Q35E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

A house, for cooling purposes, consists of two zones: the attic area zone A and the living area zone B (see Figure 5.4). The living area is cooled by a 2 – ton air conditioning unit that removes 24,000 Btu/hr. The heat capacity of zone B is $\frac{\mathbf{1}}{\mathbf{2}}{}^{\mathbf{\circ }}\mathbf{F}$ per thousand Btu. The time constant for heat transfer between zone A and the outside is 2 hr, between zone B and the outside is 4 hr, and between the two zones is 4 hr. If the outside temperature stays at $\mathbf{100}\mathbf{°}\mathbf{F}$, how warm does it eventually get in the attic zone A? (Heating and cooling buildings was treated in Section 3.3 on page 102.)

 

Step-by-Step Solution

Answer

Therefore, the warm of zone A is eventually got $\mathbf{90}\mathbf{.}\mathbf{4}\mathbf{°}\mathbf{F}$. 

Step 1: General form

Elimination Procedure for 2 × 2 Systems:

 

To find a general solution for the system

$$\begin{gathered} {\mathbf{L}}_{\mathbf{1}}\left[\mathbf{x}\right]\mathbf{+}{\mathbf{L}}_{\mathbf{2}}\left[\mathbf{y}\right]\mathbf{=}{\mathbf{f}}_{\mathbf{1}}\mathbf{,} \\ {\mathbf{L}}_{\mathbf{3}}\left[\mathbf{x}\right]\mathbf{+}{\mathbf{L}}_{\mathbf{4}}\left[\mathbf{y}\right]\mathbf{=}{\mathbf{f}}_{\mathbf{2}}\mathbf{,} \end{gathered}$$ 

Where ${\mathbf{L}}_{\mathbf{1}}\mathbf{,}{\mathbf{L}}_{\mathbf{2}}\mathbf{,}{\mathbf{L}}_{\mathbf{3}}\mathbf{,}$ and L₄ are polynomials in $\mathbf{D}\mathbf{=}\frac{\mathbf{d}}{\mathbf{dt}}.$

 

1. Make sure that the system is written in operator form.

 

1. Eliminate one of the variables, say, y, and solve the resulting equation for x(t). If the system is degenerating, stop! A separate analysis is required to determine whether or not there are solutions.

 

1. (Shortcut) If possible, use the system to derive an equation that involves y(t) but not its derivatives. [Otherwise, go to step (d).] Substitute the found expression for x(t) into this equation to get a formula for y(t). The expressions for x(t), and y(t) give the desired general solution.

 

1. Eliminate x from the system and solve for y(t). [Solving for y(t) gives more constants----in fact, twice as many as needed.]

 

1. Remove the extra constants by substituting the expressions for x(t) and y(t) into one or both of the equations in the system. Write the expressions for x(t) and y(t) in terms of the remaining constants.

 

Vistas’ formulas for finding roots:

 

For function y(t) to be bounded when $\mathbf{t}\to \mathbf{+}\infty$ we need for both roots of the auxiliary equation to be non-positive if they are reals and, if they are complex, then the real part has to be non-positive. In other words,

 

1. If ${\mathbf{r}}_{\mathbf{1}}\mathbf{,}{\mathbf{r}}_{\mathbf{2}}\in \mathbf{R}$, then ${\mathbf{r}}_{\mathbf{1}}·{\mathbf{r}}_{\mathbf{2}}⩾\mathbf{0}\mathbf{,}{\mathbf{r}}_{\mathbf{1}}\mathbf{+}{\mathbf{r}}_{\mathbf{2}}⩽\mathbf{0}$,
2. If ${\mathbf{r}}_{\mathbf{1}}\mathbf{,}{\mathbf{r}}_{\mathbf{2}}\mathbf{=}\mathbf{\alpha }\pm \mathbf{\beta i}$, $\mathbf{\beta }\ne 0$ , then $\mathbf{\alpha }\mathbf{=}\frac{{\mathbf{r}}_{\mathbf{1}}\mathbf{+}{\mathbf{r}}_{\mathbf{2}}}{\mathbf{2}}⩽0$.

