2. Predict the volumetric flow rate of the siphon (cm³/s), assuming that L = 100 cm, Az = 80cm, and that there is friction in the tube (Wɛgiven by equation above). %3D

Structural Analysis
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Chapter2: Loads On Structures
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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
bottle such that the corresponding volume between the two
AZ
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
expressed as wf = kFLuavg
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 yovg is the average velocity in
the tube. You may take the water density to be 999kg/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 (w = 0). In order to get the
volumetric flow rate, you must first calculate ygvg:
2. Predict the volumetric flow rate of the siphon (cm3/s), assuming that L = 100cm,
Az = 80cm, and that there is friction in the tube (wi given by equation above).
3. Repeat the flow rate calculation from number 2 with the following change: assume a
half-length of tubing is used, with L = 50 cm and Az = 33 cm
1
Transcribed Image 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 bottle such that the corresponding volume between the two AZ 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 expressed as wf = kFLuavg 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 yovg is the average velocity in the tube. You may take the water density to be 999kg/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 (w = 0). In order to get the volumetric flow rate, you must first calculate ygvg: 2. Predict the volumetric flow rate of the siphon (cm3/s), assuming that L = 100cm, Az = 80cm, and that there is friction in the tube (wi given by equation above). 3. Repeat the flow rate calculation from number 2 with the following change: assume a half-length of tubing is used, with L = 50 cm and Az = 33 cm 1
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