ABSTRACT

In this paper, the group proposes an analytical representation for the occurrence of hydraulic jump flow. The experiment showed that hydraulic jumps happen when a high velocity liquid enters a zone of lower velocity. The approach used by the group is controlled volume method, as it is the most commonly used approach in analyzing hydraulic jumps. Using the Reynolds Transport Theorem and with the aid of some very helpful assumptions, the group found a relationship between the characteristics of the jump upstream and jump downstream, as well as the entrainment. Unfortunately, the figures that were observed experimentally were not close enough to theoretical values. But the group has seen the accuracy and performance of the experiment apparatuses – scales, sluice gate – will affect mainly the data that will be observed.

INTRODUCTION

Hydraulic Jump occurs when an open channel flows at a high velocity and low depth then changes instantly to a high depth with low velocity. [2] In this experiment, letting water flow from a tank and then through a small opening of a gate, we create a rapid low-depth flow. By controlling a certain apparatus on the far side, a hydraulic jump is created. The hydraulic jump should be stabilized at one point so that a productive investigation is achieved The rapidly flowing liquid is abruptly slowed and increases in height, converting some of the flow's initial kinetic energy into an increase in potential energy, with some energy irreversibly lost through turbulence to heat. In an open channel flow, this manifests as the fast flow rapidly slowing and piling up on top of itself similar to how a shockwave forms. [3]

THEORETICAL BACKGROUND

The hydraulic jump is a phenomenon that happens in the flow of water. It reduces the flow with high velocity to low velocity while making the water level higher. In this part, we will derive the equations that were used in the experiment.

Figure 1 (The Hydraulic Jump) Figure 2 (Cross sectional Area of the flow)

Hydraulic jump’s characteristic still follows the Reynolds Transport Theorem (RTT). In that case, the equation of the discharge is Q=2/3 √ (2g) LH3/2 (1)

Where Q= discharge

L= width of the gate

H= height of the flume

Also, the equation for the flow just in the front of the opening of the gate is Q=AV(2)

Where Q=discharge

A=cross-sectional area

V1=velocity1

Substituting Eq. (2) to Eq. (1), we will have

AV1=2/3 √ (2g) LH3/2 (3)

However, the cross-sectional area A is the area of the gate opening. If the assumption is that there exists contraction of the cross-sectional area of the flow, the area of the flow would be A=Y1 * B (4)

Where Y1= height of the flow

B= width of the gate

Replacing A of the Eq. (3) with the variables of Eq. (4), we will have V1Y1 B =2/3 √ (2g) LH3/2 (5)

Evaluating the equation to get the value of V1, we will get

V 1=2/3 √ (2g) LH3/2 (6)

Y1 B

Since the gate of the setup is rectangular, the width of the flume is just equal to width of the gate.

L=B (7)

Thus, the equation of velocity1 of the flow is equal to

V1 =2/3 √ (2g) H3/2 / Y1 (8)

Given the momentum equation,

F = (gh12/2) – (gh22/2) = -v12h1 + v22h2 (9)

And the continuity equation,

v1h1=v2h2 (10)

We can eliminate V2 by substituting the continuity equation to the momentum equation. (gh12/2) – (gh22/2) = -v12h1 + v22h12 /h2 (11)

Evaluating the expression and isolating the h2, we will have, h12 – h22 = 2v12h1/g (-1 + h1/h2)

(h1 – h2) (h1 + h2) = (2v12h1/g) (h1 – h2)(1/h2) (h1 + h2)h2 = 2v12h1/g

h1 h2+ h22 = 2v12h1/g

Adding h12/4 at both sides,

h12/4+ h1 h2+ h22 = 2v12h1/g + h12/4

(h2 +...

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