CIRCLE YOUR LECURE BELOW: Div. 1 10:30 am Div. :30 m Div. 3 4:30 m Prof. Gore Prof. Udua Prof. Chen EXAM # 3 INSRUCIONS 1. his is a closed book examination. All needed roerty tables are rovided.. Do not hesitate to ask the instructor if you do not understand a roblem statement. 3. Start each roblem on the same age as the roblem statement. Write on only one side of the age. Materials on the back side of the age ill not be graded. here are blank ages rovided for your ork. 4. Put only one roblem on a age. A second roblem on the same age ill not be graded. 5. Label your system and list relevant assumtions for roblems and 3. 6. If you give multile solutions, you ill receive only a artial credit although one of the solutions is correct. Delete the solutions you do not ant. 7. For your on benefit, lease rite clearly and legibly. Maximum credit for each roblem is indicated belo. 8. After you have comleted the exam, at your seat, ut your aers in order. his may mean that you have to remove the stale and re-stale. Do not turn in loose ages. 9. Once time is called you ill have three minutes to turn in your exam. Points ill be subtracted for exams turned in after these three minutes. 10. Use calculators authorized by the School of Mechanical Engineering. Problem Possible Score 1 40 30 3 30 otal 100 1
Problem 1. (40 oints) Anser the folloing short-anser questions. You must sho ork and/or rovide an exlanation to receive credit. a) Consider to vaor-comression refrigeration cycles. he refrigerant enters the throttling valve as a saturated liquid at 30 o C in one cycle and as a subcooled liquid at 8 o C in the other. Both cycles oerate ithin the same ressure range. Which cycle ill have a higher COP? Use a -s and/or P-h diagram to exlain. You can assume that liquid enthaly only deends on temerature. 3b 3a 3b 3a 4b 4a 1 4b 4a 1 s h Cycle 1--3b-4b (ith subcooled liquid entering the throttle ill have a higher COP than the cycle 1--3a-4a because for identical amount of ork inut, the former has a higher heat removed from the refrigerated material (hich is at tem. designated by the red dashed line). b) A heat um oerates ith a condenser temerature of 7 o C and delivers heat to a house maintained at 17 o C. he evaorator of this heat um oerates at an average temerature of 3 o C and the environment is at 13 o C. Find the maximum COP and the maximum second la efficiency of the heat um. COPmax-HP = H/(H-L) = (73+7)/(7-3) = 1.5 COPmax-revHP = H-house/(H-house-Lenv) = (73+17)/(17-13) = 7.5 Second La Efficiency = COPmax-HP/COPmax-rev= (1.5/7.5) (100)= 17.4%
c) Considering the definition of molar enthaly of secies k defined in the standard equation o sheet as: h () h (98) h () h (98), state if the folloing statements are k f,k s,k s,k RUE/FALSE? Exlain. N A) he term n k h s,k() h s,k(98) of an exhaust gas mixture leaving a combustor k 1 deends on the number of moles of each secies and the temerature of the mixture. B) he term N k 1 nh (98) of an exhaust gas mixture deends on the on the number k o f,k ofmoles of each secies and the temerature of the mixture. A) RUE because sensible energy does deend on the number of mols and temerature of each secies and the temerature is equal for all secies and is equal to the mixture temerature. B) FALSE because the heat of formation does not deend on the temerature of the mixture but only on the reference temerature. d) What is the difference beteen the Engineering Equations Solver (EES) statement (A) and EES statement (B) belo? (A) hb enthaly( H O, 1 ) (B) hd enthaly(ater, 1, P P1) (A) Gives the enthaly of HO considered as an ideal gas at temerature of 1. (B) Gives the enthaly of liquid ater at a temerature of 1 and a ressure of P1. e) A mixture of O N NOin equilibrium at a certain temerature and ressure is isothermally comressed to tice the ressure. Considering only the single equilibrium reaction, O N NOill the number of moles of each individual secies change if the ressure is increased? Exlain. he number of moles of each individual secies ill not change since the temerature is constant the quantity PNOPNO/POPN = nno nno (N) (N)/nO nn (N) (N) = nno nno / no nn is constant since N the total moles cancel out from the numerator and the denominator. 3
Problem. (30 oints) A Rankine vaor oer cycle oerates ith ater as the orking fluid. Saturated liquid leaves the condenser at 60 o C, and saturated vaor enters the turbine at 00 o C. he turbine oerates in an isentroic manner, hile the um has an isentroic efficiency of 0.8. Assume the corresonding Carnot cycle oerates beteen 60 o C and 00 o C. (Given: the secific enthaly at the turbine exit is 114.3 kj/kg) Find: (1) State your assumtions and dra the cycle on a -s diagram. () Determine the thermal efficiency. (3) Determine the nd la efficiency. (1) Assumtions: Steady-state steady flo No change in kinetic energy and otential energy -s diagram: 1s 1 4 3 s () Using thermodynamic data (Provided on cover hoto) and given definition of states: o 00 C 1 1s 15.54bar, h 793.kJ/kg o 3 4 3 60 C 4 3 0.1994bar, h4 51.13kJ/kg, v 4=0.001017m /kg Pum (4 to 1): 3 3 m 100kJ/m,s v 4(1s 4) 0.001017 (15.54 0.1994)bar 1.56kJ / kg kg 1bar,s 1.56kJ / kg 1.95kJ / kg 0.8 h1 h4 h1 h 4 (1.95 51.13) 53.08kJ / kg 4
Boiler (1 to ): q h h (793. 53.08)kJ / kg 540.1kJ / kg in 1 urbine ( to 3): From given in roblem statement: h3 114.3kJ / kg h h 3 (793. 114.3)kJ/kg=678.88kJ/kg hermal efficiency: (678.88 1.95)kJ/kg th 6.6% q 540.1kJ / kg in (3) Carnot cycle efficiency: L (60 73)K C 1 1 9.6% (00 73)K H nd la efficiency: th 6.6% 89.9% 9.6% C 5
Problem 3. (30 oints) he combustion roducts from burning one mole of entane, ith ure oxygen in a stoichiometric ratio exit at 400 K, 100 kpa. (a) Determine the coefficients a, b and c in the stoichiometric reaction belo Next, consider the dissociation of only. (b) What is the equilibrium constant for the reaction shon belo at 400 K 1 (c) It is observed that 0.4 moles of are resent in the exhaust, of an actual combustor, for every mole of entane that as burnt. Write a balanced equation for the actual reaction. Ho many moles of and are resent in the exhaust? (d) If the exhaust in art (c) continues to react, hich direction is the dissociation of reaction (described in art (b) likely to roceed, forard or backard, to attain equilibrium? Solution: a. 8 5 6 b. Referring to table on cover age, the equilibrium constant is, log 1.679 0.01 c. 8 6 0.4 Balancing C: 5 0.4 4.6 Balancing O: 16 0.4 6 0. herefore, number of moles of CO is 4.6 moles and number of moles of O is 0... d. 0.0357... 0.0179 0.41. / 0.0116 Since this is less than K calculated in art b, the reaction ill move forard to roduce more CO. 6