Microsoft PowerPoint - Heat Exchanger - IX(mod)

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1 Principles f Heat Exchanger Design (ix) Design Examples Using the LMTD Methd 中興大學機械系沈君洋教授 weeks

2 熱交換器設計原理之教授內容 (1). 熱傳之基本定理 --- (i) 熱傳之簡介 (ii) 熱傳導與 Furier s law (iii) 熱對流與 Newtn s law f cling (iv) 熱輻射與 Stefan- Bltzmann law (v) 簡例說明 () 熱傳導 --- (i) 熱阻與熱阻串聯之觀念 - 直角座標 (ii) 熱阻與熱阻串聯之觀念 - 圓柱體座標 (iii) 簡例說明 (iv) 多維熱傳導公式 (3) 散熱鰭片之基本原理 --- (i) 散熱鰭片之型式 (ii) 一維長方形散熱鰭片之熱傳理論 (iii) 一維環狀散熱鰭片之熱傳理論 (iv) 長方形散熱鰭片之散熱模組 (v) 環狀散熱鰭片之散熱模組 (4) 散熱鰭片之設計與分析 (5) 外流場之熱對流 -- (i) 外流場流體之邊界層 (ii) 外流場流體之邊界層與雷諾數之關係 (iii) 摩擦係數 (C f,x ) frictin cefficient (iv) 熱邊界層之觀念與熱對流係數 (v) 紐塞數與普朗特數 (vi)reynlds Analgy and Chiltn-Clburn Analgy (vii) 二維等溫平板之熱傳 (viii) 二維等熱通量平板之熱傳 (ix) 管排 (Tube Bank) (6) 內流場之熱對流 -- (i) 內流場流體之邊界層與雷諾數 (ii) 管道內層流與紊流之流場 (iii) Darcy 摩擦係數 - --frictin factr (iv) 管道內熱對流係數與紐塞數 (v) 管內熱傳增強之方法 (7) 自然熱對流 冷凝與沸騰之熱對流 (8) 熱交換器之設計 --- LMTD 設計方法之理論基礎 (9) 以 LMTD 方法設計管殼式熱交換器之實例說明 (10) 以 LMTD 方法設計鰭管式熱交換器與其他形式熱交換器之實例說明 (11) 熱交換器之設計 --- Ntu 方法之理論基礎 (1) Ntu 方法設計熱交換器之實例說明

3 (9) 以 LMTD 方法設計熱交換器之 實例說明

4 內容 例 1: 冷凍空調中之管殼式冷凝器之設計 (Shell and Tube Heat Exchanger) 例 : 變更例 1 中冷凝器之設計 --- 冷凝溫度由 35 0 C 昇高至 45 0 C 例 3: 改良例 1 中冷凝器之設計 --- 增加管數例 4: 改良例 中冷凝器之設計 --- 增加管數例 5: 冷凍空調中之 Tw-Passes 管殼式蒸發器之設計 - Water n Tube Side 例 6: 例 5 之蒸發器之修改 -( 管徑由 1 mm 增大至 15.8 mm) 例 7: 例 5 之 Tw-Passes 管殼式蒸發器改為 Fur-Passes 之設計例 8: 液壓油冷卻器 (shell-and-tube type) 之設計 Oil n Shell Side (3/8 steel pipe) 例 9: 液壓油冷卻器 (shell-and-tube type) 之設計 例 8 之變更設計 (3/4 steel pipe) 例 10: 液壓油冷卻器 (shell-and-tube type) 之設計 Oil n Tube Side 例 11: 冷凝器之設計 ( 冷凝溫度 = 50 0 C )

5 例 1: 冷凍空調中之管殼式冷凝器之設計 (Shell and Tube Heat Exchanger) 考慮設計一組空調系統中之水冷式冷凝器 (water-cled cndenser), 此冷凝器為二通路管殼式 (tw-tube-passes shell-and-tube cndenser ( 如圖 ) 此冷凝器之負荷 ( 熱傳量 ) 為 78.3 kw, 冷凝溫度為 35 0 C, 冷媒為 R-, 試計算 (i) 冷卻水於管內之流速 (m/s)? (ii) 冷凝器中 ( 見下圖 ) 銅管所需之長度為多少 (m)? 冷媒入口 35 0 C (sat. vapr) Water cled cndenser Tw-passes shell-and-tube type 冷媒出口 35 0 C (sat. liquid) 管長 (L) 冷卻水入口 5 0 C 冷卻水出口 30 0 C

6 銅管 44 支外徑 : 16 mm 內徑 : 14mm 註 : refrigerant (R-, 35 0 C): k = (W/m-K), = (Pa-s) h fg ( 蒸發熱 ) = J/kg water(8 0 C): w = 996 (kg/m 3 ), w = (Pa-s), k w = 0.61 (W/m-K), c p,w =4190 (J/kg-K), F i = fuling factr (water t cpper) = (m -K/W) cpper: k c = 390 W/m-K

7 ΔT ΔT (35 5) (35 30) LMTD ΔTlm 7.1 ln( ) ΔT 1 0 ΔT1 ln[(35 5) /(35 30)] C 管內冷卻水側 冷卻水之質量流率 = 冷卻量 / (c p,w T) = 78.3 kw / [4.19 (kj/kg-k) x (30-5)(K) = (kg/s) 冷卻水之體積流率 = 質量流率 / 密度 = (m 3 /s) = 5.1 (L/min) 冷卻水於銅管內之流速 (V) = 體積流率 / [ ( 支管 ) x 每支管之內截面積 ] = 1.1 (m/s) Re = w Vd i / w = 996 (kg/m 3 ) x 1.1 (m/s) x (m) / (pa-s) = Dittus-Blter Equatin: Nu d =0.03 Re 0.8 Pr 0.4 (Fully Turbulent Flw) Pr = c p,w w / k w = 4190 (J/kg-K) x (pa-s) / 0.61 (W/k-K) = 5.35 h i = (k w / d i )(0.03) Re 0.8 Pr 0.4 = 5338 (W/m -K)

8 管外冷媒側 44 tubes 13 rws assume g 9.81 (m/s ) ; N 3.38 tubes/rw ; d m 假設 T 是 C 必須驗算! h d gρ h k 0.79 [ ] μ(δt)d 390 (W/m 3 fg 1/ (m/s ) (1153) (kg/m ) (J/kg) (0.083) (W/m K) 0.79[ ] K) (J s/m ) 3.7 (K) (m) 1/4 ( Heat and Mass Transfer by Y. A. Cengel, p. 586) ( Intrductin t Heat Transfer by Frank P. Incrpera, p. 586) 註 : 1 (J) = 1 (N-m) ; 1 (N) =1 (kg-m/s ) ; 1 (W) = 1 (J/s)

9

10 Staggered Arrangement: N = 3.38, 查表可得 : F 0.93 d,n d h h F h (W/m -K) 管內流體熱阻管外流體熱阻管內污垢熱阻管外污垢熱阻) h 1 )( d d ( )F d d ( ) d d )ln( k d ( F h 1 U 1 i i i i i c 代回右式, 管壁熱阻

11 1 1 d d d d 1 F ( )ln( ) ( )F i ( )( ) U h kc di di di hi [ ]ln( ) ( )( ) ( )( ) 3 ()(390) m K ( ) W 管外流體之熱阻相對最大 W Hence, U 1153 ( ) m-k Q = U A (LMTD) Hence A = Q / [U (LMTD)] = (W) / [1153 (W/m -K) x 7.1(K)] = 所有銅管之總外表面積 = 9.4 (m ) 管之長度 = A / [ 總管數 x 單支管之周長 ] = 4.6 (m) 太長!

