HT Module 2 Pper solution Qulity Solutions wwwqulitytutorilin Module 2 Q6Discuss Electricl nlogy of combined het conduction nd convection in composite wll M-16-Q1(c)-5m Ans: It is frequently convient to compre the het conduction process with the flow of electricity nd spek of quntity nlogous to electricl resistnce clled therml resistnce Corresponding to the current i nd the potentil difference in n electricl circuit the nlogous quntities in therml circuit re the het flow rte q nd the temperture difference (note: > ) Consider slb of thickness L nd hving therml conductivity K Let h be the het trnsfer coefficient on right fce s shown in fig For conduction between point 1 nd W, For convection between point W nd 2, For combined conduction nd convection: Sitution Electric Circuit Therml circuit 3 Driving force Voltge, Temp difference, Flow Current, I Het flow rte, Q Resistnce Electric resistnce R / Therml resistnce, /
Qulity Solutions HT Module 2 Pper solution wwwqulitytutorilin Q7 Derive Fourier s three dimensionl differentil eqution in the Crtesin co-ordintes nd hence deduce expression for one dimensionl stedy ste het conduction [D-14-Q-2()-10M] Ans Consider smll volume element in Crtesin co-ordintes hving smll sides dx, dy nd dz s shown in fig1 Let, the temperture t left fce is T nd is constnt through-out surfce The temperture grdient for smll distnce is / in the direction x Let, the het flow in x direction, het flux for smll time d is A d (1) Now, het out flux is, d (2) het stored within control volume when it flows in -direction d d d d d (3) d A d d Substituting Eqution (i) in Eqution (iii) d A d dz d d dz d (4) Similrly, het stored in control volume, when it flows in direction d A d dz d d dz d (5) Similrly, het stored in control volume, when it flows in z direction d Z A Z d d Z Z d d dz d (6) Let, internl het generted per unit volume is, then totl internl het, V d d d dz d (7) Also, het storge cpcity of body is, C d d d dz C V 4 d (8)
HT Module 2 Pper solution Qulity Solutions wwwqulitytutorilin For het blnce, d d d [ d d dz C C d ] d d dz d d d dz C d (conductivity is constnt in ll direction or uniform therml conductivity) [ ] C ( ) z α ( ) α If het flow is stedy stte, temperture is not the function of time z Poisson eqution If there is no internl het genertion in control volume then, Newton s eqution or Lplce eqution If het flows only in one-direction (Assuming then, nd z Known s Fourier s one dimensionl stedy stte het flow eqution without internl het genertion Three dimensionl conduction eqution for constnt therml conductivity, ( ) z α Extr ;Assumptions: i Conduction of het tkes plce under stedy stte ii The het flow is uni-directionl iii The temperture grdient is constnt nd the temperture profile is liner iv There is no internl het genertion v The boundry surfces re isotherml in chrcter vi The mteril is homogenous nd isotropic (Ie K constnt in ll directions) 5
Qulity Solutions HT Module 2 Pper solution wwwqulitytutorilin Q8 Derive the temperture profile eqution for cylindricl system from the generl differentil eqution strting the ssumptions for one dimensionl stedy stte het trnsfer [D-14-Q-2(b)-10M-CBGS] Ans: Consider long cylinder of length L, inside rdius r, nd outside rdius r0 Inner nd outer surfces re t uniform tempertures of nd respectively, see Fig 1 FIGURE 1 Het trnsfer through cylindricl shell () Cylindricl system nd the equivlent therml circuit (b) Vrition of temperture long the rdius Assumption i Stedy stte conduction ii One-dimensionl conduction, in the r direction only iii Homogeneous, isotropic mteril with constnt k iv No internl het genertion Now, this is cylindricl system; so, it is logicl tht we strt with the generl differentil eqution for one dimensionl conduction, in cylindricl coordintes So, we hve, ( ) ( ) z α In this cse: / τ, since it is stedy stte conduction / / z, since it is one-dimensionl conduction, in the r direction only,since there is no internl het genertion Therefore, the controlling differentil eqution for the cylindricl system, under the bove mentioned stipultions, becomes: (A) [Note : now, it is not prtil derivtive, since there is only one vrible,r] We hve to solve this differentil eqution to get the temperture distribution long r nd then pply Fourier s lw to clculte the het flux t ny position Multiplying eqn(a) by r, we get ( ) 6
