What is the maximum operating capacity of the fan motor? !
Thread Content
The figure shows the characteristic curve of a typical fan. The power of the motor connected to the fan is N = Q (air volume) * P (air pressure). In fact, since the fan is integrated with the piping network, the intersection point of the fan’s characteristic curve and that of the piping network constitutes the fan’s operating point. 1. The greater the pipeline resistance, the smaller the air volume, and the lower the current in the motor driving the fan. So, question (1): For a regular motor, when the pipeline resistance is high and the air volume is low, does the motor’s power decrease at this time? ! Because, N=V (voltage) * I (current); the voltage remains constant (does the voltage change?) ), the current decreased. If the power decreases, doesn’t that contradict N=Q (air volume) * P (air pressure)? ! 2. When using an inverter-driven motor, it is generally indicated that the constant-torque frequency range is 5–50 Hz, while the constant-power range is 50–100 Hz. Under normal operation, within the range of 5-50HZ, the current flowing through the motor is relatively low; when the frequency is increased to 50-100HZ, the motor current rises and the air volume increases. So, question (2): For piping systems with high resistance, does the motor not reach its full output? ! Question (3): Can the motor generate greater force by increasing the frequency, as long as it does not exceed its rated current? ! Because, in reality, the rated power of a motor represents the safe operating range; its maximum capacity to perform work exceeds that rated power, but this will result in burning out of the motor. I seek advice from experts! @arpcd
"The current has decreased. If the power decreases, doesn’t that contradict N=Q (air volume) * P (air pressure)? "I don’t see any contradiction here; a low current corresponds to a low air volume, so there’s no conflict. When the rated air volume is achieved, the motor is operating at full capacity. Increasing the load further will raise the current, which may damage the motor. It is about increasing the value that meets the requirements, not reducing the resistance of the pipelines. Some electric fans or water pumps, due to their size and properties, do not experience a continuous increase in load; as a result, their motors cannot be damaged
For example, the resistance is 1000 Pa and the air volume is 2000 CMH. If the resistance becomes 2000 Pa, the air volume will become 1000 CMH (in reality it won’t be that much); since N=Q*P, the power remains unchanged! But in reality, the current has decreased! N=V*I; the voltage remains unchanged, the current decreases, and thus the power also decreases! One remains unchanged, while the other has become smaller – there’s a contradiction!
“For example, the resistance is 1000 Pa and the air volume is 2000 CMH. If the resistance becomes 2000 Pa, the air volume will become 1000 CMH (in reality it won’t be that much); since N=Q*P, the power remains unchanged! ” This algorithm is questionable; P represents the head of the fan, which is almost a constant, and it is not the same value as what is measured by the pressure gauge at the outlet. The pressure difference when the outlet valve is opened to different degrees is very small; moreover, these values of 1000 and 2000 are not the same parameter, so the calculation is incorrect
That’s not the right way to put it, is it? If the resistances are different, is the work done by the fan exactly the same?
Let professionals handle professional tasks; this post would be more appropriate in the section dedicated to rotating equipment or pump systems. . . Personally, I think the original poster doesn’t lack actual on-site measurement data; it’s also easy to ask the manufacturer for the performance curves – resistance 1, flow rate 2, resistance 2, flow rate 1? What is this of yours? . . . Basic skills. . . :lol
Okay, thank you, expert. .