A Portable and Easily Constructed Manual Water Pump – Part 1, by G.G.


“Water, water everywhere but nary a drop to drink”

– Samuel Taylor Coleridge’s “The Rime of the Ancient Mariner”.

Finding water becomes the major problem faced by people in a survival situation. Go without water only a few days and death ensues. Storing water in the form of barrels is a logistical nightmare. They must be drained and refilled annually. Further, you only have as much water as you have stored.

Living in Idaho gives those of us who live here access to one of the world’s largest aquifers. In fact, that body of water lies only 70 feet below my house. The problem I have is getting through the layers of hard pan, lava, gravel, and dirt that lie in between. Yes, I do have a well, but when TEOTWAWKI comes, there will be no electricity to bring that water to the surface. I guess you could say I am lucky water is only 70 feet down for me. Others in the Treasure Valley have to go down 400 feet or more to reach potable water. In order to access this water source, a portable, easily constructed manual pump is needed. (What follows is a mildly technical description of the principles employed by this pump. If this does not interest you, skip down to the assembly of the pump).

Technical Description of Manual Pump

Typical hand pumps can only pump water up from about 33 feet (14.7 psi or 1 atmosphere). Pumping water further than that pulls the dissolved air out of the water and creates bubbles. An example of this can be seen when a nurse draws medication out of a vial. Air always ends up in the syringe even though the needle remains submerged in the solution in the vial. This is due to the negative pressure being applied by the syringe. Where the air in the solution is dissolved at a higher pressure than what exists inside the syringe a natural gradient causes the dissolved gases to boil out. (This is also similar to what happens when a can of soda foams up. If the pressure outside of the open can was higher than the pressure with which the gas was dissolved in the soda, no fizzing would occur.)

So, the problem remains; how do we access water deeper than 33 feet? As I have thought about this, I kept thinking that electric pumps usually push water up from the bottom of the well, since most pumps are located at the bottom of the well, rather than pull the water up (unless special pump considerations are made). Furthermore, it is easy to conclude that a pump down at 400 feet needs to generate more pressure to push the water out than a pump at 70 feet does.

The answer to this problem came partly when I was studying for my PADI Dive Master rating. In the physics portion of this course, we learned about buoyancy. This is also known as Archimedes principle. Simply put, it means that the amount of force trying to lift a submerged object is equal to the weight of the water displaced by that object. For example, each cubic foot of seawater weighs 64 pounds. If a 1 ft3 container was placed in sea water, it is pushing away 64 pounds of salt water. Now, if we were to attach a 64-pound weight to that container, the container would be neutrally buoyant. In other words, there is no force trying to move the container up or down in the water. However, if we attached a 63-pound weight, the container would float up or would be positively buoyant. If we attached a 65-pound weight, the container would sink, or would be negatively buoyant.

If we were to consider these principles as they apply to water in a pipe, we would find there is no difference between the principles of buoyancy inside the pipe versus those inside the ocean. Application of these principles implies that in order to bring water up from the bottom of a well, the weight of that column of water (the water in the pipe) would have to be overcome.

For instance, if we had a 1-inch square column of water, the weight of that water would be 14.7 pounds for every 32 feet. Therefore the amount of pressure to elevate water from 70 feet would be (70 / 32) x 14.7 = 32.16 psi. The amount of pressure to elevate water from 400 feet would be (400 / 32) x 14.7 = 183.75 psi.

Let’s shift gears for a moment and consider areas of a circle. Take two circles of different diameter and place one circle inside a larger circle. Let’s make the area of the smaller circle equal the area between the smaller circle and the larger one. Let’s make the circles such that half of the area is outside the small circle and half is inside of it. Rather than going through the algebra to figure this problem out, let’s just calculate the area of several circles:

The area of a 1½” circle: A = πR2 or, 3.14 x 0.75 x 0.75 = 1.766 square inches.

To make the area inside the smaller circle equal the area outside of the smaller circle, we divide 1.766 / 2 = 0.883 square inches

The area of a ¾” circle is 0.442 square inches.

The area of a 1” circle is 0.785 square inches

The area of a 1 1/8” circle is 0.993 square inches.

Now, if we extend these circles to a third dimension, we would essentially have one pipe inside of another. Because pipes only come in certain sizes, let’s look at this problem using what is available to us.

I feel polyethylene pipe is ideal for this application, due to its inherent flexibility. This means the pipe can be pulled by hand out of the well in one continuous length. PVC, or some other stiff pipe, must be separated length by length as the pipe is lifted out of the well. This means that each section must be connected to each other by fittings, which can fail either by separating due to faulty gluing or by leaking due to improper sealing. The most fail-safe pipe material is the flexible polyethylene pipe. I have included calculations for PVC pipe below, if this material is chosen.

The actual inside diameter (ID) of a 1.5” x 160 psi NSF polyethylene pipe is 1.61” (Home Depot Internet # 205909030). This gives an area of 2.035 in2. The actual outside diameter of a 1” x 160 psi NSF polyethylene pipe is 1.125” (Home Depot Internet # 1000030866). This gives an area of 0.994 in2. So, if we were to subtract the inside area of the larger pipe from the outside area of the smaller pipe, we would get, 2.035 in2 – 0.994 in2 = 1.041 in2. In other words, the larger pipe is 0.047 in2 greater than the smaller pipe. This difference is important

So, by plugging the bottom end of the smaller pipe, it will remain hollow as water fills the volume outside of this pipe. To use our scuba diving analogy, we want the hollow container (smaller pipe inside the 1 1/2” larger pipe) to lift with a force equal to weight of the water surrounding it.

Because the smaller pipe is 0.047 in2 smaller than the larger pipe, there will be a net force lifting the smaller pipe. This would make it difficult to push down the inner pipe to move the pump mechanism. However, if the weight of the smaller pipe is considered, the forces become negligible. Thus, this system is in a neutrally buoyant condition. It will lift the water from the bottom of the well with very little force. The depth of the well is irrelevant.

So, the final solution to our initial problem is the overwhelmingly simple principle of buoyancy.

Tomorrow, we will put this pump together.

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