Solution_Prelim I - If a ‘3 e 1 figugimxttvtk i“ leflfiwkgk WWW CornellUniversity College of Engineering EN GRD 2210 Thermodynamics

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Unformatted text preview: If a ‘3 e 1., figugimxttvtk i“ leflfiwkgk { WWW ; CornellUniversity College of Engineering EN GRD 2210: Thermodynamics Instructors: Stephen Pope, Max Zhang 1"t Prelim Thursday October 15th. 2009 — 7 :30—9z30pm. You may use a calculator and one 8.5” by 11” sheet of paper. Answer all parts of all questions. In order to obtain the maximum credit for your solutions, show all steps in your computations. The point value for each problem is given to help you allocate your time. Make sure to include units with your answers. Important: in scoring this exam, we will be looking for your approach to solving each part of the questions. If you get stuck on one part, move on and set up the equations for the other parts to get partial credit. Read each problem thoroughly and make sure you understand what is being asked of you. Make sure to attempt each problem and show What you would have done if you are unable to fully answer the problem or if you run out of time. Exam: 4 problems Given: R = 8.314 kJ/(kmol K) 0°C : 273.15 K wwwwel 1. Air is compressed at steady state from a pressure of 136 kPa and a temperature of 305 K with a volumetric flow rate of 37 m3 min'l, to a pressure of 680 kPa and a temperature of 400 K. The power required to drive the compressor is 155 kW. Heat transfer occurs to cooling water circulating in a water jacket enclosing the compressor, which enters at a temperature of TA=3 00 K. The mass flow rate of cooling water is 82 kg min“. The kinetic and potential energy changes of the air and cooling water are negligible. a. What is the mass flow rate of air in kg s'l? b. Calculate TB 9 W :1 l ‘ Cooling water, TAi“? WW SK» Air, ch p2=680 kPa, “\‘Tzz400 K w I t l W n Chg?) 21': teem? ‘ g i D f/k 3,3853;ng , 1 £ l W ,V. M a - v H“ W -‘J Air, 1312136 kPa’ Cooling water, T13 T3305 K V‘ -.: wigs”, b "1 W“ 33:: (“MR-i Q'W‘AW We 1' M421 «1:; m A p “"* (“Q m We; hm?) ‘ q p x” “mm.” «mm MVWNW.WN‘LEMW_WMWW“ ‘2‘ “WW/\ W,» G‘ 5‘ 1m ,3 it) t“: 5:; E 2% r W” A” a} ( to»; (A: ,w x « [5 MK} w»; W am; W W“ “‘“fm'g‘ Tu www-mmm.,.,, w, , “A” ,. ‘WWW,,,....,.‘.WWW "(Ml “A M W?» y 6 fi"'" m u ‘ ., a W W. “gm WWW)WWm“,WM.WwW.dw, W3 :2th . m “YMVNQXQWWLW (i W“ {mmmk‘fm-v \ ij-N’xlflw‘j, Win‘ (9; 2. The sketch shows part of a refrigeration device that is used to maintain a cooling box at a low temperature. The device and the cooling box are in a steady state. Saturated liquid, Refrigerant 22, enters at “1” at a temperature of 40°C and a flow rate of 0.02kg/s. The stream passes through a throttle and enters the chamber, which is maintained at a pressure of 0.8 bar. The refrigerant exits the chamber in two separate streams: a liquid stream “2” and a vapor stream “3”. (The temperature and pressure of the two exiting streams are the same.) a. What is the pressure p1 of the inflowing stream? At a particular operating condition, the heat transfer rate to the chamber is 1.5 kW. Determine: b. The temperature of the exiting streams. c. The mass flow rate m2 of the exiting liquid. d. Consider now different conditions to those in (b) and (0): what is the maximum possible cooling rate that can be achieved if the cooling box is to be maintained at —20°C? heat transfer CooHn box from surroundings g qmzfiglwflyg Q’Mfi Q heat transfer (\\ 1 Refrigeration 24%} , ., f» ., , chamber mp ” M2 Pa 1”; ‘ «a? {fmw ’fi’m J ‘5}: A F M :2: a)?" “*1“§ww Clix ; $‘\ A 2 I: M QCMQM “i” Q ghg‘fiflm- .ar M $1; Wig %" “:3 W i} y 3- V" {NRC} {A Q3 A , ’kstf‘ K ( ,, M , ,R a» (0&0 :2» 2%; '(Ctaéng3-~2s:a<:m;m 4;; iv x»; m w MTWWWWWMWNWmmnmwwmwmw‘Wmm.” « M 2 ‘3‘ ‘11" v J: m 1;“ we, m M5 mMW fim’r mm baa, @dffi‘fig :41 rigiwm I ‘. 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For the process 1-2 (going from State 1 to State 2) identify and quantify all of the energy interactions between the indicated system and the environment. 8 .i in Hw— 1:2; 0 L \ w” /' System ' . im‘kegug H "“ nc‘flr {A} . 5 boundary etc: ‘ ’ Electric ; a current ‘2 3» $0 with “(3 w a" m: 4 “M H WQM k ,9” T» “T: m (gawm‘ék G W “p W “W Wk nmmfwwwmmmwm x: 8 -? my, T $5? 0 R 7a "Mywxmwmwwh 5 g 9 Fm "'EébEy-Q, i, m3 «a e E K " $33 xx ::; :; OB f) 0 {in Pa 412* x; 2V3. Emit “3:” , Wu-uw_,,..w\mw«wwm: 5 x "‘ ‘ me firmlaai4锣iw “EVAN ("mgwem Cgasagfi‘wwg ‘ é 3: 21., m (311% “A t a w...“ .I , n a , n ‘ _ an , 5‘57 RDA. 1.5:: “Eh w («Lgmmmmg “23%; 0602403 "‘ in g ‘ , “'3‘ 933W fiawwfimwg meiizeaggiywmiimcx‘ P» 50ka {32". 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The piston is free to move so that the pressure is the same on both sides. The piston does NOT conduct heat. The water is slowly heated to a final state, where the temperature is Tw2=200 OC. During the process, air is compressed with pvn=constant for n = 1.4. a. Determine the total volume of the container in m3 b. Calculate the amount of work done on the air in kJ 0. Calculate the amount of heat transfer from the external source to the water in kJ /‘ may {Gian $23? LEM \ Ti,“ m (I: ll E331; Mass ~—~ * W “ i Um; QM”; Lees?) rm zxniagmj J imafiiive 01,") F9" wafi‘gd gr €‘mmz, VJQ‘ETEJ‘J“ 3"?» flailfimmé‘au a :: ioméagrflflfl m“ (95535055 m3 ‘M/I'Nwa/A-«w VVVV “Mn-NW“)! Viv-w {33?} w w ,. wwmmma WW 3% ‘ ‘ \fm 3“ mawm m 7 Nu ,._ a :3»; z n W ( 5:2. [:25 ’ gq *w; g3: “‘1mva‘mAKunmww.&umvnwmmmmmmw1M mm ( ‘“ K M _ ‘3 fa (9)64:ng M3), "aggflwf $3:va hqggfimakgfia «up «a g kwwégm z , Q Q. um WW :« in“ \ TM a“ y \ L TH; RAM 1" “xii/mt; w R? 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Solution_Prelim I - If a ‘3 e 1 figugimxttvtk i“ leflfiwkgk WWW CornellUniversity College of Engineering EN GRD 2210 Thermodynamics

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