prelim_pract3_solns - EngrD 219 Fall 2004 Prelim 3 Total...

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Unformatted text preview: EngrD 219 - Fall 2004 Prelim 3 November 17, 2004 Total Points: 100 Time: 8:35 - 9:55 (80 minutes) Name: 50pm, fl 0 M 5 (Please initial every sheet) Please write your solutions in the allocated spaces below the questions. You can use the back of the sheets for scratch work (it will not be graded). The last two sheets of the exam are blank. Write on these two sheets if you need extra room, These two sheets will be graded. If you take apart the pages, staple them when you turn in your exam, Partial credit will be given, but you must show your work to get any credit, and you must box your answer. To Leggy; appropriate credit, he sure to state where you 3;; data that you use in lotions e. . "this er is from tab] .X on a " Good Lucid! 1. The first step in making ice cream is to pasteurize (heat) the ingredients. A liquid stream of in kg/s of ingredients (milk and cream) at 10°C enters a pasteurization unit. The ingredients are combined with saturated steam at 110°C that enters the unit at in: kg/s. Pasteurized product exits the unit at 70°C at a rate of 2 kg/s. a) Draw a flow chart of the process. ”32 Eat mauve (“PM I31. b) Solve for iii} and ring. Assume the ingredients andproducr streams have the some properties as pure water and ihat the pasteurization wait is perfeciiy insulated State at! other assumpiions. ND k-E‘. {Hi-155 flower: n31, “in = 2_ 15/; m Penna. NB SHAFTWIU‘C . 4 ' A mm was .- wuH. + WI? p11 ‘ 23:75 “Imam-3 Show; high: I at 41- 0 L745 {3-5) [4‘1 s urn; M45 (3:6,mflrpoidrl) A ”P: 293 ”/9 (6-3“) 42.0%” mm»); = 20-1?) 91.0.?” + unfit-mi.) = 3‘“: r 15%.; .j‘, = - Wins “‘4’ = [2' H JNG’MHW) lit/I1, = 0J9 Sid“ gain—W S I 2. The output stream from roblem 2 is new combined with air, whipped and chilled to make delicious ice cream at C. Half of the water in the final product is ice and the other half is liquid. (The freezing point is below that of pure water due to the other ingredients.) To produce 2 kg/s of ice cream, 30 kw of power must be supplied to the whipping unit. Air 2 kgz‘s pasteurized Chiller and whipping ingredients unit 70°C 1 atm Heat removed 30 kW to by chiller whipping unit Do an energy balance on this process to find the amount of heat removed by the chiller. As in problem 2, treat the feed and output streams to have the properties of pure water. Neglect the mass and enthalpy of the injected air. State all other approximations you make. ND than? [n LIE-er.bfi, up = to — ws £1!) (2%l(l:ouwur) ' l1%)lfi,mf) : b'“ 30 to“ ? Z?- ?le ‘(1’53K like ) 1” (1%)Cfi Tm) we ar 0's. Ar 0°C. = (lgfl—uoorrflfl—‘i‘fl (”2?) + (Lfiflo) A Humor -‘- ‘2‘i3 kg:- (retail) 51¢;th info Hf) — 333 :1 -(2 elm-e“) = as {—20 5:5) “333 ‘§8fa : Q+30 3. A liquid mixture consisting of 60 mol% pentane (P) and 40 mol% hexane (H) at 75°C and 40 atm is fed to an evaporator that operales at pressure P0. The vapor and liquid streams that exit the evaporator are at 60°C and PD atm. The liquid stream is equimolar in pentane and hexane, as shown in the flow chart below. fr, molimin 60°C Po atm y mo] P(v)r'm0[ (l-y) mol H(v)lmol 100 molimin 75°C 4.0 aim 0.60 mol P(l)fmol 0.40 mol H(l).-’mol ti Incl/min 60°C Po atm 0.50 mol P(l).r‘mol Q (kas) 0.50 mol Hfly‘mol 3) Assume the vapor and liquid streams, exiting the evaporator are in equilibrium. Use RaOult’s law to find Po and y. (Vapor pressure data: P;(60°C)=1607 mm Hg, P; (60°C) 2 572.7 mm Hg) RABHLT : 0- 50 P; {5001.) = 73 Pa 0. 90 P; {‘D‘U = (ll—g.) 170 0.513 (“‘07) :— 13,P° } ADD: “390 MMH} =.:_I—P—D4] 0.510(5'723') = (f-gJPo 8:15;? = “:1?“ w mm?“ b) Find ii“ and fl}. ”Jami weft: mum: loo = n", + 734 Mm ml: (Mom! 60 = {0.737) r'lv + (0.9041 Go = (OJZYJHM' I511) + 0.513 Viz 0-237 ”1 : f3.7 Viv = 42.1953: mm ii: = {3—3: :g-j. fig; .13“! g 6) Some entries in the enthalpy table are already filled in, and you can assume they are correct. Complete the enthalpy table. References: liquid pentane and liquid hexane at 60°C. Fw Ideadme ' Cr 3 (l31-m-txro‘3J + (4018'? xw'SJT —(ZZ.1'1M0”')T2' VAfi'afl. +(S”7.uxm"’) 7'3 1. 1:”ka 3 (rW‘ 8‘ ) -' C A" '5 _ {REM-rd P (ya no ) WW 68-7!) A to . 4'- 4 [so crumb d? + AHVKHN) 561.14 QM“ T -._.___._._.—-—P'/ “—‘J \-..__.-r" Q @ (Z) ,_,or~ we U/al+5mr? Q) = 0.1lb3(eg.w_5n) = {.219 3333"“ (bar €60); 23.e5(:;:m : 285'; M Q = 19.95 1‘34»; 9n7.’£-'{6.£R+2?? 311.33 5° ‘— 29.?2 W- @ .-.- f f. 5-9 Wnutw Ill-bow. J an" AMI as rm _ —3 - [057..wa )‘n {19.v1xlo”$)T1 , ( 7.97 mod)?” 61’) + (“J-w xm‘” 7“] 63.79 = 3-959 — 10.3w = — was: ”An: A — 313' HM - AM: + 2115‘ — 3.9611 c 2.7.ng m hqmm I A Hm ., {0.1453)(7§-603 = 1m 5’5- mu! m.“ flow; Pumm HIM-7'37) = 3m “Mum 97W (A) (57.2) hr: 3 2t? mug.“ Hum (u) (V‘ZJJIJGJ) = H. r MV/NM “Win-L {1} (5,-1.3) (-5) a 1"? “‘0’th @L’AH = Z ”1'36 " Z M W “UPI-Er = (swung) +(n.;)(21.31,) - (Lo)(3.72,‘) - 6mm“) : 7%..1K} : was kid n1 EH ...
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