Chapter 2.4, Problem 57AYU

Wind Chill 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. One formula for computing the equivalent temperature is

$W=\{\begin{array}{l}t0\le v<1.79\\ 33-\frac{\left(10.45+10\sqrt{v}-v\right)\left(33-t\right)}{22.04}\text{}1.79\le v\le 20\\ 33-1.5958\left(33-t\right)v>20\end{array}$

where $v$ represents the wind speed (in meters per second) and $t$ represents the air temperature $\left(°C\right)$. Compute the wind chill for the following:

(a) An air temperature of ${10}^{\circ }\text{C}$ and a wind speed of 1 meter per second (m/sec)

(b) An air temperature of ${10}^{\circ }\text{C}$ and a wind speed of 5 m/sec

(c) An air temperature of ${10}^{\circ }\text{C}$ and a wind speed of 15 m/sec

(d) An air temperature of ${10}^{\circ }\text{C}$ and a wind speed of 25 m/sec

(e) Explain the physical meaning of the equation corresponding to $0\le v<1.79$.

(f) Explain the physical meaning of the equation corresponding to $v>20$.

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 57AYU

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 57AYU

b. Wind chill factor when an air temperature of $10°C$ and a wind speed of 5 meter per second is $W = 4℃$.

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.

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

$= 3.94 \approx 4℃$

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 57AYU

c. Wind chill factor when an air temperature of $10°C$ and a wind speed of 15 meter per second is $W = -3℃$.

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.

$W = 33 - \frac{\left(10.45 + 10\sqrt{15} - 15\right)\left(33 - 10\right)}{22.04} = 33 - \frac{34.15 \times 23}{22.04} = 33 - \frac{785.45}{22.04} = 33 - 35.64$

$= -2.64 \approx -3℃$

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 57AYU

d. Wind chill factor when an air temperature of $10°C$ and a wind speed of 25 meter per second is $W = -4℃$.

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.

$W = 33 - 1.5958\left(33 - 10\right) = 33 - 36.7034 = -3.7 = -4℃$

To determine

To find:

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

Answer to Problem 57AYU

e. $\text{Wind chill is equal to the air temperature}$.

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

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

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

e. Wind chill factor when an air temperature of 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 determine

To find:

f. The physical meaning of the equation corresponding to $v > 20$.

Answer to Problem 57AYU

f. $\text{As wind speed greater than 20m/s the wind chill factor depends only on the air temperature}$.

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

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

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

f. The equation $v > 20$ corresponds to the function $W = 33 - 1.5958\left(33 - t\right)$. This directly implies that as wind speed greater than 20m/s the wind chill factor depends only on the air temperature.