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Discussion Overview

The discussion revolves around a homework problem involving hydraulic systems and instrumentation measuring systems. Participants are addressing calculations related to force requirements for a hydraulic jack and analyzing a differential amplifier's performance, specifically focusing on common mode voltage and gain calculations.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • Post 1 presents two questions regarding a hydraulic jack and an instrumentation measuring system, including relevant equations for calculations.
  • Post 2 provides a partial solution for the second question, calculating the common mode voltage based on the given CMRR and differential gain.
  • Post 3 suggests that participants should limit each post to one question and questions the unit of the load in the first question.
  • Post 4 reiterates the suggestion to limit questions and provides an alternative stroke calculation, arriving at 65 strokes.
  • Post 5 details the calculation method for determining the number of strokes required, showing the steps and arriving at approximately 64.8 strokes.
  • Post 6 confirms the calculation method and acknowledges an error in the previous calculation, seeking feedback on the second question's approach.

Areas of Agreement / Disagreement

Participants express differing views on the number of strokes required for the hydraulic jack, with calculations yielding results of 65 strokes and approximately 64.8 strokes. There is no consensus on the approach to the second question, as one participant seeks guidance while another provides a calculation.

Contextual Notes

Participants have not fully resolved the assumptions regarding units in the first question, and there may be dependencies on the definitions of terms used in the calculations. The discussion on the second question remains open, with varying levels of understanding among participants.

Keeeen
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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.
 
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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?
 

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