Formula to calculate the average diameter of a mixture of two sizes of marbles.
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Im looking to mix two particle sizes of powder metals, and need to calculate the resultant average particle size.
O.K. More specifically: I mix 50 pounds of metal powder that has a spherical particle size of 6 microns with 50 pounds of metal powder that has a spherical particle size of 0.8 micron. What is the resultant average particle size? All have the same density. By volume, 1 large particle is 419 times larger than a small particle, so there are 419 times more small particles than large particles. D(avg) = .5D(1) + .5D(2) does not work. The mixture is much finer because of the large size difference between the two particle sizes. I need to know the average particle size.
physics word-problem chemistry
edited Dec 22 '18 at 5:19
Shaun
9,060113682
asked Aug 26 '17 at 3:18
C DorschC Dorsch
11
$endgroup$
add a comment |
$begingroup$
Im looking to mix two particle sizes of powder metals, and need to calculate the resultant average particle size.
O.K. More specifically: I mix 50 pounds of metal powder that has a spherical particle size of 6 microns with 50 pounds of metal powder that has a spherical particle size of 0.8 micron. What is the resultant average particle size? All have the same density. By volume, 1 large particle is 419 times larger than a small particle, so there are 419 times more small particles than large particles. D(avg) = .5D(1) + .5D(2) does not work. The mixture is much finer because of the large size difference between the two particle sizes. I need to know the average particle size.
physics word-problem chemistry
edited Dec 22 '18 at 5:19
Shaun
9,060113682
asked Aug 26 '17 at 3:18
C DorschC Dorsch
11
$endgroup$
$begingroup$
Do you mean: what is the expected diameter of a randomly chosen particle (particles are chosen with equal probability, regardless of diamater)? In this case, the average you want is simply the arithmetic mean of the two particle diameters ...
$endgroup$
– Zubin Mukerjee
Aug 26 '17 at 3:38
$begingroup$
You should clarify what you mean to ask! I've voted to close the question since it's unclear at the moment.
$endgroup$
– Zubin Mukerjee
Aug 26 '17 at 3:39
$begingroup$
It sounds somewhat like a "balls in an urn" problem except the goal is not to find a probability. Rather you draw out, let is say, 1000 balls at random and $40$% have diameter $d_1$ and $60$% have diameter $d_2$, what is the average diameter? Then $d_{avg}=0.4,d_1+0.6,d_2$. Is this what you have in mind?
$endgroup$
– John Wayland Bales
Aug 26 '17 at 3:46
$begingroup$
On the other hand if what you really want to know is what will be the volume of a certain quantity of the mixture, then it depends on the relative sizes of the particles, not on their average size and becomes a difficult sphere packing problem.
$endgroup$
– John Wayland Bales
Aug 26 '17 at 3:49
add a comment |
$begingroup$
Im looking to mix two particle sizes of powder metals, and need to calculate the resultant average particle size.
O.K. More specifically: I mix 50 pounds of metal powder that has a spherical particle size of 6 microns with 50 pounds of metal powder that has a spherical particle size of 0.8 micron. What is the resultant average particle size? All have the same density. By volume, 1 large particle is 419 times larger than a small particle, so there are 419 times more small particles than large particles. D(avg) = .5D(1) + .5D(2) does not work. The mixture is much finer because of the large size difference between the two particle sizes. I need to know the average particle size.
physics word-problem chemistry
edited Dec 22 '18 at 5:19
Shaun
9,060113682
asked Aug 26 '17 at 3:18
C DorschC Dorsch
11
$endgroup$
Im looking to mix two particle sizes of powder metals, and need to calculate the resultant average particle size.
O.K. More specifically: I mix 50 pounds of metal powder that has a spherical particle size of 6 microns with 50 pounds of metal powder that has a spherical particle size of 0.8 micron. What is the resultant average particle size? All have the same density. By volume, 1 large particle is 419 times larger than a small particle, so there are 419 times more small particles than large particles. D(avg) = .5D(1) + .5D(2) does not work. The mixture is much finer because of the large size difference between the two particle sizes. I need to know the average particle size.
physics word-problem chemistry
physics word-problem chemistry
edited Dec 22 '18 at 5:19
Shaun
9,060113682
asked Aug 26 '17 at 3:18
C DorschC Dorsch
11
edited Dec 22 '18 at 5:19
Shaun
9,060113682
asked Aug 26 '17 at 3:18
C DorschC Dorsch
11
edited Dec 22 '18 at 5:19
Shaun
9,060113682
edited Dec 22 '18 at 5:19
Shaun
9,060113682
edited Dec 22 '18 at 5:19
Shaun
9,060113682
9,060113682
asked Aug 26 '17 at 3:18
C DorschC Dorsch
11
asked Aug 26 '17 at 3:18
C DorschC Dorsch
11
asked Aug 26 '17 at 3:18
C DorschC Dorsch
11
11
$begingroup$
Do you mean: what is the expected diameter of a randomly chosen particle (particles are chosen with equal probability, regardless of diamater)? In this case, the average you want is simply the arithmetic mean of the two particle diameters ...
$endgroup$
– Zubin Mukerjee
Aug 26 '17 at 3:38
$begingroup$
You should clarify what you mean to ask! I've voted to close the question since it's unclear at the moment.
