Anyone any good at physics?

The Breakroom > The Water Cooler

<< < (3/4) > >>

Lew_Merrick_PE:
Come on people, think this through. (1) The cited specific gravity is 0.866. Water has a specific gravity of 1.000, so this fluid is likely to be an oil of some kind. Gasses would have a specific gravity fairly close to 0. (2) Whereas the density of the fluid may change is it is being forced through a (reducing or expanding) nozzle, the mass flow rate through the nozzle will remain constant (Green's Theorem). (3) If there is no energy gradient across the nozzle then there is no flow through the nozzle (First Law of Thermodynamics)!

I do a moderate amount of fluid flow work. I don't do enough to know all this noise of the top of what's left of my mind (though I have friends who do). Most normally, you are given: either a volumetric or mass flow requirement, required nozzle exit fluid speed, or a specific pressure delta to achieve. You then work out a pressure/thermal head needed to fit that requirement and size the pump or "drop" (etc.) to suit.

Water clocks are the classical example of this type of flow. You use a very large area, but otherwise shallow, collector at each stage of a water clock to minimum variations in pressure head. You use a (relatively) small orifice to regulate the outflow into your collection cylinders that, when you measure the height of the fluid in the cylinder, gives you the amount of that gradient of time that has passed. Such systems were common by 500 BC and were probably in existence prior to that.

andyf:

--- Quote from: Lew_Merrick_PE on December 07, 2012, 12:06:11 PM ---... Water clocks are the classical example of this type of flow. You use a very large area, but otherwise shallow, collector at each stage of a water clock to minimum variations in pressure head. You use a (relatively) small orifice to regulate the outflow into your collection cylinders that, when you measure the height of the fluid in the cylinder, gives you the amount of that gradient of time that has passed. Such systems were common by 500 BC and were probably in existence prior to that.

--- End quote ---

Ah, the clypsedra! The Babylonians and Egyptians used those around 1600 BC.

Having now exhibited my knowledge of useless trivia, all I can say about Ross's problem is that it looks like a trick question, incorporating data which is intended to confuse. Other more essential data is omitted, like the rate at which fluid is being fed into the nozzle, and its nature ("fluids" comprised both liquids and gases when I was at school).

Given this dearth of real information, I think the best that can be said is that (ignoring gases, and assuming liquids to be incompressible) the flow rate in both volume and weight terms will be the same as the supply flow rate and will be uniform all the way along the nozzle. This will hold true whatever the relative density of the fluid, though if one fluid has a relative density of 1 and the other 0.866 and the volumetric flow rate at the input is the same for both, then (ignoring any difference in friction) the mass flow rate of the latter fluid will be 0.866 that of the former.

Andy

Ross:
Thanks for your help guys, ill have a look in the morning and see what I can do.

I forgot to mention ism working this out and assuming the liquid is not compressible and the only forces that's needed to worry about is just the flow of the liquid.

Ross:
Hi guys,

I'm doing the first part (a) it seems I NEED the velocity to do anything and seeming as pressure isn't given or anything it makes it a lot harder.

I understand the flow rate will remain constant but the question asks for the midpoint which Is what I shall do. But having so little info is proving the most tricky :loco:

Lew_Merrick_PE:

--- Quote from: Ross on December 09, 2012, 07:39:04 AM ---I'm doing the first part (a) it seems I NEED the velocity to do anything and seeming as pressure isn't given or anything it makes it a lot harder.

I understand the flow rate will remain constant but the question asks for the midpoint which Is what I shall do. But having so little info is proving the most tricky
--- End quote ---

Ross,

The velocity of flow will be inversely-proportional to the cross-sectional area. For an incompressible fluid, this means that, the larger the area, the lower the velocity. Given the ø0.50 m to ø2 m ratio, the velocity of flow will be (1²/0.25²) 16X the velocity of the large end at the small end. The velocity of flow will vary along the length of the taper in proportion to the square of the change in radius (Delta-r). Assuming we go from small to large (i.e. ø0.50 m to ø2 m), then the increase in radius from X=0 (i.e. R0.25 m) to X=6 (i.e. R1 m) is given by: r(x) = 0.25 + x(0.75/6) = 0.25 + 0.125x. r²(x) = (0.25 + 0.125x)² = 0.0625 + 0.0625x + 0.015625x². The latter (r²(x)) gives you the r² at any point between x=0 and x=6. r²(x)/0.0625 = 16r²(x) will give you a plot of the local (x-based) velocity with respect to the velocity at the small (i.e. ø0.5 m) end of the taper. Plotting this out might be helpful to your understanding.

Navigation

 Message Index

Go to full version