Continuity Equation Homework: Diameter of Constriction

In summary, you should use Bernoulli's equation to find the velocity at the constriction, and then use equation of continuity to find the velocity of flow in the larger tube.
  • #1
Roger Wilco
30
0

Homework Statement


The inside diameters of the larger portions of the horizontal pipe as shown in the image (attached) are 2.50 cm. Water flows to the right at a rate of 1.80*10^4 m^3/s. What is the diameter of the constriction.
th_Photo5.jpg

Homework Equations

Continuity equation Rate of Volume flow=constant.

The Attempt at a Solution


I got a proportion worked down to
[tex]\frac{V_i}{V_o}=\frac{D_o^2}{D_i^2}[/tex]

where V is velocity and the subscripts i and o denote in and out respectively.I am having trouble figuring out how to utilize the two heights of the fluids in the vertical tubes.

RW

Edit: I am now considering using Bernoulli's equation to take into account pressure. Sound better?
 
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  • #2
Using equation of continuity find the velocity of flow in larger tube. And using Bernoulli's equation find the velocity at constriction
 
  • #3
rl.bhat said:
Using equation of continuity find the velocity of flow in larger tube. And using Bernoulli's equation find the velocity at constriction

There is also pressure in the larger tube (left hand side) though, so wouldn't it be Bernoulli's only? I must admit, I have been eating cookies for the last 20 minutes...let me have another go here. :)

RW
 
  • #4
Before asking next question you should have found the velocity of flow in the larger tube. Now a1v1 = rate of flow of water. From this find v1.
According to Bernoulli's equation P1 + 1/2dv1^2 = P2 + 1/2dv2^2. where d is the density of water. From this find v2.
 
  • #5
rl.bhat said:
Before asking next question you should have found the velocity of flow in the larger tube. Now a1v1 = rate of flow of water. From this find v1.
According to Bernoulli's equation P1 + 1/2dv1^2 = P2 + 1/2dv2^2. where d is the density of water. From this find v2.

So, do I need to even consider the flow at the left-most portion of the tube (where the 10cm vertical pipe is)?

Or can I just look at it where h= 5 cm and the rightmost portion of the tube?

I have no trouble sorting out the algebra. My trouble is with which parts of the tube I actually need to condider.

RW
 
  • #6
Wow. This one is really giving me trouble if I choose to use all three sections of the tube. Maybe I can just limit it to the constricted portion and the right-most portions?
 
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  • #7
This isn't working out at all right now.

So i have so far from continuity equation:

Subscript A for rightmost portion of tube and B for the constricted portion under the 5cm vertical tube.

[tex]R_v=constant[/tex]

[tex]\Rightarrow R_v=V_AA_A[/tex]

[tex]\Rightarrow \pi r^2_AV_A=R_v[/tex]

[tex]\Rightarrow V_A=\frac{1.8*10^{-4}}{(.025)^2\pi}=9.167*10^{-2} \frac{m}{s}[/tex]

From Bernoulli:

[tex]p_B+.5\rho V_B^2+\rho gy_B=p_A+.5\rho V_A^2+\rho gy_A[/tex] p=0, rho is factor of all

[tex]\Rightarrow V_B=\sqrt{V_A^2-2gy_B}=DNE[/tex] The quantity I get under the radical is NEGATIVE.

Do I need to include the 10cm height? If so how?

I reall need some help on this one:confused:
 
  • #8
rl.bhat said:
Before asking next question you should have found the velocity of flow in the larger tube. Now a1v1 = rate of flow of water. From this find v1.
According to Bernoulli's equation P1 + 1/2dv1^2 = P2 + 1/2dv2^2. where d is the density of water. From this find v2.

So I used this where P1= rho*g*y1 and P2=rho*g*y2 (i.e., Bernoulli's Equation)
Then instead of calculating V, I used [tex]R_v=AV[/tex]

[tex]\Rightarrow V=\frac{R_v}{A}[/tex]

[tex]\Rightarrow V= \frac{R_v}{\pi(.5 D)^2}[/tex]
RW
 
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  • #9
You have made 2 mistakes. 1) 2.5 cm is the diameter and not the radius. 2) PA is not equal to PB. PA = rho*g*0.10 and PB = rho*g*0.05. Since tube is horizontal rho*g*yA = rho*g*yB. Now try it again.
 
  • #10
rl.bhat said:
You have made 2 mistakes. 1) 2.5 cm is the diameter and not the radius. 2) PA is not equal to PB. PA = rho*g*0.10 and PB = rho*g*0.05. Since tube is horizontal rho*g*yA = rho*g*yB. Now try it again.

I don't want to seem ungrateful, but I think you should reread my response again before assuming that I have made more errors. It is even more confusing when you say that I am incorrect when I am not.
Roger Wilco said:
So I used this where P1= rho*g*y1 and P2=rho*g*y2 (i.e., Bernoulli's Equation)
Then instead of calculating V, I used [tex]R_v=AV[/tex]

[tex]\Rightarrow V=\frac{R_v}{A}[/tex]

[tex]\Rightarrow V= \frac{R_v}{\pi (.5D)^2}[/tex]
RW
As you can see from above quote: P1=rho*g*y1 and P2=rho*g*y2
which is exactly what you wrote.

And nowhere did I indicate that I planned on using 2.5cm as the Radius. I would be plugging in V=(R_v)/[pi*(D/2)^2]

Other than the miscommunication, I suspect that this is the correct approach.

Thanks for your help,
RW
 
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  • #11
I posted my responce when I received this:
Do I need to include the 10cm height? If so how? Both of us were wrighting the responce simultaneously. 7.24 and 7.32. When I was camposing and sending the responce I might have received your responce, which I don't know how to see simultaneously. Any way I am sorry.
 
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  • #12
rl.bhat said:
I posted my responce when I received this:
Any way I am sorry.


Don't be:wink: I was just trying to be sure that we were on the same page. I think I can wrap this one up now:smile: Thank you for your help rl. bhat!

RW
 

1. What is the continuity equation and how is it related to the diameter of constriction?

The continuity equation is a fundamental equation in fluid mechanics that states that the mass flow rate of a fluid must remain constant throughout a closed system. This equation is directly related to the diameter of constriction, as the constriction of a pipe or channel will affect the flow rate of fluid through it, and therefore impact the continuity equation.

2. How is the diameter of constriction calculated?

The diameter of constriction can be calculated using the continuity equation, which states that the cross-sectional area of the fluid at any point in a closed system must remain constant. Therefore, the diameter of constriction can be found by measuring the cross-sectional area of the fluid before and after the constriction, and using this information to solve for the diameter.

3. What factors can affect the diameter of constriction in a fluid system?

Several factors can impact the diameter of constriction in a fluid system, such as the velocity of the fluid, the viscosity of the fluid, and the shape and size of the constriction. Other external factors, such as temperature and pressure, can also play a role in determining the diameter of constriction.

4. How does the diameter of constriction affect the flow rate of a fluid?

The diameter of constriction has a direct impact on the flow rate of a fluid. A smaller diameter of constriction will result in a higher velocity of the fluid, leading to a higher flow rate. On the other hand, a larger diameter of constriction will result in a lower velocity and a lower flow rate.

5. Can the diameter of constriction be changed without affecting the flow rate of a fluid?

No, the diameter of constriction directly affects the flow rate of a fluid. Any change in the diameter, whether it is an increase or decrease, will result in a change in the flow rate. However, this change can be controlled and predicted by using the continuity equation and other fluid mechanics principles.

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