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LECTURE NOTES 03/07  STKM3212: FOOD PROCESSING TECHNOLOGY   MASS AND HEAT TRANSFER IN STEADY STATE  (Perpindahan Jisim dan Tenaga Haba dalam Keadaan Mantap)  SAIFUL IRWAN ZUBAIRI   PMIFT, Grad B.E.M.   B. Eng. (Chemical-Bioprocess) (Hons.), UTM M. Eng. (Bioprocess), UTM ROOM NO.: 2166, CHEMISTRY BUILDING, TEL. (OFF.): 03-89215828, FOOD SCIENCE PROGRAMME, CENTRE OF CHEMICAL SCIENCES AND FOOD TECHNOLOGY,  UKM BANGI, SELANGOR
1.1 OUTLINES ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
1.2 MASS TRANSFER EXAMPLES AND FICK’S LAW ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
CONTINUE: Schematic diagram of molecules diffusion process
CONTINUES: ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],
1.3  EXAMPLES OF THE MASS TRANSFER IN THE FOOD PROCESSING INDUSTRY (MEMBRANE SEPARATION PROCESSES): ,[object Object],[object Object],[object Object],[object Object],[object Object]
CONTINUE: ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
CONTINUE: LIQUID PERMEATION MEMBRANCE PROCESS
CONTINUE: ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
CONTINUE: ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],into {1} -------- {1} -------- {2}
CONTINUE: ,[object Object],[object Object],-------- {3} 
EXAMPLE 1:
CONTINUE:
CONTINUE:
[object Object]
1.4 BASIC MECHANISMS OF HEAT TRANSFER ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
CONDUCTION: ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
CONTINUE: ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
CONTINUE: ,[object Object],[object Object],[object Object]
CONTINUE: ,[object Object],
EXAMPLE 2:
CONVECTION: ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
CONTINUE: ,[object Object],[object Object],[object Object],[object Object]
CONTINUE: ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
RADIATION ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
CONTINUE: ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
[object Object],[object Object],1.5 CONDUCTION THROUGH HOLLOW CYLINDER  (Air-Metal-Fluid)
CONTINUE: REMEMBER ----------
CONTINUE: As if, A 2 /A 1  > 1.5 ----------- As if, A 2 /A 1  < 1.5 ----------- A 1m  = A 1  + A 2 /2 Cylinder resistance ------------ = A 2  - A 1 /In (A 2 /A 1 ) LOG MEAN AREA Unit R = K/W
EXAMPLE 3: ,[object Object],[object Object],[object Object],T 2  = 297.1 K (outer cooling coil) q = 14.65 W must be removed. T bath  = not given.  WATER BATH ROOM T 1  = 274.9 K (inside coil) Ice H 2 O Length of cooling coil tubing needed?
