Question

A new gel is being developed to use inside padding and helmets to cushion the body from impacts. The gel is stored in a 4.1 m^3 cylindrical tank with a diameter of 2 m. The tank is pressurized to 1.3 atm of surface pressure to prevent evaporation. A total pressure probe located at the bottom of the tank reads 60 ft of water. What is the specific gravity of the gel contained in the tank?

Answer 1

Answer:

$SE_{gel} = 3.75$

Explanation:

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

$V=\pi\times r \times h\\h=\frac{V}{\pi \times r}\\h=\frac{4.1m^3}{\pi \times 1m}= 1.30 m$

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:

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

$60 ft water \times \frac{0.0295 atm}{1 ft water}= 1.77 atm\\P_{bottom}=P_{surface}+P_{gel}\\P_{gel}=P_{bottom}-P_{surface}=1.77 atm - 1.3 atm\\P_{gel}= 0.47 atm\times \frac{101325Pa}{1 atm}=47622.75 Pa$

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

$P_{gel} = \rho gh\\\rho = \frac{P_{gel}}{gh}=\frac{47622.75 Pa}{9.8 m/s^2 \times 1.30m}= 3738.04 \frac{kg}{m^3}$

$SE_{gel} = \frac{\rho_{gel}}{\rho_{water}}= \frac{3738.04 kg/m^3}{997 kg/m^3} = 3.75\\SE_{gel} =3.75$

The specific gravity of the gel is 3.75.