- 19 Mar '15 14:42 / 1 editHello All,

Out of shear curiosity I'm looking for some help on validating a model equation for the interaction of the fuel(compressed air)/propellant(water) for the rocket in question. I'm trying to obtain the thrust force as a function of time, not the rockets position in time. I am also unsure of certain assumptions that I made, and am more importantly looking for clarification on how to properly apply the mechanics.

If I model the soda bottle as a cylinder with constant cross-sectional area; ignoring the bottle neck (truncated nozzle) with respect to fuel expansion (technically this would be a two stage rocket, but I'm only concerned with the first stage), and that the nozzle has appreciable volume when compared to the volume of the propellant (water).

Coordinate system:

Picture a rocket in launch position (nozzle capped) partially filled with water. Directly above the water is a cavity of compressed air.

The single coordinate will be**x**↓+, and it will be zero at the gas/water interface.**The rocket is held in place, and will be kept from accelerating (keeping the frame inertial)**

I'm only concerned with the forces acting internally (not the external holding force). I'm not sure if this is/isn't valid if I neglect that external "holding force" in the subsequent analysis.

Newtons Second Law:

Σ**F**= d(m**v**)/dt

Σ**F**= F_p + W

Where "F_p" is the force from pressure, and "W" is the weight of the propellant. They are both functions of**x**( and time)

F_p = P*A

Assuming the compressed air behaves as an Ideal Gas, expands adiabatically, and k = 1.4

Po*Vo^1.4 = P*V^1.4

P = Po*(Vo/V)^1.4

Assume:

Vo = A*Lo

V = A*(Lo+**x**)

Substitute into "P"

P = Po*(Lo/(Lo+**x**))^1.4

F_p = P*A = A*Po*(Lo/(Lo+**x**))^1.4

Moving on to "W"

m(**x**) = mo - µ*A***x**

W = (mo - µ*A***x**)*g

g = acceleration from gravity

µ = mass density of water

That takes care of the right side.

LHS:

d(m***v**)/dt = m*d(**v**)/dt +**v_e***dm/dt

The first term on the right represents the "Force" associated with the accelerated mass (water) inside the bottle (neglecting mass from air), and the second term is the "Thrust force" from ejecting mass.

Looking at the first term on the right:

m*d(**v**)/dt

The instantaneous mass, m(x) =mo - µ*A***x**

Acceleration of that mass (from gas expansion) d(**v**)/dt = d²**x**/dt²

m*d(**v**)/dt = (mo - µ*A***x**)*d²**x**/dt²

The "Thrust" (second term on the right)

with v_e = the velocity at the exit of the nozzel, from continuity

A_e***v_e**= A***v**

**v_e**= A/A_e***v**

**v_e**= A/A_e*d**x**/dt

Then dm/dt is equal to the displaced mass from compressed air expansion

dm/dt = µ*A***v**

dm/dt = µ*A*d**x**/dt

**v_e***dm/dt = µ*A²/A_e*(d**x**/dt)²

Then If everything I did was correct? Solve:

A*Po*(Lo/(Lo+**x**))^1.4 + (mo - µ*A***x**)*g = (mo - µ*A***x**)*d²**x**/dt² + µ*A²/A_e*(d**x**/dt)²

Like I said, mostly interested in the correct application of the mechanics at this point? - 20 Mar '15 15:55Before I look at your maths I want to check I've understood your model. You have a cylinder of water of cross-sectional area A. Water comes out of an aperture of cross-sectional area A' where A' ≤ A. As a simplifying assumption you have A = A'? I don't think the last is necessary, but we'll leave that for now. The gas pushes the water out of the cylinder and so you have a cylinder of gas at pressure P(t) above a cylinder of water which is forced out of the aperture. You want to know the thrust as a function of time, until the water is exhausted?

Are you assuming that there is no gas dissolved in the water? - 20 Mar '15 16:03 / 4 edits

A =/= A'.*Originally posted by DeepThought***Before I look at your maths I want to check I've understood your model. You have a cylinder of water of cross-sectional area A. Water comes out of an aperture of cross-sectional area A' where A' ≤ A. As a simplifying assumption you have A = A'? I don't think the last is necessary, but we'll leave that for now. The gas pushes the water out of the cyl ...[text shortened]... , until the water is exhausted?**

Are you assuming that there is no gas dissolved in the water?

A' = A_e. and,

A_e<A

I was neglecting the change from A→A_e as far as the final moments of the first stage are concerned, and the mass contained in the nozzle at any time, and neglecting dissolved gas(hadn't thought of it, not sure if its good or bad assumption, but it will certainly add unwanted complexity). Other than that your interpretation is correct.

