Does capillary rise violate hydrostatic paradox?Hydrostatic pressure?Question on the hydrostatic paradoxIt's about capillary rise of waterIs hydrostatic pressure independent of temperature?About hydrostatic pressure affecting measured weight on a scaleAssuming hydrostatic pressure distribution despite fluid motionPitot tube, assumption of hydrostatic pressure distributionHydrostatic pressure in a gasHydrostatic pressure: clarificationsHydrostatic Condition in Fluid
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Does capillary rise violate hydrostatic paradox?
Hydrostatic pressure?Question on the hydrostatic paradoxIt's about capillary rise of waterIs hydrostatic pressure independent of temperature?About hydrostatic pressure affecting measured weight on a scaleAssuming hydrostatic pressure distribution despite fluid motionPitot tube, assumption of hydrostatic pressure distributionHydrostatic pressure in a gasHydrostatic pressure: clarificationsHydrostatic Condition in Fluid
$begingroup$
If $p$ is a pressure and $p_A = p_textatm + hdg,,$ $p_B = p_textatm$, is hydrostatic paradox violated, shouldn't $p_A=p_B$?
fluid-statics capillary-action
edited 10 mins ago
Qmechanic♦
106k121961223
asked 5 hours ago
Lelouche LamperougeLelouche Lamperouge
604
$endgroup$
add a comment |
$begingroup$
If $p$ is a pressure and $p_A = p_textatm + hdg,,$ $p_B = p_textatm$, is hydrostatic paradox violated, shouldn't $p_A=p_B$?
fluid-statics capillary-action
edited 10 mins ago
Qmechanic♦
106k121961223
asked 5 hours ago
Lelouche LamperougeLelouche Lamperouge
604
$endgroup$
$begingroup$
So P(A) need not be equal to P(B)???
$endgroup$
– Lelouche Lamperouge
5 hours ago
$begingroup$
You must have meant hydrostatic principle rather than hydrostatic paradox. ;) A paradox cannot be "violated"--rather, it appears to violate principles. And true principles resolve the paradox rather than violating it. :P
$endgroup$
– Dvij Mankad
4 hours ago
add a comment |
$begingroup$
If $p$ is a pressure and $p_A = p_textatm + hdg,,$ $p_B = p_textatm$, is hydrostatic paradox violated, shouldn't $p_A=p_B$?
fluid-statics capillary-action
edited 10 mins ago
Qmechanic♦
106k121961223
asked 5 hours ago
Lelouche LamperougeLelouche Lamperouge
604
$endgroup$
If $p$ is a pressure and $p_A = p_textatm + hdg,,$ $p_B = p_textatm$, is hydrostatic paradox violated, shouldn't $p_A=p_B$?
fluid-statics capillary-action
fluid-statics capillary-action
edited 10 mins ago
Qmechanic♦
106k121961223
asked 5 hours ago
Lelouche LamperougeLelouche Lamperouge
604
edited 10 mins ago
Qmechanic♦
106k121961223
asked 5 hours ago
Lelouche LamperougeLelouche Lamperouge
604
edited 10 mins ago
Qmechanic♦
106k121961223
edited 10 mins ago
Qmechanic♦
106k121961223
edited 10 mins ago
Qmechanic♦
106k121961223
106k121961223
asked 5 hours ago
Lelouche LamperougeLelouche Lamperouge
604
asked 5 hours ago
Lelouche LamperougeLelouche Lamperouge
604
asked 5 hours ago
Lelouche LamperougeLelouche Lamperouge
604
604
$begingroup$
So P(A) need not be equal to P(B)???
$endgroup$
– Lelouche Lamperouge
5 hours ago
$begingroup$
You must have meant hydrostatic principle rather than hydrostatic paradox. ;) A paradox cannot be "violated"--rather, it appears to violate principles. And true principles resolve the paradox rather than violating it. :P
$endgroup$
– Dvij Mankad
4 hours ago
add a comment |
$begingroup$
So P(A) need not be equal to P(B)???
$endgroup$
– Lelouche Lamperouge
5 hours ago
$begingroup$
You must have meant hydrostatic principle rather than hydrostatic paradox. ;) A paradox cannot be "violated"--rather, it appears to violate principles. And true principles resolve the paradox rather than violating it. :P
$endgroup$
– Dvij Mankad
4 hours ago
$begingroup$
So P(A) need not be equal to P(B)???
$endgroup$
– Lelouche Lamperouge
5 hours ago
$begingroup$
So P(A) need not be equal to P(B)???
$endgroup$
– Lelouche Lamperouge
5 hours ago
$begingroup$
You must have meant hydrostatic principle rather than hydrostatic paradox. ;) A paradox cannot be "violated"--rather, it appears to violate principles. And true principles resolve the paradox rather than violating it. :P
$endgroup$
– Dvij Mankad
4 hours ago
$begingroup$
You must have meant hydrostatic principle rather than hydrostatic paradox. ;) A paradox cannot be "violated"--rather, it appears to violate principles. And true principles resolve the paradox rather than violating it. :P
$endgroup$
– Dvij Mankad
4 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The pressures at A and B are indeed equal. However, the pressure in the fluid immediately below the curved meniscus is equal to $p_atm-hdg$ as a result of surface tension. So the pressure at A is $$p_A=p_atm-hdg+hdg=p_atm=p_B$$That is, there is a discontinuous change in pressure across the meniscus as a result of the surface tension in combination with the curvature. The pressure on the upper side of the interface is $p_atm$ and the pressure on the lower side of the interface is $p_atm-hdg$.
answered 4 hours ago
Chester MillerChester Miller
15.8k2825
$endgroup$
add a comment |
$begingroup$
$p_A$ is equal to $p_B$ here. The disparity is arising due to the fact that pressure just outside the meniscus is greater than the pressure inside. This is due to the curvature of the meniscus and surface tension.
