Question

A thick oak wall, initially at 25 C, is suddenly exposed to combustion products for which T 800 C and h 20 W/m2 K. (a) Determine the time of exposure required for the surface to reach the ignition temperature of 400 C.

Answer 1

The time of exposure required for the surface to reach the ignition temperature of 400 °C is 310 seconds.

Step-by-step explanation:

For us to solve the question, if we consider the wall to be acting like a semi - infinite wall, the equation to solve it will be given by;

(T(o) - T)/(T(∞) - T(o)) = [erf x/√(2αt)] + [exp ((hx/k) + (h²αt/k²))] x [1 - erf {(h√(αt))/k} + {(x/(2√(αt))}]

Where;

T(o) = initial uniform temperature of wall

T = required ignition temperature of the wall,

T(∞) = temperature of combustion exhaust.

h = surface coefficient between the wall and combustion gas,

α = thermal diffusivity of oak,

k = thermal conductivity of oak

x = distance from the surface of wall.

From the image i attached, looking at oak under hardwood, we have the following properties ;

ρ = 545 kg/m³

c = 2385 J/Kg.k

k = 0.17 W/m.k

Now, thermal diffusivity of oak, α is given by; α = k/ρc = 0.17/(545 x 2385) = 1.308 x 10^(-7) m²/s

Now, from the question,

T(o) = 400 °C

T = 25 °C

T(∞) = 800 °C

Thus;

(T(o) - T)/(T(∞) - T(o)) will give;

(400 - 25)/(800 - 25) = 375/775 = 0.484

At the surface, x =0,we have;

(x/(2√(αt)) = 0 and erf(0) = 0 and

1 - erf(0) = 1

also, since h= 20 W/m²K, thus, (h√(αt))/k = [20 x √((1.308 x 10^(-7) x t))]/0.17 = [20 x √(1.308 x 10^(-7) x √t)]/0.17 = 0.0425√t

Putting x = 0 in the initial equation as well as the other relevant values and rearranging, we obtain ;

0.484 = [erf 0] + [((h√αt)/k)²)] x [1 - erf {(h√(αt))/k}]

Now, from earlier, (h√(αt))/k = 0.0425√t and also erf 0 = 0 and 1 - erf (0) = 1. So;

Thus, 0.484 = e^(0.0425√t)² [(1 - erf 0.0425√t)]

By using error function calculator and trial and error method, t is approximately equal to 310 seconds.

Answer 2

The time of exposure is t = 355.69 s

Step-by-step explanation:

according to the exercise we have to:

Tinf = 800°C

h = 20 W/m^2*K

The properties at 300 K are the follow:

p = 545 kg/m^3

c = 2385 J/kg*K

K = 0.17 W/m*K

α = K/(p*c) = 0.17/(545 * 2385) = 1.31x10^-7 m^2/s

the temperature distribution will be equal to:

(T(0,t) - Ti)/(Tinf - Ti) = (400-25)/(800-25) = 0.484 ς x/(2*(α*t)^1/2 =

0

Thus:

(h(α*t)^1/2)/K = 0.8

t = ((0.8*K)/(h*α^1/2))^2 = ((0.8*0.17)/(20*(1.3x10^-7)^1/2))^2 = 355.69 s