EDIT: I found an error (I think). I need to change the final result, by making the last term on the right negative, since dm/dt is inherently negative.

A*Po*(Lo/(Lo+x))^1.4 + (mo - µ*A*x)*g = (mo - µ*A*x)*d²x/dt² - µ*A²/A_e*(dx/dt)² - 20 Mar '15 21:53

I'm slightly confused about what you've done. I think you've got the velocity of the water jet confounded with the velocity of the rocket. You also seem to have neglected external air pressure.*Originally posted by joe shmo***A =/= A'.**

A' = A_e. and,

A_e<A

I was neglecting the change from A→A_e as far as the final moments of the first stage are concerned, and the mass contained in the nozzle at any time, and neglecting dissolved gas(hadn't thought of it, not sure if its good or bad assumption, but it will certainly add unwanted complexity). Other than that your interpre ...[text shortened]... ly negative.

A*Po*(Lo/(Lo+x))^1.4 + (mo - µ*A*x)*g = (mo - µ*A*x)*d²x/dt² - µ*A²/A_e*(dx/dt)²

constant = PV^γ

So P∝V^-γ

Suppose the initial volume of the gas = V(0) and when the water runs out it is V(T) = V, so I have volume as a function of time t. Suppose there is an internal partition and we can ignore gravity within the water. Then before the partition is released the pressure in the water is 1 atmosphere. When the partition is released the pressure of the water and gas are the same. The force on the water coming out of the nozzle then has two contributions one from the gas pressure the other competing one from external air pressure.

P(0)V(0)^γ = P(t)V(t)^γ

P(t) = P(0) V(0)^γ V(t)^-γ

F_gas_out(t) = P(0) V(0)^γ A_nozzle V(t)^-γ

F_gas_out(t) = k V(t)^-γ

k = P(0) V(0)^γ A_nozzle

And the overall force taking external air pressure into account is:

From now:

A = A_nozzle

P = P_a = air pressure

F = kV(t)^-γ - P A

Where F is the net thrust. We need to work out how that changes with time.

F = dp/dt = vdm/dt = ρv² A

v is the velocity of the water relative to the*instantaneous*rest frame of the rocket.

ρv(t)² = [k/A]V(t)^-γ - P

Now, v(t) = d(V(t)/A)/dt = (dV(t)/dt)/A = V'(t)/A, the velocity of the outgoing water jet is the rate of change of volume divided by the area of the*nozzle*:

x' ≝ dx/dt

so we get:

V'(t) = A √[([k/ρA]V(t)^-γ ) - P/ρ]

V(t)

∫dV /√[([k/ρA]V(t)^-γ ) - P/ρ] = At

V(0)

if 0 ≤ t ≤ T

This is a painful looking integral. It may be a standard integral, if not it looks harsh and may be worth doing numerically. Once it's done we get the thrust as a function of time. Then you can use this to find the acceleration of the rocket, you seemed to be combining this all into one thing. - 22 Mar '15 22:23 / 2 edits
*Originally posted by DeepThought***I'm slightly confused about what you've done. I think you've got the velocity of the water jet confounded with the velocity of the rocket. You also seem to have neglected external air pressure.**

constant = PV^γ

So P∝V^-γ

Suppose the initial volume of the gas = V(0) and when the water runs out it is V(T) = V, so I have volume as a function of ti ...[text shortened]... this to find the acceleration of the rocket, you seemed to be combining this all into one thing.I'm slightly confused about what you've done. I think you've got the velocity of the water jet confounded with the velocity of the rocket.

I'm taking it from this statement and analysis that you believe:

ΣF = m*dv/dt + v*dm/dt

m*dv/dt = 0 (the rocket is stationary)

ΣF = v*dm/dt

look at it in a different way (or I'm wrong); what if control volume was fixed inside the cylinder around the gas/propellant system?

Using the Momentum Equation:

Σ**F**= d/dt∫**v***ρ*(A*dx) + ∫**v***ρ***V**·d**A**

In words:

[sum of forces acting on the matter in the control volume] = [time rate of change of momentum in control volume] + [net outflow rate of momentum through control surface]

Your analysis is neglecting the first term on the RHS, by saying that the momentum in the control volume is constant in time.

I'm saying that it is a non-zero function of the coordinate**x**that I described, and the "thrust" force ( which comes from the simplification of the second term on the right) is also a function of**x**. Consequently the only way to determine the "thrust" force as a function of time is to solve the DE for x(t) and the differentiate.

Also as you noted, I was taking P_a = 0 gauge, so I ignored it.

