Question

What does the top pressure gauge read in $FIGUREEX14.30?$

Short answer

The top pressure gauge reads$110\mathrm{kPa}$.

Step-by-step solution

Definition

Use Bernoulli's equation to solve for the reading in the top pressure gauge. Thus, it states that the total energy of the oil is constant at every part of the tube.

Figure for movement of oil 

The figure which shows the movement of oil through the tube is given below.

Explanation

The total energy of the oil is equal to the sum of its pressure energy, kinetic energy and potential energy,

$p_1+\frac{1}{2}\rho v_1^2+\rho g{y}_1=p_2+\frac{1}{2}\rho v_2^2+\rho g{v}_2$

Here $\rho$is the density of the oil, $p_1$and $p_2$are the pressures at two points, $v_1$and $v_2$are the speeds, $y_1$and $y_2$are the heights of these two points respectively.

Rearrange this for the pressure of the oil at the second point,

$p_2+\frac{1}{2}\rho v_2^2+\rho g{v}_2=p_1+\frac{1}{2}\rho v_1^2+\rho g{y}_1$

$p_2=p_1+\frac{1}{2}\rho v_1^2+\rho g{v}_1-\left(p_2+\frac{1}{2}\rho v_2^2+\rho g{v}_2\right)$

$=p_1+\frac{1}{2}\rho \left(v_1^2-v_2^2\right)+\rho g\left(y_1-y_2\right)$

Calculation of pressure

Substitute $200\mathrm{kPa}$for $p_1$,$900\mathrm{kg}/{\mathrm{m}}^3$for $\rho$,$2.0\mathrm{m}/\mathrm{s}$for $v_1$,$3.0\mathrm{m}/\mathrm{s}$for $v_2$,$0\mathrm{m}$for $y_1$. $10\mathrm{m}$for $y_2$and $9.8\mathrm{m}/{\mathrm{s}}^2$$9.8\mathrm{m}/{\mathrm{s}}^2$for $g$.

$p_2=\left[\begin{array}{c}200\mathrm{kPa}+\frac{1}{2}\left(900\mathrm{kg}/{\mathrm{m}}^3\right)\left((2.0\mathrm{m}/\mathrm{s}{)}^2-(3.0\mathrm{m}/\mathrm{s}{)}^2\right)\\ +\left(900\mathrm{kg}/{\mathrm{m}}^3\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)(0\mathrm{m}-10\mathrm{m})\end{array}\right]$

$=\left[\begin{array}{c}200\mathrm{kPa}+\frac{1}{2}\left(900\mathrm{kg}/{\mathrm{m}}^3\right)\left(-5.0{\mathrm{m}}^2/{\mathrm{s}}^2\right)\\ +\left(900\mathrm{kg}/{\mathrm{m}}^3\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)(-10\mathrm{m})\end{array}\right]$

$=200\mathrm{kPa}-2250\mathrm{Pa}-88,200\mathrm{Pa}$

$=200\mathrm{kPa}-90.45\mathrm{kPa}$

role="math" localid="1648184153162" $p_2=110\mathrm{kPa}$