Mean Flow and Mixing Properties of a Vertical Round Turbulent Buoyant Jet in a Weak Crosscurrent

Abstract

This paper deals with the mean flow and mixing properties of a vertical round turbulent buoyant jet in a weak crosscurrent, through an implementation of the integral method. The phenomenon is described by the Reynolds averaged partial differential equations of continuity, momentum and conservation of tracer mass formulated in a curvilinear cylindrical coordinate system. Applying second order mathematical approximations, the equations are integrated on a reduced cross-sectional area of the jet under the similarity assumption and the boundary conditions. The reduced area by a cyclical sector provides increased entrainment due to the increase of its perimeter, which reduces the fluxes and affects the dilution, and the model predictions, mainly the dilutions, are considerably improved. A system of ordinary differential equations is produced, which is solved numerically using a 4^th order Runge-Kutta method. The results obtained for several values of the normalized ratio of ambient over buoyant-jet exit velocity are compared with experimental data of normalized trajectories and dilutions available in the literature. The satisfactory performance of the integral model for weak current velocities makes it suitable for research, for studying effluent discharges in water bodies or in the atmosphere, as well as for design purposes.

Article Highlights

• A curvilinear cylindrical coordinate system is used.

• Reduced cross-sectional area is used in the bent-over phase.

• The integral model predictions for weak currents agree well with experimental data.

Appendix A

1.1 Details in PDE Integration

PDE (1)-(4) are integrated on the cross-sectional area A under the boundary conditions and assumptions given in the main text. Fluxes are reduced by a factor $I=1-\frac{\omega }{2\uppi}$, while the cross-sectional perimeter is increased and equal to 2πR = 2π(1 + f_πI)B_w.

Continuity:

$${\int}_A\frac{\partial (rhu)}{\partial r}\mathrm{d}r\mathrm{d}\varphi +{\int}_A\frac{\partial (rw)}{\partial \xi}\mathrm{d}r\mathrm{d}\varphi =0$$

(A.1)

${A}_1={\int}_A\frac{\partial (rhu)}{\partial r}\mathrm{d}r\mathrm{d}\varphi ={\int}_0^{2\uppi}R{(hu)}_R\mathrm{d}\varphi =R{\int}_0^{2\uppi}{h}_R\left({u}_R+{u}_a\sin \theta \sin \varphi \right)\mathrm{d}\varphi =2\uppi R\left({u}_{1R}+{u}_{2R}\right)$, where u_1R = h_Ru_R, ${u}_{2R}=\frac{R}{2}{u}_a\sin \theta \frac{\mathrm{d}\theta }{\mathrm{d}\xi }$ and ${h}_R=1+R\sin \varphi \frac{\mathrm{d}\theta }{\mathrm{d}\xi }$

${A}_2=I{\int}_A\frac{\partial (rw)}{\partial \xi}\mathrm{d}r\mathrm{d}\varphi =I\frac{\mathrm{d}}{\mathrm{d}\xi}\int_0^{2\uppi}\int_0^{B_w}r\left({w}_1+{u}_a\cos \theta \right)\mathrm{d}r\mathrm{d}\varphi -I{\int}_0^{2\uppi}{B}_w{\left({u}_a\cos \theta \right)}_{B_w}\mathrm{d}\varphi \frac{\mathrm{d}{B}_w}{\mathrm{d}\xi }=\frac{\mathrm{d}\overset{\sim }{\mu }}{\mathrm{d}\xi }-2I\uppi {B}_w\frac{\mathrm{d}{B}_w}{\mathrm{d}\xi }{u}_a\cos \theta$, where the Leibniz rule has been applied; ${h}_B=1+{B}_w\sin \varphi \frac{\mathrm{d}\theta }{\mathrm{d}\xi }\ {\overset{\sim }{h}}_B=1-{B}_w\sin \varphi \frac{\mathrm{d}\theta }{\mathrm{d}\xi }$, ${h}_B{\overset{\sim }{h}}_B\approx 1$, $\overset{\sim }{\mu }=\mu +I\uppi {B}_w^2{u}_a\cos \theta$, $\frac{\mathrm{d}\mu }{\mathrm{d}\xi }=-2\uppi R{u}_{1R}$, $\mu =I\uppi {b}_w^2{w}_m$, ${u}_{1R}=-{K}_w\left\{\frac{1}{2}\left(2+\frac{\xi\ \mathrm{d}{w}_m}{w_m\ \mathrm{d}\xi}\right)\left[1-\exp \left(-{n}_R^2\right)\right]{n}_R^{-1}-n\exp \left(-{n}_R^2\right)\right\}$, n_R = R/b_w (Yannopoulos 2006).