12 驗算 T 是否接近 C Q = 78.3 kw T B T water T A =35 0 C 1/h F (d /k)ln(d /d i ) (d /d i )(F i ) (d /d i )(1/h i ) 略 Q = h A T = h A (T A T B ) 平均管外表面溫度 T = Q / (h A) = (W) / [9.4 (m ) x 3 (W/m-K)] = C 註 : 嚴格之計算必須重新假設一個 T 值, 再重覆以上之計算, 但差別應已不大

13 (i) 利用電腦解答 : 疊代法 3 fg 1/4 h = Fhd 0.79 F[ ] μ(δt)d gρ h k 由先前熱阻之關係 : Q=h A T U A (LMTD) T 代入 h 與 U 各項後, U(LMTD) h 1 (LMTD) 1 d d d d 1 F ( )ln( ) ( )F i ( )( ) h kc di di di h i T 3 gρ hfgk 1/ F[ ] μ(δt)d (LMTD) gρ h k d d d d 0.79 F[ ] [ F ( )ln( ) ( )F ( )( )] μ(δt)d h k d d d h 3 fg 1/4 1 1 i c i i i i

14 代入各數值後, T 3 fg 1/4 1 1 i c i i i i 1/4 (LMTD) gρ h k d d d d 0.79 F[ ] [ F ( )ln( ) ( )F ( )( )] μ(δt)d h k d d d h (LMTD) gρ h k d d d d F[ ] [F ( )ln( ) ( )F ( )( )] μ(δt)d k d d d h (ΔT) (ΔT) 3 fg 1/4 i c i i i i 7.1 [ ] 1/4 因此, 3/4 T (ΔT) 利用 EES 軟體疊代計算 : T = C 接著利用此 T值, 以相同之方法計算 h 與 U, 然後計算管子長度 (L)

15 (ii) 利用計算機計算 : Gauss-Seidel 疊代法 解答 T = (ΔT) 3/4 疊代次數 初始假設值 ( C) 得到值 ( C) 疊代次數 初始假設值 ( C) 得到值 ( C) 疊代計算後 : T = C

16 (iii) 利用計算機計算 : Newtn-Raphsn 疊代法 利用 Newtn-Raphsn's Iteratin Methd: 3/4 f ( T) =ΔT (ΔT) f ( T) n ( T) n+1 ( T) n ' f ( T) n (ΔT) n (ΔT) That is, ( T) n+1 ( T) n 1/4 1 (3/ 4)(1.878)( ΔT) n 3/4 n 疊代次數 1 3 初始假設值 ( C) 得到值 ( C) 疊代計算後 : T = C

17 Pressure Drp ( p, 管內水流之壓力降 ) p G v1 v A vm v ( )( )[( K c 1 ) ( 1) f ' (1 K e ) p1 p v A v v 1 (i) 入口 v v 1 = free-flw area / frntal area = 空孔率 = 0.15 K c = 0.35 ; K e =0.7 ; = (ii) Flw acceleratin Fanning frictin factr c 1 (iii) Cre frictin ~1.0 (iv) 出口 ~1.0 1 ] ( )( ) [( K 1 ) f' ( 1 K e)] p p A p G v1 c A 1 1 c ( V) v 1 A ( V) v1 4L' [ ]( ) [ f ' ] [ ]( ) [ f ' ] p A p D 1 c 1 ( V) v1 L' [ ]( ) [ f ] p D 1

18 f = Fanning frictin factr ; f = Darcy frictin factr = 4 f L = L = 8.5 (m) ; D = d i = 14 (mm) = (m) V = 1.1 (m/s) ; = 996 kg/m 3 Frm Mdy chart: Re = ; = mm = 1.5 m f = /D = 1.07(10) -4 ( V) L' 1 p [ ]( v ) [1.07 f ] D 並不大! V L' [ ] [1.07 f ] D 996(kg/m 3) [1.1( m/ s)] [1.07 (0.065)(608.6)] (kg/m-s ) (pa) (kpa) 一般水泵 (Pump) 可接受! (p < 40 kpa)

19 Mdy Diagram

20 Cntractin Lss Cefficient (K c ) and Expansin Lss Cefficient (K e )

21 討論 : 欲縮短冷凝器之長度則必須縮短圓管之長度, 有效方法有下列幾種 : (i) 增加管數 (ii) 增強管外殼側之熱傳性能, 例如採用具鰭片之熱傳增強管 (iii) 銅管經常清洗,fuling factr (F i ) 可以降低至 (m -K/W) (iv) 昇高冷凝溫度

22 計算流程圖 決定銅管大小 管數與管排 計算 LMTD 計算管內側水流之熱對流係數 (h i ) 假設管壁與冷媒蒸汽間之溫差 (T) 計算管外側冷媒冷凝之熱對流係數 (h ) 計算整體熱傳係數 (U ) 計算熱交換器所需之熱傳面積 (A ) 與管長 (L) 與原先假設值差別不大 驗證溫差 (T) 與原先假設值差別大 (T < 0.5 C) 計算水流側之壓差

23 例 : 變更例 1 中冷凝器之設計 --- 冷凝溫度由 35 0 C 昇高至 45 0 C 冷媒入口 45 0 C (sat. vapr) 冷卻水入口 5 0 C Water cled cndenser Tw-passes shell-and-tube type 冷媒出口 45 0 C (sat. liquid) 管長 (L) 冷卻水出口 30 0 C

24 銅管 44 支外徑 : 16 mm 內徑 : 14mm 重力, g 註 : refrigerant (R-, 45 0 C): k = (W/m-K), = (Pa-s) h fg = (J/kg), =1109 (kg/m 3 ) water(8 0 C): w = 996 (kg/m 3 ), w = (Pa-s), k w = 0.61 (W/m-K), c p,w =4190 (J/kg-K), F i = fuling factr (water t cpper)= (m -K/W) cpper: k c = 390 W/m-K

