Rural alpine natural water system

I visited a low-tech mountain hut during a walking tour in the Alps in France. At 1500m elevation, it can only be accessed via cable car and then a mountain walk, and it is isolated from both electricity and water networks. This means the only power comes from gas, used for cooking, and wood, burned for heating. Water must be obtained from a nearby stream, yet I found the system to be surprisingly sophisticated.

To check it out, I walked about a hundred metres from where the hut stands, to locate the stream. Basically, the system consists of two plastic tanks, each holding about 100 litres of water, sitting in the stream. Water enters the first tank directly from the stream – it is fresh as it has come straight from the glacier higher up. This tank contains a filter to catch any sediment; at present, this is improvised with an old stocking.

The water that passes through the filter is led via a plastic hose to the second tank, which is a couple of metres down into the valley and is a storage tank: when supply exceeds demand, some water can be held here until it is needed on drier days. An overspill pipe attached to the top allows water to run safely back into the stream if the tank is full.

The main pipe leads from the storage tank, downhill to the hut, where it can be used cold or heated with gas.

Some calculations for quantifying the system

We can examine two situations; the first, where the second is not in overspill so nothing flows through location 6; the second where it does. Bernoulli’s equation is useful here. It is an energy balance equation and states that [pressure + kinetic energy + gravitational potential energy] = constant in a flow system, as long as the mass flow rate is constant (there are a couple more technicalities, but I won’t dwell on them here). Bernoull’s equation allows us to consider only start and end point, ignoring what happens in between.

Where p is pressure, rho is density, v is velocity, g is gravitational acceleration and h is height from a fixed point.

Also, with a constant mass flow rate (and incompressible fluid) we can assume:

where A is pipe area.

Situation 1: At both 5 and 1, the density of the water is the same and the area of the pipe is the same. Therefore the velocity of the water entering the house is the same as the velocity of the water from the stream. To calculate the pressure change, we can cancel the second term in Bernoulli’s equation (since velocity and density don’t change), giving p1 + (rho)(g)(h1) = p2 + (rho)(g)(h2). Rearranging gives p2 = p1 – (rho)(g)(h1-h2).

Sticking into this that p1 = air pressure = 1 bar = 100kPa and h1-h2 = 30, p2 is 3.94 bar.