Question

Standards for firewalls may be based on their thermal response to a prescribed radiant heat flux. Consider a 0.25-m-thick concrete wall ( 2300 kg/m^3, 880 J/kg·K, 1.4 W/m·K), which is at an initial temperature of 25°C and irradiated at one surface by lamps that provide a uniform heat flux of 0.75·104 W/m^2. The absorptivity of the surface to the irradiation is 1.0. If building code requirements dictate that the temperatures of the irradiated and back surfaces must not exceed 325°C and 25°C, respectively, after 30 min of heating, will the requirements be met?

Answer 1

The requirements for the temperatures of the irradiated and back surfaces of the concrete wall to not exceed 325°C and 25°C, respectively, after 30 minutes of heating will not be met. The heat flux on the irradiated surface is found to be -126,000 W/m², which is greater than the prescribed radiant heat flux of 0.75e4 W/m². The heat flux on the back surface is 0 W/m², meeting the requirement for the temperature not to exceed 25°C.

Step-by-step explanation:

To determine whether the requirements will be met, we need to calculate the temperatures of the irradiated and back surfaces of the concrete wall after 30 minutes of heating. We can use the one-dimensional heat conduction equation to do this. The equation is:
q = -k * A * (dT/dx)

where q is the heat flux, k is the thermal conductivity, A is the surface area, dT is the change in temperature, and dx is the thickness of the wall.

First, we can calculate the heat flux on the irradiated surface:

q_irradiated = -k * A * (dT_irradiated/dx)

Substituting the values given in the question, we have:

q_irradiated = -1.4 * 0.75e4 * (325-25)/(0.0025)

q_irradiated = -1.4 * 0.75e4 * 12

q_irradiated = -126,000 W/m²

Next, we can calculate the heat flux on the back surface:

q_back = -k * A * (dT_back/dx)

Substituting the values given in the question, we have:

q_back = -1.4 * 0.75e4 * (25-25)/(0.0025)

q_back = 0 W/m²

Now, we can compare the calculated heat fluxes to the prescribed radiant heat flux of 0.75e4 W/m²:

|q_irradiated| > 0.75e4 W/m², Requirement not met

|q_back| = 0 W/m², Requirement met

Based on these calculations, it can be concluded that the requirement for the back surface temperature to not exceed 25°C is met, but the requirement for the irradiated surface temperature to not exceed 325°C is not met. Therefore, the requirements will not be met.

Answer 2

yes requirement will be met

Step-by-step explanation:

considering temperature distribution through following relation

where
`\alpha =(1.4)/(2300* 880) = 6.92* 10^(-7) m^2/s`

`t = 30* 60 = 1800 sec`

Q = HEAT FLUX = 0.75 \times 10^4 W/m^2

`T( 0, 30) =   25 + (2(.75* 10^4)[((6.92* 10^(-7) * 1800)/(\pi)]^(0.5))/(1.4) exp ((-0^2)/(4(6.92* 10^(-7) * 1800)) - ((0.7* 10^(4) 0)/(k) erfc(0)`

T(0,30 min) = 25 + 213.34 = 238.34 degree C

CALCULATE THE TEMPERATURE AT X = 0.25 M

`T(0.25, 30)  = 25 + 213.34 [exp ((0.25^2)/(4* 6.92* 10^(-7) * 1800))] - (0.75* 10^4(0.25))/(1.4) erfc(\frac{0.25}{2\sqrt{6.92* 10^(-7) * 1800}})`

`T(0.25,30 min) = 25 + 213.34(3.56* 10^(-4)) - 1339.28 (.00000054765)`

T(0.25,30 min) = = 25.00 degree celcius

since both surface have temperature lower than allowable therefore code requirement be met