Zahar, I know you'll find this very funny too..
I tried to simplify it further using simplified Bernoulli's equation
p1 + 1/2 * q * v1^2 = p2 + 1/2 * q * v2^2
Because in this case
v2 > v1 -> v1^2 << v2^2;
p2 < p1
We can write even more simple
p1 = 1/2 * q * v2^2
p1 - pressure of water in muzzle in front of the piston ?
v1 - speed of the piston/water being pushed by the piston = 30 m/s
p2 - pressure in space between shock absorber ond the tang of the spear p2 < p1
v2 - the speed of water leaving the muzzle = 207,27 m/s
q - (1020 - 1030) kg/m3 (for sea water)
A1 - cross section of the barrel (piston)
A2 - cross section of gap where water can exit the muzzle forward
v2 = A1/A2*v1
p1 (Pas) = 1/2 * 1020 kg/m3 * 207.27^2 m2/s2 = 21910035 Pas = 219 bar
So the pressure of water in front of the piston would be at least 219 bar (in case the mass of the piston would be 0 g.
Becauce the mass of the piston is about 10 g, the pressure of water p1 in front of the piston would be even higher...?
I tried to take into account the mass of the piston as well.The piston I was considering had a mass of 11.7 g and a volume of 4.7 cm3.The density of the piston is 11.7/4.7 = 2.48 g/cm3. The amount of water in front of the piston could be about 1 cm3, which is 1 ml. If I imagine the piston as a liquid of the same mass and volume and add the mass and volume of the water in front of the piston, then the total mass is 11.7 + 1 = 12.7 g and the volume is 4.7 + 1 = 5.7 cm3. It is as if I were applying the formula for pressure p1 to a liquid whose density is q = 12.7/5.7 = 2.23 g/cm3 = 2230 kg/m3. Now we can calculate the new value of p1:
p1 (Pas) = 1/2 * 2230 kg/m3 * 207.27^2 m2/s2 = 47901351 Pas = 479 bar
Regardless of which result is more accurate (219 bar or 479 bar) the fact is that a pulse of enormous pressure will be created in front of the piston...
Mass of the piston:
View attachment 60660
This is to determine the volume of the piston.
View attachment 60662
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