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1 Solution Key Last Name: First Name: Purdue ID: Please write your name in BLOCK letters. Otherwise Gradescope may not recognize your name. CIRCLE YOUR LECTURE BELOW: MWF 10:30 am MWF 3:30 pm TR 8:30 am Prof. Chen Prof. Meyer Prof. Jain EXAM # 1 ME 300 SP2018 INSTRUCTIONS 1. Please place all your electronics, including but not limited to cell phones, computers, watches, tablets, etc., into your bag and store your bag under your seat. 2. The only calculator that is permitted to be used during this exam is the TI-30X IIS basic scientific calculator. 3. This is a closed book and closed notes examination. You are provided with an equation sheet and all the necessary property tables. 4. Do not hesitate to ask the instructor if you do not understand a problem statement. 5. Start each problem on the same page as the problem statement. Write on only one side of the page. Materials on the back side of the page will not be graded. 6. Put only one problem on a page. Another problem on the same page will not be graded. 7. Identify your system(s), list relevant assumptions, and start with basic equations for Problem If you give multiple solutions, you will receive only a partial credit although one of the solutions might be correct. Cross out any solutions that you do not want to be graded. 9. For your own benefit, please write clearly and legibly. Maximum credit for each problem is indicated below. 10. After you have completed the exam, put your papers in order. This may mean that you have to remove the staple and then re-staple the exam. Do not turn in loose pages. 11. If you use additional blank pages, do not stable on your exam but use clip to hold it together. 12. Once time is called you will have three minutes to turn in your exam. Points will be subtracted for exams turned in after these three minutes. 13. Keep your eyes on your own paper. If you are caught cheating you will get a zero for the exam and your name will be turned over to the Dean of Students. Problem Possible score Total 100 1

2 Problem 1 (30 points). Each part is 3 points. Please fill only ONE circle for your answer to each problem. (1) What happens to the internal energy of a piston-cylinder device filled with ideal gas that undergoes an isothermal expansion process? Increase Decrease Remain the same Internal energy is only a function of temperature for an ideal gas. (2) What happens to the entropy of an ideal gas during an isothermal expansion process? Increase Decrease Remain the same ss = ss 2 (TT 2 ) ss 1 (TT 1 ) RR ln(pp 2 PP 1 ) = RR ln(pp 2 PP 1 ) > 0 (3) Is the entropy change for an irreversible process higher than that for a reversible process occurring between the same two thermodynamic equilibrium states? Entropy change depends only on the initial and final states. (4) Can exergy change in a process be negative? Exergy change can be positive or negative. (5) Is the exergy of system A (TA < T0, PA = P0) higher than that of system B (TB > T0, PB = P0)? Kinetic and potential energy effects are negligibly small for both systems. To determine the exergy of a system, a substance of the system is required additionally. (6) Is the exergy of a system different in different environments? By definition, exergy is a function of To and po (7) Is the second-law (exergetic) efficiency of a compressor higher than its isentropic efficiency? For isentropic process, both equal to one. Other conditions vary. 2

3 Problem 1 (Continued) (8) One kmol of N2 at 1 atm and 20 o C is mixed with 1 kmol of O2 at 1 atm and 20 o C. The mixture state is 1 atm and 20 o C. What happens to the specific entropy of N2? Increase Decrease Remain the same ss = ss 2 (TT 2 ) ss 1 (TT 1 ) RR ln(pp 2 PP 1 ) = RR ln PP 2,NN2 PP 1,,NN2 > 0 (9) A mixture of several gases contains an identical number of moles of each component. Will all the mass fractions be identical? For different gases, molar mass is different. (10) Can a thermal process undergo constant relative humidity? Along 100% relative humidity line under cooling and dehumidification. 3

4 Problem 2 (25 points) A tank with a volume of 1.5 m 3 contains moist air at 25 o C, 1 bar, and humidity ratio of 0.01 kgv/kga. (a) Label the state and the corresponding partial vapor pressure, pv, saturated partial vapor pressure, pg, dry bulb temperature Tdb, and dew point temperature Tdp on the T-v diagram. (b) Calculate the relative humidity, in %. (c) Find the dew point temperature, in o C. (d) Determine the mass of water vapor contained in the tank, in kg. The molar mass for air is kg/kmol and for water is kg/kmol. Do not interpolate but use the closest table values. No need to write assumptions and basic equations for this problem. Solution: (a) Pg Pv Tdb Tdp (b) Using the given information, ωω = 0.01 kkkk wwwwwwwwww kkkk aaaaaa, TT dddd = 25, aaaaaa PP = 1 bbbbbb = 100 kkkkkk ωω = PP vv ωω PP PP PP vv = vv ωω PP = kkkkkk = kkkkkk From Table A-2, PP gg = bbbbbb = φφ = PP vv kkkkkk = = oooo 49.9% PP gg kkkkkk 4

