Calculating Pressure Threshold for Passive Solar Pressure Pump Design

[Simon1989]

Hi there - this is my first post here so apologies if it is not in the correct sub-forum.

I am working on a passive solar powered pump that uses pressure built up by heat in a closed cylinder (expansion chamber) to displace water from a sealed reservoir to an elevated target tank.

I am trying to calculate the theoretical performance of this system but am struggling on the connection between the thermodynamics and hydraulics.

The pressure exchange tube (between the cylinder and the reservoir) is connected to the top of the reservoir, and forces water out of a 15mm tube connected to the bottom of the reservoir to the target tank. I need to calculate the pressure threshold at which water will begin to be displaced by the expanding air to the elevated tank.

Any ideas would be greatly appreciated.

 

[Baluncore]

Welcome to PF.
The diameter of the tube is unimportant since pressure is “force per area”.
On the other hand, the height is very important. The pressure in the “expansion” chamber will need to be equal to the head of liquid in the riser tube to the upper reservoir. The height will be measured from the water level in the lower reservoir, to the riser outlet, or if the outlet is submerged, the water level in the upper reservoir.

Given that water has a density of 1000 kg/m^3, and that the acceleration due to gravity at the surface of the Earth is about 9.8 m/s^2. A 1 metre column of water will exert a force of 9.8 kN (kilo Newton) on the 1 square metre base. The pressure due to each metre of water is therefore P = 9.8 kPa (kilo pascal)

To push water up a head of, say 10 metres, will require a pressure of 10 * 9.8 kPa = 98.0 kPa
Multiply pressure in kPa by 0.145 to convert to psi, 98.0 kPa = 14.21 psi

See: http://en.wikipedia.org/wiki/Hydraulic_head