Originally Posted By: paul
F=[m*Ve]+[(Pt-Pa)*Ae]

do you see the (+) sign in the above formula?

Yes, and had you read what I wrote above you'd understand why its there - the force generated as a compressed gas moves from a region of higher pressure to lower pressure is equal to the sum of the force generated by the expansion of the pressurized gas against a nozzle (if there is one) PLUS the force exerted due to the pressure differential.

Strangely enough, when describing sums mathematically, we use '+' signs.

I know, its shocking - using the appropriate mathematical symbol to represent its concordant mathematical procedure.

Originally Posted By: paul
all you are calculating is pressure times the nozzle area.

Which described the force produced given the pressure differential and the area connecting the high-pressure region to the low-pressure one.

If you think that is wrong, certainly you can provide the mathematical evidence it is wrong.

Originally Posted By: paul
what if the fluid is methane do you think that you will get the same result in you formula if worked the way you say it can be worked?

Assuming it is under the same pressure and doesn't go through a phase transition - yes. The force generated due to a pressure differential is always the same for a given pressure difference and cross-sectional area of the opening.

Originally Posted By: paul

you cant leave this
Quote:
[m*Ve]+

out of this
Quote:
F=[m*Ve]+[(Pt-Pa)*Ae]

or all you have is this
Quote:
F=(Pt-Pa)*Ae

But I can leave it out m*Ve specifically deals with the force generated as the gas expands and pushes against the nozzle. Since we don't have a conical nozzle to harness the energy, Ve is zero. m*Ve will always be zero if Ve itself is zero.

Basic math, the fact you are arguing about it says wonders about your understanding of what is quite simple math.

Originally Posted By: paul
suppose each of the above also have 1000 psi pressure
we get your exact result for each of the above fluids no matter what fluid we use.


Exactly - although the amount of gas required to achieve 1000psi will vary gas-to-gas.

Originally Posted By: paul
this is rocket science
F = q Ve + (Pe - Pa) Ae

where F = Thrust
q = Propellant mass flow rate
Ve = Velocity of exhaust gases
Pe = Pressure at nozzle exit
Pa = Ambient pressure
Ae = Area of nozzle exit

not what you put up.

LOL, and once again paul shows his total ignorance of math. The formula you just wrote is THE EXACT SAME ONE I HAVE BEEN WRITING ALL ALONG. Substituting one symbol for another - in your case, q for m - doesn't magically make it another formula.

If anything, your good for a morning laugh.

Oh, and by the way F is force, as in newtons (N). You need to integrate to get thrust (which is in newton-seconds [N*s]).

Originally Posted By: paul
The product of q*Ve is called the momentum, or velocity, thrust and the product (Pe-Pa)Ae is called the pressure thrust.
you are only including the pressure not the momentum of the mass , because you didn't include mass.


Nope. q*Ve (m*Ve in the way I wrote it) is the force produced by the nozzle, due to the expansion (and thus acceleration) of the gas by the conical shape of the nozzle. The actual pressure-driven force is accounted for by the [Pt-Pa]*Ae.
Ve is calculated using:


The only real elements of that you need to understand is the ratio: 1-[Pe/Po]

[Pe-Po] is the pressure difference at the entrance verses exit of the nozzle; essentially you are calculating the force generated by the change in pressure as the air expands in the nozzle. In our case there is no expansion, as the nozzle is just a hole. Thus, Pe = Po, ergo:

1-[Pe/Po] = 1-[1] = 0

Since the remainder of the formula is multiplied by that value, the Ve will also end up being zero.

Originally Posted By: paul
you left out the most important part of the formula.

Nope, I understand the formula and thus calculated it correctly. Since Ve is zero, m*Ve (or q*Ve in your case, although its the same bloody thing) is also zero, ergo that part of the math can be ignored.

Its funny though, you lecturing me on math, when you seem to have completely missed the fact that you "corrected" my math by providing the same formula and changing one letter.

LOL, stupid is funny!

Bryan
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