Qulity Solutions ie integrting, HT Module 2 Pper solution )C C, ( ) B ( Integrting gin, C n C C where, C1 nd C2 re constnts of integrtion Eq( C ) gives the temperture distribution s function of rdius C1 nd C2 re found out by pplying the two BC's:, i ii, B C e, C n C B C e, C n C b Subtrcting Eqution (b) from Eqution (): C n( /, ) C n( /, ) n / And, from Eqution () C n n ubun C nd C n E C, e e n n n n e n ( ) D n Eqution (D) is the desired eqution for temperture distribution long the rdius Note: Eqution (D) cn be written in non-dimensionl form s follows: n(/, ) n / 7 wwwqulitytutorilin
Qulity Solutions HT Module 2 Pper solution wwwqulitytutorilin Q9 Derive n expression of temperture profile for infinite plne wll with uniform het genertion Wll thickness is b m nd both surfces re mintined t constnt temperture t w C Het genertion in the wll is q W/m3 Therml conductivity of the wll mteril is k W/m C [M-13-Q3 ()-10M] Ans: Consider plne slb of thickness b 2L s shown in Fig Other dimensions of the slb re comprtively lrge so tht het trnsfer my be considered s one-dimensionl in the x-direction, s shown The slb hs, constnt therml conductivity k, nd, uniform internl het genertion rte of q g (W/m3) Both the sides of the slb re mintined t the sme, uniform temperture of T w Then, it is intuitively cler tht mximum temperture will occur t the centre line, since the het hs to flow from the centre outwrds Therefore, it is dvntgeous to select the origin of the rectngulr coordinte system on the centre line, s shown Let us nlyse this cse for temperture distribution within the slb nd the het-trnsfer to the sides Assumptions: i One-dimensionl conduction, ie thickness L is smll compred FIGURE to the dimensions in the y nd z- directions ii Stedy stte conduction, ie temperture t ny point within the Plne slb with internl het genertion both slb does not chnge with time; of course, tempertures t different sides t the sme temperture points within the slb will be different 3 iii Uniform internl het genertion rte, q g (W/m ) iv Mteril of the slb is homogeneous (ie constnt density) nd isotropic (ie vlue of k is sme in ll directions) We, know, Fourier s one dimensionl stedy stte het flux flow eqution in Crtesin co-ordintes per unit volume per unit time is, (A) Integrting Eq(A) X C Integrting gin, C C (B) Two BC's re required to solve this second order differentil eqution BC's: i t,, since temperture is mximum t the centre line ii At, iii At, Substituting BC (i) in eqution (B) C C (C) Substituting BC (iii) nd Eq(C) in eqution (B) C (D) Substituting BC (ii) in eqution (B) C (E) Subtrcting Eq (E) from Eq (D) C, C 8
Qulity Solutions HT Module 2 Pper solution wwwqulitytutorilin Substituting vlues of C nd C in Eq (B) (Required Answer) be eun, Above eqution gives the mximum temperture difference within the slb(l is the hlf-thickness, nd b 2L) Q10 write short note oncriticl rdius nd thickness of insultion Q11Criticl rdius with its pplictions Q12 Explin the criticl thickness of insultion with its significnce [ [D-14-Q-5(b)(3)-4M]] [M-13-Q7 (b)-7m] [D-13-Q1 (e)-5m] Criticl rdius of insultion: Het loss from n insulted pipe vries s rdius of insultion Het loss is minimum t criticl rdius The thickness of insultion corresponding to criticl rdius of insultion is known s criticl insultion thickness If we insulte beyond this point, het loss rte increses Insultion is pplied on metllic wires or pipes to reduce het trnsfer to surrounding As the insultion thickness increses therml resistnce due to conduction lso increses which cuses the reduction in het flow rte, but t the sme time the outer surfce re of insultion which is responsible for convection het trnsfer lso increses so the net effect of insultion is to increse the het trnsfer insted of decrese These two opposing effects leds to n optimum insultion thickness As we increses the thickness of insultion the het flow rte increses linerly nd its vlue is mximum t prticulr rdius of insultion ( rc ) Criticl rdius of insultion ( rc ) for cylinder ie pipe If we increse insultion rdius beyond criticl rdius then het flow rte decreses s shown in figb At certin rdius r the het trnsfer through insulted pipe is exctly equl to het trnsfer through bre pipe Rdius of insultion is selected bsed on ppliction(significnce:) s below1 In cse of electric wire I Q Therefore we hve pply criticl rdius of insultion (we required q qmx) 2 In cse of refrigertion pipe t entry to evportor the het trnsfer to surrounding should be minimum s possible Therefore we hve to pply insultion rdius greter thn r s shown in fig 9
