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 4.24.2 cubic meters [m cubed ]m3 cylindrical tank with a diameter of 2 meters​ [m]. The tank is pressurized to 1.41.4 atmosphere​ [atm] of surface pressure to prevent evaporation. A total pressure probe located at the bottom of the tank reads 6565 feet of water​ [ft Upper H 2 Upper OH2O​]. What is specific gravity of the gel contained in the​ tank?

Answer 1

Answer:

The specific gravity of the gel is 1470.68

Explanation:

Given:

Volume $V = 4.24$ $m^{3}$

Diameter of tank $d = 2$ m

Pressure at surface $P _{o} = 1.41$ atm

Height $h = 6565$ ft $= \frac{6565}{3.281} = 2001$ m

Pressure at bottom is given by,

 $P_{bottom} = \rho _{w} g h$

Where $\rho _{w} = 1000 \frac{kg}{m^{3} }$, $g = 9.8\frac{m}{s^{2} }$

 $P_{bottom} = 1000 \times 9.8 \times 2001$

 $P_{bottom} = 19.6 \times 10^{6}$ $\frac{N}{m^{2} }$

Here volume is 4.24 $m^{3}$

   $\pi r^{2} h = 4.24$

 $h = \frac{4.24}{3.14} = 1.35$ m

From pascal equation,

$P_{bottom} = P_{o} + \rho _{gel} g h$

Find density of gel from above equation,

$19600000 = 142865.25 + \rho_{gel} \times 9.8 \times 1.35$

$\rho_{gel} = 1470682.89$ $\frac{kg}{m^{3} }$

So specific gravity is given by,

   ζ = $\frac{\rho _{gel} }{\rho_{w} }$

   ζ = $\frac{1470682.89}{1000}$

   ζ = $1470.68$

Therefore, the specific gravity of the gel is 1470.68