R ADIATIVE H EAT T RANSFER

1.2 HEAT TRANSFER IN FOOD PROCESSING .1 M ODES OF H EAT T RANSFER.1 MODESOF HEAT TRANSFER

1.2.4 R ADIATIVE H EAT T RANSFER

Example 1.5

What is the overall heat transfer coeffi cient based on inside area and outside area for Example 1.4?

Solution 1.5

From Example 1.4, we have

− °

= × ³

i 1.27 10 C/W R

− °

= × ⁵

w 1.38 10 C/W R

− °

= × ²

o 1.06 10 C/W R

Therefore, the overall heat transfer coeffi cient based on inside area is

− − − °

= +i w+ o= × ³+ × ⁵+ × ²=

i i

1 R R R 1.27 10 1.38 10 1.06 10 0.0119 C/W

U A

= ²°

i 10.70 W/m C U

Similarly, the overall heat transfer coeffi cient based on outside area is

= ²°

o 8.92 W/m C U

Therefore, a shape factor, F, is defi ned. The radiative energy exchange between the surface, 1, of a body and the surroundings, 2, of the body can be determined by

4 4

12 1 ( K1 K 2)

q=sF Ae T −T (1.27)

If the surface, 1, is enclosed by the surroundings, 2, as shown in Figure 1.1c, then F12 = 1. Similar to the convective heat transfer coeffi cient, a radiative heat transfer coeffi cient, h_r, may be expressed as

r ( K 1 K 2)

q=h A T −T (1.28)

where

2 2

r ( K 1 K 2)( K 1 K 2)

h =se T +T T +T (1.29)

Example 1.6

A food item is heated in an oven. The oven temperature is set at 180°C. The average surface temperature of the food is 50°C. The food item has a spherical shape with a diameter of 20 cm. What is the radiative heat transfer rate? What is the radiative heat transfer coeffi cient?

Solution 1.6

The surface area of the food item is

= π = ×4 ² 4 3.14159 0.1× ²=0.1257 m²

A r

Since the food item is placed in an oven, the shape factor is F₁₂ = 1. The emissivity of the food surface is assumed to be 1. Using Equation 1.27, the radiative heat transfer rate is

( ) ( )

−

= −

⎡ ⎤

= × × × × ×⎣ + − + ⎦

=

4 4

12 1 K1 K 2

4 4

8

( )

5.669 10 1 0.1257 1 273.15 180 273.15 50 222.8 W

q sF A Te T

Using Equation 1.29, the radiative heat transfer coeffi cient is

= ²+ ² + = ² °

r (K1 K 2)(K1 K 2) 13.63 W/m C

h seT T T T

1.2.5 PHASE CHANGE HEAT TRANSFER

Most foods such as raw meats and vegetables have high moisture contents. Water itself is widely used as a processing medium in food processes. Water normally exists in one of three states: solid, liquid, and gas. The transition between two states is called a phase transformation or phase change. During phase transformation, the

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temperature of pure water remains constant with added energy because all energy is used to transform water from one state to another. As water is widely present in foods and is used as a processing medium, it is necessary to discuss heat transfer with the phase changes of water in the food industry.

During thermal food processing, the water in the food may experience phase changes. Frying and grilling of foods involve a phase change from liquid water to vapor. There is an evaporation front, which divides the food body into two parts of the outer crust and internal core regions, as shown in Figure 1.2. The evaporation front moves toward the center as frying and grilling processes proceed. If a frozen food is used during frying and grilling, there will be two moving boundaries:

the thawing front and the evaporation front. The heat transfer mechanisms across the moving boundary must account for the latent heat of phase changes of water.

The moving front of phase change in the food can be tracked by the energy balance on the front, which is given by (Wang and Singh, 2004)

1 2 w

1 2

d ( ) d

T T S t

k k X

x x ρ t

∂ ∂

⎛ ⎞ ⎛ ⎞

− ⎜⎝∂ ⎟⎠ + ⎜⎝∂ ⎟⎠ =l (1.30)

0, ( )

t> x=S t (1.31)

where

subscripts 1 and 2 denote phase 1 and phase 2, respectively

∂T/∂x is the temperature gradient k is thermal conductivity

Outer crust

Internal core region

Evaporation front, S(t)

q dT

dx

T2

T1

x1 x2

x Tb

1

dT dx ₂ T

FIGURE 1.2 Schematic of heat transfer with phase changes (frying). (Reprinted from Wang, L.J. and Sun, D.W., Thermal Food Processing: New Technologies and Quality Issues, CRC Press, Boca Raton, 2006. With permission.)

