I find that students often find pipe networks difficult because they have difficulty visualizing that the head change along two different paths that are connected at each end must be the same. One way to demonstrate this is to drain a tank through two different length pipes. You can do this with the same equipment as the siphoning experiment though you will need an extra length of tube. I use a tank with pneumatic tubing connectors in it so that you don’t have to hold the tubes at both ends.

**Equipment**

- A large tank and a table to put it on
- Two tubes or pipes of substantially different (known) lengths
- Two measuring cups
- A stop watch.

In the example in the photos there is a press-to-connect fitting in the base of the tank which I connect to a T-junction and then have the two pipes connected to that.

**Demonstration**

- Fill the tank with water and flood the tubes.
- Place the measuring cups on the floor so that there is a substantial head difference between the top of the water in the tank and the top of the measuring cups
- Lower the outlet of the tubes to the top of each measuring cup and start the water flowing
- Record the time taken for each cup to fill. The longer tube should take a longer time

**Analysis**

The analysis is very similar to the siphoning demonstration so I will only summarize it here.

The total head (H) is the distance from the water surface in the tank to the top of the measuring cups. The head is balanced by the entrance head loss (h_{en}), the head loss in the junction (hj), the head loss in the pipe (h_{p1} or h_{p2}) and the exit loss from each pipe (h_{ex1 }and h_{ex2}). That is,

H = h_{en} +h_{j} + h_{p1} + h_{ex1} = h_{en} +h_{j} + h_{p2} + h_{ex2.}

The first two loss terms cancel. Using the Darcy Weisbach equation we can re-write the remaining terms as

U_{1}^{2}/2g (1+f_{1}L_{1}/d_{1})=U_{2}^{2}/2g(1+f_{2}L_{2}/d_{2})

The ratio of the flow rates is given by

Q_{2}/Q_{1}=((1+f_{1}L_{1}/d_{1})/(1+f_{2}L_{2}/d_{2}))^{1/2}

Here is where it gets messy and, as a result, you won’t get a nice neat ratio of flow rates. In the example that I show in the video the tubes are 1/4” in diameter and 256” and 64” long respectively. Therefore, if the exit velocity head is negligible, the flow rate in the long tube should be roughly half that in the shorter tube. In the test I ran the ratio was 2.5:1. There are various explanations for this difference. First, the exit velocity head may not have been negligible (though accounting for that would have reduced the predicted flow rate ratio). Second, the friction factors for the tubes would be different due to the different Reynolds number (though again, accounting for that would have reduced the predicted flow rate ratio). The most likely explanation for the difference between the predicted and measured flow rate ratio is that the longer tube was curled up in a series of loops whereas the shorter tube was straight. There are additional measurement errors in the tube lengths, and the time for each cup to fill. For example, the times we measured were 25 seconds and 63. It does not take much of an error to get a substantial change in ratio. For example, there is a parallax error in reading the water level in the cups, and There can also be a delay between the person watching the cup calling out ans the timer registering the time. If the recorded times were each out by say plus or minus 3 seconds then the flow rate ratio would vary from 2.1 to 3. Clearly the demonstration provides a great opportunity to talk about experimental error and uncertainty.

Thank you to Kate and Austin for helping out with the video.

An index of all the demonstrations posted on this blog can be found on my website here. Don’t forget to follow @nbkaye on twitter for updates to this blog. If you have a demonstration that you use in class that you would like to share on this blog please email me (nbkaye@clemson.edu). I also welcome comments (through the comments section or via email) on improving the demonstrations.

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