Chapter 7, Problem 7.32P

To determine

(a)

The pump resistance and the steady-state height.

Answer to Problem 7.32P

The pump resistance is $600\text{N}\text{.}\text{s}/\left(\text{kg}{\text{.m}}^{\text{2}}\right)$ and the steady-state height is $1.223\text{m}$.

Explanation of Solution

Calculation:

The figure below shows the pump curve and the line for steady-state flow.

Figure-(1)

Write the expression of the two resistances in series.

$q_{m2}=\left(\frac{1}{R_1+R_2}\right)\Delta p$.... (I)

Here, the outlet flow rate of the tank is $q_{m2}$, pump resistance is $R_1$, the resistance of the outlet pipe is $R_2$ and the pressure rise in the pump is $\Delta p$.

Write the expression of mass conservation.

$q_{m1}-q_{m2}=\frac{\left(\Delta p-\rho gh\right)}{R_1}-\left(\frac{\rho gh}{R_2}\right)$.... (II)

Here, the outlet flow rate of the tank is $q_{m2}$, the inlet flow rate of the tank is $q_{m1}$, pump resistance is $R_1$, the resistance of outlet pipe is $R_2$, the density of the fluid is $\rho$, the acceleration due to gravity is $g$, the height of liquid in the tank is $h$ and the pressure rise in the pump is $\Delta p$.

Write the expression of the height of the tank.

$h=\frac{\Delta p}{\rho g}\left(\frac{R_2}{R_1+R_1}\right)$.... (III)

Here, the pump resistance is $R_1$, the resistance of outlet pipe is $R_2$, the density of the fluid is $\rho$, the acceleration due to gravity is $g$, the height of liquid in the tank is $h$ and the pressure rise in the pump is $\Delta p$.

Refer to Figure-(1) to obtain the pressure rise in the pump as $3\times {10}^4\text{N}/{\text{m}}^{\text{2}}$ and outer flow rate as $30\text{kg}/\text{sec}$.

Substitute $30\text{kg}/\text{sec}$ for $q_{m2}$

$3\times {10}^4\text{N}/{\text{m}}^{\text{2}}$ for $\Delta p$ and $400\text{l}/\left(\text{m}\text{.s}\right)$ for $R_2$ in Equation (I).

$\begin{array}{l}\left(30\text{kg}/\text{sec}\right)=\frac{1}{R_1+\left(400\text{l}/\left(\text{m}\text{.sec}\right)\right)}\times \left(3\times {10}^4\text{N}/{\text{m}}^{\text{2}}\right)\\ \left(30\text{kg}/\text{sec}\right)\times \left(R_1+400\text{l}/\left(\text{m}\text{.sec}\right)\right)=\left(3\times {10}^4\text{N}/{\text{m}}^{\text{2}}\right)\\ \left[R_1+400\text{l}/\left(\text{m}\text{.sec}\right)\right]=\frac{\left(3\times {10}^4\text{N}/{\text{m}}^{\text{2}}\right)}{\left(30\text{kg}/\text{sec}\right)}\\ R_1=\frac{\left(3\times {10}^4\text{N}/{\text{m}}^{\text{2}}\right)}{\left(30\text{kg}/\text{sec}\right)}-400\text{l}/\left(\text{m}\text{.sec}\right)\end{array}$

$R_1=\left(600\text{N}\text{.s}/\left(\text{kg}{\text{.m}}^{\text{2}}\right)\right)$

Thus, the pump resistance is $600\text{N}\text{.}\text{s}/\left(\text{kg}{\text{.m}}^{\text{2}}\right)$.

Substitute 0 for $\left(q_{m1}-q_{m2}\right)$ in Equation (II).

$\begin{array}{l}\frac{\Delta p-\rho gh}{R_1}-\frac{\rho gh}{R_2}=0\\ \frac{\Delta p}{R_1}-\frac{\rho gh}{R_1}-\frac{\rho gh}{R_2}=0\\ \frac{\Delta p}{R_1}-\rho gh\left[\frac{1}{R_1}+\frac{1}{R_2}\right]=0\\ \rho gh\left[\frac{1}{R_1}+\frac{1}{R_2}\right]=\frac{\Delta p}{R_1}\end{array}$

$\rho gh\left[\frac{R_1+R_2}{R_2}\right]=\Delta p$

Substitute $3\times {10}^4\text{N}/{\text{m}}^{\text{2}}$ for $\Delta p$, $\left(600\text{N}\text{.s}/\left(\text{kg}{\text{.m}}^{\text{2}}\right)\right)$ for $R_1$, $\left(400\text{N}\text{.s}/\left(\text{kg}{\text{.m}}^{\text{2}}\right)\right)$ for $R_2$, $1000\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$ and $9.81\text{m}/{\text{sec}}^{\text{2}}$ for $g$ in Equation (III).

$\begin{array}{c}h=\left(\frac{3\times {10}^4\text{N}/{\text{m}}^{\text{2}}}{\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\times \left(9.81\text{m}/{\text{sec}}^{\text{2}}\right)}\right)\left(\frac{\left(400\text{N}\text{.s}/\left(\text{kg}{\text{.m}}^{\text{2}}\right)\right)}{\left(600\text{N}\text{.s}/\left(\text{kg}{\text{.m}}^{\text{2}}\right)\right)+\left(400\text{N}\text{.s}/\left(\text{kg}{\text{.m}}^{\text{2}}\right)\right)}\right)\\ =\left(3.0581\text{m}\right)\left(0.4\right)\\ =1.223\text{m}\end{array}$

Thus, the steady-state height is $1.223\text{m}$.