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l is the latent heat of phase change r is the density of the product X_w is the moisture content

S(t) is the position of the moving boundary at time t

Water is also widely used as a processing medium. Boiling and condensation involve a phase change between liquid water and vapor. Boiling heat transfer is particularly important in processing operations such as evaporation in which the boiling of liquids takes place either at submerged surfaces or on the inside surface of vertical tubes as in a climbing fi lm evaporator. The heat fl ux changes dramatically as a function of the temperature difference between the surface and the boiling liquid, rising to a peak value and falling away sharply. This is caused by the strong dependence between the heat transfer coeffi cient and the temperature difference, which is shown in Figure 1.3. In order to avoid overheating and damaging the walls of the heater, equipment should ideally be operated in the nucleate boiling zone, just below the critical tem-perature difference as shown in Figure 1.3. Vapor condensation is also used in food thermal processes. Consider a sterilization process used for manufacturing canned foods. If steam is used as a heating medium, the condensation of steam on the metal surface of a can results in a signifi cantly higher heat transfer than if hot water were used to heat the cans. Vapor condenses on a cold surface in one of two distinct ways:

fi lm condensation and drop condensation. The presence of noncondensable gases in steam affects the rate of condensation, and the fi lm heat transfer coeffi cient may be reduced considerably (Wang and Sun, 2006).

The heat fl ux due to phase change of boiling and condensation can be expressed as

p ( s )

q=h A T −T_∞ (1.32)

FIGURE 1.3 Relationship between boiling heat transfer coeffi cient and temperature difference.

(Reprinted from Wang, L.J. and Sun, D.W., Thermal Food Processing: New Technologies and Quality Issues, CRC Press, Boca Raton, 2006. With permission.)

Critical temperature difference

Temperature difference between surface and liquid

Boiling heat transfer coefficient

Film boiling Nucleate boiling

Interface evaporation

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where

hp is the heat transfer coeffi cient of phase change

Ts and T_∞ are the temperatures of solid surface and bulk fl uid, respectively A is the surface area where the phase change occurs

The heat transfer coeffi cients of a phase change when a liquid is vaporized or when a vapor is condensed are considerably greater than that for heat transfer without a phase change, as shown in Table 1.3. However, it is diffi cult to measure heat transfer coeffi cients of phase changes.

The heat transfer coeffi cient in nucleate boiling may be calculated by a correla-tion. Kutateladze’s correlation is a commonly used one, which is given by (Wang and Sun, 2006):

0.7

0.5 max 0.35

v

0.0007 Pr

hb q P

k σ

⎡ ⎤ −

⎛ ⎞ = ⎢ Ψ⎥

⎜ ⎟⎝ ⎠y ⎣ ⎦

alr (1.33)

where

l v

( )

Ψ =g

− s

r r (1.34)

0.25

l v

max v 2

v

( )

0.16 g

q = ⎛⎜⎝σ − ⎞⎟⎠

r r

lr r (1.35)

where

k is the thermal conductivity (W/m°C) a is the thermal diffusivity (m²/s) P is the pressure (Pa)

l is the latent heat of phase change (J/kg) r is the density (kg/m³)

Pr is the Prandtl number

s is the Stefan–Boltzmann constant at s = 5.669 × 10⁻⁸ W/m²K⁴ g is the acceleration due to gravity (m/s²)

Subscripts l and v denote the liquid and gaseous phases, respectively

The fi lm heat transfer coeffi cient for condensation can be predicted from the Nusselt theory, which gives the mean fi lm coeffi cient by (Wang and Sun, 2006)

2 3 0.25

cd 0.943 k g , for a vertical surface

h L T

⎛ ⎞

= ⎜⎝ ∆ ⎟⎠

r l

m (1.36)

