91影视

The temperature outside the building varies as a sine wave, with a minimum of $16掳\mathrm{C}$at $2:00\mathrm{a}.\mathrm{m}.$and a maximum of $32{}^{鈭�}\mathrm{C}$at $2:00\mathrm{p}.\mathrm{m}.$If the time constants for the building are 1 hours and 5 hours, it has to find the lowest and highest temperatures inside the building.

$M=M_0-Bcos\left(蝇t\right)$ 鈥︹�� (1)

$$\begin{gathered} M_0-B=16鈥�鈥�鈥�鈥�鈥�鈥�......\left(a\right) \\ {M}_0+B=32鈥�鈥�鈥�鈥�鈥�鈥�鈥�......\left(b\right) \end{gathered}$$

At $2:00\mathrm{a}.\mathrm{m}.$t=0,

And at 2:00 p.m, t=12 hours

Adding the equations (a) and (b),

$$\begin{gathered} 2M_0=48 \\ {M}_0=24 \end{gathered}$$

The forcing function$\mathrm{Q}\left(\mathrm{t}\right)$is given by,

$$\begin{gathered} Q\left(t\right)=K\left(M_0-Bcos\left(蝇t\right)\right) \\ Q\left(t\right)=K\left(24-8cos\left(蝇t\right)\right) \end{gathered}$$

Temperature T(t) is given by

$T\left(t\right)=B_0-BF\left(t\right)+C{e}^{-kt}$ 鈥︹�� (2)

Where, $F\left(t\right)=\frac{cos\left(蝇t\right)+\left(\frac{蝇}{k}\right)sin\left(蝇t\right)}{1+\frac{{蝇}^2}{k^2}}$.

Substituting K=1,B=8 and B₀ =M₀=24 in equation (2),

$T\left(t\right)=24-8F\left(t\right)+C{e}^{-t}$

Now as the exponential term died off, therefore,

$T\left(t\right)=24-8F\left(t\right)$ 鈥︹�� (3)

Where, the value of F(t) is,

$F\left(t\right)=\frac{1}{\sqrt{1+\frac{{蝇}^2}{k^2}}}\left(\frac{cos\left(蝇t\right)}{\sqrt{1+\frac{{蝇}^2}{k^2}}}+\frac{\left(\frac{蝇}{k}\right)sin\left(蝇t\right)}{\sqrt{1+\frac{{蝇}^2}{k^2}}}\right)$

$F\left(t\right)=\frac{sin\left(蝇t+ta{n}^{-1}\left(\frac{k}{蝇}\right)\right)}{\sqrt{1+\frac{{蝇}^2}{k^2}}}$ 鈥︹�� (4)

Now as the maximum value of sin x is 1.

Therefore, from equation (4),

$F\left(t\right)=\frac{1}{\sqrt{1+\frac{{蝇}^2}{k^2}}}$

So, by substituting the value of F(t) in equation (3),

$T\left(t\right)=24-\frac{8}{\sqrt{1+\frac{{蝇}^2}{k^2}}}$

鈥︹�� (5)

When the time constant is 1 hour i.e., when$\frac{1}{\mathrm{k}}=1$

And$蝇=\frac{蟺}{12}$

Therefore, from equation (5),

Let $T_L$be the lowest temperature,

$$\begin{gathered} T_L=24-\frac{8}{\sqrt{1+\frac{{蝇}^2}{k^2}}} \\ {T}_L=24-\frac{8}{\sqrt{1+\frac{{蟺}^2}{144}}} \\ {T}_L=16.3^{0}C \end{gathered}$$

Now as the minimum value of sin x is 1.

Therefore, from equation (4),

$F\left(t\right)=-\frac{1}{\sqrt{1+\frac{{蝇}^2}{k^2}}}$

Thus, from equation (5),

Let $T_H$be the highest temperature,

$$\begin{gathered} T_H=24+\frac{8}{\sqrt{1+\frac{{蝇}^2}{k^2}}} \\ {T}_H=24+\frac{8}{\sqrt{1+\frac{{蟺}^2}{144}}} \\ {T}_H=31.7^{0}C \end{gathered}$$

Hence, if the time constant is 1 hour, the lowest temperature inside the building will reach role="math" localid="1664185402769" $\mathbf{16}\mathbf{.}\mathbf{3}\mathbf{掳}\mathbf{C}$and the highest temperature will reach role="math" localid="1664185416044" $\mathbf{31}\mathbf{.}\mathbf{7}\mathbf{掳}\mathbf{C}$.

When the time constant is 5 hours i.e., when$\frac{1}{\mathrm{K}}=\frac{1}{5}$

Hence, from equation (5),

Let $T_L$be the lowest temperature,

$$\begin{gathered} T_L=24-\frac{8}{\sqrt{1+\frac{{蝇}^2}{k^2}}} \\ {T}_L=24-\frac{8}{\sqrt{1+\frac{25{蟺}^2}{144}}} \\ {T}_L=19.1^{0}C \end{gathered}$$

Let $T_H$be the highest temperature,

So, from equation (5),

$$\begin{gathered} T_H=24+\frac{8}{\sqrt{1+\frac{{蝇}^2}{k^2}}} \\ {T}_H=24+\frac{8}{\sqrt{1+\frac{25{蟺}^2}{144}}} \\ {T}_H=28.9^{0}C \end{gathered}$$

Thereafter, if the time constant is 5 hours, the lowest temperature inside the building will reach $19.1掳\mathrm{C}$and the highest temperature will reach $28.9掳\mathrm{C}$.