Help with force/pressure calculations

Original post by pjs412

Thanks and apologies.

By the way, you're a lot more likely to get replies if you choose one part of your question to start with and explain why you're stuck and provide any working that you've done.

Original post by pjs412

Hi All

I understand this may be really simple stuff but please bear with me, I'm new to this course and really need help with the following:

a) If a cylinder is required to overcome a force of 2.5 tonne with a working pressure of 210 bar, what diameter of cylinder is required if the piston rod is 20mm in diameter?

b) if the cylinder above is to extend at a rate of 50mm per second, what flow rate will be required from the pump in litres per minute?

c) The volume of a gas is 0.48m³ at a pressure of 635 kPa, if the pressure is then decreased to 125 kPa but the temp maintained at a constant level what will its volume change to?

d) State the thermodynamic law used and also state an alternative?

e) Plot a graph with info obtained in part 'C' (I could do this if i understood where/how to get the data obviously)

f) explain 2 applications in real industrial systems of the 3 basic gas laws you have used for calculations , give examples?

If anyone is able to help with this it would be greatly appreciated, If it possible to show how you got the answers as well as i would like to learn this stuff as opposed to just using someone else's answers.

Thanks

Post your working so far so we can see what you have attempted, or where you have gone wrong, so that we can offer advice.

Reply 6

OP

Original post by Smack

Post your working so far so we can see what you have attempted, or where you have gone wrong, so that we can offer advice.

Its basically knowing what equations to use, like in a) i understand i need to find the area of the rod so i can calculate the force on one side but i just cant understand how id calculate how big the cylinder needed to be?

Original post by pjs412

Its basically knowing what equations to use, like in a) i understand i need to find the area of the rod so i can calculate the force on one side but i just cant understand how id calculate how big the cylinder needed to be?

Are you familiar with hydraulic cylinders? I think you need some understanding of hydraulics to answer questions A and B. If not, the questions can still be answered, but it'll take a little more explaining.

(edited 7 years ago)

Reply 8

OP

I have done A) & B) so far but i think there both wrong.

A)

F=2.5 Tonne/2500 KG
P=210 Bar = 21000 KPA
A=?

A=2500/21000
A=0.11904m^2

i then transposed piR^2 to become Square root 0.11904/Pi = 0.19465, i then multiplied this by 2 to get the diameter. What am i doing wrong?

B)

50mm per second = 0.05 m/s
Area = 0.11904 m^2
Flow Rate = ?

Q=AV
Q= 0.11904 x 0.05
Q= 5.952x10^-3 m^3/s

again i know this is wrong but its the best i can conjure, any help with where im going wrong would be massively appreciated.

Reply 9

A mass is not a force. You need to multiply by g (9.80665N/kg) to get the weight. You also seem to have lost the k in kPa.

Right approaches, just some small mistakes.

(edited 7 years ago)

Reply 10

Original post by pjs412

I have done A) & B) so far but i think there both wrong.

A)

F=2.5 Tonne/2500 KG
P=210 Bar = 21000 KPA
A=?

A=2500/21000
A=0.11904m^2

i then transposed piR^2 to become Square root 0.11904/Pi = 0.19465, i then multiplied this by 2 to get the diameter. What am i doing wrong?

F=mg

F=PA

\therefore PA=mg

\therefore A=\frac{mg}{P}=\frac{\pi d^2}{4}

\therefore d=2\sqrt{\frac{mg}{\pi P}}

Reply 11

Original post by RogerOxon

\therefore d=2\sqrt{\frac{mg}{\pi P}}

That gives me 0.0386m (3s.f.) or 38.6mm. (P=210bar = 210.100000 = 21000000 = 21 MPa)

Reply 12

Original post by pjs412

B)

50mm per second = 0.05 m/s
Area = 0.11904 m^2
Flow Rate = ?

Q=AV
Q= 0.11904 x 0.05
Q= 5.952x10^-3 m^3/s

again i know this is wrong but its the best i can conjure, any help with where im going wrong would be massively appreciated.

Right approach, but using the wrong area - see the posts above.

Reply 13

OP

Original post by RogerOxon

Right approach, but using the wrong area - see the posts above.

Huge help. Thank you so much for taking the time.

Original post by pjs412

Huge help. Thank you so much for taking the time.

Your methodology was correct. You just need to ensure that you keep your units consistent. It's generally easier if you convert everything to SI - e.g. m, Pa, N, etc.

And also take note of the difference between mass and weight.

Reply 15

OP

Original post by RogerOxon

F=mg

F=PA

\therefore PA=mg

\therefore A=\frac{mg}{P}=\frac{\pi d^2}{4}

\therefore d=2\sqrt{\frac{mg}{\pi P}}

why do you divide pi d^2 by 4?

Original post by pjs412

why do you divide pi d^2 by 4?

One of the formulas used to work out the area of a circle is:

A = (π * d²)/4

This formula is most typically used when it is the diameter provided, rather than the radius.

Reply 17

Original post by pjs412

why do you divide pi d^2 by 4?

A=\pi r^2=\pi (\frac{d}{2})^2=\frac{\pi d^2}{4}

as

d=2r

(edited 7 years ago)

Reply 18