Hydraulic diameter

The hydraulic diameter is a calculated variable that can be used to calculate pressure loss and throughput in pipes or channels if the cross-section of the pipe or channel deviates from the circular shape. The use of the hydraulic diameter is a good approximation for turbulent flows , but for laminar flow conditions it can lead to considerable errors. ${\displaystyle d_{h}}$

The hydraulic radius is defined similarly to the hydraulic diameter and is mainly used to calculate flows in open channels . ${\displaystyle r_{h}}$

Hydraulic diameter

definition

The flow conditions for pipes with a circular cross-section are extensively documented. The calculation of the hydraulic diameter can be understood as an attempt to determine the diameter of that pipe with a circular cross-section for a flow channel with any cross-section , which has the same pressure loss as the given flow channel for the same length and the same average flow velocity. ${\displaystyle {\bar {v}}}$ ${\displaystyle \Delta p}$

When a pipe flow is established, the shear forces on the pipe wall of a certain pipe section are in equilibrium with the compressive forces that occur on the cross-sectional areas at the inflow and outflow of this pipe section. The definition of the hydraulic diameter is based on the idea that comparable conditions exist if the cross-sectional area and the wetted circumference are in the same ratio to one another. When considering the cross section, the wetted circumference is the length of the curve at which the fluid touches the pipe wall. ${\displaystyle A}$ ${\displaystyle P}$

${\displaystyle d_{h}=4\cdot {\frac {A}{P}}}$

Examples

• For the flow in an annular gap the width between two concentric tubes with the diameters  D and  d, respectively ${\displaystyle s={\frac {D-d}{2}}}$

${\displaystyle d_{h}=4\cdot {\frac {{\tfrac {\pi }{4}}\cdot (D^{2}-d^{2})}{\pi \cdot (D+d)}}=D-d=2s}$

• For a channel with a square cross-section the length of the side the hydraulic diameter results in ${\displaystyle a}$

${\displaystyle d_{h}=4\cdot {\frac {a^{2}}{4\cdot a}}=a}$

Limited applicability

For most of the cases that exist in practice, the actually prevailing flow conditions can be determined with usable accuracy with the aid of the hydraulic diameter. However, relevant deviations can arise if the shape of the cross-section deviates from a circle in a certain way or if there are laminar flow conditions. Here are two examples:

• Shape of the considered cross-section:

For a cross-section that is composed of a circle (diameter d) and a long narrow gap (gap width b) ("circle with a narrow extension"):

${\displaystyle d_{h}\approx {\frac {\pi d^{2}}{\pi d+2b}}}$

That would be what one would use to determine a higher pressure loss than in a round pipe. In fact, if the gap is narrow enough, it has no effect on the pressure loss. ${\displaystyle d_{h}<d}$

• Laminar gap flow:

For a small, wide gap (gap width, gap height ) the hydraulic diameter is: ${\displaystyle b\gg }$ ${\displaystyle h}$

${\displaystyle d_{h}={\frac {4bh}{2b+2h}}\approx 2h}$

The pressure loss in a round pipe with laminar flow and medium velocity is with the Hagen-Poiseuille law ${\displaystyle {\bar {v}}}$

${\displaystyle \Delta p_{\text{Kreisrohr}}={\frac {32\eta l{\bar {v}}}{d^{2}}}}$

This would mean the pressure loss in the gap

${\displaystyle \Delta p_{{\text{Spalt}},dh}={\frac {8\eta l{\bar {v}}}{h^{2}}}}$.

According to Hagen-Poiseuille, an exact solution can also be given directly for the low, wide gap. This reads

${\displaystyle \Delta p_{\text{Spalt, exakt}}={\frac {12\eta l{\bar {v}}}{h^{2}}}}$.

In this case, the calculation with the aid of delivers a pressure loss that is 33% too low. ${\displaystyle d_{h}}$

Hydraulic radius

The hydraulic radius is the quotient of the flow cross-section  A and the wetted circumference : ${\displaystyle r_{h}}$ ${\displaystyle P}$

${\displaystyle r_{h}={\frac {A}{P}}={\frac {d_{h}}{4}}}$

Especially in the case of open channels is more convenient to use than . ${\displaystyle r_{h}}$ ${\displaystyle d_{h}}$

Examples of the hydraulic radius

	rectangle	Trapezoid	triangle	circle	parabola
width  ${\displaystyle B}$	${\displaystyle b}$	${\displaystyle b+2\cdot mh}$	${\displaystyle 2\cdot mh}$	${\displaystyle \left(\sin {\frac {\theta }{2}}\right)\cdot D}$ or ${\displaystyle 2{\sqrt {h\cdot (D-h)}}}$	${\displaystyle {\frac {3}{2}}{\frac {A}{h}}}$
Medium water depth	${\displaystyle h}$	${\displaystyle {\frac {(b+mh)h}{b+2\cdot mh}}}$	${\displaystyle {\frac {1}{2}}h}$	${\displaystyle \left[{\frac {\theta -\sin \theta }{\sin {\frac {\theta }{2}}}}\right]{\frac {D}{8}}}$	${\displaystyle {\frac {2}{3}}h}$
Cross sectional area  ${\displaystyle A}$	${\displaystyle b\cdot h}$	${\displaystyle (b+mh)\cdot {}h}$	${\displaystyle m\cdot h^{2}}$	${\displaystyle {\frac {1}{8}}(\theta -\sin {\theta })\cdot {}D^{2}}$	${\displaystyle {\frac {2}{3}}Bh}$
wetted perimeter ${\displaystyle P}$	${\displaystyle b+2h}$	${\displaystyle b+2\cdot {}h\cdot {}{\sqrt {1+m^{2}}}}$	${\displaystyle 2h\cdot {}{\sqrt {1+m^{2}}}}$	${\displaystyle {\frac {1}{2}}\theta \cdot {}D}$	${\displaystyle B+{\frac {8}{3}}{\frac {h^{2}}{B}}}$
Hydraulic radius  ${\displaystyle r_{h}}$	${\displaystyle {\frac {bh}{b+2h}}}$	${\displaystyle {\frac {(b+mh)\cdot {}h}{b+2h\cdot {}{\sqrt {1+m^{2}}}}}}$	${\displaystyle {\frac {mh}{2\cdot {}{\sqrt {1+m^{2}}}}}}$	${\displaystyle {\frac {1}{4}}\left[1-{\frac {\sin \theta }{\theta }}\right]D}$	${\displaystyle {\frac {2B^{2}h}{3B^{2}+8h^{2}}}}$

1. ↑ ^{a b} valid for , with . Si ${\displaystyle 0<\xi <1}$ ${\displaystyle \xi ={\tfrac {4h}{b}}}$ ${\displaystyle \textstyle \xi >1:P=\left({\frac {B}{2}}\right)\left[{\sqrt {1+\xi ^{2}}}+{\frac {1}{\xi }}\ln \left(\xi +{\sqrt {1+\xi ^{2}}}\right)\right]}$