Question

Q6. marks) Consider an electromagnetic flowmeter which is used to measure the average volumetric flow rate of a process fluid in pipe of 0.1 m diameter. The velocity profile is symmetrical and can be assumed uniform. The flux density in the liquid is 1.4 Wb/m2. The output from the fow meter is given to an amplifier of gain 10 and the internal resistance developed due to the fluid between the electrodes is found to be is 200 kf2. The input impedance of the amplifier is 2,2 MQ. Find average velocity of the liquid when the peak-to-peak output voltage at the amplifier output is a. 4 volts marks) b. Find the percentage change in the reading of the amplifier for a 10% increase in the resistivity of the flowing fluid. Note that: Induced voltage for conductor moving through a magnetic field, is expressed by the relation: Blv marks)

Answer 1

Given : CTi nPipe Diameter (D) O 1m Flux Density (B) 4 wb/m2 Gain of Amplifier (β): lo Internal Resistance of Fluid 200 k2 Input Impedance of Amplifier= 2.2 MS2 Circuit 200 K 32 Vo Peak to peak voltage o/p = 4V n series From the above circuit diagram 200 k.2 and 22 M2 are -p Vin=厂2-2 x 106 200x103+ 22x10 Also lo → Vin = 0.2 volt g Volt = 10 in Vin Using Vin 0 2 volts in U) we get 9167 x Eo 0.2 > Ed 0.218 volts B-D-9 Eo= Also 0.218 = 1.4 x 0-1 x B ab 29 = 0218.- = 1.557 m/sec. = Fluid velocity 14x0l b) Con ductivity of Fluid Ises by 10 2 Ep Resistance of 4luid noill becomes 0-909 times of previous value ie 200 K2 changec to 0909x200 131-81 K2

Vin becomes 181 8Ix103+ 22x106 x 0 218 131 81x102-2x 106 0.20126 = 10× 0.20136 20196 V 7 Change in o/p 2136-2x loo -x100 2 0.68%