Since A₁ + A₂ = 0, then

$$\frac{\mathrm{d}\overset{\sim }{\mu }}{\mathrm{d}\xi }=-2\uppi R{u}_e=\frac{\mathrm{d}\mu }{\mathrm{d}\xi }+\uppi \left(2{IB}_w\frac{\mathrm{d}{B}_w}{\mathrm{d}\xi}\cos \theta -{R}^2\sin \theta \frac{\mathrm{d}\theta }{\mathrm{d}\xi}\right){u}_a$$

(A.2)

ψ-momentum

$${\int}_A\frac{\partial }{\partial r}\left[r\sin \varphi\ h\left({u}^2+\frac{p_d}{\rho_0}\right)\right]\mathrm{d}r\mathrm{d}\varphi +{\int}_A\frac{\partial }{\partial \xi}\left[r\sin \varphi \left( uw-\frac{\tau_{r\xi}}{\rho_0}\right)\right]\mathrm{d}r\mathrm{d}\varphi +{\int}_Ar\left({w}^2+{w}^{\prime 2}+\frac{p_d}{\rho_0}\right)\frac{\mathrm{d}\theta }{\mathrm{d}\xi}\mathrm{d}r\mathrm{d}\varphi =-{\int}_A{g}_0^{\prime } rhc\ \cos \theta\ \mathrm{d}r\mathrm{d}\varphi$$

(A.3)

$${B}_1={\int}_0^{2\uppi}R\sin \varphi\ {h}_R\left[{\left({u}_R+{u}_a\sin \theta \sin \varphi \right)}^2-\frac{p_d}{\rho_0}\right]\mathrm{d}\varphi \cong {\int}_0^{2\uppi}R\sin \varphi\ {h}_R{u}_R^2\mathrm{d}\varphi +{\int}_0^{2\uppi}2R\sin \varphi\ {h}_R{u}_R{u}_a\sin \theta \sin \varphi \mathrm{d}\varphi +{\int}_0^{2\uppi}R\sin \varphi\ {h}_R{u}_a^2{\sin}^2\theta {\sin}^2\varphi \mathrm{d}\varphi =-\uppi {R}^2{\left({u}_{1R}+{u}_{2R}\right)}^2\frac{\mathrm{d}\theta }{\mathrm{d}\xi }+2\uppi R\left({u}_{1R}+{u}_{2R}\right){u}_a\sin \theta +\frac{3\uppi}{4}{R}^2{u}_a^2{\sin}^2\theta \frac{\mathrm{d}\theta }{\mathrm{d}\xi }$$

$${B}_2=\int_0^{2\uppi}\int_0^{B_w}I\frac{\partial }{\partial \xi}\left(r\sin \varphi\ uw\right)\mathrm{d}r\mathrm{d}\varphi =I\frac{\mathrm{d}}{\mathrm{d}\xi}\int_0^{2\uppi}\int_0^{B_w}r\sin \varphi \left({u}_1+{u}_2+{u}_a\sin \theta \sin \varphi \right)\left({w}_1+{u}_a\cos \theta \right)\mathrm{d}r\mathrm{d}\varphi -I{\int}_0^{2\uppi}{B}_w\frac{\mathrm{d}{B}_w}{\mathrm{d}\xi}\sin \varphi\ \left(\frac{u_{1B}+{u}_{2B}}{h_B{\overset{\sim }{h}}_B}{\overset{\sim }{h}}_B+{u}_a\sin \theta \sin \varphi \right)\left({w}_{1B}+{u}_a\cos \theta \right)\mathrm{d}\varphi =\frac{u_a}{2}\sin \theta \frac{\mathrm{d}\mu }{\mathrm{d}\xi }+\frac{\mu }{2}{u}_a\cos \theta \frac{\mathrm{d}\theta }{\mathrm{d}\xi }+\frac{\uppi}{2}I{B}_w^2{u}_a^2\left({\cos}^2\theta -{\sin}^2\theta \right)\frac{\mathrm{d}\theta }{\mathrm{d}\xi }+\uppi I{B}_w^2\frac{\mathrm{d}{B}_w}{\mathrm{d}\xi}\left({u}_{1B}+{u}_{2B}\right)\cos \theta \frac{\mathrm{d}\theta }{\mathrm{d}\xi }$$