2 3 0.25

cd 0.725 k g , for a horizontal tube

h d T

⎛ ⎞

= ⎜⎝ ∆ ⎟⎠

r l

m (1.37)

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where

r is the density (kg/m³)

k is the thermal conductivity (W/m°C) g is the acceleration due to gravity (m/s²) l is the latent heat of the phase change (J/kg) m is the viscosity (Pa s)

DT is the temperature difference for the phase change (°C) L is the length (m)

d is the diameter of the tube (m)

1.2.6 HEAT TRANSFERWITH ELECTROMAGNETIC WAVES

Microwave energy, which is an electromagnetic energy, is widely used in the food indus-try. Microwaves are transmitted as waves, which can penetrate foods and interact with the polar molecules in foods such as water to be converted to heat. An electromagnetic spectrum is normally characterized by wavelength (l) and frequency (f). Microwaves are nonionizing electromagnetic waves, and commercial microwave heating applications use frequencies of 2450 MHz, sometimes 915 MHz in the United States and 896 MHz in Europe. The depth of penetration into a food is directly related to frequency, and lower frequency microwaves penetrate more deeply. As a microwave can penetrate into foods, it can heat foods quicker than traditional thermal processing methods and transfer heat from the outer surface to the inside of foods by conduction. However, it should be noted that conduction or convection may also occur during microwave heating if a temperature difference exists in foods. In this case, the conversion rate of microwave energy per unit volume of foods can be considered as a source term in a heat transfer model.

The conversion of microwave energy to heat depends on the properties of the microwave energy source and the dielectric properties of the foodstuffs. The power dissipation or rate of energy conversion per unit volume of foods, P (W/m³), is given by (Wang and Sun, 2006)

15 2

5.56 10

P= × ⁻ E fe ″ (1.38)

and

″ = tan

e e¢ d (1.39)

where

E is the microwave electrical fi eld strength (V/m) f is the frequency of the microwave

e¢ and e¢¢ are the dielectric constant and dielectric loss, respectively d is the loss angle

However, the suitability of a food for microwave heating is crucially dependant on the penetration characteristics of microwave into the food. The distribution of micro-wave energy within a material is determined by the attenuation factor, a, which is calculated by

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(

²

)

^0.5

2 1 tan 1

2

π ⎡ ⎤

= ⎢⎣ + − ⎥⎦

a e¢ d

l (1.40)

The microwave electrical fi eld strength is a function of penetration depth, x, which can be given by

2

0e ^x

E=E ^−a (1.41)

The microwave electrical fi eld strength profi le obtained by Equation 1.41 can be used to calculate the local conversion rate of microwave energy using Equation 1.38. The power absorption rate at the penetration depth, x, can also be determined by the Lambert’s model, which is expressed as

2

0e ^x

P=P ^−a (1.42)

where P₀ is the incident power on the surface (W/m²).

1.2.7 UNSTEADY-STATE HEAT TRANSFER

For steady-state heat transfer, the temperature of a food does not change with time.

However, in the majority of thermal food processes, the temperature of a food product changes continuously and unsteady-state heat transfer is more commonly found. The generalized governing equation of unsteady-sate heat transfer can be expressed as

Source

Accumulation Convection Diffusion

x y z

T T T T T T T

c u u u k k k S

t x y z x x y y z z

⎛ ⎞

⎛ ⎞

∂ ∂ ∂ ∂ ∂ ⎛ ∂ ⎞ ∂ ∂ ∂ ⎛ ∂ ⎞

⎜ + + + ⎟= ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟+

⎜ ∂ ∂ ∂ ∂ ⎟ ∂ ⎝ ∂ ⎠ ∂ ⎝ ∂ ⎠ ∂ ⎝ ∂ ⎠

⎜ ⎟

⎝ ⎠ 

r (1.43)

In the above equation, the specifi c heat, c, of a food product can be calculated from the composition and temperature of the food. The temperature-dependent specifi c heat of major food components is given in Table 1.2. The power dissipation rate during microwave heating determined by Equation 1.38 or 1.42 is one example of source item, S, to be included in the above equation. In order to fi nd the solution of Equation 1.43, it is necessary to know the initial and boundary conditions. The initial conditions give what happens at the start. The initial conditions may be the same initial temperature, T|_t=0 = T₀. The initial conditions may also be an initial tempera-ture profi le, T|_t=0 = T₀(x,y,z). The boundary conditions give what happens at the boundaries of the material to be investigated. The boundary conditions may be (1) a constant, T|_G = T_s; (2) a fl ux, T|_G = q_s; (3) a convection, T|_G = h(T_s − T_∞); or (4) a com-bination of fl ux and convection, T|_G = q_s + h(T_s − T_∞).

