Q 3.3-9E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

A warehouse is being built that will have neither heating nor cooling. Depending on the amount of insulation, the time constant for the building may range from 1 to 5 hr. To illustrate the effect insulation will have on the temperature inside the warehouse, assume the outside temperature varies as a sine wave, with a minimum of $\mathbf{16}\mathbf{°}\mathbf{C}$ at $\mathbf{2}\mathbf{:}\mathbf{00}\mathbf{a}\mathbf{.}\mathbf{m}\mathbf{.}$ and a maximum of $\mathbf{32}\mathbf{°}\mathbf{C}$ at $\mathbf{2}\mathbf{:}\mathbf{00}\mathbf{p}\mathbf{.}\mathbf{m}\mathbf{.}$ Assuming the exponential term (which involves the initial temperature T0) has died off, what is the lowest temperature inside the building if the time constant is 1 hr? If it is 5 hr? What is the highest temperature inside the building if the time constant is 1 hr? If it is 5 hr?

Step-by-Step Solution

Answer

If the time constant is 1 hour, the lowest temperature inside the building will reach $16.3°\mathrm{C}$and the highest temperature will reach $31.7°\mathrm{C}$.

If the time constant is 5 hours, the lowest temperature inside the building will reach $19.1°\mathrm{C}$ and the highest temperature will reach $28.9^{\mathrm{o}}\mathrm{C}$.

Step1: Given data.

The temperature outside the building varies as a sine wave, with a minimum of $16°\mathrm{C}$ at $2:00\mathrm{a}.\mathrm{m}.$and a maximum of $32{}^{\circ }\mathrm{C}$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.

Step 2: Analyzing the given statement

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

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

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

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

Adding the equations (a) and (b),

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

Therefore, B=8.

Step3: 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 T(t) is given by

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

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=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,

$T\left(t\right)=24-8F\left(t\right)$                                                                                             …… (3)

 

Where, the value of F(t) is,

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

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

 

Step 4: To determine lowest and highest temperatures inside the building if the time constant is 1 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 F(t) in equation (3),

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

                                                                                          …… (5)

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

And $\omega =\frac{\pi }{12}$

Therefore, from equation (5),

Let $T_L$be the lowest temperature,

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

Now as the minimum value of sin x is 1.

Therefore, from equation (4),

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

Thus, from equation (5),

 

Let  $T_{H }$be the highest temperature,

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

Hence, if the time constant is 1 hour, the lowest temperature inside the building will reach $\mathbf{16}\mathbf{.}\mathbf{3}\mathbf{°}\mathbf{C}$ and the highest temperature will reach $\mathbf{31}\mathbf{.}\mathbf{7}\mathbf{°}\mathbf{C}$.

Step 5: To determine lowest and highest temperatures inside the building if the time constant is 5 hr

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,

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

Let  $T_H$be the highest temperature,

So, from equation (5),

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

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