Q37E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In Problem 36, if a small furnace that generates 1000 Btu/hr is placed in zone B, determine the coldest it would eventually get in zone B has a heat capacity of $\mathbf{2}\mathbf{°}\mathbf{F}$ per thousand Btu.

Step-by-Step Solution

Answer

Therefore, the cold of zone B is eventually got $\frac{460}{11}\approx 41.8°F$.

Step 1: General form

Elimination Procedure for 2 x 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 $D=\frac{d}{dt}$:

1. Make sure that the system is written in operator form.
2. 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.
3. (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.
4. Eliminate x from the system and solve for y(t). [Solving for y(t) gives more constants- twice as many as needed.]
5. 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.

 

Vieta’s formulas for finding roots:

For function y(t) to be bounded, we need 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 $r_1,r_2\in R$, then $r_1·r_2⩾0,r_1+r_2⩽0$,
2. If $r_1,r_2=\alpha \pm \beta i$,$\beta \ne 0$, then $\alpha =\frac{r_1+r_2}{2}⩽0$.

 

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

Step 2: Evaluate the given equation

Given that, $1000Btu/hr$ is placed in zone B.

 Referring to problem 36:

 Only zone A is heated by a furnace, which generates $80,000Btu/hr$.

 The heat capacity of zone A is $\frac{1}{4}°F$ per thousand Btu.

 

The time constant for heat transfer between zone A and the outside is 4 hours, between the unheated zone B and the outside is 5 hours, and between the two zones is 2 hours.

 

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

 

Using the given information create the system of equation.

 

Then, $\frac{dx}{dt}=\frac{1}{2}\left(y\left(t\right)-x\left(t\right)\right)-\frac{x\left(t\right)}{4}+20$and $\frac{dy}{dt}=\frac{1}{2}\left(x\left(t\right)-y\left(t\right)\right)-\frac{y\left(t\right)}{5}+2$

 

The above equations can be rewritten as,

  $$\begin{gathered} \frac{dx}{dt}=\frac{1}{2}\left(y\left(t\right)-x\left(t\right)\right)-\frac{x\left(t\right)}{4}+20 \\ 4\frac{dx}{dt}=2\left(y\left(t\right)-x\left(t\right)\right)-x\left(t\right)+80 \\ =2y\left(t\right)-2x\left(t\right)+80-x\left(t\right) \\ =2y\left(t\right)-3x\left(t\right)+80 \end{gathered}$$

$$\begin{gathered} 4\frac{dx}{dt}+3x\left(t\right)-2y\left(t\right)=80 ......\mathbf{\left(}\mathbf{1}\mathbf{\right)} \\ \frac{dy}{dt}=\frac{1}{2}\left(x\left(t\right)-y\left(t\right)\right)-\frac{y\left(t\right)}{5}+2 \\ 10\frac{dy}{dt}=5\left(x\left(t\right)-y\left(t\right)\right)-2y\left(t\right)+20 \\ =5x\left(t\right)-5y\left(t\right)-2y\left(t\right)+20 \end{gathered}$$

$$\begin{gathered} =5x\left(t\right)-7y\left(t\right)+20 \\ 10\frac{dy}{dt}-5x\left(t\right)+7y\left(t\right)=20 ......\mathbf{\left(}\mathbf{2}\mathbf{\right)} \end{gathered}$$

 

Rewrite the system in operator form:

 $$\begin{gathered} \left(4D+3\right)\left[x\right]-2y=80 ......\mathbf{\left(}\mathbf{3}\mathbf{\right)} \\ -5x+\left(10D+7\right)\left[y\right]=20 ......\mathbf{\left(}\mathbf{4}\mathbf{\right)} \end{gathered}$$

Step 3: Solve the equations

Multiply 5 on equation (3) and multiply $4D + 3$ on equation (4). Then, add them together to get.

$$\begin{gathered} 5\left(4D+3\right)\left[x\right]-10y-5\left(4D+3\right)x+\left(4D+3\right)\left(10D+7\right)\left[y\right]=400+60 \\ \left(4D+3\right)\left(10D+7\right)\left[y\right]-10y=460 \\ \left(40D^2+58D+21-10\right)\left[y\right]=460 \\ \left(40D^2+58D+11\right)\left[y\right]=460 \end{gathered}$$ 

             $\left(40D^2+58D+11\right)\left[y\right]=400 ......\mathbf{\left(}\mathbf{5}\mathbf{\right)}$

Since the auxiliary equation to the corresponding homogeneous equation is:

$40r^2+58r+11=0$. 

 

Then,

 $$\begin{gathered} r=\frac{-58\pm \sqrt{{\left(58\right)}^2-44\times 40}}{2\times 40} \\ =\frac{-29\pm \sqrt{401}}{40} \end{gathered}$$

 

So, the roots are $r=\frac{-29+\sqrt{401}}{40}$ and $r=\frac{-29-\sqrt{401}}{40}$.

 

Then, the general solution of y is $y_h\left(t\right)=A{e}^{\left(\frac{-29+\sqrt{401}}{40}\right)t}+B{e}^{\left(\frac{-29-\sqrt{401}}{40}\right)t} ......\mathbf{\left(}\mathbf{6}\mathbf{\right)}$

 

Let us assume that, $y_p\left(t\right)=C ......\mathbf{\left(}\mathbf{7}\mathbf{\right)}$

 Substitute equation (7) in equation (5).

$$\begin{gathered} \left(40D^2+58D+11\right)\left[y\right]=460 \\ \left(40D^2+58D+11\right)\left[C\right]=460 \\ 11C=460 \\ C=\frac{460}{11} \end{gathered}$$

Substitute the value of C in equation (7).

$$\begin{gathered} y\left(t\right)=y_h\left(t\right)+y_p\left(t\right) \\ =A{e}^{\left(\frac{-29+\sqrt{401}}{40}\right)t}+B{e}^{\left(\frac{-29-\sqrt{401}}{40}\right)t}+\frac{460}{11} \end{gathered}$$

So, the general solution is $y\left(t\right)=A{e}^{\left(\frac{-29+\sqrt{401}}{40}\right)t}+B{e}^{\left(\frac{-29-\sqrt{401}}{40}\right)t}+\frac{460}{11} ......\mathbf{\left(}\mathbf{8}\mathbf{\right)}$

Step 4: limit method

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

 

Implement the limits on equation (8).

  $$\begin{gathered} \underset{t\to \infty }{lim}y\left(t\right)=\underset{t\to \infty }{lim}\left[A{e}^{\left(\frac{-29+\sqrt{401}}{40}\right)t}+B{e}^{\left(\frac{-29-\sqrt{401}}{40}\right)t}+\frac{460}{11}\right] \\ =\frac{460}{11} \end{gathered}$$

So, the solution is founded