The document discusses hydraulic principles for analyzing flow through pipes in series and parallel configurations. It provides examples of calculating head loss, equivalent pipe coefficients, and determining flow rates and heads at different points in pipe systems. Sample problems are worked through step-by-step showing computations for pressure and total heads at nodes, equivalent roughness coefficients, head loss, and flow distribution between parallel pipes.

1. A series of pipes may have different pipe diameters and/or roughness parameters. The total head
loss is equal to the sum of the head loss in each pipe segment
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Hydraulics (CEC 301)
Lecture 2
Ramat
Polytechnic

2. Energy Loss in Non-Circular Pipes
Problem 2
For a flow rate of 0.04m3
/s determine the pressure and total heads at points A, B, C, and D for
series pipes shown in Figure below. Assume fully turbulent flow for all cases and the pressure
head at point A is 40m.
Fig. 2.1: Pipes in series with data

3. Solution
The pressure and total heads are computed using the energy equation along the path beginning at
point A. Given the pressure head and elevation, the total head at point A is:
Note the velocity and velocity head are:
The velocity head is four orders of magnitude less than the static head so it can be neglected.
Neglecting velocity head is a common assumption in pipe network analysis. All energy loss in
the system is due to friction. So following the path of flow the total heads at A, B, C, and D are:

4. Problem 3.
For the series pipe system in figure 2.1, find the equivalent roughness coefficient and the total
head at point D for a flow rate of 0.03 m3
/s.
Solution
The equivalent pipe loss coefficient is equal to the sum of the pipe coefficients or:
Note: An equivalent pipe is a pipe of uniform diameter having loss of equal head and discharge
of compound pipes of different lengths and diameters.
For this problem:
Note that pipe 2 has the largest loss coefficient since it has the smallest diameter and
highest flow velocity. As seen in Example 5.1, although it has the shortest length, most of
the head loss occurs in this section. The head loss between nodes A and D for Q = 0.03
m3
/s is then:

5. Considering the figure below:
The relationship that must hold is that the head loss in pipes 1, 2, and 3 must be the same.
Since all begin at a single node (A) and all end at a single node (B) and the difference in
head between those two nodes is unique, regardless of the pipe characteristics the head
loss in the pipes is the same or
Fig. 2.3: Pipes in parallel with data
Problem 4
Given the data for the three parallel pipes in Figure 2.3 above, compute:
(1) the equivalent parallel pipe coefficient,
(2) the head loss between nodes A and B,
(3) the flow rates in each pipe, and
(4) the total head at node B
Solution
The equivalent parallel pipe coefficient allows us to determine the head loss that can then
be used to disaggregate the flow between pipes. The loss coefficient for the Hazen
Williams equation for pipe 1 with English units is:
Similarly, K2
and K3
equal to 0.0286 and 0.0439, respectively. The equivalent loss
coefficient is:

6. NB: n = 1.85
Note:
(2 and 4) The head loss between nodes A and B is then:
(4) So the head at node B, HB, is:
(3) The flow in each pipe can be computed from the individual pipe head loss
equations since the head loss is known for each pipe (hL= 0.43ft).
The flows in pipes 2 and 3 can be computed by the same equation and are 4.32 and
3.43 cfs, respectively. The sum of the three pipe flows equals 10 cfs, which is same
as inflow to node A.
Problem 5.

7. Solution