Relationship between system pressure drop and production capacity
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I would like to ask the experts: 1. With the fan speed kept constant, if the pressure drop is reduced by 1000 Pa, by how much will the air volume increase? How much can the sulfuric acid production capacity be increased (assuming there is sufficient headroom in the system)? Please provide the calculation method. 2. With the air volume kept constant, if the pressure drop is reduced by 1000 Pa, how much energy can the fan save? --Request for calculation formula sheet (are the formulas for motors and turbines the same?) ）
This post was last edited by Shang Xiaoke on 2016-5-5 at 15:51. 1. (V1/V2)2 = H1/H2 + 1000 Pa. V1 and H1 represent the pressure at the fan outlet corresponding to the current air volume. In a sulfuric acid production system, 2300–2400 m3 of air is required per ton of acid produced. The additional air volume divided by the air volume needed per ton of acid gives the increase in production volume; the actual increase in production shall be determined based on statistics. 2. With the air volume unchanged, the pressure drop needs to be reduced by 1000 Pa. The pressure drop of the system equipment should have decreased. The work done by the motor itself to overcome resistance decreases, and the electricity consumption can be calculated based on the current. Current multiplied by voltage then multiplied by time, plus square root of 3, plus efficiency, plus power factor.
A relevant link is as follows: http://bbs.hcbbs.com/thread-1680372-1-1.html