principles of heat exchanger design - iv-mod

21
Principles of Heat Exchanger Design (iv) ---- Analysis and Design of Fins 中興大學 機械系 沈君洋教授

Upload: covek-vicko

Post on 24-Apr-2015

192 views

Category:

Documents


9 download

TRANSCRIPT

Principles of Heat Exchanger Design (iv) ---- Analysis and Design of Fins

: Heat Transfer with Applications by Kirk D. Hagen (1). --- (i) (ii) Fouriers law (iii) Newtons law of cooling (iv) StefanBoltzmann law (v) (2) --- (i) - (ii) - (iii) (iv) (3) --- (i) (ii) (iii) (iv) (v) (4) (5) -- (i) (ii)(iii)(Cf,x) friction coefficient (iv) (v) (vi)Reynolds Analogy and Chilton-Colburn Analogy (vii) (viii) (ix) (Tube Bank) (6) -- (i) (ii) (iii) Darcy--friction factor (iv) (v) (7) (8) --- LMTD (9) LMTD (10) LMTD (11) --- Ntu (12) Ntu

(4) (i) (ii)

(i) CASE 1 t = 0.3 mm, =12 mm, L =100 mm, kcopper= 390 W/m-K, h = 50 W/m2-K T0 = 1000C, T = 300C m = [(2h)/(kt)]0.5 = 29.2 m-1, 0.35, : = fin efficiency = 0.97

=

tanh(mb) mb

Q = kA(T0 T )m tanh(mb)

Q = (390 W/m-K) [(0.1 m)(3x10-4 m)] (100-30)(K) (29.2 m-1) tanh(0.35) = 8.044 (W)

= c1e-mx + c2 emx

(T0 T ) e bm 0 e bm = bm = 0 .668 0 c1 = bm e + e bm e + e bm (T0 T ) e bm 0 e bm c2 = = bm = 0 .332 0 e bm + e bm e + e bm

At the tip, x = 0.012 m = T(x) T = 0.668 (100-30) e -(29.2)(0.012) + 0.332 (100-30) e (29.2)(0.012) = 65.90C Ttip = 65.9 + 30 = 95.90C 1000C

CASE 2 t = 0.2 mm, =12 mm, L =100 mm, kcopper=390 W/m-K, h = 50 W/m2-K T0 = 500C, T = 5000C m = [(2h)/(kt)]0.5 = 35.8 m-1, 0.43, : = fin efficiency = 0.95

(T0 T ) e bm 0 e bm = bm = 0 .702 0 c1 = bm bm bm e +e e +e (T0 T ) e bm 0 e bm = bm = 0 .298 0 c2 = bm bm bm e +e e +e At the tip, x = 0.012 m = T(x) T = [(0.702) e -(35.8)(0.012) + (0.298) e (35.8)(0.012)] (50-500) = - 4110C Ttip = - 411 + 500 = 890C 500C !

CASE 3 t = 0.2 mm, =12 mm, L =100 mm, kcopper=390 W/m-K, h = 200 W/m2-K T0 = 500C, T = 5000C m = [(2h)/(kt)]0.5 = 71.6 m-1, 0.86, : = fin efficiency = 0.83

(T0 T )e bm 0 ebm c1 = bm bm = bm bm = 0.848 0 e +e e +e (T0 T )e bm 0 e bm c2 = bm bm = bm bm = 0.152 0 e +e e +e At the tip, x = 0.012 m = T(x) T = [(0.848) e -(71.6)(0.012) + (0.152) e (71.6)(0.012)] (50-500) = - 3230C Ttip = - 323 + 500 = 1770C 500C!

