Chapter 10, Problem 1P

A faucet supplies fluid to a container of cross-sectional area A at a volume flow rate $Q_{in}$, as shown in Fig. P10.1. At the same time, the fluid leaks out the bottom at a rate $Q_{out}=kh\left(t\right)$, where $k$ is a constant. It the container is initially empty, the fluid height $h\left(t\right)$ satisfies the following first-order differential equation and initial condition:

$A\frac{dh\left(t\right)}{dt}+kh\left(t\right)=Q_{in}$ $h\left(0\right)=0$

1. Determine the transient solution, $h_{tran}\left(t\right)$.

(b) Suppose the faucet is turned on and off in a sinusoidal fashion, so that $Q_{in}=\frac{Q}{2}\left(1-\sin \omega t\right)$. Determine the steady-state solution, $h_{ss}\left(t\right)$.

(c) Determine the total solution $h\left(t\right)$, subject to the initial condition.

To determine

(a)

The transient solution of the given system.

Answer to Problem 1P

The transient solution of the given system is $h_{tran}\left(t\right)=c{e}^{\frac{-k}{A}t}$.

Explanation of Solution

Given:

Initial fluid height $h\left(0\right)$ is $0$.

The expression for $Q_{out}$ is $Q_{out}=kh\left(t\right)$

The system equation is,

  $A\frac{dh\left(t\right)}{dt}+kh\left(t\right)=Q_{in}$ ...... (1)

Concept used:

The transient solution of the equation is in the form given as:

  $h_{tran}\left(t\right)=c{e}^{st}$ ...... (2)

Calculation:

To obtain transient solution of an equation put right hand side of the equation equal to zero.

Substitute $c{e}^{st}$ for $h\left(t\right)$ in equation (1).

  $\begin{array}{l}A\frac{d\left(c{e}^{st}\right)}{dt}+k\left(c{e}^{st}\right)=0\\ Acs{e}^{st}+kc{e}^{st}=0\\ c{e}^{st}\left(As+k\right)=0\\ As+k=0\end{array}$

Solve further for $s$.

  $s=\frac{-k}{A}$

Substitute $\frac{-k}{A}$ for $s$ in equation (2).

  $h_{tran}\left(t\right)=c{e}^{\frac{-k}{A}t}$ ...... (3)

Since, the transient solution of the equation is in the time constant form given as:

  $h_{tran}\left(t\right)=c{e}^{\frac{-k}{A}t}$

Conclusion:

Thus, the transient solution of the given system is $h_{tran}\left(t\right)=c{e}^{\frac{-k}{A}t}$.

To determine

(b)

The steady state solution of the given system.

Answer to Problem 1P

The steady state solution of the given system is $v_{ss}\left(t\right)=\frac{-Q}{2\left(A^2{\omega }^2+k^2\right)}\left(k\sin \omega t-A\omega \cos \omega t\right)$.

Explanation of Solution

Given:

The expression for $Q_{in}$ is $Q_{in}=\frac{Q}{2}\left(1-\sin \omega t\right)$

Concept used:

The steady state solution of the equation is in the form given as:

  $h_{ss}\left(t\right)=C\sin \omega t+D\cos \omega t$ ...... (4)

Calculation:

Substitute $C\sin \omega t+D\cos \omega t$ for $h\left(t\right)$ and $\frac{Q}{2}\left(1-\sin \omega t\right)$ for $Q_{in}$ in equation (1).

  $\begin{array}{c}A\frac{d}{dt}\left(C\sin \omega t+D\cos \omega t\right)+k\left(C\sin \omega t+D\cos \omega t\right)=\frac{Q}{2}\left(1-\sin \omega t\right)\\ A\left(C\omega \cos \omega t-D\omega \sin \omega t\right)+k\left(C\sin \omega t+D\cos \omega t\right)=\frac{Q}{2}\left(1-\sin \omega t\right)\end{array}$

Simplify further.

   $\cos \omega t\left(AC\omega +kD\right)+\sin \omega t\left(-AD\omega +kC\right)=\frac{Q}{2}-\frac{Q}{2}\sin \omega t$ ...... (5)

Equate coefficient of $\sin \omega t$ on both side of the equation.

   $\left(-AD\omega +kC\right)=-\frac{Q}{2}$ ...... (6)

Equate coefficient of $\cos \omega t$ on both side of the equation.

  $\left(AC\omega +kD\right)=0$

Rearrange equation for $D$.

   $D=-\frac{AC\omega }{k}$ ...... (7)

Substitute $-\frac{AC\omega }{k}$ for $D$ in equation (6).

