Ducted Fan Pressure Differential: Questions Answered

[bartrocs]

Hi guys, I recently completed a fluid dynamics practical at Uni and submitted a report on it, however, I still have some burning questions which I need to be answered.

Firstly, the data that we collected shows that as the flow rate in the duct increases, the pressure differential of the fan decreases. My understanding of this is through Bernoulli, a higher flow rate implies a faster velocity (Area is constant) which implies a higher dynamic pressure. As total pressure must remain constant, this must mean that the static pressure has gone down. Am I right in this reasoning? I just don't know if it's okay to use a streamline which goes through a fan for bernoulli.

I guess I am looking for a complete explanation on what happens to the different pressures (dynamic, static, total) as the air moves through the duct and especially through the fan.

Here is a link to our prac description:

 

[rcgldr]

Firstly, the data that we collected shows that as the flow rate in the duct increases, the pressure differential of the fan decreases. My understanding of this is through Bernoulli, a higher flow rate implies a faster velocity (Area is constant) which implies a higher dynamic pressure. As total pressure must remain constant, this must mean that the static pressure has gone down. Am I right in this reasoning?

Ignoring resistance of the ducting, Bernoulli applies to the flow before and after the area swept out by the fan, but not to the flow across the fan, because the fan performs work on the air, mostly increasing the pressure with little change in speed. The air accelerates as it's pressure decreases from ambient to below ambient as it approaches the fan intake, and accelerates again after it passes through the fan as it's pressure decreases from above ambient back to ambient. The pressure differential times the swept area essentially equals the force that the fan applies to the air, and this force times the speed of the flow at the fan equals the power exerted onto the air.

As far as flow rate versus pressure differential, that would be a limitation of a specific fan and also related to the size and effective resistance to flow of the duct tube. A different fan designed to operate at higher power levels and higher speeds could achieve the same pressure differential at a higher flow rate than the fan you used for testing. There are limitations to this due to resistance to flow by the duct tube, and even in the case of a large tube or otherwise unrestricted flow, other issues come into play at very high flow rates, starting around mach 0.3 to mach 0.5 depending on the circumstances.

 

[bartrocs]

I understand this, but what is the mechanism which causes the flow rate to be lower if the outlet area is smaller? Is it because the static pressure increases on the outlet side of the fan, so it has to push harder in order to overcome the pressure gradient, so if the fan is not powerful enough, the flow rate decreases?

 

[russ_watters]

Bernoulli's principle is, first and foremost, a conservation of energy statement. Is energy conserved in the situations you are describing?

I understand this, but what is the mechanism which causes the flow rate to be lower if the outlet area is smaller? Is it because the static pressure increases on the outlet side of the fan, so it has to push harder in order to overcome the pressure gradient, so if the fan is not powerful enough, the flow rate decreases?

That's exactly correct. So, why is that? Why doesn't Bernoulli's equation apply in the way you expected?

 

[bartrocs]

Is it because the fan does work on the flow, which is not taken into account in Bernoulli's equation?

 

[russ_watters]

Is it because the fan does work on the flow, which is not taken into account in Bernoulli's equation?

No. Bernoulli's principle (equation) is a conservation of energy statement and as with any conservation of energy statement it can easily be adapted to include other energy inputs and outputs. Indeed, it says the energy is constant -- at what value? At the value that the fan provided!

The problem is that Bernoulli's equation applies to a single flow situation at a time. It is conservation of energy along the duct for a single snapshot in time or set of conditions. When you change the setup, you need to re-do Bernoulli's equation to describe the new situation. That's why your experiment involved calculating different flow coefficients for different setups.

Or, to put it another way involving the fan: when you change the duct configuration or the rpm of the fan, you change the amount of energy input by the fan.

 

[russ_watters]

...the fan performs work on the air, mostly increasing the pressure with little change in speed. The air accelerates as it's pressure decreases from ambient to below ambient as it approaches the fan intake, and accelerates again after it passes through the fan as it's pressure decreases from above ambient back to ambient.

The first part was right ("little change in speed" = essentially no acceleration), but the second part contradicts it and is therefore not. If the flows and pressures are low -- which pretty much by definition must be true if you call it a "fan", the air is not compressible and there is no acceleration. The only place it accelerates or decelerates is where it enters and leaves the duct: once inside, continuity requires a constant flow rate/speed.

 

[bartrocs]

Ah, I am sort of beginning to understand. So say we keep the power going to the fan constant. Now we start with a fully open outlet and take our readings. Then we reduce the outlet diameter, and take a reading again .The fan is providing a constant energy to the flow, so the flow rate of the smaller outlet case is less because the static pressure across the fan is greater, which means that less of the fan's energy can go into the dynamic pressure, so the velocity that it imparts onto the fluid is less. Am I correct?

 

[russ_watters]

Ah, I am sort of beginning to understand. So say we keep the power going to the fan constant. Now we start with a fully open outlet and take our readings. Then we reduce the outlet diameter, and take a reading again .The fan is providing a constant energy to the flow, so the flow rate of the smaller outlet case is less because the static pressure across the fan is greater, which means that less of the fan's energy can go into the dynamic pressure, so the velocity that it imparts onto the fluid is less. Am I correct?

In practice, getting the fan to output the same power is not a straightforward task: it requires speed control tied to the watt-meter. But yes, if the power is the same (and the efficiency of the fan is the same...) and the static pressure is higher because of a restriction, the flow must be lower.

 

[rcgldr]

The only place it accelerates or decelerates is where it enters and leaves the duct: once inside, continuity requires a constant flow rate/speed.

True, in the situation of a constant diameter duct, the acceleration only occurs outside the duct (some turbulent flow at entry and exit points). The reference to "smaller outlet area" led me to think that one of the experiments involved a tapered duct, in which case the velocity of the air does increase as it flows through the duct, similar to the ducted fan setups used on radio control models.

 

[russ_watters]

True, in the situation of a constant diameter duct, the acceleration only occurs outside the duct (some turbulent flow at entry and exit points). The reference to "smaller outlet area" led me to think that one of the experiments involved a tapered duct, in which case the velocity of the air does increase as it flows through the duct, similar to the ducted fan setups used on radio control models.

The experiment involves some sort of variable outlet orifice or outlet venturi, so clearly there can be acceleration there if it is constricted. But the rest of the duct looked pretty much straight and you seemed to be saying there is acceleration and deceleration through the duct as the pressure changes. In a real straight duct, pressure is lost to friction, but there is no acceleration or deceleration.

 

[bartrocs]

just as a question that leads on from this, what is the mechanism by which the static pressure increases? the pressure outside the duct is atmospheric so what property of air/what physical phenomena allows for the static pressure within the duct to increase?

 

[russ_watters]

just as a question that leads on from this, what is the mechanism by which the static pressure increases? the pressure outside the duct is atmospheric so what property of air/what physical phenomena allows for the static pressure within the duct to increase?

Increase where, exactly, in response to what?

 

[bartrocs]

On the outlet side of the duct, when the outlet is narrowed.

 

[russ_watters]

On the outlet side of the duct, when the outlet is narrowed.

Narrowing the outlet forces the air to move faster, which increases the frictional losses in the duct and velocity pressure of the airstream (which is lost when the air exits the duct). Now, the fan can't maintain the same airflow in the face of more resistance, so the airflow drops and a new balance of airflow and pressure is established.

 

[rcgldr]

just as a question that leads on from this, what is the mechanism by which the static pressure increases? the pressure outside the duct is atmospheric so what property of air/what physical phenomena allows for the static pressure within the duct to increase?

The fan performs work on the air. It creates a low pressure zone on the intake side of the fan and a high pressure zone on the output side of the fan. This results in a lower than ambient pressure on the intake side of the duct, and the air accelerates and decreases in pressure as it approaches the duct. On the output side of the fan, the higher pressure air accelerates and decreases in pressure after it exits the duct.