TRANSIENT HEAT CONDUCTION, ANALOGY METHODS AND HEAT FLOW MEASURING

Transcription

1 HEA RANSFER - LAB LESSON NO. 3 RANSIEN HEA CONDUCION ANALOGY MEHODS AND HEA FLOW MEASURING Before the start of the lab you should be able to answer the followng questons: 1. Explan the Schdt s ethod. How does t wor?. Descrbe (at least) two dfferent analogy ethods whch ay be used n solvng heat transfer probles. 3. What are the benefts usng analogy ethods? 4. In these analoges what are the enttes correspondng to teperature heat transfer resstance heat capacty and heat flow? 5. Explan how a two-densonal steady state heat conductng proble can be solved usng fnte dfference equaton approach. C:\Mna douent\insttutonell tänstgörng\v-labbar\#3\lab-pe.doc 1 (16)

2 HEA RANSFER - LAB LESSON NO. 3 1 OBJECIVE In ths lab Schdt s ethod for solvng a transent heat conductng proble through a wall wll be used. Experent wll be conducted to verfy the calculated soluton. A sple coputer progra wll also be used for coparson. Further ore two analogy ethods Hydraulc and Electrcal wll be deonstrated. wo dfferent electrcal approaches wll be nvestgated one contnuous and one dscrete. Agreeent between the s nvestgated. he hydraulc analogy wll be deonstrated on a wall wth changng teperature on one sde. A MS-Excel progra s wrtten for nuercal study of the proble at hand. Also a coercal FEM software s deonstrated. Soe dfferent ethods of easurng the heat flow through a wall s deonstrated one old and two ore odern ethods. AIM Get hands on experence of soe heat flow easurng ethods. Experence of analogy ethod n solvng heat transfer probles. Introducton n solvng heat transfer proble nuercally understand the approach of FDE (Fnte Dfference Equatons). KEEP IN MIND - KEEP I SIMPLE C:\Mna douent\insttutonell tänstgörng\v-labbar\#3\lab-pe.doc (16)

3 HEORY HEA RANSFER - LAB LESSON NO Introducton he general heat transfer proble has four densons.e. three denson n space n one n te. However t s often possble to splfy to a certan lt. Often the steady-state soluton s wanted. Many proble can be treated wth only two densons n space and soetes even wth only one denson n space. he general governng equaton for heat transfer n a sold assung constant ateral propertes and no nternal heat generaton s: τ α z (1) eperature τ te x y z dstance α ( ρ c p ) Eq. (1) can be solved nuercally whch wll be showed n ths lab for two dfferent cases transent heat transfer n an one densonal wall (Schdt s ethod) and steady-state heat transfer n a two densonal wall. Another way to nvestgate the nfluence of dfferent paraeters n heat transfer probles s to use analogy ethods. Electrcal currents hydraulc flows and heat flows are governed by the sae type of dfferental equatons and thus t s possble to use electrc or hydraulc analoges n the study of heat transfer. C:\Mna douent\insttutonell tänstgörng\v-labbar\#3\lab-pe.doc 3 (16)

4 Analogy ethods HEA RANSFER - LAB LESSON NO. 3 As stated earler there are analogy between heat transfer and electrcty and hydraulc. A forer professor of our departent professor Bo Perre nvestgated the nfluence of the rbs on the nsulatng capablty of the hull of refrgerated freght shps by usng the electrc analogy. A couple of artcle were publshed on the subect (n Swedsh Spantens nveran på solerngen vd yllastfartyg belyst ed eletrsa analogförsö ) see appendx. he followng equatons are vald for the theral and electrc flows: q U A / R th () I V / R (3) hus the heat flow q correspond to the current I the teperature dfference correspond to the dfference n electrc potental V and the heat transfer resstance R th correspond to the electrc resstance. In transent probles there s an addtonal correspondence between heat capacty and electrc capactance. A hydraulc odel ay be used for vsualzng the heat flow and teperature dstrbuton through a wall. he heat flow correspond to the lqud flow the heat transfer resstance n dfferent layers n the wall correspond to the flow resstance of capllary tubes of dfferent length and daeter. he teperature n these layers correspond to the pressure nsde the tube n our case vsualzed as the water level n the vertcal tubes. In transent probles there s an addtonal correspondence between the aount of heat stored n each layer and the aount of water stored n the vertcal tubes. A large tube daeter s able to store ore water at a gven water level whch then correspond to a large aount of heat stored at a gven teperature. C:\Mna douent\insttutonell tänstgörng\v-labbar\#3\lab-pe.doc 4 (16)

