Heat Transfer. Solutions for Vol I _ Classroom Practice Questions. Chapter 1 Conduction
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2 Heat ransfer Solutions for Vol I _ lassroom Practice Questions hapter onduction r r r K K. ns: () ase (): Higher thermal conductive material is inside and lo thermal conductive material is outside K K r r r Q L ln ln K K Q.6(KL()) () From () and () L Q r r ln ln r r K K Let, r mm, r mm, r mm L Q ln ln K K Q.(KL).. () No, ase (): Higher thermal conductivity material outside, loer thermal conductivity material is inside. Q > Q Means that, loer thermal conductivity material should e used for inner layer and higher thermal conductivity material should e used for outer layer so that heat transfer ill e loer insulation is effective.. ns: (a) r 5mm r 5+5 mm r mm Q r r ln ln r r KL KL K 65 ln ln K 5K r r r K
3 : : ME GE _ Vol I _ Solutions Q K 65 ln ln K % decrease in heat transfer %.66. ns: (c) H m, L m,.5 m i,, K.5 W/mK h i.5w/m K, Q 6 h h. ns: (d) i 6 8 J K r,. 8 K r K r r 5K r r r r r Due to steady state H.. h W/m K r r r K r Kr.7 o ns: () o avoid condensation in the uilding, the inside all temp should e greater than or equal to DP o h i 8 h o W/m K i K. H. from inside air to inside all H. from inside all surface to outside air.7.7 δ 8. δ δ.7 m δ. ommon Data for Q. 6 and Q max 6 Q Q Kr r r r K rr r r x L.m
4 : : Heat ransfer Q G 8 MW/m 8 6 W/m 6, W/m K For -D steady state ith heat generation equation is d Q dx G d QG d QG x dx dx Q G x x (i) Where, & are constants that can e evaluated y oundary conditions. t, x, 6 6 t x L. m,, 6 8 Q G x x (ii) o get maximum temperature d dx Q G.. 6 x x 6 8 Q G 7. By putting the value of x 5 m in eq.(ii) ns (c) Q 9. ns: (a) all all Resistance. 5W 5 W, K W/mK, R.6 K/W Q R d cm cm R R W 5 m 5 mm
5 : : ME GE _ Vol I _ Solutions. ns: () Q i r r Kr r r ri i at radius.5 cm Q Kr r i ns: (d) K. 5 ritical radius of cylinder, r c h.5 m 5 mm ritical thicness 5 5 mm. ns: (a). ritical radius, r c h mm cm For cale as r (.5 cm) > r ( cm), the heat transfer decreases y adding insulation.. ns: (c) Q Q Q R eq R + R L L L + L L + L Q + Q L L 5. ns: () r 5 mm, h W/m K, i. W/mK Max. heat flo occurs at critical radius or diameter i d c r c h h. i.6 6 mm. ns: (c) r mm, K.75 r c ( r ). m h 5 r c. mm, s r ( mm) < r c (. mm) he heat transfer increases y adding insulation till r r c and then it decreases.
6 : 5 : Heat ransfer hapter onvection. ns: (). ns: (a). ns: (a) f., , V 5 m/sec p J/gK Sat. Stream p Pr f St Pr h f V P Sat. ater P h f V P r P W/m K. ns: () L cm, p 5,, Q 8 W Q h. gl Grashoff numer (Gr) Gr L (eeping others constant) Nusselt numer (Nu) (Gr. Pr) / for laminar flo Nu (Gr) / Nu (L ) / hl / L h L / Q h Q h Q h dl Q h L Q L / L Q L / / Q L Q L 8 L / / 8 6 L cm L 5. ns: (c) 6. ns: () thicness of hydrodynamic oundary layer.5 Pr p t.5mm 7. ns: (a) 8. ns: (d) Q t Q top + Q ottom + Q side 6 h h + h + h 6h h + h + h h h h h 6
7 : 6 : ME GE _ Vol I _ Solutions 9. ns: (). e 5y t No slip region, d dy Q Q cond h d dy y e 5y (e 5y )( ) 5 e Su () in () 5y convection (). () 5e 5y ( ) h s ( ) 5 h y. 5 e h 5 W/m K Vd Vd 5. R e < Hence flo is laminar constant Nu.66 hd h d. (i) ns: (c), (ii) ns: (d).6 W/m K g. W/m K, W/m K. K/m y s Q Q g d dy d.6 dx g. But, Q Q conv. ns: () d dy y.6 h 8 d g dx g.5 h(8 ) 75 W / m. ns: (). W/mK R e 5 means that the flo is laminar D cm.m For a fully developed laminar flo hrough pipe (i) With constant heat flux Nu.6 constant hd.6. h.6. (ii) With constant all temp hd Nu h. K 6.6 W / m W / mk K
