# Fluid pressure question!

## 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

Delphi51
Homework Helper
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.

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

NascentOxygen
Staff Emeritus
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.

NascentOxygen
Staff Emeritus
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 cant seem to derive an equation to relate the display reading (in %) to the
sensor pressure?

Any ideas?

NascentOxygen
Staff Emeritus