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Wind Chill The wind chill factor represents the equivalent 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

$$\begin{gathered} W=\{t0≤v<1.79 \\ 33-\frac{(10.45+10\sqrt{v}-v)(33-t)}{22.04},1.79≤v≤20 \\ 33-1.5958(33-t),v>20 \end{gathered}$$

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

(a) An air temperature of localid="1645962228946" $10°\mathrm{C}$and a wind speed of 1 meter per second localid="1645962235368" $(\mathrm{m}/\sec )$

(b) An air temperature of localid="1645962241584" $10°\mathrm{C}$and a wind speed of localid="1645962248218" $5\mathrm{m}/\sec$

(c) An air temperature of localid="1645962254138" $10°\mathrm{C}$and a wind speed of localid="1645962259043" $15\mathrm{m}/\sec$

(d) An air temperature of localid="1645962265610" $10°\mathrm{C}$and a wind speed of localid="1645962274629" $25\mathrm{m}/\sec$

(e) Explain the physical meaning of the equation corresponding to localid="1645962280348" $0≤v<1.79$

(f) Explain the physical meaning of the equation corresponding to localid="1645962285160" $v>20$.

Step by step solution

01

Part (a) Step 1: Given information

Given the air temperature of${10}^{∘}C$and wind speed of$1m/s$

02

Part (a) Step 2: Calculating the values

As $v=1$is between 0 and $1.79$we have

$W=t$

Substitute $10$for $t$

$⇒W=10$

03

Part (b) Step 1: Given information

Given the air temperature of${10}^{∘}C$and wind speed of$15m/s$

04

Part (b) Step 2: Substitute t=10,v=5 and calculate the value

Here $v=5$is between $1.79$ and $20$ , so we have

$$\begin{gathered} W=33-\frac{(10.45+10\sqrt{v}-v)·(33-t)}{22.04} \\ W=33-\frac{(10.45+10\sqrt{5}-5)·(33-10)}{22.04} \\ W=33-\frac{(5.45+22.36)·23}{22.04} \\ W=33-\frac{27.81·23}{22.04} \\ W=33-29.02 \\ W=3.98 \end{gathered}$$Substituting, we get

Here $v=5$is between $1.79$and $20$, so we have

$W=33-\frac{(10.45+10\sqrt{v}-v)·(33-t)}{22.04}$

05

Part (c) Step 1: Given information

Given the air temperature of${10}^{∘}C$and wind speed of$25m/sec$

06

Part (c) Step 2: Substitute t=10,v=15 and calculate the value

Substituting, we get

$v=15$is between $1.79$and $20$, so we have

$W=33-\frac{(10.45+10\sqrt{v}-v)·(33-t)}{22.04}$

$$\begin{gathered} W=33-\frac{(10.45+10\sqrt{v}-v)·(33-t)}{22.04} \\ W=33-\frac{(10.45+10\sqrt{15}-15)·(33-10)}{22.04} \\ W=33-\frac{(-4.55+38.72)·23}{22.04} \\ W=33-\frac{34.17·23}{22.04} \\ W=33-35.66 \\ W=-2.66 \end{gathered}$$

07

Part (d) Step 1: Given information

Given the air temperature of${10}^{∘}C$and wind speed of$25m/sec$

08

Part (d) Step 2: Substitute t=10  and solve

Calculating, we get

As $25>20$

$W=33-1.5958(33-t)$

$$\begin{gathered} W=33-1.5958(33-10) \\ W=33-36.7 \\ W=-3.7 \end{gathered}$$

09

Part (e) Step 1: Given information

Given the equation corresponding to$0≤v<1.79$

10

Part (e) Step 2: Checking the required equation

Here for$0≤v<1.79$we have

$W=t$

Therefore, the wind chill is equal to the air temperature for$0≤v<1.79$

11

Part (f) Step 1: Given information

Given the equation corresponding to$v>20$

12

Part (f) Step 2: Checking the required equation 

Here for

From the function for $v>20$we have

$W=33-1.5958(33-t)$

Therefore, for $v>20$the wind chill depends only on the air temperature.

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