Hydraulic

Written by Jerry Ratzlaff on . Posted in Fluid Dynamics

Hydraulics is the force or motion applied on a confined liquid.

Hydraulic Conductivity

Hydraulic conductivity ( $$k$$ ) is the ease with which a fluid can move through porous spaces or fractures.  See the Darcy's Law.

Hydraulic Diameter

Hydraulic diameter ( $$d_h$$ ) is normally used when the flow is in non-circular pipe or tubes and channels.  Circular pipe has the same pressure drop of a rectangular channel but a greater average velocity.  Square or rectangular pipes have a greater weight and a greater pressure drop compared with a circular pipe with the same section.

Hydraulic Diameter FORMULA

$$\large{ d_h = \frac { 4 A } { P } }$$

Where:

$$\large{ d_h }$$ = hydraulic diameter

$$\large{ A }$$ = cross section area

$$\large{ P }$$ = cross section wetting perimeter

Hydraulic Diameter of a tube with a tube on the Inside

Hydraulic diameter ( $$d_h$$ ) of tube in tube is when the flow is between the inside tube and the outside tube.

Hydraulic Diameter of a tube with a tube on the Inside FORMULA

$$\large{ d_h = \frac { 4 \left( \pi r_{o}{^2} - \pi r_{i}{^2} \right) } { 2 \pi r_o + 2 \pi r_i } }$$

$$\large{ d_h = \frac { 4 \pi \left( r_{o}{^2} - r_{i}{^2} \right) } { 2 \pi \left( r_o + r_i \right) } }$$

Where:

$$\large{ d_h }$$ = hydraulic diameter

$$\large{ r_i }$$ = pipe outside radius of the inside tube

$$\large{ r_o }$$ = pipe inside radius of the outside tube

Hydraulic Diameter of an Isosceles Triangle

Hydraulic Diameter of an Isosceles Triangle FORMULA

$$\large{ d_h = \frac { w\; sin \; \theta } { 1 + sin \frac {\theta}{2} } }$$

Where:

$$\large{ d_h }$$ = hydraulic diameter

$$\large{ w }$$ = length of side

$$\large{ \theta }$$   (Greek symbol theta) = degree

Hydraulic Diameter of a rectangular Tube

Hydraulic diameter ( $$d_h$$ ) of a rectangular tube is when the flow is within the tube.

Hydraulic Diameter of a rectangular Tube FORMULA

$$\large{ d_h = \frac { 4wh } { 2 \left( w + h \right) } }$$

$$\large{ d_h = \frac { 2wh } { w + h } }$$

Where:

$$\large{ d_h }$$ = hydraulic diameter

$$\large{ h }$$ = height of tube

$$\large{ w }$$ = width of tube

Hydraulic Diameter of a Right Triangle

Hydraulic Diameter of a Right Triangle FORMULA

$$\large{ d_h = \frac { 2wh } { w + h + \left( w^2 + h^2 \right) ^ \frac{1}{2} } }$$

Where:

$$\large{ d_h }$$ = hydraulic diameter

$$\large{ h }$$ = height of tube

$$\large{ w }$$ = width of tube

Hydraulic Diameter of a Square Tube

Hydraulic diameter ( $$d_h$$ ) of a square tube is when the flow is within the tube.

Hydraulic Diameter of a Square Tube FORMULA

$$\large{ d_h = w }$$

Where:

$$\large{ d_h }$$ = hydraulic diameter

$$\large{ w }$$ = width of tube

Hydraulic gradient ( $$i$$ ) (dimensionless number) is the change in height (pressure) to length.

$$\large{ i = \frac { h_1 \;-\; h_2} { l} }$$

Where:

$$\large{ i }$$ = hydraulic gradient

$$\large{ h_1 }$$ = pressure head at point 1

$$\large{ h_2 }$$ = pressure head at point 2

$$\large{ l }$$ = length of column

Solve for:

$$\large{ h_1 = i l \;+\; h_2 }$$

$$\large{ h_2 = h_1 \;-\; i l }$$

$$\large{ l = \frac { h_1 \;-\; h_2} { i} }$$

Hydraulic head ( $$h$$ ) is the measurement mechanical energy due to pressure of a fluid from a higher elevation to a lower elevation.

$$\large{ h = \frac {p}{g \rho} }$$

$$\large{ h = h_1 \;-\; h_2 }$$

Where:

$$\large{ h }$$= head

$$\large{ g }$$ = gravitational acceleration

$$\large{ h_1 }$$ = pressure head at point 1

$$\large{ h_2 }$$ = pressure head at point 2

$$\large{ p }$$ = pressure

$$\large{ \rho }$$   (Greek symbol rho) = density

Hydraulic radius ( $$r_h$$ ) is the cross section area of water in a pipe or channel divided by the wetting perimeter.

$$\large{ r_h = \frac { A } { P } }$$

Where:

$$\large{ r_h }$$ = hydraulic radius

$$\large{ A }$$ = cross section flow area

$$\large{ P }$$ = wetted perimeter

Solve for:

$$\large{ A = r_h P }$$

$$\large{ P = \frac { A } { r_h} }$$

Hydraulic Radius of a Pipe

Hydraulic radius ( $$r_h$$ ) is the cross section area of water in a pipe or channel divided by the wetting perimeter.

