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kamran
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I am going to address stratified layers of liquid and for simplicity assume your gate is a square, not a disc.

First off the gravity will always place the liquids layered, heavy to light from the bottom to top. So the mercury will always settle to the bottom.

Let's call the thickness of mercury $H_M< D/2 sin\theta$ and the thickness of water $H-H_M$.

Then the hydrostatic distribution from the bottom will be two triangles with different slopes. We set the datum at the interface between the water and mercury and ignore the atmospheric pressure.

Then we have two trapezoids of pressure acting on the gate.

From the datum up to the level of the top of the gate, we have $P= D/2*sin\theta*1$ to $P=H-H_M*1$ and we know the pressure acts at the mid-height of this trapezoid.

From the datum to the bottom of the tank we have the pressure of the water above the datum plus the hydrostatic pressure of mercury.

So the top and bottom of this trapezoid are

$$P_{top}=(H-H_M)*1\ and\ the bottom \ P_{bottom}=P_{top}+ H_M*13.5$$$$P_{top}=(H-H_M)*1\ and\quad the bottom\ pressure= \ P_{bottom}=P_{top}+ H_M*13.5$$

And the resultant of these two vectors (P_w+P_m)falls at the height of

$$ H_{effective}= \frac { (P_{W}*d_{W from datum} + P_{M}*d_{mercuty from datum})}{(P_{W}+ P_{M}) } $$

Now we can apply this effective pressure at its resultant height to the gate.

I am going to address stratified layers of liquid and for simplicity assume your gate is a square, not a disc.

First off the gravity will always place the liquids layered, heavy to light from the bottom to top. So the mercury will always settle to the bottom.

Let's call the thickness of mercury $H_M< D/2 sin\theta$ and the thickness of water $H-H_M$.

Then the hydrostatic distribution from the bottom will be two triangles with different slopes. We set the datum at the interface between the water and mercury and ignore the atmospheric pressure.

Then we have two trapezoids of pressure acting on the gate.

From the datum up to the level of the top of the gate, we have $P= D/2*sin\theta*1$ to $P=H-H_M*1$ and we know the pressure acts at the mid-height of this trapezoid.

From the datum to the bottom of the tank we have the pressure of the water above the datum plus the hydrostatic pressure of mercury.

So the top and bottom of this trapezoid are

$$P_{top}=(H-H_M)*1\ and\ the bottom \ P_{bottom}=P_{top}+ H_M*13.5$$

And the resultant of these two vectors (P_w+P_m)falls at the height of

$$ H_{effective}= \frac { (P_{W}*d_{W from datum} + P_{M}*d_{mercuty from datum})}{(P_{W}+ P_{M}) } $$

Now we can apply this effective pressure at its resultant height to the gate.

I am going to address stratified layers of liquid and for simplicity assume your gate is a square, not a disc.

First off the gravity will always place the liquids layered, heavy to light from the bottom to top. So the mercury will always settle to the bottom.

Let's call the thickness of mercury $H_M< D/2 sin\theta$ and the thickness of water $H-H_M$.

Then the hydrostatic distribution from the bottom will be two triangles with different slopes. We set the datum at the interface between the water and mercury and ignore the atmospheric pressure.

Then we have two trapezoids of pressure acting on the gate.

From the datum up to the level of the top of the gate, we have $P= D/2*sin\theta*1$ to $P=H-H_M*1$ and we know the pressure acts at the mid-height of this trapezoid.

From the datum to the bottom of the tank we have the pressure of the water above the datum plus the hydrostatic pressure of mercury.

So the top and bottom of this trapezoid are

$$P_{top}=(H-H_M)*1\ and\quad the bottom\ pressure= \ P_{bottom}=P_{top}+ H_M*13.5$$

And the resultant of these two vectors (P_w+P_m)falls at the height of

$$ H_{effective}= \frac { (P_{W}*d_{W from datum} + P_{M}*d_{mercuty from datum})}{(P_{W}+ P_{M}) } $$

Now we can apply this effective pressure at its resultant height to the gate.

Source Link
kamran
  • 19.7k
  • 2
  • 15
  • 33

I am going to address stratified layers of liquid and for simplicity assume your gate is a square, not a disc.

First off the gravity will always place the liquids layered, heavy to light from the bottom to top. So the mercury will always settle to the bottom.

Let's call the thickness of mercury $H_M< D/2 sin\theta$ and the thickness of water $H-H_M$.

Then the hydrostatic distribution from the bottom will be two triangles with different slopes. We set the datum at the interface between the water and mercury and ignore the atmospheric pressure.

Then we have two trapezoids of pressure acting on the gate.

From the datum up to the level of the top of the gate, we have $P= D/2*sin\theta*1$ to $P=H-H_M*1$ and we know the pressure acts at the mid-height of this trapezoid.

From the datum to the bottom of the tank we have the pressure of the water above the datum plus the hydrostatic pressure of mercury.

So the top and bottom of this trapezoid are

$$P_{top}=(H-H_M)*1\ and\ the bottom \ P_{bottom}=P_{top}+ H_M*13.5$$

And the resultant of these two vectors (P_w+P_m)falls at the height of

$$ H_{effective}= \frac { (P_{W}*d_{W from datum} + P_{M}*d_{mercuty from datum})}{(P_{W}+ P_{M}) } $$

Now we can apply this effective pressure at its resultant height to the gate.