J.P. Holman: 3.09) T sur := Use table 3-1 to determine the shape factor for this problem. 4π r S := T sphere := 30K r 1. S = m k := 1.

Transcription

1 .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

2 .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 Temp

3 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 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 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

4 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 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 Figure give Fo : 0 Thu the time for thi i Fo r τ : τ 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 T can : T 00 + ( T 0 T 00 ) ϑ 0 can T can 0.56 C

5 .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 a + b T m The boundary layer become turbulent at the critical Reynoold number Re cr : 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 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 u oo x δ otherwie 0.0 δ( x) x

6 .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: g ρ : ρ g c p : g c p g m : m : m m µ : ρ µ 0. Pa µ c p Pr : Pr.99 0 α : α 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 d i Re Pr Nu :.66 + Nu 9.6 d i 0.0 Re Pr +

7 a impler eq. i w m : w m 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 ( 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 g d i µ w : w ρ w µ w 0.06 Pa Re Pr > 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 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!!!!!!!!!!

8 .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 : µ µ : Pa : m g ρ 0.06 : α : α m µ c p Pr : Pr m ρ c p u D H Re : Re Highly turbulent! Dittu Boelter, eq. 6- Nu : 0.0 Re 0.8 Pr 0. Nu 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 µ w f 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

9 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 ε ε : ρ ice x r ice A m µ µ air 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 The flow i firt laminar and air at the end of the plate it become turbulent! τ a 9. 0 Ue eq 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 m τ b : ε h b T 00 T ice τ b τ b min

10 .P. Holman, 7.09) : m C : T : 00 C C T air : 5 C C T + T air : H: m C.5 Table A-5: ρ : g ρ 0.77 g m - 50 c p : 500 µ : : g m Pa c p.07 0 g µ Pa 0.09 m α : ρ c p µ c p µ Pr : : α m ρ Pr m β : β g β T T air H Gr : Gr Gr Pr Tab (7-) > C : 0.0 n : Equation 7-5, > Nu : C ( Gr Pr) n Nu Nu h : h 6.5 H m q: h H T T air q. 0

11 .P. Holman, 7.7) :.m C : T : 55 C C T air : C C T + T air : 0.65 Table A-5: ρ g : H: 7m 00 c p : 50 µ : : g 0.06 α m Pr 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 µ 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 : Ra 0.9 Pr q q: h H T T air q.50 0 Nu.8 0 Nu h : h.85 H m