Question

4. (15) A capacitive fluid level sensor is designed using two parallel plates immersed in the water whose level is to be measured in direct sensing mode. Each plate is $l$ cm by w mm, where $l = 5$ cm, and $w = 20$ mm. The plates are separated by a gap of 3 mm, which is air in the absence of fluid. The water level, $h$, may vary from 0 to $l$. a) (11) Calculate the capacitance of the sensor when the tank is half full (fluid is at mid point), and when the tank is quarter full ($h = \frac{1}{4}l$). (Use the table provided for dielectric constants on the last page. Permittivity of vacuum, $\epsilon_o$, is $8.854 \times 10^{-12}$ F/m. State your assumptions) b) (4) Compare and comment on the two results for the two fluid levels. fs Go to last page and look at water in table. Use glycerin as an example

          4. (15) A capacitive fluid level sensor is designed using two parallel plates immersed in
the water whose level is to be measured in direct sensing mode. Each plate is $l$ cm by
w mm, where $l = 5$ cm, and $w = 20$ mm. The plates are separated by a gap of 3 mm,
which is air in the absence of fluid. The water level, $h$, may vary from 0 to $l$.
a) (11) Calculate the capacitance of the sensor when the tank is half full (fluid is at
mid point), and when the tank is quarter full ($h = \frac{1}{4}l$).
(Use the table provided for dielectric constants on the last page. Permittivity of
vacuum, $\epsilon_o$, is $8.854 \times 10^{-12}$ F/m. State your assumptions)
b) (4) Compare and comment on the two results for the two fluid levels.
fs
Go to last page and look at water in table.
Use glycerin as an example

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4. (15) A capacitive fluid level sensor is designed using two parallel plates immersed in
the water whose level is to be measured in direct sensing mode. Each plate is l cm by
w mm, where l = 5 cm, and w = 20 mm. The plates are separated by a gap of 3 mm,
which is air in the absence of fluid. The water level, h, may vary from 0 to l.
a) (11) Calculate the capacitance of the sensor when the tank is half full (fluid is at
mid point), and when the tank is quarter full (h = (1)/(4)l).
(Use the table provided for dielectric constants on the last page. Permittivity of
vacuum, , is 8.854 × 10^-12 F/m. State your assumptions)
b) (4) Compare and comment on the two results for the two fluid levels.
fs
Go to last page and look at water in table.
Use glycerin as an example

Added by Victor K.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Madhur L

08/31/2023

Step 1

Given that the length of each plate is l = 5 cm and the width is w = 20 mm, we need to convert the width to centimeters. 1 mm = 0.1 cm So, the width in centimeters is w = 20 mm * 0.1 cm/mm = 2 cm. The area of each plate is given by: Area = length * width = l * w Show more…

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Texts: Sensors, Actuators, and Their Interfaces 4. (15) A capacitive fluid level sensor is designed using two parallel plates immersed in the water whose level is to be measured in direct sensing mode. Each plate is l cm by w mm, where l = 5 cm, and w = 20 mm. The plates are separated by a gap of 3 mm, which is filled with air in the absence of fluid. The water level, h, may vary from 0 to l. a) (11) Calculate the capacitance of the sensor when the tank is half full (fluid is at the midpoint), and when the tank is quarter full (h = 1/4 * l). (Use the table provided for dielectric constants on the last page. Permittivity of vacuum, ε₀, is 8.854 x 10⁻¹² F/m. State your assumptions.) b) (4) Compare and comment on the two results for the two fluid levels. Go to the last page and look at the water in the table. Use glycerin as an example.

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Transcript

00:01 Hi, this is a question based on capacitance.

00:05 So, capacitance so here to find the capacitance of the transducer.

00:10 So, here we have as a function of the liquid at a depth of h.

00:16 So, depth is h therefore, we have the capacitance is given by epsilon not epsilon a divided by d where epsilon is the dielectric constant this is the permittivity of the free space a is the area of the overlap of the transducer and d is the so d is the separation between the plates here a can be written as so liquid depth into so liquid depth into width.

00:46 So, we can write the value of capacitance as so value of capacitance at the function of h can be written as epsilon not epsilon w into h divided by d.

00:57 So, now for half width.

00:58 So, for part b we have half full liquid h is equal to 1 meter.

01:05 So, we have epsilon is equal to 3 .8 epsilon not is 8 .85 picofarad per meter and w is 0 .05 meter and d is equal to 6 meter.

01:19 So, 6 millimeter that is 0 .06 meter.

01:22 So, we have the value of c half is given by so it is given by 3 .8 into 8 .85 into 10 power minus 12 into 0 .05 into 1 divided by d is 6 into 10 power minus 3.

01:42 So, we have the value of capacitance is given by 280 picofarad to determine the sensitivity.

01:52 So, sensitivity k is given by the rate of change in the capacitance at the depth h is h divided by d h.

02:03 So, this is the sensitivity given...

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