# Sensors and transducers, transmission, signal conditioning

1. Jun 24, 2017

### Keeeen

1. The problem statement, all variables and given/known data
Q1. A hydraulic jack has a load piston diameter of 150 mm, and a plunger diameter of 25 mm
a) Ignoring loses, calculate the force needed on the plunger to raise a load of 2500 kg
b) Calculate how many stroke of the plunger will be required to raise the load 180 mm, if each stroke of the plunger is 100 mm long.

Q2. An instrumentation measuring system uses a differential amplifier have a CMRR of 120 dB and a differential gain of $10^5$. The maximum differential input signal is $60\mu V$. If the amplified noise voltage is specified to be not more than 1% of the maximum output voltage, calculate the maximum common mode voltage that can be present in the input to the amplifier.

2. Relevant equations
$A = \frac {\pi D} {4}$
$Effort = Load (\frac {D2}{D1})^2$
$number of strokes = \frac {A1 * Distance}{A2 * Length}$
$CMRR = 20Log10 \frac {Adiff}{Acm}$

3. The attempt at a solution

6a. $Area of D1 = \frac {\pi * 0.15}{4}$
$A1 = 0.018m^2$

$Area of D2 = \frac {\pi * 0.025}{4}$
$A2 = 0.00049m^2$

$Effort = 2500 (\frac {0.025}{0.15})^2$
$= 69.4N$

6b. $Strokes = \frac {0.018*0.18}{0.00049*0.1}$
$=66.12$
$=66 strokes$

How's this looking?

As for Q2 I'm a bit uncertain and would just like a bit of guidance of where i can get started or someone to point me in the right direction where i can find the information to answer the question.
Cheers.

Last edited: Jun 24, 2017
2. Jun 24, 2017

### Keeeen

For Q2 i've gotten this far.

$120 = 20log10 \frac{10^5}{Acm}$
$6 = log10 \frac {10^5}{Acm}$
$1*10^6 = \frac {10^5}{Acm}$
$Acm = \frac {10^5}{1*10^6}$
$Acm = 0.1$

3. Jun 24, 2017

### CWatters

Usually best to limit each post to just one question, especially if they are very different.

For Q1
A).Check the units of the load. Is it kg or Newton's?
B). I made it 65 strokes.

4. Jun 24, 2017

### Keeeen

Ah Kg.
How did you get 65 strokes?

5. Jun 25, 2017

### CWatters

Volume of load stroke/Volume of plunger stroke

= {Stroke of Load * Area of Load piston} / {Stroke of plunger * Area of plunger}

= {180*π*(150/2)2} / {100*π*(25/2)2}

π and 22 cancels

= {180*1502} / {100*252}

= 4,050,000 / 62,500

= 64.8 strokes

6. Jun 25, 2017

### Keeeen

Right ok i see where i've gone wrong now. thanks. how about Q2 does it look like i'm on the right lines?