Q 3.3-10E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Early Monday morning, the temperature in the lecture hall has fallen to $\mathbf{40}\mathbf{°}\mathbf{F}$, the same as the temperature outside. At $\mathbf{7}\mathbf{:}\mathbf{00}\mathbf{a}\mathbf{.}\mathbf{m}\mathbf{.}$, the janitor turns on the furnace with the thermostat set at $\mathbf{70}\mathbf{°}\mathbf{F}$. The time constant for the building is $\frac{\mathbf{1}}{\mathbf{k}}\mathbf{=}\mathbf{2}\mathbf{ }\mathit{h}\mathit{r}$ and that for the building along with its heating system is $\frac{\mathbf{1}}{{\mathbf{k}}_{\mathbf{1}}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{ }\mathit{h}\mathit{r}$. Assuming that the outside temperature remains constant, what will be the temperature inside the lecture hall at $\mathbf{8}\mathbf{:}\mathbf{00}\mathbf{a}\mathbf{.}\mathbf{m}\mathbf{.}$? When will the temperature inside the hall reach $\mathbf{65}\mathbf{°}\mathbf{F}$?

Step-by-Step Solution

Answer

At $8:00\mathrm{a}.\mathrm{m}.$, the temperature inside the lecture will reach $98°\mathrm{F}$and the temperature inside will reach $65°\mathrm{F}$after $29.64 \mathrm{minutes}$.

Step1: Given data.

The temperature inside and outside the lecture hall is $40°\mathrm{F}.-$At $7:00\mathrm{a}.\mathrm{m}.$, the janitor turns on the furnace with the thermostat set at $70°\mathrm{F}$. The time constant for the building is $\frac{1}{\mathrm{k}}=2 \mathrm{hr}$ and that for the building along with its heating system is $\frac{1}{{\mathrm{k}}_1}=\frac{1}{2} \mathrm{hr}$.

Step 2: Analyzing the given statement

Assuming that

Here, temperature inside the lecture hall, $T_{in}={40}^{0}F$.

Temperature outside the hall, $T_{out}={40}^{0}F$.

Temperature value on furnace, $T_f={70}^{0}F$.

The time constant for the building is $\frac{1}{\mathrm{k}}=2\mathrm{hr}$.

The time constant for the building with its heating system is $\frac{1}{{\mathrm{k}}_1}=\frac{1}{2} \mathrm{hr}$.

It will use the following formula to find the solution,

 $\frac{dT}{dt}=K_1\left(T_{out}-T\right)+K_u\left(T_f-T\right)$

Step 3: To find the value of K u

As it knows that,

 $K_1+K_u=K$

Using values from step1,

 $$\begin{gathered} 2+K_u=\frac{1}{2} \\ K_u=\frac{1}{2}-2 \\ K_u=\frac{1-4}{2} \\ K_u=\frac{-3}{2} \end{gathered}$$

One will use this value in equation (1).

Step 4: To determine the temperature inside the lecture hall at 8:00 a.m.

Now from equation (1),

$$\begin{gathered} \frac{dT}{dt}=2\left(40-T\right)-\frac{3}{2}\left(70-T\right) \\ \frac{dT}{dt}=\frac{-50-T}{2} \\ \frac{dT}{dt}=-25-\frac{T}{2} \end{gathered}$$ 

i.e., $\frac{dT}{dt}+\frac{T}{2}=-25$               …… (2)

 

Integrating factor = ${\mathit{e}}^{\mathbf{\int }\frac{\mathbf{1}}{\mathbf{2}}\mathbf{d}\mathbf{t}}\mathbf{=}{\mathit{e}}^{\frac{\mathbf{1}}{\mathbf{2}}\mathbf{t}}$

Multiplying both sides of (2) by $e^{\frac{1}{2}t}$,

 $$\begin{gathered} e^{\frac{1}{2}t}·\frac{dT}{dt}+e^{\frac{1}{2}t}·\frac{T}{2}=-25·e^{\frac{1}{2}t} \\ \frac{d}{dt}\left(T·e^{\frac{1}{2}t}\right)=-25·e^{\frac{1}{2}t} \end{gathered}$$

Integrating both sides,

$T·e^{\frac{1}{2}t}=-50e^{\frac{1}{2}t}+C$ 

Where, C is an arbitrary constant.

 

When t=0, $\mathrm{T}=40°\mathrm{F}$

$$\begin{gathered} 40=-50+C \\ C=90 \end{gathered}$$

Therefore,   $\mathit{T}\mathbf{=}\mathbf{-}\mathbf{50}\mathbf{+}\mathbf{90}{\mathit{e}}^{\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{t}}$     …… (3)

Temperature in the lecture hall when t=1 hour,

$$\begin{gathered} T=-50+90e^{-\frac{1}{2}} \\ T={98}^{0}F \end{gathered}$$

Hence, the temperature inside the lecture will reach $\mathbf{98}\mathbf{°}\mathbf{C}$ at  $\mathbf{8}\mathbf{:}\mathbf{00}\mathbf{a}\mathbf{.}\mathbf{m}\mathbf{.}$

Step 5: To determine the time when the temperature inside the lecture hall will reach 65 ∘ F

Now from equation (3),

Substituting $\mathrm{T}=65°\mathrm{F}$

$$\begin{gathered} T=-50+90e^{-\frac{1}{2}t} \\ 65=-50+90e^{-\frac{1}{2}t} \\ t=2ln\left(1.28\right) \\ t=0.494hour \\ t=29.64min \end{gathered}$$

 

Hence, the temperature inside will reach $\mathbf{65}\mathbf{°}\mathbf{F}$ after  $\mathbf{29}\mathbf{.}\mathbf{64} \mathbf{minutes}$.