This difference is compensated by $hdg$ to make $p_A=p_B$.
edited 32 mins ago
Sebastiano
306119
answered 4 hours ago
himanshuhimanshu
353
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
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active
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$begingroup$
The pressures at A and B are indeed equal. However, the pressure in the fluid immediately below the curved meniscus is equal to $p_atm-hdg$ as a result of surface tension. So the pressure at A is $$p_A=p_atm-hdg+hdg=p_atm=p_B$$That is, there is a discontinuous change in pressure across the meniscus as a result of the surface tension in combination with the curvature. The pressure on the upper side of the interface is $p_atm$ and the pressure on the lower side of the interface is $p_atm-hdg$.
answered 4 hours ago
Chester MillerChester Miller
15.8k2825
$endgroup$
add a comment |
$begingroup$
The pressures at A and B are indeed equal. However, the pressure in the fluid immediately below the curved meniscus is equal to $p_atm-hdg$ as a result of surface tension. So the pressure at A is $$p_A=p_atm-hdg+hdg=p_atm=p_B$$That is, there is a discontinuous change in pressure across the meniscus as a result of the surface tension in combination with the curvature. The pressure on the upper side of the interface is $p_atm$ and the pressure on the lower side of the interface is $p_atm-hdg$.
answered 4 hours ago
Chester MillerChester Miller
15.8k2825
$endgroup$
add a comment |
$begingroup$
The pressures at A and B are indeed equal. However, the pressure in the fluid immediately below the curved meniscus is equal to $p_atm-hdg$ as a result of surface tension. So the pressure at A is $$p_A=p_atm-hdg+hdg=p_atm=p_B$$That is, there is a discontinuous change in pressure across the meniscus as a result of the surface tension in combination with the curvature. The pressure on the upper side of the interface is $p_atm$ and the pressure on the lower side of the interface is $p_atm-hdg$.
answered 4 hours ago
Chester MillerChester Miller
15.8k2825
$endgroup$
The pressures at A and B are indeed equal. However, the pressure in the fluid immediately below the curved meniscus is equal to $p_atm-hdg$ as a result of surface tension. So the pressure at A is $$p_A=p_atm-hdg+hdg=p_atm=p_B$$That is, there is a discontinuous change in pressure across the meniscus as a result of the surface tension in combination with the curvature. The pressure on the upper side of the interface is $p_atm$ and the pressure on the lower side of the interface is $p_atm-hdg$.
answered 4 hours ago
Chester MillerChester Miller
15.8k2825
answered 4 hours ago
Chester MillerChester Miller
15.8k2825
answered 4 hours ago
Chester MillerChester Miller
15.8k2825
answered 4 hours ago
Chester MillerChester Miller
15.8k2825
15.8k2825
add a comment |
add a comment |
$begingroup$
$p_A$ is equal to $p_B$ here. The disparity is arising due to the fact that pressure just outside the meniscus is greater than the pressure inside. This is due to the curvature of the meniscus and surface tension.
This difference is compensated by $hdg$ to make $p_A=p_B$.
edited 32 mins ago
Sebastiano
306119
answered 4 hours ago
himanshuhimanshu
353
$endgroup$
add a comment |
$begingroup$
$p_A$ is equal to $p_B$ here. The disparity is arising due to the fact that pressure just outside the meniscus is greater than the pressure inside. This is due to the curvature of the meniscus and surface tension.
This difference is compensated by $hdg$ to make $p_A=p_B$.
edited 32 mins ago
Sebastiano
306119
answered 4 hours ago
himanshuhimanshu
353
$endgroup$
add a comment |
$begingroup$
$p_A$ is equal to $p_B$ here. The disparity is arising due to the fact that pressure just outside the meniscus is greater than the pressure inside. This is due to the curvature of the meniscus and surface tension.
This difference is compensated by $hdg$ to make $p_A=p_B$.
edited 32 mins ago
Sebastiano
306119
answered 4 hours ago
himanshuhimanshu
353
$endgroup$
$p_A$ is equal to $p_B$ here. The disparity is arising due to the fact that pressure just outside the meniscus is greater than the pressure inside. This is due to the curvature of the meniscus and surface tension.
This difference is compensated by $hdg$ to make $p_A=p_B$.
edited 32 mins ago
Sebastiano
306119
answered 4 hours ago
himanshuhimanshu
353
edited 32 mins ago
Sebastiano
306119
edited 32 mins ago
Sebastiano
306119
edited 32 mins ago
Sebastiano
306119
306119
answered 4 hours ago
himanshuhimanshu
353
answered 4 hours ago
himanshuhimanshu
353
answered 4 hours ago
himanshuhimanshu
353
353
add a comment |
add a comment |
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$begingroup$
So P(A) need not be equal to P(B)???
$endgroup$
– Lelouche Lamperouge
5 hours ago
$begingroup$
You must have meant hydrostatic principle rather than hydrostatic paradox. ;) A paradox cannot be "violated"--rather, it appears to violate principles. And true principles resolve the paradox rather than violating it. :P
$endgroup$
– Dvij Mankad
4 hours ago