Flow rate

Aj83

For a fixed pressure differential will the flow rate remain constant or will it increase as the restriction decreases. So imagine certain volumetric flow rate through partially opened valve, will the flow rate increase or will it remain constant as the valves is opened further? Assume density is constant and no pressure losses. Thanks

russ_watters

Welcome to PF!

Do you have an idea about what might happen?

Aj83

I think the velocity will decrease and flow rate will remain same due to continuity?

Aj83

Anyone?

cwolfe

For sizing valves we use Q=Cv x pressure drop. Cv is a dimensionless number indicating the orifice size. Q=flow. This simple calculation can grow to be very complex when variables like piping, temperature, viscosity, critical pressure drops and many others are included. But the base is fairly accurate for cold water at low pressure drops (200 or so). What it tells you is that if the inlet pressure and Delta-P remain constant, the flow will vary with orifice size.

russ_watters

Minus losses, the velocity should stay about the same, but yes, flow rate will increase because the orifice gets larger.

Aj83

Right, Let me put it in another way. If you put your finger in front of a pipe outlet, lets say while watering your garden, so that you are blocking the outlet partially. When you do that, will there be a reduction in flow rate or will the velocity increase to keep the flow rate same?

I am trying to get my head around continuity equation (mass flow rate)in = (mass flow rate)out or considering density remains constant (Volume flow rate)in = (Volume flow rate)out.

Also q=vA (q=volume flow rate, v=velocity, A=cross sectional area)

So my understanding is that keeping the pressure differential same,if you change the cross sectional area, the velocity will adjust accordingly to keep the flow rate same and vice versa. Which is what we see in a nozzle, when the area decreases the velocity increases. So why does flow rate increases when you open the valve further (increasing the exit cross section), shouldn't velocity decrease as the area increases while flow rate remaining constant?

P.S To keep things simple, let's say we don't take the pressure losses in to account due to viscosity or any changes in temperature.

jakestenson

Information about orifice calculations here

Alkim

Probably you should consult Bernoulli´s principle.

Aj83

Thanks for your reply. I understand the Bernoulli's principle which says with increase in flow velocity,the static pressure decreases and vice versa. So in this case as the restriction increases the static pressure drops further across it which will lead to an increase in flow velocity(increase in dynamic pressure). Now I am inclined to think that this change in static pressure and velocity will be such that the volumetric flow rate remain same to satisfy continuity. So as the restriction increase or the corss sectional area "A" decreases the velocity "V" will increase such that volumetric flow rate "q=VA" remains constant. Is that correct?

P.S As I have been saying so far, ignore the 'pressure loss' (which is quite significant across sharp edged orifice) so even though static pressure 'drops' there is no static pressure 'loss' therefore 'total pressure' remains same throughout the flow.

harrylin

Isn't that the situation of a water tap? Increase the opening and the volume flow rate increases.

The continuity equation merely tells you that the in-flow equals the out-flow at a certain valve opening.

Hkwoo123

You should consult the poiseuille flow equation.
It shows the relationship between pressure difference, flow rate and radius.

Aj83

So say tap is open to some specific flow rate, if you put a nozzle (a converging shape) at its end why does the velocity increase? or why does velocity increase if you put your finger such that its blocking the exit partially?

jbriggs444

There are (at least) three effects that may apply.

1. Water hammer.

When you squeeze down the nozzle or put your thumb over the end of the hose the water in the hose has momentum -- it cannot slow down instantly. So there is a momentary pressure increase across the orifice.

This will explain a momentary increase in exit velocity.

2. Bernoulli.

For a fixed pressure drop there is a fixed increase in kinetic energy per unit volume. Drop the orifice size while maintaining a fixed pressure drop and nozzle velocity does not increase at all.

By itself, this will predict that squeezing down the nozzle will not result in a long-term increase nozzle velocity. But it will predict a reduction in flow rate.

3. Where's the bottleneck.

If there is no constriction at the end of the hose then there is no pressure drop at the end of the hose. If the water company is supplying water at (for instance) 60 psi and the water pressure on egress from the hose is 0 psi then there has to be a pressure drop somewhere.

Bernoulli can explain some of this pressure drop. The unrestricted flow velocity out the end of the hose with the faucet wide open can be respectable. But in my gardening experience, it's nowhere near as high as the stream velocity you get with a nozzle.

Reduce the orifice size and you reduce the flow rate (as above). This, in turn reduces the pressure drop at all the other bottlenecks in the system. You have less pressure drop within the length of the hose, less pressure drop at the faucet, less pressure drops in the pipes in your house and less pressure drop at every elbow, tee and water meter.

Now you have closer to 60 psi at the nozzle and closer to 60 psi worth of velocity at the nozzle.

This predicts the long term increase in nozzle velocity that is seen.

Aj83

1. Water Hammer.

This effect will occur only momentarily as soon as lets say the thumb is placed on the end but once the fluid settles to that change it will disappear i,e no surge or hydraulic shock. However velocity increase is not momentary, it remains increased (try doing that to your tap). In fact nozzle is a shape which is used to increase the flow velocity (jet engines, garden hose etc). So imagine nozzle as a 'control volume', lets say 1 kg/s enters the wider end then according to law of conservation of mass 1 kg/s must exit the narrow side too which makes sense because as the area is reduced the velocity increases to keep the flow rate same due to reduction in area (mass flow rate = density * velocity * area "density remains constant for incompressible flow")

2. Bernoulli.

Yes you are right, with a drop in 'static' pressure there is an increase in 'dynamic' pressure aka kinetic energy per unit volume or in simple words increase in flow velocity. I am not aware of any situation where you can keep the pressure drop same by changing the restriction, perhaps you can elaborate on that.

3. Where's the bottleneck.

Why does there have to be a pressure drop? 60 psi is pressure energy contained by the fluid and as it comes out of the end where pressure is 0 psi it still has the same amount of energy in form of kinetic/pressure/internal energy so the energy is conserved. There will however be a minor 'pressure loss' in form of viscous dissipation due to viscosity or in bends due to flow separation (turbulence) but as i have been saying from start lets just ignore the pressure losses and gravity effects.

The reason why velocity increases when there is a nozzle attached to the end is that flow area is reduced so the flow velocity increases to satisfy the law of conservation of mass. Now why don't we see the same effect when we open and close the tap? Thats is the dilemma I have.

jbriggs444

You should not with one breath say "let's ignore pressure losses" and with the next breath ask "why do we see what we see in the real world". I maintain that real-world pressure losses added up from the water company to the end of the hose are non-negligible.

The evidence for this is what I've already given: The arc of water from the end of a hose running wide open and pointed at an upward angle does not reach very high. The arc of water from the end of a hose with a nozzle reaches higher.

Ideally -- ignoring pressure losses due to viscosity the height of a stream pointed upward at a steep angle should be roughly equal to the height of the local water tower, with or without a nozzle.

You asked for a situation where you can keep the pressure drop the same "by changing the restriction". If we take you at your literal word, there is a common device that does exactly that. It is called a pressure regulator. When output pressure is low, it opens a valve wider. When output pressure is high it closes the valve tighter.

Aj83

jbriggs, I appreciate your responses but this doesn't answer the question I asked in this thread. For simplicity, lets assume flow is 1 dimensional (horizontal), pipe's surface is very smooth, length is very short so the viscous losses are very small.

In your regulator example, what does the output pressure mean?