8 : 7 : Heat ransfer hapter Extended Surfaces FINS 6 osh(.6) o. ns: (c) o,, 6. ns: (d). ns: () cm 55 5 In case of rod of length L ith insulated tip, the temp at the end of tip osh m L x osh (ml) osh ml osh osh ml 6 5 ml ml osh - ().6.6 m L In case of rod of length L ith oth ends maintained at same temp of, at the center the temp gradient is zero, means that from one side upto the center of long rod, it can e treated as short fin of length L L, for hich temp distriution is osh(m(l x)) 6 osh() osh(ml) osh(ml) 6.6 osh L L L L + D m +.5 cm.5 m h d cosh m L x cosh ml cosh cosh cosh m L x cosh ml cosh cosh cm cm cm
9 : 8 : ME GE _ Vol I _ Solutions. ns: (d) cm () 5 long fin 5 t x 5 mm x cm () 6at x L Given, W/mK,? x cm, x cm e e mx mx m x mx e e m x m x x m h x m m d x x x x cosh(ml) osh(ml).57 ml.99 L L.99 h d 6. ns: (c).99 m m cm 9 cm K W/mK 5W / mk mm 5. ns: () If, x L cosh m L x cosh ml cosh ml (cosh ) h 8.5 W/m K Q loss hpk P d..68 m d... 8
10 : 9 : Heat ransfer Q loss W h W/m K Q sin glefin hp s tanh ml 7. ns: (c) e mx m ln x 5 8 ln W/m.K h d L m m 6 P. m.9 6 m Q sin gle.58 W 5.8 otal heat generated n n.8 W Q dissipatedinsin glefin.6 8. ns: () 5 Effective length, L c.5mm L c.5 m Q hp tanh ml c. ns: (c) h m 8.9 d 5 Q Q 5W 5 5 tanh ns: Q gen 8 W 7, l 7 W/mK,
11 : : ME GE _ Vol I _ Solutions hapter ransient Heat onduction. ns: (c) D mm m K 5 W/mK, 8 Kg/m G J/gK, h 56 W/m K h V e t t t t Given, t t. (t t ) e e..6 sec. ns: () d Qentering dx x 5 5x + x + 5x 5x d d d dx dx dx 96x x.5. ns: (c) 5, m 5 gm ime sec m m.5 V V 9 ccording to the Neton s la of cooling the rate of cooling is directly proportional to difference in temperature. d ( ) d d K d d K d d 5 + x + 5x 6x dx d + 9x 8x dx d 96x dx d ooling rate is d ooling rate or heating rate to e maximum d d or minimum dx d ln K e Kt e K 5 K./sec No next sec means that at t emp? e.() 5 5.9
12 : : Heat ransfer. ns: (a) For lumped analysis to e used Bi. hl a. (L for solid cue is ) 6 Where, a is a side of cue) ha max. 6 a max. 6 h m 5 5. ns: (d) d 5 cm.5 m i 5, 9, 5 Lumped capacity analysis can e applied e cause internal temperature gradients are neglected. i e 5 9 e 5 9 h Vp 5 8 V e sec 7.7 min V d [ (for sphere) ] 6 6. ns: (a) r 5 r r i.5 m L.5 m. m /hr d. dt r d dt d dt 7. ns: (d) inside outside 8. ns: (c) d dr r o. m 5.5 W/m.K d r. dr...5 r i d dr d dr r d ri dr 8 85 i r i.6 (8 7 r i ).6 [8 7 (.5)] 5.5/ hr r o r o d. dr d dr d ro. dr 8r o 85r o /hr 9. ns:.5 (range. to.6) It is given in the question that at all times the surface is remaining at temperature of amient.
13 : : ME GE _ Vol I _ Solutions (F O ) (F O ) B L c B L c c B 6 c B B.. B.5 hours 5 6 hapter 5 Radiation. ns: () 8 K, 5 K.8,. 6 ( ) Q (8 5 ) W/m.8, K, 6K s.5 Q s Q s s s s 57K q Q s s s s.7w / m s. ns: ().5.5
14 : : Heat ransfer.5 Q ns: () F +F F F F F F DL D R W / m F +F F ns: (d) K, 5 K,.7 Irradiation of ody W/m.5 6. ns: (d) K, 5 K.7,.8 Q.7 W/m 7. ns: (c) Q Q S W.S 79 Q WS 79Q S E E Given S.5.8 N.8 79 E E N S N ns: (c) % Reduction 75% Q S.75 Q Q Q S WS W.S n n.5
15 : : ME GE _ Vol I _ Solutions 9. ns: (a).9 c recorded temperature K all temperature K fluid temperature? h 8. W/m K Q conv Q radation h c c c.59 K ns: () F + F + F + F F ( ).75.5 F F 8. F.5.5 F d pipe cm, pipe.8, L p m Q pipe pipe P p all all s room area is very large compared to pipe, hence Q d p plp p all pipe d p p pipe p Q L (67 ) W. ns: (d). ns: (c). ns: () () () p K
16 : 5 : Heat ransfer. ns: (). ns: (a) hapter 6 Heat Exchangers oil ater 7 c 8 max h 5.5 min.5m, U W/m NU U min ns: (c) NU.5, h c For counter flo NU.5 NU.5. ns: (d) Given,, NU c NU NU m 7 m No Parallel flo Heat exchanger m m W W m m 8 5. ns: (c) Parallel flo Heat exchanger P c p e e NU () () max 58 7 () min 6 5 5
17 : 6 : ME GE _ Vol I _ Solutions 5 LMD 5 n n NU 57.9 max 5 8. LMD ns: () NU NU NU.8 NU NU 7. ns: () h, h 5, U W/m K, 5m, parallel flo H.E exp ( NU) U 5 NU min exp.5 c c min h c c 5 5 c Because the hot fluid is steam and hose temp remains constant, the flo is immaterial 8 9, 8 9 m n m m P U m ns: (a), m c 5 6 h, P 87 J/g K 8, U W/m K.7 g / sec
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