91Ó°ÊÓ

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.

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

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

$p_1+\frac{1}{2}\mathrm{ÒÏ}{v}_1^2+\mathrm{ÒÏ}g{y}_1=p_2+\frac{1}{2}\mathrm{ÒÏ}{v}_2^2+\mathrm{ÒÏ}g{v}_2$

Here $\mathrm{ÒÏ}$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,

$p_2+\frac{1}{2}\mathrm{ÒÏ}{v}_2^2+\mathrm{ÒÏ}g{v}_2=p_1+\frac{1}{2}\mathrm{ÒÏ}{v}_1^2+\mathrm{ÒÏ}g{y}_1$

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

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

Substitute $200\mathrm{kPa}$for $p_1$,$900\mathrm{kg}/{\mathrm{m}}^3$for $\mathrm{ÒÏ}$,$2.0\mathrm{m}/\mathrm{s}$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 $g$.

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

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

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

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