similar problems at other sites.) Name: Email: To receive full credit all work must be clearly provided. Please use units in all answers. Problem 1: Air expands through a turbine operating at steady state on an instrumented test stand. At the inlet, P 1 = 160 lbf/in. 2, T1 = 1560 R, and at the exit, P 2 = 14.8 lbf/in. 2 The mass flow rate of air entering the turbine is 10.5 lb/s, and the power developed is measured as 2580 horsepower. Neglecting heat transfer and kinetic and potential energy effects, determine the exit temperature, T 2, in R. Δ KE +Δ PE +Δ H = Qnet Wcv Δ KE =Δ PE = Q net = 0!m(h 1! h 2 ) =W cv = 2580hp Btu 1hp = 2545 =.707 h Table A-22E for T 1 gives h 1 = 385.08 Btu/lbm Btu / s 2580 hp(.707 ) hp Btu Btu Btu = + h = 173.72 + 385.08 = 211.36 lbm 10.5 lbm lbm lbm s 2 1 Table A-22E for h 2 gives T 2 = 880 0 R hr Btu s 1
Problem 2: Air enters the compressor of a gas turbine power plant operating at steady state at 290K, 100 kpa and exits at 420K, 330 kpa. Stray heat transfer and kinetic and potential energy effects are negligible. Using the ideal gas model for air, determine the isentropic compressor efficiency. Be sure to show all work.! c = h 2s! h 1 h 2! h 1 to find h 2s, use eqn 6.41 p r2 = p r1 ( p 2 p 1 ) =1.2311(330 /100) = 4.0626 Using Table A-22 h 2s interpolates to 408.5 kj/kg and! c = h 2s! h 1 h 2! h 1 = 408.5! 290.16 421.26! 290.16 =.903 (90.3%) 2
Exam is Open Textbook, Open Class Notes, Computers can be used (Computer limited to class notes, lectures, homework, book material, calculator, conversion utilities, etc. No searching for similar problems at other sites.) Name: Email: To receive full credit all work must be clearly provided. Please use units in all answers and do not pull- apart the exam. 1.) (70%) Two kilograms of water at 25 C are placed in a piston cylinder device under 100 kpa absolute pressure as shown in the diagram (State(1)). Heat is added to the water at constant pressure until the piston reaches the stops at a total volume of 0.4 m 3 (State (2)). More heat is then added at constant volume until the temperature of the water reaches 300 C (State (3)). ES 3001 Thermodynamics Exam- 1 (Sullivan, A- Term 2012) 1
i) Draw a P-v and a T-v diagram of the states and processes of the problem and include all the relevant information on the diagram. In this case there are three states and two processes (state 1 to state 2 and state 2 to state 3). The diagrams do not have to be drawn to scale. It is not known where quality x2 is, just somewhere on the horizontal line in the wet region. P CR line is not necessary, nor are the blue lines necessary. 1 3 2 Constant Temp. ES 3001 Thermodynamics Exam- 1 (Sullivan, A- Term 2012) 2
ii) Determine the quality (x) of the fluid and the mass of the vapor at state (2). (Schematic is not necessarily to scale.) Calculate the specific volume, specific internal energy, and specific enthalpy of the state.! =! =!! +!!!!! =!! = 0.4 2 = 0.2!"!!! (!!!! ) =. 2.0010432 (1.694.0010432) =.117! =!! +!!!!! = 417.36 +.117(2506.1-417.36) = 661.74 h = h! +! h! h! = 417.46 +.117(2675.5-417.46) = 681.65! =!!! (!!!! ) =. 2.0010432 (1.694.0010432) =.117 mass of vapor = x (mass ) =.117 * 2kg =.234 kg iii) Determine the pressure of the fluid at state (3). Calculate the specific volume, specific internal energy, and specific enthalpy of the state. Double interpolation for v, u, and h.! = 0.2!" volume is constant find in super heat region where v =.2 kg somewhere between 10 bar and 15 bar (Table A- 4) Need: P = 10 bar P=15 bar T = 280 u=.2480 u=.1627 T = 300 u=(.2480+.2678)/2!!!"! = u=(.1627+.1765)/2 u=.2579!!"!!" u=.1696 T = 320 u=.2678 u=.1765 X =.6557 P = 10bar + x(15bar- 10bar) = 10 +.6557*5 = 13.28 bar Need to do similar double interpolation for both u and h as done for T ES 3001 Thermodynamics Exam- 1 (Sullivan, A- Term 2012) 3
