Why does the current of the fan decrease when the pipeline resistance increases? !
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As the pipeline resistance increases, the fan current decreases. According to N=U*I, as the current decreases, the motor power also decreases. However, the resistance of the fan increases and the air volume decreases, but its work done does not decrease. Is static pressure SP2/SP1=（rpm2/rpm1）^2 and breaking horsepower BHP2/BHP1=（rpm2/rpm1）^3 an explanation of these two formulas? I would appreciate your advice. Thank you. . .
Using I=U/R, resistance and flow rate are inversely proportional.
The power output by the motor is converted into the kinetic energy and pressure of the wind. The increase in pressure results in a small increase in energy, but the kinetic energy of the wind decreases significantly; as a result, the motor’s output power drops!
The dynamic pressure isn’t that high actually; a wind speed of 10 meters results in just over 100 Pascals, so it’s the static pressure that plays the main role. N = Q * P.
As the operating point shifts to the left, the current naturally decreases. The intersection of the two curves represents their actual operating point.
How can it be explained from the perspective of work done?
Put simply, an increase in pipeline resistance raises the wind pressure; higher wind pressure increases the load on the fan. However, an increase in resistance also leads to a decrease in flow rate, and a reduced flow rate lowers the fan’s load. Moreover, the reduction in load caused by lower flow rate is greater than the increase in load caused by higher pressure, so the overall load decreases. It’s so hard to say!
From the formula for work done: work done equals wind pressure multiplied by air volume. As the resistance in the ductwork increases, wind pressure rises while the air volume decreases. However, the decrease in air volume is greater than the increase in wind pressure, resulting in a lower total value when these two values are multiplied together. This is the same principle as that of centrifugal fans and centrifugal pumps. It is also the most significant characteristic that distinguishes it from positive-displacement machines and pumps.
I see, I understand now. Thank you, expert.
So can I use an inverter to increase the frequency and thus the current, when the resistance increases, in order to boost the air volume?