How Gibbs module outlet work in the canal outlet system? describe the working principal.
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Gibb’s Rigid Module
Gibb’s module has an inlet pipe below the distributary bank. The pipe takes water from the distributary to a rising spiral, which is connected to the eddy chamber. This produces free vortex motion owing to which there is heading up of water (due to smaller velocity at larger radius – a feature of vortex motion) near the outer wall of the rising pipe. The water surface thus slopes towards the inner wall. A series of baffle plates of appropriate size are attached to the roof of the eddy chamber such that their lower ends slope against the direction of flow. As the head increases, water banks up at the outer wall at the eddy chamber and strikes against the baffles and spins round in the compartment between two adjacent baffle plates. This results in the dissipation of excess energy an2 release of a constant discharge. The outlet is relatively more expensive, and its sediment withdrawing characteristic is also not good.
The following discharge formula was given by Gibbs:
Q = ro √2g (d1 + ho)1.5 [(m2 – 1)/m3 x loge m + 1/m loge m – (m2 – 1)/2m2]
where Q = discharge passing down the module,
ro = radius of the outer semicircle of the eddy chamber,
m = ro/ r1 = ratio of outer radius to inner radius,
r1 = radius of inner semicircle,
d1 = depth of water at inner circumference,
ho = head loss in inlet pipe.
The formula is based on the free vortex flow in which the velocity at any point varies inversely as the radius, and by Bernoulli’s theorem, the total energy of all filaments is constant. Gibb’s formula holds good only for his standard design in which m or (r0/r1) = 2 and (ho/D) = 1/7, where D is the difference of level measured from the minimum water level in the parent channel to the floor of eddy chamber.
can you send working animation GIF or any video ?
because i have doubt that can we lift water without pump? vertex flow is sufficient to lift the water of that quantity ?