ME 300 Thermodynamics II Spring 2015 Exam 3. Son Jain Lucht 8:30AM 11:30AM 2:30PM

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1 NAME: PUID#: ME 300 Thermodynamics II Spring 05 Exam 3 Circle your section (-5 points for not circling correct section): Son Jain Lucht 8:30AM :30AM :30PM Instructions: This is a closed book/note exam. You may use the equation sheet available from the class website. You may use an approved calculator (see syllabus). You must start from the basic form of the governing equations and simplify accordingly where necessary to solve each problem. Hence, you must show all your work for full credit. 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. Question Total Score Total (out of 00)

2 . True / False [ pt each NO PARTIAL CREDIT] a) T The back work ratio for a Rankine cycle is lower than that for a Brayton cycle. b) F Lowering the average temperature of heat addition increases efficiency in a Rankine cycle. c) T The throttling process in heat pumps can be modeled as isenthalpic. d) F The Clausius-Clapeyron equation applies to phase change between any two phases. e) F Different refrigerants cannot be used in a cascade refrigeration cycle. f) F In turbojets, propulsive thrust force is proportional to the product of the inlet and exit air velocities. g) T In a Brayton refrigeration cycle, the expansion process is achieved using a turbine. h) T The use of heat pumps is limited in very cold environments because of the need to expand the refrigerant to a colder temperature than the ambient environment. i) F The maximum theoretical COP of a vapor compression system is infinite. j) T Intercooling in a Brayton cycle requires a multi-stage compressor.

3 k) (0 points) You are experimenting with ammonia and need to know the saturated pressure at 30 C, but you only have data for hfg, vfg, and psat at 4 C. Using either the Clapeyron or the Clausius Clapeyron equation, find an estimate for psat at 30 C using only hfg, vfg, and psat at 4 C from the tables provided. (Note: The molecular weight of ammonia is 7.04 /kmol.) Table A-3: hfg = 69.6 / vf =.6547e-3 vg = 0.30 psat= bar Since vg >> vf and assuming that ammonia in its gaseous phase can be treated as an ideal gas, we can use the Clausius Clapeyron equation to find psat at 30 C: p h g hf ln = p R T T sat p 69.6 ln = p 8.34 = sat p p sat p, sat 7.04 =.733 =.49 bar sat Alternatively the Clapeyron equation can be used: dp dt hg = T v ( v ) sat g f dp = dt T h h g f h ( vg v f ) f Assuming that the ratio of hfg to vfg doesn t change much with temperature, we can integrate and arrive at the following expression: hg h ( ) f T p p = ln sat v v T p, sat ( g f ) 69.6 = = 79.4 kpa 3 = 5.4 kpa =.5 bar 303 ln 97

4 l) (0 points) Develop an expression for the specific entropy change, s( v, T ) s( v, T ), using the Van der Waals equation of state. v p s ( v, T ) s ( v, T ) = v dv T v for Van der Waals equation, p T v R =. v b ' Therefore R s d b b v ( ') ( ') v b ' v s = v = R ln v b ' = R ln v ln v = R ln. v v v b ' v b ' An alternative solution would be as follows: v p s ( v, T ) s ( v, T ) = v dv T v for a different form of the Van der Waals equation, p T v R =. v b Since this partial derivative is in terms of v, we need to modify the integrand so that it is in terms o f v : R R s s = dv = ln ( Mv b) v Mv b M R s s = ln ( Mv b) ln ( Mv b) M Mv b v b s s = R ln = R ln. Mv b v b v v v