$${B}_3=\int_0^{2\uppi}\int_0^{B_w} Ir\left[{\left({w}_1+{u}_a\cos \theta \right)}^2+\frac{p_d}{\rho_0}\right]\frac{\mathrm{d}\theta }{\mathrm{d}\xi}\mathrm{d}r\mathrm{d}\varphi =\overset{\sim }{m}\frac{\mathrm{d}\theta }{\mathrm{d}\xi }$$

$${B}_4=\int_0^{2\uppi}\int_0^{B_w}I{g}_0^{\prime } rhc\ \cos \theta\ \mathrm{d}r\mathrm{d}\varphi =\zeta \cos \theta$$

$${B}_5=\int_0^{2\uppi}\int_0^{B_w}I\frac{\partial }{\partial r}\left(\frac{p_d}{\rho_0} rh\ \sin \varphi\ \right)\mathrm{d}r\mathrm{d}\varphi \cong 0$$

where $\overset{\sim }{m}=m+2\mu\ {u}_a\cos \theta +I\uppi {B}_w^2{u}_a^2{\cos}^2\theta$, $\zeta =I\uppi {g}_0^{\prime }{b}_c^2{c}_m$, ${u}_{2B}=\frac{B_w}{2}{u}_a\sin \theta \frac{\mathrm{d}\theta }{\mathrm{d}\xi }$, w_1B ≅ 0 and u_1B is calculated using the relationship for u_1R and substituting R = B_w.

Since B₁ + B₂ − B₃ =  − B₄ − B₅, after substitution and solution with respect to $\frac{\mathrm{d}\theta }{\mathrm{d}\xi }$, the following equation is derived:

$$\frac{d\theta}{d\xi}=\frac{\zeta cos\theta +\pi R{u}_1{u}_a sin\theta}{D}$$

(A.4)

where $D=\overset{\sim }{m}+\uppi {R}^2\left[{\left({u}_{1R}+{u}_{2R}\right)}^2-\frac{7}{4}{u}_a^2{\sin}^2\theta \right]-\left[\frac{\mu }{2}+\uppi I{B}_w^2\frac{\mathrm{d}{B}_w}{\mathrm{d}\xi}\left({u}_{1B}+{u}_{2B}\right)\right]{u}_a\cos \theta -\frac{\uppi}{2}I{B}_w^2{u}_a^2\cos 2\theta$.

ξ – momentum

$${\int}_A\frac{\partial }{\partial r}\left[ rh\left( uw-\frac{\tau_{r\xi}}{\rho_0}\right)\right]\mathrm{d}r\mathrm{d}\varphi +{\int}_A\frac{\partial }{\partial \xi}\left[r\left({w}^2+{w}^{\prime 2}+\frac{p_d}{\rho_0}\right)\right]\mathrm{d}r\mathrm{d}\varphi +{\int}_Ar\sin \varphi \left( uw-\frac{\tau_{r\xi}}{\rho_0}\right)\frac{\mathrm{d}\theta }{\mathrm{d}\xi}\mathrm{d}r\mathrm{d}\varphi ={\int}_A{g}_0^{\prime } rhc\ sin\theta\ \mathrm{d}r\mathrm{d}\varphi$$