HT Module 2 Pper solution Qulity Solutions wwwqulitytutorilin Q13 A pipe, 2 cm dimeter,t 400C is plced in (i)n ir flow t 500C with h 20 w/m2k or in (ii) wter t 300C with h 70 w/m2k Find the het trnsfer per unit length of the pipe nd comment on the results in both cses Ans: [D-14-Q-1(c)-5M-CBGS] For ir t 500C: Since we know tht, Het trnsfer is given by, A / For wter t 300C Het trnsfer is given by, A / Q14 Find the het flux cross composite slb of width 025m nd 015 m of conductivity 388 W/m-K nd 250 W/ m-k respectively when its one surfce is t 150 C nd the other surfce is t 40 C Also find the temperture t the joint of two mterils of the slb [D-13-Q1 ()-5M] Ans: Given: Figure with given dt is s shown in fig,,, / To find: (i) het flux (ii) temperture t the joint of two mterils of the slb Assumption : stedy stte het trnsfer Anlysis: Het trnsfer to composite wll by conduction is given by, e A, e e A, (1) o (2) (3) 10 A A From eqution (1), un / / /
HT Module 2 Pper solution Qulity Solutions wwwqulitytutorilin Now putting vlue of in Eq(2), C Q15 A rectngulr slb (k 10 w/m-k) of thickness 15 cm nd inside temperture of 4000C is insulted by mterils of thickness 10 cm ( K 30 w/m-k) The mbient ir is t 280C nd the outside convective het trnsfer coefficient is 15 w/m2k Determine the stedy stte het trnsfer per init surfce re nd the temperture of outside surfce of the slb nd the insultion (D-14-Q1 ()-5M-CBGS) Ans: Given: Figure with given dt is s shown in fig Referring to fig c c, /, / 15 w/m2k Assumption : stedy stte het trnsfer Anlysis: Het trnsfer to composite wll by conduction is given by, e A, e e A, (1) o (2) (3) A, A (4) From eqution (1), / Now putting vlue of in Eq(2), A un / C Now, putting vlue of in Eq(3) or Eq(4), we get, 11 / C
HT Module 2 Pper solution Qulity Solutions wwwqulitytutorilin Q16 The composite wll of furnce is mde up with 120 mm of fire cly [K 025(100009 t)w/m C] nd 600 mm of red brick (k08 W/m C)The inside surfce temperture is 1250 C nd the outside temperture is 40 C Determine: [M-14-Q4 ()-10M] i The temperture t the lyer interfce nd 2 ii The het loss for 1 m of surfce wll Ans: Refering to fig [ /; ] To find: eeue e nefce,, e f f e funce, Assumption: stedy stte het trnsfer Anlysis: eeue e nefce, : i Aee/en e cnduc f fe c, [ [ ] ] e ence f e fe c, [, e ence f ed bc, ] e f f funce, unde ed e cndn e e un f e f u ec e en cnden e f u e ed bc, e e f eun nd, e bn 12
HT Module 2 Pper solution Qulity Solutions wwwqulitytutorilin b n, e e e, : e f f e funce, Q17 An electric hot plte is mintined t temperture of 3500C nd is used to keep solution boiling t 950C The solution is contined in cst iron vessel of wll thickness 25 mm which is enmeled inside to thickness of 08 mm The het trnsfer coefficient for the boiling solution is 55 kw/m2k nd therml conductivities of cst iron nd enmel re 50 nd 105 w/m2k respectivelyclculte 1) the overll het trnsfer coefficient2) the rte of the het trnsfer per unit re [D-14-Q2 (b)-10m] Ans: Given: e e e nfe ceffcen, ; I I / e e f e nfe e un e, A 13 /
