.P. Holman:.09) T ur : 0 Ue table - to determine the hape factor for thi problem. D :.m r : 0.5m π r S : T phere : 0 r D S 7.0 m :.7 m Ue eq. - to calculate the heat lo. q : S T phere T ur q 57.70
.P. Holman:.5) Generally, -D with no internal heat gereration and teady tate: dt dx dt + 0 dy T B : 700 Here, x y and hence T A : 00 T C : 00 T + T + T + T i+, j i, j+ i, j i, j T i, j So, Fourier law becomme a ytem of linear equation. T D : 500 Set a gue value for all temperature T A + T B + T C + T D T : T : T T : T T : T Given T C + T B + T + T T T + T + T A + T D T T C + T + T + T D T T + T B + T A + T T Temp : Find T, T, T, T Temp 87.5 0 Temp 87.5 6.5 7.5.5
Exam problem P.6) A urf 0.006m T i : 65 : h urf : 50 m dl : 0.liter T end : 50 T 00 : 0 V: dl ma : 0.g ma ρ : ρ 000 g V c p : 70 g h l : 8 m A l : 0.0m In the text, it i aumed that inner thermal reitance i neglected! > umped Capacity!!!!, Eq. -5. T T 00 T i T 00 ha τ ρcv e Since we have heat tranfer at everal interface, the uperpoition principal i to be ued, ee page in Handboo by Granryd. ρ c p V τ 0_cylinder : h l A l τ 0_cylinder. 0 ρ c p V τ 0_urf : τ 0_urf.78 0 h urf A urf Now, according to the handboo, page, and the textboo, ection -5. T T 00 T i T 00 τ τ τ 0_cylinder τ 0_urf e e which can be implified to τ T T 00 τ 0_tot e T i T 00 where τ 0_tot : + τ 0_tot.695 0 τ 0_cylinder τ 0_urf which alo could be expreed a ρ c p V ρ c p V τ 0_tot τ 0_tot : τ 0_tot.695 0 Σ( h A) h l A l + h urf A urf So, olving for the time, for which T: T end τ τ 0_tot ln T T 00 : T i T 00 Thu, τ 687. τ.55 min
Exam problem P.7) T 0 : 0 C T 00 : 0 C h out : 00 m Heat tranfer by conduction inide the can! C : d Coe : 65mm d Coe r : : 5mm coe 0.58 : c p : 00 ρ : 000 g m g a) Calculate how long tim it tae to cool the Coe to T: C if the can i conidered to be long. b) Calculate the temperature the coe will have at the calculated intance if the edge are conidered. a) h out r Bi : Bi 5.60 coe α τ coe Fo : where r α : ρ c p α.8 0 7 m In fig.88 c, the mean temperature of a infinite cylinder i already calculated. ϑ m T T 00 ϑ 0 T 0 T 00 T T 00 T 0 T 00 0.6 Figure give Fo : 0 Thu the time for thi i Fo r τ : τ 76.87 τ.78 min α b) If conideration i taen to the edge, what i the temperature? h out See the edge a the pecial cae of plate Bi : Bi 9.9 coe Figure.78-c in handboo. α τ Fo : Fo 0.0 ϑ m 0.88 Hence, the average temperature in the can i ϑ 0 plate ϑ m 0.6 0.88 T can : T 00 + ( T 0 T 00 ) 0.6 0.88 ϑ 0 can T can 0.56 C
.P. Holman, 5.) T oo : 90 C p : bar u oo : 0 m bar : 0 5 Pa C : x δ :.5cm δ Chec if the boundary layer i laminar or turbulent at the location. u oo x δ Re x : where Re x.9 0. 0 5 a + b T 90 + 7.5 6.5 50 00 m The boundary layer become turbulent at the critical Reynoold number Re cr : 5 0 5 50 00 0.76 0 6 5.90 6.5 0.76 5.90 m. 6.5 (Table A5) Re x > Re cr 0 > No, the boundary layer i till laminar. So, eq. 5-a i ued: δ x 5.0 Re x and 5.0 x δ δ : δ 0.679 mm Re x How doe the entire layer loo lie? x cr : Re cr u oo x cr 0.69 m δ( x) : 5 x u oo x δ if x < x cr 5 u oo x δ x 0.8 0.56 u oo x δ otherwie 0.0 δ( x) 0.0 0.0 0 0 0. 0. 0. 0. 0.5 x
.P. Holman, 6.09) C : Engine Oil a Fluid d i :.5cm T in : 8 C u: 0 cm : m T wall : 65 C Etimate the total heat tranfer and the outgoing oil temperature! T in + T wall Aume the average bul temperature of the oil for propertie! T bul : T bul 5.5 C Set T bul : 50 C for implicity! Table A-, page 657: 0 876.05 g ρ : 60 86.0 50 ρ 870.05 g c p : 0 60 96 07 50 g c p.006 0 g 0 0.000 m : 60 0.89 0.69 0 m 50 0 0. : 0. 60 0.0 50 m m µ : ρ µ 0. Pa µ c p Pr : Pr.99 0 α : α 8.8 0 8 m ρ c p Recalculate the propertie if neceary when the outlet temperature i nown! u d i Re : Re.55 AMINAR FO Contant wall temperature, ue eq. 6-9. d i 0.0668 Re Pr Nu :.66 + Nu 9.6 d i 0.0 Re Pr +
a impler eq. i 6-0 60 w 0.89 0 m : 80 0.75 0 w 7. 0 5 m 65 60 ρ w : 80 ( T wall T out ) T wall T in + ϑ m T T out T in A : π d i A 0.8 m Q m u π d i : ρ Q m 0.0 g So, inerted in energy balance h π Solving for 86.0 85.0 ( T wall T out ) T wall T in + d i u π d i ρ c p T out T in 65 g ρ w 86.05 g d i µ w : w ρ w µ w 0.06 Pa Re Pr 9.998 > 0, and can be ued d 0. i Nu.86 ( Re Pr) µ : µ w Nu.0 Nu So, the heat tranfer coefficient i h : h 6.665 d i m Energy balance over the tube q h A ϑ m Q m c p T where h T wall h T in + u d i ρ c p T in T out : T ( h + u d i ρ c p ) out.0 and q: Q m c p T out T in q 86.0 T in + T out Controll the aumed Bul temperature T bul : T bul.007 C Redo all the calculation with new thermal propertie of the OI!!!!!!!!!!
