some question about the velocity of fluid

dzsz99185 · November 8, 2015, 21:50

Why the square root of X Y Z velocity is not equal to velocity magnitude?

LuckyTran · November 9, 2015, 00:30

The root of sums of squares of x,y,z velocity components does give the velocity magnitude, but that's not what you're computing.

You are computing the average face values of x,y,z velocities and comparing the root of sum of squares of these to the average face value of the velocity magnitude. These two quantities aren't equivalent and are not equal in general.

You need to be very careful and make sure you are using the right type of weights to compute the quantity you are interested in. There are simple arithmetic averages, area-weighted averages, mass-flow weighted averages, etc.

As a simple example consider two velocity vectors:
(1,-1,1) and (1,1,1)
The local velocity magnitudes are sqrt(3) each

The arithmetic average of these two local velocities is the average of sqrt(3) and sqrt(3), which is simply sqrt(3). Hence the arithmetic averaged velocity magnitude is sqrt(3)

The average of the individual velocities components are:
(1,0,1)
The root of sum of square of this average velocity is sqrt(2) which is different than sqrt(3). Hence taking the square root of sum of squares of averaged velocity components does not give the average velocity magnitude.

It helps to avoid confusion if you say out in very long sentences exactly what you are computing so that you do not confuse a quantity for another quantity. If you start calling both of these "average velocity magnitude" then it's very easy to fool yourself into thinking that you should get the same result. The error begins with fooling yourself into thinking that they are the same quantity in the first place.

flotus1 · November 9, 2015, 01:52

Or in short: $\overline{\phi^2} \neq \overline{\phi}^2$

dzsz99185 · November 9, 2015, 07:04

Thank u very much! Very useful!

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November 9, 2015, 00:30		#2
LuckyTran Senior Member Lucky Join Date: Apr 2011 Location: Orlando, FL USA Posts: 5,674 Rep Power: 65	The root of sums of squares of x,y,z velocity components does give the velocity magnitude, but that's not what you're computing. You are computing the average face values of x,y,z velocities and comparing the root of sum of squares of these to the average face value of the velocity magnitude. These two quantities aren't equivalent and are not equal in general. You need to be very careful and make sure you are using the right type of weights to compute the quantity you are interested in. There are simple arithmetic averages, area-weighted averages, mass-flow weighted averages, etc. As a simple example consider two velocity vectors: (1,-1,1) and (1,1,1) The local velocity magnitudes are sqrt(3) each The arithmetic average of these two local velocities is the average of sqrt(3) and sqrt(3), which is simply sqrt(3). Hence the arithmetic averaged velocity magnitude is sqrt(3) The average of the individual velocities components are: (1,0,1) The root of sum of square of this average velocity is sqrt(2) which is different than sqrt(3). Hence taking the square root of sum of squares of averaged velocity components does not give the average velocity magnitude. It helps to avoid confusion if you say out in very long sentences exactly what you are computing so that you do not confuse a quantity for another quantity. If you start calling both of these "average velocity magnitude" then it's very easy to fool yourself into thinking that you should get the same result. The error begins with fooling yourself into thinking that they are the same quantity in the first place.

November 9, 2015, 01:52		#3
flotus1 Super Moderator Alex Join Date: Jun 2012 Location: Germany Posts: 3,399 Rep Power: 46	Or in short: $\overline{\phi^2} \neq \overline{\phi}^2$

November 9, 2015, 07:04		#4
dzsz99185 New Member dzs Join Date: Jul 2015 Posts: 2 Rep Power: 0	Thank u very much! Very useful!