To determine

(b)

The linearized dynamic model of the height deviation in the tank.

Answer to Problem 7.32P

The linearized dynamic model of the height deviation in the tank is $A\frac{d}{dt}\delta h=-0.0595\times \delta h$.

Explanation of Solution

Calculation:

Write the expression of the linear model.

$\delta q_{m1}=-\frac{1}{r}\delta \left(\Delta p\right)$.... (IV) Here, the small change in flow rate is $\delta q_{m1}$, the slope of the line is $\left(\frac{1}{r}\right)$ and the small change in pressure is $\delta \left(\Delta p\right)$.

Write the expression of mass conservation.

$\rho gh\left[\frac{R_1+R_2}{R_2}\right]=\Delta p$.... (V)

Here, the pump resistance is $R_1$, the resistance of outlet pipe is $R_2$, the density of the fluid is $\rho$, the acceleration due to gravity is $g$, the height of liquid in the tank is $h$ and the pressure rise in the pump is $\Delta p$.

Write the expression of small deviation in mass flow rates of the mass conservation.

$\rho A\frac{d}{dt}\delta h=\delta q_{m1}-\delta q_{m2}$.... (VI)

Here, the area of the tank is $A$, the density of the fluid is $\rho$, the deviation in height is $\delta h$ and the small change in flow rate is $\delta q_{m1}$ and $\delta q_{m2}$.

Write the expression of the outlet flow rate of the tank.

$q_{m2}=\frac{\rho gh}{R_2}$.... (VII) Here, the outlet flow rate of the tank is $\delta q_{m2}$. the resistance of outlet pipe is $R_2$, the density of the fluid is $\rho$, the acceleration due to gravity is $g$ and the height of liquid in the tank is $h$.

Write the expression of the deviation.

$\begin{array}{c}\rho A\frac{d}{dt}\delta h=\left(-\frac{1}{r}\right)\left(\frac{R_1+R_2}{R_2}\rho g\delta h\right)-\frac{\rho g}{R_2}\delta h\\ A\frac{d}{dt}\delta h=-\left(\frac{1}{r}\left(\frac{R_1+R_2}{R_2}\right)+\frac{1}{R_2}\right)g\delta h\end{array}$.... (VIII).

Here, the area of the tank is $A$, the pump resistance is $R_1$, the resistance of outlet pipe is $R_2$, the slope is $\frac{1}{r}$, the acceleration due to gravity is $g$, the time interval is $t$ and the deviation in height is $\delta h$.

Substitute $\frac{R_1+R_2}{R_2}\rho g\delta h$ for $\delta \left(\Delta p\right)$ in Equation (IV).

$\delta q_{m1}=\left(-\frac{1}{r}\right)\left(\frac{R_1+R_2}{R_2}\rho g\delta h\right)$

Substitute $\left(-\frac{1}{r}\right)\left(\frac{R_1+R_2}{R_2}\rho g\delta h\right)$ for $\delta q_{m1}$ and $\frac{\rho g}{R_2}\delta h$ for $\delta q_{m2}$ in Equation (VI).

$\begin{array}{c}\rho A\frac{d}{dt}\delta h=\left(-\frac{1}{r}\right)\left(\frac{R_1+R_2}{R_2}\rho g\delta h\right)-\frac{\rho g}{R_2}\delta h\\ A\frac{d}{dt}\delta h=-\left(\frac{1}{r}\left(\frac{R_1+R_2}{R_2}\right)+\frac{1}{R_2}\right)g\delta h\end{array}$

Substitute $\left(400\text{N}\text{.s}/\left(\text{kg}{\text{.m}}^{\text{2}}\right)\right)$ for $R_2$, $\left(600\text{N}\text{.s}/\left(\text{kg}{\text{.m}}^{\text{2}}\right)\right)$ for $R_1$, $700$ for $r$ and $9.81\text{m}/{\text{sec}}^{\text{2}}$ for $g$ in Equation (VIII).

$\begin{array}{c}A\frac{d}{dt}\delta h=\left[\begin{array}{l}-\left(\frac{1}{700}\right)\left(\frac{\begin{array}{l}\left(600\text{N}\text{.s}/\left(\text{kg}{\text{.m}}^{\text{2}}\right)\right)\\ +\left(400\text{N}\text{.s}/\left(\text{kg}{\text{.m}}^{\text{2}}\right)\right)\end{array}}{\left(400\text{N}\text{.s}/\left(\text{kg}{\text{.m}}^{\text{2}}\right)\right)}\right)\\ +\left(\frac{1}{\left(400\text{N}\text{.s}/\left(\text{kg}{\text{.m}}^{\text{2}}\right)\right)}\right)\end{array}\right]\times \left(9.81\text{m}/{\text{sec}}^{\text{2}}\right)\times \delta h\\ =-\left(0.00357+0.0025\right)\times \left(9.81\text{m}/{\text{sec}}^{\text{2}}\right)\times \delta h\\ =-0.595\times \delta h\end{array}$

Thus, the linearized dynamic model of the height deviation in the tank is $A\frac{d}{dt}\delta h=-0.0595\times \delta h$.