EDIT: Btw, how do you bring in ∫, ρ into this format (I copied yours). I use alt-codes but they are somewhat limited for symbols - 23 Mar '15 05:29

So you have the rocket clamped in place and not accelerating?*Originally posted by joe shmo*I'm slightly confused about what you've done. I think you've got the velocity of the water jet confounded with the velocity of the rocket.

I'm taking it from this statement and analysis that you believe:

ΣF = m*dv/dt + v*dm/dt

m*dv/dt = 0 (the rocket is stationary)

ΣF = v*dm/dt

look at it in a different way (or I'm wrong); wha ...[text shortened]... ρ into this format (I copied yours). I use alt-codes but they are somewhat limited for symbols

I'm wary of neglecting air pressure, the pressure of the gas will at most be a few atmospheres and the formula C = PV^γ, uses the absolute pressure of the gas, you can't subtract anything off it.

You shouldn't have that mdv/dt term, roughly speaking the water comes out of the nozzle at a constant velocity, all the acceleration has happened inside your control volume. Besides it's worth neglecting anyway to avoid the resultant non-linear differential equation.

As an alternative approach to the problem let's think about it from an energy point of view:

The Work done on an element of fluid as it is emitted is PdV

P = pressure

dV = volume of element of fluid

Power = P dV/dt = PAv

v = nozzle velocity.

The Work done of the volume of water emitted = ½ mv² = ½ ρ v² dV

Power supplied to fluid = ½ ρ v² dV/dt = ½ ρ A v³

Equating them we have:

½ ρ A v³ = PAv

v² = 2P/ ρ

P = (P_gas - P_atmos) = kV^-γ - P_a

dV/dt = A √[2(kV^-γ - P_a)/ρ]

Which, apart from a factor of two which I can't account for, is the same as the formula I got earlier. I'm worried about the factor of two, as it rather clearly indicates a mistake in one or other set of working. I'll work out where it came from tomorrow.

The symbols are in unicode. To enter unicode on a Linux machine press Ctrl + shift + u, you'll see an underlined u, release the keys and then type in the hexadecimal code, and then press space. So for □ you need u25a1. For Windows machines you need to look at [1]. There's a search function and list of maths symbols in [2].

[1] http://www.fileformat.info/tip/microsoft/enter_unicode.htm

[2] http://www.fileformat.info/info/unicode/block/mathematical_operators/images.htm - 24 Mar '15 01:15
*Originally posted by DeepThought***So you have the rocket clamped in place and not accelerating?**

I'm wary of neglecting air pressure, the pressure of the gas will at most be a few atmospheres and the formula C = PV^γ, uses the absolute pressure of the gas, you can't subtract anything off it.

You shouldn't have that mdv/dt term, roughly speaking the water comes out of the nozzle at a ...[text shortened]... unicode.htm

[2] http://www.fileformat.info/info/unicode/block/mathematical_operators/images.htmWhich, apart from a factor of two which I can't account for, is the same as the formula I got earlier. I'm worried about the factor of two, as it rather clearly indicates a mistake in one or other set of working. I'll work out where it came from tomorrow.

I'm not exactly sure what has happened, but at this point:

v² = 2P/ ρ

You have effectively developed a pre-reduced version of the Bernoulli Equation Along a Streamline, which is valid given your assumption below:

roughly speaking the water comes out of the nozzle at a constant velocity, all the acceleration has happened inside your control volume

However, when you vary the pressure the flow is not steady (dv/dt =/= 0)

I may be missing something, but it seems like a contradiction? - 24 Mar '15 02:08
*Originally posted by DeepThought***So you have the rocket clamped in place and not accelerating?**

I'm wary of neglecting air pressure, the pressure of the gas will at most be a few atmospheres and the formula C = PV^γ, uses the absolute pressure of the gas, you can't subtract anything off it.

You shouldn't have that mdv/dt term, roughly speaking the water comes out of the nozzle at a ...[text shortened]... unicode.htm

[2] http://www.fileformat.info/info/unicode/block/mathematical_operators/images.htmThe Work done of the volume of water emitted = ½ mv² = ½ ρ v² dV

Power supplied to fluid = ½ ρ v² dV/dt = ½ ρ A v³

I think I see the problem...You used kinetic energy for a constant mass.

The power supplied to the fluid = F·v = v*dm/dt·v = ρ A v³

ρ A v³ = PAv

v² = P/ ρ - 24 Mar '15 02:21
*Originally posted by DeepThought***So you have the rocket clamped in place and not accelerating?**

I'm wary of neglecting air pressure, the pressure of the gas will at most be a few atmospheres and the formula C = PV^γ, uses the absolute pressure of the gas, you can't subtract anything off it.