In the modeling of an unsteady-state thermal process, the values of temperature depend on the time and position in the material. The equation governing the unsteady-state heat transfer is thus of a partial differential type, as shown in Equation 1.43.

Numerical methods have been widely used to solve the partial differential equation.

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Numerical methods can generate discretized solutions to the partial differential equa-tion (Wang and Sun, 2002, 2003; Wang and Singh, 2004; Amezquita et al., 2005).

An analytical solution of the partial differential equation is continuous. The possibility of an analytical solution is restricted to rather simple forms of the gov-erning equations, and the boundary and initial conditions. Sometimes, depending on the geometry of the product to be studied, it is useful to consider alternative coordinate systems such as the cylindrical coordinate and spherical coordinate sys-tems. However, the intrinsic mechanisms and physical laws of heat transfer remain the same, irrespective of the system used. For one-dimensional heat transfer, Equa-tion 1.43 can be rewritten in a general format as (Singh and Heldman, 2001)

1 _n

n

T T

c k r

t r r r

∂∂ = ∂∂ ⎛⎜⎝ ∂∂ ⎞⎟⎠

r (1.44)

where

r is the distance from the center (m) n = 0 for an infi nite slab

n = 1 for an infi nite cylinder n = 2 for a sphere

Suppose the initial condition and boundary conditions are

0 0

Tt₌ =T (1.45)

0

0

r

k T r ₌

− ∂ =

∂ (1.46)

( _s )

r R

k T h T T

r ₌ ^∞

− ∂ = −

∂ (1.47)

To simplify the analysis, two parameters are defi ned as (thermal diffusivity) k

= c

a r (1.48)

Bio hD(Biot number)

= k (1.49)

Biot number is the ratio of the internal resistance to heat transfer in the solid to the external resistance to heat transfer in the fl uid. There are

Bio

• £ 0.1, negligible internal resistance to heat transfer

0.1 < Bio < 40, fi nite internal and external resistance to heat transfer

• Bio

• ³ 40, negligible external resistance to heat transfer (Singh and Heldman, 2001)

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If Bio £ 0.1, the internal resistance can be neglected, and the heat balance equation becomes

(

^a

)

cV T hA T T

∂ =t −

r ∂ (1.50)

Integrating Equation 1.44, we obtain

( )

a

a i

e ^{hA cV t}

T T

T T

− = −

−

r (1.51)

If 0.1 < Bio < 40, both internal and external resistances should be considered. The solutions to Equation 1.44 with initial and boundary conditions are

Sphere:

( )

¹ ( ^{2 2 / 2})

( )

a a i 1

2 1

e sin

n

n t D

n

T T D

n r D

T T r n

∞ + −

−

− = ⎛ ⎞⎜ ⎟⎝ ⎠ − π

− π

∑

^{p a (1.52)}

Root equation,

1 1

Bio= − b1 cotb (1.53)

Infi nite cylinder:

( )

( )

( )

2 2

a 0

a i 1 1

2 e

n t D n

n n n

J r D

T T

T T J

∞ −

−

− =

−

∑

^{l a}_{l l}^{l (1.54)}

Root equation,

( ) ( )

1 1 1

0 1

Bio J

= bJ b

b (1.55)

Infi nite slab:

( )

¹ ( ^{2 2})

( )

a

a i 1

2 1 e ⁿ cos

n

tD n

n n

T T

T T r D

∞ +

−

−

− = −

−

∑

_l ^{l a l (1.56)}

Root equation,

1 1

Bio= b tanb (1.57)

If Bio ³ 40, the external resistance can be neglected, and the boundary condition becomes constant ambient temperature or it is assumed that the surface heat transfer coeffi cient is infi nitely big.

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1.3 FLUID MECHANICS IN FOOD PROCESSING