lecture 03: stkm3212

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.), UTMM. 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 OUTLINES1.2 MASS TRANSFER EXAMPLES AND FICK’S LAW.1.3 EXAMPLES OF THE MASS TRANSFER IN THE

FOOD PROCESSING INDUSTRY (MEMBRANE SEPARATION PROCESSES):

(a) Liquid Permeation Membrane (Membran Peresapan Cecair)(b) Reverse Osmosis (Keterbalikkan Osmosis)

1.4 BASIC MECHANISMS OF HEAT TRANSFER.1.5 CONDUCTION THROUGH HOLLOW CYLINDER.1.6 CONDUCTION THROUGH MULTILAYER

CYLINDER.1.7 COMBINED CONVECTION AND CONDUCTION

HEAT TRANSFER.

1.2 MASS TRANSFER EXAMPLES AND FICK’S LAW Mass transfer occurs when a component in a mixture MIGRATES in

the same phase or from phase to phase because of a difference in concentration between 2 points

EXAMPLES IN NATURE:1. Liquid in an open pail of water – evaporates – air – REASON: Difference in

concentration of water vapor at water surface & surrounding air.2. A piece of sugar added to a cup of coffee – dissolves – by itself – diffuses to the

surrounding solution. This mass transfer can be interpreted using: FICK’S LAW FICK’S LAW: “A random path molecule A might take diffusing

through B molecules from point (1) to (2) IF there are a greater number of A molecules near point (1) than at (2), and since molecules diffuse randomly in both directions, MORE A molecules will diffuse from (1) to (2) than (2) to (1) So, the net diffusion of A is HIGH-TO-LOW CONCENTRATION regions.

CONTINUE:

Schematic diagram of molecules diffusion process

CONTINUES:The general Fick’s Law equation:

J*Az = Molar flux of component A in the z direction due to molecular diffusion in (kg mol A/s.m2)

DAB = The molecular diffusivity of the molecules A in B in (m2/s)

CA = The concentration of A in (kg mol/m3)

z = The distance of diffusion in (m)AT STEADY STATE: J*Az & DAB = constant ------------ SO,

After rearranging and integrating the equation above:

1.3 EXAMPLES OF THE MASS TRANSFER IN THE FOOD PROCESSING INDUSTRY (MEMBRANE SEPARATION PROCESSES):

Separations by use of membranes are becoming important in the food process industries.

The membrane acts - semipermeable/semiporous barrier - separation occurs by the membrane controlling the rate of movement of various molecules - between - 2 liquid phases - 2 gas phases or liquid gas phase.

Method of concentrating food substances HEATING. Using HEAT will resulted a changes in product nutrients, active

ingredient and rheology. Therefore = MEMBRANCE SEPARATION IS COMMONLY USED

to concentrated the food substances without involving HEAT during the process FOOD QUALITY CAN BE ASSURED.

CONTINUE: PRINCIPLES OF MEMBRANE SEPARATION:

Fluids that contain 2 or more component inside will be channelled through membrane.

Membrane will allowed only certain component to pass through compared to the other components (SELECTIVITY).

This selectivity properties will be affected the separation. The factors that effecting the selectivity is PORES SIZE OF THE MEMBRANE

NORMAL PARTICLES FILTRATION > 1m pores size MICRO FILTRATION 0.05 m - 1.3m pores size ULTRA FILTRATION 0.005 m - 0.3m pores size NANO FILTRATION 0.001 m - 0.01m pores size REVERSE OSMOSIS < 0.001m pores size NOTE THAT: The smaller the pores size, the higher pressure

needed to run the separation process.

CONTINUE:

LIQUID PERMEATION MEMBRANCE PROCESS

CONTINUE: In membrane processes with liquids, the solute molecules

must first be transported or diffuse through the liquid film of the first liquid phase on one side of the solid membrane, through the membrane itself, and then through the film of the second liquid phase.

C1 = the bulk liquid-phase concentration of the diffusing solute A in (Kg mol A/m3).

C1i = the concentration of A in the fluid just adjacent to the solid.