CONTINUE: ,[object Object],[object Object],[object Object],[object Object],= 0.0680 m 2 = 0.1257 – 0.0314  In (0.1257/0.0314 ) --------------------------------
CONTINUE: ,[object Object],[object Object],[object Object],[object Object],[object Object]
1.6 CONDUCTION THROUGH MULTILAYER CYLINDER   (Air-Insulation-Metal-Fluid) ,[object Object],[object Object],[object Object],[object Object]
CONTINUE: Radial heat flow through multiple cylinders in series  METAL TUBE WALL = A INSULATED 1 = B INSULATED 2 = C
CONTINUE: Where: COMBINE the equations to eliminate T 2  and T 3 . SO, THE FINAL EQUATION     R = The overall  resistance is the  SUM  of individual resistances in series
EXAMPLE 4: ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
CONTINUE: ,[object Object],[object Object],[object Object],[object Object],[object Object],= = = = 0.0351 m 2 0.0703 m 2 0.0487 - 0.024  In (0.0487/0.024 ) -------------------------------- 0.0974 - 0.0487  In (0.0974/0.0487 ) --------------------------------
CONTINUE: . Asbestos (B) Stainless Steel (A) Inside the pipe (fluid or gas flow)  r 1 r 2 r 3 0.0254 m q
CONTINUE: ,[object Object],[object Object],[object Object],= = = 811 - 310.8 -------------------- 0.01673 + 1.491 =  331.7 W = 811 - T 2 ------------ 0.01673  =  T 2  =  805.5 K   “ Only small temperature drop occurs across the metal wall because of its high thermal conductivity” R A  = r 2  - r 1 ---------- k A  . A AIm 0.0125 ------------------- 21.63(0.0351) =  0.01673 K/W R B  = r 3  - r 2 ---------- k B   . A BIm 0.0254 ------------------- 0.2423(0.0703) =  1.491 K/W q = T 1  - T 2 ------------ R A   q = T 1  - T 3 ------------ R A  + R B
1.7 COMBINED CONVECTION AND CONDUCTION HEAT TRANSFER  (Fluid-Insulation-Metal-Fluid) ,[object Object],[object Object],[object Object]
CONTINUE: METAL TUBE WALL
CONTINUE: The  HEAT TRANSFER RATE  is using the combination of  CONVECTIVE  and  CONDUCTION Expressing  1/ h i A ,     xA /k A A and 1/h o A  = R (resistances)  and combine:
CONTINUE: ,[object Object],[object Object],[object Object],[object Object],L = Length of a pipe  q =  T 1  - T 4 1/ h i  . A i  +  (r o  - r i )/k A  . A Im  + 1/h o  . A o   ------------------------------------------------- T 1  - T 4    R  ----------- = REMEMBER ----------
CONTINUE: The overall heat transfer by combined  CONDUCTION & CONVECTIVE  is often expressed in terms of an  OVERALL HEAT TRANSFER COEFFICIENT (U) defined by :   T overall  = T 1  - T 4 A more important application is heat transfer from a fluid outside a cylinder through a METAL WALL and to a fluid inside the TUBE, as often occurs in  “ HEAT EXCHANGER ” ------ REFER FIG 4.3-3 (b)
CONTINUE: ,[object Object],[object Object],REMEMBER ---------- 1/ h i  +  (r o  - r i )  A i  /k A  . A Im  +  A i  / A o  . h o   U i U o A o  / A i  . h i   +  (r o  - r i )  A o  /k A  . A Im  + 1/h o   = = ------------------------------------------------ ------------------------------------------------ 1 1 1 T 1  - T 4    R  ----------- = 1 = --------- A i .   R = --------- A o .   R
EXAMPLE 5: ,[object Object],[object Object],[object Object],[object Object]
CONTINUE: Lagging (B) Stainless Steel (A) Inside the pipe (saturated steam flow), T i  = 267  0 F  T 0  = 80  0 F Surrounding Fluid A  r i r 1 r 0 1.5 in. q
CONTINUE: ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
CONTINUE: ,[object Object],[object Object],--------- = = A AIm  = A 1  - A i ------------- = In   (A 1 /A i ) 0.2750 - 0.2157 ------------------------ In(0.2750/0.2157) =  0.245 ft 2 A BIm  = A 0  - A 1 ------------- = In   (A 0 /A 1 ) 1.060 - 0.2750 ------------------------ In(1.060/0.2750) =  0.583 ft 2 R i  =  h i .A i 1 = 1 1000(0.2157 ) ------------------ 0.00464 R A  =  k A .A AIm r 1  - r i   ------------ 0.044 - 0.034  --------------------- 26(0.245) =  0.00148
CONTINUE: = SO: = q = 29.8 btu/hrs  R B  =  k B .A BIm r 0  - r 1 ----------- 0.169 - 0.044  --------------------- 0.037(0.583) =  5.80 R 0  =  h 0 .A 0 1 = 1 2(1.060) ------------ = 0.472 ------- q =  T i  - T 0 R i  + R A  + R B  + R 0 ------------------------- 267 - 80 --------------------------------------------------- 0.00464 + 0.00148 + 5.80 + 0.472
CONTINUE: ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],T i  - T o    R  ----------- = 1 = --------- A i .   R U i

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Lecture 03: STKM3212

  • 1. LECTURE NOTES 03/07 STKM3212: FOOD PROCESSING TECHNOLOGY MASS AND HEAT TRANSFER IN STEADY STATE (Perpindahan Jisim dan Tenaga Haba dalam Keadaan Mantap) SAIFUL IRWAN ZUBAIRI PMIFT, Grad B.E.M. B. Eng. (Chemical-Bioprocess) (Hons.), UTM M. Eng. (Bioprocess), UTM ROOM NO.: 2166, CHEMISTRY BUILDING, TEL. (OFF.): 03-89215828, FOOD SCIENCE PROGRAMME, CENTRE OF CHEMICAL SCIENCES AND FOOD TECHNOLOGY, UKM BANGI, SELANGOR
  • 2.
