Conduction in the Cylindrical Geometry

Transcription

1 Cnductin in the Cylindrical Gemetry R. Shankar Subramanian Department f Chemical and Bimlecular Engineering Clarksn University Chemical engineers encunter cnductin in the cylindrical gemetry when they analyze heat lss thrugh pipe walls, heat transfer in duble-pipe r shell-and-tube heat exchangers, heat transfer frm nuclear fuel rds, and ther similar situatins. Unlike cnductin in the rectangular gemetry that we have cnsidered s far, the key difference is that the area fr heat flw changes frm ne radial lcatin t anther in the cylindrical gemetry. This affects the temperature prfile in steady cnductin. s an example, recall that the steady temperature prfile fr nedimensinal cnductin in a rectangular slab is a straight line, prvided the thermal cnductivity is a cnstant. In the cylindrical gemetry, we find the steady temperature prfile t be lgarithmic in the radial crdinate in an analgus situatin. T see why, let us cnstruct a mdel f steady cnductin in the radial directin thrugh a cylindrical pipe wall when the inner and uter surfaces are maintained at tw different temperatures. Steady Cnductin Thrugh a Straight Cylindrical Pipe Wall Cnsider a straight circular pipe f inner radius r, uter radius r and length L depicted belw. T r r z r T L Let the steady temperature f the inner surface be T and that f the uter surface be T, as shwn n the sketch. The temperature varies nly in the radial directin. Fr mdeling heat flw thrugh the pipe wall, it is cnvenient t use the end view shwn n the fllwing page.

2 ( + r) Q r r r Q( r) T r T r We use a shell balance apprach. Cnsider a cylindrical shell f inner radius r and uter radius r+ r lcated within the pipe wall as shwn in the sketch. The shell extends the entire length Q r be the radial heat flw rate at the radial lcatin r within the pipe wall. L f the pipe. Let ( ) Then, in the end view shwn abve, the heat flw rate int the cylindrical shell is Q( r ), while the heat flw rate ut f the cylindrical shell is Q( r+ r). t steady state, Q( r) Q( r+ r). Rearrange this result after divisin by r as shwn belw. ( + ) ( ) Q r r Q r r 0 Nw take the limit as r 0. This leads t the simple differential equatin dq dr 0 Integratin is straightfrward, and leads t the result Q cnstant independent f radial lcatin. The heat flw rate Q q r, where is the area f the cylindrical surface nrmal t the r directin, and q r is the heat flux in the radial directin. The area f the cylindrical surface is dt π rl, where L is the length f the pipe. Frm Furier s law, qr k. Therefre, we dr find that

3 dt k cnstant dr dt Substituting fr the area, we can write π rlk cnstant. Because π Lk is cnstant, dr this leads t a simple differential equatin fr the temperature distributin in the pipe wall. dt r C dr Here, C is a cnstant that needs t be determined later. Rearrange this equatin as dr dt C r and integrate bth sides t yield T C ln r + C where C is anther cnstant that needs t be determined. Thus, we find that the steady state temperature distributin in the pipe wall fr cnstant thermal cnductivity is lgarithmic, in cntrast t cnductin thrugh a rectangular slab in which case, the steady temperature distributin was fund t be linear. T find the tw cnstants in the slutin, we must use bundary cnditins n the temperature distributin. The temperature is specified at bth the inner and uter pipe wall surfaces. Thus, we can write the bundary cnditins as fllws. ( ) T T( r ) T r T By substituting these tw bundary cnditins in the slutin fr the temperature field in turn, we btain tw equatins fr the undetermined cnstants C and C. T C ln r + C T C ln r + C These tw linear equatins can be slved fr the values f the cnstants C and C in a straightfrward manner. We find T T C ln / ( r r ) T T C T ln r ln ( r/ r) Substituting the results fr the tw cnstants in the slutin fr the temperature prfile, fllwed by rearrangement, yields the fllwing result fr the temperature distributin in the pipe wall. ( r r ) ( ) T ln / T T T ln r / r 3

4 sample sketch f the steady temperature prfile in the pipe wall is shwn belw. T T r r We are ready t evaluate the heat flw rate Q. dt Q qr k π rl dr Recall that dt / dr C / r. Substituting in the result fr Q leads t T T T T Q ( kc ) πl πlk πlk ln / ln / ( r r ) ( r r ) This result is a bit hard t remember. Let us try recasting it in a frm similar t that we used fr steady cnductin in a rectangular slab. Let us write T Q where T T T is the driving frce, and R is the resistance t heat flw. We find R R ln r / r / π Lk. Let us rearrange this result. that the resistance is ( ) ( ) 4

