Q 3.3-13E - Fundamentals Of Differential Equations And Boundary Value Problems

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?

Step-by-Step Solution

Answer

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

Step1: 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 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,

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

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

Step2: 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.

Step 3: Determining the temperature in the tank after 12 hr

The differential equation obtained in 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 $\mathbf{148}\mathbf{.}\mathbf{6}\mathbf{°}\mathbf{F}$.