How to find the resultant force in a T tube section

[lucasLima]

Homework Statement

So, I need to find the X and Y components of the force in the T section

Homework Equations

Due the fact that the problem dosn't ask for a specific force I couldn't think in any other equation to use, because even tough there is Pressure and Gravity acting I need to decompose the overall sum of it, that as far as I know should be given by the equation above.

The Attempt at a Solution

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[Chestermiller]

You have already determined the three volumetric flow rates, presumably correctly. Before starting to apply momentum balances, the next thing you need to do is determine the pressure at the very junction of the two pipes, P_j. You can do this using a modified version of the Bernoulli Equation. Do you know how to do this? Knowing the pressures at the exits of the tubes will be necessary in order to apply the momentum balance correctly.

 

[lucasLima]

You have already determined the three volumetric flow rates, presumably correctly. Before starting to apply momentum balances, the next thing you need to do is determine the pressure at the very junction of the two pipes, P_j. You can do this using a modified version of the Bernoulli Equation. Do you know how to do this? Knowing the pressures at the exits of the tubes will be necessary in order to apply the momentum balance correctly.

(V₁²)/2 + g*z1 + P₁/p = (V_j²)/2 + g*z_j + P_j/ 1000

z1=z2 V1= 12 P1=500 kPa and p = 1000

72+500 = (V_j²)/2 + P_j/ 1000
572= (V_j²)/2 + P_j/ 1000

That's what I can come up with. But I still can't understand what I should do with it and where the pressures in the exits of the tube will be necessary.

And after further reading I'm questioning if the result I've got isn't Fx , which would be equal to Rx + Fp = Rx + 500 kPa* A1

Thanks for the help!
/

 

[Chestermiller]

(V₁²)/2 + g*z1 + P₁/p = (V_j²)/2 + g*z_j + P_j/ 1000

z1=z2 V1= 12 P1=500 kPa and p = 1000

72+500 = (V_j²)/2 + P_j/ 1000
572= (V_j²)/2 + P_j/ 1000

That's what I can come up with. But I still can't understand what I should do with it and where the pressures in the exits of the tube will be necessary.

And after further reading I'm questioning if the result I've got isn't Fx , which would be equal to Rx + Fp = Rx + 500 kPa* A1

View attachment 204935

Thanks for the help!
/

To get the pressure at the junction, you use the following equation:

$$Q_1\left(\frac{P_1}{\rho}+\frac{v^2_1}{2}\right)=Q_2\left(\frac{P_j}{\rho}+\frac{v^2_2}{2}\right)+Q_3\left(\frac{P_j}{\rho}+\frac{v^2_3}{2}\right)$$
From this equation, what value do you get for the pressure at the junction (in kPa)?

 

[lucasLima]

To get the pressure at the junction, you use the following equation:

$$Q_1\left(\frac{P_1}{\rho}+\frac{v^2_1}{2}\right)=Q_2\left(\frac{P_j}{\rho}+\frac{v^2_2}{2}\right)+Q_3\left(\frac{P_j}{\rho}+\frac{v^2_3}{2}\right)$$
From this equation, what value do you get for the pressure at the junction (in kPa)?

I used the above equation and arrived in the result of 540 046 PA .

Should I decompose Pj in x and y to know the amount of pressuire in each axis so I can put in the linear momentum equation (in the Pressure Force)? If so, how should I do that?

 

[Chestermiller]

I used the above equation and arrived in the result of 540 046 PA .

Should I decompose Pj in x and y to know the amount of pressuire in each axis so I can put in the linear momentum equation (in the Pressure Force)? If so, how should I do that?

First let me check to see if I can confirm your result. Then, I'll show you what to do next. Hopefully we can get the same answers

 

[Chestermiller]

I don't confirm your answer for the pressure at the junction. I get $P_j=87.5\ kPa$. Please check your arithmetic.

 

[Chestermiller]

There is a more accurate way of doing what we've been trying to accomplish (rather than by trying to estimate the pressure at the junction), by treating the inlet stream as being comprised of two entirely separate flow streams of volume flow rates of $Q_2$ and $Q_3$. This approach will immediately enable us to determine the pressures at the exit cross sections 2 and 3. Applying the Bernoulli equation to these two separate flow streams, we have $$\frac{P_1}{\rho}+ \frac{v^2_1}{2}=\frac{P_2}{\rho}+\frac{v^2_2}{2}\tag{1}$$$$\frac{P_1}{\rho}+ \frac{v^2_1}{2}=\frac{P_3}{\rho}+\frac{v^2_3}{2}+gy_3\tag{2}$$where $-y_3$ is that depth of the exit at cross section 3 (and is a positive number, so $y_3$ is negative). From Eqn. 1, I calculate that $P_2=86.5\ kPa$ and from Eqn. 2, I calculate that $P_3=(90.0-9.81y_3)\ kPa$. What do you obtain?

The next step is to carry out momentum balances in the horizontal and vertical directions to get the reaction forces.