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    #1

    maths question- increase with the square root

    Flow increases with the square root of the pressure gradient, so quadrupling the pressure gradient merely doubles the flow.

    I am translating the sentence into my native language. Is it correct to say : Flow is directly proportional to the square root ... ?

  1. konungursvia's Avatar
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    #2

    Re: maths question- increase with the square root

    Yes, that's right: Flow is directly proportional to the square root of the pressure.


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    #3

    Re: maths question- increase with the square root

    Quote Originally Posted by alla View Post
    Flow increases with the square root of the pressure gradient, so quadrupling the pressure gradient merely doubles the flow.

    I am translating the sentence into my native language. Is it correct to say : Flow is directly proportional to the square root ... ?
    Gas dynamics?
    "grad p" is a vector. It can't be proportionate to a scalar quantity.
    Say this:
    proportional to the square root of pressure drop.

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    #4

    Re: maths question- increase with the square root

    I don't agree. An increase in pressure also increases the flow. Both are in the same vectorial direction, so they can be proportional. So it's not a proportionality to the pressure "drop", but to the square root of the pressure gradient (increase). Flow is inversely proportional to its drop or decrease. Less pressure, less flow.


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    #5

    Re: maths question- increase with the square root

    Hello Konun,

    Quote Originally Posted by konungursvia View Post
    I don't agree. An increase in pressure also increases the flow.
    Incorrect. Flow has nothing to do with pressure. :)

    I ≠ I(p)
    I = I(|∇p|)!

    Quote Originally Posted by konungursvia View Post
    Both are in the same vectorial direction, so they can be proportional.
    Pressure is an isotropic scalar field, so it has no direction. P is scalar, on the other hand grad p is vectorial.

    I ~ -sqrt|∇p|.


    Quote Originally Posted by konungursvia View Post
    So it's not a proportionality to the pressure "drop", but to the square root of the pressure gradient (increase) decrease - that is why the minus sign above.
    Quote Originally Posted by konungursvia View Post
    Flow is inversely proportional to its drop or decrease. Less pressure, less flow.
    Konun, particles take the line of least resistance. They move towards lower pressure, counter to gradient direction. The difference in pressure that counts, not the pressure itself.
    Last edited by svartnik; 01-Jun-2009 at 18:20.

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    #6

    Re: maths question- increase with the square root

    Okay, I accept your authoritative knowledge; I did a Bachelor's in Chemistry, but I remember my physics far less than you do.

    Having said that, an increase in the pressure gradient (obviously that of a fluid in a pipe) would not decrease flow, would it?

    I am taking for granted that the OP was talking about the amount of flow, not its direction relative to the gradient (a convention):

    "Flow increases with the square root of the pressure gradient, so quadrupling the pressure gradient merely doubles the flow."

    It seems that if this is correct, then his idea of expressing the magnitude of the quantity as proportional to the square root of the change in pressure gradient seems a good strategy, linguistically, doesn't it?


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    #7

    Re: maths question- increase with the square root

    Quote Originally Posted by konungursvia View Post
    Having said that, an increase in the pressure gradient (obviously that of a fluid or gas in a pipe) would not decrease flow, would it?
    Yes, it would. The magnitude of the gadient vector in the vector field at any point tells us how quickly pressure changes in the gradient's direction. If the pace of change increases, intensity of flow increases. That is what proportionality means in this case. That negative symbol in that scalar equation in my previous post is meant to express that the direction of I is opposite to ∇p.

    Quote Originally Posted by konungursvia View Post
    I am taking for granted that the OP was talking about the amount of flow, not its direction relative to the gradient (a convention):
    So am I.

    Quote Originally Posted by konungursvia View Post
    "Flow increases with the square root of the pressure gradient, so quadrupling the pressure gradient merely doubles the flow."

    It seems that if this is correct, then his idea of expressing the magnitude of the quantity as proportional to the square root of the change in pressure gradient seems a good strategy, linguistically, doesn't it?
    I ~ -sqrt|∇p| - This is what we have to translate from mathematics into English.

    The square root of the magnitude of the pressure gradient at any point is proportionate to the amount of flow.

    Does this sound English? You are a more reliable authority than me to answer my question.

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    #8

    Re: maths question- increase with the square root

    Sounds okay, but I'd use "the magnitude of the flow" rather than the amount. Nice work, Svartnik, I've learnt a lot.

    How about this for a last question: if the pressure gradient were decreased, would this increase flow? If so, would a decrease to zero gradient mean that flow is maximized?

    Also, am I wrong in understanding there is a contradiction here: "quadrupling the pressure gradient merely doubles the flow"; and yet, you tell me I am wrong in understanding that increasing the pressure gradient increases the flow's magnitude. I remain, a bit confused.
    Last edited by konungursvia; 01-Jun-2009 at 19:34.


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    #9

    Re: maths question- increase with the square root

    alla: can you clear this up.
    Are you translating a section on turbulent flow as opposed to laminar flow?

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    #10

    Re: maths question- increase with the square root

    Hi folks.... I've thought about it a good deal, and am convinced there are some errors all around. Although Svartnik is a physicist, I recall I often make elementary errors like mistaking east for west, a decrease for a negative, a linear for a polar coordinate, and so on.

    So here's my thinking: we have all accepted the OP's original statement: "Flow increases with the square root of the pressure gradient, so quadrupling the pressure gradient merely doubles the flow."

    Here I interpret the pressure gradient to be inside pipe of undefined length with a higher fluid (gas) pressure toward one end, let's call it the East or the Russian end, and a lower pressure toward the other end, let's call it the West or the European Union end. So, since the pressure is higher at the Russian end, and is lower at the EU end, there is a net flow toward the low pressure end, toward the lucky Europeans.

    My point is this: given that the flow is a result of the gradient (presumably logarithmic), a propane molecule (for instance) being more likely to be pushed away from the Russian end toward the Euro-end than pushed away from the Euro-zone back toward Russia, it follows that decreasing the pressure gradient (the "potential difference" if you will), which renders the two ends nearer to each other in pressure, will decrease the tendency for net flow to occur.

    So I think the negative you were pointing out is merely the result of the vector direction convention, according to which the higher pressure end is considered positive in terms of the vector for the pressure gradient, whereas the flow goes in the opposite direction, toward the negative end of the pressure gradient.

    It does not follow that an increase in the magnitude of the gradient (an increase in the pressure differential from one end to the other) would result in a lesser magnitude of flow; rather, it would result in a greater flow, albeit in the direction opposite the gradient.

    Doesn't that make sense? Of course, perhaps you mean the same thing, only you're calling a larger "negative" flow a decrease in flow, which is also a possible translation of math into natural language.

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