Average velocity of fluid flow












3












$begingroup$


Washburn equation for fluid flow in horizontal capillary is given by :
$frac{dL}{dt} = frac{γR}{4µL}$



Find the time dependencies of length of travel, $L(T)$ and average velocity.



I can find the first part by integrating the Washburn eqn but I am a little confused as to how to find the average velocity as a function of t.






average fluid-dynamics







share|cite|improve this question











$endgroup$












  • $begingroup$
    Perhaps you could ask this in the physics stack exchange?
    $endgroup$
    – cheesyfluff
    Nov 7 '15 at 4:19










  • $begingroup$
    Nevertheless, I came up with the best answer I could. Tell me if this helped.
    $endgroup$
    – cheesyfluff
    Nov 7 '15 at 4:31










  • $begingroup$
    You have the Washburn equation wrong. $L^2 = frac{γRt}{4µ}$. This gives distance from starting point and starting time. So at t=0, then L=0.
    $endgroup$
    – MaxW
    Nov 7 '15 at 7:40












  • $begingroup$
    If you integrate the differential equation,you get your form of the Washburn eqn.
    $endgroup$
    – Nayak
    Nov 7 '15 at 18:14










  • $begingroup$
    Velocity of what? What is the difference between $dot{L}$ and $v$?
    $endgroup$
    – mvw
    Aug 25 '18 at 16:04
















3












$begingroup$


Washburn equation for fluid flow in horizontal capillary is given by :
$frac{dL}{dt} = frac{γR}{4µL}$



Find the time dependencies of length of travel, $L(T)$ and average velocity.



I can find the first part by integrating the Washburn eqn but I am a little confused as to how to find the average velocity as a function of t.






average fluid-dynamics







share|cite|improve this question











$endgroup$












  • $begingroup$
    Perhaps you could ask this in the physics stack exchange?
    $endgroup$
    – cheesyfluff
    Nov 7 '15 at 4:19










  • $begingroup$
    Nevertheless, I came up with the best answer I could. Tell me if this helped.
    $endgroup$
    – cheesyfluff
    Nov 7 '15 at 4:31










  • $begingroup$
    You have the Washburn equation wrong. $L^2 = frac{γRt}{4µ}$. This gives distance from starting point and starting time. So at t=0, then L=0.
    $endgroup$
    – MaxW
    Nov 7 '15 at 7:40












  • $begingroup$
    If you integrate the differential equation,you get your form of the Washburn eqn.
    $endgroup$
    – Nayak
    Nov 7 '15 at 18:14










  • $begingroup$
    Velocity of what? What is the difference between $dot{L}$ and $v$?
    $endgroup$
    – mvw
    Aug 25 '18 at 16:04














3












3








3





$begingroup$


Washburn equation for fluid flow in horizontal capillary is given by :
$frac{dL}{dt} = frac{γR}{4µL}$



Find the time dependencies of length of travel, $L(T)$ and average velocity.



I can find the first part by integrating the Washburn eqn but I am a little confused as to how to find the average velocity as a function of t.






average fluid-dynamics







share|cite|improve this question











$endgroup$




Washburn equation for fluid flow in horizontal capillary is given by :
$frac{dL}{dt} = frac{γR}{4µL}$



Find the time dependencies of length of travel, $L(T)$ and average velocity.



I can find the first part by integrating the Washburn eqn but I am a little confused as to how to find the average velocity as a function of t.






average fluid-dynamics




average fluid-dynamics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 7 '15 at 4:02









Don

440415




440415










asked Nov 7 '15 at 2:54









NayakNayak

283




283












  • $begingroup$
    Perhaps you could ask this in the physics stack exchange?
    $endgroup$
    – cheesyfluff
    Nov 7 '15 at 4:19










  • $begingroup$
    Nevertheless, I came up with the best answer I could. Tell me if this helped.
    $endgroup$
    – cheesyfluff
    Nov 7 '15 at 4:31










  • $begingroup$
    You have the Washburn equation wrong. $L^2 = frac{γRt}{4µ}$. This gives distance from starting point and starting time. So at t=0, then L=0.
    $endgroup$
    – MaxW
    Nov 7 '15 at 7:40












