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Stream power, originally derived by R. A. Bagnold in the 1960s, is the amount of energy the water in a river or stream is exerting on the sides and bottom of the river. [1] Stream power is the result of multiplying the density of the water, the acceleration of the water due to gravity, the volume of water flowing through the river, and the ...
This is an average measure. For measuring the discharge of a river we need a different method and the most common is the 'area-velocity' method. The area is the cross sectional area across a river and the average velocity across that section needs to be measured for a unit time, commonly a minute.
The term stream power law describes a semi-empirical family of equations used to predict the rate of erosion of a river into its bed. These combine equations describing conservation of water mass and momentum in streams with relations for channel hydraulic geometry (width-discharge scaling) and basin hydrology (discharge-area scaling) and an assumed dependency of erosion rate on either unit ...
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It uses a combination of the energy, momentum, and continuity equations to determine water depth with a given a friction slope (), channel slope (), channel geometry, and also a given flow rate. In practice, this technique is widely used through the computer program HEC-RAS , developed by the US Army Corps of Engineers Hydrologic Engineering ...
The graph takes sediment particle size and water velocity into account. [2] The upper curve shows the critical erosion velocity in cm/s as a function of particle size in mm, while the lower curve shows the deposition velocity as a function of particle size. Note that the axes are logarithmic.
The discharge formula, Q = A V, can be used to rewrite Gauckler–Manning's equation by substitution for V. Solving for Q then allows an estimate of the volumetric flow rate (discharge) without knowing the limiting or actual flow velocity. The formula can be obtained by use of dimensional analysis.
The discharge may also be expressed as: Q = − dS/dT . Substituting herein the expression of Q in equation (1) gives the differential equation dS/dT = A·S, of which the solution is: S = exp(− A·t) . Replacing herein S by Q/A according to equation (1), it is obtained that: Q = A exp(− A·t) .