# Hydraulic Diameter – Calculation Of Hydraulic Diameter of tubes of different geometrical shape

**Hydraulic Diameter **is the characteristics length that is used to calculate the Reynolds Number (*R*_{e}) to determine whether a flow is turbulent or laminar.

It is used when handling flow in non-circular tubes and channels.

Hydraulic diameter converts a cross-section of any geometrical shape into a circular cross-section.

Using the Reynold Number (*R*_{e}), it can be determined whether the flow is turbulent or laminar, or transient based on the following conditions.

# If **R**_{e} < 2300, the flow is laminar

# If 2300 < **R**_{e} < 4000, the flow is transient

# If **R**_{e} > 4000, the flow is turbulent.

The formula of hydraulic diameter is:**D**_{H} = 4**R**_{H} =4 A/P

where,**A** is the cross-sectional area of the flow**P** is the wetted perimeter of the cross-section.

Here, the wetted parameter includes all the surfaces acted upon by the shear stress from the fluid.

For complicated shapes, hydraulic radius **R**_{H} is given by,**R**_{H} = A/P

Where, **A **is the area of the geometric figure.**P** is the perimeter of the geometric figure.

Now the hydraulic diameter **D**_{H} is 4 times the hydraulic radius,

Therefore, **D**_{H} =4R**D**_{H} = 4 A/P

For circular pipe,**D _{H}**= 4πR²/2πR= 2R

Hydraulic diameter is mainly used for calculations including turbulent flow.

Hydraulic diameter is also used to calculate heat transfer in internal-flow problems.

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**Now we will see the hydraulic diameter of tubes of different geometrical shapes.**

**i) Hydraulic Diameter Of Circular Tube:**

In this circular tube, water flows through the whole cross-section of the circular tube.

For full-flowing circular section, geometric diameter and hydraulic diameter are the same.

Hydraulic Radius (**R**_{H}) = A/P = π(d^{2}/4) / πd = d/4

Now, Hydraulic Diameter (**D**_{H}) = 4 **R**_{H} = 4 * d/4 = d

So, hydraulic diameter is equal to geometric diameter in case of circular tube.**ii) Hydraulic diameter of Square Tube:**

Let the length each side of the square tube be a.

Then, Hydraulic Radius (**R**_{H}) = A/P = a^{2}/4a = a/4

Now, Hydraulic Diameter (**D**_{H}) = 4 * R_{H} = 4 * a/4 =a

Therefore, hydraulic diameter of square tube is equal to length of each side of the square.**iii) Hydraulic Diameter Of Rectangular Tube:**

Let the length and breadth of the rectangular tube be l and b.

Now, Hydraulic Radius (**R**_{H}) = A/P = l*b /2(l+b)

and Hydraulic Diameter (**D**_{H}) = 4 * **R**_{H} = 4 * l*b / 2 ( l + b) = 2 x lxb / l+b

So for full flowing Rectangular Tube,**Hydraulic Diameter (D**_{H}**) = 4 * R**_{H}** = 4 * l*b / 2 ( l + b) = 2 x lxb / l+b****iv) Hydraulic Diameter Of Circular Tube with Circular Tube on the Inside:**

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Let r_{o} be the inner radius of the outer tube and r_{i }be the outer radius of the inner tube.

Then Hydraulic Radius (**R**_{H}) = A/P = (πr_{o}^{2}– πr_{i}^{2}) / ( 2πr_{o }+ 2πr_{i})

And Hydraulic Diameter (**D**_{H}) = 4 * (πr_{o}^{2}– πr_{i}^{2}) / ( 2πr_{o }+ 2πr_{i}) = 2 ( r_{o }-r_{i})

Above we have calculated the hydraulic diameter of four **closed channels**. Now we will see how to find the hydraulic diameter of the **open channel**.

**Difference between open channel and closed channel:**

Whenever the flow of water is under influence of atmospheric pressure, it is called open channel. Eg-river.

In case of water flow in a pipe, if the pipe is full of water it is called closed channel and if the water is flowing without filling the whole pipe i.e flowing partially full, it is called open channel.

**Now let us see the calculation of Hydraulic Diameter of some open channel:**

**i) Hydraulic Diameter of rectangular tube ( Open Channel ):**

Let the depth of flow in rectangular tube is represented by** y** and **b** be the bottom width of the tube.

Now, Hydrauic Radius ( **R**_{H} )

= Area / Wetted Perimeter

= y x b / 2y + b

And Hydraulic Diameter ( **D**_{H} )

= 4 x **R**_{H}

=4 x y x b/ 2y +b

= 4yb/ 2y+b**ii) Hydraulic Diameter of Trapezoidal Tube (Open Channel):**

List of parameters for calculation of Hydraulic Diameter of Trapezoidal Tube in Open Channel are:

a) **y** is the depth of the lqiuid

b) **b** is the bottom width

c) **Q** is the width of the liquid surface

d) **W** is the wetted length measured along the slope side

e) **α** is the angle of the slope side from vertical

f) the slope is specified as horizontal:vertical = **z:1**

Now Hydraulic Radius ( **R**_{H} ) = A/P

Area Of cross-section = y (b + Q)/2 .. (1)

Now as Q = b + 2zy

We can write eq (1) as…

A = y/2 (b +b +2zy)

=> A = y/2 (2b +2zy)

=> A = y (b+zy)

Now, the Wetted Parameter (P) = b + 2W

But by Pythagoras formula, we have

W^{2} = y^{2} + (yz)^{2}

or W = (y^{2} + (yz)^{2} )^{1/2}

Hence, Wetted Paramater P = b + 2y (1 + z^{2})^{1/2}

So, the Hydraulic Radius ( **R**_{H} ) = A/P = y (b+zy) / [b + 2y (1 + z^{2})^{1/2}]

And Hydraulic Diameter ( **D**_{H} ) = 4 x **R**_{H} = 4 x y (b+zy) / [b + 2y (1 + z^{2})^{1/2}]**iii) Hydraulic Diameter of Triangular Tube (Open Channel):**

List of parameters for calculation of Hydraulic Diameter of Triangular Tube in Open Channel are:

a) **B** is the width of the liquid flow

b) **Q** is the sloped length of the triangle

c) **y** is the depth of the liquid measured from the vertex of triangle

d) and the side slope specification is horizontal : vertical = **z:1**

Area of cross-section (A) = By/2

From the figure, we have

B=2yz

So Area (A) = y^{2}z

and Wetted Parameter (P) = 2Q

By Pythagoras theorem on triangle , we have

Q^{2} = y^{2} + (yz)^{2}

So Parameter (P) becomes 2[ y^{2} (1 + z^{2}) ]^{1/2}

Hence, the Hydraulic Radius ( **R**_{H} )

= A/P

= y^{2}z / 2[ y^{2} (1 + z^{2}) ]^{1/2}

And Hydraulic Diameter ( **D**_{H} )

= 4 x **R**_{H}

= 4 x y^{2}z / 2[ y^{2} (1 + z^{2}) ]^{1/2}

= 2 x y^{2}z / [ y^{2} (1 + z^{2}) ]^{1/2}