5 ( r r ) ( r r ) ln / r r ln / r r r r R πlk r r πlk k [ πlr πlr] klm ln ( πlr / πlr) where we have intrduced a new symbl, which stands fr lg mean area. It is defined as lm where π Lr ln ( / ) lm is the inner surface area f the pipe wall, and π Lr is the uter surface area f the pipe wall. The lg mean always lies between the tw values being + averaged. Try cmparing it with the mre cmmn arithmetic average, which is. Yu /, the will find that the lg mean is always smaller than the arithmetic average. Fr ( ) difference between the tw averages is less than 4%. Steady Cnductin Thrugh Multiple Layers in the Cylindrical Gemetry It is straightfrward t extend ur analysis f steady state cnductin in a pipe wall t multiple layers in the cylindrical gemetry. Cnsider, fr example, a pipe f length L carrying ht r cld fluid that needs t be insulated frm the surrundings. We add an insulatin layer t the utside f the pipe. Here is an end view f the setup. r 3 T 3 T r r T ir B In the abve sketch, regin is the pipe wall, and regin B is the insulatin layer. The pipe is surrunded by air in this example. Based n ur earlier analysis, we can immediately write the steady state heat flw rate Q frm the interir wall f the pipe t the utside surface f the insulatin in terms f the driving frce fr cnductin in each f the tw layers and the resistance f each layer. 5

6 Q T T T T 3 where Rpipe Rpipe Rins kpipe lm, pipe r r r3 r and Rins are the resistances t k heat flw in the pipe wall and the insulatin layer, respectively. The tw lg mean areas are defined as fllws. lm, pipe and ln ( / ) 3 lm, ins, and π Lr ln ( 3 / ) ins lm, ins, π Lr, 3 π Lr3. We can cmbine the resistances and use an verall driving frce, just as we did in the case f steady cnductin thrugh a cmpsite slab. Because T T QRpipe T T3 QRins, adding the tw left sides, we btain ( ) T T Q R + R 3 pipe ins s that we can write Q T T 3 ( Rpipe + Rins ) It is pssible t include additinal cnvective resistances t heat transfer in the interir f the pipe and frm the utside surface t the air surrunding the insulated pipe. If we define a heat transfer cefficient h i t describe cnvective heat transfer between the fluid flwing thrugh the pipe at a temperature T i and the pipe wall, whse interir surface is at a temperature T, we can write Ti T Q hi i( Ti T ) / ( h) i i where i π Lr is the inside area fr heat transfer, and is the same as defined earlier. Thus, the cnvective heat transfer resistance n the inside f the pipe is / ( h i i). Likewise, if the cnvective heat transfer cefficient between the utside surface f the insulated pipe and the surrunding air is h, and the temperature f the air is T, we can write T Q h ( T3 T) / 3 T ( h) where π Lr3 is the utside surface area fr heat transfer, and is the same as 3. We see that the cnvective heat transfer resistance n the utside f the pipe is / ( h ). 6

7 Nw, we can write the fllwing result fr the heat flw rate. T T T T T T T T ( ) ( ) i 3 3 Q Uii Ti T U Ti T / ( h i i ) Rpipe Rins / ( h ) where U i is the verall heat transfer cefficient based n the inside heat transfer area, and U is the verall heat transfer cefficient based n the utside area. Nte that in the cylindrical gemetry, we have t specify the area upn which the definitin f the verall heat transfer cefficient is based, unlike in the rectangular gemetry where the area fr heat flw did nt change acrss the path. It is straightfrward t see that the verall heat transfer cefficients can be btained frm the fllwing result. r r r r U U h k k h 3 Rk i i k i i pipe lm, pipe ins lm, ins This equatin states that the verall resistance t heat transfer, signified by either / ( U i i) r / ( U ) is cmprised f cntributins frm each individual resistance t heat transfer in series. Other resistances can be added as needed, fr example when there is a thermal cntact resistance between the pipe wall and the insulatin layer, r when fuling ccurs in an industrial heat transfer situatin. Yu ll find additinal infrmatin in yur textbk. ls, tw gd references are the texts by Mills () and Hlman (). S far, we have assumed the thermal cnductivity t be a cnstant. If the temperature variatin acrss a layer is sufficiently large that the thermal cnductivity changes significantly, then the simplest apprach is t use the thermal cnductivity evaluated at the arithmetic average f the temperatures at the inside and utside surfaces f that layer. In prblems invlving cnductin thrugh an insulated pipe wall, the tw surface temperatures f the insulatin layer may nt be knwn initially. In such cases, yu can begin by making an educated guess f the temperatures invlved and estimate the thermal cnductivity. fter calculatins are made using this guessed thermal cnductivity, the temperatures f the tw surfaces f the insulatin layer can be calculated. These temperatures are then used in a secnd iteratin t make a mre accurate estimate f the average temperature f the insulatin layer, and the thermal cnductivity at that temperature. Usually, ne r tw iteratins f this type will suffice, and the thermal cnductivity f the layer will n lnger change appreciably at the next iteratin. Steady Cnductin with Heat Generatin in a Cylinder When a current passes thrugh an electrical wire, because f the resistance t the flw f electricity in the wire, heat is generated. The heat generatin can be characterized as a vlumetric 3 surce f heat S. In SI units, S wuld be measured in W / m. t steady state, this heat must flw radially ut f the wire int the surrundings either directly r thrugh an insulatin layer. Likewise, in nuclear fuel rds, a nuclear prcess within the rd leads t a surce f heat. 7

