ch11 - Problem 11.1 Given: Rectangular channel flow Find:...

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
Problem 11.1 [1] Given: Rectangular channel flow Find: Discharge Solution: Basic equation: Q 1 n A R 2 3 S 0 1 2 = Note that this is an "engineering" equation, to be used without units! For a rectangular channel of width B w 2m = and depth y 1.5 m = we find from Table 11.2 AB w y = A 3.00 m 2 = R B w y B w 2y + = R 0.600 m = Manning's roughness coefficient is n 0.015 = and S 0 0.0005 = Q 1.49 n A R 2 3 S 0 1 2 = Q 3.18 m 3 s =
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Problem 11.2 [3] Given: Data on rectangular channel Find: Depth of flow Solution: Basic equation: Q 1 n A R 2 3 S 0 1 2 = Note that this is an "engineering" equation, to be used without units! For a rectangular channel of width B w 2.5 m = and flow rate Q3 m 3 s = we find from Table 11.2 AB w y = R B w y B w 2y + = Manning's roughness coefficient is n 0.015 = and S 0 0.0004 = Hence the basic equation becomes Q 1 n B w y B w y B w + 2 3 S 0 1 2 = Solving for y y B w y B w + 2 3 Qn B w S 0 1 2 = This is a nonlinear implicit equation for y and must be solved numerically. We can use one of a number of numerical root finding techniques, such as Newton's method, or we can use Excel 's Solver or Goal Seek , or we can manually iterate, as below, to make the left side evaluate to B w S 0 1 2 0.900 = . For y1 = m () y B w y B w + 2 3 0.676 = For y 1.2 = m y B w y B w + 2 3 0.865 = For y 1.23 = m y B w y B w + 2 3 0.894 = For y 1.24 = m y B w y B w + 2 3 0.904 = The solution to three figures is y 1.24 = (m)
Background image of page 2
Problem 11.3 [3] Given: Data on trapzoidal channel Find: Depth of flow Solution: Basic equation: Q 1.49 n A R 2 3 S 0 1 2 = Note that this is an "engineering" equation, to be used without units! For the trapezoidal channel we have B w 8f t = z2 = Q 100 ft 3 s = S 0 0.0004 = n 0.015 = Hence from Table 11.2 AB w zy + () y = 82 y + y = R B w + y B w 2y 1z 2 + + = y + y y 5 + = Hence Q 1.49 n A R 2 3 S 0 1 2 = 1.49 0.015 y + y y + y y 5 + 2 3 0.0004 1 2 = 100 = (Note that we don't use units!) Solving for y y + y [] 5 3 y 5 + 2 3 50.3 = This is a nonlinear implicit equation for y and must be solved numerically. We can use one of a number of numerical root finding techniqu such as Newton's method, or we can use Excel 's Solver or Goal Seek , or we can manually iterate, as below. For y2 = ft y + y 5 3 y 5 + 2 3 30.27 = For y3 = ft y + y 5 3 y 5 + 2 3 65.8 = For y 2.6 = ft y + y 5 3 y 5 + 2 3 49.81 = For y 2.61 = ft y + y 5 3 y 5 + 2 3 50.18 = The solution to three figures is y 2.61 = (ft)
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Problem 11.4 [3] Given: Data on trapezoidal channel Find: Depth of flow Solution: Basic equation: Q 1 n A R 2 3 S 0 1 2 = Note that this is an "engineering" equation, to be used without units!
Background image of page 4
Image of page 5
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 62

ch11 - Problem 11.1 Given: Rectangular channel flow Find:...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online