(ii) Case 1:t = 0.3 mm, Ro = 20 mm, Ri = 8 mm, kcopper= 390 W/m-K, h = 50 W/m2-K m = [(2h)/(kt)]0.5 = 29.2 m-1, : = fin efficiency = 0.93 = m(Ro-Ri) = 0.35, Ri/Ro = 0.4

Case 2:, If h = 200 W/m2-K (increased by fourfold) m = [(2h)/(kt)]0.5 = 58.5 m-1, Ri/Ro = 0.4 : = m(Ro-Ri) = 0.702,

= Q/Qmax = fin efficiency = 0.82 h Qmax

Qmax ! !

fin !

h h() () h ()

500C

1500C 500C 1500C

: (i) (ii) fin (iii) 1500C

(1) () (400C) (copper, 0.4 mm )

40 mm

10000C

82 mm

Plate Fin (k = 390 W/m-K, t = 0.4 mm )

(T2 = 400C) h2 = 10 W/m2-K2L =

40 m m

500C

(T1 = 10000C) h1 = 20 W/m2-K

(Tb) 800C ?

:

T ( x) T ,mean Tb T ,mean

cosh[m( L x)] = cosh(mL)

where T ,mean

h1T1 + h2T2 = mean fluid temperature = 6800 C h1 + h2

m

h1 + h2 = 13.87 (m 1 ) , cosh(mL) = 1.0387 kt

on the center line : x = L cosh[m( L x)] T ( L) = T ,mean + (Tb T ,mean ) cosh(mL) = 1020 C !

() Q = (Ti-T1)/[1/(hiAi)] = (Ti-T1)/[1/(2r1Lhi)] (Ti-T1) = Q [1/(2r1Lhi)] = Q Rth,conv1 Q = (T1-T2)/[ln(r2/r1)/2k1L] (T1-T2) = Q [ln(r2/r1)/2k1L] = Q Rth,1 Q Ti Tth,conv1 T1 Tth,1 T2 Tth,conv2 To 2r2 2r1 1 2 To (gas)

Q = 0 At ho (T2 To )

(T2 T0 ) = Q [1 /( 0 At ho )] Q R th , conv 2 Ti To = Q [Rth,conv1 + Rth,1 + Rth,conv2] = Q Rth,total Q = (Ti To)/Rth,total Rth,total = Rth,conv1 + Rth,1 + Rth,conv2

Ti (liquid)

Overall heat transfer coefficient (U) - Q = (T) / Rth,total = UoAt (T) = UiAi (T) Ai = 2r1L At = total external surface area (fin area+ base area)

Ui = Overall heat transfer coefficient based on Ai = 1/(Rth,totalAi) = 1/[(Rth,conv1 + Rth,1 + Rth,conv2)Ai] = =1 /[(1 /[

1 ln( r2 / r1 ) 1 + + ) Ai ] o ho At hi Ai 2k1 L

A 1 Ai ln( r2 / r1 ) 1 + + ( i )] 2 k1 L hi o ho At

Uo = Overall heat transfer coefficient based on At = 1/(Rth,totalAt) = =1 /[( 1 ln( r2 / r1 ) 1 + + ) At ] hi Ai o ho At 2 k 1 L A ln( r2 / r1 ) 1 A 1 1 /[ ( t ) + t + ] 2 k 1 L hi Ai o ho

U !

hi ho () ( )hi ho

(h0)T2 < 1000C P2

T1 ,P1 m

1000C

1000C

Q = () = ( m : ) x (cp: ) x (T2 - T1) Qmax = = () x () x (1000C - T1) Effectiveness () = Q / Qmax , U0 = Ntu ( m cp)min/At

() Ntu = At U0/(m cp)min h0

m m h0

U0 h0 :

At ln( r2 / r1 ) 1 At 1 U o = 1 /[ ( ) + + ] hi Ai 2 k 1 L o ho 1 1 /[ ] o ho o hoNote: 1 At ( ) hi Ai

At ln( r2 / r1 ) 2k1 L

!

8 fins / in

1f/4

2f/4

7.75 fins / in

5/8

St Pr

2/3

Nu =( )Pr 2/3 Re Pr h = Pr 2/3 c p U

(U)