Using the cylindrical view factor what would be the heat flux to person at ground level 3.5 m from a 4m diameter kerosene fire with a average flame height using would be 3.5 m. Include the recommended safety factor. Round to the nearest kW/m2. Provide numerical answer with units of seconds, do not include units in input, submit number only.
I apologize for the confusion earlier. In order to calculate the heat flux to a person at ground level 3.5 m from a 4m diameter kerosene fire with an average flame height of 3.5 m and including the recommended safety factor, we need to use the following formula:
q = (sigma * F * (Ts^4 - Ta^4)) / r
where q is the heat flux in kW/m², sigma is Stefan-Boltzmann constant (5.67 × 10^(-8) W/(m²K⁴)), F is the view factor determined by the cylindrical view factor formula, Ts is the surface temperature of the fire (assume 1000 K for kerosene fire), Ta is the ambient temperature (assume 300 K) and r is the safety factor (assume 2).
First, we need to calculate the view factor (F) using the cylindrical view factor formula:
F = [(H / D) * atan(1 / sqrt((H / D)^2 + (R / D)^2))] / pi
where H is the height of the flame (3.5 m), D is the diameter of the fire (4 m), R is the distance from the fire (3.5 m), and atan is the arctangent function.
F = [(3.5 / 4) * atan(1 / sqrt((3.5 / 4)^2 + (3.5 / 4)^2))] / pi
F ≈ 0.1638
Now, we can calculate the heat flux (q) using the formula:
q = (sigma * F * (Ts^4 - Ta^4)) / r
q = (5.67 × 10^(-8) * 0.1638 * (1000^4 - 300^4)) / 2
q ≈ 29.602 kW/m²
Therefore, the heat flux to the person at ground level 3.5 m from the 4m diameter kerosene fire with a flame height of 3.5 m and including the recommended safety factor is approximately 30 kW/m².