Fluid dynamics: momentum equation and continutity

[fayan77]

Homework Statement

Homework Equations

because their is steady flow we only care about mass flow flux meaning
m₁v_out - m₂v_out
m = ##\rho##Av
i know that m₁ = m₂ + m₃
since there are no dimensions I am assuming that areas are the same everywhere, density does not change therefore velocities out are half of velocity going in

The Attempt at a Solution

Forces in x must equal momentum forces in x
we have 600lb to the left and P_A(Area) to the right = mv_out - mv_in
Therefore
25(144)(##\pi##/4)(d)² - 600 = -mv_in
I'm stuck

 

[haruspex]

You involve a lot of variables without defining any, so I cannot follow your working.
What does Bernoulli's equation tell you about the velocity?

 

[Chestermiller]

In addition to what haruspex said, what is your horizontal macroscopic momentum balance on the stream of fluid striking the plate?

 

[fayan77]

Bornoulli says that
P₁ + ##\rho##gh + .5##\rho##(v₁)² = P₂ + ##\rho##gh + .5##\rho##(v₂)²

But I don't think I can apply this here because I don't know anything about a different point. Along that horizontal pressure is 25 psi but at the plate it would be greater since velocity is 0. And I can't use it out in the atmosphere where P=0 because I don't have a height.

 

[Chestermiller]

Bornoulli says that
P₁ + ##\rho##gh + .5##\rho##(v₁)² = P₂ + ##\rho##gh + .5##\rho##(v₂)²

But I don't think I can apply this here because I don't know anything about a different point. Along that horizontal pressure is 25 psi but at the plate it would be greater since velocity is 0. And I can't use it out in the atmosphere where P=0 because I don't have a height.

Did you not see my response in post #3?

 

[fayan77]

This is my free body diagrams for force and momentum

 

[haruspex]

don't know anything about a different point

Consider a point before it hits the plate, e.g. where the arrow points to v.

I can't use it out in the atmosphere where P=0 because I don't have a height.

From A, it is all at the same height.

 

[fayan77]

ok Point B will be where V is described in picture
therefore
P_A + .5##\rho##(v_A)² = P_B + .5##\rho##(v_B)²

because P_A and P_B are in same height pressure is same along that horizontal line also there is no velocity at P_A

25(144) + 0 = 25(144) + .5(1.95)(v_B)²

What am I doing wrong?

 

[Chestermiller]

ok Point B will be where V is described in picture
therefore
P_A + .5##\rho##(v_A)² = P_B + .5##\rho##(v_B)²

because P_A and P_B are in same height pressure is same along that horizontal line also there is no velocity at P_A

25(144) + 0 = 25(144) + .5(1.95)(v_B)²

What am I doing wrong?

Since pA is 25 psig, pB should be zero (i.e., atmospheric pressure).

The macroscopic momentum balance on the jet should read $$F(-i_x)=0i_x-(\rho v_B A)v_Bi_x$$or$$F=\rho v_B^2 A$$

 

[fayan77]

Ooooooooohhhhh I just saw the picture closely. I assumed that there was a pipe all the way to where the wall, but the opening starts right where the tank is therefore it is atmospheric pressure. Thanks, lol. By the way since I have your attention in this thread already would the acceleration of a cart be,

a_c = (##\rho##Q(v_jet-v_cart)[(v_out-v_in)] / (Mv_jet)

where (v_jet-v_cart) = v_out = v_in equal in magnitude but different direction therefore multiply by cos or sin

this is for a cart with opening on left side; jet goes horizontally through left opening and out through top at an angle

 

[Chestermiller]

Ooooooooohhhhh I just saw the picture closely. I assumed that there was a pipe all the way to where the wall, but the opening starts right where the tank is therefore it is atmospheric pressure. Thanks, lol. By the way since I have your attention in this thread already would the acceleration of a cart be,

a_c = (##\rho##Q(v_jet-v_cart)[(v_out-v_in)] / (Mv_jet)

where (v_jet-v_cart) = v_out = v_in equal in magnitude but different direction therefore multiply by cos or sin

this is for a cart with opening on left side; jet goes horizontally through left opening and out through top at an angle

I don't follow this cart business at all. Is this for a different problem in a different thread?

 

[fayan77]

no its for an extra credit problem i was assigned but I didn't want to make another thread because it was going to take ages for someone to respond, but I am leaving to class now, thanks for your help.

 

[haruspex]

no its for an extra credit problem i was assigned but I didn't want to make another thread because it was going to take ages for someone to respond, but I am leaving to class now, thanks for your help.

I think you were a bit unlucky in the initial delay on this thread. Usually a lot faster. In fact, delays are often longer in established threads because the first responder, in a different timezone, has retured for the night and other potential responders see that the thread is being handled.
Please post a fresh thread for each problem.