HT Module 2 Pper solution Qulity Solutions wwwqulitytutorilin Q18 A stndrd cst iron pipe (inner dimeter 50mm nd outer dimeter 55mm) is insulted with 85 % mgnesium insultion ( K002W/m0C) Temperture t the interfce between the pipe nd insultion is 3000C The llowble het loss through the pipe is 600W/m length of pipe nd for the sfety, the temperture of the outside surfce of insultion must not exceed 1000C Determine minimum thickness of insultion required Given: efen fue (D-15-Q2(b)-08M) e e ee en f e { } To find: ℎ Anlysis: F ed e e f, π ℓn π ℓn π ℓn π ℓn, b eun be e nu cne f nun Q19 A refrigertion suction line hving outer dimeter 30 mm is required to be thermlly insulted The outside ir film coefficient of het trnsfer is 12 / The therml conductivity of insultion is 03 w/mk i) Determine whether the insultion will be effective; ii) Estimte the mximum vlue of therml conductivity of insulting mteril to reduce het trnsfer; iii) Determine the thickness of cork insultion to reduce the het trnsfer to 22 % if the therml conductivity of cork is 0038 W/mk M-16-Q3(b)-06M Given: Objective: decee e nfe e To find: ee e nun be effece u ue f educe e nfe cne f c nun, f 14 nd
HT Module 2 Pper solution Qulity Solutions Anlysis: Cc du, <, wwwqulitytutorilin e en nun n be effece e n ee e ue f F nun be effece: ence, e u ue f iii), e cn ue educe e nfe nce nn en bu neecn nde cnnece f ence nd be e e ccu A 15 A du f c nun e e ccu A ℓn A
HT Module 2 Pper solution Qulity Solutions Given: ℓn π A ℓn π wwwqulitytutorilin π A cne f c nun π Q20 The inner nd outer rdii of hollow cylinder re 50mm nd 100mm respectively The inside surfce is mintined t 300 C nd the outer surfce t 100 C The therml conductivity vries with the temperture over this temperture rnge s k(t) 05 05X10-3T, where T is in C nd k(t) is in W/m C Determine (i) het flow rte per meter length of cylinder [D-13-Q4 (b)-10m] (ii) Temperture t mid thickness of the shell Given: Referring to figure c c [ ] To find: I Het flow rte per meter length of cylinder II Temperture t mid thickness of the shell Assumption: stedy stte het trnsfer Anlysis: [, [ ] ] cn [ β] e cn e, e nfe e, nd β F fndn nd en eefe, [ β ] [ 16 ] / en eeue en eeue
HT Module 2 Pper solution Qulity Solutions e ence f cndc, n un: π n π / un: eeue d cne f e, e From Fourier s eqution, we hve A π β π en e u nd nen f nd d π π o d π o n [ β [ β { π o 17 ( )] β β n π β d ] { β, β d n } } wwwqulitytutorilin
HT Module 2 Pper solution Qulity Solutions wwwqulitytutorilin Q21 An insulted het pipe of 16 cm dimeter is covered with 4 cm thick lyer of insultion (k 09 W/m C) nd crries process stem Determine the percentge chnge in the rte of het loss if n extr 2 cm thick lyer of lgging (k 125 W/m C) is provided Given tht surrounding temperture remins constnt nd the outer surfce het trnsfer coefficient for both the configurtions is 12 W/m2 C [M-13-Q2 ()-10M] Given: Figure with given dt is s shown in figure FIGURE Exmple () Pipe with one lyer of lgging FIGURE Exmple (b) Pipe with two lyer of lgging Dt: : ℎ : Since this is cse of stedy stte, one-dimensionl conduction with no internl het genertion, therml resistnce concept is pplicble To find: % chnge in het loss Anlysis: In cse (i): Therml resistnce is the sum of conduction resistnce in lgging lyer number 1 nd convective resistnce over its surfce Conduction resistnce of the pipe mteril nd the convective resistnce between stem nd inner surfce of pipe re neglected, since no dt is given See Fig Exmple In cse (ii): Therml resistnce is the sum of conduction resistnces in lgging lyers number 1 nd number 2 nd the convective resistnce over the surfce of lgging lyer number 2 Obviously, Rtotl for cse (ii) is more thn tht for cse (i); ccordingly, het trnsfer rte for the second cse, Q2 is less thn tht for first cse, Q1 F n O, e e: o nd o ee e e eeue dffeence, c e e f b ce eefe, / 18 /
HT Module 2 Pper solution Qulity Solutions And, % cne n e f e ( ) : n ] un: / / π un: / / n [ un: / wwwqulitytutorilin Q22 A slb of 12 cm thickness nd generting het uniformly t 106 W/m3 hs therml conductivity of 200 W/m C Both surfces of the slb re mintined t 150 C Determine i) the temperture, temperture grdients nd het flow rte t plnes 3 cm nd 9 cm from one of the two plnes, ii) mximum temperture nd its loction : A figure with given dt is s shown in figure efe f cne f b, e e f e nfe, / e cnduc f b, / e eeue f ec ufce ee eeue f e ufce [M-13-Q 3(b)-10M] To find: e eeue e d nd ue ne: Anlysis: e eeue e d nd ue ne: e eeue dbun e b: A d ne: ( ) ( A ue ne / nd / 19 L ) An
HT Module 2 Pper solution Qulity Solutions, ( ) ( An ) An e e f e nd eeue den e d nd ue ne: F un e, e e f / A / eeue den A / / 20 d d / / / d d A An An An An wwwqulitytutorilin