Calculating Volume of Water Flow in a 14mm Diameter Pipe at 2m/s

[mathdad]

Water in a 14mm diameter pipe flows at 2m/s. How many liters flow along the pipe in 1 minute?

 

[Greg]

Hi RTCNTC. Any thoughts on how to begin?

 

[mathdad]

Hi RTCNTC. Any thoughts on how to begin?

How to begin?

How about using conversion factors?

We must convert units to mm, right?

 

[MarkFL]

How to begin?

How about using conversion factors?

We must convert units to mm, right?

Yes, converting all measures of length to the same units would be a good start. I think I would convert everything to cm since 1 liter is 1000 cm³. I would also convert all measures of time to minutes.

Now we must determine the volume of water that flows along the pipe in one minute...what shape can we use? What are its dimensions?

 

[HallsofIvy]

14 mm is 1.4 cm so the cross-section area of the pipe is \(\displaystyle \pi (1.4)^2= 1.96\pi\) square cm. Since the water flows through the pipe at 2 m/s, in one second, the water that flows through the pipe can be thought of as a cylinder with that area and length 2 meters. What is that volume? So how much flows through the pipe in 1 min= 60 seconds?

 

[mathdad]

Yes, converting all measures of length to the same units would be a good start. I think I would convert everything to cm since 1 liter is 1000 cm³. I would also convert all measures of time to minutes.

Now we must determine the volume of water that flows along the pipe in one minute...what shape can we use? What are its dimensions?

My Work:

d: Pipe Inner Diameter (m)
Ｑw : Water Flow Rate (m^3/h)
v: Water Velocity (m/s)

v = Qw / (3600 * π * (d / 2)^2 )

2 = Qw / (3600 * π * (0.014 / 2)^2 )
2 = Qw / (3600 * π * (0.014 / 2)^2 )
2 = Qw / 0.55417
Q = 1.10835 m^3/h

1 dm^3 = 1 liter
1.10835 * 1000 = 1108.35 liters/hour
1108.35 / 60 = 18.4725 liters(minute)

Is this right?

 

[MarkFL]

The volume $V$ of a cylinder in terms of its diameter $D$ and height $h$ is:

\(\displaystyle V=\frac{\pi}{4}D^2h\)

We are given the diameter, and to determine the height of the cylindrical volume of water, we may use the kinematic relationship between distance $d$, average speed $v$ and time $t$ to get its height:

\(\displaystyle d=vt\)

And so, we have:

\(\displaystyle V=\frac{\pi}{4}D^2vt\)

Putting in the given values, and converting units, we obtain:

\(\displaystyle V=\frac{\pi}{4}\left(14\text{ mm}\frac{1\text{ cm}}{10\text{ mm}}\right)^2\left(2\,\frac{\text{m}}{\text{s}}\cdot\frac{100\text{ cm}}{1\text{ m}}\cdot\frac{60\text{ s}}{1\text{ min}}\right)\left(1\text{ min}\right)\cdot\frac{1\text{ L}}{1000\text{ cm}^3}=\frac{147}{25}\pi\text{ L}\approx18.47\text{ L}\quad\checkmark\)

 

[mathdad]

The volume $V$ of a cylinder in terms of its diameter $D$ and height $h$ is:

\(\displaystyle V=\frac{\pi}{4}D^2h\)

We are given the diameter, and to determine the height of the cylindrical volume of water, we may use the kinematic relationship between distance $d$, average speed $v$ and time $t$ to get its height:

\(\displaystyle d=vt\)

And so, we have:

\(\displaystyle V=\frac{\pi}{4}D^2vt\)

Putting in the given values, and converting units, we obtain:

\(\displaystyle V=\frac{\pi}{4}\left(14\text{ mm}\frac{1\text{ cm}}{10\text{ mm}}\right)^2\left(2\,\frac{\text{m}}{\text{s}}\cdot\frac{100\text{ cm}}{1\text{ m}}\cdot\frac{60\text{ s}}{1\text{ min}}\right)\left(1\text{ min}\right)\cdot\frac{1\text{ L}}{1000\text{ cm}^3}=\frac{147}{25}\pi\text{ L}\approx18.47\text{ L}\quad\checkmark\)

Ok. I was right.