# Control volumes and conservation of ping-pong balls

This is a slightly silly demonstration that I use to illustrate control volume analysis and conservation of volume (or mass). I have the students do calculations during the demonstration and get their answers using the IClicker system. While the demonstration is simple in concept, the execution is complex and there are usually a few false starts.

Equipment

1. Three buckets
2. Stop watch
4. six or seven student volunteers

Demonstration

Line all but one of the students up along the front of the class room facing the class and give each student a ping-pong ball. Place the buckets at the far left, far right, and middle of the line of students. The middle and left hand buckets should be empty and the right hand bucket should be full of ping-pong balls. The student who is not lined up is the timer. The timer uses the stop watch to call out each second. The demonstration has three parts.

Part 1: ping-pong ball flow rate

1. Every two seconds have the students pass their ball to the person standing next to them toward the empty bucket at the far end of the line. The person at the end of the line simply puts their ball into the bucket every two seconds. There should be a continuous line of balls flowing into the bucket. Run the demonstration for 30 seconds.
2. Have the students calculate the ball flow rate in balls per second (should be 1/2  ball/sec).
3. Have the students calculate the number of balls that were placed in the bucket (#balls=flow rate x time=1/2 x 30= 15 balls).
4. Put all the balls back to where they started

Part 2: pipe junction

∑Qout-∑Qin=0

1. Every two seconds have the students pass their ball just like before. Only this time the student standing in front of the middle bucket (the junction student) puts one ball in that bucket every 6 seconds. Again run for 30 seconds.
2. Have the students calculate the rate at which balls are being added to the bucket at the far end. This requires using conservation of ping pong balls (volume). The sum of the flow rates into the junction (1/2  balls/sec) equals the sum of the flow rates out (1/6 balls/sec into the middle bucket and the ball flow rate to the end bucket). Therefore the flow rate into the far bucket is 1/2 – 1/6=1/3 balls/sec. This can be confirmed by seeing that there should be roughly 10 balls in the far end bucket.
3. Put all the balls back to where they started

Part 3: change in storage

dV/dt+∑ Qout-∑Qin=0

1. Every two seconds have the students pass their balls just like last time. Again have the middle student put a ball in the middle bucket every 6 seconds.
2. Every six seconds have a student take a ball out of the far end bucket. Again run for 30 seconds.
3. Have the students calculate the number of balls in the far bucket. The time rate of change of balls in the control volume (far bucket) plus the sum of the flow rates out minus the sum of the flow rates in equals zero. Therefore, dB/dt+1/3-1/6=0 or dB/dt=1/6. The change in number of balls is dB=∫(dB/dt)dt=1/6 x 30=5 balls.

Analysis

After the demonstration is complete I go over all the calculations on the board drawing up the three cases with control volumes where appropriate. The diagrams I use are shown below the descriptions of the individual cases.

An index of all the demonstrations posted on this blog can be found 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.