Pressure difference

Storyboard

When two columns of liquid with different heights are connected, it can generate a pressure difference that leads to a flow of liquid from the taller column to the shorter one. This movement continues until both columns reach the same height, eliminating any pressure difference.

>Model

ID:(1608, 0)



Mechanisms

Iframe

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Code
Concept

Mechanisms

ID:(15478, 0)



Connecting two liquid columns

Concept

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If two columns of water with different heights at their bases are connected, a situation arises where there is a pressure difference along the connecting tube.

This setup allows us to study how the pressure difference generates a liquid flow along the tube. We can consider an element of liquid with a certain length and a section equal to that of the tube, and estimate the corresponding mass using the density. With the section, we can also convert the pressure difference into a force difference and, ultimately, study how volumes in liquids are accelerated due to pressure differences.

ID:(933, 0)



Pressure difference between columns

Concept

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If there is the pressure difference ($\Delta p$) between two points, as determined by the equation:

$ \Delta p = p_2 - p_1 $



we can utilize the water column pressure ($p$), which is defined as:

$ p_t = p_0 + \rho_w g h $



This results in:

$\Delta p=p_2-p_1=p_0+\rho_wh_2g-p_0-\rho_wh_1g=\rho_w(h_2-h_1)g$



As the height difference ($\Delta h$) is:

$ \Delta h = h_2 - h_1 $



the pressure difference ($\Delta p$) can be expressed as:

$ \Delta p = \rho_w g \Delta h $

ID:(15704, 0)



Model

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Parameters

Symbol
Text
Variable
Value
Units
Calculate
MKS Value
MKS Units
$g$
g
Gravitational Acceleration
m/s^2
$\rho_w$
rho_w
Liquid density
kg/m^3
$\Delta p$
Dp
Variación de la Presión
Pa

Variables

Symbol
Text
Variable
Value
Units
Calculate
MKS Value
MKS Units
$\Delta h$
Dh
Height of liquid column
m
$h_1$
h_1
Height or depth 1
m
$h_2$
h_2
Height or depth 2
m
$p_1$
p_1
Pressure in column 1
Pa
$p_2$
p_2
Pressure in column 2
Pa

Calculations


First, select the equation: to , then, select the variable: to

Calculations

Symbol
Equation
Solved
Translated

Calculations

Symbol
Equation
Solved
Translated

Variable Given Calculate Target : Equation To be used




Equations

#
Equation

$ \Delta h = h_2 - h_1 $

Dh = h_2 - h_1


$ \Delta p = p_2 - p_1 $

Dp = p_2 - p_1


$ \Delta p = \rho_w g \Delta h $

Dp = rho_w * g * Dh


$ p_1 = \rho_w g h_1 $

p = rho_w * g * h


$ p_2 = \rho_w g h_2 $

p = rho_w * g * h

ID:(15479, 0)



Height difference

Equation

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When two liquid columns are connected with the height of liquid column 1 ($h_1$) and the height of liquid column 2 ($h_2$), a the height difference ($\Delta h$) is formed, which is calculated as follows:

$ \Delta h = h_2 - h_1 $

$\Delta h$
Height of liquid column
$m$
5819
$h_1$
Height or depth 1
$m$
6259
$h_2$
Height or depth 2
$m$
6260



the height difference ($\Delta h$) will generate the pressure difference that will cause the liquid to flow from the higher column to the lower one.

ID:(4251, 0)



Pressure difference

Equation

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When two liquid columns are connected with the pressure in column 1 ($p_1$) and the pressure in column 2 ($p_2$), a the pressure difference ($\Delta p$) is formed, which is calculated according to the following formula:

$ \Delta p = p_2 - p_1 $

$p_1$
Pressure in column 1
$Pa$
6261
$p_2$
Pressure in column 2
$Pa$
6262
$\Delta p$
Variación de la Presión
$Pa$
6673



the pressure difference ($\Delta p$) represents the pressure difference that will cause the liquid to flow from the taller column to the shorter one.

ID:(4252, 0)



Pressure of a column (1)

Equation

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If we consider the expression of the column force ($F$) and divide it by the column Section ($S$), we obtain the water column pressure ($p$). In this process, we simplify the column Section ($S$), so it no longer depends on it. The resulting expression is:

$ p_1 = \rho_w g h_1 $

$ p = \rho_w g h $

$h$
$h_1$
Height or depth 1
$m$
6259
$g$
Gravitational Acceleration
9.8
$m/s^2$
5310
$\rho_w$
Liquid density
$kg/m^3$
5407
$p$
$p_1$
Pressure in column 1
$Pa$
6261

As the the column force ($F$) generated by a column of liquid of the column height ($h$), the column Section ($S$), the liquid density ($\rho_w$), and the gravitational Acceleration ($g$) is

$ F = S h \rho_w g $



and the the water column pressure ($p$) is then defined as

$ p \equiv\displaystyle\frac{ F }{ S }$



we have that the the water column pressure ($p$) generated by a column of liquid is

$ p = \rho_w g h $

ID:(4249, 1)



Pressure of a column (2)

Equation

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If we consider the expression of the column force ($F$) and divide it by the column Section ($S$), we obtain the water column pressure ($p$). In this process, we simplify the column Section ($S$), so it no longer depends on it. The resulting expression is:

$ p_2 = \rho_w g h_2 $

$ p = \rho_w g h $

$h$
$h_2$
Height or depth 2
$m$
6260
$g$
Gravitational Acceleration
9.8
$m/s^2$
5310
$\rho_w$
Liquid density
$kg/m^3$
5407
$p$
$p_2$
Pressure in column 2
$Pa$
6262

As the the column force ($F$) generated by a column of liquid of the column height ($h$), the column Section ($S$), the liquid density ($\rho_w$), and the gravitational Acceleration ($g$) is

$ F = S h \rho_w g $



and the the water column pressure ($p$) is then defined as

$ p \equiv\displaystyle\frac{ F }{ S }$



we have that the the water column pressure ($p$) generated by a column of liquid is

$ p = \rho_w g h $

ID:(4249, 2)



Pressure difference between columns

Equation

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The height difference, denoted by the height difference ($\Delta h$), implies that the pressure in both columns is distinct. In particular, the pressure difference ($\Delta p$) is a function of the liquid density ($\rho_w$), the gravitational Acceleration ($g$), and the height difference ($\Delta h$), as follows:

$ \Delta p = \rho_w g \Delta h $

$g$
Gravitational Acceleration
9.8
$m/s^2$
5310
$\Delta h$
Height of liquid column
$m$
5819
$\rho_w$
Liquid density
$kg/m^3$
5407
$\Delta p$
Variación de la Presión
$Pa$
6673

If there is the pressure difference ($\Delta p$) between two points, as determined by the equation:

$ \Delta p = p_2 - p_1 $



we can utilize the water column pressure ($p$), which is defined as:

$ p_t = p_0 + \rho_w g h $



This results in:

$\Delta p=p_2-p_1=p_0+\rho_wh_2g-p_0-\rho_wh_1g=\rho_w(h_2-h_1)g$



As the height difference ($\Delta h$) is:

$ \Delta h = h_2 - h_1 $



the pressure difference ($\Delta p$) can be expressed as:

$ \Delta p = \rho_w g \Delta h $

ID:(4345, 0)