5 Problem # [35 points] Given: An ideal gas turbine cycle with an overall pressure ratio of 64 uses a two- stage compressor with an inter-cooler and a two-stage turbine with a reheat and regeneration. The inlet temperature for both stages of the compressor is 300K whereas that for the turbine stages is 000K. Neglect kinetic energy, stray heat losses, pressure drops across the intercooler, combustors and transition ducts. Using the ideal gas assumption, complete the table and calculate ideal cycle thermal efficiency of the plant operating at sea level. You don t need to interpolate properties in the ideal air table. Hint: T = T4 and T6 = T8. Also, note that for a two-stage compression and expansion the work is maximized when both stages of the compressor and the turbine have the same pressure ratio. PUT YOUR ANSWERS IN BOXES PROVIDED. Assumptions: +3 Neglect kinetic energy, stray heat losses, pressure drops across the intercooler, combustors and transition ducts. Assume air standard assumptions (ideal gas, SSSF) DO NOT ASSUME constant heat capacities. Basic equations: +4 pr ( T ) = ; η w net p p p T r Solution: q in Note that h = h 3,h = h 4,h 6 = h 8, h 5 = h 7 Energy conservation reduces to the following for the compressors and turbines: w c,in = ( h h ) = ( h 4 h 3 ) w = h h = h h t

6 State P T h pr atm k KJ/ For Table: +3 (- for each wrong entry, minimum is zero, max is +3) To get h and h6: p = p = p T r p p r ( T ) p T p r p r T p p 6 r 6 ( T6 ) p6 = pr T = pr T p p T p r 7 7 7

7 For a two-stage compression and expansion the work is maximized when both stages of the compressor and the turbine have the same pressure ratio. Optimum pressure ratio is: Pressure Ratio = Total Pressure Ratio = 64 = 8 = ( ) = w c,in = h h w = h h = = 90. t 6 7 w net = w t w c = = ( 6 5 ) ( 8 7 ) = 90. q = h h = h h +5 in q in = η w net = 43.9 = 0.74 = 74% q in 90. η = 74% +

8 Problem #3 (35 points) Given: A vapor compression refrigeration cycle operates with propane as the refrigerant. The system is used to keep a refrigerated space at -0 C, and the heat input rate to the evaporator is 4000 kw. The measured state data for the cycle are indicated in the diagram. Find: (a) Indicate the assumptions for your analysis. (b) Determine the pressure at the compressor exit. (c) Calculate the mass flow rate (/s) of propane for the cycle. (d) Calculate the COP for the cycle. (e) Calculate the isentropic efficiency η C of the compressor. In the propane SLVM and SHV tables (A-7, A-8), pick the closest values, do not interpolate. Assumptions: + 3 () Steady state () Uniform flow (3) Adiabatic compressors, throttling valve (4) No pressure drop across condenser, evaporator

9 Basic Equations: Qɺ = Qɺ = mɺ h h in C evap in c 4 + h Wɺ = mɺ h h + = Qɺ COP = β = Wɺ s ηc + h h + h Solution: Find properties at all states. State p =.0 bar, T = 0 C SHV h = 464.9, s = K State 4 p =.0 bar, x = 0.46 SLVM 4 4 h = x h + x h = = 0.7 T f 4 4g 4 = 5.43 C + 4 State 3 x = 0.0, h = h = p = p = 6.0 bars, T = C 3 3 p3 = p = 6.0 bars + 4 State p = 6.0 bar, T = 70 C h = SHV + 3 State s ps = p = 5(.0) = 6.0 bar, ss = s =.877 K SHV T s = 70 C h s =

10 Solution: η ( h h ) s c = = = h h η c = COP, mass flow rate, compressor work 4000 Qɺ in Qɺ in = Qɺ evap = mɺ ( h h4 ) = 4000 mɺ = = s s ( h h4 ) mɺ = 6.4 s COP β Qɺ mɺ ( h h ) ( h h ) Wɺ c mɺ ( h h ) ( h h ) in 4 4 = = = = = COP =

ME 200 Final Exam December 12, :00 a.m. to 10:00 a.m.

CIRCLE YOUR LECTURE BELOW: First Name Last Name 7:30 a.m. 8:30 a.m. 10:30 a.m. 1:30 p.m. 3:30 p.m. Mongia Abraham Sojka Bae Naik ME 200 Final Exam December 12, 2011 8:00 a.m. to 10:00 a.m. INSTRUCTIONS