ChE 548 Final Exam Spring, 2004
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1 . Keffer, eprtment of Chemil Engineering, University of ennessee ChE 58 Finl Em Spring, Problem. Consider single-omponent, inompressible flid moving down n ninslted fnnel. erive the energy blne for this system. Show ll work involved in eh step of the derivtion. Epress the energy blne in sh form tht the left-hnd-side ontins only the time derivtive of the tempertre. Stte ny ssmptions tht yo mke. Introde vribles sh s the density, het pity, therml ondtivity, et s neessry. he ft tht the flid is inompressible n be epressed by mking the veloity fntion of il position; do so. ssme the srrondings re hotter thn the flid inside the fnnel. Qlittively sketh the stedy stte profile for two vles of the het trnsfer oeffiient, ero (inslted nd non-ero for yor bondry onditions. For the inslted se, one n obtin n nlytil soltion for the stedy stte profile. ime permitting, obtin it.
2 . Keffer, eprtment of Chemil Engineering, University of ennessee Problem. Soltion: mss blne: ρ t for inompressible flid. density is onstnt. enthlpy ssme onstnt het pity ssme het pity is given on per mss bsis H C ref p length of fnnel d C dimeter of the retor ross-setionl re π p ( ref π for n inompressible flid, the volmetri flowrte is onstnt F ons tn t v herefore, the veloity s fntion of position is v F volme element ssme no vrition in rdil or nglr dimensions V
3 . Keffer, eprtment of Chemil Engineering, University of ennessee energy blne ρh V t onv ρvh ρvh ond q q loss s ( π h( h srr where srr is the tempertre of the srrondings nd h is the het trnsfer oeffiient. srr VρH ρvh ρvh q q π ( h srr divide by inrementl volme ρh t ρvh ρvh q q h ( srr tke limit s goes to ero. ρh t ρvh insert Forier s w q h ( srr ρh t ρvh k h ( srr eliminte the enthlpy in fvor of the tempertre ρh ρcp t t ρvh ρv H v H ρc p v v ( ref ssme onstnt therml ondtivity. se the ft tht we hve n epression for the veloity s fntion fo il position. 3
4 . Keffer, eprtment of Chemil Engineering, University of ennessee F C vh p ρ ρ srr p p h k k v C t C ρ ρ divide by ρc p srr p h C ln v t ρ α α where α is the therml diffsivity p C k ρ α ln π π srr p h C v t ρ α α his is the evoltion eqtion for tempertre.
5 . Keffer, eprtment of Chemil Engineering, University of ennessee Now, let s solve for the stedy stte profile for the inslted se, where h. Cse. he tempertre nd the tempertre grdient t re known. F α α (. F α α (. F α α π (.3 let (. F α α (.5 π his OE is of the form: ( (.6 where F ( (.7 his OE hs the soltion: (o ep ( d (.8 o (.9 d d (. 5
6 . Keffer, eprtment of Chemil Engineering, University of ennessee 6 F ln d F d F d F (d o o (. F ep ( F ln ep ( o o (. F ep (.3 integrte gin: o ( ( d F ep d (. o d F ep ( ( (.5 d F ep ( d F ep ( ( (.6
7 . Keffer, eprtment of Chemil Engineering, University of ennessee his integrl is of the form: ep bd (.7 o Use the sbstittion / nd we hve ep( b d (.8 o his integrl n be evlted nlytilly to yield ep( b d ep( b ep( (.9 o o So for or se (. F b (. F (. So we hve ep o ( b d ep( b ep o (.3 ( ( ep( b ep (. 7
8 . Keffer, eprtment of Chemil Engineering, University of ennessee his soltion ssmes tht we know the tempertre nd the tempertre grdient t the inlet of the fnnel. We old lso work the problem ot where or onstnts of integrtion re determined by tempertre t eh bondry. here yo hve the nlytil soltion. et s mke ople plots of the nlytil soltion. First, we write qik little Mtlb ode to evlte the soltion. 8
9 . Keffer, eprtment of Chemil Engineering, University of ennessee ler ll; lose ll; nint ; np nint ; eros(,np; eros(,np; ; o ; f o ; d /nint; Cp ; ro ; r ; s (r - ro/; F ; rho ; k ; lph k/(rho*cp; ddo ; o 3; -*F/(lph*pi*s; b *F/(lph*pi*s*ro; epb ep(b; term ep(/ro*(/ro^ - /(*ro /^; for i ::np (i (i-*d o; r ro s*(i; pi/*r*r; v F/; term ep(/r*(/r^ - /(*r /^; (i o ddo*ro^/s*epb*(term - term; end plot(,; 9
10 . Keffer, eprtment of Chemil Engineering, University of ennessee Plot for bse se:
11 . Keffer, eprtment of Chemil Engineering, University of ennessee plot for bse se eept we inrese the volmetri flowrte to F
12 . Keffer, eprtment of Chemil Engineering, University of ennessee plot for bse se eept we inrese the therml ondtivity to k
13 . Keffer, eprtment of Chemil Engineering, University of ennessee 3 Cse B. he tempertre t nd the tempertre t re known. Eqtions (. to (. remin the sme. Eqtion (. n be rewritten in terms of generl nknown onstnt of integrtion. ep ( o F (. F ep (B. F ep (B.3 integrte gin: o d F d ep ( ( (B. o d F ep ( ( (B.5 d F d F ep ( ep ( ( (B.6 his integrl is of the form: o d b ep (.7 Use the sbstittion / nd we hve o d b ep (.8
14 . Keffer, eprtment of Chemil Engineering, University of ennessee his integrl n be evlted nlytilly to yield o o ep ep b d b ep (.9 So for or se (. b (B. F (. So we hve o o d b ep ep (B.3 ep ( ( (B. Finlly, we evlte the nknown onstnt, by foring it to stisfy the seond bondry ondition ep ( ( (B.5 ep ( ( (B.6 here yo hve the nlytil soltion. et s mke ople plots of the nlytil soltion. First, we write qik little Mtlb ode to evlte the soltion.
15 . Keffer, eprtment of Chemil Engineering, University of ennessee ler ll; lose ll; nint ; np nint ; eros(,np; eros(,np; ; o ; f o ; d /nint; Cp ; ro ; r ; s (r - ro/; F ; rho ; k ; lph k/(rho*cp; o 3; ; -*F/(lph*pi*s; term ep(/ro*(/ro^ - /(*ro /^; term3 ep(/r*(/r^ - /(*r /^; (-o*s/(term3-term; for i ::np (i (i-*d o; r ro s*(i; term ep(/r*(/r^ - /(*r /^; (i o /s*(term - term; end plot(,; dd_o /ro^*ep(/ro 5
16 . Keffer, eprtment of Chemil Engineering, University of ennessee bse se F nd k 6
17 . Keffer, eprtment of Chemil Engineering, University of ennessee F nd k 7
18 . Keffer, eprtment of Chemil Engineering, University of ennessee F nd k 8
19 . Keffer, eprtment of Chemil Engineering, University of ennessee For the se where the het trnsfer oeffiient is non-ero, the OE is more diffilt to solve nlytilly. he oeffiients of the terms re fntions of il position,. I m pretty sre tht n nlytil soltion does eist, bt I hven t derived it myself. nywy, the problem doesn t sk for n nlytil derivtion, only sketh. So, one yo hve estblished wht the inslted se shold look like, then yo hve to modify tht to ont for het loss. he het loss will be greter t the wider end of the fnnel so the differene between the inslted n ninslted ses ot to be greter t the wider end of the fnnel. Seeing s we re on pge 9 of the soltions, I m going to leve it t tht. 9
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