CHAPTER 5
HEAT TRANSFER THEORY (cont'd)

SURFACE HEAT TRANSFER

Newton found, experimentally, that the rate of cooling of the surface of a solid, immersed in a colder fluid, was proportional to the difference between the temperature of the surface of the solid and the temperature of the cooling fluid. This is known as Newton's Law of Cooling, and it can be expressed by the equation, analogous to eqn. (5.2),

q = h_sA(T_a– T_s)                                                        (5.4)

where h_s is called the surface heat transfer coefficient, T_a is the temperature of the cooling fluid and T_s is the temperature at the surface of the solid. The surface heat transfer coefficient can be regarded as the conductance of a hypothetical surface film of the cooling medium of thickness x_f such that
      h_s  = k_f /x_f  
where k_f is the thermal conductivity of the cooling medium.

Following on this reasoning, it may be seen that h_s can be considered as arising from the presence of another layer, this time at the surface, added to the case of the composite slab considered previously. The heat passes through the surface, then through the various elements of a composite slab and then it may pass through a further surface film. We can at once write the important equation:

q = ADT[(1/h_s1) + x₁/k₁ + x₂/k₂ + .. + (1/h_s2)]                                                           (5.5)

   = UADT

where 1/U = (1/h_s1) + x₁/k₁ + x₂/k₂ +.. + (1/h_s2)
and h_s1, h_s2 are the surface coefficients on either side of the composite slab, x₁, x₂ ...... are the thicknesses of the layers making up the slab, and k₁, k₂... are the conductivities of layers of thickness x₁, ..... . The coefficient h_s is also known as the convection heat transfer coefficient and values for it will be discussed in detail under the heading of convection. It is useful at this point, however, to appreciate the magnitude of h_s under various common conditions and these are shown in Table 5.1.

TABLE 5.1
APPROXIMATE RANGE OF SURFACE HEAT TRANSFER COEFFICIENTS

EXAMPLE 5.4. Heat transfer in jacketed pan
Sugar solution is being heated in a jacketed pan made from stainless steel, 1.6 mm thick. Heat is supplied by condensing steam at 200 kPa gauge in the jacket. The surface transfer coefficients are, for condensing steam and for the sugar solution, 12,000 and 3000 J m^-2 s^-1 °C^-1 respectively, and the thermal conductivity of stainless steel is 21 J m^-1 s^-1 °C^-1.
Calculate the quantity of steam being condensed per minute if the transfer surface is 1.4 m² and the temperature of the sugar solution is 83°C.

From steam tables, Appendix 8, the saturation temperature of steam at 200 kPa gauge(300 kPa Absolute) = 134°C, and the latent heat = 2164 kJ kg^-1.

For stainless steel x/k = 0.0016/21 = 7.6 x 10^-5

DT = (condensing temperature of steam) - (temperature of sugar solution)
      = 134 - 83 = 51°C.
From eqn. (5.5)

1/U = 1/12,000 + 7.6 x 10^-5 + 1/3000
   U = 2032 J m^-2 s^-1 °C^-1

and since A = 1.4 m²

   q = UADT
      = 2032 x 1.4 x 51
      = 1.45 x 10⁵ J s^-1

Therefore steam required

      = 1.45 x 10⁵ / (2.164 x 10⁶) kg s^-1
      = 0.067 kg s^-1
      = 4 kg min^-1
      

Heat-Transfer Theory > UNSTEADY STATE HEAT TRANSFER

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