Question

The absolute pressure in water at a depth of 5m is read to be 145 kPa. Determine (a) the local atmospheric pressure, and (b) the absolute pressure at a depth of 5 m in a liquid whose specific gravity if 0.85 at the same location. (density of water is 1000 kg/m3)

Answer 1

Answer:

a) 95950 pascals

b) 137642.5 pascals

Explanation:

The absolute pressure (Pabs) on a fluid is:

$P_{abs}=P_{gauge}+P_{atm}$ (1)

With Pgauge the pressure due depth on the fluid and Patm the atmospheric pressure. Pgauge is equal to:

$P_{gauge}=\rho gh$ (2)

with ρ the fluid density, g the gravitational acceleration and h the depth on the fluid. Using (2) on (1) and solving for Patm:

$P_{atm}=P_{abs}-P_{gauge}=P_{abs}-\rho_{water} gh$

$P_{atm}=(145000Pa)-(1000\frac{kg}{m^{3}})(9.81\frac{m}{s^{2}})(5m)$

$P_{atm}=95950Pa$

b) Here we're going to use again (1) but now we have another value of density because it's other liquid, to know that value we should use the fact that specific gravity (S.G) for liquids is the ratio between fluid density and water density:

$S.G=\frac{\rho_{fluid}}{\rho_{water}}$

$\rho_{liquid}=S.G*\rho_{water}$

$\rho_{liquid}=(0.85)*(1000\frac{kg}{m^{3}})=850\frac{kg}{m^{3}}$

so:

$P_{abs}=\rho_{liquid} gh+P_{atm}=(850\frac{kg}{m^{3}})(9.81\frac{m}{s})(5m)+95950Pa$

$P_{abs}=137642.5 Pa$