Borda–Carnot equation

Relation to the total head and Bernoulli's principle

The Borda–Carnot equation gives the decrease in the constant of the Bernoulli equation. For an incompressible flow the result is – for two locations labelled 1 and 2, with location 2 downstream to 1 – along a streamline:

${\displaystyle p_1+\frac{1}{2}\rho v_1^2+\rho g{z}_1=p_2+\frac{1}{2}\rho v_2^2+\rho g{z}_2+\mathrm{\Delta }E,}$

with

p₁ and p₂ the pressure at location 1 and 2,

z₁ and z₂ the vertical elevation (above some reference level) of the fluid particle,

g the gravitational acceleration.

The first three terms on either side of the equal sign are respectively the pressure, the kinetic energy density of the fluid and the potential energy density due to gravity. As can be seen, pressure acts effectively as a form of potential energy.

In case of high-pressure pipe flows, when gravitational effects can be neglected, ΔE is equal to the loss Δ(p + ρv²/2):

${\displaystyle \mathrm{\Delta }E=\mathrm{\Delta }\left(p+\frac{1}{2}\rho v^2\right).}$

For open-channel flows, ΔE is related to the total head loss ΔH as

${\displaystyle \mathrm{\Delta }E=\rho g\mathrm{\Delta }H,}$

with H the total head:

${\displaystyle H=h+\frac{v^2}{2g},}$

where h is the hydraulic head – the free surface elevation above a reference datum: h = z + p/(ρg).

Examples

Sudden expansion of a pipe

The Borda–Carnot equation is applied to the flow through a sudden expansion of a horizontal pipe. At cross section 1, the mean flow velocity is equal to v₁, the pressure is p₁ and the cross-sectional area is A₁. The corresponding flow quantities at cross section 2 – well behind the expansion (and regions of separated flow) – are v₂, p₂ and A₂, respectively. At the expansion, the flow separates and there are turbulent recirculating flow zones with mechanical energy losses. The loss coefficient ξ for this sudden expansion is approximately equal to one: ξ ≈ 1.0. Due to mass conservation, assuming a constant fluid density ρ, the volumetric flow rate through both cross sections 1 and 2 has to be equal:

${\displaystyle A_1{v}_1=A_2{v}_2}$ so ${\displaystyle v_2=\frac{A_1}{A_2}{v}_1.}$

Consequently – according to the Borda–Carnot equation – the mechanical energy loss in this sudden expansion is:

${\displaystyle \mathrm{\Delta }E=\frac{1}{2}\rho {\left(1-\frac{A_1}{A_2}\right)}^2{v}_1^2.}$

The corresponding loss of total head ΔH is:

${\displaystyle \mathrm{\Delta }H=\frac{\mathrm{\Delta }E}{\rho g}=\frac{1}{2g}{\left(1-\frac{A_1}{A_2}\right)}^2{v}_1^2.}$

For this case with ξ = 1, the total change in kinetic energy between the two cross sections is dissipated. As a result, the pressure change between both cross sections is (for this horizontal pipe without gravity effects):

${\displaystyle \mathrm{\Delta }p=p_1-p_2=-\rho \frac{A_1}{A_2}\left(1-\frac{A_1}{A_2}\right)v_1^2,}$

and the change in hydraulic head h = z + p/(ρg):

${\displaystyle \mathrm{\Delta }h=h_1-h_2=-\frac{1}{g}\frac{A_1}{A_2}\left(1-\frac{A_1}{A_2}\right)v_1^2.}$

The minus signs, in front of the right-hand sides, mean that the pressure (and hydraulic head) are larger after the pipe expansion. That this change in the pressures (and hydraulic heads), just before and after the pipe expansion, corresponds with an energy loss becomes clear when comparing with the results of Bernoulli's principle. According to this dissipationless principle, a reduction in flow speed is associated with a much larger increase in pressure than found in the present case with mechanical energy losses.

Sudden contraction of a pipe

In case of a sudden reduction of pipe diameter, without streamlining, the flow is not able to follow the sharp bend into the narrower pipe. As a result, there is flow separation, creating recirculating separation zones at the entrance of the narrower pipe. The main flow is contracted between the separated flow areas, and later on expands again to cover the full pipe area.