C1iS = the concentration of A in the solid at the surface and is in equilibrium with C1i.

kc1 and kc2 = the mass transfer coefficient. K’ = the equilibrium distribution coefficient =

CONTINUE: So, MASS TRANSFER FLUX (PROFIL PEMINDAHAN JISIM) through each phase are

EQUAL to each other.

Where; PM = permeability in solid (m/s) L = thickness in (m)DAB = Diffusivity of A in the solid (m2/s)

into {1}

-------- {1}

-------- {2}

CONTINUE:SIMPLIFY THE EQUATION:

Adding all equation {3} in ONE equation, the internal concentration c1i and c2i drop out, the FINAL EQUATION :

-------- {3}

EXAMPLE 1:

CONTINUE:

CONTINUE:

**LOGIN THE eLEARNING SYSTEM TO GET THE ADDITIONAL INFO ON

THE REVERSE OSMOSIS APPLICATION**

1.4 BASIC MECHANISMS OF HEAT TRANSFER Food processing always involve with a lot of heat transfer.

e.g.: Heat intake process: Pasteurization, sterilization & concentration.

e.g.: Heat removal process: Freezing and Cooling. Heat transfer always occur from HOT MEDIUM (Medium

panas) TO FOOD MATERIAL OR FOOD MATERIAL TO COOL MEDIUM (medium sejuk).

3 ways of heat transfer : CONDUCTION = Pengaliran CONVENCTION = Perolakan RADIATION = Sinaran

CONDUCTION:CONDUCTION (Pengaliran): Heat transfer through conduction involves energy transfer at

the molecules level. When a molecule were given an energy, it will VIBRATE at its

location. AMPLITUDE OF THE VIBRATION will increased heat energy. This vibration will transfer from one molecule to the other

molecule WITHOUT molecules translation movement (TIDAK MELIBATKAN PERUBAHAN PADA PERGERAKAN MOLEKUL ASAL Only vibration).

If there is a temperature gradient in the substances, the heat transfer will occurred from the HIGH TEMPERATURE REGION TO LOW TEMPERATURE REGION.

CONTINUE: The RATE OF HEAT TRANFER (q)/SURFACE AREA (A) is called = HEAT

TRANSFER FLUX (Profil Pemindahan Haba). THE HEAT FLUX for the heat transfer through CONDUCTION is

PROPORTIONAL with the TEMPERATURE GRADIENT.

“Fourier’s First Law of Heat Transfer”Where:q = Heat transfer rate in direction x (W)A = Surface area (m2)

(TABLE 4.1-1) --- k = Thermal conductivity for materials (W/m. K)dT/dx = The temperature gradient in the x direction (K/m)

CONTINUE:

(-) -ve sign shows that the heat is conduct (mengalir) from the high temperature to the low temperature.

dT/dx This Gradient Temperature is the driving force (daya penggerak) for the heat transfer.

The heat conduction can be in the STEADY STATE OR IN UNSTEADY STATE.

CONTINUE:Under the steady state condition: (INTEGRATING AND REARRANGING THE FOURIER’S FIRST LAW):

EXAMPLE 2:

CONVECTION:CONVECTION (Perolakan): The way of heat transfer through CONVECTION is involving

the molecules movement from ONE LOCATION TO ANOTHER LOCATION.

It also involves heat energy exchanged (pertukaran tenaga haba) to other molecules at a new location.

2 types of CONVECTION: Free/natural convection (Perolakan bebas/tabii): a) Occurred on its own without outside force.b) It is because of the changes of DENSITY and TEMPERATURE DIFFERENTIAL.c) e.g.: Air with high temperature becoming less dense and eventually flow upwards.

CONTINUE: Forced convection (Perolakan paksa)

a) Occurred when there was an OUTSIDE FORCE (DAYA LUARAN) to the system.

b) The fluid is forced to flow by pressure differences = e.g.: pump, fan etc.

