Problem #1. Known: All required parameters. Schematic: Find: Depth of freezing as function of time. Strategy:

Transcription

1 BEE Prelm Soluton Problem #1 Known: All requred parameter. Schematc: Fnd: Depth of freezng a functon of tme. Strategy: In thee mplfed analy for freezng tme, a wa done n cla for a lab geometry, we equate the rate of heat tranfer through the already frozen layer to the rate at whch heat need to be removed to freeze an adonal layer. The analy here exactly ame a that done for a lab, wth the only dfference beng the retance here not jut of that of the frozen layer but alo the platc wrap n ere wth the frozen layer. Soluton: 1) Tm T dx H f A Lplatc x 1 k A k A ha platc frozen Tm T dx Hf Lplatc x 1 k k h platc frozen 1

2 BEE Prelm Soluton Tm T Lplatc x 1 dx H f k platc k frozen h t Lbo Tm T Lplatc x 1 dx H 0 f k 0 platc k frozen h T L m T platc x x t x H k k h f platc frozen T T L L L t m platc bo bo t Lbo H f k platc k frozen h freeze H f Lplatc Lbo L bo Lbo Tm T k platc k frozen h ) It wll add an extra layer of conductve retance that ncreae the tme to freeze.

3 BEE Prelm Soluton Problem # Known: Reflectvty = 0; T 1 =303 K; T =88 K; T g =93K; T =83K; I = 1000 W/m ; h o =5 W/m K Schematc: Fnd: h, T Strategy: There are two eparate problem here. Frt, we need to fnd the nde heat tranfer coeffcent. To calculate convectve heat tranfer coeffcent, we follow the tep provded n the courenote for th chapter. Although the urface nclned, equaton for vertcal urface can be ued, a mentoned. For the econd part of the problem, follow the tep provded n the courenote for olvng a radatve energy balance problem. Note the tranmvty gven for gla whch would be needed for all radatve exchange term. Soluton: 1) 3

4 BEE Prelm Soluton Nu Ra L L hl 0.387Ra 0.85 k / Pr Gr Pr Gr 3 g L T T K K L 4m 1/6 L 8/7 8/7 Properte hould be evaluated at ½(30+0) = 5, (.e. 98 K). Aume properte at 300 K for mplcty. Pr kg m 5 kg m W k 0.06 mk Gr Ra L h / W h.91 m K /16 1/ 6 8/ ) 5 F F F0 848 F , 1 1 Hence, the gla can be condered a black body 4

5 BEE Prelm Soluton An alternate way to ay th that max T=897.6 µmk. Thu, for the nde urface temperature around 303 K, max = 9.56 µm for whch the gla ha zero tranmvty and therefore an aborptvty of 1. 3) Solar radaton In-Out+Generaton-Conumpton=Storage h T T h T T F1 Tgla T3 F1 3 Tgla T forced gla ar natural gla ar F F Angle olar Gen Change n radaton make Fracton that Convectve lo Convectve lo Radatve lo Radatve lo torage wth gla get aborbed to outde to nde to ground to other gla urface In Out F F 1 n( ) F F Tar ) T=349.4K=76.4 C 5) Update the properte baed on the new flm temperature, ( )/=53. C, and then recalculate natural convecton heat tranfer coeffcent and then update new ar temperature. Repeat untl no change. We dd not expect you to do any calculaton at th tep. 5

6 BEE Prelm Soluton Problem #3 Known: Varou rate of generaton and elmnaton Schematc: Fnd: M n tomach, M n ntetne, M b (jut equaton for M b ) a functon of tme Strategy: Ue ma balance Soluton: Rate n Rate out + Rate of Generaton = Rate of change n Storage 1) 0 km 0 ln M k t C 0 M t 0 M M M exp k t ) 0 6

7 BEE Prelm Soluton km km kam 0 km 0 exp kt k ka M 0 k ka M km 0 exp kt c M P k k Q k M k t exp a 0 c exp P Q exp P D exp P P k ka k ka t P k ka k ka t exp exp exp exp exp exp M exp k ka t Q exp k ka t D exp k ka t a 0 a km 0 exp k ka k t km 0 exp k ka k t Q exp k k t k M exp k t exp k k t M k k k a km 0 exp k ka k t M exp k ka t D exp k ka t k ka k km 0 exp k t M k ka k D km 0 k k k a km 0 k k k D D exp k ka t a 0 exp 0 k k k k k k k M k t km M exp k ka t a a km k t exp k k t 0 exp a k ka k 7

8 BEE Prelm Soluton 3) b FkaM kel M b 0 k M Fk exp k t exp k k t k M 0 0 k k k 0 a a el b a b 8

9 BEE Prelm Soluton Problem #4 #1 Henry # At hgher alttude, there le oxygen nce lower preure. Therefore there lower amount of oxygen n lung due to a mall ambent concentraton #3 A the temperature ncreae, the olublty of carbon doxde decreae. Therefore, the CO come out of the oluton and the concentraton buld up n the head pace, ncreang the preure untl the can explode. #4 Hydrophobc nce Partton coeffcent = C drug n old (tue) /C drug n water #5 There le moture n the ar n the wnter o the kn urface comng to equlbrum wth ar alo ha le water. #6 E E C f C0 exp k0 exp t1 C0 exp k0 exp t RT1 RT E E exp t exp t RT 1 1 RT t E E E E exp / exp exp t1 RT1 RT RT1 RT t E 1 1 ln t1 R T1 T #7 p H x O o o p H x CO co co Compare xo and xco to ee whch one more oluble. 9

10 BEE Prelm Soluton #8 At the nterface of the two phae, the equlbrum reached frt. Th really fat, can be aumed ntantaneou for our purpoe. #9 c p RT #10 b) harder 10