  • $begingroup$
    If you integrate the differential equation,you get your form of the Washburn eqn.
    $endgroup$
    – Nayak
    Nov 7 '15 at 18:14










  • $begingroup$
    Velocity of what? What is the difference between $dot{L}$ and $v$?
    $endgroup$
    – mvw
    Aug 25 '18 at 16:04


















  • $begingroup$
    Perhaps you could ask this in the physics stack exchange?
    $endgroup$
    – cheesyfluff
    Nov 7 '15 at 4:19










  • $begingroup$
    Nevertheless, I came up with the best answer I could. Tell me if this helped.
    $endgroup$
    – cheesyfluff
    Nov 7 '15 at 4:31










  • $begingroup$
    You have the Washburn equation wrong. $L^2 = frac{γRt}{4µ}$. This gives distance from starting point and starting time. So at t=0, then L=0.
    $endgroup$
    – MaxW
    Nov 7 '15 at 7:40












  • $begingroup$
    If you integrate the differential equation,you get your form of the Washburn eqn.
    $endgroup$
    – Nayak
    Nov 7 '15 at 18:14










  • $begingroup$
    Velocity of what? What is the difference between $dot{L}$ and $v$?
    $endgroup$
    – mvw
    Aug 25 '18 at 16:04
















$begingroup$
Perhaps you could ask this in the physics stack exchange?
$endgroup$
– cheesyfluff
Nov 7 '15 at 4:19




$begingroup$
Perhaps you could ask this in the physics stack exchange?
$endgroup$
– cheesyfluff
Nov 7 '15 at 4:19












$begingroup$
Nevertheless, I came up with the best answer I could. Tell me if this helped.
$endgroup$
– cheesyfluff
Nov 7 '15 at 4:31




$begingroup$
Nevertheless, I came up with the best answer I could. Tell me if this helped.
$endgroup$
– cheesyfluff
Nov 7 '15 at 4:31












$begingroup$
You have the Washburn equation wrong. $L^2 = frac{γRt}{4µ}$. This gives distance from starting point and starting time. So at t=0, then L=0.
$endgroup$
– MaxW
Nov 7 '15 at 7:40






$begingroup$
You have the Washburn equation wrong. $L^2 = frac{γRt}{4µ}$. This gives distance from starting point and starting time. So at t=0, then L=0.
$endgroup$
– MaxW
Nov 7 '15 at 7:40














$begingroup$
If you integrate the differential equation,you get your form of the Washburn eqn.
$endgroup$
– Nayak
Nov 7 '15 at 18:14




$begingroup$
If you integrate the differential equation,you get your form of the Washburn eqn.
$endgroup$
– Nayak
Nov 7 '15 at 18:14












$begingroup$
Velocity of what? What is the difference between $dot{L}$ and $v$?
$endgroup$
– mvw
Aug 25 '18 at 16:04




$begingroup$
Velocity of what? What is the difference between $dot{L}$ and $v$?
$endgroup$
– mvw
Aug 25 '18 at 16:04










1 Answer
1






active

oldest

votes


















0












$begingroup$

Remember that the average velocity is equal to
$$frac{Delta L}{Delta t}=frac{L_2-L_1}{t_2-t_1}$$
Which can be found by integrating the velocity over a certain interval and dividing it by the length of that interval.



The expression $frac{dL}{dt}$ is equal to the instantaneous velocity, so integrating the function gives
$$L(t) = sqrt{frac{gamma Rt}{2mu}}$$



and so the average velocity from $t=a$ to $t=b$ is



$$frac{L(b)-L(a)}{b-a}
= frac{sqrt{frac{gamma Rb}{2mu}}-sqrt{frac{gamma Ra}{2mu}}}{b-a}
=sqrt{frac{gamma R}{2mu}}frac{sqrt{b}-sqrt{a}}{b-a}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks. That clears it up a bit. But a and b are not given quantities in the question. I will ask in the physics stack forum. :)
    $endgroup$
    – Nayak
    Nov 7 '15 at 4:55











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









0












$begingroup$

Remember that the average velocity is equal to
$$frac{Delta L}{Delta t}=frac{L_2-L_1}{t_2-t_1}$$
Which can be found by integrating the velocity over a certain interval and dividing it by the length of that interval.