SO, THE HEAT TRANSFER THROUGH CONVECTION can be represent using the “Newtonian Cooling Law”

CONTINUE:

q = hA(Tm - Ts)Where: q = Heat transfer rate (W)

h = Convective coefficient (W/m2.K)A = Surface area (m2)Tm = Bulk of average temperature of the fluid (K)

Ts = Temperature of the wall in contact with fluid (K) The higher h value means the higher of the heat transfer rate of the convective system.

RADIATION

RADIATION (Sinaran): In radiant heat transfer, the medium through which the

heat is transferred usually not HEATED. Radiation heat transfer is the transfer of heat by

electromagnetic radiation. Medium to transfer heat is not applicable. No involvement of the molecules. Can occur in the vacuum. e.g.: x-rays, light waves, gamma rays e.g.: Furnace with boiler tubes, microwave oven baking.

CONTINUE: FLUX HEAT TRANSFER by radiation from solid surface can be

represent as:

q = TA4

Where: q = Heat transfer rate (W)A = Surface area of the body (m2) = Constant Stefan-Boltzmann (5.669 x 10-8W/m2.K4) = Emissivity (kepancaran)TA = Temperature of the solid body (K)

Emissivity for the PERFECT BLACK BODY = 1.0 All real materials have an emissivity < 1.0

In many examples in the process industries, HEAT IS TRANSFERRED through the walls of a thick-walled cylinder as in a pipe that may or may not be insulated.

Figure below = HEAT (q) is flowing radially from inside to outside.

1.5 CONDUCTION THROUGH HOLLOW CYLINDER (Air-Metal-Fluid)

CONTINUE:

REMEMBER ----------

CONTINUE:

As if, A2/A1 > 1.5 -----------

As if, A2/A1 < 1.5 -----------

A1m = A1 + A2/2

Cylinder resistance ------------

= A2 - A1/In (A2/A1)

LOG MEAN AREA

Unit R = K/W

EXAMPLE 3:LENGTH OF TUBING FOR COOLING COILA thick-walled cylindrical tubing of hard rubber having an inside radius of 5 mm and an outside radius of 20 mm is being used as a temporary cooling coil in a bath. Ice water is flowing rapidly inside and the inside wall temperature is 274.9 K. The outside surface temperature cooling coil is 297.1 K. A total of 14.65 W must be removed from the bath by cooling coil. How many (m) of tubing are needed?ANS:

T2 = 297.1 K (outer cooling coil)

q = 14.65 W must be removed.Tbath = not given.

WATER BATH ROOM

T1 = 274.9 K (inside coil)Ice H2O

Length of cooling coil tubing needed?

CONTINUE:From TABLE 4.1.1: The thermal conductivity at 0 0C (273 K) is k = 0.151 W/m.K. Since data at the other temperature (Tbath) not given, assume that the outer temperature cooling coil (T2) that contact with the wall of the bath room is same: T2 = Tbath

r1 = 5/1000 = 0.005 m; r2 = 20/1000 = 0.02 mThe calculation will be done first for a LENGTH OF 1.0 m of tubing (Assumption):A1 = 2Lr1 = 2(1.0m)(0.005m) = 0.0314 m2; A2 = 0.1257 m2

=0.1257 – 0.0314 In (0.1257/0.0314)-------------------------------- 0.0680 m2=

CONTINUE:Substituting into:

q = 0.151(0.0682)[(274.9 - 297.1)/(0.02 - 0.005)]

q = - 15.2 W in 1-m assumption length of cooling tube (Negative sign indicates that the heat flow from r2 on the outside and contact the bath room wall TO r1 on the inside. Since 15.2 W is removed for a 1-m length (ASSUMPTION), the ACTUAL length needed is:

Length = 14.65 W/15.2 (W/m) = 0.964 m **Note that the thermal conductivity of rubber is quite small.

Generally, metal cooling coils are used, BUT the thermal conductivity of metals is quite high NOT SUITABLE FOR SMALL COOLING CONDITION**

1.6 CONDUCTION THROUGH MULTILAYER CYLINDER (Air-Insulation-Metal-Fluid) In the process industries, HEAT TRANSFER OFTEN occurs

through MULTILAYERS OF CYLINDERS. e.g.: When heat is being transferred through the walls of an

insulated pipe. Figure below shows a PIPE WITH TWO LAYERS OF

INSULATION AROUND IT, SO, we have 3 concentric hollow cylinders.

