Question

Annealing is a process by which steel is reheated and then cooled to make it less brittle. Consider the reheat stage for a 100-mm-thick steel plate ( 7830 kg/m3, 550 J/kg·K, 48 W/m·K), which is initially at a uniform temperature of 170°C and is to be heated to a minimum temperature of 550°C. Heating is effected in a gas-fired furnace, where products of combustion at [infinity] 800°C maintain a convection coefficient of 250 W/m2·K on both surfaces of the plate. How long should the plate be left in the furnace?

Answer 1

Answer:

The answer to the question is;

The plate be left in the furnace for 905.69 seconds.

Explanation:

To solve the question, we have to check the Bi number as follows

Bi = $\frac{hL}{k} = \frac{250\frac{W}{m^{2} K} *0.05 m}{48\frac{W}{mK} } = 0.2604$

As the Bi number is > 0.1 we have to account for the variation of temperature with location in the mass.

We perform nonlumped analysis

The relation for heat transfer given by

Y =  $\frac{T_f-T_{inf}}{T_i- T_{inf}}$

=$\frac{550-800}{170- 800}$ = 0.3968 = C₁ exp (ζ₁² F₀)  

where

C₁ and ζ₁ are coefficients of a series solution

We therefore look for the values of C₁ and ζ₁ from Bi tables to be

ζ₁ = 0.4801 +(0.26-0.25) (0.5218-0.4801)/(0.3-0.25) ≈ 0.4884 and

C₁ = 0.4801 +(0.26-0.25) (1.0450 - 1.0382)/(0.3-0.25) ≈ 1.03956 and  

This gives the relation

0.3968 = 1.03956 exp (ζ₁² F₀)  

or ζ₁² $(\frac{\alpha t}{L^2})$

where

α = Thermal diffusivity of solid = k/(ρ·c$_p$) = $\frac{48}{7830*550}$ = 1.1146×10⁻⁵

c$_p$ = Specific heat capacity of solid at constant pressure = 550 J/kg·K

ρ = Density of the solid = 7830 kg/m³

=㏑$(\frac{0.3968 }{1.03956 })$ = -0.9631 from where we have

t = $\frac{0.9631 *0.05^{2} }{0.4884^2*1.11*10^{-5}}$  = 905 seconds.