25 ΔT ΔT (45 5) (45 30) LMTD ΔTlm ln( ) ΔT 管內冷卻水側 1 0 ΔT1 ln[(45 5) /(45 30)] 冷卻水之質量流率 = 冷卻量 / (c p,w T) = 78.3 kw / [4.19 (kj/kg-k) x (30-5)(K) = (kg/s) 冷卻水之體積流率 = 質量流率 / 密度 = (m 3 /s) = 5.1 (L/min) 冷卻水於銅管內之流速 (V) = 體積流率 / [ ( 支管 ) x 每支管之內截面積 ] = 1.1 (m/s) C Re = w Vd i / w = 996 (kg/m 3 ) x 1.1 (m/s) x (m) / (pa-s) = Dittus-Blter Equatin: Nu d =0.03 Re 0.8 Pr 0.4 (Fully Turbulent Flw) Pr = c p,w w / k w = 4190 (J/kg-K) x (pa-s) / 0.61 (W/k-K) = 5.35 h i = (k w / d i )(0.03) Re 0.8 Pr 0.4 = 5338 (W/m -K)

26 管外冷媒側 i c i i i i 1 1 d d d d 1 F ( )ln( ) ( )F ( )( ) U h k d d d h 管內流體熱阻管外流體熱阻管內污垢熱阻管外污垢熱阻管壁熱阻

27 Staggered Arrangement: N = 3.38, 查表可得 : F 0.93 assume T 10K ; g 9.81 (m/s ) ; N 3.38 tubes/rw ; h d 44 tubes 13 rws d m gρ h k 0.79 [ ] μ(δt)d 3 fg 1/ (m/s ) (1153) (kg/m ) (J/kg) (0.078) (W/m 0.79[ 1775 (W/m K) 必須驗算! (J s/m ) 10 (K) (m) 3 K) ] 1/4 h h d,n F h d (W/m -K)

28 1 1 d d d d 1 F ( )ln( ) ( )F ( )( ) U h k d d d h i c i i i i [ ]ln( ) ( )( ) ( )( ) 1651 ()(390) m K ( ) W 管外流體之熱阻相對最大 W Hence, U 977 ( ) m-k Q = U A (LMTD) Hence A = Q / [U (LMTD)] = (W) / [977 (W/m -K) x (K)] = 所有銅管之總外表面積 = 4.61 (m ) 管之長度 = A / [ 總管數 x 單支管之周長 ] =.09 (m)

29 驗算 T 是否接近 10 0 C Q = 78.3 kw T B Q = 78.3 kw T water T A =45 0 C 1/h F (d /k)ln(d /d i ) (d /d i )(F i ) (d /d i )(1/h i ) 略 Q = h A T = h A (T A T B ) 平均管外表面溫度 T = Q / (h A) = (W) / [4.61 (m ) x 1651 (W/m-K)] = C 註 : 嚴格之計算必須重新假設一個 T 值, 再重覆以上之計算, 但差別應已不大

30 討論 : 冷凝溫度由 35 0 C 昇高至 45 0 C, 冷凝器中之管長可大幅縮短, 亦即所需之熱傳面積可大幅減少, 但冷凝溫度之昇高會造成蒸發溫度之上昇, 進而造成冷媒壓縮機耗電量之增加 冷凝溫度由 35 0 C 昇高至 45 0 C, 管長大幅縮短, 因此冷卻水側之壓降亦會大幅下降 一般國外系統之冷凝溫度之設計為 40 ~ 4 C, 冷卻水入口溫度之設計為 9 ~30 C, 冷卻水出口溫度之設計為 35 C

31 例 3: 改良例 1 中冷凝器 ( 冷凝溫度 35 0 C) 之設計 --- 增加管數 冷媒入口 35 0 C (sat. vapr) 冷卻水入口 5 0 C Water cled cndenser Tw-passes shell-and-tube type 冷媒出口 35 0 C (sat. liquid) 冷卻水出口 30 0 C 管長 (L)

32 紅銅管 d i = 14 mm d = 16 mm 總共 74 支管 冷卻水隔板 重力, g 註 : refrigerant (R-, 35 0 C): k = (W/m-K), = (Pa-s) water(8 0 C): w = 996 (kg/m 3 ), w = (Pa-s), k w = 0.61 (W/m-K), c p,w =4190 (J/kg-K) F i = fuling factr (water t cpper)= (m -K/W) cpper: k c = 390 W/m-K

33 管內冷卻水側 ΔT ΔT (35 5) (35 30) LMTD ΔTlm 7.1 ln( ) ΔT 1 0 ΔT1 ln[(35 5) /(35 30)] 冷卻水之質量流率 = 冷卻量 / (c p,w T) = 78.3 kw / [4.19 (kj/kg-k) x (30-5)(K) = (kg/s) 冷卻水之體積流率 = 質量流率 / 密度 = (m 3 /s) = 5.1 (L/min) 水於銅管內之流速 (V) = 體積流率 / [37 ( 支管 ) x 每支管之內截面積 ] = (m/s) C Re = w Vd i / w = 996 (kg/m 3 ) x (m/s) x (m) / (pa-s) = 1169 Dittus-Blter Equatin: Nu d =0.03 Re 0.8 Pr 0.4 (Fully Turbulent Flw) Pr = c p,w w / k w = 4190 (J/kg-K) x (pa-s) / 0.61 (W/k-K) = 5.35 h i = (k w / d i )(0.03) Re 0.8 Pr 0.4 = 350 (W/m -K)

34 管外冷媒側 assume T 3.3 K ; g 9.81 (m/s ) ; N 4.35 tubes/rw ; d m gρ h k hd 0.75 [ ] μ(δt)nd 0.75[ 3 fg 1/4 74 tubes 17 rws 3 K) 3 ] (J s/m ) 3.3 (K) (m) (m/s ) (1153) (kg/m ) (J/kg) (0.083) (W/m 445 (W/m K) 必須驗算! 1/4 Staggered Arrangement: N = 4.35, 查表可得 : F 0.9 h h d,n F h d (W/m -K)

35 1 1 d d d d 1 F ( )ln( ) ( )F i ( )( ) U h kc di di di hi [ ]ln( ) ( )( ) ( )( ) 50 ()(390) m K ( ) W 管外流體之熱阻相對最大 W Hence, U 108 ( ) m-k Q = U A (LMTD) Hence A = Q / [U (LMTD)] = (W) / [108 (W/m -K) x 7.1(K)] = 所有銅管之總外表面積 = (m ) 管之長度 = A / [ 總管數 x 單支管之周長 ] =10.56 / (74 x 3.14 x 0.016) =.84 (m) 略長

36 驗算 T 是否接近 C Q = 78.3 kw T B T water T A =35 0 C 1/h F (d /k)ln(d /d i ) (d /d i )(F i ) (d /d i )(1/h ) 略 Q = h A T = h A (T A T B ) 平均管外表面溫度 T = Q / (h A) = (W) / [10.56 (m ) x 50 (W/m-K)] = C 註 : OK

37 討論 : 增加冷凝器中之管數, 亦可大幅縮短管長 但冷凝器中所需之總熱傳面積並不會減少 增加冷凝器中之管數, 管長縮短, 同時水流於管中流動之速率下降, 因此冷卻水側之壓降亦會大幅下降