(A.5)

$${\varGamma}_1={\int}_0^{2\uppi}R\ {h}_R\left[\left({u}_R+{u}_a\sin \theta \sin \varphi \right)\left({w}_{1R}+{u}_a\cos \theta \right)-\frac{\tau_{R\xi}}{\rho_0}\right]\mathrm{d}\varphi =2\uppi R\left({u}_{1R}+{u}_{2R}\right){u}_a\cos \theta +\uppi {R}^2{u}_a^2\sin \theta \cos \theta \frac{\mathrm{d}\theta }{\mathrm{d}\xi }$$

$${\varGamma}_2=\int_0^{2\uppi}\int_0^{B_w}I\frac{\partial }{\partial \xi}\left\{r\left[{\left({w}_1+{u}_a\cos \theta \right)}^2+{\mathrm{w}}^{\prime 2}+\frac{p_d}{\rho_0}\right]\right\}\mathrm{d}r\mathrm{d}\varphi =\frac{\mathrm{d}\overset{\sim }{m}}{\mathrm{d}\xi }-\frac{\mathrm{d}{B}_w}{\mathrm{d}\xi }{\int}_0^{2\uppi}I{B}_w\left[{\left({w}_{1R}+{u}_a\cos \theta \right)}^2+{w}_{1R}^{\prime 2}+{\left(\frac{p_d}{\rho_0}\right)}_R\right]\mathrm{d}\varphi =\frac{\mathrm{d}\overset{\sim }{m}}{\mathrm{d}\xi }-2\uppi I{B}_w\frac{\mathrm{d}{B}_w}{\mathrm{d}\xi }{u}_a^2{\cos}^2\theta$$

$${\varGamma}_3=\frac{\mathrm{d}\theta }{\mathrm{d}\xi}\int_0^{2\uppi}\int_0^{B_w} Ir\ \sin \varphi \left[\left(\frac{u_1+{u}_2}{h}+{u}_a\sin \theta \sin \varphi \right)\left({w}_1+{u}_a\cos \theta \right)-\frac{\tau_{r\xi}}{\rho_0}\right]\mathrm{d}r\mathrm{d}\varphi =\frac{\mathrm{d}\theta }{\mathrm{d}\xi}\left\{\int_0^{2\uppi}\int_0^{B_w} Ir\ \sin \varphi \left(\frac{u_1+{u}_2}{h\overset{\sim }{h}}\overset{\sim }{h}{w}_1-\frac{\tau_{r\xi}}{\rho_0}\right)\mathrm{d}r\mathrm{d}\varphi +\int_0^{2\uppi}\int_0^{B_w} Ir\ \sin \varphi \left(\frac{u_1+{u}_2}{h\overset{\sim }{h}}\overset{\sim }{h}{u}_a\cos \theta \right)\mathrm{d}r\mathrm{d}\varphi +\int_0^{2\uppi}\int_0^{B_w} Ir\ {w}_1{u}_a\sin \theta {\sin}^2\varphi \mathrm{d}r\mathrm{d}\varphi +\int_0^{2\uppi}\int_0^{B_w} Ir\ {u}_a^2\sin \theta \cos \theta {\sin}^2\varphi \mathrm{d}r\mathrm{d}\varphi \right\}=\frac{\mathrm{d}\theta }{\mathrm{d}\xi}\left[-\uppi I{b}_w^3{w}_m^2\frac{\mathrm{d}\theta }{\mathrm{d}\xi }{I}_1-\frac{\uppi}{2}I{u}_a\sin \theta {\left(\frac{\mathrm{d}\theta }{\mathrm{d}\xi}\right)}^2{b}_w^4{w}_m{I}_2-\uppi I{u}_a\cos \theta \frac{\mathrm{d}\theta }{\mathrm{d}\xi }{b}_w^3{w}_m{I}_3-\frac{\uppi}{8}I{u}_a^2\sin \theta \cos \theta {\left(\frac{\mathrm{d}\theta }{\mathrm{d}\xi}\right)}^2{B}_w^4+\frac{\uppi}{2}\mu {u}_a\sin \theta +\frac{\uppi}{2}I{B}_w^2{u}_a^2\sin \theta \cos \theta \right]=\frac{\mathrm{d}\theta }{\mathrm{d}\xi}\left[\frac{\uppi}{2}{u}_a\sin \theta \left(\mu +I{B}_w^2{u}_a\cos \theta \right)-\frac{\uppi}{8}I{u}_a\sin {\theta b}_w^4\left(4{w}_m{I}_2+{n}_B^4{u}_a\cos \theta \right){\left(\frac{\mathrm{d}\theta }{\mathrm{d}\xi}\right)}^2\right]$$

