Conservation of Momentum Fluids Question

[Chestermiller]

Momentum leaving would be $$\iint_{A_2} \rho \vec{V} (\vec{V} \cdot \hat{n})\, dS = \rho \pi R^2 V^2 = \frac{\rho Q^2}{\pi R^2}$$. The backward pressure force would be $$-\iint_{A_2}P_2 \, \vec{dS} = -P_2 \pi r^2\hat{z}$$ where $r$ is the radius of the hemisphere.

Do you agree? Sorry for the late response; I'd been busy the last few days.

Yes, I do agree. So now, what is the momentum balance?

 

[member 428835]

Yes, I do agree. So now, what is the momentum balance?

The momentum balance is $$-\pi r^2 V^2 \rho+\frac{\rho Q^2}{\pi R^2} = -P_2 \pi r^2+ \left(P_1-\frac{\rho}{2}\left(\frac{Q}{2 \pi r^2}\right)^2\right)\pi r^2 - F_{wall}$$. How does that look?

 

[Chestermiller]

The momentum balance is $$-\pi r^2 V^2 \rho+\frac{\rho Q^2}{\pi R^2} = -P_2 \pi r^2+ \left(P_1-\frac{\rho}{2}\left(\frac{Q}{2 \pi r^2}\right)^2\right)\pi r^2 - F_{wall}$$. How does that look?

This looks good. But in the "momentum-in" term, I would have replaced V with $\frac{Q}{2\pi r^2}$.

Now, to complete the solution to this problem, you need one additional relationship to eliminate $Q^2$ and replace it with something involving $(p_1-p_2)$. Any ideas?

Chet

 

[member 428835]

This looks good. But in the "momentum-in" term, I would have replaced V with $\frac{Q}{2\pi r^2}$.

Now, to complete the solution to this problem, you need one additional relationship to eliminate $Q^2$ and replace it with something involving $(p_1-p_2)$. Any ideas?

Chet

Let's apply Bernoulli at $x\to-\infty$ and $x=0$ along the x-axis. Then we have $$P_1 + \lim_{r\to\infty}\frac{\rho Q}{4 \pi r^2} = P_2 + \frac{\rho Q}{4 \pi R^2} \implies\\ Q = \frac{(P_1-P_2)4\pi R^2}{\rho}$$

think we're done now, right?

 

[Chestermiller]

Let's apply Bernoulli at $x\to-\infty$ and $x=0$ along the x-axis. Then we have $$P_1 + \lim_{r\to\infty}\frac{\rho Q}{4 \pi r^2} = P_2 + \frac{\rho Q}{4 \pi R^2} \implies\\ Q = \frac{(P_1-P_2)4\pi R^2}{\rho}$$

think we're done now, right?

I get $$P_1=P_2+\frac{\rho}{2}\left(\frac{Q}{\pi R^2}\right)^2$$

 

[member 428835]

I get $$P_1=P_2+\frac{\rho}{2}\left(\frac{Q}{\pi R^2}\right)^2$$

Oh my gosh, can't believe I didn't write that! It was late and I was sleepy. Yes, of course! So we substitute that into the mass balance for $Q$ and we're set, right?

 

[Chestermiller]

Oh my gosh, can't believe I didn't write that! It was late and I was sleepy. Yes, of course! So we substitute that into the mass balance for $Q$ and we're set, right?

You mean you substitute that into the momentum balance, not the mass balance. Hopefully, that will lead to a result which matches the given answer.

 

[member 428835]

You mean you substitute that into the momentum balance, not the mass balance. Hopefully, that will lead to a result which matches the given answer.

Yeeeeees, this is what I meant! Hahahaha evidently I'm still tired. Thanks a ton!