Fluid Mechanics: Momentum Equation- When to include p_atm in equation?

[Master1022]

Homework Statement

General problem. For example, air enters a wind tunnel (or a jet; basically something that has a larger area of entry than exit) and the total pressure is atmospheric pressure. Write out the momentum equation for this control volume.

Relevant Equations

Bernoulli's Equation
Momentum Equation

So if we define point 1 at the entrance and point 2 at the exit, then we can write out Bernoulli's equation along a horizontal streamline as such: p_1 + \frac{1}{2}\rho v_{1}^2 = p_2 + \frac{1}{2}\rho v_{2}^2 = p_{atm}

One question is: won't there be p_atm also contributing to the static pressure at p1 and p2 (i.e. p_atm + p_1 or p_atm + p_2)? Mathematically I see that there would be problems if we had this in the equation, but I cannot understand why. Does atmospheric pressure not contribute to the static pressure inside an object (e.g. this wind tunnel or a pipe)?

Then later on, when we write the momentum equation for our control volume, we get (Force = Change in Momentum Flux R(-->)):
p_1 A_1 - p_2 A_2 - p_{atm}(A_1 - A_2) + F = \rho u_2 ^ 2 A_2 - \rho u_1 ^ 2 A_1

However, once again, I cannot understand why there isn't a p_atm term for A1 & A2?

Apologies for this, as I understand there is some basic error.

Thanks in advance

 

[Chestermiller]

In the Bernoulli equation, it doesn't matter (since it cancels).

It isn't clear to me what you are doing in the momentum equation. You can work in terms of absolute pressures or you can work in terms of gauge pressures. But your choice dictates your interpretation of the force F.

 

[Master1022]

In the Bernoulli equation, it doesn't matter (since it cancels).

Thank you for your response. So if we had a p_{atm} term, the equation would read p_{atm} = p_{atm} + \frac{1}{2}\rho u^2 + p_1 If we were to consider pressure this way, would this mean that the p_1 is negative if u > 0?

 

[Chestermiller]

Thank you for your response. So if we had a p_{atm} term, the equation would read p_{atm} = p_{atm} + \frac{1}{2}\rho u^2 + p_1 If we were to consider pressure this way, would this mean that the p_1 is negative if u > 0?

Please tell me what you mean by this equation. You seem to be using gauge pressures. Is that correct. If so, then $\p_1$ is sub-atmospheric (i.e., partial vacuum).

So, in the Bernoulli equation, you can use either gauge pressures or absolute pressures as long as you are consistent.

In the momentum equation, you need to be careful what you are doing, particularly if you are trying to find the net force exerted on the control volume (by the combination of fluid and surrounding atmosphere).

 

[Chestermiller]

Thank you for your response. So if we had a p_{atm} term, the equation would read p_{atm} = p_{atm} + \frac{1}{2}\rho u^2 + p_1 If we were to consider pressure this way, would this mean that the p_1 is negative if u > 0?

In your equation, I'm guessing that $p_1$ is gauge pressure. If that interpretation is correct, then the equation is telling us that $p_1$ is less than atmospheric pressure (negative gauge pressure).

In the Bernoulli equation, it doesn't matter if you use gauge pressure or absolute pressure, as long as you are consistent.

In the momentum equation, you need to be careful and be aware of which you are using, particularly if you want to use the equation to determine the net force of the atmosphere and the fluid on the body of the control volume.

 

[Master1022]

Please tell me what you mean by this equation. You seem to be using gauge pressures. Is that correct. If so, then $\p_1$ is sub-atmospheric (i.e., partial vacuum).

So, in the Bernoulli equation, you can use either gauge pressures or absolute pressures as long as you are consistent.

In the momentum equation, you need to be careful what you are doing, particularly if you are trying to find the net force exerted on the control volume (by the combination of fluid and surrounding atmosphere).

Thank you for your response. Yes, you are correct, that is what I meant. It now makes more sense.

 

[Chestermiller]

Thank you for your response. Yes, you are correct, that is what I meant. It now makes more sense.

If you have any additional issues, please get back with me.