Chapter 2.4, Problem 58AYU

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 58AYU

Solution:

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

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

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

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 $-{10}^{\circ }\text{C}$ and a wind speed of 5 meter per second.

Answer to Problem 58AYU

Solution:

b) Wind chill factor when an air temperature of $-{10}^{\circ }\text{C}$ and a wind speed of 5 meter per second is $W=-{21}^{\circ }\text{C}$

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

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

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

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. $W=33-\frac{(10.45+10\sqrt{5}-5)(33-(-10))}{22.04}=33-\frac{27.85\times 43}{22.04}=33-54.335$

  $=-21.335\approx -{21}^{°}\text{C}$

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 58AYU

Solution:

c) Wind chill factor when an air temperature of $-{10}^{°}\text{C}$ and a wind speed of 15 meter per second is $W=-{32}^{°}\text{C}$

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.

  $\begin{array}{c}W=33-\frac{(10.45+10\sqrt{15}-15)(33-(-10))}{22.04}=33-\frac{34.15\times 43}{22.04}\\ =33-\frac{1468.45}{22.04}=33-66.63\end{array}$

  $=-32.64\approx -{32}^{\circ }\text{C}$

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 58AYU

Solution:

d) Wind chill factor when an air temperature of $-{10}^{°}\text{C}$ and a wind speed of 25 meter per second is $W=-{36}^{°}\text{C}$.

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

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

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

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

  $t=-10$ and $v=25$ meter per second.

  $\begin{array}{c}W=33-1.5958(33-(-10))=33-1.5958\times 43=33-68.6194\\ =-35.619=-{36}^{°}\text{C}\end{array}$