Background Siphoning, the process of using a tube to carry a liquid over a barrier and to a lower location, can be described well bythe mechanical energy equation. In this assignment you will analyse...

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Extracted text: Background Siphoning, the process of using a tube to carry a liquid over a barrier and to a lower location, can be described well bythe mechanical energy equation. In this assignment you will analyse siphoning behaviour theoretically, and then will carry out an experiment using a plastic bottle and length of tubing provided to you. The plastic bottle has a nominal volume of one litre. However, two marks have been placed on the side of the AZ bottle such that the corresponding volume between the two marks is approximately AV = 650 cm3. The plastic tubing given to you has an inside diameter of D = 2 mm and a length of L = 100 cm. In the picture to the right, Az is the variable distance between the liquid level in the bottle and the tube outlet level. The diameter of the tube is small enough that the flow inside will be laminar, an orderly state of flow in which a = 2 in the mechanical energy equation. Furthermore, the friction (wterm can be expressed as wf= kehlava where k = 9.5s-1 is a friction constant for room-temperature laminar flow of water in this particular tubing. L is end-to-end length of tube and yayg is the average velocity in the tube. You may take the water density to be 999 kg/m3. Analysis Forthis experiment the analysis will be performed first. 1. Calculate the volumetric flow rate of the siphon (cm³/s), assuming that L=100cm, Az = 80 cm, and that there is no friction in the tube (wf =0). In order to get the volumetric flow rate, you must first calculate Uavg: