CHAPTER 5
HEAT TRANSFER THEORY (cont'd)

CONVECTION HEAT TRANSFER

Natural Convection
Natural Convection Equations
Forced Convection
Forced convection Equations


Convection heat transfer is the transfer of energy by the mass movement of groups of molecules. It is restricted to liquids and gases, as mass molecular movement does not occur at an appreciable speed in solids. It cannot be mathematically predicted as easily as can transfer by conduction or radiation and so its study is largely based on experimental results rather than on theory.
The most satisfactory convection heat transfer formulae are relationships between dimensionless groups of physical quantities. Furthermore, since the laws of molecular transport govern both heat flow and viscosity, convection heat transfer and fluid friction are closely related to each other.

Heat transfer by natural convection occurs when a fluid is in contact with a surface hotter or colder than itself. As the fluid is heated or cooled it changes its density. This difference in density causes movement in the fluid that has been heated or cooled and causes the heat transfer to continue.

There are many examples of natural convection in the food industry. Convection is significant when hot surfaces, such as retorts which may be vertical or horizontal cylinders, are exposed with or without insulation to colder ambient air. It occurs when food is placed inside a chiller or freezer store in which circulation is not assisted by fans. Convection is important when material is placed in ovens without fans and afterwards when the cooked material is removed to cool in air.

It has been found that natural convection rates depend upon the physical constants of the fluid, density r, viscosity m, thermal conductivity k, specific heat at constant pressure c_p and coefficient of thermal expansion b (beta) which for gases = l/T by Charles' Law. Other factors that also affect convection-heat transfer are, some linear dimension of the system, diameter D or length L, a temperature difference term, DT, and the gravitational acceleration g since it is density differences acted upon by gravity that create circulation. Heat transfer rates are expressed in terms of a convection heat transfer coefficient h_c, which is part of the general surface coefficient h_s, in eqn. (5.5).

Experimentally, if has been shown that convection heat transfer can be described in terms of these factors grouped in dimensionless numbers which are known by the names of eminent workers in this field:

Nusselt number (Nu) = (h_cD/k)
Prandtl number (Pr)  = (c_pm/k)
Grashof number (Gr) = (D³r²g b DT/m²)

Experimental work has evaluated K, k, m, n, under various conditions. For a discussion, see for example McAdams ( 1954) . Once K, k, m, n, are known for a particular case, together with the appropriate physical characteristics of the fluid, the Nusselt number can be calculated. From the Nusselt number we can find h_c and so determine the rate of convection-heat transfer by applying eqn. (5.5). In natural convection equations, the values of the physical constants of the fluid are taken at the mean temperature between the surface and the bulk fluid. The Nusselt and Biot numbers look similar: they differ in that for Nusselt, k and h both refer to the fluid, for Biot k is in the solid and h is in the fluid.

Natural Convection Equations

These are related to a characteristic dimension of the body (food material for example) being considered, and typically this is a length for rectangular bodies and a diameter for spherical/cylindrical ones.

(1) Natural convection about vertical cylinders and planes, such as vertical retorts and oven walls

(Nu) = 0.53(Pr.Gr)^0.25 for 10⁴ < (Pr.Gr) < 10⁹                                                            (5.15)

(Nu) = 0.12(Pr.Gr)^0.33 for 10⁹ < (Pr.Gr) < 10¹²                                                          (5.16)

For air these equations can be approximated respectively by:

   h_c = 1.3(DT/L)^0.25                                                                                               (5.17)

   h_c = 1.8(DT)^0.25                                                                                                  (5.18)

Equations (5.17) and (5.18) are dimensional equations and are in standard units (DT in °C and L (or D) in metres and h_c in J m^-2 s^-1 °C^-1). The characteristic dimension to be used in the calculation of (Nu) and (Gr) in these equations is the height of the plane or cylinder.

(2) Natural convection about horizontal cylinders such as a steam pipe or sausages lying on a rack

(Nu) = 0.54(Pr.Gr)^0.25         for laminar flow in range 10³ < (Pr.Gr) < 10⁹.                    (5.19)

Simplified equations can be employed in the case of air, which is so often encountered in contact with hotter or colder foods giving again:
For    10⁴ < (Pr.Gr) < 10⁹

   h_c = 1.3(DT/D)^0.25                                                                                                (5.20)

and for 10⁹< (Pr.Gr) < 10¹²

   h_c = 1.8(DT)^0.33                                                                                                   (5.21)

(3) Natural convection from horizontal planes, such as slabs of cake cooling

The corresponding cylinder equations may be used, employing the length of the plane instead of the diameter of the cylinder whenever D occurs in (Nu) and (Gr). In the case of horizontal planes, cooled when facing upwards, or heated when facing downwards, which appear to be working against natural convection circulation, it has been found that half of the value of h_c in eqns. (5.19) - (5.21) corresponds reasonably well with the experimental results.

Note carefully that the simplified equations are dimensional. Temperatures must be in °C and lengths in m and then h_c will be in J m^-2 s^-1 °C^-1. Values for s, k and m are measured at the film temperature, which is midway between the surface temperature and the temperature of the bulk liquid.

EXAMPLE 5.7. Heat loss from a cooking vessel
Calculate the rate of convection heat loss to ambient air from the side walls of a cooking vessel in the form of a vertical cylinder 0.9 m in diameter and 1.2 m high. The outside of the vessel insulation, facing ambient air, is found to be at 49°C and the air temperature is 17°C.

First it is necessary to establish the value of (Pr.Gr).
From the properties of air, at the mean film temperature, (49 + 17)/2, that is 33°C,
m = 1.9 x 10^-5 N s m^-2, c_p = 1.0 kJ kg^-1°C^-1, k = 0.025 J m^-1 s^-1° C^-1, b = 1/308, r =1.12 kg m^-3.
From the conditions of the problem, characteristic dimension = height = 1.2 m, DT = 32°C.

Therefore           (Pr.Gr) = (c_pm /k) (D³r²g b DT /m²)
                                   = (L³r²g b DT c_p) / (mk)

                                   = [(1.2)³ x (1.12)² x 9.81 x 32 x 1.0 x 10³ )/(308 x 1.9 x 10^-5 x 0.025)

                                   = 5 x 10⁹

Therefore eqn. (5.18) is applicable.

and so                     h_c = 1.8DT^0.25 = 1.8(32)^0.25
                                   = 4.3 J m^-2 s^-1 °C^-1

Total area of vessel wall = pDL = p x 0.9 x 1.2 = 3.4 m²
                              DT = 32°C.

Therefore heat loss rate = h_c A(T₁ - T₂)

                                   = 4.3 x 3.4 x 32
                                   = 468 J s^-1

Forced Convection

When a fluid is forced past a solid body and heat is transferred between the fluid and the body, this is called forced convection heat transfer. Examples in the food industry are in the forced-convection ovens for baking bread, in blast and fluidized freezing, in ice-cream hardening rooms, in agitated retorts, in meat chillers. In all of these, foodstuffs of various geometrical shapes are heated or cooled by a surrounding fluid, which is moved relative to them by external means.

The fluid is constantly being replaced, and the rates of heat transfer are, therefore, higher than for natural convection. Also, as might be expected, the higher the velocity of the fluid the higher the rate of heat transfer. In the case of low velocities, where rates of natural convection heat transfer are comparable to those of forced convection heat transfer, the Grashof number is still significant. But in general the influence of natural circulation, depending as it does on coefficients of thermal expansion and on the gravitational acceleration, is replaced by dependence on circulation velocities and the Reynold’s number.

As with natural convection, the results are substantially based on experiment and are grouped to deal with various commonly met situations such as fluids flowing in pipes, outside pipes, etc.

Forced convection Equations

(1) Heating and cooling inside tubes, generally fluid foods being pumped through pipes

In cases of moderate temperature differences and where tubes are reasonably long, for laminar flow it is found that:

(Nu) = 4                                                                                                                   (5.22)

and where turbulence is developed for (Re) > 2100 and (Pr) > 0.5

(Nu) = 0.023(Re)^0.8 (Pr)^0.4                                                                                       (5.23)

For more viscous liquids, such as oils and syrups, the surface heat transfer will be affected, depending upon whether the fluid is heating or being cooled. Under these cases, the viscosity effect can be allowed, for (Re) > 10,000, by using the equation:

(Nu) = 0.027(m/m_s)^0.14(Re)^0.8 (Pr)^0.33                                                                      (5.24)

In both cases, the fluid properties are those of the bulk fluid except for m_s, which is the viscosity of the fluid at the temperature of the tube surface.

Since (Pr) varies little for gases, either between gases or with temperature, it can be taken as 0.75 and eqn. (5.23) simplifies for gases to:

(Nu) = 0.02(Re)^0.8.                                                                                                  (5.25)

In this equation the viscosity ratio is assumed to have no effect and all quantities are evaluated at the bulk gas temperature. For other factors constant, this becomes h_c = k' v^0.8, as in equation (5.28)

(2) Heating or cooling over plane surfaces

Many instances of foods approximate to plane surfaces, such as cartons of meat or ice cream or slabs of cheese. For a plane surface, the problem of characterizing the flow arises, as it is no longer obvious what length to choose for the Reynolds number. It has been found, however, that experimental data correlate quite well if the length of the plate measured in the direction of the flow is taken for D in the Reynolds number and the recommended equation is:

(Nu) = 0.036 (Re)^0.8(Pr)^0.33 for (Re) > 2 x 10⁴                                                           (5.26)

For the flow of air over flat surfaces simplified equations are:

   h_c = 5.7 + 3.9v for v < 5 m s^-1                                                                               (5.27)

   h_c = 7.4v^0.8 for 5 < v < 30 m s^-1                                                                            (5.28)

These again are dimensional equations and they apply only to smooth plates. Values for h_c for rough plates are slightly higher.