 

Section 3.3: Heat transfer, it is modelled by the following equation $\frac{\mathbf{dT}}{\mathbf{dt}}\mathbf{=}\mathbf{K}\left(\mathbf{M}\left(\mathbf{t}\right)\mathbf{-}\mathbf{T}\left(\mathbf{t}\right)\right)\mathbf{+}\mathbf{H}\left(\mathbf{t}\right)\mathbf{+}\mathbf{U}\left(\mathbf{t}\right)$ where $\frac{\mathbf{1}}{\mathbf{K}}$ is the time constant for the building given in hours, M(t) is the outside temperature, T(t) is the inside temperature, H(t) is the heating in the building, U(t) is the cooling.

 

Step 2: Evaluate the given equation

Given that, the living area is cooled by a 2 – ton air conditioning unit that removes 24,000 Btu/hr. 

 

The heat capacity of zone B is $\frac{\mathbf{1}}{\mathbf{2}}{}^{\mathbf{\circ }}\mathbf{F}$ per thousand Btu. 

 

The time constant for heat transfer between zone A and the outside is 2 hr, between zone B and the outside is 4 hr, and between the two zones is 4 hr.

 

Let x(t) be denoted as the temperature in A at time t and y(t) denoted as the temperature in B at time t.

 

Using the given information create the system of equation.

 

Then , $\frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}}\left(\mathbf{y}\left(\mathbf{t}\right)\mathbf{-}\mathbf{x}\left(\mathbf{t}\right)\right)\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\left(\mathbf{100}\mathbf{-}\mathbf{x}\left(\mathbf{t}\right)\right)$and $\frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}}\left(\mathbf{x}\left(\mathbf{t}\right)\mathbf{-}\mathbf{y}\left(\mathbf{t}\right)\right)\mathbf{+}\frac{\mathbf{1}}{\mathbf{4}}\left(\mathbf{100}\mathbf{-}\mathbf{y}\left(\mathbf{t}\right)\right)\mathbf{-}\mathbf{12}$

 

The above equations can be rewritten as,

$$\begin{gathered} \frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}}\left(\mathbf{y}\left(\mathbf{t}\right)\mathbf{-}\mathbf{x}\left(\mathbf{t}\right)\right)\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\left(\mathbf{100}\mathbf{-}\mathbf{x}\left(\mathbf{t}\right)\right) \\ \mathbf{4}\frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{=}\left(\mathbf{y}\left(\mathbf{t}\right)\mathbf{-}\mathbf{x}\left(\mathbf{t}\right)\right)\mathbf{+}\mathbf{2}\left(\mathbf{100}\mathbf{-}\mathbf{x}\left(\mathbf{t}\right)\right) \\ \mathbf{=}\mathbf{y}\left(\mathbf{t}\right)\mathbf{-}\mathbf{x}\left(\mathbf{t}\right)\mathbf{+}\mathbf{200}\mathbf{-}\mathbf{2}\mathbf{x}\left(\mathbf{t}\right) \\ \mathbf{=}\mathbf{y}\left(\mathbf{t}\right)\mathbf{-}\mathbf{3}\mathbf{x}\left(\mathbf{t}\right)\mathbf{+}\mathbf{200} \\ \mathbf{4}\frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{+}\mathbf{3}\mathbf{x}\left(\mathbf{t}\right)\mathbf{-}\mathbf{y}\left(\mathbf{t}\right)\mathbf{=}\mathbf{200} \dots \left(1\right) \end{gathered}$$

$$\begin{gathered} \frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}}\left(\mathbf{x}\left(\mathbf{t}\right)\mathbf{-}\mathbf{y}\left(\mathbf{t}\right)\right)\mathbf{+}\frac{\mathbf{1}}{\mathbf{4}}\left(\mathbf{100}\mathbf{-}\mathbf{y}\left(\mathbf{t}\right)\right)\mathbf{-}\mathbf{12} \\ \mathbf{4}\frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{=}\left(\mathbf{x}\left(\mathbf{t}\right)\mathbf{-}\mathbf{y}\left(\mathbf{t}\right)\right)\mathbf{+}\left(\mathbf{100}\mathbf{-}\mathbf{y}\left(\mathbf{t}\right)\right)\mathbf{-}\mathbf{48} \\ \mathbf{=}\mathbf{x}\left(\mathbf{t}\right)\mathbf{-}\mathbf{y}\left(\mathbf{t}\right)\mathbf{+}\mathbf{100}\mathbf{-}\mathbf{y}\left(\mathbf{t}\right)\mathbf{-}\mathbf{48} \\ \mathbf{=}\mathbf{x}\left(\mathbf{t}\right)\mathbf{-}\mathbf{2}\mathbf{y}\left(\mathbf{t}\right)\mathbf{+}\mathbf{52} \\ \mathbf{4}\frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{-}\mathbf{x}\left(\mathbf{t}\right)\mathbf{+}\mathbf{2}\mathbf{y}\left(\mathbf{t}\right)\mathbf{=}\mathbf{52} \cdots \left(2\right) \end{gathered}$$

