wind facotors and chill

Question

1.)    Vis windspeeds in m)sec, find wind chill tept given

 

2.)T=10(small zero) C,V =9,/sec

 

3.)T=0(small zero)C, V=15m/sec

Answer 1

Hi Sherry - This problem requires plugging your given values into a formula that relates the wind speed to the temperature.  Were you given this equation? Since your given values are all in metric units (meters and degrees Celsius), you'll need to make sure that the formula you use states the inputs in metric units also. 

 

Here's one that I found: 

 

T_chill=13.12 + 0.6215(T_{a) -} 11.37(V^0.16) + 0.3965(T_a)(V^0.16)

 

T_{a = actual temperature from thermometer}

V = Wind speed in Km/hr.

 

First you'll first have to convert your speed (the "V" in the equation) from m/sec to Km/hr:

 

9m/sec x (1Km/1000m) x (3600 sec/1 hr) = 9 x .001 x 3600 = 32.4 Km/Hr.

 

You can build your equation by putting this number in for V, but you'll have to use a calculator to raise 32.4 to a 0.16 exponent. Do you know how to do this with your calculator? 

 

Here's my final equation when wind speed = 9m/sec or (32.4 Km/hr) and T_a = 10^oC

 

T_chill = 13.12 + 0.6215(10) - 11.37(32.4)^0.16 + 0.3965(10)(32.4)^0.16

 

13.12 + 6.215 - 19.84 + 6.92 = 6.42 degrees C

 

This agrees with the value given on a conversion application I found elsewhere online. 

 

Can you do this calculation for the second temperature above? 

 

Good luck