91Ó°ÊÓ

Given that the outside temperature varies as a sine wave with a minimum of $50°\mathrm{F}$at $2:00\mathrm{a}.\mathrm{m}.$and a maximum of $80°\mathrm{F}$at $2:00\mathrm{p}.\mathrm{m}.$, It has to determine the times at which the building reaches its lowest temperature and its highest temperature, assuming the exponential term has died off.

Now, the value of M is,

$M=M_0-Bcos\left(\mathrm{Ó\neg }t\right)$ …… (1)

Here,$\mathrm{Ó\neg }=\frac{2\mathrm{Ï€}}{24}=\frac{\mathrm{Ï€}}{12}$

$$\begin{gathered} M_0-B=50        ......\left(a\right) \\ {M}_0+B=80        ......\mathbf{\left(}\mathbf{b}\mathbf{\right)} \end{gathered}$$

At 2:00 a.m.,t=0

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

Adding the equations (a) and (b),

$$\begin{gathered} 2M_0=130 \\ {M}_0=65 \end{gathered}$$

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

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

Temperature $\mathrm{T}\left(\mathrm{t}\right)$is given by

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

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

Substituting K=1, B=15 and B₀=M₀=65in equation (2),

$T\left(t\right)=65-15F\left(t\right)+C{e}^{-t}$

Now as the exponential term died off, therefore,

$T\left(t\right)=65-15F\left(t\right)$ …… (3)

Where, the value of F(t) is,

$$\begin{gathered} F\left(t\right)=\frac{1}{\sqrt{1+\frac{{\mathrm{Ó\neg }}^2}{k^2}}}\left(\frac{cos\left(\mathrm{Ó\neg }t\right)}{\sqrt{1+\frac{{\mathrm{Ó\neg }}^2}{k^2}}}+\frac{\left(\frac{\mathrm{Ó\neg }}{k}\right)sin\left(\mathrm{Ó\neg }t\right)}{\sqrt{1+\frac{{\mathrm{Ó\neg }}^2}{k^2}}}\right) \\ F\left(t\right)=\frac{1}{\sqrt{1+\frac{{\mathrm{Ó\neg }}^2}{k^2}}}\left(\frac{cos\left(\mathrm{Ó\neg }t\right)}{\sqrt{1+\frac{{\mathrm{Ó\neg }}^2}{k^2}}}+\frac{\left(\frac{\mathrm{Ó\neg }}{k}\right)sin\left(\mathrm{Ó\neg }t\right)}{\sqrt{1+\frac{{\mathrm{Ó\neg }}^2}{k^2}}}\right) \end{gathered}$$

$F\left(t\right)=\frac{sin\left(\mathrm{Ó\neg }t+ta{n}^{-1}\left(\frac{k}{\mathrm{Ó\neg }}\right)\right)}{\sqrt{1+\frac{{\mathrm{Ó\neg }}^2}{k^2}}}$ …… (4)

Now as the maximum value of sin x is 1.

Therefore, from equation (4),

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

So, by substituting the value of $\mathrm{F}\left(\mathrm{t}\right)$in equation (3),

$T\left(t\right)=65-\frac{15}{\sqrt{1+\frac{{\mathrm{Ó\neg }}^2}{k^2}}}$

…… (5)

When the time constant is 2 hours i.e., when $\frac{1}{\mathrm{K}}=\frac{1}{2}$

And$\mathrm{Ó\neg }=\frac{\mathrm{Ï€}}{12}$

Thus, from equation (5),

Let $T_L$be the lowest temperature,

$$\begin{gathered} T_L=65-\frac{15}{\sqrt{1+\frac{{\mathrm{Ó\neg }}^2}{k^2}}} \\ {T}_L=65-\frac{15}{\sqrt{1+\frac{4{\mathrm{Ï€}}^2}{144}}} \\ {T}_L={61}^{0}F \end{gathered}$$

Now as the minimum value of sin x is 1.

Hence, from equation (4),

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

So, from equation (5),

Let $T_H$be the highest temperature,

$$\begin{gathered} T_H=65+\frac{15}{\sqrt{1+\frac{{\mathrm{Ó\neg }}^2}{k^2}}} \\ {T}_H=65+\frac{15}{\sqrt{1+\frac{4{\mathrm{Ï€}}^2}{144}}} \\ {T}_H=68.9^{0}F \end{gathered}$$

Thereafter, if the time constant is 2 hours, the lowest temperature inside the building will reach 61°F and the highest temperature will reach 68.9°F.

Newton’s Law of cooling is,

$T\left(t\right)=M_0+\left(T_0-M_0\right)e^{-kt}$

When ,$\mathrm{T}\left(\mathrm{t}\right)=61^{\mathrm{o}}\mathrm{F}$

$$\begin{gathered} 61=65+\left(15-65\right)e^{-\frac{t}{2}} \\ -4=-50e^{-\frac{t}{2}} \\ 4=50e^{-\frac{t}{2}} \\ {e}^{\frac{t}{2}}=\frac{50}{4} \\ \frac{t}{2}=ln\left(12.5\right) \\ t=2ln\left(12.5\right) \\ t=5.05hr \end{gathered}$$

Accordingly, the building reaches its lowest temperature at7:05 a.m.

When ,$\mathrm{T}\left(\mathrm{t}\right)=68.9^{\mathrm{o}}\mathrm{F}$

$$\begin{gathered} 68.9=65+\left(15-65\right)e^{-\frac{t}{2}} \\ 3.9=-50e^{-\frac{t}{2}} \\ 3.9=-50e^{-\frac{t}{2}} \\ {e}^{\frac{t}{2}}=\frac{-50}{3.9} \\ \frac{t}{2}=-ln\left(12.8\right) \\ t=-2ln\left(12.8\right) \end{gathered}$$

Therefore, the building reaches its highest temperature at 8:51 p.m.