Rewrite the system in operator form:

 

 $\left(\mathbf{4}\mathbf{D}\mathbf{+}\mathbf{3}\right)\left[\mathbf{x}\right]\mathbf{-}\mathbf{y}\mathbf{=}\mathbf{200}$ …… (3)

$\mathbf{-}\mathbf{x}\mathbf{+}\left(\mathbf{4}\mathbf{D}\mathbf{+}\mathbf{2}\right)\left[\mathbf{y}\right]\mathbf{=}\mathbf{52}$  …… (4)

Step 3: Solve the equations

Multiply 4D+2 on equation (3) and add with equation (4).

$$\begin{gathered} \left(\mathbf{4}\mathbf{D}\mathbf{+}\mathbf{2}\right)\left(\mathbf{4}\mathbf{D}\mathbf{+}\mathbf{3}\right)\left[\mathbf{x}\right]\mathbf{-}\left(\mathbf{4}\mathbf{D}\mathbf{+}\mathbf{2}\right)\left[\mathbf{y}\right]\mathbf{-}\mathbf{x}\mathbf{+}\left(\mathbf{4}\mathbf{D}\mathbf{+}\mathbf{2}\right)\left[\mathbf{y}\right]\mathbf{=}\left(\mathbf{4}\mathbf{D}\mathbf{+}\mathbf{2}\right)\mathbf{200}\mathbf{+}\mathbf{52} \\ \left(\mathbf{4}\mathbf{D}\mathbf{+}\mathbf{2}\right)\left(\mathbf{4}\mathbf{D}\mathbf{+}\mathbf{3}\right)\left[\mathbf{x}\right]\mathbf{-}\left[\mathbf{x}\right]\mathbf{=}\mathbf{452} \\ \left(\mathbf{16}{\mathbf{D}}^{\mathbf{2}}\mathbf{+}\mathbf{20}\mathbf{D}\mathbf{+}\mathbf{6}\mathbf{-}\mathbf{1}\right)\left[\mathbf{x}\right]\mathbf{=}\mathbf{452} \\ \left(\mathbf{16}{\mathbf{D}}^{\mathbf{2}}\mathbf{+}\mathbf{20}\mathbf{D}\mathbf{+}\mathbf{5}\right)\left[\mathbf{x}\right]\mathbf{=}\mathbf{452} \\ \left(\mathbf{16}{\mathbf{D}}^{\mathbf{2}}\mathbf{+}\mathbf{20}\mathbf{D}\mathbf{+}\mathbf{5}\right)\left[\mathbf{x}\right]\mathbf{=}\mathbf{452} \cdots \left(5\right) \end{gathered}$$ 

 Since the auxiliary equation to the corresponding homogeneous equation is $16{\mathbf{r}}^{\mathbf{2}}\mathbf{+}20\mathbf{r}\mathbf{+}5\mathbf{=}\mathbf{0}$.

 Then,

$$\begin{gathered} \mathbf{r}\mathbf{=}\frac{\mathbf{-}20\pm \sqrt{{\left(20\right)}^{\mathbf{2}}\mathbf{-}20\mathbf{\times }16}}{\mathbf{2}\mathbf{\times }\mathbf{16}} \\ \mathbf{=}\frac{\mathbf{-}20\pm \sqrt{400\mathbf{-}320}}{32} \\ \mathbf{=}\frac{\mathbf{-}20\pm \sqrt{80}}{3\mathbf{2}} \\ \mathbf{=}\frac{4\left(\mathbf{-}5\pm \sqrt{5}\right)}{32} \\ \mathbf{=}\frac{\mathbf{-}5\pm \sqrt{5}}{8} \end{gathered}$$

Hence, the roots are $\mathbf{r}\mathbf{=}\frac{\mathbf{-}5\mathbf{+}\sqrt{5}}{8}$ and $\mathbf{r}\mathbf{=}\frac{\mathbf{-}5\mathbf{-}\sqrt{5}}{8}$.

