# Fluid Pressure Equation for 150 m3 Silo | Sensor Calibration and Display Reading

• joe98
In summary, the conversation discusses a 150 m3 silo used to store cream, with an electronic pressure sensor mounted 0.6 m up from the base. The display is calibrated to show the level of cream based on the sensor's gauge pressure, but the display can be adjusted with a constant offset to account for extra fluid. An equation is derived to relate the display reading to the sensor pressure, and it is determined that the silo must be 10.2 m high for the display to read 100% when full of cream. If water is used instead of cream, the silo must be higher than 12 m for the display to read 100%.
joe98

## Homework Statement

A 150 m3 silo- a cylinder (12 m high) is used to store cream (density 850 kg m-3). An electronic pressure sensor (measures gauge pressure) is mounted 0.6 m up from the base of the tank, and a 0 to 100% display is used to indicate the level. It is not possible to sense
the level once it is below the sensor, but the display can be calibrated with a constant
offset to account for this extra fluid (for example, the sensor can show 10% when
there is no pressure). Derive an equation to relate the display reading (in %) to the
sensor pressure. What offset should be used on the display to account for the 0.6 m
mounting height? If water is used instead of cream, how high must the silo be in
order for the display (calibrated for cream) to read 100% when the silo is full?

P=ρgh
m=ρv

## The Attempt at a Solution

Pressure(at the base of cylinder)=850x9.8x12=99960Pa
Pressure at sensor=1000x9.8x0.6=5880Pa

how high must the silo be in
order for the display (calibrated for cream) to read 100% when the silo is full?

99960=pgh
99960=1000x9.8xh
h=10.2meters

derive an equation
i got (9.8hx850/99960)x100

not sure of the offset but i got 0.6/12=0.05 not sure with this

any suggestion guys, much appreciated

Derive an equation to relate the display reading (in %) to the
sensor pressure.
This means you need a formula to relate the display Reading (R) to the actual Pressure (P). For instance, if the sensor was at the bottom, it would be
R = P/(ρg12)*100%
You'll have to figure out how to incorporate the sensor height of 0.6 into that formula. One method is to do a couple of examples, say cream height 0.6 and cream height 10.4. Then fiddle with a linear formula until it fits.

If water is used instead of cream, how high must the silo be in
order for the display (calibrated for cream) to read 100% when the silo is full?
The pressure would have to be as high as it was with cream but water is less dense, so according to P=ρgh the height would have to be greater than 12 m to make up for the smaller density.

Delphi51 said:
This means you need a formula to relate the display Reading (R) to the actual Pressure (P). For instance, if the sensor was at the bottom, it would be
R = P/(ρg12)*100%

so would R=P/(pg0.6)*100

Delphi51 said:
The pressure would have to be as high as it was with cream but water is [strike]less[/strike] more dense,
Cream floats on the surface of milk.

joe98 said:
Pressure(at the base of cylinder)=850x9.8x12=99960Pa
Pressure at sensor=1000x9.8x0.6=5880Pa
Why did you use ρ of H₂O when this is offset for cream?

that was for a different part of the question where "If water is used instead of cream, how high must the silo be in
order for the display (calibrated for cream) to read 100% when the silo is full"

but i can't seem to derive an equation to relate the display reading (in %) to the
sensor pressure?

Any ideas?

joe98 said:
that was for a different part of the question where "If water is used instead of cream, how high must the silo be in
order for the display (calibrated for cream) to read 100% when the silo is full"
The nature of the fluid below the level of the probe cannot affect the display reading because the probe cannot react to it. The display reading is calibrated with that level of cream as an offset, and it remains that way for all fluids.

wouldnt the overall pressure difference=850x1000x(12-0.6)=94962Pa

so the reading could be 0.05+(pressure/94962)

## 1. What is fluid pressure?

Fluid pressure is the force exerted by a fluid per unit area. It is measured in units of force per unit area, such as pounds per square inch (psi) or pascals (Pa).

## 2. How is fluid pressure calculated?

Fluid pressure is calculated by dividing the force exerted by the fluid by the area over which it is exerted. The formula for fluid pressure is P = F/A, where P is pressure in units of force per unit area, F is the force exerted by the fluid, and A is the area over which the force is exerted.

## 3. What factors affect fluid pressure?

The factors that affect fluid pressure include the density of the fluid, the depth or height of the fluid, and the acceleration due to gravity. Other factors, such as the shape and size of the container holding the fluid, can also impact fluid pressure.

## 4. How does fluid pressure change with depth?

Fluid pressure increases with depth due to the weight of the fluid above pushing down on the fluid below. This is known as hydrostatic pressure. The deeper you go in a fluid, the greater the pressure becomes.

## 5. What is the difference between static and dynamic fluid pressure?

Static fluid pressure refers to the pressure exerted by a fluid that is not moving. Dynamic fluid pressure, on the other hand, is the pressure exerted by a fluid that is in motion. This type of pressure is also known as fluid velocity pressure and is often seen in applications such as fluid flow in pipes.

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