An electric furnace is to be used for batch heating a certain material with specific heat of 670 J/(kg K) from 20 to 760°C. The material is placed on the furnace floor which is 2m x 4m in area as shown in the accompanying sketch. The side walls of the furnace are made of a refractory material. Parallel to the plane of the roof, but several inches below it, a grid of round resistor rods is installed. The resistors are 13 mm in diameter and are spaced 5 cm center to center. The resistor temperature is to be maintained at 1100°C, under these conditions the emissivity of the resistor surface is 0.6. If the top surface of the stock is assumed to have an emissivity of 0.9, estimate the time required for heating a 6 metric ton batch. External heat losses from the furnace may be neglected, the

Question

temperature gradient through the stock can be considered negligibly small, and steady-state conditions can be assumed.





GIVEN

- Batch heating of material in the furnace shown above

- Specific heat of material (c) = 670 Jkg K

- Material temperatures

- Initial

- Final

- Furnace dimensions: 2 m x 4 m x 1.3 m high

- Side walls are refractory material

- Resistor rod diameter (Dr) = 13 mm = 0.013 m

- Resistor center to center distance (s) = 5 cm = 0.05 m

- Resistor temperature (T2) = 1100°C = 1373 K

- Emissivity of the resistor surface (?2) = 0.6

- Emissivity of the material surface (?1) = 0.9

- Mass of material (m) = 6 metric tons = 6000 kg

FIND

- The time required (t) for heating the 6 metric ton batch

ASSUMPTIONS

- Quasi-steady state conditions

- External heat losses are negligible

- Temperature gradient through the material is negligible (negligible internal thermal resistance)

- Material is gray

- Convective heat transfer is negligible

PROPERTIES AND CONSTANTS

the Stephan-Boltzmann constant (?) = 5.67 x 10–8 W/(m2 K4

Answer 1

The shape factor F21 can be read off. For s/D = 50/13 = 3.85 and one row: F21 = 0.60.

Note that A1 = A2, therefore, F21 = F12.

The sum of the shape factors from a given surface must sum to unity



The rate of radiative heat transfer, between two gray surfaces connected by re-radiating surfaces is



where A2 f21 is note that A1 = A2 = A



The temperature changes in the material is given by



As T1 increases, the rate of heat transfer will decrease. Therefore, the equation above will be solved for

a chosen time increment and the temperature T1 will then be updated. This procedure will be repeated

until T1 = 760°C = 1033 K.

Let ?t = 5 min = 300 s initially