# Hydraulic

Written by Jerry Ratzlaff on . Posted in Fluid Dynamics

## 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 Depth

Hydraulic depth ( $$d_h$$ ) (also called hydraulic mean depth) is the cross section area of water in a pipe or channel divided by the wetting perimeter.

### Hydraulic Depth formula

$$d_h = \frac { A } { w }$$          $$hydraulic \; depth \;=\; \frac { cross \; section \; flow \; area } { top \; width \; of \; flow }$$

Where:

$$d_h$$ = hydraulic depth

$$A$$ = area of section flow

$$w$$ = top width of fluid

$$P$$ = cross section wetting perimeter

Solve for:

$$A = d_h w$$

$$w = \frac { A } { d_h}$$

## 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

$$d_h = \frac { 4 A } { P }$$          $$hydraulic \; diameter \;=\; \frac { 4 \;\;x\;\; cross \; section \; area } { cross \; section \; wetting \; perimeter }$$

Where:

$$d_h$$ = hydraulic diameter

$$A$$ = cross section area

$$P$$ = cross section wetting perimeter

## Hydraulic Diameter of an Ellipse

### Hydraulic Diameter of an Ellipse FORMULA

$$d_h = \frac { 4wh \left( 64 - 16 e^2 \right) } { \left( w + h \right) \left( 64 - 3 e^4 \right) }$$          $$hydraulic \; diameter \;=\; \frac { 4 \;\;x\;\; width \;\;x\;\; hight \; \left( \; 64 \;-\; 16 \;\;x\;\; e^2 \; \right) } { \left( \; width \;+\; height \; \right) \; \left( \; 64 \;-\; 3 \;\;x\;\; e^4 \; \right) }$$

$$e = \frac { w - h } { w + h }$$          $$e \;=\; \frac { width \;-\; height } { width \;+\; height }$$

Where:

$$d_h$$ = hydraulic diameter

$$h$$ = height of ellipse

$$w$$ = width of ellipse

## Hydraulic Diameter of a Duct, Pipe or tube

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

### Hydraulic Diameter of a Duct, Pipe or tube FORMULA

$$d_h = \frac { 4 \pi r^2 } { 2 \pi r }$$          $$hydraulic \; diameter \;=\; \frac { 4 \;\;x\;\; \pi \;\;x\;\; radius^2 } { 2 \;\;x\;\; \pi \;\;x\;\; radius }$$

Where:

$$d_h$$ = hydraulic diameter

$$r$$ = pipe inside radius

## 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

$$d_h = \frac { 4 \left( \pi r_o^2 - \pi r_i^2 \right) } { 2 \pi r_o + 2 \pi r_i }$$          $$hydraulic \; diameter \;=\; \frac { 4 \;\;x\;\; \left( \pi \;\;x\;\; radius_o^2 \;-\; \pi \;\;x\;\; radius_i^2 \right) } { 2 \;\;x\;\; \pi \;\;x\;\; radius_o \;+\; 2 \;\;x\;\; \pi \;\;x\;\; radius_i }$$

$$d_h = \frac { 4 \pi \left( r_o^2 - r_i^2 \right) } { 2 \pi \left( r_o + r_i \right) }$$          $$hydraulic \; diameter \;=\; \frac { 4 \;\;x\;\; \pi \;\;x\;\; \left( radius_o^2 \;-\; radius_i^2 \right) } { 2 \;\;x\;\; \pi \;\;x\;\; \left( radius_o + radius_i \right) }$$

Where:

$$d_h$$ = hydraulic diameter

$$r_i$$ = pipe outside radius of the inside tube

$$r_o$$ = pipe inside radius of the outside tube

## Hydraulic Diameter of an Isosceles Triangle

### Hydraulic Diameter of an Isosceles Triangle FORMULA

$$d_h = \frac { w\; sin \theta } { 1 + sin \frac {\theta}{2} }$$          $$hydraulic \; diameter \;=\; \frac { length \;\;x\;\; sin \; \theta } { 1 \;+\; sin \; \frac {\theta} {2} }$$

Where:

$$d_h$$ = hydraulic diameter

$$w$$ = length of side

$$\theta$$ = 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

$$d_h = \frac { 4wh } { 2 \left( w + h \right) }$$          $$hydraulic \; diameter \;=\; \frac { 4 \;\;x\;\; width \;\;x\;\; height } { 2 \;\;x\;\; \left( width \;+\; height \right) }$$

$$d_h = \frac { 2wh } { w + h }$$          $$hydraulic \; diameter \;=\; \frac { 2 \;\;x\;\; width \;\;x\;\; height } { width \;+\; height }$$

Where:

$$d_h$$ = hydraulic diameter

$$h$$ = height of tube

$$w$$ = width of tube

## Hydraulic Diameter of a Right Triangle

### Hydraulic Diameter of a Right Triangle FORMULA

$$d_h = \frac { 2wh } { w + h + \left( w^2 + h^2 \right) ^ \frac{1}{2} }$$          $$hydraulic \; diameter \;=\; \frac { 2 \;\;x\;\; width \;\;x\;\; height } { width \;+\; height \;+\; \left( \; width^2 \;+\; height^2 \; \right) ^ \frac{1}{2} }$$

