water pressure
water pressure
(OP)
Hi there !
This time I've got a odd question.
Seems banal, but I can't work it out.
Issue: A sealed tank is completely filled with water @ 72.5 PSI and a known outside temperature, (ex 77 F).
The tank is not insulated with respect of outside temperature
Let's suppose that, after one day, pressure changes to 50 PSI.
Question 1: Is this pressure variation physically possible ? The sole affecting variable would be outside temperature
Question 2: Should it be the case, how can I calculate new outside temperature ? It seems to me it is not possible to utilize saturation properties of water. At described pressures we got temperatures over 212 F.
Many thanks
This time I've got a odd question.
Seems banal, but I can't work it out.
Issue: A sealed tank is completely filled with water @ 72.5 PSI and a known outside temperature, (ex 77 F).
The tank is not insulated with respect of outside temperature
Let's suppose that, after one day, pressure changes to 50 PSI.
Question 1: Is this pressure variation physically possible ? The sole affecting variable would be outside temperature
Question 2: Should it be the case, how can I calculate new outside temperature ? It seems to me it is not possible to utilize saturation properties of water. At described pressures we got temperatures over 212 F.
Many thanks





RE: water pressure
I think you must have your sums wrong. You say the temperature rises from 77 F to 212 F. This cannot be right. If the pressure decays then the temperature must drop not increase.
RE: water pressure
A liquid full vessel will change pressure about 100 psi for ever 1°F temperature change, so if it cooled of a fraction of a degree the change from 72.5 to 50 psi(g?) is very possible.
For a partially full vessel you would have go give us a lot more information, but like D5B123 says the temperature would have had to have dropped, not risen
David
RE: water pressure
Please take a look at this FAQ
http://www.eng-tips.com/faqs.cfm?fid=1339
Then rearrange the formula you'll find there solving to calculate the temperature reduction ΔT that will lead to a pressure reduction ΔP of 22.5 psi.
αfΔT-βΔP=αvΔT+ΔPD/tE
RE: water pressure