5 HEA RANSFER - LAB LESSON NO. 3 o evaluate the results of the analogy test a nuber of scale factors are needed. hese are shown n table 1. able 1: Analogy odels analogous enttes and scale factors. Analogous quanttes heral Hydraulc Electrc Heat Q eperature e τ Lqud volue v Water level h v e Electrc charge Q el I dτel Electrc potental V e τ h τ el Relatons: Q / τ K t v / τ h C h h v Q el / τ h I V / R ransport functon Conducton: C h Capllary tube (correspondng to heat transfer) K A / x Surface heat transfer: constant R h 1 / C h K A h Conservaton functons Q C v A r h v Q el C el V (correspondng to heat C c p A x ρ c p A r tube area C el capactance balance) Scales: Q v h hv n Q h Q nel Qel n n el V τ τ τ nh τh nel τel Relatons between scale factors n n n Q h Q h τ h K C h C A r n n C C K A h h h r n τ h n n n Q el Q el τ el K R C C el el n el el n el C 1 K R C n el τ el C:\Mna douent\insttutonell tänstgörng\v-labbar\#3\lab-pe.doc 5 (16)

6 Fnte Dfference Equatons HEA RANSFER - LAB LESSON NO. 3 An analytcal soluton for eq.(1) can be attaned for soe sple geoetres however for the general case no such soluton has been reported. o be able solve the proble at hand soe other ethod ust be used. Wth the fast and cheap coputers of today t s now possble for alost anyone to solve eq.(1) by usng nuercal ethods. -1-1/ 1/ 1 x Fgure 1: eperature as a functon of space Fro eq.(1) t s apparent that we ust fnd a way of expressng the dervatves n an sply way. We are nterested n solvng the teperature n poston. If we only study the teperature dependency of the x coordnate realzng that the dependency n y- and z- denson s treated n an analogous anner the teperature gradents can be expressed as: x x ( ) ( ) ( ) 1 ( ) ( ) ( ) 1 1 x x 1 where denotes ponts n x y z denson respectvely and () denotes te step. (4) C:\Mna douent\insttutonell tänstgörng\v-labbar\#3\lab-pe.doc 6 (16)

7 HEA RANSFER - LAB LESSON NO. 3 C:\Mna douent\insttutonell tänstgörng\v-labbar\#3\lab-pe.doc 7 (16) he second dervatve of teperature n pont s ( ) ( ) ( ) x x x x 1 1 (5) Insertng eq.(4) nto eq.(5) yelds ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) x x x x x (6) and for y- and z-denson ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) z z z z z y y y y y (7) he te dervatve can be expressed as ( ) ( ) τ τ 1 (8) Insertng eqs.(6) - (8) nto eq.(1) yelds ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) α τ z y x (9) whch s the fnte dfference equaton of eq.(1). 3.4 Steady-state two densonal heat transfer In a steady state analyss the teperature at any gven pont s constant wth respect to te 0 τ. In eq.(9) the left hand sde s then zero. For a two densonal analyss (xy)

8 HEA RANSFER - LAB LESSON NO. 3 the teperature n z-drecton s constant 0. Eq.(9) can then be splfed assung z that the coputatonal doan s spaced equal n both drectons (y x) as: whch easly can be solved ransent one densonal heat transfer Schdt s ethod Agan we use eq.(9) as a startng pont. Now we have a transent probles whch eans that the left hand sde s not equal to zero. We are only nterested n one space denson say x. Eq.(9) then splfes to: ( 1) ( ) ( ) ( ) ( ) τ α 1 1 x Collectng ters wth teperature of the new te step on the left hand sde and the old te step on the rght hand sde. ( ) ( ) ( 1) α τ ( ) ( ) ( ) 1 1 x whch can be rearranged even ore α τ x ( ) ( ) ( 1) ( ) ( ) 1 1 Introducng M defned as x M α τ and nsertng n eq.(13) yelds ( 1) ( ) ( ) 1 M 1 ( ) 1 M 1 α τ x By ang the clever observaton and settng M.e. the relaton between step n space and step n te Schdt attaned an equaton whch s easy to calculate. It can be seen that the second ter on the rght hand sde vanshes and eq.(15) splfes to (10) (11) (1) (13) (14) (15) C:\Mna douent\insttutonell tänstgörng\v-labbar\#3\lab-pe.doc 8 (16)

