Answers

From Build a Level Sensor

1. Gauge pressure is a pressure measured relative to the surrounding atmosphere. In other words, the amount to which the measured pressure is higher than atmospheric pressure. The atmospheric pressure is constantly varying so a value of 101,325 Pa is used as a standard atmospheric pressure.

Absolute pressure is measured relative to absolute zero or total vacuum.

Differential pressure is the difference between two pressures. This is often measured. A typical application would be a Venturi or Orifice Plate flow meter where differential pressure is an indication of fluid velocity or flow rate.

Go to Pressure Measurement for more information on pressure measurement.

Pressure due to a height or head of water: P = ρgh

Height of water above sensor = 5 – 0.3 m = 4.7 m

P = 1000 x 9.81 x 4.7

P = 46107 Pa

This is a gauge pressure as it does not include the column of air in the atmosphere sitting on top of the tank. Add 1 atmosphere to get the absolute pressure:

P = 101325 +46107

P = 147432 Pa absolute

2. 8 mbar = 800 Pa (1 bar = 100,000 Pa so 1 mbar = 100 Pa)

There are 6895 Pa in a psi. Divide the pressure in Pa by 6895 to get the answer in psi:

P = 46107/6895

P= 6.7 psig

P = 147432/6895

P= 21.4 psia

3. A piezoresistor is a device whose resistance changes when it is pressed. Piezo is the Greek word for Press. Silicon doped with small amounts of N or P type material exhibits a large piezoresistive effect. It is used to make piezoresistors or strain gauges with very high sensitivities or gauge factors.

It is very similar to the electrical strain gauge in that it’s resistance changes as it bends. However, it has a much higher sensitivity – of the order of 100 for the piezoresistor compared to 2 for the electrical strain gauge. The electrical strain gauge is made from a very long thin wire (an alloy of 54% copper, 44% nickel and 1% manganese is an example) whereas the piezoresistor is made from semiconductor material. The piezoresistor is more easy to connect to a diaphragm as it can be deposited on the diaphragm during construction. The diaphragm is then exposed directly to the pressure to be measured or where there is a concern about damaging the diaphragm it is connected to another diaphragm via a silicone oil filled tube. See following description from http://www.burkert.com/COM/599.html

A chemical seal consists of a chamber filled with a transfer fluid (oil-based mixture), closed on one side by a pressure sensor and closed off from the process on the other by means of a diaphragm (stainless steel or plastic). The pressure bends the diaphragm and is transferred from the fluid to the pressure sensor.
Pressure transmitters are used if the process conditions no longer allow direct attachment of a pressure sensor (extreme temperatures, chemical resistance, hygiene requirements or media which form coatings or are very viscous ...). A wide range of process connections also allows use in special applications.

Plot the data as follows. Make sure the input is on the x axis and the output is on the y axis. Make sure the axes are scaled and labelled. Make sure the graph has a title.

Sensitivity = slope (if input is on x).

Sensitivity = 0.025 mV/mm

Offset is when x = 0

Offset = 0.59 mV

Max non linearity is point that is furthest from the line. This occurs at 82 mm where an output of 2.4 mV was measured. We can work out what we should have measured at 82 mm by using the straight line equation above.

Vout = 0.0249 x Level + 0.5883

Vout = 0.0249 x 82 + 0.5883

Vout = 2.63 mV

Difference = 2.63 – 2.4 = 0.23 mV

Max non linearity = 0.23 mV

The stated sensitivity is 10 mV/psi and we have measured 0.025 mV/mm. To compare, we must convert psi to mm. In this case mm means mm of H₂O which has a density of 1000 kg/m³.

1 mm H₂O exerts a pressure of 1000 x 9.81 x 0.001 = 9.81 Pa.

1 psi = 6895 Pa = 6895/9.81 mm H₂O = 703 mm H₂O

10 mV/psi becomes 10 mV/703 mm = 0.014 mV/mm

The stated sensitivity is 0.014 mV/mm and we measured 0.025 mV/mm. The reason for the difference may be the excitation or supply voltage to the bridge. The stated sensitivity applies only if the excitation voltage is 10 V. A different supply may have been used.

4. (i) The gain needed is to amplify 0.025 mV/mm to 1 mV/mm. Divide the overall sensitivity by the sensitivity of the amplifier: 1 / 0.025 = 40

A gain of 40 is needed.

(ii) Circuit diagram:

(iii) Resistor sizes. Gain of instrumentation amp needs to be 40. Equation for gain of the instrumentation amp is:

40 = Gain = V_out/(V₂ – V₁) = 1 + 2R/R_gain

If we choose a value of 1 kΩ for R, then R_gain works out to be:

40 = 1 + 2R/R_gain = 1 + 2(1000)/R_gain

R_gain = 2(1000)/(40 – 1) = 51.3 Ω

This is a very small resistor. All resistors in the amplifier circuit should be in the kΩ range for it to work well. Therefore, R_gain should be twenty times bigger (50 x 20 = 1000) so that means that R should be twenty times bigger too. Redo the calculation for R = 20 kΩ:

40 = 1 + 2R/R_gain = 1 + 2(20000)/R_gain

R_gain = 2(20000)/(40 – 1) = 1025 Ω

This is a more acceptable value. Summary: R = 20 kΩ and R_gain = 1.025 kΩ

In practice, R_gain would be adjustable so the gain can be changed to give the exact sensitivity required.

5. Start block diagram. From question we can get the following pieces:

To get the rest we convert the sensitivity of 10 mV/psi to mV/mmH2O. 1 psi = 6895 Pa and 1 mmH2O = ρgh = 1000 x 9.81 x 0.001 = 9.81 Pa. Therefore, 1 psi = 6895 Pa = 6895/9.81 mm H₂O = 703 mm H₂O

10 mV/psi becomes 10 mV/703 mm = 0.014 mV/mm

If the input is 200 mm, the output = 200 x 0.014 = 2.8 mV. The block diagram can now be completed:

Check above – all inputs and outputs have ranges and units, all sensitivities calculated.

Check sensitivities: Since 0 to 200 mmH2O goes in at start and comes out at end, the overall sensitivity is 1. Therefore, multiply all the individual sensitivities by each other and we should get 1.

Check: 0.014 x 1.78 x 51 x 0.78 = 0.99 which is almost 1 so this passes the check test.