You shouldn't have that mdv/dt term, roughly speaking the water comes out of the nozzle at a ...[text shortened]... unicode.htm

[2] http://www.fileformat.info/info/unicode/block/mathematical_operators/images.htmI'm wary of neglecting air pressure, the pressure of the gas will at most be a few atmospheres and the formula C = PV^γ, uses the absolute pressure of the gas, you can't subtract anything off it.

I read in my fluids text that external pressure on a jet can be neglected if the flow is subsonic. Perhaps that was a bad interpretation under these particular circumstances.

The symbols are in unicode. To enter unicode on a Linux machine press Ctrl + shift + u, you'll see an underlined u, release the keys and then type in the hexadecimal code, and then press space. So for □ you need u25a1. For Windows machines you need to look at [1]. There's a search function and list of maths symbols in [2].

Thanks for that, I'll give them a read. - 24 Mar '15 05:31

The mass in the fluid element doesn't change. A fluid element is also known as a fluid parcel, see [1] below. So I'm still confused about the discrepancy. The energy version considering the final energy of the fluid element and the energy supplied to it gives a variant of Bernoulli's equation.*Originally posted by joe shmo*I'm wary of neglecting air pressure, the pressure of the gas will at most be a few atmospheres and the formula C = PV^γ, uses the absolute pressure of the gas, you can't subtract anything off it.

I read in my fluids text that external pressure on a jet can be neglected if the flow is subsonic. Perhaps that was a bad interpretation under t ...[text shortened]... ch function and list of maths symbols in [2].[/quote]

Thanks for that, I'll give them a read.

Problem is my fluid mechanics lectures were 23 years ago and they insisted on putting them at 9am and presented this really bitty course with 4 different lecturers giving about 5 lectures each so I tended to lie in instead and did a bit of question spotting for the exam. So I'm having problems working out what the assumptions should be. Since both derivations are pretty convincing to me.

There's a fairly standard physics problem along the lines of a container of water has a hole drilled in the side near the bottom, the distance between the top of the water and the hole is h, what rate does the water flow out at. Apart from the driver in this case being the pressure in the water from gravity, rather than a gas, it's basically the same problem. I'm wondering if it's in your fluids book as that'll tell you which version of the result is needed.

[1] http://en.wikipedia.org/wiki/Fluid_parcel - 25 Mar '15 01:38 / 1 edit

Ok, I suppose my conclusion about the KE was wrong. I found the problem in question, with the pressurized tank!*Originally posted by DeepThought***The mass in the fluid element doesn't change. A fluid element is also known as a fluid parcel, see [1] below. So I'm still confused about the discrepancy. The energy version considering the final energy of the fluid element and the energy supplied to it gives a variant of Bernoulli's equation.**

Problem is my fluid mechanics lectures were 23 years ag ...[text shortened]... tell you which version of the result is needed.

[1] http://en.wikipedia.org/wiki/Fluid_parcel

They give the exit velocity explicitly in the problem statement.

v = √( 2gh + 2P/ρ )

Which is certainly derived from Bernoulli's Equation; and exactly what you have in your energy approach if pressure contribution from the fluids own weight is neglected.

Also, the pressure is allowed to vary as well, and it is also explicitly given:

(P_o + P_atm)(L-h_o)/(L-h) - P_atm

I see that it is the ideal gas law (γ=1), and absolutes are needed from the relationship.

The question remains why your original approach missing that factor of 2? - 25 Mar '15 03:36

In the second calculation I was considering a fluid element properly. In the first calculation I was considering some fixed volume of the exit nozzle which had fluid flowing through it and assumed that the water had a fixed velocity there (for a small enough length of pipe this is true), but the water has to accelerate at some point. I now think you were right to be using F = ma rather than F = v dm/dt, but there was something wrong with how you had it applied.*Originally posted by joe shmo***Ok, I suppose my conclusion about the KE was wrong. I found the problem in question, with the pressurized tank!**

They give the exit velocity explicitly in the problem statement.

v = √( 2gh + 2P/ρ )

Which is certainly derived from Bernoulli's Equation; and exactly what you have in your energy approach if pressure contribution from the fluids own wei ...[text shortened]... m the relationship.

The question remains why your original approach missing that factor of 2?I see that it is the ideal gas law (γ=1), and absolutes are needed from the relationship.

In their example they will be assuming isothermal expansion, so given that PV = NRT, if they have T constant then Boyle's law applies. If your set up is adiabatic then the temperature of the gas will change during the expansion and you need PV^γ. So in practice it depends on how rapidly the gas drives the water out of the bottle and how insulating the bottle is.