Flow in pressurized conduits 67
for k=1:nb
% Flow velocities in the various branches;
u(k)=q(k)/1000*1.273/d(k)^2;
% Head loss in the various branches;
dh(k)=r(k)*q(k)^2/1000000;
end
plot(1:n,diff,'k','Linewidth',1.5)
xlabel('Number of iterations'),ylabel('Max discharge
difference between successive iterations')
PROBLEM 4.1
Solve the same problem by making the following suggested changes while
keeping the rest of the data constant:
1. Since both the resistance coefficient and the pipe diameter are given,
the quantity fL
g RD
=
π
2
5
8
can be estimated for each branch. Based on
the data given, find out whether a constant value for the friction coefficient (f) was used. If yes, what was that value?
2. Calculate the discharge in all branches by considering an inflow of
500 m 3 /s at the junction (1–2) (i.e. connecting branches 1 and 2) and
the same amount of outflow at junction (12–13).
3. What kind of changes on pipe diameter would you implement on
the given network in order to keep velocity in all branches less than
1.5 m/s?
4. If the network pipes were to be replaced with pipes of the same material but of constant diameter D = 0.2 m, what changes in the computer program would you implement to solve the problem?
5. Solve the same pipe network problem as in Example 4.1 but during your initial selection of discharge values take Q 1 = 450 m 3 /s and
Q 2 = 50 m 3 /s. The rest of the initial discharges should be selected in
such a way so that continuity is preserved at each junction. After how
many iteration steps is the correction ΔQ l reduced to less than 0.001?
Compare the simulation data obtained by running the modifications, derive
conclusions and discuss the significance of the various variables involved to
the problem of the pipe network and the Hardy Cross method.
4.3 UNSTEADY FLOW IN A CLOSED CONDUIT
For most practical situations in pipe hydraulics, the water is considered
as incompressible and the pipe as rigid. However, those assumptions are
not always true. For instance, when there is a sudden change of the flow
for k=1:nb
% Flow velocities in the various branches;
u(k)=q(k)/1000*1.273/d(k)^2;
% Head loss in the various branches;
dh(k)=r(k)*q(k)^2/1000000;
end
plot(1:n,diff,'k','Linewidth',1.5)
xlabel('Number of iterations'),ylabel('Max discharge
difference between successive iterations')
PROBLEM 4.1
Solve the same problem by making the following suggested changes while
keeping the rest of the data constant:
1. Since both the resistance coefficient and the pipe diameter are given,
the quantity fL
g RD
=
π
2
5
8
can be estimated for each branch. Based on
the data given, find out whether a constant value for the friction coefficient (f) was used. If yes, what was that value?
2. Calculate the discharge in all branches by considering an inflow of
500 m 3 /s at the junction (1–2) (i.e. connecting branches 1 and 2) and
the same amount of outflow at junction (12–13).
3. What kind of changes on pipe diameter would you implement on
the given network in order to keep velocity in all branches less than
1.5 m/s?
4. If the network pipes were to be replaced with pipes of the same material but of constant diameter D = 0.2 m, what changes in the computer program would you implement to solve the problem?
5. Solve the same pipe network problem as in Example 4.1 but during your initial selection of discharge values take Q 1 = 450 m 3 /s and
Q 2 = 50 m 3 /s. The rest of the initial discharges should be selected in
such a way so that continuity is preserved at each junction. After how
many iteration steps is the correction ΔQ l reduced to less than 0.001?
Compare the simulation data obtained by running the modifications, derive
conclusions and discuss the significance of the various variables involved to
the problem of the pipe network and the Hardy Cross method.
4.3 UNSTEADY FLOW IN A CLOSED CONDUIT
For most practical situations in pipe hydraulics, the water is considered
as incompressible and the pipe as rigid. However, those assumptions are
not always true. For instance, when there is a sudden change of the flow