9 ( 1) ( ) ( ) 1 1 HEA RANSFER - LAB LESSON NO. 3 (16) whch says that the teperature at node and te 1 s the arthetc ean of the surroundng nodes 1 and -1 at the prevous te step. Because of t s splcty eq.(16) can be solved graphcally wthout any calculatons requred. At the pre-coputer era ths was a necessty. By settng M we have lned the ncreent n te together wth the ncreent n space and the propertes of the ateral. We ust fulfll ths relaton f Schdt s ethod s used.e. eq.(16). However for the odern engneer calculaton capablty s no proble and the requreent of settng M s soewhat obsolete. But for nuercal stablty reasons t can be shown that M should be equal to or greater than. 4 Experent 4.1 Analogy ethod 1: he nfluence of the rbs on the nsulatng capablty of the hull of refrgeratng freght shps he apparatus conssts of a plastc tray wth alunu rals at two sdes and a nuber of loose peces of ral. On the botto of the tray there s a plastc sheet sutable for drawng lnes wth a pencl. he tray s flled wth slghtly salt water. Wth the loose rals a profle odel of the shp s rbs s bult see fgure. o calculate the factor y by whch the heat transfer n to the refrgerated roo s ncreased wth the rbs copared to wthout the rbs s calculated as: δ z l y (17) δ z h z where d and l s defned as C:\Mna douent\insttutonell tänstgörng\v-labbar\#3\lab-pe.doc 9 (16)

10 HEA RANSFER - LAB LESSON NO. 3 1 δ δ λ α nt h z 0.90 l 0.99 h 1 α ext δ λ z h b b b (18) δ b δ z b h Fgure : Scheatc odel of the shp s rbs. An electrcal potental s appled between the rbs and a parallel ral and ths potental corresponds to the teperature dfference between the outsde wall of the hull and the nsde wall of the refrgerated roo. A volt-eter s suppled for readng the potental at dfferent postons n the water and a Apére-eter s connected n the crcut so that the current through the odel can be read. ES PROCEDURE A. Use the volt eter to fnd fve ponts of equal potental n between the rbs and the nsde wall. Connect the ponts to a curve. Draw such lnes for three dfferent potentals. C:\Mna douent\insttutonell tänstgörng\v-labbar\#3\lab-pe.doc 10 (16)

11 HEA RANSFER - LAB LESSON NO. 3 B. Read the current through the odel wth the rbs connected. hen dsconnect the rbs and reove the fro the tray. Read the new current wthout the rbs. By what factor y s the heat flow (current) ncreased wth the rbs copared to wthout the rbs? C. Copare the results fro B wth the equaton of Perre eq.(17) usng the followng data: α ext α ext δ b 0 l 0.04 W/( K). z b h and δ are found by easurng the odel. D. What nds of splfcatons have we done n the odel? How do you thn that these splfcatons affect the accuracy of the results? 4. Steady state two-densonal heat transfer A slar odel as above s bult by usng a esh of electrc resstances. In ths case the potental ay only be found n a fnte nuber of dscrete ponts. Note the slarty between the esh odel and the nuercal odel later used n the coputer sulaton. ES PROCEDURE A. Measure the electrcal potental n two nodal ponts n the esh. Note whch nodes and the results n suppled fgure fgure 3. B. Measure the potental of the four surroundng nodes for each of the two nodes easured n A. C. Copare the average of the four surroundng nodes wth the potental of the node tself. D. Measure the current through the odel wth and wthout the connectons representng the rb. Copare the result to that fro the odel n secton 4.1. C:\Mna douent\insttutonell tänstgörng\v-labbar\#3\lab-pe.doc 11 (16)

