There are some inconsistencies in the question as it is asking for the heat flux, then wants the answer in seconds. Also, the recommended safety factor is not provided.
Assuming the question is seeking heat flux in kW/m2, I can continue with this clarification and the given information:
Given:
- Distance from fire (ground level): H = 3.5m
- Diameter of the kerosene fire: D = 4m
- Flame height: L = 3.5m
We can calculate the view factor (F) between a point and a circular area using the following formula:
F = 1/4 * (1 - cos(B)), where B is the angle formed between the observer and the edges of the circle, in radians
To find B, we can use the tangent function:
tan(B) = D/2 / (H + L)
B = arctan( (D/2) / (H+L))
B = arctan( (4m/2) / (3.5m+3.5m)) = arctan(2/7)
B ≈ 0.28 rads
Now calculate F:
F = 1/4 * (1-cos(0.28)) ≈ 0.122
Next, you will need to find the radiant heat flux (q) emitted by the fire. However, this is not possible without knowing the radiative properties of the flame and surroundings (emissivity, reflectivity, etc.). Therefore, it is impossible for me to provide a numerical answer in kW/m2 for the heat flux without making additional assumptions.
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.
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