#### 2.1.1. Existing Form of the Division Line in Rectangular Channels

Consider the case of steady, uniform flow in a smooth rectangular channel with an aspect ratio of

b/

h, where

h = water depth and

b = channel width. Due to existence of a secondary current, the division line is actually a zero Reynolds shear stress line in the flow area, and also a line without energy cross-transfer [

21]. Yang et al. [

22] have proven the physical existence of division lines by using experimental data recorded at the Hydraulics Laboratory, University of Wollongong, as well as the experimental data recorded previously by Melling and Whitelaw [

23] and that of Tracy [

24].

Einstein [

25] proposed that the flow cross-sectional area can be divided into three sub-sections that correspond to the side wall and channel bed. However, he did not propose any method of determining the exact location of the division line. Chien and Wang [

26] performed an in-depth study of Einstein’s idea. They gave the total overall boundary shear stress,

${\tau}_{0}$, as:

where

$\rho $ = fluid density;

g = gravitational acceleration;

R = hydraulic radius; and

S_{e} = energy slope.

Then, they divided

${\tau}_{0}$ into two parts corresponding to the bed,

$\overline{{\tau}_{b}}$, and side wall,

$\overline{{\tau}_{w}}$. They also defined two hydraulic radii, one for the wall

${R}_{w}$, and one for the bed

${R}_{b}$. Thus:

where

$\overline{{\tau}_{b}}$ = mean bed shear stress;

$\overline{{\tau}_{w}}$ = mean wall shear stress;

A_{b} = flow area corresponding to channel bed;

A_{w} = flow area corresponding to channel side wall;

R_{w} = hydraulic radius corresponding to side wall; and

R_{b} = hydraulic radius corresponding to bed.

According to Yang [

27], the mechanical energy contained in any flow will be transmitted to the boundary nearest to the “relative distance” and dissipated. The dimensionless relative distance Φ is defined as the ratio of geometrical length between a point in the flow field and the characteristic boundary length, representing the energy dissipation capacity of the boundary (for a smooth boundary channel, this would be the viscous layer thickness

D_{s}).

The side wall and bed will each have their own fair share of surplus energy transferred from the main flow depending on the minimum relative distance Φ. Accordingly, for any unit volume in the flow field, there are two possible ways to transfer the surplus energy: either toward the bed (${\mathsf{\Phi}}_{b}$) or toward the side wall (${\mathsf{\Phi}}_{w}$). If ${\mathsf{\Phi}}_{w}\le {\mathsf{\Phi}}_{b}$, the energy will be transferred toward the side wall. Conversely, if ${\mathsf{\Phi}}_{w}\ge {\mathsf{\Phi}}_{b}$, then the energy will be transported to the bed. It follows that the condition where ${\mathsf{\Phi}}_{w}={\mathsf{\Phi}}_{b}$ will define a division line by which the flow region near the corner is divided into two sub-flow sections.

Therefore, the division line (

${\mathsf{\Phi}}_{w}={\mathsf{\Phi}}_{b}$) in

Figure 1a can be expressed as:

where

z and

y denote the horizontal and vertical axes in the Cartesian coordinates system, respectively;

y = vertical direction normal to bed;

z = transverse flow direction;

${D}_{sw}$ = characteristic length for smooth side wall; and

${D}_{sb}$ = characteristic length for smooth channel bed.

Substituting

${D}_{sw}={C}_{s}\nu /\overline{{u}_{\ast w}}$ and

${D}_{sb}={C}_{s}\nu /\overline{{u}_{\ast b}}$ into Equation (5) gives:

where

${C}_{s}$ = constant;

$\overline{{u}_{\ast w}}$ = mean side wall shear velocity; and

$\overline{{u}_{\ast b}}$ = mean bed shear velocity. The geometric significance of

k in

Figure 1a is the reciprocal of the slope of the line OD. Point D is the intersection of the division line and the water surface line.

Then, it can be obtained that:

Using Equations (3)–(8),

k can be evaluated, yielding:

Using a similar method, Yang and Lim [

27] also gave the evaluation of

k_{1} in

Figure 1b:

where

k_{1} represents the slope of the division line OD in

Figure 1b.

Highly inspired by the derivation of Yang and Lim [

27], we also take the two cases of

Figure 1 to analyze. It has been proven that the critical aspect ratio

α is equal to 2 in smooth rectangular channels by Yang and Lim [

27]. As the rectangle is axisymmetric, we take the vertical line of the channel section as the axis and only analyze half of it.