  $\begin{array}{c}\left(-A\left(-\frac{AC\omega }{k}\right)\omega +kC\right)=-\frac{Q}{2}\\ \left(\left(\frac{A^{2}C{\omega }^2}{k}\right)+kC\right)=-\frac{Q}{2}\\ C\left(\frac{A^2{\omega }^2}{k}+k\right)=-\frac{Q}{2}\\ C=-\frac{Qk}{2\left(A^2{\omega }^2+k^2\right)}\end{array}$

Substitute $-\frac{Qk}{2\left(A^2{\omega }^2+k^2\right)}$ for $C$ in equation (7).

  $\begin{array}{c}D=-\frac{A\left(-\frac{Qk}{2\left(A^2{\omega }^2+k^2\right)}\right)\omega }{k}\\ =\frac{AQ\omega }{2\left(A^2{\omega }^2+k^2\right)}\end{array}$

Substitute $-\frac{Qk}{2\left(A^2{\omega }^2+k^2\right)}$ for $C$ and $\frac{AQ\omega }{2\left(A^2{\omega }^2+k^2\right)}$ for $D$ in equation (4).

  $\begin{array}{c}{h}_{ss}\left(t\right)=\left(-\frac{Qk}{2\left(A^2{\omega }^2+k^2\right)}\right)\sin \omega t+\left(\frac{AQ\omega }{2\left(A^2{\omega }^2+k^2\right)}\right)\cos \omega t\\ =\frac{-Q}{2\left(A^2{\omega }^2+k^2\right)}\left(k\sin \omega t-A\omega \cos \omega t\right)\end{array}$

Conclusion:

Thus, the steady state solution of the given system is $h_{ss}\left(t\right)=\frac{-Q}{2\left(A^2{\omega }^2+k^2\right)}\left(k\sin \omega t-A\omega \cos \omega t\right)$.

To determine

(c)

The total solution of the given system.

Answer to Problem 1P

The total solution of the given system is $h\left(t\right)=\frac{-Q}{2\left(A^2{\omega }^2+k^2\right)}\left(A\omega e^{\frac{-k}{A}t}+k\sin \omega t-A\omega \cos \omega t\right)$.

Explanation of Solution

Concept used:

The total solution of the equation is given as:

  $h\left(t\right)=h_{tran}\left(t\right)+h_{ss}\left(t\right)$ ...... (8)

Calculation:

Substitute $\frac{-Q}{2\left(A^2{\omega }^2+k^2\right)}\left(k\sin \omega t-A\omega \cos \omega t\right)$ for $h_{ss}\left(t\right)$ and $c{e}^{\frac{-k}{A}t}$ for $h_{tran}\left(t\right)$ in equation (8).

   $h\left(t\right)=c{e}^{\frac{-k}{A}t}\text{+}\frac{-Q}{2\left(A^2{\omega }^2+k^2\right)}\left(k\sin \omega t-A\omega \cos \omega t\right)$ ...... (9)

Substitute $0$ for $t$ and $0$ for $h\left(0\right)$ in equation (9).

  $\begin{array}{c}0=c{e}^{\frac{-k}{A}\left(0\right)}\text{+}\frac{-Q}{2\left(A^2{\omega }^2+k^2\right)}\left(k\sin \omega \left(0\right)-A\omega \cos \omega \left(0\right)\right)\\ 0=c+\frac{-Q}{2\left(A^2{\omega }^2+k^2\right)}\left(-A\omega \right)\\ 0=c+\frac{QA\omega }{2\left(A^2{\omega }^2+k^2\right)}\\ \end{array}$

Rearrange for $c$.

  $c=-\frac{QA\omega }{2\left(A^2{\omega }^2+k^2\right)}$

Substitute $-\frac{QA\omega }{2\left(A^2{\omega }^2+k^2\right)}$ for $c$ in equation (9).

  $\begin{array}{c}h\left(t\right)=\left(-\frac{QA\omega }{2\left(A^2{\omega }^2+k^2\right)}\right)e^{\frac{-k}{A}t}\text{+}\frac{-Q}{2\left(A^2{\omega }^2+k^2\right)}\left(k\sin \omega t-A\omega \cos \omega t\right)\\ =\frac{-Q}{2\left(A^2{\omega }^2+k^2\right)}\left(A\omega e^{\frac{-k}{A}t}+k\sin \omega t-A\omega \cos \omega t\right)\end{array}$

Conclusion:

Thus, the total solution of the given system is $h\left(t\right)=\frac{-Q}{2\left(A^2{\omega }^2+k^2\right)}\left(A\omega e^{\frac{-k}{A}t}+k\sin \omega t-A\omega \cos \omega t\right)$.