Would a 2-holed watergun shoot with reduced velocity?

[iamwearingpants]

I've been arguing with my dad about whether drilling a 2nd hole on our PVC water gun (which has 1 hole) would reduce the velocity of the water coming out.

He thinks that 2 holes will halve the velocity of the water.

I think that when putting the same energy into the gun, it will shoot the same as the 1-holed variant we have now because more energy will be put into the water moving, instead of being held up against the walls of the gun.

Think of it this way, a crowd of people is moving out of a theater. There are 100 They are going towards 1 door, they move at the same speed, but they're kinda stuck, and the people moving out of the room is 10 people/s
Now, a week later the managers of the theater make a second exit. The people travel at the same speed, with the same energy, there's still 100, but they're less stuck and now they move out of the room at 20 people/s

So, will the water leave at the same velocity with a 2-holed gun?

(BTW is this the appropriate forum? The Space math/physics forum is for space math and physics, and I don't think waterguns are space physics )

 

[RisingFury]

The pressure drives the water. Initially the water velocity will be the same as in a 1 holed gun, but the pressure will drop sooner.

You'd expand the same amount of energy, just a two holed gun would do it quicker.

 

[iamwearingpants]

The pressure drives the water. Initially the water velocity will be the same as in a 1 holed gun, but the pressure will drop sooner.

You'd expand the same amount of energy, just a two holed gun would do it quicker.

Ok, I can live with that, I'll go tell him to drill the 2nd hole.


EDIT: Hmmm...
After a quick test fire, the velocity is constant, and is entirely dependent on the force I put into it. Also, is there a formula for this?

 

[JEL]

Here's a water-'gun' with 2 holes in it... looks like fun

 

[Eccentrus]

I believe the formula would be the Bernoulli's one, just search for that in the wiki, I don't know how to put greek characters here, but the concept is of the debit and the velocity, and the surface area of the holes, so in principle, to get the water jets into the same velocity, you'll need twice the debit, so then, the water gun would empty twice as fast as in a single-holed water gun, much the same with what RF said.

 

[jedidia]

pressure is force over surface area. You double the surface area, you half the pressure, unless you're putting more force into it. As such, there are two possible outcomes: same force, same firing time, reduced exhaust velocity. or: double force, same velocity, reduced firing time.

It's a very similiar relation as thrust and ISP in a rocket drive...

 

[iamwearingpants]

pressure is force over surface area. You double the surface area, you half the pressure, unless you're putting more force into it. As such, there are two possible outcomes: same force, same firing time, reduced exhaust velocity. or: double force, same velocity, reduced firing time.

It's a very similiar relation as thrust and ISP in a rocket drive...

Ok. The surface area is going to be almost unchanged, because the 2-3 mm holes are 0.01% of the total surface area of the gun's interior.
What's confusing me is the quantity of the water being shot. Will this remain constant with 2 holes and the same pressure? Or will it stay the same with half the velocity?

 

[C3PO]

Ok. The surface area is going to be almost unchanged, because the 2-3 mm holes are 0.01% of the total surface area of the gun's interior.
What's confusing me is the quantity of the water being shot. Will this remain constant with 2 holes and the same pressure? Or will it stay the same with half the velocity?

The surface area of 2 holes should be 200% of 1 hole

 

[jedidia]

What's confusing me is the quantity of the water being shot. Will this remain constant with 2 holes and the same pressure? Or will it stay the same with half the velocity?

It's a bit more complicated, as we're not dealing with constant pressure. If the pressure would be constant (i.e. you have a pump continiously keeping pressure at the same level despite the significant increase of volume that happens as soon as you pull the trigger), the velocity would reduce in proportion with the increase of surface area, if I'm not mistaken, and the flowrate would increase by the same ratio.

However, in your case, pressure is constantly dropping once you pull the trigger, which means that both velocity and flowrate will decrease over time. drilling a second whole changes your initial values of flowrate and velocity, but since flowrate determines the speed at which the pressure will drop, it also means that your pressure will drop faster. The gun therefore wouldn't empty double as fast with two holes, since flowrate drops faster than if you had only one hole. The beam will just lose power a lot faster than with one hole (which is basically what rising fury said earlier).

