# Conversion between Power and Flow rate of Fan taking into consideration the geometry of the fan

I need to find the relation between two properties of a fan, that is between power and flow-rate. I expect such a formula to depend on the pressure and on the geometry of the fan.

I have found in here: https://www.axair-fans.co.uk/news/applications/understanding-basic-fan-laws/ a relation between rotational velocity of a fan with each of pressure, power and flow rate. Then following those relations, a relation between power and flow rate could be build easily. But this relation is unuseful since I don't need to convert between two states of each same property.

By the way, the kind of fan is one with slots or holes. I just need the method of building such a relation no matter how the fan looks like then I could apply that to the fan I have.

Thank you

• Cross posted here : engineering.stackexchange.com/q/29381/10902 – user207455 Jul 29 '19 at 7:08
• @SolarMike Hi. just to make sure about the rules, is it ok to put my question in more than one site? – user65035 Jul 29 '19 at 7:11
• It is, or seems to be, considered good manners to inform people that the question is cross posted so people don't waste effort on answers that are duplicates through no fault of their own. But you just deleted the post on the Engineering stack... – user207455 Jul 29 '19 at 7:13
• I haven't thought about that so thanks for letting me know. As the question is related to both fields of Physics and engineering, so I have decided to keep it here. – user65035 Jul 29 '19 at 7:25
• So, further to my other comment you could check out the dimensionless groups - many textbooks cover those. – user207455 Jul 29 '19 at 7:26

I calculated the flow parameters through the oval opening of the air purifier Dyson pure cool . I wondered how the air flow is created through the opening when blowing a jet through a thin slot around the perimeter of the oval. I compiled a mathematical model and wrote code for Mathematica 12. Figure 1 shows the flow lines and the magnitude of speed when blowing jets on the perimeter of the oval (shows one section across the oval).

The second question that interested me was how the stream penetrates the filter. Here is a simple example where a stream is filtered through a thin channel system. Figure 2 shows the magnitude of velocity (left), the velocity profile at the outlet (in the center) and the longitudinal component of velocity in a flow behind the filter.

Finally, the third question is how air is distributed through the ventilation system. Fig. 3 shows a ventilation duct with several outputs: at the top, the distribution of velocity magnitude is shown, in the center these are velocity profiles in each branch, and below is the pressure distribution in several sections.

To maintain air circulation in the system, it is necessary to install a fan of a certain power, which is able to develop the required pressure and provide the required air flow. The standard formula relating fan power ($$P_e$$), air flow rate ($$q_v$$), pressure drop ($$P_r$$) and system efficiency ($$\eta$$) is $$P_e=\frac {q_v P_r}{\eta}$$ The international standard uses specific fan power (SFP): $$SFP=\frac {\sum P_e}{q_v}$$ The efficiency of converting electrical power to pressure and air flow rate depends on the type of fan: $$\eta =\frac {P_{out}}{P_{in}}=\eta _{vsd} \eta _{motor} \eta _{transf} \eta _{fan}$$

$$P_{out}$$ is useful output of the fan in [W]=$$\Delta P_rq_v$$, $$\Delta P_r$$ is pressure rise in [Pa], $$q_v$$ is flow rate in $$m^3/s$$;

$$P_{in}$$ is power input [W]; $$\eta _{vsd}$$ is variable speed drive efficiency; $$\eta _{motor}$$ is motor efficiency; $$\eta _{transf}$$ is power transfer efficiency; $$\eta _{fan}$$ fan aerodynamic efficiency.

Blade speed and pressure distribution (my calculations for viscous flow)

Fan efficiency depends on flow and diameter. In fig. Figure 5 shows the dependence of the efficiency of five types of fans on $$q_v$$ and diameter. In most cases, the pressure in the system decreases with increasing flow rate, but the efficiency increases, so the power increases with increasing flow rate. Typical dependencies are shown in Fig. 6. for a real fan (left) and in theory (right) The theory that I am developing is very simple. It is assumed that the pressure drop in the system is related to the flow rate by a quadratic dependence: $$\Delta p=p_0(1-aq_v^2)$$ In fig. 7 shows the dependence of fan pressure on flow rate for several types of fans. There are two types of system efficiency - with reaching a constant as in Fig. 5 or with a local maximum as in Fig. 6: $$\eta =\frac {kq_v}{1+bq_v+cq_v^2}$$ Then the electric power depends on the flow rate as $$P_e=\frac { q_v\Delta p}{\eta}=(p_0/k)(1-aq_v^2)(1+bq_v+cq_v^2)$$ Here the parameters $$a, b, c, p_0/k$$ depend on the geometry of the fan.This dependence is shown in Fig. 6. Qualitatively, this dependence is consistent with experiment.

• Thank you, but do you consider this as an answer to my question or more like an advertisement for the product? – user65035 Jul 31 '19 at 7:09
• @hellothere I did the calculations solely for scientific purposes, and not for advertising or business. If you have a question about the flow through the oval, then I can do the calculations. – Alex Trounev Jul 31 '19 at 8:00
• It doesn't answer my question at all. My question is to find a relation (like a formula) between power and flowrate of a fan. – user65035 Jul 31 '19 at 8:28
• @hellothere This is not a correctly posed problem. Each fan will have its own formula. Describe your fan or submit a photo. – Alex Trounev Jul 31 '19 at 8:56
• @hellothere you're welcome! This is an interesting topic, I will add experimental data and theory. – Alex Trounev Aug 6 '19 at 9:38

the power delivered by the fan blades to the air moving through it will be equal to the mass flow rate of air through the fan times the pressure difference across the fan. it will also be equal to the power delivered by the motor times the efficiency of the fan.

the power delivered by the motor to the fan will be equal to the torque developed by the motor times its RPM. it will also be equal to the power delivered by the AC mains times the efficiency of the motor.

the power delivered by the AC mains to the motor will be equal to the current flow through the motor times the mains voltage.

• Thank you. So the relation between power of the fan and its flow rate will not depend on the geometry of the fan ? And are these two expressions the same: power of fan, power delivered by the fan to the air? – user65035 Aug 2 '19 at 9:34
• all those things will sensitively depend on the details of the fan's geometry but they are hard to quantify experimentally or theoretically. they are generally lumped together into the efficiency number for the fan, where the more artful the design, the higher the efficiency. the fan power is what it absorbs from the motor. the power delivered to the air is less than this by the efficiency number. – niels nielsen Aug 2 '19 at 18:18

From the link you posted you can derive the equation $$\frac{P_2}{P_1}=\left(\frac{\dot V_2}{\dot V_1}\right)^3$$ where I have changed their notation slightly. I will use $$P$$ for power and $$p$$ for pressure, and $$\dot V$$ for the volumetric flow rate and $$V$$ for volume and $$v$$ for flow velocity, and $$A$$ for area.

As you said, you do not want to compare two points, so we can collect terms to get $$P=k \ \dot V^3$$ which is your desired relationship between power and volumetric flow rate.

Now, $$k$$ is a characteristic of the fan, and must be calculated or measured at some given power level for the fan. Ideally, a smaller $$k$$ will mean less power for a given volumetric flow rate, so we would like a fan that minimizes $$k$$. Generally this will be difficult to calculate theoretically, but it should be something that can be determined by the specifications for any given fan.

However, since you are specifically interested in the geometry, we can do a little rearranging to get a sense of how the geometry may affect things: $$k=\frac{P}{\dot V^3}=\frac{p\ \dot V}{\dot V^3}=\frac{p}{\dot V^2}=\frac{p}{(A\ v)^2}$$

So, to minimize $$k$$ you want to have a low pressure and high velocity over a large area. However, realistically pressure and velocity and area are all related, so you cannot usually just increase one without affecting the others. However, this does give you a general idea that increasing the area will reduce the power for a fixed flow rate.