Question

A liquid level control system linearly converts a displacement of 2-3 meters into a 4-20 mA control signal. A relay closes at 12 mA and opens at 10 mA. Find (a) the relation between displacement level and current, and (b) the neutral zone (between 12mA and 10 mA) or displacement gap in meters.

Answer 1

Answer:

a)H = 0.0625 I + 1.75

b)Neutral zone=0.125 m

Explanation:

Given that

$H_{min}= 2 m$

$H_{max}= 3 m$

$I_{min}=4\ mA$

$H_{max}= 20\ mA$

The relationship between current and displacement is given as follows

H= K I + H0

Now by putting the values

3 = K x 20 + H0

2 = K x 4 + H0

Therefore

1 = 16 K

K=0.0625

H0=2- 0.0625 x 4

H0=1.75

The relationship is given as follows

H = 0.0625 I + 1.75

Now ,

$I_H=12\ mA$

$H_H=0.0625\times 12+1.75$

$H_H=2.5$

$I_L=10\ mA$

$H_L=0.0625\times 10+1.75$

$H_L=2.375$

Neutral zone is given as follows

Neutral zone=2.5-2.375

Neutral zone=0.125 m

Answer 2

The relationship between the displacement level and current is $I_x=(16x-28)mA$ and the neutral zone is between 2.375m to 2.5m.

Given that

• I$_$$_1$ = 4mA
• I$_2$ = 20mA

when

$x_1=2m, I_x_1=4mA\\x_2=3m, I_x_2=20mA$

Using equation of a line

$\frac{y_2-y_1}{y-y_1} =\frac{x_2-x_1}{x-x_1}$

substituting the values into the above equation,

$\frac{20-4}{I_x_1-4}=\frac{3-2}{x-2} \\\frac{16}{I_x_1-4} =\frac{1}{x-2}$

cross multiply both sides

$16(x-2)=(I_x-4)(1)\\16x-32=I_x-4$

Make $I_x_1$ the subject of formula

$I_x=(16x-28)mA$

From the calculation above, the relationship between displacement level and current is $I_x=(16x-28)mA$

b) The neutral zone or displacement gap.

Given that the range is 12mA to 10mA,

$I_x=12mA\\Ix=(16x_3-28)$

This relationship comes from above.

$12=16x_3-28\\x_3=\frac{12+28}{16} \\x_3=2.5m$

For 10mA

$I_x=10mA\\10=16x_4-28\\x_4=\frac{10+28}{16}\\x_4=2.375m$

From the calculation above, the neutral zone is between 2.375m to 2.5m

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