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 which is initially at a uniform temperature of Ti = 200 degree C and is to be heated to a minimum temperature of 550 degree C. Heating is effected in a gas-fired furnace, where products of combustion at T infinity = 800 degree C maintain a convection coefficient of h = 250 W/m2 K on both surfaces of the plate How long should the plate be left in the furnace?

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Complete question is;

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 plate (ρ = 7830 kg/m3, Cp = 550 J/kg K, k = 48 W/m K). The plate initially is at 200 °C and is to be heated to a minimum temperature of 550 °C. Heating is effected in a gas-fired furnace where the products of combustion at T∞ = 800 °C maintain a convection heat transfer coefficient of h = 250 W/m.K on both surfaces of the plate. How long should the plate be left in the furnace?

Answer:

860 seconds

Explanation:

We are given;

Initial Temperature; Ti = 200 °C

Minimum Temperature; T_i = 550 °C

T∞ = 800 °C

convection coefficient; h = 250 W/m².K

ρ = 7830 kg/m³

Cp = 550 J/kg K

k = 48 W/m K

Plate thickness = 100mm

Thus,L = 100/2 = 50mm = 0.05 m

Let's find the biot number from the formula;

Bi = hL/K

Bi = (250 × 0.05)/48

Bi = 0.2604

Now, lowest temperature in the slab is given as;

θ_o = (T_o - T∞)/(T_i - T∞)

θ_o = (550 - 800)/(200 - 800)

θ_o = 0.4167

Now, from online tables calculation, we can find the root of the biot number.

Thus, root of the biot number Bi = 0.2604 is;

ζ1 = 0.488 rad

Also, C1 is gotten as 1.0396

Now,formula for thermal diffusivity is;

α = k/ρc

α = 48/(7830 × 550)

α = 1.115 × 10^(-5) m²/s

Also, from online tables, f(ζ1) = 0.401

Thus, we can find the time the plate should the plate be left in the furnace from;

-(ζ1)²(αt/L²) = In 0.401

-(ζ1)²(αt/L²) = -0.9138

t = (-0.9138 × 0.05²)/-(0.488² × 1.115 × 10^(-5))

t ≈ 860 s

In this exercise we want to calculate the time, in seconds, of the time left on the plate in the furnace, like this:

860 seconds

Organizing the information given in the statement we find that:

  • Initial Temperature; Ti = 200 °C
  • Minimum Temperature; T_i = 550 °C
  • Infinity Temperature: T=800°C
  • convection coefficient; h = 250 W/m².K
  • ρ = 7830 kg/m³
  • Cp = 550 J/kg K
  • k = 48 W/m K
  • Plate thickness = L = 0.05 m

Using the formula given below, we will have how to calculate the number of biot, like this:

[tex]B = hL/K\\B = (250 * 0.05)/48\\B = 0.2604[/tex]

calculating the angle that corresponds to the temperature difference as:

[tex]\theta_o = (T_o - T)/(T_i - T)\\\theta_o = (550 - 800)/(200 - 800)\\\theta_o = 0.4167[/tex]

Using the formula below, we will have:

[tex]\alpha = k/\rho c\\\alpha = 48/(7830 * 550)\\\alpha = 1.115 * 10^{-5}[/tex]

Combining all the information from the previous calculations, we have that the time will be calculated as:

[tex]-(\phi)^2(\alpha t/L^2) = In 0.401\\-(\phi )^2(\apha t/^2) = -0.9138\\t = (-0.9138 * 0.05^2)/-(0.488^2 * 1.115 * 10^{-5})\\t = 860 s[/tex]

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