The heat transfer rate (q) = is same for each layer ----- STEADY STATE CONDITION

CONTINUE:

Radial heat flow through multiple cylinders in series

METAL TUBE WALL = AINSULATED 1 = B

INSULATED 2 = C

CONTINUE:

Where:

COMBINE the equations to eliminate T2 and T3. SO, THE FINAL EQUATION

R = The overall resistance is the SUM of individual resistances in series

EXAMPLE 4:HEAT LOSS FROM AN INSULATED PIPEA thick-wall tube of stainless steel (A) having a k = 21.63 W/m.K with dimensions of 0.0254 m ID and 0.0508 m OD is covered with a 0.0254 m layer of asbestos (B) insulation, k = 0.2423 W/m.K. The inside wall temperature of the pipe is 811 K and the outside surface of the insulation is at 310.8 K. For a 0.305 m length of pipe, calculate the heat loss and also the temperature at the interface between the metal and the insulation.ANS: Calling T1 = 811 K, T2 (the interface) = ?, T3 = 310.8 K, the dimensions are:r1 = 0.0254/2 = 0.0127m; r2 = 0.0508/2 = 0.0254m; r3 = (r2-r1) + r1 + 0.0254 m (layer of asbestos) = 0.0508m

CONTINUE:The areas are as follows for L = 0.305 mA1 = 2Lr1 = 2(0.305)(0.0127) = 0.024 m2

A2 = 2Lr2 = 2(0.305)(0.0254) = 0.0487 m2

A3 = 2Lr3 = 2(0.305)(0.0508) = 0.0974 m2

The log mean areas for the stainless steel (A) and asbestos (B) are:

0.0487 - 0.024 In (0.0487/0.024)--------------------------------= =

0.0974 - 0.0487 In (0.0974/0.0487)--------------------------------= =

0.0351 m2

0.0703 m2

CONTINUE:

.

Asbestos (B)

Stainless Steel (A)

Inside the pipe(fluid or gas flow) r1

r2r3

0.0254 m

q

CONTINUE:The resistances (R) are:

Hence, the HEAT TRANSFER RATE IS:

To calculate temperature T2:

=RA =r2 - r1----------kA . AAIm

0.0125-------------------21.63(0.0351)

= 0.01673 K/W

RB =r3 - r2----------kB . ABIm

0.0254-------------------0.2423(0.0703)

= 1.491 K/W=

q =T1 - T2------------ RA

=811 - 310.8

--------------------0.01673 + 1.491

= 331.7 Wq =

T1 - T3------------RA + RB

=811 - T2------------0.01673

= T2 = 805.5 K

“Only small temperature drop occurs across the metal wall because of its high thermal conductivity”

1.7 COMBINED CONVECTION AND CONDUCTION HEAT TRANSFER (Fluid-Insulation-Metal-Fluid) In many practical situations the surface temperature (or

boundary conditions at the surface) are not known, but there is a FLUID on both sides of the solid surfaces.

Consider the plane wall in fig. 4.3-3a with HOT FLUID in T1 on the inside surface and COOL FLUID at T4 on the outside surface.

The outside CONVECTIVE COEFFICIENT is ho (W/m2.K) and hi

on the inside.

CONTINUE:METAL TUBE WALL

CONTINUE:

The HEAT TRANSFER RATE is using the combination of CONVECTIVE and CONDUCTION

Expressing 1/hiA, xA/kAA and 1/hoA = R (resistances) and combine:

CONTINUE:SO, THE OVERALL HEAT TRANSFER RATE through the cylinder will be:

Where: Ai = 2..L.ri (The inside area of the metal tube)

Aim = Log mean Area of the metal tube.

Ao = The outside area.