$${\varGamma}_4=\int_0^{2\uppi}\int_0^{B_w}I{g}_0^{\prime } rhc\ \sin \theta\ \mathrm{d}r\mathrm{d}\varphi =\zeta \sin \theta$$

where ${I}_1={\int}_0^{n_B}{n}^2\frac{u_1}{w_m}\frac{w_1}{w_m}\mathrm{d}n$ and −0.00427 ≤ I₁ ≤ 0.00844 for K_w = 0.11, $-1\le {s}_1=\frac{\xi }{w_m}\frac{\mathrm{d}{w}_m}{\mathrm{d}\xi}\le -\frac{1}{3}$ (Yannopoulos 2006); thus I₁ ≅ 0; ${I}_2={\int}_0^{n_B}{n}^3\frac{w_1}{w_m}\mathrm{d}n\cong \frac{1-3{\mathrm{e}}^{-2}}{2}=0.2970$; ${I}_3={\int}_0^{n_B}{n}^2\frac{u_1}{w_m}\mathrm{d}n=-\frac{K_w}{4}\left[{s}_1+\left(8+{s}_1\right){\mathrm{e}}^{-2}\right]$ and −0.0194 ≤ I₃ ≤ 0.00145 for K_w = 0.11, ${n}_B=\sqrt{2}$, $-1\le {s}_1\le -\frac{1}{3}$; thus I₃ ≅ 0.

Since Γ₁ + Γ₂ + Γ₃ = Γ₄, after substitution and solution with respect to $\frac{\mathrm{d}\overset{\sim }{m}}{\mathrm{d}\xi }$, the following equation is derived:

$$\frac{\mathrm{d}\overset{\sim }{m}}{\mathrm{d}\xi }=\zeta cos\theta -\uppi {u}_a\left\{2\cos \theta \left(R{u}_{1R}-I{B}_w\frac{\mathrm{d}{B}_w}{\mathrm{d}\xi }{u}_a\cos \theta \right)+\frac{sin\theta}{2}\left[\mu +\left(I{B}_w^2+4{R}^2\right){u}_a\cos \theta -I{b}_w^4\left({I}_2{w}_m+\frac{1}{4}{n}_B^4{u}_a\cos \theta \right){\left(\frac{\mathrm{d}\theta }{\mathrm{d}\xi}\right)}^2\right]\frac{\mathrm{d}\theta }{\mathrm{d}\xi}\right\}$$

(A.6)

Tracer conservation

$${\int}_A\frac{\partial \left[ rh\left( uc+{u}^{\prime }{c}^{\prime}\right)\right]}{\partial r}\mathrm{d}r\mathrm{d}\varphi +{\int}_A\frac{\partial \left[r\left( wc+{w}^{\prime }{c}^{\prime}\right)\right]}{\partial \xi}\mathrm{d}r\mathrm{d}\varphi =0$$

(A.7)

$${\varDelta}_1={\int}_0^{2\uppi}R\ {h}_R\left[{u}_R{c}_R+{\left({u}^{\prime }{c}^{\prime}\right)}_R\right]\mathrm{d}\varphi =0$$

$${\varDelta}_2=\frac{\mathrm{d}}{\mathrm{d}\xi}\int_0^{2\uppi}\int_0^{B_w} Ir\left[\left({w}_1+{u}_a\cos \theta \right)c+{w}^{\prime }{c}^{\prime}\right]\mathrm{d}r\mathrm{d}\varphi -\frac{\mathrm{d}{B}_w}{\mathrm{d}\xi }{\int}_0^{2\uppi}I{B}_w\left[\left({w}_{1B}+{u}_a\cos \theta \right){c}_B+{\left({w}^{\prime }{c}^{\prime}\right)}_B\right]\mathrm{d}\varphi =\frac{1}{g_0^{\prime }}\frac{\mathrm{d}\overset{\sim }{\beta }}{\mathrm{d}\xi }=0$$

where c_R = c_B = (u^′c^′)_R = (w^′c^′)_B = 0; $\overset{\sim }{\beta }=\beta +\zeta {u}_a\cos \theta$ and, thus,

$$\beta ={\beta}_0-{u}_a\left(\zeta \cos \theta -{\zeta}_0\cos {\theta}_0\right)$$

(A.8)