- (a)
Determination of the location of division line in wide–shallow channel (b/h ≥ 2)

The division line can be determined by Equation (10) when

b/

h ≥ 2 (

Figure 1a). Since

$2h/b$ is a constant, solving Equation (10) will give the value of

k.

Referring to the value of k, a comparison relationship between

$1/k$ and

$2h/b$ is as follows:

It can be shown from Inequality (13), so

That is to say, when b/h ≥ 2, the slope of the division line is always greater than 2h/b. The intersection point of the division lines is on or above the water surface line.

- (b)
Determination of the location of division line in narrow–deep channel (b/h ≤ 2)

The division line can be determined by Equation (11) when

b/

h ≤ 2 (

Figure 1b). Since

$b/2h$ is a constant, the value of

k_{1} can be obtained by solving Equation (11).

Referring to the value of

k_{1}, a comparison relationship between

k_{1} and

$2h/b$ is as follows:

It can be shown from Inequality (15), so

That is to say, when b/h ≤ 2, the slope of the division line is always greater than 2h/b. The intersection point of the division lines is on or above the water surface line, the same as b/h ≥ 2.

By combining cases (

**a**) and (

**b**), we know that

Figure 1b will not occur. Whether in a wide–shallow or in a narrow–deep channel, the existing form of the division line is always shown as

Figure 1a.

#### 2.1.2. CPL of Lines in Rectangular Channel

As shown in

Figure 2, taking the left half cross-section of the channel as an example, the dividing line divides the cross-section of the channel into regions I (ABCD), II (OAD) and III (ODE).

In region I, the discharge of the shadow rectangle area can be expressed using Equation (16):

where

dQ is the discharge of the shadow rectangle area;

dA is the area of shadow rectangle; and

u is the longitudinal velocity of any point in the area.

By integrating

y, the discharge

Q can be obtained as follows:

As the rectangular area can be expressed as

$A={h}_{I}dz$, the average velocity of this flow area can be obtained as:

According to the log-law of vertical velocity in channel section:

Substituting Equation (19) to (18) gives:

Combining Equations (19) and (20) gives:

$y$ is expressed as:

where

${y}_{I}$ represents the location of depth average velocity in region I, and

${h}_{I}\left(=h\right)$ is the water depth of the channel.

In region II,

${y}_{II}$ represents the CPL of depth average velocity, which is similar to region I:

where

${y}_{II}$ represents the location of depth average velocity in region II, and

${h}_{II}$ is the distance perpendicular to the channel bed to the division line.

In region III, as shown in

Figure 2, the discharge of the shadow rectangle area can be expressed as:

Integrating

z along the direction of

z axis, the discharge can be determined using Equation (25):

where

${h}_{III}$ is the horizontal distance from the side wall of the channel to the division line.

As the rectangular cross-sectional area can be expressed as

$A={h}_{III}dy$, the average velocity of this flow area can be obtained as shown as Equation (26):

According to the log-law of vertical velocity in channel section:

Using Equations (26) and (27),

$z$ can be evaluated using:

Giving:

where

${z}_{III}$ represents the location of the depth average velocity in region III.

#### 2.1.3. CPL of Regions in Rectangular Channel

For region I, according to Equation (22), since the water depth

h in the channel is constant, measuring the velocity of the CPL of depth average velocity can provide the average velocity for the whole section of region I. Therefore, the average discharge can be obtained by multiplying the average cross-section velocity by the area of the region. Similarly, the flow in regions II and III can be obtained using the same method.

Figure 3 shows that

P_{I}, P_{II}, and

P_{III} are the characteristic points that represent the mean velocities in regions I, II, and III, respectively.

For region II, the discharge

Q can be expressed using Equation (30):

where

e is a constant;

S is the slope of channel; and

z_{0} is shown in

Figure 3.

The mean velocity can be calculated from the value of

Q/A, which can be expressed using Equation (31):

Using Equations (6), (19) and (31), Equation (32) can be obtained:

Therefore, the

y and

z value of the point

P_{II} can be calculated using Equation (33):

where

$\overline{{z}_{II}}$ and

$\overline{{y}_{II}}$ represent the coordinates of

P_{II}.

For region III, the analysis method is similar to that of region II. The

z and

y value of the point

P_{III}, which can be used to calculate the value of

Q of region III, can be expressed using Equation (34):

where

$\overline{{z}_{III}}$ and

$\overline{{y}_{III}}$ represent the coordinates of

P_{III}.