91Ó°ÊÓ

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

Given thata 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,

$$\begin{gathered} \frac{dT}{dt}=Rateofchangeintemperatureofthetank-Rateofchangeintemperatureduetosolar \\ heater \end{gathered}$$

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.

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.=$e^{∫\frac{1}{72}dt}=e^{\frac{1}{72}t}$

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 localid="1664176196883" $\mathrm{t}=0$, $\mathrm{T}=110°\mathrm{F}$,

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

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 ofsunlight will be localid="1664176516004" $\mathbf{127}\mathbf{.}\mathbf{5}\mathbf{°}\mathbf{F}$.