2.) (30%) A pressure cooker allows much faster (and more tender) cooking by maintaining a higher boiling temperature of the water inside. It is well sealed, and steam can only escape through an opening on the lid, on which sits a metal petcock. When the pressure overcomes the weight of the petcock, the steam escapes, maintaining a constant high pressure while the water boils. Assume the opening under the petcock has an area of 8 mm 2 and the atmospheric pressure is constant at 101 kpa. a) Draw a free body diagram of the petcock. F = mg F = P gage Area b) Determine the mass of the petcock required in order to maintain an operating pressure of 99 kpa gage. F = P gage Area = mg 99kPa 8mm 2 (1mm 2 /1000 2 m 2 ) =.792 N = m (9.8m/s 2 ) m = 0.0808kg = 80.8 grams c) Determine the corresponding temperature of the boiling water. Solution: P abs = P atm + 99kPa (gage) = 200 kpa = 2 bar T sat at 2 bar = 120.2 o C (Table A-3) ES 3001 Thermodynamics Exam- 1 (Sullivan, A- Term 2012) 4
Exam is Open Textbook, Open Class Notes, Computers can be used (Computer limited to class notes, lectures, homeworks, book material, calculator, conversion utilities, etc. No searching for similar problems at other sites.) Name: Email: To receive full credit all work must be clearly provided. Please use units in all answers. Problem 1: Consider a system similar to what we ve had in class. 1 2 4 3 A fluid (water) travels around the cycle in the schematic. Two state variables are given at each location in the cycle. Complete the table (all squares filled in). Note if a value is not applicable for a given state, just enter NA (for Not Applicable). Assume the following properties exist: P (bar) T (deg. C) v m 3 /kg h kj/kg x 1 100 95 100.0013897.001385 397.96 416.12 NA 2 100 600.03837 3625.3 NA 3.5 81.33 2.916 2415.4.90 4.5 81.33.00103 340.49 0.0
a) determine State 1 missing properties (show all work) for P1, v1: v1=vf(t) = therefore in Table A2 find where vf =.0010397 (which occurs at T = 95C) x= NA, h1=h f (95) = 397.96 T1 = 95C b) determine State 2 missing properties (show all work) for P2, T2: Superheated region (T2>Tsat) x = NA Directly read from table A-4 h2 = 3625.3 v2 =.03837 c) determine State 3 missing properties (show all work) for P3, x: Wet region at phase change Temp T3 = Tsat = 81.33C v3 = v f + x(v g v f ) =.00103 + 0.9 (3.24-.00103)=2.916 h3 = hf + x(hg hf) = 340.49 + x(2645.9-340.49)=2415.4 d) determine State 4 missing properties (show all work) for P4, v4: v4 = vf =.00103 x = 0 Pure Liquid, T4 = Tsat = 81.33C h4 = hf = 340.49 e) Determine the work per unit mass (kj/kg) W net = W turbine + W pump = (h2-h3) + (h4-h1) = (3625.3-2415.4) + (340.49-397.96) = 1,152.43 kj/kg = (h2-h3) + (h4-h1) = (3625.3-2415.4) + (340.49-416.12) = 1,134.27 kj/kg (Depending which exam you were given.) f) Determine the efficiency: Efficiency = Wnet/Qin = 1152.43/(h2-h1) = 35.71% Or if using h1 at 416.12; η = 35.34%
2.) What is the First Law of Thermodynamics? (10 pts) a) In English words Conservation of Energy b) As an equation including appropriate terms!"!" =!! +!!" +!!!" +!!!" (!!"# +!!!"# +!!!"# ) c) For each of the terms in part (b) provide their units. For example if (which it is not) Power was a term identified in (b), then part (c) would state that Power = Work per unit time or Force * distance / time.!!" +!!!" +!!!" =!"!!"!" =!"! ;! =!"! ;! =!"! ; ;!"#"#$"!!""!"#$%!"#$!"#$h!"#$%&'.
3.) Consider the figure below of a perfect gas situation. One kilogram of nitrogen fills the cylinder of a piston- cylinder assembly. There is no friction between the piston and the cylinder walls, and the surroundings are at 1 atm. The initial volume and pressure in the cylinder are 1 m 3 and 1 atm, respectively. Heat transfer to the nitrogen occurs until the volume is doubled. A) Determine the work for the process, in kj. W = Integral(Pdv), but P is constant = P(V2 V1) = [1.01325x10 5 N/m 2 ][1m 3 ] W = 1.01325 x 10 5 N m = 101.325 kj B) Determine the heat transfer for the process, in kj, assuming the specific heat (0.742 kj/(kgk) is constant. Molecular weight of N 2 gas is M N2 = 28.01kg/kmol R = R(universal constant)/m (Molecular weight) = 8.314/28.01 =.2968 kj/(kg K) PV = mrt (Know P 1, V 1, m 1, R) Solve for T 1 = 341.4K PV = mrt (Know P 2, V 2, m 2, R) Solve for T 2 (But P 2 =P 1, V 2 = 2V 1 ), therefore T 2 = 2*T 1 =682.8K Q = m(u 2 -u 1 ) + W du/dt = c v Or (u 2 -u 1 ) = c v (T 2 -T 1 ) =.742(341.4K)= 253.32 kj/kg Q = 1kg(253.32kJ/kg) + 1.01325x10 5 N/m 2 (2m 3-1m 3 )(1kJ/(10 3 N/m 2 )) = 354.6 kj