(3) Heating and cooling outside tubes

Typical examples in food processing are water chillers, chilling sausages, processing spaghetti.
Experimental data in this case have been correlated by the usual form of equation:

(Nu) = K (Re)ⁿ(Pr)^m                                                                                                 (5.29)

The powers n and m vary with the Reynolds number. Values for D in (Re) are again a difficulty and the diameter of the tube, over which the flow occurs, is used. It should be noted that in this case the same values of (Re) cannot be used to denote streamline or turbulent conditions as for fluids flowing inside pipes.

For gases and for liquids at high or moderate Reynolds numbers:

(Nu) = 0.26(Re)^0.6(Pr)^0.3                                                                                          (5.30)

whereas for liquids at low Reynolds numbers, 1 < (Re) < 200:

(Nu) = 0.86(Re)^0.43(Pr)^0.3                                                                                        (5.31)

As in eqn. (5.23), (Pr) for gases is nearly constant so that simplified equations can be written. Fluid properties in these forced convection equations are evaluated at the mean film temperature, which is the arithmetic mean temperature between the temperature of the tube walls and the temperature of the bulk fluid.

EXAMPLE 5.8. Heat transfer in water flowing over a sausage
Water is flowing at 0.3 m s^-1 across a 7.5 cm diameter sausage at 74°C. If the bulk water temperature is 24°C, estimate the heat-transfer coefficient.

Mean film temperature = (74 + 24)/2 = 49°C.

Properties of water at 49°C are:
                             c_p  = 4.186 kJ kg^-1°C^-1, k = 0.64 J m^-1 s^-1°C^-1, m = 5.6 x 10^-4 N s m^-2, r = 1000 kg m^-3.

Therefore             (Re)  = (Dvr/µ)
                                  = (0.075 x 0.3 x 1000)/(5.6 x 10^-4)
                                  = 4.02 x 10⁴
                       (Re)^0.6 = 580

                           (Pr)  = (c_p m/k)
                                  = (4186 x 5.6 x 10^-4)/0.64
                                  = 3.66.
                        (Pr)^0.3 = 1.48

                           (Nu) = (h_cD/k)
                                  = 0.26(Re)^0.6(Pr)^0.3

Therefore               h_c  = k/D x 0.26 x (Re)^0.6(Pr)^0.3
                                  = (0.64 x 0.26 x 580 x 1.48)/0.075
                                  = 1904 J m^-2 s^-1 °C^-1

EXAMPLE 5.9. Surface heat transfer to vegetable puree
Calculate the surface heat transfer coefficient to a vegetable puree, which is flowing at an estimated
3 m min^-1 over a flat plate 0.9 m long by 0.6 m wide. Steam is condensing on the other side of the plate and maintaining the surface, which is in contact with the puree, at 104°C. Assume that the properties of the vegetable puree are, density 1040 kg m^-3, specific heat 3980 J kg^-1 °C^-1, viscosity 0.002 N s m^-2, thermal conductivity 0.52 J m^-1 s^-1 °C^-1.
                              v   =   3m min^{-1 =   3/60 ms-1}

                           (Re) = (Lvr/m)
                                  = (0.9 x 3 x 1040)/(2 x 10^-3 x 60)
                                  = 2.34 x 10⁴
Therefore eqn. (5.26) is applicable and so:
                       (h_cL/k) = 0.036(Re)^0.8(Pr)^0.33

                              Pr = (c_pm/k)
                                  = (3980 x 2 x 10^-3)/0.52
                                  = 15.3
and so

                       (h_cL/k) = 0.036(2.34 x 10⁴)^0.815.3^0.33

                             h_c = (0.52 x 0.036) (3.13 x 10³)(2.46)/0.9
                                  = 160 J m^-2 s^-1°C^-1

EXAMPLE 5.10. Heat loss from a cooking vessel
What would be the rate of heat loss from the cooking vessel of Example 5.7, if a draught caused the air to move past the cooking vessel at a speed of 61 m min^-1

Assuming the vessel is equivalent to a flat plate then from eqn. (5.27)
                               v   = 61/60 = 1.02 m s^-1, that is v < 1.02 m s^-1

Heat-Transfer Theory > OVERALL HEAT TRANSFER COEFFICIENTS

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