Is the solution to this problem correct? !
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This post was last edited by Wang Tianze on 2016-10-25 at 00:58. There is a pneumatic differential pressure transmitter with a range of 20 KPa, corresponding to a flow rate of 40 t/h. What is the output of this transmitter when the flow rate is 30 t/h? Answer: Since there is a square root relationship between flow rate and differential pressure, to determine the differential pressure from the known flow rate, squaring is required: (30/40)²×100% = 56.25%. The differential pressure output signal is: 56.25%×80 + 20 = 65 Kpa
There is a pneumatic differential pressure transmitter with a range of 20 KPa, corresponding to a flow rate of 40 t/h. What is the output of this transmitter when the flow rate is 30 t/h? Answer: Since there is a square root relationship between flow rate and differential pressure, to find the differential pressure given the flow rate, one uses the square of (30/40)²×100% = 56.25%. The differential pressure output signal is: 56.25%×80 + 20 = 65 Kpa.
The range is 20 KPA; getting a value of 65 KPA means it’s beyond the range, right? I really can’t understand the questions in the question bank
I think it’s (Q_current/Q_MAX) squared = P_current/PMAX; (30/40) squared = P/20. So P=11.25, right? Please point out any mistakes. Thank you
Did he ask you about the output current of the regulator?
Yes, it must be 11.25kPA
This question is very simple: the square of Q is proportional to the pressure difference!
This post was last edited by YuYuDeYuYu on 2016-10-25 at 09:37. Instrument Question Bank: There is a pneumatic differential pressure transmitter with a range of 20 KPa, corresponding to a flow rate of 40 t/h. When the flow rate is 30 t/h, what will be the output of the differential pressure transmitter? Answer: Since flow rate and differential pressure are related by a square root relationship, to find the differential pressure given the flow rate, one needs to apply the square operation: (30/40)² × 100% = 56.25%. The differential pressure output signal is then 56.25% × 80 + 20 = 65 Kpa. There’s something wrong with this question; I don’t understand what the questioner is trying to express. Thank you
It seems that he must have made a mistake; if the calculated output current is also incorrect, then his second calculation is completely unreadable! Think about it carefully again
(30/40)^2=(p-20)/100-20; p=80×(30/40)^2+20=64.80