Pressure and flow

votes1

In the case of centrifugal pumps, as pressure decreases flow increases and cice verse.

In the case Positive Displacement pumps Flow and pressure are independant of each other. Hence you can get various flow rates at specific pressures by adjusting the speed of the motor.

votes0

Hi,

Flow expresses the volume of a liquid moved in a given unit of time.

Example: a pump pushes 10 m3 of water per hour through a pipe.

Pressure is a force applied by water to a surface.

Example: in your home, tap pressure is determined by the difference in height between the water stored in your utility’s tank and the elevation height. Pressure is measured in bars.

votes0

In order to determine the relationship between pressure and flow, you need to follow the Bernoulli equation that links the three types of energy—pressure, kinetic and potential—from a point of departure to a point of arrival. By omitting energy pressure, therefore calculating in terms of absolute energy, you get the following equation: V²/2g +Z, where Z is pressure and Q=V.S so V=Q/S will give you P=Q²/2gS²+Z

votes0

Hi,

In order for everyone to understand (including me):

Q = flow rate

V = velocity

S = surface

Z = potential energy

P = pressure

g = gravity or approx. 1kg

Thanks.

votes0

For my part I would simply answer that the pressure decreases as your flow increases.

If you put a gauge on a pipe with the valve closed, your pressure level is the static pressure. It's the maximum pressure.

If you open the valve a little and create a flow, the pressure on the gauge decreases a little.

If the valve is wide open, there's a big draw and the pressure decreases substantially.

Is this the type of answer you were looking for?

I suggest you use a nomograph to obtain the calculations for head loss (and therefore pressure) according to the flow and other parameters. It's less complicated than the formulas and works just as well.

votes0

Hi,

I don't know what your level of knowledge is regarding hydraulics/fluid mechanics, but simply put, in order to understand the relationship between flow and pressure, we need to go back to the concept of "potential energy" or "head" found in Bernoulli's equation:

H = z + P/(Ro*g) + V²/(2g), where H represents the head in height (m) of water column. That is to say the energy of the water.

The term Z represents the height (m).

P/(Ro*g) means the pressure expressed in m of water column (P is the pressure in pascals, Ro the density of water in kg/m3, and g the acceleration due to gravity. (As mentioned previously, the pressure is the force exerted by the water on the walls of the pipe.)

The term v²/(2g) represents the kinetic energy of water expressed in meters of column water.

So, the head (the energy expressed in meters water column) is the sum of the pressure and kinetic energy.

Therefore, under a constant head, if you increase the flow rate (and thus the speed in your pipe), the kinetic energy increases and so the pressure decreases and vice versa. That's what cd2016 is getting at in the answer to your query.

NB. I am deliberately avoiding head loss so as not to complicate matters.

votes0

Hi,

I don't know what your level of knowledge is regarding hydraulics/fluid mechanics, but simply put, in order to understand the relationship between flow and pressure, we need to go back to the concept of "potential energy" or "head" found in Bernoulli's equation:

H = z + P/(Ro*g) + V²/(2g), where H represents the head in height (m) of water column. That is to say the energy of the water.

The term Z represents the height (m).

P/(Ro*g) means the pressure expressed in m of water column (P is the pressure in pascals, Ro the density of water in kg/m3, and g the acceleration due to gravity. (As mentioned previously, the pressure is the force exerted by the water on the walls of the pipe.)

The term v²/(2g) represents the kinetic energy of water expressed in meters of column water.

So, the head (the energy expressed in meters water column) is the sum of the pressure and kinetic energy.

Therefore, under a constant head, if you increase the flow rate (and thus the speed in your pipe), the kinetic energy increases and so the pressure decreases and vice versa. That's what cd2016 is getting at in the answer to your query.

NB. I am deliberately avoiding head loss so as not to complicate matters.

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