  • 3.
  • 4. CONTINUE: Schematic diagram of molecules diffusion process
  • 5.
  • 6.
  • 7.
  • 8. CONTINUE: LIQUID PERMEATION MEMBRANCE PROCESS
  • 9.
  • 10.
  • 11.
  • 15.
  • 16.
  • 17.
  • 18.
  • 19.
  • 20.
  • 22.
  • 23.
  • 24.
  • 25.
  • 26.
  • 27.
  • 29. CONTINUE: As if, A 2 /A 1 > 1.5 ----------- As if, A 2 /A 1 < 1.5 ----------- A 1m = A 1 + A 2 /2 Cylinder resistance ------------ = A 2 - A 1 /In (A 2 /A 1 ) LOG MEAN AREA Unit R = K/W
  • 30.
  • 31.
  • 32.
  • 33.
  • 34. CONTINUE: Radial heat flow through multiple cylinders in series METAL TUBE WALL = A INSULATED 1 = B INSULATED 2 = C
  • 35. CONTINUE: Where: COMBINE the equations to eliminate T 2 and T 3 . SO, THE FINAL EQUATION  R = The overall resistance is the SUM of individual resistances in series
  • 36.
  • 37.
  • 38. CONTINUE: . Asbestos (B) Stainless Steel (A) Inside the pipe (fluid or gas flow) r 1 r 2 r 3 0.0254 m q
  • 39.
  • 40.
  • 42. CONTINUE: The HEAT TRANSFER RATE is using the combination of CONVECTIVE and CONDUCTION Expressing 1/ h i A ,  xA /k A A and 1/h o A = R (resistances) and combine:
  • 43.
  • 44. CONTINUE: The overall heat transfer by combined CONDUCTION & CONVECTIVE is often expressed in terms of an OVERALL HEAT TRANSFER COEFFICIENT (U) defined by :  T overall = T 1 - T 4 A more important application is heat transfer from a fluid outside a cylinder through a METAL WALL and to a fluid inside the TUBE, as often occurs in “ HEAT EXCHANGER ” ------ REFER FIG 4.3-3 (b)
  • 45.
  • 46.
  • 47. CONTINUE: Lagging (B) Stainless Steel (A) Inside the pipe (saturated steam flow), T i = 267 0 F T 0 = 80 0 F Surrounding Fluid A r i r 1 r 0 1.5 in. q
  • 48.
  • 49.
  • 50. CONTINUE: = SO: = q = 29.8 btu/hrs R B = k B .A BIm r 0 - r 1 ----------- 0.169 - 0.044 --------------------- 0.037(0.583) = 5.80 R 0 = h 0 .A 0 1 = 1 2(1.060) ------------ = 0.472 ------- q = T i - T 0 R i + R A + R B + R 0 ------------------------- 267 - 80 --------------------------------------------------- 0.00464 + 0.00148 + 5.80 + 0.472
  • 51.