 

[iamwearingpants]

It's a bit more complicated, as we're not dealing with constant pressure. If the pressure would be constant (i.e. you have a pump continiously keeping pressure at the same level despite the significant increase of volume that happens as soon as you pull the trigger), the velocity would reduce in proportion with the increase of surface area, if I'm not mistaken, and the flowrate would increase by the same ratio.

However, in your case, pressure is constantly dropping once you pull the trigger, which means that both velocity and flowrate will decrease over time. drilling a second whole changes your initial values of flowrate and velocity, but since flowrate determines the speed at which the pressure will drop, it also means that your pressure will drop faster. The gun therefore wouldn't empty double as fast with two holes, since flowrate drops faster than if you had only one hole. The beam will just lose power a lot faster than with one hole (which is basically what rising fury said earlier).

There is no trigger on my gun, the water is being pushed out directly, like a syringe type thing. The pressure is constant throughout the firing.

 

[jedidia]

If it's a syringe type thing, then that means that the ratio between the area of your holes and the surface area of the pressure vessel will go up, which in turn means that presure is not constant if the force pushing the piston doesn't increase, if I'm not mistaken. But it's been a while since I did this kind of calculations, so I might be wrong.

The effect in a water gun would be rather negligible though, I guess (I assume you're pushing the piston by hand, which means there's no constant aplication of force anyways).

The only effect you'll probably notice is that you'll empty your gun faster than before.

 

[HarvesteR]

Hmm, then again, if you're pushing the water out by hand, there is probably a maximum peak amount of pressure you can develop... If one hole made the gun too 'stiff', a second hole will have the effect of enlarging the (no pun intented) bottleneck, meaning you can drive the piston in faster, effectively making the gun more efficient overall... More volume delivered in a shorter time.

About range and muzzle velocity, I'd say you wouldn't really notice it, since it's hand pumped and all...

This would be more of a concern if you were using some pneumatic firing device, as in those big spud cannons... Then the laws of physics are easier to notice...

Cheers

 

[martins]

Since we can consider water to be incompressible, the water pressure will remain constant while you move the piston at constant speed, and the velocity of the water emerging is a linear function of the piston speed and the ratio of the piston cross section and the area of the hole. If you add a second hole, you have to push the piston at twice the speed in order to obtain the same water velocity.

 

[HAL9001]

I tzhink: yes it would, It would shoot half speeded...
but I'm not really shure.

 

[RisingFury]

pressure is force over surface area. You double the surface area, you half the pressure, unless you're putting more force into it. As such, there are two possible outcomes: same force, same firing time, reduced exhaust velocity. or: double force, same velocity, reduced firing time.

It's a very similiar relation as thrust and ISP in a rocket drive...

Partially incorrect...

If you have a pressurized tank and you poke two holes into it, the pressure at t=0 will be the same as in the tank with one hole. The difference is that in a one hole system the firing time will be longer and in a two hole system the mass flow will be greater.

Do this: Take two plastic bottles, at the bottom poke one hole into the first bottle and two holes into the other. Seal the holes with sticky tape and fill both bottles with the same amount of water. Then pull off the sticky tape on both bottles at the same time. You'll see that initially both bottles will shoot water equally far away (same velocity), but the 2 hole bottle will empty first. As the water level drops, the pressure will drop. That simulates a pressurized tank...

 

[iamwearingpants]

The 2nd gun was built, and with 3 holes. It shoots half as far as the new 1 holed variant we also built. Both guns were improved designs so the 3-holed 'shotgun' shoots just as far as the old 1-holed design.

:thankyou: for the help and explanations guys, Now I have a better understanding of water physics.

 

[JEL]

Consider squeezing the end of a water-hose for your garden. The more you squeeze it (making the hole smaller basically) the farther the water will usually spray.

Fun home-science at its best