# Hydraulics II Report 2 S1 2023 Final

Hydraulics II Report 2 S1 2023 Final :

#### Question 1 – Pipe Network (100 Marks)

A pipe network as shown in Figure 1 has been constructed in order to convey water from a reservoir A to a number of delivery points.

The details of each pipe are given in the table below. You may also neglect all minor losses that may occur in the system.

Figure 1 – Pipe network for Q1

The head added by the pump (P) is a constant 20m. The pressure head elevation at point A is 78 m.

The elevation of each node of the pipe network is given below.

1. Use the linearisation method to solve for the unknown discharges in each pipe of the network.
• Accounting for the elevation of each node, estimate the pressure head in metres at each pipe junction (A, B, C, D, E, F)

HINT: The pump has the opposite effect (opposite direction) to the friction loss in pipe BC

Marking Scheme: Question 1 – Pipe Network

#### Question 2 – Vegetative Lined Channels (60 Marks)

Design the broad shallow grassed waterway for the transmission of flood flows in a compound channel illustrated below:

Design Data:

• The soil type is erosion resistant
• The channel is currently planted with Rhodes grass
• The Bed slope is 3%

The channel consists of:

• a narrow concrete lined section (n = 0.014) designed to carry the normal low streamflow; and
• a broad shallow grassed waterway for the transmission of flood flows. Your design must satisfy two main criteria:

The velocity of flow does not exceed the permissible velocity nominated for the particular

grass and soil in the waterway, that is, the channel must be stable.

The depth of flow does not exceed the height of the channel banks, that is, the channel must have sufficient capacity.

#### Marking Scheme: Question 2

Question 3 – Control Structure (70 marks)

An irrigation scheme is fed from a river via a diversion channel. The irrigation channel is 2 m wide and is constructed of rough concrete with an estimated value of 0.025 for the Manning

n. The bed slope is 0.0025.

The discharge into the channel is controlled by a vertical sluice gate (Cc = 0.61). The depth upstream of the gate is a constant 3.0 m, and the maximum discharge is 7 m3 /s.

The designer of the gate has prepared a rating curve for the sluice gate (yg vs Q) for free- flowing conditions as shown below in the table.

The issue is that the designer has not considered those cases where the gate may be drowned by the hydraulic jump downstream of the gate. The depth on the downstream side of the gate is equal to the normal flow depth.

Independently you have determined the normal flow depths over the operating range of discharges, which is also included in the table below.

Figure 2 – Gate rating curve assuming free flowing conditions

• Determine at what discharge the gate changes from freely flowing to submerged conditions (to the nearest m3 /s)?
• Calculate the new gate opening (yg) for those discharges for where the gate is submerged by the depth downstream of the gate.
• Plot the new rating curve (alter the part where gate is submerged)

#### Marking Scheme Question 3 – Control Gate

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