There is not much head loss between cross section 1, before the contraction, and cross section 3, the vena contracta at which the main flow is contracted most. But there are substantial losses in the flow expansion from cross section 3 to 2. These head losses can be expressed by using the Borda–Carnot equation, through the use of the coefficient of contraction μ:

${\displaystyle \mu =\frac{A_3}{A_2},}$

with A₃ the cross-sectional area at the location of strongest main flow contraction 3, and A₂ the cross-sectional area of the narrower part of the pipe. Since A₃ ≤ A₂, the coefficient of contraction is less than one: μ ≤ 1. Again there is conservation of mass, so the volume fluxes in the three cross sections are a constant (for constant fluid density ρ):

${\displaystyle A_1{v}_1=A_2{v}_2=A_3{v}_3,}$

with v₁, v₂ and v₃ the mean flow velocity in the associated cross sections. Then, according to the Borda–Carnot equation (with loss coefficient ξ=1), the energy loss ΔE per unit of fluid volume and due to the pipe contraction is:

${\displaystyle \mathrm{\Delta }E=\frac{1}{2}\rho {\left(v_3-v_2\right)}^2=\frac{1}{2}\rho {\left(\frac{1}{\mu }-1\right)}^2{v}_2^2=\frac{1}{2}\rho {\left(\frac{1}{\mu }-1\right)}^2{\left(\frac{A_1}{A_2}\right)}^2{v}_1^2.}$

The corresponding loss of total head ΔH can be computed as ΔH = ΔE/(ρg).

According to measurements by Weisbach, the contraction coefficient for a sharp-edged contraction is approximately:

${\displaystyle \mu =0.63+0.37{\left(\frac{A_2}{A_1}\right)}^3.}$

Derivation from the momentum balance for a sudden expansion

For a sudden expansion in a pipe, see the figure above, the Borda–Carnot equation can be derived from mass- and momentum conservation of the flow. The momentum flux S (i.e. for the fluid momentum component parallel to the pipe axis) through a cross section of area A is – according to the Euler equations:

${\displaystyle S=A\left(\rho v^2+p\right).}$

Consider the conservation of mass and momentum for a control volume bounded by cross section 1 just upstream of the expansion, cross section 2 downstream of where the flow re-attaches again to the pipe wall (after the flow separation at the expansion), and the pipe wall. There is the control volume's gain of momentum S₁ at the inflow and loss S₂ at the outflow. Besides, there is also the contribution of the force F by the pressure on the fluid exerted by the expansion's wall (perpendicular to the pipe axis):

${\displaystyle F=(A_2-A_1)p_1,}$

where it has been assumed that the pressure is equal to the close-by upstream pressure p₁.

Adding contributions, the momentum balance for the control volume between cross sections 1 and 2 gives:

${\displaystyle S_1-S_2+F=A_1\left(\rho v_1^2+p_1\right)-A_2\left(\rho v_2^2+p_2\right)+(A_2-A_1)p_1=0.}$

Consequently, since by mass conservation ρ A₁ v₁ = ρ A₂ v₂:

${\displaystyle \mathrm{\Delta }p=p_1-p_2=-\rho \left(\frac{A_1}{A_2}{v}_1^2-v_2^2\right)=-\rho \frac{A_1}{A_2}\left(1-\frac{A_1}{A_2}\right)v_1^2,}$

in agreement with the pressure drop Δp in the example above.

The mechanical energy loss ΔE is:

${\displaystyle \begin{array}{cc}\mathrm{\Delta }E& =\left(p_1+\frac{1}{2}\rho v_1^2\right)-\left(p_2+\frac{1}{2}\rho v_2^2\right)=\mathrm{\Delta }p+\frac{1}{2}\rho \left(v_1^2-v_2^2\right)\\ & =-\rho v_2\left(v_1-v_2\right)+\frac{1}{2}\rho \left(v_1^2-v_2^2\right)=\frac{1}{2}\rho \left(v_1^2-2v_1{v}_2+v_2^2\right)\\ & =\frac{1}{2}\rho {\left(v_1-v_2\right)}^2,\end{array}}$

which is the Borda–Carnot equation (with ξ = 1).