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Overview

I am trying to find out what happens to the air flow within a pipe if a fan cannot generate the required pressure to overcome the friction within the pipe.

After much research I am aware that air moves from a point of high pressure to a point of low pressure. Also as the length of pipe increases, the pressure drop due to the friction of the pipe on the air also increases.

So as a working example, assume that I have a fan that has an airflow rating of 1 m3/s and generates a static pressure of 50 Pa. The fan and pipe have a diameter of 30cm and finally the pipe has a pressure drop of 7 Pa/m.


Situation 1

If the pipe has a length of 1 meter, then the air pressure at the end of the pipe would be:

$$50\, Pa - ( 7\, Pa\cdot 1 ) = 43\, Pa.$$

So the air flow would continue past the end of pipe into the surrounding environment.


Situation 2

If the pipe had a length of say 10 meters, then this image gives a clearer idea of what I'd imagine happening with the air pressure:

Piping

As each meter of pipe drops the air pressure by 7 Pa, then just after 7 meters of pipe the air pressure would be 0 - essentially neutral to the surrounding air.

And because air flow moves from a point of high pressure to a point of low pressure, then there would be no reason for air to flow within the pipe after 7 meters. Since the fan will still be generating 50 Pa, then air will still flow through the fan into the end of the pipe connected directly to the fan.

So, what happens then to the air that is already within the pipe but has no pressure difference to move and what happens to the air that continues being pulled into the pipe?

Clearly I'm not much up on fluid dynamics and physics, so would appreciate a simplistic answer to the question. What am I missing here?

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  • $\begingroup$ You realise that no matter what you do, air is going to come out the end of your pipe, yes? Why do you think pressure drop is linear? $\endgroup$ – Rory Alsop Jun 23 '16 at 22:30
  • $\begingroup$ Oh I'm not trying to not get air out, I'm just struggling to understand why it would. The pressure drop being linear I got from a chart such as this one (arca53.dsl.pipex.com/index_files/619.gif) that states the pressure drop of a pipe per meter (with an assumption of a constant friction factor). $\endgroup$ – R4D4 Jun 23 '16 at 22:34
  • $\begingroup$ I read from that link that pressure drop per unit length is not a static number, but depends on flow volume. As the flow drops, the pressure drop per length does as well. $\endgroup$ – BowlOfRed Jun 23 '16 at 23:58
  • $\begingroup$ @R4D4 , the fan will produce a pressure on one end of the pipe. Assuming that the other end of the pipe is at ambient pressure, the pressure drop across the pipe (of whatever length) will be a known number. The flow rate will then depend on the length of the pipe and the pressure drop per meter. In other words, the flow rate will vary depending on the specifics of your problem, and it will match the conditions that you have. $\endgroup$ – David White Jun 24 '16 at 1:58
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The pressure drop shown in the pipe in situation 2 is based on an assumed flow rate, but that flow rate can't be achieved when the pressure is only 50 Pa. You cannot deliver a fixed flow rate and a fixed static pressure: when the flow rate is kept constant, like produced by volumetric pumps, the pressure will vary with resistance: if you block the exit the pressure will rise until something breaks (often the pump itself). If you keep the pressure constant at 50 Pa, with a supply of compressed air for example, the flow rate in the pipe will be the one corresponding to a pressure drop of 5 Pa/m.

For a fan, neither flow rate nor static pressure is fixed: the flow rate depends on the static pressure, which itself is determined by the resistance the air flow encounters at the outlet. Without a load (ie nothing connected to the outlet) the static pressure is zero, because you can't build up pressure in open air. This also applies to the end of the pipe in situation 1: the pressure at the exit is zero, not 43 Pa, and the static pressure at the outlet of the fan will be 7 Pa. Which is too low for a flow rate of 1m³/s (assuming the fan can deliver that rate at 50 Pa): the flow rate will increase, causing the static pressure to rise due to increased resistance in the pipe. Eventually the system will reach a new equilibrium with $Q>1m³/s$ and $P_s<50Pa$.

(note btw that fan specifications can be rather misleading: unless they explicitly state otherwise, the flow rate and static pressure values listed are usually maximum values: the flow rate measured at zero static pressure, and the static pressure measured at zero flow rate)

For most calculations one needs the operating curve (static pressure versus flow rate) of the fan, or the fan performance tables (listing flow rates for a range of static pressure values)

operating curve of a fan, with resistance curves of pipes

The green line shows the operating curve of a (hypothetical) fan with flow rate of 1m³/s at 50 Pa static pressure. The red and blue lines are the resistance curves of 30 cm pipes with lengths of respectively 10 m and 7.14 m. (based on http://www.arca53.dsl.pipex.com/index_files/619.gif). The intersection with the operating curve gives you the operating point of the system.

For the 10 m pipe, let's assume that the flow rate in the system is 0.8 m³/s. Then pressure drop in the pipe is 50 Pa, and since pressure at the pipe end is zero, the static pressure at the fan outlet is also 50 Pa. At that pressure, the fan can deliver a flow of 1 m³/s, so the flow will increase. Conversely, when we assume an initial flow rate to the right of the intersection point, the pressure required is more than fan can deliver at the given flow rate, so the flow will decrease. In both cases, the system will end up at the intersection point.

Notice that in the region between 0 and 0.5m³/s, the static pressure rises with increasing flow rate. This can potentially cause unstable operation, and fans aren't (or shouldn't be) run in that region. The shape of the curve shown is typical for centrifugal fans with backward curved blades, other types (centrifugal with forward curved blades, axial fans, propeller blades) can have an s-shaped curve (first down, then up, then down again).

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  • $\begingroup$ Thank you! What resources and keywords should I be looking at to understand this better? $\endgroup$ – R4D4 Jun 28 '16 at 12:32
  • $\begingroup$ Actually, regarding the last paragraph: I'm not seeing how it is even possible to operate the fan in that region? Also, did you draw those graphs or is there a nice tool/website to create them? $\endgroup$ – R4D4 Jun 28 '16 at 12:58
  • $\begingroup$ @R4D4 for ideal flow, Bernoulli equation is important, for viscous flow (with friction losses) Reynold's number, laminar and turbulent flow. For real systems, maybe this HVAC system design guide, chapter 5 (page 95): waterandfire.ir/Down_En/HVAC_Systems_Design_Handbook.pdf Examples and calculations of ventilation systems may be helpful. Pipe flow examples, formulas: web2.clarkson.edu/projects/subramanian/ch330/notes/… $\endgroup$ – Previous Jun 29 '16 at 1:24
  • $\begingroup$ @R4D4 If you restrict the outlet enough, a fan will operate in that region (or when blocking the inlet, a vacuum cleaner for example). It's mainly the air flow around the blades that becomes unstable aerovent.com/docs/fan-engineering-topics/… I made the graphs in a paint program, only realised later I could have put the values in a spreadsheet and let it draw the curves, would have been easier. $\endgroup$ – Previous Jun 29 '16 at 2:25
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At that exact moment, when there is no pressure drop at point A, air will not move at point A. But just before that point there is a point with a slightly higher pressure, so air (and pressure!) will move from there towards point A.

The first comment was "why do you presume pressure drop is linear". Id rather ask "why do you presume the system is static?". It will not be until the system has reached equilibrium, when pressure is even in the whole pipe.

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