Average velocity of fluid flow
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
edited Nov 7 '15 at 4:02
Don
435415
asked Nov 7 '15 at 2:54
Nayak
283
bumped to the homepage by Community♦ 21 hours ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
add a comment |
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
edited Nov 7 '15 at 4:02
Don
435415
asked Nov 7 '15 at 2:54
Nayak
283
bumped to the homepage by Community♦ 21 hours ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
Perhaps you could ask this in the physics stack exchange?
– cheesyfluff
Nov 7 '15 at 4:19
Nevertheless, I came up with the best answer I could. Tell me if this helped.
– cheesyfluff
Nov 7 '15 at 4:31
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.
– MaxW
Nov 7 '15 at 7:40
If you integrate the differential equation,you get your form of the Washburn eqn.
– Nayak
Nov 7 '15 at 18:14
Velocity of what? What is the difference between $dot{L}$ and $v$?
– mvw
Aug 25 '18 at 16:04
add a comment |
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
edited Nov 7 '15 at 4:02
Don
435415
asked Nov 7 '15 at 2:54
Nayak
283
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
edited Nov 7 '15 at 4:02
Don
435415
asked Nov 7 '15 at 2:54
Nayak
283
edited Nov 7 '15 at 4:02
Don
435415
asked Nov 7 '15 at 2:54
Nayak
283
edited Nov 7 '15 at 4:02
Don
435415
edited Nov 7 '15 at 4:02
Don
435415
edited Nov 7 '15 at 4:02
Don
435415
435415
asked Nov 7 '15 at 2:54
Nayak
283
asked Nov 7 '15 at 2:54
Nayak
283
asked Nov 7 '15 at 2:54
Nayak
283
283
bumped to the homepage by Community♦ 21 hours ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
bumped to the homepage by Community♦ 21 hours ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
Perhaps you could ask this in the physics stack exchange?
– cheesyfluff
Nov 7 '15 at 4:19
Nevertheless, I came up with the best answer I could. Tell me if this helped.
– cheesyfluff
Nov 7 '15 at 4:31
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.
– MaxW
Nov 7 '15 at 7:40
If you integrate the differential equation,you get your form of the Washburn eqn.
– Nayak
Nov 7 '15 at 18:14
Velocity of what? What is the difference between $dot{L}$ and $v$?
– mvw
Aug 25 '18 at 16:04
add a comment |
Perhaps you could ask this in the physics stack exchange?
– cheesyfluff
Nov 7 '15 at 4:19
Nevertheless, I came up with the best answer I could. Tell me if this helped.
– cheesyfluff
Nov 7 '15 at 4:31
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.
– MaxW
Nov 7 '15 at 7:40
If you integrate the differential equation,you get your form of the Washburn eqn.
– Nayak
Nov 7 '15 at 18:14
Velocity of what? What is the difference between $dot{L}$ and $v$?
– mvw
Aug 25 '18 at 16:04
Perhaps you could ask this in the physics stack exchange?
– cheesyfluff
Nov 7 '15 at 4:19
Perhaps you could ask this in the physics stack exchange?
– cheesyfluff
Nov 7 '15 at 4:19
Nevertheless, I came up with the best answer I could. Tell me if this helped.
– cheesyfluff
Nov 7 '15 at 4:31
Nevertheless, I came up with the best answer I could. Tell me if this helped.
– cheesyfluff
Nov 7 '15 at 4:31
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.
– MaxW
Nov 7 '15 at 7:40
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.
– MaxW
Nov 7 '15 at 7:40
If you integrate the differential equation,you get your form of the Washburn eqn.
– Nayak
Nov 7 '15 at 18:14
If you integrate the differential equation,you get your form of the Washburn eqn.
– Nayak
Nov 7 '15 at 18:14
Velocity of what? What is the difference between $dot{L}$ and $v$?
– mvw
Aug 25 '18 at 16:04
Velocity of what? What is the difference between $dot{L}$ and $v$?
– mvw
Aug 25 '18 at 16:04
add a comment |
1 Answer
1
active
oldest
votes
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
$$frac{L(b)-L(a)}{b-a}
= frac{sqrt{frac{gamma Rb}{2mu}}-sqrt{frac{gamma Ra}{2mu}}}{b-a}
=sqrt{frac{gamma R}{2mu}}frac{sqrt{b}-sqrt{a}}{b-a}$$
answered Nov 7 '15 at 4:30
cheesyfluff
1,472520
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. :)
– Nayak
Nov 7 '15 at 4:55
add a comment |
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1 Answer
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1 Answer
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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
$$frac{L(b)-L(a)}{b-a}
= frac{sqrt{frac{gamma Rb}{2mu}}-sqrt{frac{gamma Ra}{2mu}}}{b-a}
=sqrt{frac{gamma R}{2mu}}frac{sqrt{b}-sqrt{a}}{b-a}$$
answered Nov 7 '15 at 4:30
cheesyfluff
1,472520
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. :)
– Nayak
Nov 7 '15 at 4:55
add a comment |
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
$$frac{L(b)-L(a)}{b-a}
= frac{sqrt{frac{gamma Rb}{2mu}}-sqrt{frac{gamma Ra}{2mu}}}{b-a}
=sqrt{frac{gamma R}{2mu}}frac{sqrt{b}-sqrt{a}}{b-a}$$
answered Nov 7 '15 at 4:30
cheesyfluff
1,472520
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. :)
– Nayak
Nov 7 '15 at 4:55
add a comment |
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
$$frac{L(b)-L(a)}{b-a}
= frac{sqrt{frac{gamma Rb}{2mu}}-sqrt{frac{gamma Ra}{2mu}}}{b-a}
=sqrt{frac{gamma R}{2mu}}frac{sqrt{b}-sqrt{a}}{b-a}$$
answered Nov 7 '15 at 4:30
cheesyfluff
1,472520
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
$$frac{L(b)-L(a)}{b-a}
= frac{sqrt{frac{gamma Rb}{2mu}}-sqrt{frac{gamma Ra}{2mu}}}{b-a}
=sqrt{frac{gamma R}{2mu}}frac{sqrt{b}-sqrt{a}}{b-a}$$
answered Nov 7 '15 at 4:30
cheesyfluff
1,472520
answered Nov 7 '15 at 4:30
cheesyfluff
1,472520
answered Nov 7 '15 at 4:30
cheesyfluff
1,472520
answered Nov 7 '15 at 4:30
cheesyfluff
1,472520
1,472520
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. :)
– Nayak
Nov 7 '15 at 4:55
add a comment |
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. :)
– Nayak
Nov 7 '15 at 4:55
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. :)
– Nayak
Nov 7 '15 at 4:55
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. :)
– Nayak
Nov 7 '15 at 4:55
add a comment |
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Perhaps you could ask this in the physics stack exchange?
– cheesyfluff
Nov 7 '15 at 4:19
Nevertheless, I came up with the best answer I could. Tell me if this helped.
– cheesyfluff
Nov 7 '15 at 4:31
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.
– MaxW
Nov 7 '15 at 7:40
If you integrate the differential equation,you get your form of the Washburn eqn.
– Nayak
Nov 7 '15 at 18:14
Velocity of what? What is the difference between $dot{L}$ and $v$?
– mvw
Aug 25 '18 at 16:04