Question

A wind chill factor is defined as the temperature in still air required for a human to suffer the same heat loss as he does for the actual air temperature with the wind blowing. On a very cold morning on the ski slopes at Big Bear, the outside temperature is 15 ℉ and the wind chill factor is-30 °F. Find the wind speed in miles/hour. Assumptions: For a person fully clothed in ski gear, assume that temperature on the outside surface of the clothes is 40 °F and the heat transfer coefficient in sl air is 4 Btu/hr ft2 °F. For simplicity, model the person as a cylinder 1 ft in diameter by 6 ft tall. Use the correlation for forced convection past an upright cylinder: Nu 0.0239 ReD 0805 . Properties of air: thermal conductivity 0.0 134 Btu/hrft。F, density 0.0845 lbm/ft, dynamic viscosity 3.996×10-2 bm/ft hr 2.0

Answer 1

Answer: V = 208514.156 ft/hr

Explanation:

we will begin by giving a step by step order of answering.

given that

A = area available for convection which will be same for both cases

h (still) = heat transfer coefficient in still air

h(blowing) = heat transfer coefficient in blowing air.

Therefore,

h(still) A [40-(-30)] = h(blowing) A (40-15)

canceling out we have

h(still) (70) = h(blowing) (25)

where h(still) = 4 Btu/hr.ft².°F

4 × 70 =  h(blowing) (25)

h (blowing) = 11.2 Btu/hr.ft².°F

Also, we have that NUD = 0.0239 ReD 0805

h (blowing) D / k = 0.0239 (ρVD/μ)˄0.805

where from our data,

D = diameter = 1 ft

ρ = density = 0.0845 lbm/ft

μ = viscosity = 3.996×10-2 bm/ft hr

K = 0.0134 Btu/hr.ft²°F

So from

h (blowing) D / k = 0.0239 (ρVD/μ)˄0.805

we have;

11.2 × 1 / K =  0.0239 (0.0845×V×1 / 3.996×10-2 )˄0.805

where K = 0.0134

V = 208514.156 ft/hr

cheers i hope this helps