Q 3.3-8E - Fundamentals Of Differential Equations And Boundary Value Problems

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.

 

Step-by-Step Solution

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 $8:51\mathrm{a}.\mathrm{m}.$

Step1: 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.

Step 2: 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.

Step 2: 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=65 in 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)

Step 3: 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.

Step 4: 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 at 7:05 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.