5 Problem 2 (Continued) (c) Saturation temperature corresponding to PP vv is 14 without interpolation. Thus, TT dddd = TT vv = 14 (d) Water vapor in moist air is treated as an ideal gas, then, we can apply ideal gas law. mm vv = PP vvvv RR wwwwwwwwww TT = kkkkkk 1.5mm 3 ( 8.314kkkk/kkkkkkkkkk 18.02kkkk/kkkkkkkk ) ( )KK = kkkk 5

6 Problem 3 (45 points) A cooker with a volume of 0.5 L is boiling on a stove with a stove-top temperature of 800 o C. The pressure in the cooker is 100 kpa. The environment temperature is 20 o C. The heat loss from the cooker to the environment through the cooker wall and lid is 100 W. Initially one-half of the volume of the cooker is filled with liquid and the other half with vapor. Neglect kinetic energy and potential energy. Assume that the temperature and pressure are uniform throughout the cooker and heat is added at a fixed rate from the stove. If the cooker runs out of liquid water after 1000 seconds, determine: (a) Draw the system sketch and the control volume you use. (b) The total mass of the steam entering the environment, in kg. (c) The rate of heat transfer from the source (stove), in kw. (d) The total entropy production in the process, in kj/k. (e) The total exergy destroyed in the process, in kj. Assumptions: System sketch: - Quasi-equilibrium - Steady flow - Uniform state throughout the cooker QQ loss=100j/s - No work - Heat addition at a fixed rate from the stove CV for (c) and (d) Basic equations: tt = 1000ss CV for (b) V=0.5 L TT 0 = 20 ddmm cccc dddd = mm ii mm ee mm cccc = mm ee TT ssssssssss = 800 Qstove ddee cccc dddd = QQ cccc WW cccc + mm ii h ii + vv ii 2 ii + ggzz 2 ii mm ee(h ee + vv ee 2 ee + ggzz 2 ee) UU cccc = QQ cccc mm ee h ee ddss cccc dddd = QQ cccc TT bb + mm iiss ii mm eess ee + σσ CCCC SS cccc = QQ cccc TT bb mm ee ss ee + σσ CCCC Solution: (a) State 1: 0.25 L of saturated water 1bar L of saturated water 1bar From Table A-3, vv ff = mm 3 /kkkk uu ff = kkkk/kkkk h ff = kkkk/kkkk ss ff = kkkk/kkkkkk 6 State 2: 0.5 L of saturated water 1bar vv gg = mm 3 /kkkk uu gg = kkkk/kkkk h gg = kkkk/kkkk ss gg = kkkk/kkkkkk

7 Problem 3 (Continued) Using the mass balance equation, mm cccc = mm ee mm ee = mm 1 mm 2 mm 1 = VV 2 + VV 2 = vv ff vv gg mm mm3 kkkk = kkkk = kkkk mm 2 = VV mm3 = = kkkk vv ff mm3 kkkk mm mm3 kkkk Therefore, mm ee = mm 1 mm 2 = kkkk = kkkk (b) Using the energy balance equation, UU cccc = QQ cccc mm ee h ee UU 2 UU 1 = QQ ssssssssss QQ llllllll mm ee h ee UU 1 = mm 1ff uu ff + mm 1gg uu gg = kkkk kkkk/kkkk kkkk kkkk/kkkk = kkkk UU 2 = mm 2 uu gg = kkkk kkkk/kkkk = kkkk QQ llllllll = QQ llllllll tt = 100JJ 1000ssssss 0.001kkkk/JJ = 100kkkk Therefore, QQ ssssssssss = UU 2 UU 1 + QQ llllllll + mm ee h ee = kkkk = kkkk QQ ssssssssss = QQ ssssssssss = kkkk = 0.641kkkk tt 1000ss 7

8 Problem 3 (Continued) (c) Using the entropy balance equation, SS tttttttttt = QQ jj TT bb,jj mm ee ss ee + σσ tttttttttt SS 2 SS 1 = QQ ssssssssss QQ llllllll mm TT ssssssssss TT ee ss ee + σσ tttttttttt 0 SS 1 = mm 1ff ss ff + mm 1gg ss gg = kkkk kkkk/kkkkkk kkkk kkkk/kkkkkk = kkkk/kk SS 2 = mm 2 ss gg = kkkk kkkk/kkkkkk = kkkk/kk QQ llllllll = 100kkkk Therefore, σσ tttttttttt = SS 2 SS 1 QQ ssssssssss + QQ llllllll + mm TT ssssssssss TT ee ss ee 0 = = kkkk/kk (d) EE dd,tttttttttt = TT 0 σσ tttttttttt = ( )KK kkkk/kk = kkkk 8

9 Problem 3 (Continued) 9

10 Problem 3 (Continued) 10