Q 3.3-14E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In Problem 13, if a larger tank with a heat capacity of $\mathbf{1}\mathbf{°}\mathbf{F}$ per thousand Btu and a time constant of 72 hr is use instead (with all other factors being the same), what will be the temperature in the tank after 12 hr?

Step-by-Step Solution

Answer

The temperature inside the tank after 12 hr of sunlight will be $\mathbf{127}\mathbf{.}\mathbf{5}\mathbf{°}\mathbf{F}$

Step1: Important formula.

$\frac{dT}{dt}$= Rate of change in temperature of the tank  -  Rate of change in temperature due to solar heater

Step 2: Analyzing the given statement

Given that a solar hot-water-heating system consists of a hot-water tank and a solar panel.

Here, the heat generated by solar panel is  $2000\mathrm{Btu}/\mathrm{hr}$

Temperature outside the tank, $T_{out}={80}^{0}F$

Heat capacity of the tank is  $1°\mathrm{F}$per thousand Btu

The time constant for the tank is $\frac{1}{\mathrm{k}}=72 \mathrm{hr}$.

Initially, the temperature of water in the tank $=110°\mathrm{F}$

We have to find the temperature in the tank after 12 hr of sunlight.

We will use the following formula to find the solution,

$$\begin{gathered} \frac{dT}{dt}=Rate of change in temperature of the tank-Rate of change in temperature due to solar \\ heater \end{gathered}$$

$\frac{dT}{dt}=K\left(T_{out}-T\right)+\left(2000Btu/hr\right)·\left(\frac{2^{0}F}{1000Btu}\right) ......\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

Step3: Formation of the differential equation using equation (1)

Substituting the values of K and in $T_{out}$ equation (1),

$$\begin{gathered} \frac{dT}{dt}=\frac{1}{72}\left(80-T\right)+\left(2000Btu/hr\right)·\left(\frac{1^{0}F}{1000Btu}\right) \\ \frac{dT}{dt}=\frac{1}{72}\left(80-T\right)+2 \\ \frac{dT}{dt}=\frac{224}{72}-\frac{T}{72} \\ \frac{dT}{dt}+\frac{T}{72}=\frac{224}{72} ......\mathbf{\left(}\mathbf{2}\mathbf{\right)} \end{gathered}$$

We will use this differential equation to find the temperature in the tank after 12 hr.

 

Step 4: Determining the temperature in the tank after 12 hr

The differential equation obtained in step1 is,

 $\frac{dT}{dt}+\frac{T}{72}=\frac{224}{72} ......\mathbf{\left(}\mathbf{3}\mathbf{\right)}$       

Integrating factor, I.F.=$e^{\int \frac{1}{72}dt}=e^{\frac{1}{72}t}$

Multiplying both sides of (3) by $e^{\frac{1}{72}t}$,

$$\begin{gathered} e^{\frac{1}{72}t}·\frac{dT}{dt}+e^{\frac{1}{72}t}·\frac{T}{72}=\frac{224}{72}·e^{\frac{1}{72}t} \\ \frac{d}{dt}\left(T·e^{\frac{1}{72}t}\right)=\frac{224}{72}·e^{\frac{1}{72}t} \end{gathered}$$

Now, integrating both sides,

$T=224+C·e^{-\frac{1}{72}t} ......\mathbf{\left(}\mathbf{4}\mathbf{\right)}$         

Initially, when $\mathrm{t}=0$, $\mathrm{T}=110°\mathrm{F}$,

$$\begin{gathered} 110=224+C \\ C=-114 \end{gathered}$$

Using this value of C in equation (4),

 $T=224-114·e^{-\frac{1}{72}t} ......\mathbf{\left(}\mathbf{5}\mathbf{\right)}$ 

When the time t = 12 hr

$$\begin{gathered} T=224-114·e^{-\frac{12}{72}t} \\ T=127.5^{0}F \end{gathered}$$

Hence, the temperature inside the tank after 12 hr of sunlight will be $\mathbf{127}\mathbf{.}\mathbf{5}\mathbf{°}\mathbf{F}$.