Hydraulic Engineering Essay Sample

8 August 2017

Flow under a perpendicular penstock gate: Separate A: Free discharge:
Part A: Free discharge
y1=0. 3680±0. 0005m
y2=0. 0280±0. 001m
y3=0. 0310±0. 003m
y4=0. 1640±0. 010m

x=0. 045m
Time for 1 regular hexahedron of H2O to go through the channel =47. 2s

Part A: Q1: eight )
Theoretical y2 = 0. 0262m
Measured y2 = 0. 0280±0. 001m
Energy disagreement = 0. 0443J
Based on the hydraulic theory velocity=discharge / tallness. with the same discharge. lower tallness of watercourse flows with higher speed than watercourse with greater tallness. The energy equation K. E. =mv22 and E=y+v22g besides show Measured y2 contains lower energy. in other words. energy is dissipated by: * Frictional opposition along the hydraulic construction such as channel and air. this consequences lessening in speed and addition in the deepness of H2O. therefore energy is dissipated. The size of clash depends on the channel surface. force per unit area and speed of H2O. and besides the viscousness of flow stuff. Although the clash to the hydraulic construction can non be eliminated wholly. but minimised by cut downing the coefficient of clash.

Part A: Q2: V )
Compare Fh and Fg ; explicate any difference – the existent force per unit area distribution is reduced as the speed additions ( energy conversation ) .

From the computation it seen that the ratio of Fg/Fh attacks of one of several depth ratios. which implies that the differences between the existent force on gate can be represented by the hydrostatic force moving on the gate.

Water deepnesss can be measured upstream and downstream of the penstock gate to cipher the hydrostatic force per unit area force. By mensurating the flow rate. the impulse equation can be applied to cipher the horizontal force moving on the penstock gate. The deliberate force can be compared to the existent force moving on the penstock gate by utilizing piezometers to mensurate force per unit area on the upstream face of the penstock gate.

Part A: Q3: four )
Theoretical y4 = 0. 1819m
Measured y4 = 0. 1640±0. 01m
The general account of disagreement was discussed in the cherished. this portion will coerce on the disagreement specific for this state of affairs. Immediately after the procedure of hydraulic jumping. the hydraulic flow performs disruptive flow. which indicates the atoms are likely to be traveling in irregular waies. which increases the force per unit area to the boundary. and hence the energy dissipated is greater.

Part A: Q3: seven )
Discuss briefly the usage of a hydraulic leap for energy dissipation. * Its purpose is to execute as an energy-dissipating device to cut down energy degree of H2O flows. * It reduces uplift force per unit area under the foundations of hydraulic constructions. * It raises the H2O degree on the downstream side of a measurement gulch and maintains high H2O degree in the side of a measurement gulch and besides maintains high H2O degree in the channel for irrigation or other water-distribution intent. * It reduces pumping caputs.

* It used to for blending of certain chemicals like in instance of H2O
intervention workss.

Part A: Q1: seven )
Theoretical y2 = 0. 0897m
Measured y2 = 0. 0960±0. 008m
There are troubles in entering y2 as y2 is located within hydraulic leap. Water flow transforms to turbulent flow at the point which hydraulic leap occurs. therefore the possible mistake in deepness is expected to be great.

Flow under a perpendicular penstock gate: Part B: Drowned Discharge:

y1=0. 4260±0. 0001m
y2=0. 0960±0. 0008m
y1=0. 1980±0. 0005m

x=0. 045m
Time for 1 regular hexahedron of H2O to go through the channel =47. 2s

Part B: Q1: nine )
Theoretical y1 = 0. 4137m
Measured y1 = 0. 4260±0. 001m
y1 holding greater deepness than expected is a phenomenon of drowned out on penstock gate by downstream. For H2O flows though sluice gate with great speed and fixed sum of H2O fluxing. H2O is stored in upstream to supply considerable force per unit area at the penstock gate. as the deepness of downstream additions. force per unit area incurs to defy fluxing from upstream. hence y1 demands to increase to supply sufficient force per unit area.

Part B: Q2: six )
Theoretical discharge Q = 0. 07155m3s-1
Measured discharge Q = 0. 0706m3s-1
Percentage of mistake = 1. 35 %
The mistake in discharge due to:
* Human mistake
Due to dispatch is calculated utilizing reading of clip spent for 1 regular hexahedron of H2O pumped into the channel. and the measuring is taken by human opinion which may non be every bit accurate as computing machine.

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