104 Computational Modelling in Hydraulic and Coastal Engineering
plot(1:nx,h3t,'r','Linewidth',1.5)
plot(1:nx,h5t,'g','Linewidth',1.5)
xlabel('Longitudinal distance x 1000 [m]')
ylabel('Water depth [m]')
axis([0,47,0,4.5])
legend('t=Td/25','t=Td/5','t=Td')
PROBLEM 5.3
Solve the same problem by making the following suggested changes while
keeping the rest of the data constant:
1. Change the width linearly from 100 m at the upstream end to 2000
m at the downstream boundary. Then set a 2000 m upstream width,
linearly reducing to 100 m at the downstream end. Run the program
and compare the results of the two simulations.
2. Change the inflow hydrograph to a constant Q = 2000 m 3 /s lasting
from t = 0 to t = 3600 seconds. Run the program and comment on the
simulation results.
3. By using the appropriate expression, change the downstream boundary to a zero-flux (reflecting) boundary. Run the program and comment on the simulation results.
4. Modify the code as needed to simulate a constant base flow of
1000 m 3 /s. (Hint: Set the upstream boundary hydrograph at a constant
inflow of 1000 m 3 /s until steady-steady conditions are established.)
5. For a constant flow of 1000 m 3 /s, make the appropriate changes to the
downstream boundary condition to accommodate tidal disturbances.
(Hint: Set Q out = Q end + Q T sin(σt) where Q out is the total outflow, Q end
is the base flow discharge, Q T is the tidal flux and σ is the tidal frequency.) Select values for Q T and σ, run the simulation for 10 tidal
cycles and comment on the results.
5.3 TWO-DIMENSIONAL HORIZONTAL
FREE SURFACE FLOWS
For two-dimensional free surface flows, the solution domain is usually a
large water basin (e.g. open sea, lake or reservoir) characterised by low
water velocities. If the utilized coordinate system is not an inertial one but
one following the radial acceleration of earth, the effect of the rotation of
the earth, whenever important, appears indirectly. Thus, in the momentum
equations for a non-inertial system, Coriolis effects are introduced in the
form of an internal distributed force expressed as
C
U
= −
×
2ρΩ
(5.33)
plot(1:nx,h3t,'r','Linewidth',1.5)
plot(1:nx,h5t,'g','Linewidth',1.5)
xlabel('Longitudinal distance x 1000 [m]')
ylabel('Water depth [m]')
axis([0,47,0,4.5])
legend('t=Td/25','t=Td/5','t=Td')
PROBLEM 5.3
Solve the same problem by making the following suggested changes while
keeping the rest of the data constant:
1. Change the width linearly from 100 m at the upstream end to 2000
m at the downstream boundary. Then set a 2000 m upstream width,
linearly reducing to 100 m at the downstream end. Run the program
and compare the results of the two simulations.
2. Change the inflow hydrograph to a constant Q = 2000 m 3 /s lasting
from t = 0 to t = 3600 seconds. Run the program and comment on the
simulation results.
3. By using the appropriate expression, change the downstream boundary to a zero-flux (reflecting) boundary. Run the program and comment on the simulation results.
4. Modify the code as needed to simulate a constant base flow of
1000 m 3 /s. (Hint: Set the upstream boundary hydrograph at a constant
inflow of 1000 m 3 /s until steady-steady conditions are established.)
5. For a constant flow of 1000 m 3 /s, make the appropriate changes to the
downstream boundary condition to accommodate tidal disturbances.
(Hint: Set Q out = Q end + Q T sin(σt) where Q out is the total outflow, Q end
is the base flow discharge, Q T is the tidal flux and σ is the tidal frequency.) Select values for Q T and σ, run the simulation for 10 tidal
cycles and comment on the results.
5.3 TWO-DIMENSIONAL HORIZONTAL
FREE SURFACE FLOWS
For two-dimensional free surface flows, the solution domain is usually a
large water basin (e.g. open sea, lake or reservoir) characterised by low
water velocities. If the utilized coordinate system is not an inertial one but
one following the radial acceleration of earth, the effect of the rotation of
the earth, whenever important, appears indirectly. Thus, in the momentum
equations for a non-inertial system, Coriolis effects are introduced in the
form of an internal distributed force expressed as
C
U
= −
×
2ρΩ
(5.33)