L = Length of a pipe

q = T1 - T4

1/hi . Ai + (ro - ri)/kA . AIm + 1/ho . Ao -------------------------------------------------

T1 - T4

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:

Toverall = T1 - T4

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: The OVERALL HEAT TRANSFER COEFFICIENT (U) for

the cylinder may be based on the INSIDE AREA (Ai) or OUTSIDE AREA (Ao) OF THE TUBE: So -----------

q = Ui.Ai (T1 - T4) = Uo.Ao (T1 - T4) T1 - T4

R -----------=REMEMBER

----------

1/hi + (ro - ri) Ai /kA . AIm + Ai /Ao . ho Ui

UoAo /Ai . hi + (ro - ri) Ao /kA . AIm + 1/ho

=

=

------------------------------------------------

------------------------------------------------

1

1

1= ---------

Ai. R

= ---------Ao. R

1

EXAMPLE 5:Heat Loss by Convection and Conduction and Overall USaturated steam at 267 0F is flowing inside a ¾-in. steel pipe having an ID of 0.824 in. and an OD 1.050 in. The pipe is insulated with 1.5 in. of insulation on the outside. The convective coefficient for the inside steam surface of the pipe is estimated as hi = 1000 btu/h.ft2.0F and the convective coefficient on the outside of the lagging (pembalut) is estimated as ho = 2 btu/h.ft2.0F. The mean thermal conductivity of the metal is 45 W/m.K or 26 btu/h.ft.0F and 0.064 W/m.K or 0.037 btu/h.ft.0F for the insulation.

a) Calculate the heat loss for 1 ft of pipe using resistances if the surrounding fluid outside the insulated pipe is at 80 0F.

b) Repeat using the overall Ui based on the inside area Ai.

CONTINUE:

ri

r1r0

1.5 in.

q

Lagging (B)

Stainless Steel (A)

Inside the pipe (saturated steam flow), Ti = 267 0F

T0 = 80 0F Surrounding Fluid A

CONTINUE:ANS: Calling ri the inside radius of the pipe steel pipe, r1 the outside radius of the pipe & r0 the outside radius of the lagging (pembalut), then:ri = 0.0824/2 = 0.412 in. ---- convert to ft = 0.412/12 = 0.034 ft r1 = 1.050/2 = 0.525 in. ---- convert to ft = 0.525/12 = 0.044 ft ro = ri + (r1 - ri) + 1.5 in. = 0.412 + 0.113 + 1.5 = 2.025 in. ----convert to ft = 2.025/12 = 0.169 ft

For 1 ft of pipe, the area are as follows: Ai = 2Lri = 2(1)(0.034) = 0.2157 ft2

A1 = 2Lr1 = 2(1)(0.044) = 0.2750 ft2

A0 = 2Lr0 = 2(1)(2.025) = 1.060 ft2

CONTINUE:The log mean areas for the STEEL PIPE (A) & LAGGING (B)

are:

The various resistances are:

AAIm =A1 - Ai------------- =

In (A1/Ai)

0.2750 - 0.2157------------------------In(0.2750/0.2157)

= 0.245 ft2

ABIm =A0 - A1------------- =

In (A0/A1)

1.060 - 0.2750------------------------In(1.060/0.2750)

= 0.583 ft2

--------- =Ri = hi.Ai

1 = 11000(0.2157)

------------------ 0.00464

=RA = kA.AAIm

r1 - ri ------------ 0.044 - 0.034 ---------------------26(0.245) = 0.00148

CONTINUE:

RB = kB.ABIm

r0 - r1----------- 0.169 - 0.044 ---------------------0.037(0.583) = 5.80=

R0 = h0.A0

1 = 12(1.060)------------ = 0.472-------

SO:=q = Ti - T0

Ri + RA + RB + R0-------------------------

267 - 80---------------------------------------------------0.00464 + 0.00148 + 5.80 + 0.472

q = 29.8 btu/hrs

CONTINUE:Part (b): q = Ui.Ai (Ti – T0)

So; Ui = 1/0.2157(6.278) = 0.738 btu/h.ft2.0F

Then, to calculate q = q = Ui.Ai (Ti – T0) = (0.738)(0.2157)(267-80) q = 29.8 btu/hrs (873 W)

Ti - To

R -----------=

1= ---------

Ai. RUi