Question

A pressure of elevation 4.8m on a tank containing oil read 34.7kpa.Another gauge at elevation 2.2m read 57 5kpa.Cumpute specific weight density and specific gravity of the liguid

Answer 1

Step-by-step explanation:

From the given information:

Pressure at the elevation 4.8 m = 34.7 kPa

Pressure at the elevation 2.2 m = 57.5 kPa

Recall that:

At the pressure of depth "h";

p = ρgh

where;

ρgh = 34.7 × 10 ³

ρ × 9.81 × h = 34.7 × 10³

ρh = 34.7 × 10³/9.81

ρh = 3.54 × 10³ ----- equation (1)

ρg(h + 4.8 - 2.2) = 57.5 × 10³

ρ × 9.81 × (h + 2.6) = 57.5 × 10³

ρ (h + 2.6) = 57.5 × 10³ / 9.81

ρ (h + 2.6) = 5.86 × 10³ ------ equation (2)

From equation (1) and (2);

The density of the oil ρ is determined to be = 0.8939 × 10³ kg/m³

≅ 893.9 kg/m³

The specific weight w = ρg

w = 0.8939 × 10³ × 9.81

w = 8769.159

w = 8.75 × 10³ kg/m²/s²

The specific gravity SG =
`(\rho _(oil))/(\rho_(water))`

=
`(0.8939 * 10^3)/(10^3)`

= 0.8939

≅ 0.894