Where:

$$d_h$$ = hydraulic diameter

$$h$$ = height of tube

$$w$$ = 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

$$d_h = w$$

Where:

$$d_h$$ = hydraulic diameter

$$w$$ = width of tube

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

$$i = \frac { h_1 \;-\; h_2} { l}$$          $$hydraulic \; gradient \;=\; \frac { pressure \; head \; at \; point \; 1 \;-\; pressure \; head \; at \; point \; 2 } { length \; of \; column }$$

Where:

$$i$$ = hydraulic gradient

$$h_1$$ = pressure head at point 1

$$h_2$$ = pressure head at point 2

$$l$$ = length of column

Solve for:

$$h_1 = i l \;+\; h_2$$

$$h_2 = h_1 \;-\; i l$$

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

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

$$h = \frac {p}{g \rho}$$          $$head \;=\; \frac { pressure } { gravitational acceleration \;\;x\;\; density }$$

$$h = h_1 \;-\; h_2$$          $$hydraulic \; gradient \;=\; pressure \; head \; at \; point \; 1 \;-\; pressure \; head \; at \; point \; 2$$

Where:

$$h$$= head

$$p$$ = pressure

$$g$$ = gravitational acceleration

$$\rho$$ = density

$$h_1$$ = pressure head at point 1

$$h_2$$ = pressure head at point 2

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

$$r_h = \frac { A } { P }$$          $$hydraulic \; radius \;=\; \frac { cross \; section \; flow \; area } { wetted \; perimeter }$$

Where:

$$r_h$$ = hydraulic radius

$$A$$ = cross section flow area

$$P$$ = wetted perimeter

Solve for:

$$A = r_h P$$

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

## 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

$$r_h = \frac { A } { P }$$          $$hydraulic \; radius \;=\; \frac { cross \; section \; flow \; area } { wetted \; perimeter }$$

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

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

Where:

$$r_h$$ = hydraulic radius

$$d$$ = diameter ( $$2r$$ )

$$A$$ = cross section flow area

$$P$$ = wetted perimeter

Solve for:

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

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

$$P = \theta d$$

## 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

$$r_h = \frac { A } { P }$$          $$hydraulic \; radius \;=\; \frac { cross \; section \; flow \; area } { wetted \; perimeter }$$

$$r_h = \frac { b h } { b + 2h }$$          $$hydraulic \; radius \;=\; \frac { bottom \; width \; of \; fluid \;\;x\;\; depth \; of \; fluid } { bottom \; width \; of \; fluid \;+\; \left( \; 2 \;\;x\;\; depth \; of \; fluid \; \right) }$$

Where:

$$r_h$$ = hydraulic radius

$$b$$ = bottom width of fluid

$$h$$ = depth of fluid

$$w$$ = top width of fluid

$$A$$ = cross section flow area

$$P$$ = wetted perimeter

Solve for:

$$A = b h$$

$$P = b + 2h$$

## 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

$$r_h = \frac { A } { P }$$          $$hydraulic \; radius \;=\; \frac { cross \; section \; flow \; area } { wetted \; perimeter }$$

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

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

Where:

$$r_h$$ = hydraulic radius

$$b$$ = bottom width of fluid

$$h$$ = depth of fluid

$$z$$ = width of channel slope

$$w$$ = top width of fluid

$$A$$ = cross section flow area

$$P$$ = wetted perimeter

Solve for:

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

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

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

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

## 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

$$r_h = \frac { A } { P }$$          $$hydraulic \; radius \;=\; \frac { cross \; section \; flow \; area } { wetted \; perimeter }$$

$$r_h = \frac { \frac {h^2} {2} \left( z_1 + x_2 \right) + hg } { b + h \left( \sqrt { 1 + z_1^2 } + \sqrt { 1 + z_2^2 } \right) }$$

Where:

$$r_h$$ = hydraulic radius

$$b$$ = bottom width of fluid

$$h$$ = depth of fluid

$$z$$ = width of channel slope

$$w$$ = top width of fluid

$$A$$ = cross section flow area

$$P$$ = wetted perimeter

Solve for:

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

$$P = b + h \left( \sqrt { 1 + z_1^2 } + \sqrt { 1 + z_2^2 } \right)$$

## Hydraulic Radius of a Triangular 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 Triangular Channel formula

$$r_h = \frac { A } { P }$$          $$hydraulic \; radius \;=\; \frac { cross \; section \; flow \; area } { wetted \; perimeter }$$

$$r_h = \frac { zh^2 } { 2h \sqrt {1 + z^2 } }$$

Where:

$$r_h$$ = hydraulic radius

$$b$$ = bottom width of fluid

$$h$$ = depth of fluid

$$z$$ = width of channel slope

$$w$$ = top width of fluid

$$A$$ = cross section flow area

$$P$$ = wetted perimeter

Solve for:

$$A = zh^2$$

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

## Flow Area

Flow area refers to the cross section area of the flow within the channel.

## Wetted Perimeter

Wetting perimeter is the portion of the channel that is in contact with the fluid flowing.