 

Then, the general solution of y is ${\mathbf{x}}_{\mathbf{h}}\left(\mathbf{t}\right)\mathbf{=}{\mathbf{Ae}}^{\left(\frac{\mathbf{-}\mathbf{5}\mathbf{+}\sqrt{\mathbf{5}}}{\mathbf{8}}\right)\mathbf{t}}\mathbf{+}{\mathbf{Be}}^{\left(\frac{\mathbf{-}\mathbf{5}\mathbf{-}\sqrt{\mathbf{5}}}{\mathbf{8}}\right)\mathbf{t}}$  …… (6)

 

Let us assume that, ${\mathbf{x}}_{\mathbf{p}}\left(\mathbf{t}\right)\mathbf{=}\mathbf{C}$  …… (7)

 

Substitute the equation (7) in equation (5).

$$\begin{gathered} \left(\mathbf{16}{\mathbf{D}}^{\mathbf{2}}\mathbf{+}\mathbf{20}\mathbf{D}\mathbf{+}\mathbf{5}\right)\left[\mathbf{x}\right]\mathbf{=}\mathbf{452} \\ \left(\mathbf{16}{\mathbf{D}}^{\mathbf{2}}\mathbf{+}\mathbf{20}\mathbf{D}\mathbf{+}\mathbf{5}\right)\left[\mathbf{C}\right]\mathbf{=}\mathbf{452} \\ \mathbf{5}\mathbf{C}\mathbf{=}\mathbf{452} \\ \mathbf{C}\mathbf{=}\mathbf{90}\mathbf{.}\mathbf{4} \end{gathered}$$ 

Substitute the value of C in equation (7).

 $$\begin{gathered} \mathbf{x}\left(\mathbf{t}\right)\mathbf{=}{\mathbf{x}}_{\mathbf{h}}\left(\mathbf{t}\right)\mathbf{+}{\mathbf{x}}_{\mathbf{p}}\left(\mathbf{t}\right) \\ \mathbf{=}{\mathbf{Ae}}^{\left(\frac{\mathbf{-}\mathbf{5}\mathbf{+}\sqrt{\mathbf{5}}}{\mathbf{8}}\right)\mathbf{t}}\mathbf{+}{\mathbf{Be}}^{\left(\frac{\mathbf{-}\mathbf{5}\mathbf{-}\sqrt{\mathbf{5}}}{\mathbf{8}}\right)\mathbf{t}}\mathbf{+}\mathbf{90}\mathbf{.}\mathbf{4} \end{gathered}$$

 

So, the general solution is $\mathbf{x}\left(\mathbf{t}\right)\mathbf{=}{\mathbf{Ae}}^{\left(\frac{\mathbf{-}\mathbf{5}\mathbf{+}\sqrt{\mathbf{5}}}{\mathbf{8}}\right)\mathbf{t}}\mathbf{+}{\mathbf{Be}}^{\left(\frac{\mathbf{-}\mathbf{5}\mathbf{-}\sqrt{\mathbf{5}}}{\mathbf{8}}\right)\mathbf{t}}\mathbf{+}\mathbf{90}\mathbf{.}\mathbf{4}$…… (8)

 

Step 5: limit method

To find: $\underset{t\to \infty }{lim}x$.

 Implement the limits on equation (8).

$$\begin{gathered} \underset{\mathbf{t}\mathbf{\to }\mathbf{\infty }}{\mathbf{lim}}\mathbf{x}\left(\mathbf{t}\right)\mathbf{=}\underset{\mathbf{t}\mathbf{\to }\mathbf{\infty }}{\mathbf{l}\mathbf{i}\mathbf{m}}\left[{\mathbf{Ae}}^{\left(\frac{\mathbf{-}\mathbf{5}\mathbf{+}\sqrt{\mathbf{5}}}{\mathbf{8}}\right)\mathbf{t}}\mathbf{+}{\mathbf{Be}}^{\left(\frac{\mathbf{-}\mathbf{5}\mathbf{-}\sqrt{\mathbf{5}}}{\mathbf{8}}\right)\mathbf{t}}\mathbf{+}\mathbf{90}\mathbf{.}\mathbf{4}\right] \\ \mathbf{=}\mathbf{90}\mathbf{.}\mathbf{4} \end{gathered}$$ 

So, the solution is founded.