Chapter 2.4, Problem 57AYU

To determine

To find: a) Wind chill factor when an air temperature of ${10}^{°}\text{C}$ and a wind speed of 1 meter per second.

Answer to Problem 57AYU

Solution:

a) Wind chill factor when an air temperature of ${10}^{°}\text{C}$ and a wind speed of 1 meter per second is $W={10}^{°}\text{C}$

Explanation of Solution

Given:

It is given that the wind chill factor represents the air temperature at a standard wind speed that would produce the same heat loss as the given temperature and wind speed.

Formula for computing the equivalent temperature is

  $W=\{\begin{array}{ccc}t\text{ if }& 0\le v\le 1.79& \\ 33-\frac{(10.45+10\sqrt{v}-v)(33-t)}{22.04}& \text{ if }& 1.79\le v<20\\ 33-1.5958(33-t)& \text{ if }& v>20\end{array}$

Where v represents the wind speed and t represents the air temperature.

a)Wind chill factor when an air temperature of ${10}^{°}\text{C}$ and a wind speed of 1 meter per second.

  $t=10$ and $v=1$ meter per second.

  $W=t={10}^{°}\text{C}$ .

To determine

To find: b) Wind chill factor when an air temperature of ${10}^{°}\text{C}$ and a wind speed of 5 meter per second.

Answer to Problem 57AYU

Solution:

b)Wind chill factor when an air temperature of ${10}^{°}\text{C}$ and a wind speed of 5 meter per second is $W=4^{°}\text{C}$.

Explanation of Solution

Given:

It is given that the wind chill factor represents the air temperature at a standard wind speed that would produce the same heat loss as the given temperature and wind speed.

Formula for computing the equivalent temperature is

  $W=\{\begin{array}{ccc}t\text{ if }& 0\le v\le 1.79& \\ 33-\frac{(10.45+10\sqrt{v}-v)(33-t)}{22.04}& \text{ if }& 1.79\le v<20\\ 33-1.5958(33-t)& \text{ if }& v>20\end{array}$

Where V represents the wind speed and t represents the air temperature.

b)Wind chill factor when an air temperature of ${10}^{°}\text{C}$ and a wind speed of 5 meter per second.

t =10 and $v=5$ meter per second.

  $W=33-\frac{(10.45+10\sqrt{5}-5)(33-10)}{22.04}=33-\frac{27.85\times 23}{22.04}=33-29.06$

  $=3.94\approx 4^{\circ }\text{C}$

To determine

To find: c) Wind chill factor when an air temperature of ${10}^{°}\text{C}$ and a wind speed of 15 meter per second.

Answer to Problem 57AYU

Solution:

c) Wind chill factor when an air temperature of ${10}^{°}\text{C}$ and a wind speed of 15 meter per second is $W=-3^{°}\text{C}$

Explanation of Solution

Given:

It is given that the wind chill factor represents the air temperature at a standard wind speed that would produce the same heat loss as the given temperature and wind speed.

Formula for computing the equivalent temperature is

  $W=\{\begin{array}{ccc}t\text{ if }& 0\le v\le 1.79& \\ 33-\frac{(10.45+10\sqrt{v}-v)(33-t)}{22.04}& \text{ if }& 1.79\le v<20\\ 33-1.5958(33-t)& \text{ if }& v>20\end{array}$

Where $v$ represents the wind speed and $t$ represents the air temperature.

c)Wind chill factor when an air temperature of ${10}^{°}\text{C}$ and a wind speed of 15 meter per second.

  $t=10$ and $v=15$ meter per second.

  $\begin{array}{l}W=33-\frac{(10.45+10\sqrt{15}-15)(33-10)}{22.04}=33-\frac{34.15\times 23}{22.04}=33-\frac{785.45}{22.04}\hfill \\ =-2.64\approx -3^{°}\text{C}\hfill \end{array}$

To determine

To find: d) Wind chill factor when an air temperature of ${10}^{°}\text{C}$ and a wind speed of 25 meter per second.

Answer to Problem 57AYU

Solution:

Wind chill factor when an air temperature of ${10}^{°}\text{C}$ and a wind speed of 25 meter per second is $W=-4^{°}\text{C}$

Explanation of Solution

Given:

It is given that the wind chill factor represents the air temperature at a standard wind speed that would produce the same heat loss as the given temperature and wind speed.

Formula for computing the equivalent temperature is

  $W=\{\begin{array}{ccc}t\text{ if }& 0\le v\le 1.79& \\ 33-\frac{(10.45+10\sqrt{v}-v)(33-t)}{22.04}& \text{ if }& 1.79\le v<20\\ 33-1.5958(33-t)& \text{ if }& v>20\end{array}$

Where v represents the wind speed and t represents the air temperature.

d)Wind chill factor when an air temperature of ${10}^{°}\text{C}$ and a wind speed of 25 meter per second.

t =0 and v =25 meter per second.

  $W=33-1.5958(33-10)=33-36.7034=-3.7=-4^{°}\text{C}$

To determine

To find: e) The physical meaning of the equation corresponding to $0\le v<1.79$ .

Answer to Problem 57AYU

Solution:

Wind chill is equal to the air temperature.

Explanation of Solution

Given:

It is given that the wind chill factor represents the air temperature at a standard wind speed that would produce the same heat loss as the given temperature and wind speed.

Formula for computing the equivalent temperature is

  $W=\{\begin{array}{ccc}tif& 0\le v\le 1.79& \\ 33-\frac{(10.45+10\sqrt{v}-v)(33-t)}{22.04}& \text{ if }& 1.79\le v<20\\ 33-1.5958(33-t)& \text{ if }& v>20\end{array}$

Where $v$ represents the wind speed and $t$ represents the air temperature.

e) The equation $0\le v\le 1.79$ corresponds to the function $W=t.$ This directly implies that the wind chill depends on the temperature. Wind chill is equal to the air temperature.

To find: f) The physical meaning of the equation corresponding to $v>20$ .