Is the total pressure inside a negative-pressure chimney generally greater than atmospheric pressure?

Thread Content

This post was last edited by *nshiji168 on 2009-12-2 at 12:51. Today, a new differential pressure transmitter was installed; its range is 0–6.22 Kpa, and it is used for measuring flow rate. It operates on the principle of a pitot tube, with the static pressure connected to terminal L and the total pressure to terminal H. Both L and N are equipped with solenoid valves for backwashing, but the backwashing air supply is turned off, meaning backwashing is not in use. After installation, it was found that the current was below zero. It was later discovered that the backflow solenoid valve connected to terminal L had some air leakage, which reduced the negative pressure at terminal L (for example, if it is -400 PA under normal conditions, the transmitter at terminal L might only detect -100 PA). During the processing, the backflow solenoid valve at the L end was removed, allowing that end to be in direct contact with the atmosphere; the current remained below 4 mA. It can be inferred that the H end, which is used for measuring the total pressure, must also be under negative pressure (for example, -200 PA). Therefore, it is likely that the current was below 4 mA right after the transmitter was installed, as H (-200) – L (-100) = -100 PA, and since this value is lower than the default zero point of 0 PA, the current ended up being below 4 MA. I’m not sure if this inference makes sense There was no other choice; I didn’t have a backflow solenoid valve available. Since the receiving party was going to conduct an inspection, my supervisor suggested that I use fake values. So I connected the L end directly to the atmosphere, while keeping the H end in a state where it measured the total pressure. I adjusted the transmitter’s zero point, and then connected the static pressure (which, due to air leakage, probably wasn’t the accurate static pressure inside the chimney) to the L end. As a result, the current exceeded 4 mA, and the computer displayed the flow rate. For now, I’ll just use this setup. I used to think that the total pressure inside the chimney was higher than atmospheric pressure, but when I connected the L end directly to the atmosphere and the H end to the total pressure, the pressure difference was negative. I wonder if my approach and analysis above make sense?