Is this the correct way to calculate for a differential pressure transmitter?
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There is a pneumatic differential pressure transmitter with a measurement range of 2500 Pa; the corresponding maximum flow rate is 50 t/h. The process requirement is that an alarm be triggered at a flow rate of 40 t/h. Question: (1) What is the transmitter differential pressure without a square rooter? (2) What is the transmitter differential pressure with a square rooter? Solution: 1. Without a square root function, the differential pressure corresponding to a flow rate of 40 t/h is ΔP1=2500×(40/50)²=1600 Pa. 2. With a square root function, since ΔQ=K×ΔP, the differential pressure for a flow rate of 40 t/h is ΔP2=2500×(40/50)=2000 Pa. My understanding is that without the square root function, ΔP1=2500×(40/50)=2000 Pa; while with it, ΔP2=2500×(40/50)²=1600 Pa. Could someone please tell me which answer is correct?
We also often test this type of question, and the answer is the second one
The magnitude of the differential pressure is related to flow rate and has nothing to do with anything else. Whether to include a square root function depends on the calculation logic of the digital indicator, such as in DCS systems and similar ones
Could you explain why? :lol
Just remember that the differential pressure is proportional to the square of the flow rate
There is a pneumatic differential pressure transmitter with a measurement range of 25,000 Pa; the corresponding maximum flow rate is 50 t/h. The process requirement is that an alarm be triggered at a flow rate of 40 t/h. Question: (1) What should the alarm value be set at without a square root calculator? (2) When a square root calculator is used, what should the alarm value be set at? Solution: 1. Without a square root calculator, the differential pressure corresponding to a flow rate of 40 t/h is ΔP1 = 25000 × (40/50)² = 16000 Pa. The output corresponding to this flow rate is P_out1 = (16000/25000) × 80 + 20 = 71.2 KPa. Therefore, the alarm value S = 71.2 KPa. 2. With a square root calculator, since ΔQ = K × ΔP, the differential pressure corresponding to a flow rate of 40 t/h is ΔP2 = 25000 × (40/50) = 20000 Pa. The output corresponding to this flow rate is P_out2 = (20000/25000) × 80 + 20 = 84 KPa. Hence, the alarm value S = 84 KPa. What does *80 + 20 in the formula mean?
Does the measured differential pressure change with or without a square root calculator? Does your throttle device change depending on whether you have a square root function or not? :L
This post was last edited by jack_gl on 2016-2-17 at 11:46. I saw this problem in a *exercise set but never understood it, as I had never encountered an integrator that can change the differential voltage output; I’ve learned *now!
The origin of 80+20 in the formula is as follows: According to international standards, the transmission signal for pneumatic instruments is a pressure signal ranging from 20KPa to 100KPa. In other words, the lower range limit of the transmitter is 20 KPa, and the upper range limit is 100 KPa. The range of the transmitter is therefore 100 KPa – 20 KPa = 80 KPa; this is where the value 80 comes from. The zero point is 20 KPa, which is why we have the value 20.