8 nther example is the heating f fd in a micrwave ven, wherein the micrwave radiatin absrbed by the fd leads t heating within it. In all these cases, the center f the wire r fuel rd r the fd item wuld be ht, and the surface wuld be cler, because f the need fr heat t flw ut f the bject and int the surrundings. It is useful t be able t estimate the temperature difference between the center and the surface, and the difference in temperature between the surface and the surrunding fluid, typically air, at steady state. Therefre, next we analyze steady radial heat cnductin in a cylinder cntaining a unifrm vlumetric heat surce. similar analysis can be carried ut fr a sphere, but is mitted here. Cnsider a cylinder f length L and radius R cntaining a unifrm vlumetric heat surce S, and surrunded by air at a temperature T, as shwn in the sketch belw. ir at T T s R L The cylinder is assumed t be lng with insulated ends, s that heat transfer thrugh the cylinder and t the air ccurs nly in the radial directin. Let the heat transfer cefficient between the surface f the cylinder and the surrunding air be h, and the steady state temperature at the surface f the cylinder be T s. First, we shall use an verall energy balance at steady state t establish the relatinship between the surface temperature T s and the surrunding air temperature T. t steady state, all the heat generated within the cylinder must be transferred t the surrunding air. Thus, the rate f heat generatin within the cylinder, which is the prduct f the vlume f the cylinder and the heat generatin rate per unit vlume S, must be equal t the heat flux thrugh cnvective heat transfer t the surrunding air multiplied by the surface area f the cylinder. ( ) Q πr LS h Ts T πrl 8

9 Therefre, π R LS R Ts T S r π RLh h R Ts T + S h Nw, let us investigate the temperature distributin within the cylinder at steady state. Fr this purpse, we begin with a sketch f a differential sectin f the cylinder lcated at a radius r and f width r, and extending the entire length L f the cylinder, as shwn in the sketch. ( + r) Q r r r Q( r) T s R The steady state energy balance fr the cylindrical shell shwn in the sketch can be written as fllws. ( ) + π ( + ) Q r r rls Q r r In wrds, the heat flwing ut f the shell at r+ r is the sum f the heat flwing in at the radial lcatin r and the heat generated within the vlume f the shell. Rewrite the abve equatin as ( ) ( ) Q r+ r Q r π r rls Nw, divide thrugh by r, and take the limit as r 0. Limit r 0 ( + ) ( ) Q r r Q r r π rls This results in the fllwing rdinary differential equatin. 9

10 dq dr π rls The heat flw rate Q qr qrπ rl, and q r is the heat flux in the radial directin. Frm Furier s law, qr k ( dt / dr), and using this result yields Q π rlk ( dt / dr). Substitute fr Q in the differential equatin, treating the thermal cnductivity k as a cnstant, and rearrange t btain the fllwing secnd rder rdinary differential equatin fr the steady temperature distributin in the cylinder. d dt S r r dr dr k It is straightfrward t integrate this equatin if we keep it intact withut simplifying the differentiatins using the prduct rule. First, we rewrite it as d dt S r dr dr k r which can be integrated immediately, because the left side is just the derivative f respect t r. Thus, integratin leads t dt dr S k r r + C dt r dr with where we have intrduced an arbitrary cnstant f integratin C. Rearrange this equatin as dt S C r + dr k r and it can be integrated nce again! The result fr the temperature distributin in the rd is S T( r) r + C ln r + C 4k where a secnd arbitrary cnstant f integratin has been intrduced. Clearly, the temperature has t be finite everywhere, including at r 0. T satisfy this cnstraint, we can see that the cnstant C must be zer. Thus, S T( r) r + C 4k 0

11 T find the remaining arbitrary cnstant, we use the bundary cnditin that the temperature at the surface r R is T s. This cnditin is written frmally as fllws. ( ) Ts T R Evaluating bth sides f the result fr the temperature field at the surface r S Ts R + C 4k S s that C T R 4k s +. Substitute this result fr R, C in the temperature field t btain S T Ts + R r 4k ( ) Can yu guess where the temperature will be a maximum? Frm the slutin, we can see that this will ccur when r 0, which is the axis f the rd. This maximum value is fund t be S Tmax T 0 Ts + R 4k ( ) sample sketch f the radial temperature distributin in the rd is given n the fllwing page.

12 Temperature versus Radial Psitin in a Cylinder with a Unifrm Heat Surce T max Temperature T s 0 Radial Psitin References..F. Mills, Heat Transfer, Secnd Editin, Prentice-Hall, New Jersey, J.P. Hlman, Heat Transfer, Tenth Editin, McGraw-Hill, New Yrk, 00.