12 HEA RANSFER - LAB LESSON NO A B C D E F G H I J K L M N O P Q 1 C C Fgure 3: Electrcal resstance esh. USING MS-EXCEL he nodal esh of the electrc odel ay also be constructed n a spread sheet e.g. MS- Excel. o do ths let the cells of the spread sheet represent the nodal ponts specfy the boundary teperatures n the approprate cells and let the progra calculate the teperatures n the rest of the cells as the average of the four surroundng cells. An teratve procedure s needed. he calculatons should be contnued untl no further change n the teperatures s found. he odel s then sad to be relaxed. A odel teplate slar to fgure 3 s suppled. Copare the result fro the excel odel to those fro the esh! An exaple of soluton s gven n fgure 4. C:\Mna douent\insttutonell tänstgörng\v-labbar\#3\lab-pe.doc 1 (16)

13 HEA RANSFER - LAB LESSON NO. 3 A B C D E F G H I J K L M N O P Q C 0 C Fgure 4: A soluton of the two densonal wall. It s of course also possble to wrte a coputer progra n any prograng language to fnd the relaxed teperatures of the nodal esh. A separate progra has the advantage of lettng you steer the teraton process. For exaple you could change only one teperature n each step the teperature n the pont where the dfference between the nodal value and the average of the surroundng values s the largest. hs procedure s sad to ensure a convergent soluton. It s however consderably slower than ust calculatng new averages for every pont. he lab assstant wll deonstrate a coercal software (ANSYS) by sulatng the above proble. 4.3 Heat flow eterng he lab assstance wll talng about ethods of easurng the heat flow. Soe sall experent wll be conducted to let the student wth both odern and hstorcal equpent. C:\Mna douent\insttutonell tänstgörng\v-labbar\#3\lab-pe.doc 13 (16)

14 HEA RANSFER - LAB LESSON NO Electrcal and Hydraulc analogy for transent one densonal heat transfer through a wall wo odels of transent heat transfer through a wall wll be deonstrated by the lab assstant one electrcal and one hydraulc see also table 1. For both odels: A. ry to fgure out what each part of the odel corresponds to n a real wall. B. Pont out whch part of the wall that has the lowest theral conductvty and whch has the hghest heat capacty. C. In whch part of the wall s the phase delay the largest? Why? 4.5 ransent one densonal heat transfer through a wall Schdt s ethod he apparatus n ths test conssts of a ple of fve plates ade of an nsulatng ateral. he teperatures n between the plates and on both sdes of the ple are easured by therocouples. he ple s placed on top of a copper plate ept at roo teperature by water coolng. An electrcally heated copper plate ept at approxately 40 C can be placed on top of the ple. the teperatures are read by a data acquston syste and transferred to a coputer where the teperature can be read. At any nstance the students can store the teperature n MS-Excel. he nsulatng ateral has the followng propertes: ρ5 g/³ W/( K) c p J/(g K) and plate thcness x10. ES PROCEDURE A. Usng Schdt s ethod calculate the te step τ and the absolute te correspondng to sx te steps. τ sec. B. Place the heated plate on top of the ple and start the ter n the coputer progra. C. Save the teperatures n MS-Excel for te step 0 4 and 6. hen note the n the table below. D. Calculate the teperature for sx te steps usng Schdt s ethod. E. Copare your calculated result wth experent. Is the agreeent good? If not why? C:\Mna douent\insttutonell tänstgörng\v-labbar\#3\lab-pe.doc 14 (16)

15 HEA RANSFER - LAB LESSON NO. 3 F. Use Schdt s graphcal ethod to estate the teperature dstrbuton for sx te steps. G. Copare your graphcal soluton wth your nuercal soluton. H. ogether wth the lab assstant copare your result wth a coputer progra whch uses Schdt s ethod. See how the choce of plate thcness nfluence the accuracy of the result and also ncrease the nuber of te step needed to reach the sae endng te. Measured teperatures n experent Poston: e step Calculated teperature usng Schdt s ethod Poston: e step C:\Mna douent\insttutonell tänstgörng\v-labbar\#3\lab-pe.doc 15 (16)

16 HEA RANSFER - LAB LESSON NO. 3 Schdt s ethod graphcally (See Holan): ( C) Poston x C:\Mna douent\insttutonell tänstgörng\v-labbar\#3\lab-pe.doc 16 (16)

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