Answer:
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 = [tex]\dfrac{\rho _{oil}}{\rho_{water}}[/tex]
= [tex]\dfrac{0.8939 \times 10^3}{10^3}[/tex]
= 0.8939
≅ 0.894
