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 a. Wind chill factor when an air temperature of 10 ° C and a wind speed of 1 meter per second is W = 10 ℃ . Explanation 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 = { t i f 0 ≤ v ≤ 1.79 33 − ( 10.45 + 10 v − v ) ( 33 − t ) 22.04 i f 1.79 ≤ v < 20 33 − 1.5958 ( 33 − t ) i f v > 20 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 b. Wind chill factor when an air temperature of 10 ° C and a wind speed of 5 meter per second is W = 4 ℃ . Explanation 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 = { t i f 0 ≤ v ≤ 1.79 33 − ( 10.45 + 10 v − v ) ( 33 − t ) 22.04 i f 1.79 ≤ v < 20 33 − 1.5958 ( 33 − t ) i f v > 20 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 − ( 10.45 + 10 5 − 5 ) ( 33 − 10 ) 22.04 = 33 − 27.85 × 23 22.04 = 33 − 29.06 = 3.94 ≈ 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 c. Wind chill factor when an air temperature of 10 ° C and a wind speed of 15 meter per second is W = − 3 ℃ . Explanation 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 = { t i f 0 ≤ v ≤ 1.79 33 − ( 10.45 + 10 v − v ) ( 33 − t ) 22.04 i f 1.79 ≤ v < 20 33 − 1.5958 ( 33 − t ) i f v > 20 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 − ( 10.45 + 10 15 − 15 ) ( 33 − 10 ) 22.04 = 33 − 34.15 × 23 22.04 = 33 − 785.45 22.04 = 33 − 35.64 = − 2.64 ≈ − 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 d. Wind chill factor when an air temperature of 10 ° C and a wind speed of 25 meter per second is W = − 4 ℃ . Explanation 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 = { t i f 0 ≤ v ≤ 1.79 33 − ( 10.45 + 10 v − v ) ( 33 − t ) 22.04 i f 1.79 ≤ v < 20 33 − 1.5958 ( 33 − t ) i f v > 20 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 ( 33 − 10 ) = 33 − 36.7034 = − 3.7 = − 4 ℃ To determine To find: e. The physical meaning of the equation corresponding to 0 ≤ v < 1.79 . Answer e. Wind chill is equal to the air temperature . Explanation 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 = { t i f 0 ≤ v ≤ 1.79 33 − ( 10.45 + 10 v − v ) ( 33 − t ) 22.04 i f 1.79 ≤ v < 20 33 − 1.5958 ( 33 − t ) i f v > 20 Where v represents the wind speed and t represents the air temperature. e. Wind chill factor when an air temperature of The equation 0 ≤ v ≤ 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 f. As wind speed greater than 20m/s the wind chill factor depends only on the air temperature . Explanation 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 = { t i f 0 ≤ v ≤ 1.79 33 − ( 10.45 + 10 v − v ) ( 33 − t ) 22.04 i f 1.79 ≤ v < 20 33 − 1.5958 ( 33 − t ) i f v > 20 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 ( 33 − t ) . This directly implies that as wind speed greater than 20m/s the wind chill factor depends only on the air temperature.

9th Edition

ISBN: 9780321716835

Author: Michael Sullivan

Publisher: Addison Wesley

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Textbook Question

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

$W = {\begin{cases} t 0 \leq v < 1.79 \\ 33 - \frac{(10.45 + 10 \sqrt{v} - v) (33 - t)}{22.04} 1.79 \leq v \leq 20 \\ 33 - 1.5958 (33 - t) v > 20 \end{cases}$

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

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

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

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

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

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

Expert Solution

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

$W = {\begin{cases} t i f 0 \leq v \leq 1.79 \\ 33 - \frac{(10.45 + 10 \sqrt{v} - v) (33 - t)}{22.04} i f 1.79 \leq v < 20 \\ 33 - 1.5958 (33 - t) i f v > 20 \end{cases}$

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.

$W = t = 10 ℃$

Expert Solution

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

$W = {\begin{cases} t i f 0 \leq v \leq 1.79 \\ 33 - \frac{(10.45 + 10 \sqrt{v} - v) (33 - t)}{22.04} i f 1.79 \leq v < 20 \\ 33 - 1.5958 (33 - t) i f v > 20 \end{cases}$

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.

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

$= 3.94 \approx 4 ℃$

Expert Solution

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

$W = {\begin{cases} t i f 0 \leq v \leq 1.79 \\ 33 - \frac{(10.45 + 10 \sqrt{v} - v) (33 - t)}{22.04} i f 1.79 \leq v < 20 \\ 33 - 1.5958 (33 - t) i f v > 20 \end{cases}$

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.

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

$= - 2.64 \approx - 3 ℃$

Expert Solution

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

$W = {\begin{cases} t i f 0 \leq v \leq 1.79 \\ 33 - \frac{(10.45 + 10 \sqrt{v} - v) (33 - t)}{22.04} i f 1.79 \leq v < 20 \\ 33 - 1.5958 (33 - t) i f v > 20 \end{cases}$

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.

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

Expert Solution

To determine

To find:

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

Answer to Problem 57AYU

e. $Wind chill is equal to the air temperature$ .

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

$W = {\begin{cases} t i f 0 \leq v \leq 1.79 \\ 33 - \frac{(10.45 + 10 \sqrt{v} - v) (33 - t)}{22.04} i f 1.79 \leq v < 20 \\ 33 - 1.5958 (33 - t) i f v > 20 \end{cases}$

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

e. Wind chill factor when an air temperature of The equation $0 \leq v \leq 1.79$ 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.

Expert Solution

To determine

To find:

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

Answer to Problem 57AYU

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

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

$W = {\begin{cases} t i f 0 \leq v \leq 1.79 \\ 33 - \frac{(10.45 + 10 \sqrt{v} - v) (33 - t)}{22.04} i f 1.79 \leq v < 20 \\ 33 - 1.5958 (33 - t) i f v > 20 \end{cases}$

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

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

Chapter 2.4, Problem 56AYU

Chapter 2.4, Problem 58AYU

Chapter 2 Solutions

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