ME 200 Exam 2 October 16, :30 p.m. to 7:30 p.m.

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1 CIRCLE YOUR LECTURE BELOW: First Name Solution Last Name 7:30 am 8:30 am 10:30 am 11:30 am Joglekar Bae Gore Abraham 1:30 pm 3:30 pm 4:30 pm Naik Naik Cheung ME 200 Exam 2 October 16, :30 p.m. to 7:30 p.m. INSTRUCTIONS 1. This is a closed book and closed notes examination. You are provided with an equation sheet and all the needed property tables. 2. Do not hesitate to ask the instructor if you do not understand a problem statement. 3. 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. There are blank pages following problem 2 for your work. 4. Put only one problem on a page. Another problem on the same page will not be graded. 5. Show your system and list relevant assumptions and basic equations for problem If you give multiple solutions, you will receive only a partial credit although one of the solutions might be correct. Delete the solution you do not want graded. 7. For your own benefit, please write clearly and legibly. Maximum credit for each problem is indicated below. 8. After you have completed the exam, at your seat put your papers in order. This may mean that you have to remove the staple and re-staple. Do not turn in loose pages. 9. 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. Problem Possible Score Total 100 1

2 Problem 1 (45 points) Answer the following questions. (a) Consider an unknown ideal gas having specific internal energy of 150 / at a temperature of 27C. The molecular weight of the ideal gas is /kmol. Calculate the specific enthalpy (/) of the ideal gas at 27C. Show work with all necessary equation(s). (12 points) Specific enthalpy: hu Pv For ideal gas: Pv RT h u RT ; +4 h K -K h R kmol-k R 1 M K kmol (b) Liquid water is compressed from 1 bar and 20C (State 1) to 10 bar (State 2) with negligible change in temperature during the process. Calculate the change in specific enthalpy (/) of liquid water. Show work with all necessary equation(s). (10 points) Change in specific enthalpy for liquid: huvp ct vphv P +4 v v T Using compressed liquid approximation: Table A-2: v m 3 3 m h kpa f h

3 Problem 1 (continued) (c) Air has a critical temperature of 133 K and critical pressure of 38 bar. A certain chemical process uses air at 150 K and 76 bar. Can air be considered ideal gas at 150 K and 76 bar? Justify your answer. (8 points) P 76 bar Reduced pressure: PR 2 Pcritical 38 bar T 150 K Reduced temperature: TR 1.12 T 133 K critical Figure A-2: Z = 0.45 Z 1 air cannot be considered ideal gas (d) Consider adiabatic compression of 2 of air in a piston-cylinder assembly from 1 bar and 330 K (State 1) to 14 bar and 700 K (State 2). Air can be considered an ideal gas with variable specific heats at these conditions. Calculate the work () during the compression process. Show work with all necessary equation(s) and state any assumptions. (15 points) No inlets/no exits Considering energy balance: +4 de dt 2 2 CV i i Q CV W CV m i hi gzi me hi gzi i 2 e 2 Adiabatic Integrating: E U KEPE QW U mu u W W mu u Ignore KE and PE changes (for assumptions) Note: Full credit even if started Table A-22: u and u directly with Q W = E W W ; negative sign indicates work done on the system by the surroundings +3 3

4 Problem 2 (55 points) A refrigeration cycle using R-22 as the working fluid is shown below. Refrigerant 22 enters the compressor at 4 bar as saturated vapor (State 1) and it is compressed to 12 bar and 50C (State 2). After condensation R-22 leaves as saturated liquid at 12 bar (State 3) and it is throttled to the evaporator pressure (State 4) in the expansion valve. Evaporator removes 50 kw of heat from the space to be cooled. CV I CV III +3 (for control volumes) CV II (a) Complete the following table with nine missing entries. Clearly show work for determination of phase (saturated liquid, saturated vapor, saturated liquid vapor mixture (SLVM), compressed/sub-cooled liquid (CL), superheated vapor (SHV)) for states 2 and 4. Show any necessary calculations with basic equations for missing property values. Pressure Temperature Specific Enthalpy Phase State (bar) ( C) (/) 1 4 Sat. Vapor Sat. Liquid 4 4 (b) Considering mass and energy interactions of the evaporator, calculate the mass flow rate (/s) of R-22. (c) Find the coefficient of performance (COP) of the refrigeration cycle. Identify your system(s), list assumptions, and start with basic equations. 4

5 Problem 2 (continued) Assumptions - Steady state - One-dimensional (uniform) flow +5 - Ignore changes in KE and PE - Expansion (throttling) valve and compressor: No heat transfer Q CV 0 - Expansion (throttling) vale and evaporator: No work W 0 CV Basic Equation(s) dmcv mi me m1 m2 m3 m4 m +1 R22 dt i e 2 2 de CV i i Q CV W CV mi hi gzi me hi gzi dt i 2 e 2 Solution (a) State 1: P 1 = 4 bar, saturated vapor Table A-8: Since saturated vapor, T 1 = T sat (P 2 ) T C h 1 = h g (P 1 ) h State 2: P 2 = 12 bar and T 2 = 50C (phase explanation) Table A-8: T sat (P 2 ) = 30.25C T 2 > T sat (P 2 ) superheated vapor (SHV) +3 Table A-9: h State 3: P 3 = 12 bar, saturated liquid Table A-8: Since saturated liquid, T 3 = T sat (P 3 ) T C h 3 = h f (P 3 ) h State 4: Considering energy balance for CV I (expansion or throttling valve): h 3 = h 4 P 4 = 4 bar and h 4 = 81.9 / Table A-8: h f (P 4 ) < h 4 < h g (P 4 ) SLVM +3 (phase explanation) Since SLVM, T 4 = T sat (P 4 ) T C 5 +5

6 Problem 2 (continued) (b) Considering energy balance for CV II (evaporator): Q Q m h h Mass flow rate of R-22: Q 50 s h1 h4 CV in R in R 22 mr 22 m s +5 Q Q in evaporator (c) Coefficient of performance of the refrigeration cycle: COPRF +4 W in W compressor Considering energy balance for CV III (compressor): W CV W in m R22 h1 h2 +5 W compressor kw ; negative sign indicates work input s 50 kw COPRF COPRF 5.8 Note: 8.66 kw Compressor power need not be calculated for full credit 6

ME 200 Exam 2 October 22, :30 p.m. to 7:30 p.m.

ME 200 Exam 2 October 22, :30 p.m. to 7:30 p.m. CIRCLE YOUR LECTURE BELOW: First Name Solution Last Name 7:0 a.m. 8:0 a.m. 10:0 a.m. 11:0 a.m. Boregowda Boregowda Braun Bae :0 p.m. :0 p.m. 4:0 p.m. Meyer Naik Hess ME 00 Exam October, 015 6:0 p.m. to