38 例 4: 改良例 中冷凝器之設計 --- 增加管數 下圗中冷凝器之負荷 ( 熱傳量 ) 為 78.3 kw, 冷凝溫度為 45 0 C, 冷媒為 R-, 試計算 (i) 冷卻水於管內之流速 (m/s)? (ii) 冷凝器中銅管所需之長度為多少 (m)? 冷媒入口 45 0 C (sat. vapr) 冷卻水入口 5 0 C Water cled cndenser Tw-passes shell-and-tube type 冷媒出口 45 0 C (sat. liquid) 管長 (L) 冷卻水出口 30 0 C

39 紅銅管 d i = 14 mm d = 16 mm 總共 74 支管 冷卻水隔板 重力, g 註 : refrigerant (R-, 45 0 C): k = (W/m-K), = (Pa-s) h fg = (J/kg) water(8 0 C): w = 996 (kg/m 3 ), w = (Pa-s), k w = 0.61 (W/m-K), c p,w = 4190 (J/kg-K), F i = fuling factr (water t cpper) = (m -K/W) cpper: k c = 390 W/m-K

40 ΔT ΔT (45 5) (45 30) LMTD ΔTlm ln( ) ΔT 1 0 ΔT1 ln[(45 5) /(45 30)] C 管內冷卻水側 冷卻水之質量流率 = 冷卻量 / (c p,w T) = 78.3 kw / [4.19 (kj/kg-k) x (30-5)(K) = (kg/s) 冷卻水之體積流率 = 質量流率 / 密度 = (m 3 /s) = 5.1 (L/min) 水於銅管內之流速 (V) = 體積流率 / [37 ( 支管 ) x 每支管之內截面積 ] = (m/s) Re = w Vd i / w = 996 (kg/m 3 ) x (m/s) x (m) / (pa-s) = 1169 Dittus-Blter Equatin: Nu d =0.03 Re 0.8 Pr 0.4 (Fully Turbulent Flw) Pr = c p,w w / k w = 4190 (J/kg-K) x (pa-s) / 0.61 (W/k-K) = 5.35 h i = (k w / d i )(0.03) Re 0.8 Pr 0.4 = 350 (W/m -K)

41 管外冷媒側 i c i i i i 1 1 d d d d 1 F ( )ln( ) ( )F ( )( ) U h k d d d h 管內流體熱阻管外流體熱阻管內污垢熱阻管外污垢熱阻管壁熱阻

42 assume T ; g 9.81 (m/s ) ; N 4.35 tubes/rw ; h d d m gρ h k 0.79 [ ] μ(δt)d 0. 79[ 10 K 3 fg 1/4 74 tubes 17 rws (m/s ) (1153) (kg/m ) (J/kg) (0.078) (W/ m K) 1/4 3 ] 1775 (W/m K) (J s/m ) 10 (K) (m) Staggered Arrangement: N = 4.35, 查表可得 : F 0.9 h h d,n F h d (W/m -K)

43 1 1 d d d d 1 F ( )ln( ) ( )F ( )( ) U h k d d d h i c i i i i [ ]ln( ) ( )( ) ( )( ) 1633 ()(390) m K ( ) W 管外流體之熱阻相對最大 W Hence, U 971 ( ) m-k Q = U A (LMTD) Hence A = Q / [U (LMTD)] = (W) / [971 (W/m -K) x (K)] = 所有銅管之總外表面積 = 4.64 (m ) 管之長度 = A / [ 總管數 x 單支管之周長 ] = 1.5 (m)

44 驗算 T 是否接近 10 0 C Q = 78.3 kw T B Q = 78.3 kw T water T A =45 0 C 1/h F (d /k)ln(d /d i ) (d /d i )(F i ) (d /d i )(1/h i ) 略 Q = h A T = h A (T A T B ) 平均管外表面溫度 T = Q / (h A) = (W) / [4.64 (m ) x 1633 (W/m-K)] = C 註 : 嚴格之計算必須重新假設一個 T 值, 再重覆以上之計算, 但差別應已不大

45 例 5: 冷凍空調中之 Tw-Passes 管殼式蒸發器之設計 - Water n Tube Side ( 滿液式蒸發器 ) R vapr R liquid R- 飽和冷媒蒸氣 (x=100%) t =5 0 C T 1 =16 0 C 冰水入口 T = 8 0 C 冰水出口 t 1 =5 0 C R- 液氣態 (x=0.1) 冷媒 L 設計之熱傳負荷 = 98 (kw) 銅管 : d = 1 (mm) ; d i = 10.4 (mm) wall thickness = 0.8 (mm) ; k c = 390 W/m-K

46 0. m 液面 總共 56 支管 S T S S S T L D 0 (mm) 0 (mm).4 (mm) S D S L 註 : refrigerant (R-, 5 0 C): k f = 0.1 (W/m-K), f = (Pa-s) k g = (W/m-K), g = (Pa-s) h f = (kj/kg) h g = (kj/kg) water (~10 0 C): w = 1000 (kg/m 3 ), w = (Pa-s), k w = 0.58 (W/m-K), c p,w =4194 (J/kg-K), Pr =9.45 F = fuling factr (water t cpper)= (m -K/W)

47 Water ( 管側 ): Q m c (T T ) m (kg/s) 4194(J/kg K) (16 8)(K) w p,w 1 w (W) -3 3 m w.91 (kg/s) (m /s).91 (L/s) Re 4(m w / tube number) 4(.91/8)(kg/s) π d μ (0.0104)(m)( )(pa-s) i w Dittus-Belter Equatin: Nu D 0.03ReD Pr 87.9 k Hence, h ( )Nu 4903 (W/m K) w w D di ( L/min) 9777 Refrigerant ( 殼側 ): Q m [(1 x)h ] ref fg Q 98(kW) m ref (kg/s) (1x)h (10.1)( )(kJ/kg) fg

48 Heat Transfer: q μ h g(ρ ρ ) c ΔT h ΔT f g p,f e w f fg n b e σ Csf hfgpr f 3 h fg = 0144 (J/kg) ; f = (Pa-s) = 表面張力 = 10.8 x 10-3 (N/m) ; Pr f =.49 ; c p,f = 1170 (J/kg-K) ; f - g = = 141(kg/m 3 ) g = 9.81 (m/s ) ; T e 先估算 3 (K) ; C sf ; n 1.7 ( 經驗公式中之常數, 查表估算,R- 冷媒 ) 1 g(ρf ρ g) cp,fδte h b ( ) μfhfg n ΔTe σ CsfhfgPrf 3 1 (9.81)(141) (1170)(3) ( ) (0144) (0.007)(0144)(.49) (0.333)(49143)(0.1475) 414 (W/m K) 3

