Fluid Mechanics Q: Find Vol. Flow Rate & Head Loss at C

In summary, Fred is calculating the volumetric flow rate into a bottom tank while working with Bernoullis equation. He is unclear on the steps he needs to take and is looking for help.
  • #1
morry
136
0
Ok guys, here's a q from a past exam and I am not sure how to tackle it.
At point C, there is a leak and flow rate is given in L/s. I am also given the friction factor. I also know the diameter of the pipe.

The question asks for the vol. flow rate into the bottom tank. And then the head loss in BC and CD

What I am unsure of is the steps I need to take with Bernoullis equation. ie apply at AB then BC etc. I don't get that bit.

So if someone could point me in the right direction, Id be very happy.

Thanks.
 

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  • #2
Ok, so I think I've calculated the volumetric flow rate, but it came out to be 3535 L/s, which is a fair bit for a 22cm diameter pipe.

I now have to find the head loss in BC and CD, but I don't know how to do this since I don't know the pressure at C.

Finally I have to sketch the hydraulic gradient, which I've never even heard of before.

Anyone have any ideas about these last bits?
 
  • #3
Is the top reservior an open reservoir? Do you know the height at point C? That would mean that the pressure at C should be a combintation of the velocity head and the potential head at point C while the pressures at A and E are 0.

http://www.engineeringtoolbox.com/energy-hydraulic-grade-line-21_613.html

also, check out the definition on page 47 here:
http://www.engineeringtoolbox.com/energy-hydraulic-grade-line-21_613.html
 
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  • #4
Both reservoirs are open to the atmosphere.

I don't know the height of C. Normally, finding pressure at C wouldn't be a worry, but this time there is a leak at C.

So would the pressure be the velocity head (With leak taken into account) plus potential head?

Thanks Fred!
 
  • #5
Since it appears that that is the only information you have to go on, then I would say that you are correct. You should have some potential head, but if you are not given that information, then it might as well be considered insignificant. Velocity head is all you have left.
 
  • #6
Cheers Fred.
 
  • #7
Did it come out OK? Hope I helped.
 
  • #8
Yeah came out eventually. :)

I think I've been letting the algebra get in my way, instead of letting my head look at the particular question. Hope there's an easy example of this question on the exam. :)
 

Q: What is fluid mechanics?

Fluid mechanics is a branch of physics that deals with the study of fluids (liquids and gases) and the forces that act on them.

Q: What is the volume flow rate?

Volume flow rate is the amount of fluid that passes through a given cross-sectional area in a specific amount of time. It is commonly measured in liters per second or cubic meters per second.

Q: How do you calculate the volume flow rate?

The volume flow rate can be calculated by multiplying the cross-sectional area of the pipe or channel by the fluid velocity. The formula is Q = A x v, where Q is the volume flow rate, A is the cross-sectional area, and v is the fluid velocity.

Q: What is head loss in fluid mechanics?

Head loss is the decrease in the total energy of a fluid as it flows through a pipe or channel due to friction, changes in elevation, or other factors. It is commonly measured in units of length, such as meters or feet.

Q: How do you calculate head loss?

The formula for head loss is hL = f x (L/D) x (v^2/2g), where hL is the head loss, f is the friction factor, L is the length of the pipe, D is the diameter of the pipe, v is the fluid velocity, and g is the acceleration due to gravity. This is known as the Darcy-Weisbach equation.

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