About PVC drying radiator resistance and blower air volume
Thread Content
My unit uses PVC drying radiators with a single area of 1,000 square meters and a single-piece resistance of 500 PA, causing the air volume of the blower to be too low. Should we increase the air pressure of the blower or reduce the resistance of the heat sink? This post was last edited by edc1233 on 2008-9-16 16:50 ]
The resistance of the radiator is normal. It may be that the resistance of the entire system is relatively large. It is recommended to increase the fan pressure. There is an empirical formula for calculating the resistance of the radiator. It is related to the structure and cross-sectional weight and flow rate.
The same problems as yours have also occurred in my unit. The radiator resistance is large, the air volume is insufficient, and there is accumulation of material in the air flow tower, which affects the product quality and contains many impurities. We negotiated with the fan manufacturer and made a new impeller to increase the air volume while keeping the original equipment unchanged. If the heat sink is reduced, there may be insufficient heat and wet material will be produced. Because in normal production, the heat sink will leak internally due to stress corrosion of steam and other reasons. If it cannot be dismantled at this time, it has to be closed and waited for overhaul. Therefore, the heat sink cannot be reduced. This post was last written by ysy_Edited by jing at 2008-5-18 10:30 ]
Thank you for the guidance on the 2nd and 3rd floors. Where can I find the empirical formula for calculating radiator resistance? To make a new impeller without changing the original equipment, the only choice is to reduce the wind pressure and increase the air volume, or to increase the wind pressure while keeping the air volume as low as possible.
Do not move the radiator. The radiator has a small cross-sectional area, high air flow velocity, good heat exchange effect, and fast steam condensation.； The calculation formula of radiator resistance is based on standard conditions and may not be accurate. Solution: Replace the fan motor, e.g.: The motor speed increases to 1.2 times the original, then: Air volume increased to 1.2 times ； All-in pressure increased to 1.44x (1.2 squared) ； The power increases to 1.728 times (i.e. 1.2 cubed)