Sensors and transducers, transmission, signal conditioning

AI Thread Summary
The discussion focuses on solving two engineering problems related to hydraulic systems and instrumentation. For Q1, the calculations for the force needed on the plunger and the number of strokes required to raise a load of 2500 kg are discussed, with one participant arriving at approximately 66 strokes after correcting their calculations. In Q2, participants are working through the common mode voltage calculations for a differential amplifier with a specified CMRR and gain, with one user providing a partial solution and seeking clarification. There is an emphasis on ensuring unit consistency and the importance of limiting questions to maintain clarity in discussions. Overall, participants are collaboratively refining their understanding of hydraulic mechanics and amplifier behavior.
Keeeen
Messages
23
Reaction score
0

Homework Statement


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.

Homework 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}##

The Attempt at a Solution


[/B]
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:
Physics news on Phys.org
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##
 
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.
 
CWatters said:
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.

Ah Kg.
How did you get 65 strokes?
 
Keeeen said:
How did you get 65 strokes?

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
 
CWatters said:
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

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

Similar threads

Back
Top