49 內插

50 1 1 d d d d 1 F ( )ln( ) ( )F i ( )( ) U h kc di di di hi [ ]ln( ) ( )( ) ( )( ) 414 ()(390) 管 (.10 ) 內流 m K 體 ( ) W 熱阻Imprtant! W Hence, U 1170 ( ) m-k

51 Q = U A (LMTD) LMTD (16 5) (8 5) 16 5 ln( ) C A = Q / [U (LMTD)] = (W) / [1170 (W/m -K) x 6.16 (K)] = 所有銅管之總外表面積 = 13.6 (m ) 管之長度 = A / [ 總管數 x 單支管之周長 ] = 13.6 /[(56)(3.14)(0.01)] = 6.45 (m) 太長

52 驗算 T e 是否接近 3 0 C Q = 98 kw T B Q = 98 kw T water T A = 5 0 C 1/h F (d /k)ln(d /d i ) (d /d i )(F i ) (d /d i )(1/h i ) 略 Q = h A T = h A (T A T B ) 平均管外表面溫度 T = Q / (h A) = (W) / [13.6 (m ) x 414 (W/m-K)] =.99 0 C 註 : 嚴格之計算必須重新假設一個 T e 值, 再重覆以上之計算, 但差別應已不大

53 例 6: 空調中之 Tw-Passes 管殼式蒸發器之設計 - Water n Tube Side ( 增加管數 ) t 1 =5 0 C R vapr R liquid R- 液態與蒸氣 (x=0.1) 冷媒 L R- 飽和冷媒蒸氣 (x=100%) t =5 0 C T 1 =16 0 C 冰水入口 T = 8 0 C 冰水出口 設計之熱傳負荷 = Q = 98 (kw) 銅管 : d = 1 (mm) ; d i = 10.4 (mm) wall thickness = 0.8 (mm) ; k c = 390 W/m-K

54 液面 總共 146 支管 註 : water(10 0 C): w = 1000 (kg/m 3 ), w = (pa-s), k w = 0.58 (W/m-K), c p,w =4194 (J/kg-K), Pr =9.45 F = fuling factr (water t cpper)= (m -K/W)

55 Water: Q m c (T T ) m (kg/s) 4194(J/kg K) (16 8)(K) w p,w 1 w (W) -3 3 m w.91 (kg/s) (m /s).91 (L/s) Re 4(m w / tube number) 4(.91/73)(kg/s) π d μ (0.0104)(m)( )(pa-s) i w (L/min) 3750 Gnielinski Equatin: f 0.04 (smth pipe) ; Pr = 9.45 (f/8)(re 1000)Pr D 11.7(f/8) (Pr 1) L /3 Nu D [1 ( ) ] / /3 Usually /3 L D, hence (D/L) can be neglected! k Hence, h ( )Nu D 1759 (W/m K) hc di 太小!

56 Refrigerant: 註 : refrigerant (R-, 5 0 C): k f = 0.1 (W/m-K), f = (Pa-s) k g = (W/m-K), g = (Pa-s) h f = (kj/kg), h g = (kj/kg) Q m [(1 x)h ] ref fg Q m ref = 冷媒質量流率 (1 x)h fg 98(kW) (10.1)( )(kJ/kg) (kg/s)

57 實驗之經驗公式 : q μ h g(ρ ρ ) c ΔT h ΔT f g p,f e w f fg n b e σ Csf hfgpr f 3 h fg = 0144 (J/kg) ; f = (Pa-s) ; = 10.8 x 10-3 (N/m ) Pr f =.49 ; c p,f = 1170 (J/kg-K) ; f - g = = 141(kg/m 3 ) g = 9.81 (m/s ) ; assume T e.5 (K) ; C sf ; n g(ρf ρ g) cp,fδte h b ( ) μfhfg n ΔTe σ CsfhfgPrf 3 1 (9.81)(141) (1170)(.5) ( ) (0144) (0.007)(0144)(.49) (0.4)(49143)(0.0854) 1678 (W/m K) 3

58 內插

59 1 1 d d d d 1 F ( )ln( ) ( )F i ( )( ) U h kc di di di hi [ ]ln( ) ( )( ) ( )( ) 1678 ()(390) (.10 ) m K ( ) W W Imprtant! Hence, U 686 ( ) m-k LMTD (16 5) (8 5) 16 5 ln( ) C 管數越多, 管內水流速度越慢, h i 值亦越小! 熱傳面積之增加趨於無效!

60 Q = U A (LMTD) 太小! Hence A = Q / [U (LMTD)] = (W) / [686 (W/m -K) x 6.16 (K)] = 所有銅管之總外表面積 = 3.18 (m ) 管之長度 = A / [ 總管數 x 單支管之周長 ] = 8.5 /[(146)(3.14)(0.01)] = 4.1 (m)

61 驗算 T e 是否接近.5 0 C Q = 98 kw T B Q = 98 kw T water T A = 5 0 C 1/h F (d /k)ln(d /d i ) (d /d i )(F i ) (d /d i )(1/h i ) 略 Q = h A T = h A (T A T B ) 平均管外表面溫度 T = Q / (h A) = (W) / [3.18 (m ) x 1678 (W/m-K)] =.5 0 C 註 : 嚴格之計算必須重新假設一個 T e 值, 再重覆以上之計算, 但差別應已不大

62 例 7: 前例之 Tw-Passes 管殼式蒸發器改為 Fur-Passes 之設計 ( 加以管內經常清洗 ) 隔板 入口 出口 Rear View Frnt View 設計之熱傳負荷 = 98 (kw) 銅管 : d = 1 (mm) ; d i = 10.4 (mm) wall thickness = 0.8 (mm) ; k c = 390 W/m-K

63 總共 148 支管 每一個區域 37 支管 液面 Frnt View 註 : water(10 0 C): w = 1000 (kg/m 3 ), w = (Pa-s), k w = 0.58 (W/m-K), c p,w =4194 (J/kg-K), Pr =9.45 F = fuling factr (water t cpper)= (m -K/W)

64 Water: Q m c (T T ) m (kg/s) 4194(J/kg K) (16 8)(K) w p,w 1 w (W) -3 3 m w.91 (kg/s) (m /s).91 (L/s) Re 4(m w / tube number) 4(.91/37)(kg/s) π d μ (0.0104)(m)( )(pa-s) i w (L/min) 7399 Gnielinski Equatin: f 0.04 (smth pipe) ; Pr = 9.45 Usually (f/8)(re 1000)Pr D 11.7(f/8) (Pr 1) L /3 Nu D [1 ( ) ] / /3 /3 L D, hence (D/L) can be neglected! k w Hence, h ( )Nu D 4094 (W/m K) hc di 每支管之內截面積 = d i /4 = 8.49 x 10-5 (m ) 冷卻水於銅管內之流速 (V) = 體積流率 / [37 ( 支管 ) x d i /4] = 0.93 (m/s)

