It is well-known that cold air feels much colder

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It is well-known that cold air feels much colder in windy weather than what the thermometer reading indicates because of the “chilling effect"" of the wind. This effect is due to the increase in the convection heat transfer coefficient with increasing air velocities. The equivalent wind chill temperature in oF is given by (all variables with an asterisk superscript (*) are dimensional quantities)

T*equiv,F (oF) = T*adj,F (oF) + (T*ambient,F (oF) – T* adj,F (oF)) f(vb)

where f is the dimensionless function (which must have dimensionless arguments in order to be dimensionally consistent! - for any smooth function this can be proved via a Taylor series expansion)

f(vb) = 0.475 – 0.0203vb + 0.304vvb

and T*adj,F = 91.4oF is a temperature adjustment term, vb is a dimensionless wind speed that has been non-dimensionalized by miles per hour (i.e. v*b = vb, (miles=hour and T*ambient is the ambient air temperature in oF in calm air, which is taken to be light winds at speeds of no more than 4 mph. The constant 91.4oF in the above equation is the mean temperature of a resting person in a comfortable environment. Windy air at temperature Tambient and velocity v will feel as cold as the calm air at temperature Tequiv.

Notice that the units of the coefficients in the second factor of the second term are, in order: dimensionless, hours per mile, and the square root of hours per mile so that the resulting second factor is dimensionless after the dimensional variables are substituted into the equation.

Using proper conversion factors, obtain an equivalent relation in SI units where vSI is the dimensionless wind speed based on m/s and Tambient is ambient air temperature in oC.

Hint: When working with physical equations all terms must be dimensionally consistent.

To preserve the dimensional consistency whatever we do to one side of the equation, we must do to the other, just as you learned in algebra.

Reference no: EM13319748

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