$endgroup$
– Zubin Mukerjee
Aug 26 '17 at 3:39
$begingroup$
It sounds somewhat like a "balls in an urn" problem except the goal is not to find a probability. Rather you draw out, let is say, 1000 balls at random and $40$% have diameter $d_1$ and $60$% have diameter $d_2$, what is the average diameter? Then $d_{avg}=0.4,d_1+0.6,d_2$. Is this what you have in mind?
$endgroup$
– John Wayland Bales
Aug 26 '17 at 3:46
$begingroup$
On the other hand if what you really want to know is what will be the volume of a certain quantity of the mixture, then it depends on the relative sizes of the particles, not on their average size and becomes a difficult sphere packing problem.
$endgroup$
– John Wayland Bales
Aug 26 '17 at 3:49
add a comment |
$begingroup$
Do you mean: what is the expected diameter of a randomly chosen particle (particles are chosen with equal probability, regardless of diamater)? In this case, the average you want is simply the arithmetic mean of the two particle diameters ...
$endgroup$
– Zubin Mukerjee
Aug 26 '17 at 3:38
$begingroup$
You should clarify what you mean to ask! I've voted to close the question since it's unclear at the moment.
$endgroup$
– Zubin Mukerjee
Aug 26 '17 at 3:39
$begingroup$
It sounds somewhat like a "balls in an urn" problem except the goal is not to find a probability. Rather you draw out, let is say, 1000 balls at random and $40$% have diameter $d_1$ and $60$% have diameter $d_2$, what is the average diameter? Then $d_{avg}=0.4,d_1+0.6,d_2$. Is this what you have in mind?
$endgroup$
– John Wayland Bales
Aug 26 '17 at 3:46
$begingroup$
On the other hand if what you really want to know is what will be the volume of a certain quantity of the mixture, then it depends on the relative sizes of the particles, not on their average size and becomes a difficult sphere packing problem.
$endgroup$
– John Wayland Bales
Aug 26 '17 at 3:49
$begingroup$
Do you mean: what is the expected diameter of a randomly chosen particle (particles are chosen with equal probability, regardless of diamater)? In this case, the average you want is simply the arithmetic mean of the two particle diameters ...
$endgroup$
– Zubin Mukerjee
Aug 26 '17 at 3:38
$begingroup$
Do you mean: what is the expected diameter of a randomly chosen particle (particles are chosen with equal probability, regardless of diamater)? In this case, the average you want is simply the arithmetic mean of the two particle diameters ...
$endgroup$
– Zubin Mukerjee
Aug 26 '17 at 3:38
$begingroup$
You should clarify what you mean to ask! I've voted to close the question since it's unclear at the moment.
$endgroup$
– Zubin Mukerjee
Aug 26 '17 at 3:39
$begingroup$
You should clarify what you mean to ask! I've voted to close the question since it's unclear at the moment.
$endgroup$
– Zubin Mukerjee
Aug 26 '17 at 3:39
$begingroup$
It sounds somewhat like a "balls in an urn" problem except the goal is not to find a probability. Rather you draw out, let is say, 1000 balls at random and $40$% have diameter $d_1$ and $60$% have diameter $d_2$, what is the average diameter? Then $d_{avg}=0.4,d_1+0.6,d_2$. Is this what you have in mind?
$endgroup$
– John Wayland Bales
Aug 26 '17 at 3:46
$begingroup$
It sounds somewhat like a "balls in an urn" problem except the goal is not to find a probability. Rather you draw out, let is say, 1000 balls at random and $40$% have diameter $d_1$ and $60$% have diameter $d_2$, what is the average diameter? Then $d_{avg}=0.4,d_1+0.6,d_2$. Is this what you have in mind?
$endgroup$
– John Wayland Bales
Aug 26 '17 at 3:46
$begingroup$
On the other hand if what you really want to know is what will be the volume of a certain quantity of the mixture, then it depends on the relative sizes of the particles, not on their average size and becomes a difficult sphere packing problem.
$endgroup$
– John Wayland Bales
Aug 26 '17 at 3:49
$begingroup$
On the other hand if what you really want to know is what will be the volume of a certain quantity of the mixture, then it depends on the relative sizes of the particles, not on their average size and becomes a difficult sphere packing problem.
$endgroup$
– John Wayland Bales
Aug 26 '17 at 3:49
add a comment |
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$begingroup$
Do you mean: what is the expected diameter of a randomly chosen particle (particles are chosen with equal probability, regardless of diamater)? In this case, the average you want is simply the arithmetic mean of the two particle diameters ...
$endgroup$
– Zubin Mukerjee
Aug 26 '17 at 3:38
$begingroup$
You should clarify what you mean to ask! I've voted to close the question since it's unclear at the moment.
$endgroup$
– Zubin Mukerjee
Aug 26 '17 at 3:39
$begingroup$
It sounds somewhat like a "balls in an urn" problem except the goal is not to find a probability. Rather you draw out, let is say, 1000 balls at random and $40$% have diameter $d_1$ and $60$% have diameter $d_2$, what is the average diameter? Then $d_{avg}=0.4,d_1+0.6,d_2$. Is this what you have in mind?
$endgroup$
– John Wayland Bales
Aug 26 '17 at 3:46
$begingroup$
On the other hand if what you really want to know is what will be the volume of a certain quantity of the mixture, then it depends on the relative sizes of the particles, not on their average size and becomes a difficult sphere packing problem.
$endgroup$
– John Wayland Bales
Aug 26 '17 at 3:49