Parallel type electromagnetic flowmeter with dual excitations
An electromagnetic flowmeter, double excitation technology, applied in the application of electromagnetic flowmeter to detect fluid flow, volume/mass flow generated by electromagnetic effects, etc., can solve the problem of dynamic monitoring of fluid conductivity or internal resistance. There is no ideal method, etc. question
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Embodiment 1
[0068] Embodiment one: see figure 1 , the parallel dual-excitation electromagnetic flowmeter includes a sensor 1 with an excitation magnetic field B to form an induced potential generated by fluid cutting magnetic force lines, a differential amplifier 3 with an input internal resistance R3 and a signal amplification factor K1, and a signal processing unit 4, the sensor 1 There are two measuring electrodes C and D for the fluid, the two measuring electrodes C and D are connected to the input of the differential amplifier 3, and the output U of the differential amplifier 3 is connected to the signal processing unit 4; the conductivity of the fluid makes the two electrodes C and D have a fluid internal resistance R1, and the electrodes C and D of the sensor 1 have an induced potential E1=K0×B×V at a fluid flow rate V, and K0 is a coefficient of the sensor (1); it is characterized in that an additional excitation 2 is connected in parallel with the sensor (1) two ends of electrode...
Embodiment 2
[0089] Embodiment 2: This embodiment is basically the same as the above-mentioned embodiment, the difference is: see Figure 4 and Figure 5 , the above additional excitation 2 is composed of the equivalent internal resistance R2 and the controllable potential E2 connected in series, and the output U of the corresponding differential amplifier 3 is:
[0090] U = K 1 × E 1 × R 50 R 1 + R 50 + E 2 × R 40 R 2 + R 40 , in R 40 = R 1 × ...
Embodiment 3
[0105] Embodiment three: this embodiment is basically the same as embodiment one, the difference is: see Figure 6 and Figure 7 , the above additional excitation 2 is only composed of controllable internal resistance R2, corresponding to the output U of the differential amplifier 3:
[0106] U = K 1 × E 1 × R 50 R 1 + R 50 , in R 50 = R 2 × R 3 R 2 + R 3 ;
[0107] The signal processing unit 4 has a control output G3 to control the controllable internal resistance R2 of the additional...
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