Pressure VS Flow rate

Do you require the relationship for a pipeline, orifice in pipeline or an outlet (nozzle)?

Dear sir,

If i am correct,you are talking about pumps.Then pressure (head) is inversely proportional to flow.That curve is called drooping curve or pump performance curve.Formula for this is pump power calculation.

sankar

Look here.

http://www.pipeflowcalculations.com/naturalgas/theory.htm

Success to you

Pump performance curves are one part of the deal. The other part is the system characteristic curve.

Overlaying one to the other at the same scale will produce an intersection between the two curves, and that intersection will be the operating point of the system in terms of flow and pressure.

As a general rule, operating two centrifugal pumps in parallel does not give twice the flow; the system characteristic curve sees to that.

Sir,

compliments,well am an engineer by profession,i like to be part of your engineering team,infact to register with you to enable me have first hand information on the latest development in various topics that makes up mechanical engineering .

Regards.

omatseye tuasor.

There are quite a number of applications involving a flow-pressure relationship.

The 4 post above has at least 5 of them in mind.

Please supply more info!

Please validate whether you want it in water, oil, air, gas or semi-fluid (molten plastic).

Any hydraulic circuit is equal to electric one as hydraulic pressure is the volts flow rate is the current and the resistance of the hydraulic circuit is equal to the electric one and so you can estimate what ever you like by using the ohm's law.

PLease clarify in layman's terms. People who ask the simple questions should also get simple answers. My aplogies as I did not intend to offend any one.

Regards;

Nadeem

07062007

Let's not forget our friend Reynold! I'm no pro but seem to remember doing labs in school on pipe flow and diameter vs flow and turbulent resistance along the pipe lengths.

Engineering : Pressure is force applied over area - Pounds per square inch - PSI

Flowrate is quanity during a specific time - gallions per min GPM or Cubic feet per second - CFS

Layman's terms: Think of it like a faucet. Even with the faucet turned off you have pressure. The vale holds the pressure back. When you turn on the faucet you have flow. The pressure makes the water flow... er ... sort of ....

JR Peck

What you need is the bernoulli equation, and the continuity equation to solve fluid flow problems:

Continuity: Q1=Q2=Q3 The flow is not changing from one point in the system to another. Friction will slow the entire system, but from one point to another, flow rate is constant (unless parallel piping system obviously where flow splits into more than one path)

Bernoulli: I just don't have time to write this one. This is your homework

oh yea, Q=AV so you can relate the two

There is this rule that you may use:

Q=k(dP/SG)^1/2

Where,

Q: Flow

dP: Pressure drop

SG: Specific Gravity

k: Constant (depend on system and elements on piping)

This is often used in Cv valves calculations.

SaC.

is this question a joke?

Possibly how abt a try at the following eqn.

q=a*v

and v=square root of (2*g*h) considering the pr.head

a= c/s area of addressed part

g=whatever gravitational const at ur location (it varies at the equator (hi..hi..hi.. ))

pr.=w*h (frm wich u find h )

w= sp.wt.of the fluid

or

u can try (can b used only if u no Cd)

q=(Cd) (a) (root of (delta p))

Cd= coe. of discharge

a= c/s area

delta p = pr. drop

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b a PERMUTATION OF GOOD AND A COMBINATION OF BAD

Please use the Bernoulli's equation for fluid flow. The equation given below is true for only incompressible fluids i.e. liquids. For two points 1 & 2 in the flow of the liquid

Z1 + P1/ρg + V1²/2g = Z2 + P2/ρg + V2²/2g where Z = Elevation of the Point in Meters ; P = Pressure in Newtons / M² ; ρ = density in Kg / M³ ; V = velocity in M/sec.; g = Acceleration due to gravity = 9.81 M/sec²

All units can be in FPS system i.e. Z= feet ; P = Poundals / ft²; ρ = lbs./ft³; V= ft/sec. g = 32.2 ft./sec²

Also being incompressible, Mass flow rate 'm' = A1 X V1 X ρ = A2 X V2 X ρ where A is the cross sectional area of flow.

By knowing the values Z1, A1, V1 and P1 at one point and Z2, A2 at the other P2 and V2 can be found.

kasharma

Your formula (Z1 + P1/ρg + V1²/2g ≠ Z2 + P2/ρg + V2²/2g) actually an inequality, where the left part more right. Look textbooks. Success to you!