65 Refrigerant: 註 : refrigerant (R-, 5 0 C): k f = 0.1 (W/m-K), f = (Pa-s) k g = (W/m-K), g = (Pa-s) h f = (kj/kg), h g = (kj/kg) Q m [(1 x)h ] m ref ref fg Q (1 x)h fg 98(kW) (10.1)( )(kJ/kg) (kg/s)

66 實驗之經驗公式 : q μ h g(ρ ρ ) c ΔT h ΔT 因此, f g p,f e w f fg n b e σ Csf hfgpr f g(ρf ρ g ) cp,f h b=h (ΔT e) μ f h fg n σ Csf hfgpr f 1 1 d d d d 1 F ( )ln( ) ( )F ( )( ) U h k d d d h i c i i i i Hence, U h d d d d 1 1 [F ( )ln( ) ( )F ( )( )]h i kc di di di hi It can be knwn ( 由熱阻之關係 ), 3 3 T U(LMTD) e h LMTD d d d 1 [F ( )ln( ) ( k d d c i i d 1 )F i ( )( )]h di hi

67 Substituting back ( 將經驗公式 h 之結果代入 ) and F = 0, T e LMTD d d d d 1 g(ρ ρ ) c f g p,f 1 ( T e) [( )ln( ) ( )F i ( )( )] μ f hfg n kc di di di hi σ Csf hf gprf 3 Hence, 3 d d d d 1 g(ρf ρ g) c p,f ( T e) [( )ln( ) ( ) F ( )( )] i μ f hfg n e kc di di di hi σ Csfh ( T ) fgpr f LMTD 3

68 h fg = 0144 (J/kg) ; f = (Pa-s) ; = 10.8 x 10-3 (N/m ) Pr f =.49 ; c p,f = 1170 (J/kg-K) ; f - g = = 141(kg/m 3 ) g = 9.81 (m/s ) ; C sf ; n 1.7 代入數值後 3, ΔT e (ΔT e) 6.16 (i) 利用電腦解答 : 疊代法 需要電腦解答 3 ΔT e (ΔT e) 6.16 (!) Using the "Interactive Heat Transfer" sftware, the slutin If s, is, ΔT e C 3 g(ρf ρ g ) cp,f h (ΔT e) μ f h fg 78 (W/m -K) n σ Csf hfgpr f

69 (ii) 利用計算機計算 : Newtn-Raphsn 疊代法 解答 ΔT = e Gauss-Seidel 疊代法會發散 3 (ΔT e)! 利用 Newtn-Raphsn's Iteratin Methd: 3 f ( T e) =ΔT e (ΔT e) f ( T e) n ( T e) n+1( T e) n ' f ( T e) n (ΔT e) n (ΔT e) That is, ( T e) n+1( T e) n 1 (3)( )(ΔT ) e n 3 n 疊代次數 初始假設值 ( C) 得到值 ( C) 疊代計算後 : T = C

70 內插

71 1 1 d d d d 1 F ( )ln( ) ( )F i ( )( ) U h kc di di di hi [ ]ln( ) ( )( ) ( )( ) 78 ()(390) (.10 ) m K ( ) W W Hence, U 141 ( ) m-k 管內經常清洗 Imprtant! 管內水流速度越快, h i 值亦越大! LMTD (16 5) (8 5) 16 5 ln( ) C

72 Q = U A (LMTD) Hence A = Q / [U (LMTD)] = (W) / [141 (W/m -K) x 6.16 (K)] = 所有銅管之總外表面積 = 11.7 (m ) 管之長度 = A / [ 總管數 x 單支管之周長 ] = 11.7 /[(148)(3.14)(0.01)] =.0 (m) Perfect!

73 驗算 T e 是否為 C Q = 98 kw T B Q = 98 kw T water T A = 5 0 C 1/h F (d /k)ln(d /d i ) (d /d i )(F i ) (d /d i )(1/h i ) 略 Q = h A T = h A (T A T B ) 平均管外表面溫度 T = Q / (h A) = (W) / [11.7(m ) x 78 (W/m-K)] = C 註 : 等於原先之假設值

74 Pressure Drp ( p, 管內水流之壓力降 ) p G v1 v A vm v ( )( )[( K c 1 ) ( 1) f ' (1 K e ) p1 p v A v v 1 (i) 入口 v v 1 = free-flw area / frntal area = 空孔率 = 0.3 K c = 0.4 ; K e =0.45 ; = (ii) Flw acceleratin Fanning frictin factr c 1 (iii) Cre frictin ~1.0 (iv) 出口 ~1.0 1 ] ( )( ) [( K 1 ) f ' (1 Ke)] p p A p G v1 A c 1 1 c ( V) v1 A ( V) v1 4 L' [ ]( ) [0.85 f ' ] [ ]( ) [0.85 f ' ] p A p D 1 c 1 ( V) v1 L' [ ]( ) [0.85 f ] p D 1

75 L = 4 L = 4(.0) = 8.08 (m) = 流道總長 V = 0.93 (m/s) ; D = d i = 10.4 (mm) = (m) ; = 996 kg/m 3 Re = 7399 ; = mm f = (0.79 lnre ) - f = ( V) L' 1 p [ ]( v ) [0.85 f ] D V L' [ ] [0.85 f ] D 3 996(kg/m ) [0.93( m/ s)] [0.85 (0.0343)(777)] 並不大! (kg/m-s ) (Pa) (kpa) 水泵 (Pump) 可接受!

76 Cntractin Lss Cefficient (K c ) and Expansin Lss Cefficient (K e )

77 例 8: 液壓油冷卻器 (shell-and-tube type) 之設計 Oil n Shell Side 冷卻水 L / 9 L T =50 0 C t 1 =30 0 C t =34 0 C T 1 =80 0 C 油 0.15 m 3/8 steel pipe (40 schedule) wall thickness = 3. mm d i = 10.7 mm ; d = 17.1 mm 管排 S L S T 5 (mm) ; S L 0 (mm) S 3.6 (mm) D S D S T

78 m 1 (kg/min) 0. (kg/s) il (L/min) (m /s) 熱交換器之總熱傳量 Q Q m c (T T ) 0.(kg/s) 1800(J/kg K) (80 50)(K) il p,il (W) 流體之物理特性 : Fr il : 10 (m /s) ; Pr 100 ; k 0.15 (W/m-K) -5 il il il 3 90 (kg/m ) ; c 1800 (J/kg-K) il p,il Fr water : (m /s) ; Pr 4 ; k 0.65 (W/m-K) -7 w w w (kg/m ) ; c 4180 (J/kg-K) w p,w

79 (i) Water Side (tube side) : m water Q (W) (kg/s) c (t t ) 4(K) 4180 (J/kg-K) p,w 1 water 4 3 water vlumetric flwrate (V) (m /s) w V / 11 (tubes) flw velcity in a tube (U ) (m/s) ( d /4) i i Re 10 m i i Ud 756 (turbulent flw) ν Dittus-Belter Equatin: Nu 0.03Re Pr 67. k Hence, h ( )Nu 4085 (W/m K) w w D di D D

