Question

Open tank has a three liquids with total pressure equals Qi to 200 K pa, and height of water equals to oil. Given that the height of tank is 10 meter and mercury has a 4 meters. Specific weight of water and mercury are 9810 N/m3 and 13600 N/m3 respectively 1 Water 3 Mercury Calculate the density of oil and the pressure at the top of liquid 3 (bottom of oil).

please answer

Answer 1

Given,

Total pressure at the bottom of the tank = 200 kPa

Height of the tank (H) = 10 m

Height of mercury (H_m) = 4 m

Height of water (H_w) = Height of oil (H_o)

Specific weight of water ($\gamma$_w) = 9810 N/m³

Specific weight of mercury ($\gamma$_w) = 13600 N/m³

Let us assume, Atmospheric pressure = 101.325 kPa

Therefore, Gauge pressure (P_gauge) = Total pressure - Atmospheric pressure

.(P_gauge) = 200 - 101.325 = 98.675 kPa = 98.675 * 1000 N/m² [$\because$ 1 Pa = 1 N/m² ]

As the total height of the tank is 10 m,

H = (H_m) + (H_w) + (H_o)

$\Rightarrow$ 10 = 4 + (H_w) + (H_w) [ $\because$ (H_w) = (H_o)]

$\Rightarrow$ (H_w) = 3 m

$\therefore$ (H_o) = 3 m

We know that,

Hydrostatic pressure (P) is given as, P = $\gamma$ * Z ............................................................(1)

Where, $\gamma$ = Specific weight of fluid

Z = Depth of fluid

In the given problem as the fluids in the tank are static, hydrostatic pressure formula can be applied.

Therefore total gauge pressure at the bottom of the tank is calculated as follows:

P_gauge = ($\gamma$ * Z)_mercury + ($\gamma$ * Z)_water + ($\gamma$ * Z)_oil  

$\Rightarrow$ 98.675 * 1000 = 13600 * 4 + 9810 * 3 + $\gamma$_oil * 3

$\Rightarrow$$\gamma$_oil = 4948.33 N/m³

Therefore, density of oil is 4948.33 N/m³ = 4948.33/ 9.81 = 504.42 kg/m³ [ $\because$ 1 kg = 9.81 N]

Also,

Pressure at the bottom of the oil layer = ($\gamma$ * Z)_water + ($\gamma$ * Z)_oil

= 9810 * 3 + 4948.33 * 3

= 44275 N/m²

= 44.275 kPa

Summery:

Density of liquid = 4948.33 N/m³

Pressure at the bottom of the oil layer = 44.275 kPa