namenerk

2022-01-09

The gage pressure in a liquid at a depth of 3 m is read to be 42 kPa. Determine the gage pressure in the same liquid at a depth of 9 m.

Donald Cheek

Expert

Given that the gage pressure at depth of 3 m is 42 kPa. To determine the gage pressure at depth 9 m we can consider that the pressure of the first case ${P}_{1}$ is the gage pressure at depth of ${h}_{1}=3m$ and similarly we can define ${P}_{2}$ as the gage pressure at depth of ${h}_{2}=9m$.
Then we could obtain as:
(1)
(2)
Dividing equ. (2) by equ. (1)
$\frac{{P}_{2}}{{P}_{1}}=\frac{\rho Lg{h}_{2}}{\rho Lg{h}_{1}}$
$\therefore {P}_{2}={P}_{1}\frac{{h}_{2}}{{h}_{1}}$
$=42\left(kPa\right)\cdot \frac{9}{3}$
$=126kPa$

encolatgehu

Expert

Given,
Depth $1={h}_{1}=3m$
Gage pressure at depth $1={P}_{g1}=42kPa$
Depth $2={h}_{2}=9m$
Total pressure in liquid is given by,
$P={P}_{a}+\rho gh$
Where,
${P}_{a}=$ Atmoshpheric pressure
$\rho =$ Density of liquid
$g=$ Gravitational acceleration
$h=$ depth
Gage pressure is given by,
${P}_{g}=P-{P}_{a}=\rho gh$
Step 2
Gage pressure at depth 1 is given by,
${P}_{g1}=\rho g{h}_{1}\left(eq.i\right)$
Gage pressure at depth 2 is given by,
${P}_{g2}=\rho g{h}_{2}\left(eq.ii\right)$
Taking ration of equaion (i) and (ii),
$\therefore \frac{{P}_{g1}}{{P}_{g2}}=\frac{\rho g{h}_{1}}{\rho g{h}_{2}}$
$\therefore \frac{{P}_{gq}}{{P}_{g2}}=\frac{{h}_{1}}{{h}_{2}}$
$\therefore \frac{3}{9}$
$\therefore {P}_{g2}=42×\frac{9}{3}$
$\therefore {P}_{g2}=126kPa$
Gage pressure at depth of $9m=126kPa$