80 先行假設熱交換器之長度為 L = 3.6 (m), 則每片 baffle 間之距離為 L / 9 = 0.4 (m) (ii) Oil Side (shell side) : Tube bank 之型式 m ρu A il 3 U m il /(ρila) 0.(kg/s)/[(90)(kg/m ) (0.15)(m) 0.4(m)] (m/s) S T U max ( ) U (m/s) ST d U d max ReD,max 9.8 ; νil Frm Table: C 1.04; m 0.4; n 0.36; F 0.89 Pr m n 1/4 NuD F C Re D,max Pr ( ) 1 Prw 略 k h ( ) Nu 88.5 ( W/m -K) d il D 小! (Oil 之 值較大 ).1 小! (Oil 之 k 值較小 )

81 1 U 1 h F d ( k d )ln( d m K ( ) W W Hence, U 8 ( ) m - K c i ) [ ]ln( ) ()(40) d ( d i )F i d ( d i )( 1 h i ) 17.1 ( )( ) ( )( ) A = x d x L x ( 支管 ) = 1.18 L LMTD (80 34) (50 30) ln( ) C

82 查表 : R T t 1 T t Crrectin Factr (F') ; P t T 1 t t Q = F U A (LMTD) = (0.98)(8)(1.18 L)(31.) Hence, L = 3.64 (m) 與原來之假設相差不多! 太長! 總熱傳面積 = A = x d x L x ( 支管 ) = 1.18 L = 4.31 (m )

83

84 高熱傳面積之液壓油冷卻器 (Tw-tube pass) Baffle

85 液壓油側 間隙 1 mm 之 perfrated 鰭片

86 冷卻水側 (Tw-tube pass)

87 例 9: 液壓油冷卻器 (shell-and-tube type) 之設計 ---- 例 8 之變更設計 (3/4 steel pipe 與增加管數 ) 冷卻水 t 1 =30 0 C L / 9 L T =50 0 C t =34 0 C T 1 =80 0 C 油 0.0 m 3/4 steel pipe (40 schedule) wall thickness =.9 mm d i = 0.9 mm ; d = 6.7 mm S L S T 3 (mm) ; S L 30 (mm) S 34 (mm) D S D S T

88 m 1 (kg/min) 0. (kg/s) il (L/min) (m /s) 熱交換器之總熱傳量 Q Q m c (T T ) 0.(kg/s) 1800(J/kg K) (80 50)(K) il p,il (W) 流體之物理特性 : Fr il : 10 (m /s) ; Pr 100 ; k 0.15 (W/m-K) -5 il il il 3 90 (kg/m ) ; c 1800 (J/kg-K) il p,il Fr water : (m /s) ; Pr 4 ; k 0.65 (W/m-K) -7 w w w (kg/m ) ; c 4180 (J/kg-K) w p,w

89 (i) Water Side (tube side) : m water Q (W) (kg/s) c (t t ) 4(K) 4180 (J/kg-K) p,w water vlumetric flwrate (V) m water / w (m /s) V / 15 (tubes) flw velcity in a tube (U ) 0.16 (m/s) ( d /4) i i Ud i i Re 4040 (transitinal flw) ν Gnielinski Equatin: f 0.04 (smth pipe) (f/8)(re 1000)Pr D 11.7(f/8) (Pr 1) L /3 Nu D [1 ( ) ] 5.6 1/ /3 /3 Usually L D, hence (D/L) can be neglected k Hence, h ( )Nu 798 (W/m K) h w w D i di!

90 先行假設熱交換器之長度為 L = 1.35 (m), 則每片 baffle 間之距離為 L / 9 = 0.15 (m) = 0.75 D (ii) Oil Side (shell side) : Tube bank 之型式 m ρu A il 3 U m il /(ρila) 0.(kg/s)/[(90)(kg/m ) (0.15)(m) 0.(m)] (m/s) S T U max ( ) U (m/s) ST d U d max ReD,max 58 ; νil Frm Table: C 1.04; m 0.4; n 0.36; F 0.94 Pr m n 1/4 NuD F C Re D,max Pr ( ) 6.1 Prw 略 k h ( ) Nu 13 (W/m -K) d il D

91 1 U 1 h 1 13 F d ( k d )ln( d m K ( ) W W Hence, U 98 ( ) m - K c i ) [ ]ln( ) ()(40) 0.9 (810 d ( d -5 ) i )F i d ( d i )( 1 h i ) 6.7 ( )( ) ( )( 較大! 必須利用熱傳增強方式加大 h 值或增加外部熱傳面積 ) A = x d x L x (30 支管 ) =.5 L LMTD (80 34) (50 30) ln( ) C

92 查表 : R Crrectin Factr ; (F' ) 34 P Q = F U A (LMTD) = (0.98)(98)(.5 L)(31.) Hence, L = 1.44 (m) 與原來之假設 1.35 m 相差不多! 總熱傳面積 = A = p x d x L x (30 支管 ) =.5 L = 3.63 (m )

93 例 10: 液壓油冷卻器 (shell-and-tube type) 之設計 Oil n Tube Side (i) 試求所需圓管之總外表面積? (ii)tube passes (n) 之數目? Water: ne shell pass Oil: n tube passes Water, T 1 = 35 0 C Oil, t =50 0 C il = 866 (kg/m3) c w = 4180 (J/kg-K) il flwrate = 0 (L/min) Water, T = 40 0 C Oil t 1 =55 0 C 0 Q m ilcilδt il [( )(L/s) (0.866)(kg/L)](000)(J/kg-K)(55 50) (W).887 (kw) m Q/(c ΔT ) 887 (W) / [(4180)(J/kg-K)(40 35)(K)] w w w (kg/s) (m /s) (L/s) 8.88 (L/min) L

94 考慮為 cunterflw arrangement 以計算 LMTD 值! Fr cunterflw arrangement: (55 40) (50 35) LMTD ln[(55 40)/(50 35)] 0 =? 0 Nte : T ΔT 15 C 1 Oil t 1 =55 0 C Water T = 40 0 C cunter-flw Oil, t =50 0 C Water T 1 = 35 0 C LMTD 1 ' F(x) F (x) x=1 x1 ' Use L'Hspital's rule: y(x) = lim = G(x) Here, x = T/ΔT G (x) ( T1ΔT ) ΔT [( T/ΔT 1 ) 1] ΔT = = = ln[ T /ΔT ] ln[ T/ΔT ] 1/ ( T/ΔT ) ΔT T /ΔT 1 15 C x=1 1

95 t t T T T t t t p 0.5 ; R Frm Fig., F 0.99 (crrectin factr) 假設 U i = 180 (W/m -K) 對液體而言, 此值較小 (i.e. laminar flw) Q = U i A i F (LMTD), A i = Q / [U i F (LMTD)] = (887) / [(180)(0.99)(15)] = 1.08 (m ) Hence, A i = 總熱傳面積 = 1.08 (m )