Hydraulic Radius of a Pipe formula

$$\large{ r_h = \frac { A } { P } }$$

$$\large{ r_h = \frac { \frac { d^2 } { 4 \left( \theta - sin \; \left( 2 \theta \right) \right) } } { \theta d } }$$

$$\large{ r_h = \frac {d} {4} \; \frac { 1 - sin \; \left( 2 \theta \right) } { 2 \theta } }$$

Where:

$$\large{ r_h }$$ = hydraulic radius

$$\large{ d }$$ = diameter ( $$2r$$ )

$$\large{ A }$$ = cross section flow area

$$\large{ P }$$ = wetted perimeter

$$\large{ \theta }$$   (Greek symbol theta) = degree

Solve for:

$$\large{ \theta = cos^{-1} \; \left( 1 - \frac {h} {r} \right) }$$

$$\large{ A = \frac { d^2 } { 4 \left( \theta - sin \; \left( 2 \theta \right) \right) } }$$

$$\large{ P = \theta d }$$

Hydraulic Radius of a Rectangular Channel

Hydraulic radius ( $$r_h$$ ) is the cross section area of water in a pipe or channel divided by the wetting perimeter.

Hydraulic Radius of a Rectangular Channel formula

$$\large{ r_h = \frac { A } { P } }$$

$$\large{ r_h = \frac { b h } { b + 2h } }$$

Where:

$$\large{ r_h }$$ = hydraulic radius

$$\large{ A }$$ = cross section flow area

$$\large{ b }$$ = bottom width of fluid

$$\large{ h }$$ = depth of fluid

$$\large{ P }$$ = wetted perimeter

$$\large{ w }$$ = top width of fluid

Solve for:

$$\large{ A = b h }$$

$$\large{ P = b + 2h }$$

Hydraulic Radius of a Trapezoidal Channel (Equal Side Slopes)

Hydraulic radius ( $$r_h$$ ) is the cross section area of water in a pipe or channel divided by the wetting perimeter.

Hydraulic Radius of a Trapezoidal Channel formula

$$\large{ r_h = \frac { A } { P } }$$

$$\large{ r_h = \frac { bh + \left( z h \right) 2 } { b + 2h \left( bh + z h 2 \right) \frac {1}{2} } }$$

$$\large{ r_h = \frac { h \left( b + z h \right) } { b + 2h \sqrt { 1 + z^2 } } }$$

Where:

$$\large{ r_h }$$ = hydraulic radius

$$\large{ A }$$ = cross section flow area

$$\large{ b }$$ = bottom width of fluid

$$\large{ h }$$ = depth of fluid

$$\large{ P }$$ = wetted perimeter

$$\large{ w }$$ = top width of fluid

$$\large{ z }$$ = width of channel slope

Solve for:

$$\large{ A = bh + \left( z h \right) 2 }$$

$$\large{ P = b + 2h \left( bh + z h 2 \right) \frac {1}{2} }$$

$$\large{ A = h \left( b + z h \right) }$$

$$\large{ P = b + 2h \sqrt { 1 + z^2 } }$$

Hydraulic Radius of a Trapezoidal Channel (Unequal Side Slopes)

Hydraulic radius ( $$r_h$$ ) is the cross section area of water in a pipe or channel divided by the wetting perimeter.

Hydraulic Radius of a Trapezoidal Channel formula

$$\large{ r_h = \frac { A } { P } }$$

$$\large{ r_h = \frac { \frac {h^2} {2} \left( z_1 + x_2 \right) + hg } { b + h \left( \sqrt { 1 + z_{1}{^2} } + \sqrt { 1 + z_{2}{^2} } \right) } }$$

Where:

$$\large{ r_h }$$ = hydraulic radius

$$\large{ A }$$ = cross section flow area

$$\large{ b }$$ = bottom width of fluid

$$\large{ h }$$ = depth of fluid

$$\large{ P }$$ = wetted perimeter

$$\large{ w }$$ = top width of fluid

$$\large{ z }$$ = width of channel slope

Solve for:

$$\large{ A = \frac {h^2} {2} \left( z_1 + x_2 \right) + hg }$$

$$\large{ P = b + h \left( \sqrt { 1 + z_{1}{^2} } + \sqrt { 1 + z_{2}{^2} } \right) }$$