Question

A garage with no heating or cooling has a time constant of 2 hr. If the outside temperature varies as a sine wave with a minimum of $\mathbf{50}\mathbf{°}\mathbf{F}$at$\mathbf{2}\mathbf{:}\mathbf{00}\mathbf{a}\mathbf{.}\mathbf{m}\mathbf{.}$and a maximum of$\mathbf{80}\mathbf{°}\mathbf{F}$at$\mathbf{2}\mathbf{:}\mathbf{00}\mathbf{p}\mathbf{.}\mathbf{m}\mathbf{.}$, determine the times at which the building reaches its lowest temperature and its highest temperature, assuming the exponential term has died off.

Short answer

The building reaches its lowest temperature ${61}^{\mathrm{o}}\mathrm{F}$at $7:05\mathrm{a}.\mathrm{m}.$and the building reaches its highest temperature $68.9°\mathrm{F}$at$8:51\mathrm{a}.\mathrm{m}.$

Step-by-step solution

Given information.

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.

Find the value of B.

Now, the value of M is,

$M=M_0-Bcos\left(\omega t\right)$ …… (1)

Here,$\omega =\frac{2\pi }{24}=\frac{\pi }{12}$

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

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

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

Adding the equations (a) and (b),

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

Therefore,B=15.

To determine the temperature at time t.

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

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

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

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,

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

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

To determine lowest and highest temperatures inside the building if the time constant is 2 hr

Now as the maximum value of sin x is 1.

Therefore, from equation (4),

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

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

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

…… (5)

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

And$\omega =\frac{\pi }{12}$

Thus, from equation (5),

Let $T_L$be the lowest temperature,

$$\begin{gathered} T_L=65-\frac{15}{\sqrt{1+\frac{{\omega }^2}{k^2}}} \\ {T}_L=65-\frac{15}{\sqrt{1+\frac{4{\pi }^2}{144}}} \\ {T}_L={61}^{0}F \end{gathered}$$

Now as the minimum value of sin x is 1.

Hence, from equation (4),

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

So, from equation (5),

Let $T_H$be the highest temperature,

$$\begin{gathered} T_H=65+\frac{15}{\sqrt{1+\frac{{\omega }^2}{k^2}}} \\ {T}_H=65+\frac{15}{\sqrt{1+\frac{4{\pi }^2}{144}}} \\ {T}_H=68.9^{0}F \end{gathered}$$

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.

To determine times at which the temperature inside the building reaches its lowest and highest temperature

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.