.P. Holman, 6.) Air duct H: 5cm u : 7.5 m T : 00 B : 90cm p : atm : m A: H B A 0.05 m P: ( H + B) P.7 m A D H : D H 0.6 m P Table A-5 ρ.77 g : c p : 005.7 µ.86 0 5 µ : Pa :.568 0 5 m g ρ 0.06 : α : α.6 0 5 m µ c p Pr : Pr 0.708 m ρ c p u D H Re : Re.87 0 5 Highly turbulent! Dittu Boelter, eq. 6- Nu : 0.0 Re 0.8 Pr 0. Nu 65.508 Nu h : h 0.58 D H m Ue Petuhov equation, eq. 6-7 & 6-8. n: 0 for gae! µ w : µ f : (.8 log( Re).6) f 0.05 f 8 Re Pr n µ Nu : Nu 85.07 µ w f.07 +.7 Pr 8 Nu h : h 6.89 D H m p: f D H ρ u p 0.80 Pa m Figure 6- give f : 0.05 p: f D H ρ u p 0.8 Pa m
Exam problem P.) C : Given r ice : 5000 T ice : 0 C g ρ ice : 900 g T 00 : C To melt ice x: mmit tae Formelamling: E ρ ice A x r ice E M r ice ρ ice x r ice A A ρ air :.9 g air. 0 6 m : E ε.05 0 5 ε ε : ρ ice x r ice A m µ µ air.7 0 5 air : air ρ air Pa The time required i given by t c ε air : 0.0 p_air : 006 E q τ > ε q'' τ τ d q τ g m 0 q'' µ air c p_air Pr : Pr 0.78 air Heat tranfer to the ice q h A T 00 T ice > q'' h T 00 T ice Thu, τ ε h T 00 T ice b) a) h a : 8 m : 5m h b h b_conv ( + 0.5) u oo : 5 m ε τ a : u oo h a ( T 00 T ice ) Re : Re.88 0 6 The flow i firt laminar and air at the end of the plate it become turbulent! τ a 9. 0 Ue eq. 5-85 τ a 57.0 min 0.8 Nu Nu.68 0 : Pr 0.07 Re 87 τ a.67 hr F.S. 0.8 Nu 0.06 Re 85 Pr : Nu.6 0 Nu air h b_conv : h b : h b_conv ( + 0.5) h b_conv.68 m h b 8.957 m τ b : ε h b T 00 T ice τ b.976 0 τ b 66.68 min
.P. Holman, 7.09) : m C : T : 00 C + 7.5 C T air : 5 C + 7.5 C T + T air : H: m 85.65 7.5 C.5 Table A-5: ρ : 50 500.78 0.708 g ρ 0.77 g m - 50 c p : 500 µ : : 50 500 50 500 00.7 09.5.8 0 5.67 0.0707 0.008 g m Pa c p.07 0 g µ.67 0 5 Pa 0.09 m α : ρ c p µ c p µ Pr : : α 5.79 0 5 m ρ Pr 0.68.599 0 5 m β : β.059 0 g β T T air H Gr : Gr 5.88 0 9 Gr Pr.986 0 9 Tab (7-) > C : 0.0 n : Equation 7-5, > Nu : C ( Gr Pr) n Nu 58.558 Nu h : h 6.5 H m q: h H T T air q. 0
.P. Holman, 7.7) :.m C : T : 55 C + 7.5 C T air : C + 7.5 C T + T air : 0.65 Table A-5: 00.77 ρ g : 50 0.9980 H: 7m 00 c p : 50 µ : : 00 50 00 50 005.7 009.0.86 0 5.075 0.06 0.000 g 0.06 α.5 0 5 m Pr 0.707.59 0 5 m m β : β.0 0 g β ( T T air ) H Gr : Gr.9 0 Gr Pr.58 0 Tab (7-) > C : 0.0 n : Equation 7-5, > Nu : C ( Gr Pr) n Nu.65 0 m Pa µ c p µ α : Pr : : ρ.68 g m - c p.006 0 µ.858 0 5 Pa ρ c p ρ g Nu h : h.0 H m q: h H T T air OR Ue eq. 7-9 Ra Gr Pr : Nu : 0.85 + + 6 0.87 Ra 0.9 Pr 9 6 8 7 q.086 0 q: h H T T air q.50 0 Nu.8 0 Nu h : h.85 H m