Chapter 2, Problem 2.1CP

To determine

(a)

The expression for gage pressure in the reservoir.

Answer to Problem 2.1CP

If the movement of the surface of the reservoir is considered, the expression for $p_{1gage}$ is ${\rho }_{m}gh\left(1+\frac{d^2}{D^2}h\right)$.

Explanation of Solution

Given:

Concept Used:

Force acting on the panel is given by

$F=p.A$

Where, $p$ is the pressure in N/m² and $A$ is the area in m².

Pressure $p$ is given by

$p=\rho gh$

Where, $\rho$ is the density in kg/m³ and $h$ is the height in m.

Calculation:

Let, $d$ and $D$ be the diameter of open tube and the reservoir respectively.

The volume of the liquid raised in the manometer open tube is equal to the volume of liquid lowered down in the reservoir.

$\begin{array}{l}\frac{\pi }{4}{d}^{2}h=\frac{\pi }{4}{D}^{2}H\\ H=\frac{d^2}{D^2}h\end{array}$

Neglecting the effect of surface tension and the density of air, the pressure difference can be written as

$p_{1gage}={\rho }_{m}gh+{\rho }_{m}gH$

Substituting the value of $H$, we get

$p_{1gage}={\rho }_{m}gh\left(1+\frac{d^2}{D^2}h\right)$

Thus, the expression for $p_{1gage}$ is ${\rho }_{m}gh\left(1+\frac{d^2}{D^2}h\right)$.

Conclusion:

The expression for $p_{1gage}$ is ${\rho }_{m}gh\left(1+\frac{d^2}{D^2}h\right)$.

To determine

(b)

The expression for the gage pressure in the reservoir if the movement of the surface of the reservoir is neglected.

Answer to Problem 2.1CP

If the movement of the surface of the reservoir is neglected, the expression for $p_{1gage}$ is ${\rho }_{m}gh$.

Explanation of Solution

Given:

Concept Used:

Force acting on the panel is given by

$F=p.A$

Where, $p$ is the pressure in N/m² and $A$ is the area in m².

Pressure $p$ is given by

$p=\rho gh$

Where, $\rho$ is the density in kg/m³ and $h$ is the height in m.

Calculation:

Let, $d$ and $D$ be the diameter of open tube and the reservoir respectively.

The volume of the liquid raised in the manometer open tube is equal to the volume of liquid lowered down in the reservoir.

$\begin{array}{l}\frac{\pi }{4}{d}^{2}h=\frac{\pi }{4}{D}^{2}H\\ H=\frac{d^2}{D^2}h\end{array}$

Neglecting the effect of surface tension and the density of air, the pressure difference can be written as

$p_{1gage}={\rho }_{m}gh+{\rho }_{m}gH$

Substituting the value of $H$, we get

$p_{1gage}={\rho }_{m}gh\left(1+\frac{d^2}{D^2}h\right)$

If the movement of the surface of the reservoir is neglected, $H=0$.

Thus, the expression for $p_{1gage}$ is ${\rho }_{m}gh$.

Conclusion:

The expression for $p_{1gage}$ is ${\rho }_{m}gh$.

To determine

(c)

The ratio of $\frac{d}{D}$ to keep the error of the result in part (b) and part (a) within $1\%$ and $0.1\%$.

Answer to Problem 2.1CP

The ratio of $\frac{d}{D}$ for the error to be less than or equal to $1\%$ and $0.1\%$ is $9.95$ and $31.6$ respectively.

Explanation of Solution

Given:

Concept Used:

Force acting on the panel is given by

$F=p.A$

Where, $p$ is the pressure in N/m² and $A$ is the area in m².

Pressure $p$ is given by

$p=\rho gh$

Where, $\rho$ is the density in kg/m³ and $h$ is the height in m.

Calculation:

Let, $d$ and $D$ be the diameter of open tube and the reservoir respectively.

The volume of the liquid raised in the manometer open tube is equal to the volume of liquid lowered down in the reservoir.

$\begin{array}{l}\frac{\pi }{4}{d}^{2}h=\frac{\pi }{4}{D}^{2}H\\ H=\frac{d^2}{D^2}h\end{array}$

Neglecting the effect of surface tension and the density of air, the pressure difference can be written as

$p_{1gage}={\rho }_{m}gh+{\rho }_{m}gH$

Substituting the value of $H$, we get

$p_{1gage}={\rho }_{m}gh\left(1+\frac{d^2}{D^2}h\right)$

In fact, the above expression is

${\left(\Delta p\right)}_{actual}=p_1-p_a=p_{1gage}={\rho }_{m}gh\left(1+\frac{d^2}{D^2}h\right)$

If the movement of the surface of the reservoir is neglected, $H=0$, the above expression becomes

${\left(\Delta p\right)}_{approx}=p_1-p_a=p_{1gage}={\rho }_{m}gh$

It is clear from the above tow expressions, the values of ${\rho }_m, g$ and $h$ does not affect the % error. The % error is determined only by the ratio of $\frac{d}{D}$.

The percentage error is given by

$\left(\%\right)E=\frac{{\left(\Delta p\right)}_{actual}-{\left(\Delta p\right)}_{approx}}{{\left(\Delta p\right)}_{actual}}$

$\left(\%\right)E=\frac{\frac{d^2}{D^2}}{1+\frac{d^2}{D^2}}$

$\frac{d}{D}=\sqrt{\frac{1-E}{E}}$

If the error is less than or equal to $1\%$ :

$\frac{d}{D}=\sqrt{\frac{1-0.01}{0.01}}\ge 9.9498\approx 9.95$

If the error is less than or equal to $0.1\%$ :

$\frac{d}{D}=\sqrt{\frac{1-0.001}{0.001}}\ge 31.6069\approx 31.6$

Conclusion:

The ratio of $\frac{d}{D}$ for the error to be less than or equal to $1\%$ and $0.1\%$ is $9.95$ and $31.6$ respectively.