P in pipe that goes up 18' then back down...

[jamie]

So this is a system with two tanks, the tanks will be connected with a 2 inch pipe, the pipe has to go from ground level, up 18 feet over garage door etc. etc. then back down 18 feet into another lab/tank. What issues do you see developing from this set-up? Is there going to be cavitations at the top of this thing? Do the pressures on either side just cancel each other out because the inlet/outlet is at the same height? What is the pressure at the top of the 18 foot riser? Sorry, I'm rusty on this stuff. Thanks!

[admin]

For that height the pressure will be above evaporation point for the normal temperatures (20-30 OC).

In order to start the circulation energy (head) on one side must be higher than energy (head) on the other. It can be achieved with pump or with level difference between two tanks. In the second case you have somehow to start the flow in the connecting pipe. If tanks are on the same level and you have no pump, flow is not possible.

Pressure on top in the case of zero velocity will be p=pa - rho x g x h = 101325 Pa - 1000 kg/m3 x 9.81 m/s2 x 5.48 m = 47,566 Pa = 0.47 bar = 6.8 psi - absolute

[jamie]

Thanks! I thought it was just rgh - the fluid that is going to be in the pipes is glycol, one tank is open to atm and has a pump, the other is closed. The tank open to the atm will be filled approx 6 feet high - so does that make the riser effectively 18-6 = 12 feet high? Most of the time the glycol will be cold, in a refrigerated room...

if it does get up to roomt temp... let's see, the vapor pressure of glycol at 20°C is about .005 bar (water 0.023 bar), glycol is about the same density as water... so it should be ok - it won't convert to a vapor right?

[admin]

Yes, you are right on both...

[jamie]

thanks again!