The expression $frac{dL}{dt}$ is equal to the instantaneous velocity, so integrating the function gives
$$L(t) = sqrt{frac{gamma Rt}{2mu}}$$



and so the average velocity from $t=a$ to $t=b$ is



$$frac{L(b)-L(a)}{b-a}
= frac{sqrt{frac{gamma Rb}{2mu}}-sqrt{frac{gamma Ra}{2mu}}}{b-a}
=sqrt{frac{gamma R}{2mu}}frac{sqrt{b}-sqrt{a}}{b-a}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks. That clears it up a bit. But a and b are not given quantities in the question. I will ask in the physics stack forum. :)
    $endgroup$
    – Nayak
    Nov 7 '15 at 4:55
















0












$begingroup$

Remember that the average velocity is equal to
$$frac{Delta L}{Delta t}=frac{L_2-L_1}{t_2-t_1}$$
Which can be found by integrating the velocity over a certain interval and dividing it by the length of that interval.



The expression $frac{dL}{dt}$ is equal to the instantaneous velocity, so integrating the function gives
$$L(t) = sqrt{frac{gamma Rt}{2mu}}$$



and so the average velocity from $t=a$ to $t=b$ is



$$frac{L(b)-L(a)}{b-a}
= frac{sqrt{frac{gamma Rb}{2mu}}-sqrt{frac{gamma Ra}{2mu}}}{b-a}
=sqrt{frac{gamma R}{2mu}}frac{sqrt{b}-sqrt{a}}{b-a}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks. That clears it up a bit. But a and b are not given quantities in the question. I will ask in the physics stack forum. :)
    $endgroup$
    – Nayak
    Nov 7 '15 at 4:55














0












0








0





$begingroup$

Remember that the average velocity is equal to
$$frac{Delta L}{Delta t}=frac{L_2-L_1}{t_2-t_1}$$
Which can be found by integrating the velocity over a certain interval and dividing it by the length of that interval.



The expression $frac{dL}{dt}$ is equal to the instantaneous velocity, so integrating the function gives
$$L(t) = sqrt{frac{gamma Rt}{2mu}}$$



and so the average velocity from $t=a$ to $t=b$ is



$$frac{L(b)-L(a)}{b-a}
= frac{sqrt{frac{gamma Rb}{2mu}}-sqrt{frac{gamma Ra}{2mu}}}{b-a}
=sqrt{frac{gamma R}{2mu}}frac{sqrt{b}-sqrt{a}}{b-a}$$






share|cite|improve this answer









$endgroup$



Remember that the average velocity is equal to
$$frac{Delta L}{Delta t}=frac{L_2-L_1}{t_2-t_1}$$
Which can be found by integrating the velocity over a certain interval and dividing it by the length of that interval.



The expression $frac{dL}{dt}$ is equal to the instantaneous velocity, so integrating the function gives
$$L(t) = sqrt{frac{gamma Rt}{2mu}}$$



and so the average velocity from $t=a$ to $t=b$ is



$$frac{L(b)-L(a)}{b-a}
= frac{sqrt{frac{gamma Rb}{2mu}}-sqrt{frac{gamma Ra}{2mu}}}{b-a}
=sqrt{frac{gamma R}{2mu}}frac{sqrt{b}-sqrt{a}}{b-a}$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 7 '15 at 4:30









cheesyfluffcheesyfluff

1,482520




1,482520












  • $begingroup$
    Thanks. That clears it up a bit. But a and b are not given quantities in the question. I will ask in the physics stack forum. :)
    $endgroup$
    – Nayak
    Nov 7 '15 at 4:55


















  • $begingroup$
    Thanks. That clears it up a bit. But a and b are not given quantities in the question. I will ask in the physics stack forum. :)
    $endgroup$
    – Nayak
    Nov 7 '15 at 4:55
















$begingroup$
Thanks. That clears it up a bit. But a and b are not given quantities in the question. I will ask in the physics stack forum. :)
$endgroup$
– Nayak
Nov 7 '15 at 4:55




$begingroup$
Thanks. That clears it up a bit. But a and b are not given quantities in the question. I will ask in the physics stack forum. :)
$endgroup$
– Nayak
Nov 7 '15 at 4:55


















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