When a variable-frequency motor is used in the fan, how does the air pressure change as the speed increases?
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I am designing the piping system for an industrial furnace. Since the air volume that needs to be regulated by the combustion fan varies over a wide range, variable frequency speed control is being used. If the motor’s rotation speed increases from 1450 revolutions per minute to 2900 revolutions per minute, will the pressure of the fan increase or decrease? Is there a formula that can be used to calculate the fan’s pressure? Has anyone used this approach before?
Regarding the use of variable frequency speed control for the combustion air fans, although we have not measured the air pressure at low speeds, we believe that: 1. The air pressure of the fan depends on the back pressure; if the output cross-section is larger than the input cross-section, then the pressure cannot increase. 2. If the inlet and outlet cross-sections are fixed, a reduction in the input volume will inevitably lead to a decrease in the back pressure
Systems that use frequency converters to adjust the fan speed and thereby control the air volume generally come in two types: constant power and constant torque. If it is at a constant power level, as the speed increases, the air volume necessarily increases while the air pressure decreases. If it is a constant-torque type, as the speed increases, the air volume rises, and the air pressure also increases (for the same fan).
If, in a piping system, two fans are to be used simultaneously to increase the air volume, how can I ensure that both fans function properly when they have different capacities? If the pressures of the two fans differ, could the air from the fan with higher pressure flow back towards the fan with lower pressure, causing the latter to not only fail to deliver air but also to experience reverse airflow? How can I ensure that the outlet pressures of the two fans are exactly equal?
I also want to ask whether the air pressure will decrease after using the frequency converter?
For centrifugal fans, the total pressure of the fan is proportional to the square of the rotational speed, while the shaft power of the fan is proportional to the cube of the rotational speed. If the fan speed is doubled, the total pressure will become 4 times the original value, and the shaft power will become 8 times the original value. If variable frequency is used to adjust the air volume, the motor power must be specified based on the power at the maximum speed; otherwise, the motor will be damaged. Furthermore, when making significant adjustments to the air volume, it is also necessary to consider the stable range of the fan’s characteristic curve; otherwise, the fan may enter its surge zone and experience severe vibrations.
The wind pressure needs to be increased. The calculation formula is: P2 = (2900/1450)^2 × P1 = 4 × P1; in other words, the wind pressure should be four times the original value.
How is the relationship between wind pressure, shaft power, and rotational speed determined?
The fans in our company are equipped with variable frequency technology; this depends on the control mode used. 1. When pressure control is employed, the valve closes slightly (for a short period) when pressure increases, and then it gradually opens again based on the required air volume. 2. With volume control, pressure decreases when air is needed, and then it rises step by step to meet the load requirements.
It seems that for the wind pressure and shaft power to be proportional to the square and cube of the rotational speed, the rotational speed must remain within ±20% of its value in order for the velocity triangle to be satisfied (at least this is the case for centrifugal pumps; it’s not clear whether centrifugal fans also meet this condition). If the rotational speed changes too much, it seems that this proportional relationship is no longer maintained