Fluid hydraulics

  • #1
I am trying to determine the entrance loss benefits of changing the entrance conditions from water run-off (gravity flow) from a smooth (steel plate) surface into a rectangular channel box that is under induced suction. Currently, I believe that I am restricting water removal-flow through a rectangular, slotted entrance but am not sure how to calculate the potential benefits of increasing slot size, changing shape, etc. Attempting to remove approximately 3,000 US gallons per minute through a slot of 2.75-inches x 48-inches. I need to improve loss conditions by minimum of 12-inches of head in order to gain an appreciable benefit. HELP PLEASE!
 

Answers and Replies

  • #2
688
1
Let me get back to you tomorrow. I have some good reference material at work that should help us solve this problem.
 
  • #3
688
1
OK, looked into this - here is an approximate method.

The pressure drop across your rectangular entrance is the usual form:

DP = K * rho * vel ^2 / 2

DP = pressure drop
rho = density
vel = velocity = volumetric flow / cross sectional area

For now, use an entrance loss coefficient value of K = 0.32 for your channel.

Now, your velocity will decrease as you increase the height or width (height is better) of the rectangular entrance. This will decrease your velocity and pressure drop.

Determine your new entrance dimensions to reduce the pressure drop to the number you need.

Typed in a hurry - let me know if you have any questions.
 
  • #4
Units? When I use units for rho=lbs/ft^3 (water = 62.48), and vel=ft/sec (mine = 7.3), I calculated a DP of 532. I'm not sure what my units are, but 532 sounds like I may be off by a factor of at least 100.
 
  • #5
By the way edgepflow, I really appreciate your help!
 
  • #6
688
1
Units? When I use units for rho=lbs/ft^3 (water = 62.48), and vel=ft/sec (mine = 7.3), I calculated a DP of 532. I'm not sure what my units are, but 532 sounds like I may be off by a factor of at least 100.

You can make the units work by introducing the gc factor:

DP = K * rho * vel ^2 / ( 2 * gc )

where,

gc = 32.2 lbm-ft / (lbf-sec^2)

Multiply out your units or you can use MathCAD, programmable calculator, or some other digital package if you have access to it.
 
  • #7
688
1
I also figured a K = 0.9 is a better number.
 
  • #8
688
1
Some additional improvement in the calculation may be obtained by using the "hydraulic diameter" of the rectangular opening:

Dh = 2bh / (b+h) = 2h / (1 + AR)

where,

h = height of rectangular channel = 2.75 in.
b = width = 48 in

AR = aspect ration = height / width = 2.75 / 48 in = 0.057

Usually, the AR should be > 0.25, so there is some error here.

As you can see, as b and AR -> infinity, then Dh -> 2h. So it is best to increase height only to lower pressure drop. You could use Dh in your velocity calculation for better accuracy.
 

Related Threads on Fluid hydraulics

  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
5
Views
5K
  • Last Post
Replies
13
Views
4K
Replies
1
Views
3K
Replies
3
Views
7K
Replies
10
Views
5K
Replies
1
Views
2K
Replies
2
Views
3K
  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
18
Views
3K
Top