Environmental Fluid Mechanics III Model Applications · 2 20.04.2010 Institute for Hydromechanics...
Transcript of Environmental Fluid Mechanics III Model Applications · 2 20.04.2010 Institute for Hydromechanics...
KIT – University of the State of Baden-Wuerttemberg and
National Research Center of the Helmholtz Association
INSTITUTE FOR HYDROMECHANICS
www.kit.edu
Environmental Fluid Mechanics III Model Applications
Dong-Guan Seol
Integral Plume Modeling
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Agenda
Definition of jets and plumes
Challenges in modeling
Density stratification
Integral plume model
Numerical solution of the integral plume model
Problem definition
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Definition of jets and plumes
Jets: source of energy - momentum
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Plumes: source of energy - buoyancy
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CO2 injection:
Norway Sleipner Oil drilling rig
Oil well blowout:
IXTOC I Oil well
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Challenges in modeling
Cross-flow
Density-stratification
Buoyant Jet
in Crossflow
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Challenges in modeling (continued)
Density-stratification
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Density stratification
Temperature, salinity, particular material
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• Buoyancy force
• Buoyancy frequency
• Reduced gravity
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Integral plume model
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Basic assumptions
• Buossinesq approximation
• Self-similarity of the mean vertical velocity (w) and buoyancy force (g’)
• Entrainment hypothesis
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Integral plume model
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Buossinesq approximation:
: Plume density
: Ambient density
: Reference density
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Integral plume model
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Self-similarity: Gaussian profile
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Integral plume model
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Entrainment hypothesis
Entrainment velocity:
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Integral plume model
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Governing equations
• Mass conservation
• Momentum conservation
• Buoyancy conservation
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Integral plume model
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Mass conservation
• [Mass flux exiting control volume]
=[Mass entering control volume]+[Mass entraining from side]
Introducing Buossinesq approximation
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Integral plume model
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Momentum conservation conservation
• [Momentum flux exiting control volume]
=[Momentum entering control volume] + [Momentum entraining from side]
+ [Upward buoyancy]
[Upward buoyancy] =[weight of displaced water]-[actual weight of plume segment]
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Integral plume model
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Momentum conservation conservation (continued)
• [Momentum flux exiting control volume]
=[Momentum entering control volume] + [Momentum entraining from side]
+ [Upward buoyancy]
Introducing Buossinesq approximation
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Integral plume model
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Buoyancy conservation
[Buoyancy flux exiting control volume]
=[Buoyancy flux entering control volume]
+[Influx of buoyancy entraining from side]
(from the mass conservation)
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Integral plume model
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Buoyancy conservation (continued)
Multiply both sides with g/ r and introduce buoyancy frequency N2
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Integral plume model
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Governing equations
• Mass conservation
• Momentum conservation
• Buoyancy conservation
For simplicity, let us put
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Numerical solution
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Model setup
where
Initial conditions for pure plume (only buoyancy) at the plume bottom (z=0)
Assume linear stratification,
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Numerical solution
Matlab imprementation
clear all, close all, clc,
% define constants global alpha N;
alpha = 0.125; %entrainment coefficietn N =1.; % Buoyancy frequency
dz= 0.1; % integration interval Z = 5.2; % Total height of integration
z=[0:dz:Z]; % height increment
% specify initial value W0=0;
V0=0.00001; % non-zero for non-trivial solution F0=1;
Nz = length(z); % number of integration steps
% initialize the numerical integration nz=1; W=W0;
V=V0; F=F0;
while (nz < Nz)
zi=z(nz);
% non-dimensionalize the variables using F0 and N zp=zi/(F0^0.25/N^(3/4));
Wp=W/(F0^0.5/N^(5/2)); Vp=V^0.25/(F0^(3/4)/N^(5/4));
Fp=F/F0;
plot(Wp,zp,'.g‘‚Vp,zp,'.r‘,Fp,zp,'.b');
axis([-1 3 0 6]) legend('volume flux','momentum flux','buoyancy flux',4);
ylabel('Height z'); xlabel('Plume parameter');
[Wn,Vn,Fn]=func_rk4(W,V,F,dz);
nz = nz + 1;
W = Wn; V = Vn; F = Fn;
end
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Matlab imprementation (functions)
function [Wn,Vn,Fn]=func_rk4(W,V,F,dz)
k1=dz*myfunc1(W,V,F); l1=dz*myfunc2(W,V,F);
m1=dz*myfunc3(W,V,F);
k2=dz*myfunc1(W+k1/2,V+l1/2,F+m1/2);
l2=dz*myfunc2(W+k1/2,V+l1/2,F+m1/2); m2=dz*myfunc3(W+k1/2,V+l1/2,F+m1/2);
k3=dz*myfunc1(W+k2/2,V+l2/2,F+m2/2); l3=dz*myfunc2(W+k2/2,V+l2/2,F+m2/2);
m3=dz*myfunc3(W+k1/2,V+l1/2,F+m2/2);
k4=dz*myfunc1(W+k3,V+l3,F+m3); l4=dz*myfunc2(W+k3,V+l3,F+m3); m4=dz*myfunc3(W+k3,V+l3,F+m3);
Wn = W + (k1 + 2*k2 + 2*k3 + k4)/6;
Vn = V + (l1 + 2*l2 + 2*l3 + l4)/6; Fn = F + (m1 + 2*m2 + 2*m3 + m4)/6;
return
Runge-Kutta 4th order method: A numerical technique used to solve ordinary differential
equation of the form
where
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Matlab imprementation (functions)
function [dWdz] = myfunc1(W,V,F)
global alpha N;
dWdz=2*alpha*V^0.25;
return
function [dVdz] = myfunc2(W,V,F)
global alpha N;
dVdz=2*F*W;
return
function [dFdz] = myfunc3(W,V,F)
global alpha N;
dFdz=-N^2*W;
return
Mass conservation
Momentum conservation
Buoyancy conservation
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Problem definition By using a blower and some preheating, one can adjust both the upward
velocity and buoyancy of fumes exiting from the top of a
smokestack. Specifically, two scenarios are being considered, on with
more velocity and one one with more buoyancy, as follows:
Scenario 1: Average exit vertical velocity = 12 m/s
Average exit buoyancy = 0.01 m/s2
Scenario 2: Average exit vertical velocity = 1 m/s
Average exit buoyancy = 0.12 m/s2
In each case, the exit diameter is 1.5 m and the entrainment coefficient
is taken as 0.115. Assume linear stratification (N=1 s-1).
• Which of the two scenarios gives the highest vertical velocity at the
center of the plume 20 m above the smokestack?
• Briggs (1969) proposed following formula for the rise of the plume:
Compare the model results with above formula for different stratification
conditions (N=[0:0.1:1] 1/s). Assume pure plume of scenario 2.
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References
Briggs, G. A. (1965). “A plume rise model compared with observations.”
J. Air Poll. Control Assoc., 15(9), 433-438.
Lee, J. H. W., and Chu, V. H. (2003). Turbulent jets and plumes.
Springer.
Morton, B. R., Taylor, G., and Turner, J. S. (1956). “Turbulent
Gravitational Convection from Maintained and Instantaneous Sources.”
Proceedings of the Royal Society of London. Series A, Mathematical
and Physical Sciences, 234(1196), 1-23.
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