In-compressible Liquid Flow

GA Del, per usual.

GA. Nice reply by Hendrik & TrevorM.

Ok I understand that, thanks for clarifying. One thing im also trying to understand is pressure recovery.

If you measure pressure along a piece of pipe or across a restriction orifice etc. the upstream pressure will obviously be greater than the downstream pressure. This difference in pressure to me is the permanent pressure loss of the system. Why do they say there is some pressure recovery through a restriction orifice? ie. the pressure within the orifice is lower than the downstream pressure after the orifice plate?

You may get by fine with an electrical equivalent. It is exactly correct, if properly applied. Do your analogy, it is not rocket science. Use it to your advantage.

Best regards

In your hypothesis of incompressible fluid flow, the velocity increases when the cross section area decreases.

There is no such thing as absolute incompressibility, although liquids come really, really close. When you put a liquid under pressure, it decreases in volume a very minute amount. Therefore, as the liquid passes the restrictions, since mass flow is constant, the velocity increases by a very minute amount. For any real world calculations, liquid can be considered incompressible without any significant error.

I've read all the answers, and still come away with the nagging idea that in the OP scenario, it is a trick question. There is no flow, or velocity, in a hydraulic line that supplies a pressurized vessel, unless the vessel volume changes (as in a cylinder/piston) Although there will be some very small compressibility in the liquid, and any air pockets provide opportunities for compression, the velocity will be nil very quickly, won't it?

Why you think somebody would gives the drop pressure of each component in the line like valves,elbows,joints,etc.? Of course is not needed to solve this problem consider at all any air pocket or compessiblity of the fluid!!As was said uphere each component is pressure drop,let's say V1,V2,..Vn,constant flow is "I",so the energy losses mentioned by Trevor should be (V1+V2+...+Vn)•I,now the rest of energy should be converted in kinetic energy plus pot.energy.Remember,the Bernouilli eq. is just a conservation law you should adapt the idea to solve your own problems.-

I do not know why, do you?

"Why you think somebody would gives the drop pressure of each component in the line like valves,elbows,joints,etc.?"

You firstly need the maximum total head at the desired flow to select the pump and driver.

The maximum pressures at each point in the line can be used to select a lighter class of pipe or different material to make a huge difference in the capital outlay and even running costs. (ID of lower class may be bigger or the other type may be smoother)

How does this apply to the OP?

thanks guys, you've solved my problem!

Dear Mr. Polerz,

The Velocity WILL REMAIN CONSTANT which is dependant on cross-sectional area only and if there is change in cross sectional area, the velocity will inversely vary/

DHAYANANDHAN.S