9.8. COMBINED CONVECTION, RADIATION, AND CONDUCTION HEAT TRANSFER IN CONSTANT-RATE PERIOD

9.8A. Introduction

In Section 9.6B an equation was derived for predicting the rate of drying in the constant-rate period. Equation (9.6-7) was derived assuming heat transfer to the solid by convection only from the surrounding air to the drying surface. Often the drying is done in an enclosure, where the enclosure surface radiates heat to the drying solid. Also, in some cases the solid may be resting on a metal tray, and heat transfer by conduction through the metal to the bottom of the solid may occur.

9.8B. Derivation of Equation for Convection, Conduction, and Radiation

In Fig. 9.8-1 a solid material being dried by a stream of air is shown. The total rate of heat transfer to the drying surface is

Equation 9.8-1

where q_C is the convective heat transfer from the gas at T°C to the solid surface at T_S°C in W (J/s), q_R is the radiant heat transfer from the surface at T_R to T_S in W (J/s), and q_K is the rate of heat transfer by conduction from the bottom in W. The rate of convective heat transfer is similar to Eq. (9.6-3) and is as follows, where (T − T_S)°C = (T − T_S) K:

Equation 9.8-2

where A is the exposed surface area in m². The radiant heat transfer is

Equation 9.8-3

where h_R is the radiant-heat-transfer coefficient defined by Eq. (4.10-10):

Equation 4.10-10

Note that in Eq. (4.10-10) T_R and T_S are in K. For the heat transfer by conduction from the bottom, the heat transfer is first by convection from the gas to the metal plate, then by conduction through the metal, and finally by conduction through the solid. Radiation to the bottom of the tray is often small if the tray is placed above another tray, and it will be neglected here. Also, if the gas temperatures are not too high, radiation from the top surface to the tray will be small. Hence, the heat by radiation should not be overemphasized. The heat by conduction is

Equation 9.8-4

Equation 9.8-5

where z_M is the metal thickness in m, k_M the metal thermal conductivity in W/m · K, z_S the solid thickness in m, and k_S the solid thermal conductivity. The value of h_C in Eq. (9.8-4) is assumed to be the same as in Eq. (9.8-2).

The equation for the rate of mass transfer is similar to Eq. (9.6-5) and is

Equation 9.8-6

Also, rewriting Eq. (9.6-6),

Equation 9.8-7

Combining Eqs. (9.8-1), (9.8-2), (9.8-3), (9.8-4), (9.8-6), and (9.8-7),

Equation 9.8-8

This equation can be compared to Eq. (9.6-7), which gives the wet bulb temperature T_W when radiation and conduction are absent. Equation (9.8-8) gives surface temperature T_S greater than the wet bulb temperature T_W. Equation (9.8-8) must also intersect the saturated humidity line at T_S and H_S, and T_S > T_W and H_S > H_W. The equation must be solved by trial and error.

To facilitate solution of Eq. (9.8-8), it can be rearranged (T1) in the following way:

Equation 9.8-9

The ratio h_C/k_yM_B was shown in the wet bulb derivation of Eq. (9.3-18) to be approximately c_S in Eq. (9.3-6):

Equation 9.3-6

EXAMPLE 9.8-1. Constant-Rate Drying When Radiation, Conduction, and Convection Are Present An insoluble granular material wet with water is being dried in a pan 0.457 × 0.457 m and 25.4 mm deep. The material is 25.4 mm deep in the metal pan, which has a metal bottom of thickness zM = 0.610 mm having a thermal conductivity kM = 43.3 W/m · K. The thermal conductivity of the solid can be assumed as kS = 0.865 W/m · K. Heat transfer is by convection from an air stream flowing parallel to the top drying surface and the bottom metal surface at a velocity of 6.1 m/s and having a temperature of 65.6°C and humidity H = 0.010 kg H2O/kg dry air. The top surface also receives direct radiation from steam-heated pipes whose surface temperature TR = 93.3°C. The emissivity of the solid is є = 0.92. Estimate the rate of drying for the constant-rate period. Solution: Some of the given values are as follows: The velocity, temperature, and humidity of the air are the same as in Example 9.6-3 and the convective coefficient was predicted as hC = 62.45 W/m2 · K. The solution of Eq. (9.8-9) is by trial and error. The temperature TS will be above the wet bulb temperature of TW = 28.9°C and will be estimated as TS = 32.2°C. Then λS = 2424 kJ/kg from the steam tables. To predict hR from Eq. (4.10-10) for є = 0.92, Tl = 93.3 + 273.2 = 366.5 K, and T2 = 32.2 + 273.2 = 305.4 K, Using Eq. (9.8-5), From Eq. (9.3-6), This can be substituted for (hC/kyMB) into Eq. (9.8-9). Also, substituting other knowns, Equation 9.8-10 For TS assumed as 32.2°C, λS = 2424 × 103 J/kg. From the humidity chart for TS = 32.2°C, the saturation humidity HS = 0.031. Substituting into Eq. (9.8-10) and solving for TS, For the second trial, assuming that TS = 32.5°C, λS = 2423 × 103, and HS from the humidity chart at saturation is 0.032. Substituting into Eq. (9.8-10) while assuming that hR does not change appreciably, a value of TS = 32.8°C is obtained. Hence, the final value is 32.8°C. This is 3.9°C greater than the wet bulb temperature of 28.9°C in Example 9.6-3, where radiation and conduction were absent. Using Eq. (9.8-8), This compares favorably with 3.39 kg/h · m2 for Example 9.6-3 for no radiation or conduction.