A new gel is being developed to use inside padding and helmets to cushion the body from impacts. The gel is stored in a 4.5 cubic meters [m cubed ]cylindrical tank with a diameter of 2 meters [m]. The tank is pressurized to 1.2 atmosphere [atm] of surface pressure to prevent evaporation. A total pressure probe located at the bottom of the tank reads 58 feet of water [ft H2O ]. What is specific gravity of the gel contained in the tank?

The specific gravity of the gel is 3.696.

Explanation:

Given

V = 4.5 m³

D = 2 m ⇒ R = D/2 = 2 m/2 = 1m

Psurface = 1.2 atm

First, we have to calculate the gel's column height using the cylinder's volume, as follows:

V = π*R²*h ⇒ h = V / (π*R²)

⇒ h = 4.5 m³ / (π * (1 m) ²) = 1.4324 m

Then, as the pressure given at the bottom of the tank is the sum of the surface pressure and the gel's column pressure, we need to calculate only the gel's column pressure.

1 ft of water is a unit of pressure, but we need to convert it to atm and then to Pa, in order to calculate our results in the correct units. Therefore, the conversion factor is:

1 ft of water (4°C) = 0.0295 atm

58 ft of water * (0.0295 atm / 1 ft of water) = 1.711 atm

Pbottom = Psurface + Pgel

Pgel = Pbottom - Psurface

⇒ Pgel = 1.711 atm - 1.2 atm = 0.511 atm

Then, we apply the conversion factor as follows

Pgel = 0.511 atm * (101325 Pa / 1 atm) = 51777.075 Pa

Now, to calculate the specific gravity, we need to find first the gel's density:

Pgel = ρ*g*h ⇒ ρgel = Pgel / (g*h)

⇒ ρgel = 51777.075 Pa / (9.81 m/s²*1.4324 m) = 3684.72 Kg/m³

SEgel = ρgel/ρH₂0

⇒ SEgel = 3684.72 Kg/m³ / 997 Kg/m³ = 3.696

The specific gravity of the gel is 3.696.