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In some cases of level measurement in open tanks, the level transmitter has to be installed some distance below the base of the tank as shown below:

From the diagram above, it can be seen that when the tank level is minimum (H = 0), the liquid in the high pressure impulse line exerts a constant pressure (P = X*S.G) on the high pressure side.

When the liquid level is at a maximum, H meters, pressure on the high pressure side of the transmitter will be:

P(high) = S.G*H + S.G*X + P(atm)

P(low)= P(atm)

With this, the transmitter output is now proportional to both S.G*H and S.G* X. To make the transmitter output proportional to only S.G*H, we suppress the S.G*X part of the transmitter output by negatively biasing the transmitter by –S.G*X.

This implies that the range of level measurement will be between S.G*X to (S.G*H + S.G*X.) This procedure is called Zero Suppression and is done when calibrating the level or DP transmitter.

Consider the example below:

Suppose the maximum level of liquid in the tank is 40ft, the Specific gravity of the liquid, S.G = 0.9. If the level transmitter for level measurement is mounted 6ft below the base of the tank then we can calibrate the level transmitter as follows:

H = 40ft = 40*12 = 480in

X = 6ft = 6*12 = 72in

S.G = 0.9

Differential pressure between the level transmitter HP and LP when tank is full is:

= S.G*H + S.G*X = 0.9*480 + 0.9*72 = 496.8 in wc

When tank is empty (H = 0),

Differential pressure = S.G*X = 0.9*72 = 64.8 in wc.

The transmitter is then calibrated as shown below:

<![endif]-->At 0% minimum tank level (H = 0) 4mA = 64.8in wc and at 100% maximum tank level (H = 40ft) 20mA = 496.8 in wc. Note that the same calibration method above is used if the installation is a dry leg.

Continue to the next sections on Level Measurement:

Zero Suppression in Open Tank Level Measurement Installation |

From the diagram above, it can be seen that when the tank level is minimum (H = 0), the liquid in the high pressure impulse line exerts a constant pressure (P = X*S.G) on the high pressure side.

When the liquid level is at a maximum, H meters, pressure on the high pressure side of the transmitter will be:

P(high) = S.G*H + S.G*X + P(atm)

P(low)= P(atm)

Differential Pressure, ΔP = P(high) – P(low) = S.G * H + S.G * X |

With this, the transmitter output is now proportional to both S.G*H and S.G* X. To make the transmitter output proportional to only S.G*H, we suppress the S.G*X part of the transmitter output by negatively biasing the transmitter by –S.G*X.

This implies that the range of level measurement will be between S.G*X to (S.G*H + S.G*X.) This procedure is called Zero Suppression and is done when calibrating the level or DP transmitter.

Consider the example below:

Suppose the maximum level of liquid in the tank is 40ft, the Specific gravity of the liquid, S.G = 0.9. If the level transmitter for level measurement is mounted 6ft below the base of the tank then we can calibrate the level transmitter as follows:

H = 40ft = 40*12 = 480in

X = 6ft = 6*12 = 72in

S.G = 0.9

Differential pressure between the level transmitter HP and LP when tank is full is:

= S.G*H + S.G*X = 0.9*480 + 0.9*72 = 496.8 in wc

When tank is empty (H = 0),

Differential pressure = S.G*X = 0.9*72 = 64.8 in wc.

The transmitter is then calibrated as shown below:

<![endif]-->At 0% minimum tank level (H = 0) 4mA = 64.8in wc and at 100% maximum tank level (H = 40ft) 20mA = 496.8 in wc. Note that the same calibration method above is used if the installation is a dry leg.

Continue to the next sections on Level Measurement: