Siphon Pressure with Closed Tube ``` Name: Jasmine Status: student Grade: 9-12 Country: Jamaica Date: Spring 2013 ``` Question: How would you calculate the pressure at different points of a siphon tube, when one end of the tube is closed? Replies: Hi Jasmine, Your question omits several critical details, so I will assume that, by "siphon tube", you mean an inverted U-shaped tube, one end of which is immersed into a container of some kind of liquid which I will assume is water. The tube rises up out of the water, over the edge of the container and hangs down one side, to some point below the level of the water in the container. I assume that the tube is filled with water and closed at the lower end to prevent water from flowing out. The pressure at any point along the tube is simply calculated using the measurement of the vertical distance from any point of interest along the siphon tube that hangs down beside the container, up to the surface of the water in the container. Note that the part of the tube that is above the surface of the water has no influence on the water pressure. Note also that even if the tube is on an angle, ONLY the vertical distance from the point of interest on the tube, up to the water surface, matters. Here are the steps (note that I will use English/American units like feet, inches and pounds that you are probably more familiar with than metric units)..... First, you need to know that one cubic foot of water weighs 62.4 pounds. This means that a container that is 1' x 1' x 1' weights 62.4 lb. Since the bottom face is 1 square foot in area (1' x 1') it can therefore be said that this is a 1 foot high column of water that exerts a downward pressure of 62.4 pounds per square foot. But pounds per square foot is not a common way to express pressure. Lets see what this pressure is in pounds per square inch (commonly abbreviated as PSI). Since there are 144 square inches per square foot, 62.4 pounds per square foot is the same as 0.433 PSI (I just divided 62.4 by 144). So what all this says, is that a column of water that is 1 foot in height, has a pressure of 0.433 PSI at the bottom. This pressure rises by 0.433 PSI for every additional vertical foot of water added to the column. So knowing the above, we can now answer your original question. To determine the pressure at any point you choose (let us call it point X) along a siphon tube with a closed end, that hangs over the edge of the water container, you simply measure the distance (in feet) from Point X on the siphon, vertically up to the level of the water in the container, then multiply this distance (in feet) by 0.433. The result will be the pressure in PSI at your Point X along the siphon. So let us take an example. Suppose you have a container of water that has a siphon hanging over the side. The siphon hangs 6 feet below the container, is full of water and is blocked at the lower end. What is the pressure in the siphon at a point exactly 3.5 feet below the level of water in the container? The answer is simply the distance of 3.5 feet (from your point of interest up to the water surface), multiplied by 0.433, that is, 1.515 PSI. All that matters is the distance from the water surface down to the point on the siphon tube you are interested in. The additional siphon length below the point you are interested in, as well as the part of the siphon that is above the water level, are both not important and have no effect on the pressure. I realize this was a rather tedious explanation, but I wanted to take you through the problem, step by step, rather than just give you a formula, without any explanation where it came from. Regards, Bob Wilson Hi Jasmine, Thanks for the question. You can use the equation P = rho*g*h to calculate the pressure at different points in the siphon. P is the pressure, rho is the density of water (1000 kg/m^3), g is the gravitational constant (9.8 m/s^2) and h is the height difference between two points in the siphon. I hope this helps. Please let me know if you have more questions. Thanks Jeff Grell Click here to return to the Engineering Archives

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