Thread Content
A high-pressure blower is used as the sealing air; the blower has a pressure of 13 kPa, a flow rate of 3.7 cubic meters per minute, and a power rating of 1.3 kW. However, after installation it kept tripping frequently, and the reason for the tripping was the operation of the thermal relay. I installed a pressure gauge behind the three-way valve, and it showed only 7 kPa. The manufacturer said that the pressure of the device without any additional components is 18 kPa – how can there be such a large loss in pressure? Is it because the tee is too close to the outlet? How should this be handled?
The diameter of the fan outlet pipe has been changed, from 50 to 40
It is recommended to check whether the overcurrent setting of the thermal relay is too low; The pressure gauge has an excessive range, resulting in measurement errors ; Open the outlet fully without applying pressure and check whether the ammeter shows an overcurrent, using a clamp ammeter for measurement. Also, it is necessary to check whether there are any blockages in the pipelines, and whether the valves and stopcock switches are in the fully open position. It’s best to view the results of operation with the terminal fully open; if everything is normal, gradually increase the pressure. Give it a try and wait for a reply.
Where does the signal source for measuring the thermal relay come from?
Thermal relay activation indicates that the current value exceeds the thermal relay’s set value (the thermal relay’s rated current is used to protect the motor, and this value is determined based on the motor’s settings). If the current is too high, it means the motor is operating beyond its limits. Try reducing the air intake volume of the fan (the outlet pressure gauge shows that the outlet pressure isn’t high); see if the current level drops. If the current remains within normal limits, the thermal relay will not activate ; Also, the pressure gauge shows static pressure; it’s possible that the dynamic pressure is high, which leads to a higher load and thus a higher current. It’s just my own analysis; I’m not sure if it will be useful.