Impulse thrust calculations from pressure bottle

[NZ_Ben]

Hi,

Am wondering how to calculate the thrust and impulse thrust for gas escaping a pressure bottle - ie a scuba tank or the like. My maths is a lil shady so go easy on integration and such.

Like say if I wanted 1MJ of energy output - what size bottle at what pressure with what nozzle diameter would be required.

Thanks in advance
Ben

 

[thermodynamicaldude]

P = F/A;

Change in momentum = Impulse = Force * time

Change in energy = Work = (Force)(distance)(cos (theta))...since this a rocket..the force will most likely be in the same direction as the movement...so...cos(0) = 1...and you have...

Work = Force(distance).

Or, since you know the momentum, you know the velocity; you can find energy, since KE = (1/2) mv^2

For future reference, in case your force is not constant or something like that...you may need to use calculus...

...then... Fdt = dP...and F *(dot product) dr = dW
(note that r = distance)

Essentially, this means that to find Impulse, you have to integrate Fdt, and to find work, you would have to integrate the dot product of F and dr.

 

[Clausius2]

Hi,

Am wondering how to calculate the thrust and impulse thrust for gas escaping a pressure bottle - ie a scuba tank or the like. My maths is a lil shady so go easy on integration and such.

Like say if I wanted 1MJ of energy output - what size bottle at what pressure with what nozzle diameter would be required.

Thanks in advance
Ben

Very easy. You have to use the 2nd law of motion. But there is an special reformulation of that by means of the integral momentum equation of fluid dynamics:

$$\int_S \rho v v dS=-\int_S P dS$$

which means in first quasy-steady aproximation that the force produces with the flow of gas outwards is balanced with pressure forces:

If the hole of exhaust has an area of A, and the flow is exhausted at velocity U, and the external pressure is P_a and the internal one P_o:

the exhaust velocity can be estimated as:

$$U\approx \sqrt{\frac{P_o-P_a}{\rho_o}}$$

With that, you can make some figures. A more detail analysis will employ the unsteady process of exhausting, but the rocket equation can be employed by you to have a decent solution. I have made only a rude estimation.

 

[dextercioby]

Very easy. You have to use the 2nd law of motion. But there is an special reformulation of that by means of the integral momentum equation of fluid dynamics:

$$\int_S \rho v v dS=-\int_S P dS$$

With a small adaggio.It should be 2 around here
$$\int_S \frac{\rho v^2}{2} dS=-\int_S P dS$$

 

[Clausius2]

With a small adaggio.It should be 2 around here
$$\int_S \frac{\rho v^2}{2} dS=-\int_S P dS$$

Of course That's false. The complete momentum integral equation is:

$$\frac{d}{dt}\int_V \rho \overline {v} dV +\int_S \rho \overline {v} \overline {v}\cdot \overline{dS}=-\int_S P\overline{dS} + \int_S \tau \overline{dS} + \int_V \rho f_m dV$$

It is impossible I'm wrong here boy. I have used this equation a million of times. :!)

Take a look at some Fluid Mechanics book. What you posted have no sense.

 

[dextercioby]

Sorry,my mistake,the integral i posted appears in the enregy law.I mixed up equations.Damn,i was stupid...

 

[Clausius2]

Sorry,my mistake,the integral i posted appears in the enregy law.I mixed up equations.Damn,i was stupid...

It doesn't matter. If someone gives me one euro for each stupidity I have said here, I would be the richest guy on the world.