Chapter 2.4, Problem 58AYU

Wind Chill Redo Problem 61(a)-(d) for an air temperature of - $10°C$.

To determine

To find:

a. Wind chill factor when an air temperature of $-10°C$ and a wind speed of 1 meter per second.

Answer to Problem 58AYU

a. Wind chill factor when an air temperature of $-10°C$ and a wind speed of 1 meter per second is $W = 10℃.$

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

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

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

a. Wind chill factor when an air temperature of $-10°C$ and a wind speed of 1 meter per second.

$t = -10$ and $v = 1$ meter per second.

$W = t = -10℃$

To determine

To find:

b. Wind chill factor when an air temperature of $-10°C$ and a wind speed of 5 meter per second.

Answer to Problem 58AYU

b. Wind chill factor when an air temperature of $-10°C$ and a wind speed of 5 meter per second is $W = -21℃.$

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

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

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

b. Wind chill factor when an air temperature of $-10°C$ and a wind speed of 5 meter per second.

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

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

$= -21.335 \approx -21℃$

To determine

To find:

c. Wind chill factor when an air temperature of $-10°C$ and a wind speed of 15 meter per second.

Answer to Problem 58AYU

c. Wind chill factor when an air temperature of $-10°C$ and a wind speed of 15 meter per second is $W = -32℃.$

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

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

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

c. Wind chill factor when an air temperature of $-10°C$ and a wind speed of 15 meter per second.

$t = -10$ and $v = 15$ meter per second.

$W = 33 - \frac{\left(10.45 + 10\sqrt{15} - 15\right)\left(33 - \left(-10\right)\right)}{22.04} = 33 - \frac{34.15 \times 43}{22.04} = 33 - \frac{1468.45}{22.04} = 33 - 66.63$

$= -32.64 \approx -32℃$

To determine

To find:

d. Wind chill factor when an air temperature of $-10°C$ and a wind speed of 25 meter per second.

Answer to Problem 58AYU

d. Wind chill factor when an air temperature of $-10°C$ and a wind speed of 25 meter per second is $W = -36℃.$

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

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

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

d. Wind chill factor when an air temperature of $-10°C$ and a wind speed of 25 meter per second.

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

$W = 33 - 1.5958\left(33 - \left(-10\right)\right) = 33 - 1.5958 \times 43 = 33 - 68.6194$

$= -35.619 = -36℃$