96

97 略! 1 d 1 d d d 1 ( )( ) ( )F ( )ln( ) F U d h d k d h 略! 略! i i i i i c i i 相對小! 相對大! 管外水流速度大至一定程度時, h 0 > > h i, 上式中右邊第一項亦可忽略! h m = Nu m k / d i = (6)(0.14) / (0.0) = 18 (W/m -K) = h i U i h i 證實原先假設之值接近此值, 計算無誤! Fr il ( 物理特性 ): k = 0.14 (W/m-K), = 9.6 x 10-5 (m -s), = 866 (kg/m 3 ) c p = (kj/kg-k), Pr = 100, = (N-s/m )

98 假設圓管之內徑 = d i =0.0 (m) ; d = 0.06 (m) laminar flw design A i = (tube length) ( d i ) ; nl = 1.08 / [(3.14)(0.0)] = 17. (m) 若選擇 L = (m), tube passes = n = 18 U i 值大小之印證 : 考慮為 m (fr ne tube pass), 每個 tube pass 結束時, 因彎管之緣故, 速度分佈被破壞, 速度場與溫度場重新發展 4m (4)(0.887) πd μ (3.14)(0.0)(0.0831) il Re 1 i (x/d ) (0.956/0.0) Re Pr (1)(100) 1 i Gz laminar flw Nu m = 6.0

99 Mean Nusselt Number fr Thermally and Hydrdynamically Develping Laminar Flw Cnstant Wall Temperature Circular Tube

100 Pressure Drp ( p, 管內液壓油之壓力降 ) Re = 1 ; laminar flw ; (i) 因管甚長可考慮均處於完全發展區域 ; f = 64/Re = 0.9 V = vlumetric flwrate / crss sectinal area = (m 3 /s) / 0.0) /4](m ) = 0.44 (m/s) p V L' [ ][ f ] D 3 866(kg/m ) [0.44(m/s)] [(0.9)(18 / 0.0)] 188 (kg/m -s ) 188(pa) 1.88 (kpa)

101 (ii) 考慮為每個 tube pass (L=0.956 m ) 結束時, 因彎管之緣故, 速度分佈被破壞, 速度場重新發展 : Re / (L/D) = 4.6, 由圖 : cf Re 17 app Hence, f Re 4 f Re 68, f = 0 0 c.3 8 app p V L [ ] [ f ] [number f passes] D 3 866(kg/m ) [0.44(m/s)] [(0.308)(0.956/ 0.0)][18] 15 (kg/m-s ) 15 (pa).15 (kpa) 差別不大!

102 Pressure Drp frictin plus flw acceleratin Re / (L /D) = 0.45 Δp 4 c f app ρv ( L' )( ) D f ρv ( L' ) ( ) D 64 ρv ( )( Re L' )( ) D

103

104 例 11 (Hmewrk): 冷凝器之設計 ( 冷凝溫度 = 50 0 C ) 下圗中冷凝器之負荷 ( 熱傳量 ) 為 78.3 kw, 冷凝溫度為 50 0 C, 冷媒為 R-, 試計算 (i) 冷卻水於管內之流速 (m/s)? (ii) 冷凝器中銅管所需之長度為多少 (m)? 冷媒入口 50 0 C (sat. vapr) 冷卻水入口 30 0 C Water cled cndenser Tw-passes shell-and-tube type 冷媒出口 50 0 C (sat. liquid) 管長 (L) 冷卻水出口 35 0 C

105 紅銅管 d i = 14 mm d = 16 mm 總共 74 支管 冷卻水隔板 重力, g 註 : refrigerant (R-, 50 0 C): k = (W/m-K), = (pa-s) h fg = (J/kg) water(3 0 C): w = 996 (kg/m 3 ), w = (pa-s), k w = 0.6 (W/m-K), c p,w = 4180 (J/kg-K), F i = fuling factr (water t cpper) = (m -K/W) cpper: k c = 390 W/m-K

106 ΔT ΔT (50 30) (50 35) LMTD ΔTlm ln( ) ΔT 1 0 ΔT1 ln[(50 30) /(50 35)] C 管內冷卻水側 冷卻水之質量流率 = 冷卻量 / (c p,w T) = 78.3 kw / [4.18 (kj/kg-k) x (35-30)(K) = (kg/s) 冷卻水之體積流率 = 質量流率 / 密度 = (m 3 /s) = 5.7 (L/min) 水於銅管內之流速 (V) = 體積流率 / [37 ( 支管 ) x 每支管之內截面積 ] = 0.66 (m/s) Re = w Vd i / w = 996 (kg/m 3 ) x 0.66 (m/s) x (m) / (pa-s) = 171 Dittus-Blter Equatin: Nu d =0.03 Re 0.8 Pr 0.4 (Fully Turbulent Flw) Pr = c p,w w / k w = 4180 (J/kg-K) x (pa-s) / 0.6 (W/k-K) = 4.88 h i = (k w / d i )(0.03) Re 0.8 Pr 0.4 = 380 (W/m -K)

107 管外冷媒側 i c i i i i 1 1 d d d d 1 F ( )ln( ) ( )F ( )( ) U h k d d d h 管內流體熱阻管外流體熱阻管內污垢熱阻管外污垢熱阻管壁熱阻

108 assume T ; g 9.81 (m/s ) ; N 4.35 tubes/rw ; h d d m gρ h k 0.79 [ ] μ(δt)d 0. 79[ 11 K 3 fg 1/4 74 tubes 17 rws (m/s ) (1085) (kg/m ) (J/kg) (0.075) (W/ m K) 1/4 3 ] 168 (W/m K) (J s/m ) 11 (K) (m) Staggered Arrangement: N = 4.35, 查表可得 : F 0.9 h h d,n F h d (W/m -K)

109 1 1 d d d d 1 F ( )ln( ) ( )F i ( )( ) U h kc di di di hi [ ]ln( ) ( )( ) ( )( ) 1498 ()(390) m K ( ) W 管外流體之熱阻相對最大 W Hence, U 974 ( ) m-k Q = U A (LMTD) Hence A = Q / [U (LMTD)] = (W) / [974 (W/m -K) x (K)] = 所有銅管之總外表面積 = 4.6 (m ) 管之長度 = A / [ 總管數 x 單支管之周長 ] = 1.4 (m)

110 驗算 T 是否接近 11 0 C Q = 78.3 kw T B Q = 78.3 kw T water T A =50 0 C 1/h F (d /k)ln(d /d i ) (d /d i )(F i ) (d /d i )(1/h i ) 略 Q = h A T = h A (T A T B ) 平均管外表面溫度 T = Q / (h A) = (W) / [4.6 (m ) x 1498 (W/m-K)] = C 註 : 嚴格之計算必須重新假設一個 T 值, 再重覆以上之計算, 但差別應已不大