Rocket thrust equation -- What is velocity V in mass flow rate formula?

[user079622]

What is velocity V in mass flow rate formula, velocity of inlet ,outlet, velocity of rocket speed in relation to freestream?

 

[russ_watters]

V is just velocity in the generalized force equation. Inlet and outlet velocity are clearly labeled, and the rocket is at rest always under this equation (velocity of the gases is measured relative to the engine)..

 

[user079622]

V is just velocity in the generalized force equation. Inlet and outlet velocity are clearly labeled, and the rocket is at rest always under this equation (velocity of the gases is measured relative to the engine)..

I dont understand..
which V I must use to calculate mas flow rate?

 

[russ_watters]

I dont understand..
which V I must use to calculate mas flow rate?

I don't understand either; mass flow rate is an input, not an output from the equation. What, exactly, are you trying to do and what information do you have to start with?

 

[user079622]

I don't understand either; mass flow rate is an input, not an output from the equation. What, exactly, are you trying to do and what information do you have to start with?

so F= Vo x Vo x r x Ao - Ve x Ve x r x Ae ?

 

[russ_watters]

so F= Vo x Vo x r x Ao - Ve x Ve x r x Ae ?

Where did you get that? Again, what are you trying to do and what information do you have to start with? You seem to be plugging things in without a goal.

 

[user079622]

Where did you get that? Again, what are you trying to do and what information do you have to start with?

I am trying to calculate rocket thrust, but what V I must plug in in mass flow formula if I have just label Vo and Ve????

I want to say that mass flow rate at exit depend on Ve, on input depend on Vo

 

[russ_watters]

Ok, the original equation is for jets. Rockets don't have initial gas velocity, so you zero that out and just calculate from exit velocity.

 

[user079622]

Ok, the original equation is for jets. Rockets don't have initial gas velocity, so you zero that out and just calculate from exit velocity.

mass flow of rocket depend on Ve

Ve=100m/s
A=4m2
r= 1.2kg/m3

mdot= 100 x 1.2 x 4

F= mdot x v
=100^2 x 1.2 x 4Second thing, I dont understand part (pe-po)A, this is just force express through pressure times surface, that part must be equal as mdot x v..
dont make sense formula has part of pressure contribution and momentum mdotxv contribution to thrust

This like you calculate lift at wing from integration of static pressure around wing and then from how much wing air push down momentum (mdot x v) and then add this together, this is wrong

 

[russ_watters]

Second thing, I dont understand part (pe-po)A, this is just force express through pressure times surface, that part must be equal as mdot x v..
dont make sense formula has part of pressure contribution and momentum mdotxv contribution to thrust

Again, that equation is for a jet (or fan), where there may be a pressure change through the jet.

I'm not sure this works well for a rocket: you sure about r?

 

[user079622]

Again, that equation is for a jet (or fan), where there may be a pressure change through the jet.

I'm not sure this works well for a rocket: you sure about r?

 

[erobz]

View attachment 337009

The gage pressure of the exhaust jet integrated over the area of the jet is an external force on the rocket.

 

[user079622]

The gage pressure of the exhaust jet integrated over the area of the jet is an external force on the rocket.

Yes I agree and this is (pe-po)Ae, but dont make any sense to add this to mdot x v

 

[erobz]

Yes I agree an this is (pe-po)Ae, but dont make any sense to add this to mdot x v

It makes sense in the context of Newtons Second Law for control volumes " a.k.a. "The Momentum Equation" in fluid mechanics. A control volume surrounds the rocket, slicing through the exhaust jet.

$$ \sum \boldsymbol F = \frac{d}{dt}\int_{cv} \boldsymbol v \rho ~dV\llap{-} + \int_{cs} \boldsymbol v \rho \boldsymbol V \cdot d \boldsymbol A \tag{1}$$

Turing your rocket vertical:

The LHS side of $(1)$ is the sum of the external forces acting on the control volume, Drag, Weight, Gage Exhaust Pressure. The rest comes from evaluating the RHS under certain simplifying assumptions.

 

[user079622]

The gage pressure of the exhaust jet integrated over the area of the jet is an external force on the rocket.

This integration get same result as (pe-po)Ae?

 

[erobz]

This integration get same result as (pe-po)Ae?

Its a simplification. Uniform pressure distribution. In the diagram I showed the force as the effect of a uniformly distributed pressure integrated over $A_e$... i.e. $P_e A_e$

Note that the symbols are used differently. Your $p_e$ is an absolute pressure, mine ($P_e$) is a gage pressure, but the main idea is the same.

 

[erobz]

In practice the goal is to maximize thrust, and believe it or not... the pressure term exists, but it is actually a detriment to the thrust, because there is a dependency of the exhaust velocity $V_e$ on the nozzle exit pressure $P_e$, and in optimization they try to make $P_e$ as close to ambient pressure as possible ( bringing the pressure term to zero) to maximize $V_e$.

https://www.ijsr.net/archive/v8i12/ART20203435.pdf

 

[user079622]

Its a simplification. Uniform pressure distribution. In the diagram I showed the force as the effect of a uniformly distributed pressure integrated over $A_e$... i.e. $P_e A_e$

Note that the symbols are used differently. Your $p_e$ is an absolute pressure, mine ($P_e$) is a gage pressure, but the main idea is the same.

My (pe-po) is your gage pressure Pe.
So if you just integrate gage pressure over nozzle area ,like you said, that you still missing part mass flow x v ? so your post #12 is not correct?

 

[erobz]

My (pe-po) is your gage pressure Pe.
So if you just integrate gage pressure over nozzle area ,like you said, that you still missing part mass flow x v ? so your post #12 is not correct?

I said the rest of it ( meaning the momentum rate accumulation and efflux terms ) comes from evaluating the two integrals on the right hand side (RHS) of the equation $(1)$.