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Ac 1/3 (10.37a) gAc
b0 1/2 (10.37b) For a given channel shape A(y) and b0(y) and a given Q, Eq. (10.37) has to be solved by
trial and error or by EES to find the critical area Ac, from which Vc can be computed.
2 | v v This is the basis of the water-channel analogy for supersonic gas-dynamics experimentation [14, chap. 11]. | e-Text Main Menu | Textbook Table of Contents | Study Guide 674 Chapter 10 Open-Channel Flow By comparing the actual depth and velocity with the critical values, we can determine the local flow condition
y Vc: subcritical flow (Fr y Critical Uniform Flow:
The Critical Slope yc, V 1) yc, V Vc: supercritical flow (Fr 1) If a critical channel flow is also moving uniformly (at constant depth), it must correspond to a critical slope Sc, with yn yc. This condition is analyzed by equating Eq.
(10.37a) to the Chézy (or Manning) formula:
C2Ac RhSc n 2 4/3
Ac Rh Sc n2g P
Rhc b0 n2gAc
b0Rhc Sc 2 fP
8 b0 (10.38) where 2 equals 1.0 for SI units and 2.208 for BG units. Equation (10...
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This note was uploaded on 10/27/2009 for the course MAE 101a taught by Professor Sakar during the Spring '08 term at UCSD.
- Spring '08