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Dynamic equation of GVF, wide rectangular form

Short answer

From dH/dx = -Sf and dH = dz + dy + d(V²/2g), derive dy/dx = (S₀ - Sf)/(1 - Fr²). For a wide rectangle, using Manning’s equation, Sf/S₀ = (yₙ/y)^(10/3) and Fr² = (yc/y)³, giving dy/dx = S₀[1 - (yₙ/y)^(10/3)] / [1 - (yc/y)³].

Derive the dynamic equation of Gradually Varied Flow (GVF) and convert the derived equation for the case of wide rectangular channel, using Manning's equation, into following form: dydx=So[1−(yny)10/3]1−(ycy)3\frac{dy}{dx} = \frac{S_o\left[1 - \left(\frac{y_n}{y}\right)^{10/3}\right]}{1 - \left(\frac{y_c}{y}\right)^3} where SoS_o = bed slope, yny_n = normal depth, ycy_c = critical depth.

Dynamic equation of GVF

Assumptions: steady flow; hydrostatic pressure distribution (small slope); the channel is prismatic; the energy loss is given by the uniform-flow formula using the local depth and velocity; α=1\alpha = 1.

Take a short reach dxdx of an open channel. The total head above a datum is

H=z+y+V22gH = z + y + \frac{V^2}{2g}

where zz is the bed level, yy the depth and V=Q/AV = Q/A. The bed slope is S0=−dz/dxS_0 = -dz/dx and the energy slope is Sf=−dH/dxS_f = -dH/dx. Differentiating with respect to xx:

dHdx=dzdx+dydx+ddx(V22g)\frac{dH}{dx} = \frac{dz}{dx} + \frac{dy}{dx} + \frac{d}{dx}\left(\frac{V^2}{2g}\right) −Sf=−S0+dydx+ddy(Q22gA2)dydx-S_f = -S_0 + \frac{dy}{dx} + \frac{d}{dy}\left(\frac{Q^2}{2gA^2}\right)\frac{dy}{dx}

Since dA=T dydA = T\,dy:

ddy(Q22gA2)=−Q2gA3dAdy=−Q2TgA3=−Fr2\frac{d}{dy}\left(\frac{Q^2}{2gA^2}\right) = -\frac{Q^2}{gA^3}\frac{dA}{dy} = -\frac{Q^2 T}{gA^3} = -Fr^2

Therefore

−Sf=−S0+dydx(1−Fr2)-S_f = -S_0 + \frac{dy}{dx}\left(1 - Fr^2\right) dydx=S0−Sf1−Fr2Fr2=Q2TgA3\boxed{\frac{dy}{dx} = \frac{S_0 - S_f}{1 - Fr^2}}\qquad Fr^2 = \frac{Q^2 T}{gA^3}

Physical meaning: the numerator is the difference between the bed slope and the slope of the energy line. If S0>SfS_0 > S_f, depth tends to increase in the flow direction (for subcritical flow); the denominator decides the sign by flow regime (1−Fr2>01 - Fr^2 > 0 subcritical, <0< 0 supercritical). When y=yny = y_n, Sf=S0S_f = S_0 and dy/dx=0dy/dx = 0 (uniform flow). When y=ycy = y_c, Fr=1Fr = 1 and dy/dx→∞dy/dx \to \infty (the surface meets the critical depth line steeply).

Wide rectangular form using Manning's equation

For a wide rectangular channel, R≈yR \approx y, A=byA = by and q=Q/bq = Q/b.

Friction slope from Manning's equation V=1nR2/3Sf1/2V = \frac{1}{n}R^{2/3}S_f^{1/2}:

Sf=n2V2R4/3=n2q2y10/3S_f = \frac{n^2 V^2}{R^{4/3}} = \frac{n^2 q^2}{y^{10/3}}

At normal depth Sf=S0S_f = S_0 and y=yny = y_n:

S0=n2q2yn10/3  ⇒  SfS0=(yny)10/3S_0 = \frac{n^2 q^2}{y_n^{10/3}} \;\Rightarrow\; \frac{S_f}{S_0} = \left(\frac{y_n}{y}\right)^{10/3}

Froude number: at critical depth q2=g yc3q^2 = g\,y_c^3, so

Fr2=q2gy3=(ycy)3Fr^2 = \frac{q^2}{g y^3} = \left(\frac{y_c}{y}\right)^3

Substituting into the dynamic equation:

dydx=S0[1−(yny)10/3]1−(ycy)3\boxed{\frac{dy}{dx} = \frac{S_0\left[1 - \left(\dfrac{y_n}{y}\right)^{10/3}\right]}{1 - \left(\dfrac{y_c}{y}\right)^3}}

✓ Answer checked by Claude Opus 5.5 (high effort)

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