Question

A solar hot-water-heating system consists of a hot-water tank and a solar panel. The tank is well insulated and has a time constant of 64 hr. The solar panel generates 2000 Btu/hr during the day, and the tank has a heat capacity of $\mathbf{2}\mathbf{°}\mathbf{F}\mathbf{}$per thousand Btu. If the water in the tank is initially$\mathbf{80}\mathbf{°}\mathbf{F}$and the room temperature outside the tank is, what will be the temperature in the tank after 12 hr of sunlight?

Short answer

The temperature inside the tank after 12 hr of sunlight will be$148.6°\mathrm{F}.$

Step-by-step solution

Analyzing the given statement

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 Btu/hr

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

Heat capacity of the tank is $2°\mathrm{F}$per thousand Btu

The time constant for the tank is $\frac{1}{\mathrm{k}}=64 \mathrm{hr}$.

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,

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

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

Formation of the differential equation using equation (1)

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

$$\begin{gathered} \frac{dT}{dt}=\frac{1}{64}\left(80-T\right)+\left(2000Btu/hr\right)·\left(\frac{2^{0}F}{1000Btu}\right) \\ \frac{dT}{dt}=\frac{1}{64}\left(80-T\right)+4 \\ \frac{dT}{dt}=5.25-\frac{T}{64} \\ \frac{dT}{dt}+\frac{T}{64}=5.25 ......\mathbf{\left(}\mathbf{2}\mathbf{\right)} \end{gathered}$$

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

Determining the temperature in the tank after 12 hr

The differential equation obtainedin step1 is,

$\frac{dT}{dt}+\frac{T}{64}=5.25 ......\left(3\right)$

Integrating factor, I.F.=$e^{\int \frac{1}{64}dt}=e^{\frac{1}{64}t}$

Multiplying both sides of (3) by $e^{\frac{1}{64}t}$,

$$\begin{gathered} e^{\frac{1}{64}t}·\frac{dT}{dt}+e^{\frac{1}{64}t}·\frac{T}{64}=\left(5.25\right)·e^{\frac{1}{64}t} \\ \frac{d}{dt}\left(T·e^{\frac{1}{64}t}\right)=\left(5.25\right)·e^{\frac{1}{64}t} \end{gathered}$$

Now, integrating both sides,

$T·e^{\frac{1}{64}t}=\left(5.25\right)\left(64\right)·e^{\frac{1}{64}t}+C ......\left(4\right)$

Initially, when t =0, T=110°F,

$$\begin{gathered} 110=\left(5.25\right)\left(64\right)+C \\ C=-226 \end{gathered}$$

Using this value of C in equation (4),

$$\begin{gathered} T·e^{\frac{1}{64}t}=\left(5.25\right)\left(64\right)·e^{\frac{1}{64}t}-226 \\ T=\left(5.25\right)\left(64\right)-226·e^{-\frac{1}{64}t} ......\mathbf{\left(}\mathbf{5}\mathbf{\right)} \end{gathered}$$

When the time t=12 hr

$$\begin{gathered} T=\left(5.25\right)\left(64\right)-226·e^{-\frac{12}{64}} \\ T=148.6^{0}F \end{gathered}$$

Hence, the temperature inside the tank after 12 hr of sunlight will be role="math" localid="1664177809777" $\mathbf{148}\mathbf{.}\mathbf{6}\mathbf{°}\mathbf{F}$.