College Question

[UpTheCov]

Hi, I've reccently started my city and guilds level 2 and have been asked to work out the pressure of the bottom box (all will be explained in the link) Could someone explain this to me as to what formula I have to use, Thanks

Diagram [DLMURL]http://img530.imageshack.us/img530/3959/diagram.png[/DLMURL]

 

[REDSAW]

Hi, I've reccently started my city and guilds level 2 and have been asked to work out the pressure of the bottom box (all will be explained in the link) Could someone explain this to me as to what formula I have to use, Thanks

Diagram [DLMURL]http://img530.imageshack.us/img530/3959/diagram.png[/DLMURL]

.3 bar

 

[stevetheplumber]

.3 bar

well im not sure your right red but cant do it in metric but 10 ft head is roughly 5psi so only a third of a bar ?

 

[jpm]

head in m x 0.1bar =0.3bar

 

[helpsy]

3x9.81=29.43Kn/m2
1 bar=100,000N/m2
14.7psi=1 bar
This is intensity of pressure
3x9.81x0.5mx0.3m=4.41kN
This is total pressure

 

[bobble]

Hi,

The last reply's misleading, in that he's multiplied a pressure by the area upon which that pressure is being brought to bear. The result of that calculation is a FORCE. Mind you, so many C&G questions are v. badly written, and you may be marked correct for this answer. 0.3 bar is the answer, 4.41 psi in old money.
Bobble.

 

[Johnk78]

Hi Upthecov

I take it you have both the JTL and Treloar (3rd ed)?

I too am currently in year two at College and have both of the above but the latter also covers level 3 and explains all this sort of stuff alot better and in depth...

hope this helps.

J

 

[UpTheCov]

Hi all, many thanks for the replies. This is what we were told today;
To work out the total pressure you need to know the IP (Intensity of pressure)
which is head x 9.81 kPa

Therfore the sum is 3 x 9.81 = 29.43

then times this by the area of the required box, so 29.43 x 3 =
88.29 kPa

Not sure if this is the same as the previous answers in a differnt form?