16 Computational Modelling in Hydraulic and Coastal Engineering
2. Execution of the computations for the estimation of the series of z n
values (from n = 1 to n = N max ) and estimation of dependent parameters such as reservoir surface (S) and outflow discharge (Q out )
3. Storing of z, S and Q out values in proper output data files
4. End of the algorithm
The following two practical problems involving the continuity equation
(ODE) are solved numerically.
Example 2.1
A cylindrically shaped tank filled with water is emptying through a
circular orifice at the bottom. Given the following data, the purpose
of this exercise is to estimate the time required for the tank to empty:
Tank height = 10.0 m
Tank horizontal surface area = 5.0 m 2
Orifice cross-section area = 0.005 m 2
Discharge coefficient = 0.7
The numerical scheme is given by Equation 2.9 with no inflow
Q in
n
=
(
)
0 . For the numerical solution the tank is discretized vertically
into N = 10 layers of equal thickness. Each layer (n) is confined between
Observed values
Inflow
hydrograph
Δt Δt Δt
Inflow
discharge
Q in (t m+1 )
t m+1
t n+1
t m
Q in (t n+1 )
Q in (t m )
Inflow
hydrograph
time
λ 2
λ 1
Q in
δt
δt
δt
t = m·δt
Computational
time
t = n·Δt
Figure 2.3 Interpolation of the inflow hydrograph.
2. Execution of the computations for the estimation of the series of z n
values (from n = 1 to n = N max ) and estimation of dependent parameters such as reservoir surface (S) and outflow discharge (Q out )
3. Storing of z, S and Q out values in proper output data files
4. End of the algorithm
The following two practical problems involving the continuity equation
(ODE) are solved numerically.
Example 2.1
A cylindrically shaped tank filled with water is emptying through a
circular orifice at the bottom. Given the following data, the purpose
of this exercise is to estimate the time required for the tank to empty:
Tank height = 10.0 m
Tank horizontal surface area = 5.0 m 2
Orifice cross-section area = 0.005 m 2
Discharge coefficient = 0.7
The numerical scheme is given by Equation 2.9 with no inflow
Q in
n
=
(
)
0 . For the numerical solution the tank is discretized vertically
into N = 10 layers of equal thickness. Each layer (n) is confined between
Observed values
Inflow
hydrograph
Δt Δt Δt
Inflow
discharge
Q in (t m+1 )
t m+1
t n+1
t m
Q in (t n+1 )
Q in (t m )
Inflow
hydrograph
time
λ 2
λ 1
Q in
δt
δt
δt
t = m·δt
Computational
time
t = n·Δt
Figure 2.3 Interpolation of the inflow hydrograph.