Assuming the threshold for pain is 1.7kW/m2 and a fire with a diameter of 8 ft a

Question

Assuming the threshold for pain is 1.7kW/m2 and a fire with a diameter of 8 ft and a heat release rate of heat release rate of 7825 kW (30 points - mathematical calculation) Using the Shorki and Beyler's heat flux correlation, what is the minimum distance a person should stand away from this fire to avoid pain from radiation? Based on experimental data from large-scale pool fire experiments, Shokri and Beyler's developed a simple correlation of radiant heat flux of ground level as a function of the radial position of a vertical target. The incident heat flux correlation is given by q^n = 15.4 (L/D)^-1.59 [kW/m^2] where D = the diameter of the pool fire L = the distance from the center of the pool fire to the target edge A.~9.75m B.~15.75m C. ~20.75m D ~25.75m

Explanation / Answer

Height of the flame is given by,

H = 0.23Q^2/5 -1.02D

where Q is the heat release rate = 7825 kW

and D is the diameter of the pool fire = 8*0.3048= 2.438 m

So, H = 0.23(7825)^2/5-1.02(2.438)

= 5.8137 m

So, Area of the pool fire = 5.8137*2.438 m^2

= 14.176 m^2

Now, the actual heat release rate,

q = 7825/14.176 = 551.98 kW/m^2

According to S-B radiant flux correlation,

q =15.4(L/D)^-1.59

So, 551.98 = 15.4(L'/8*0.3048)^-1.59

(D/L')^1.59 = 35.843

so. L' = 25.60 m

Minimum threshold radiant flux = 1.7 kW/m^2

then, 1.7 = 15.4(L''/D)^-1.59

(D/L'')^1.59 = 0.11039

2.432/L'' = 0.25007

L'' = 9.753 m

So, the minimum distance a person should stand from this fire to avoid pain from heat radiation is given by,

L = L''-L'

= 25.60-9.753

So, L = 15.75 m

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