HYDROLOGY - TUTORIAL 2 TRAPEZOIDAL CHANNELS I

1 HYDROLOGY - TUTORIAL 2. TRAPEZOIDAL CHANNELS . In this TUTORIAL you will Derive equations associated with flow in a TRAPEZOIDAL channel. Derive equations for optimal dimensions. Solve slope of bed using Chezy and manning formulae. Solve questions from past papers. This TUTORIAL is a continuation of TUTORIAL 1 which should be studied first. 1. TRAPEZOIDAL SECTION. This topic occurs regularly in the Engineering Council Exam. The TRAPEZOIDAL section is widely used in canals to accommodate the shape of boats and reduce the erosion of the sides.

2 BEST DIMENSION Figure 1. The channel dimensions that give the maximum flow rate for a fixed cross sectional area is the one with the least amount of friction. This means that it must have the minimum wetted surface area and hence the minimum wetted perimeter P. If this value is then used in any formulae for the flow rate, we will have the maximum discharge possible. Using the notation shown on the diagram we proceed as follows. Area A = (B + b) hb from which B = (A/ hb) b = (A/ hb) hb/tan . Wetted Perimeter P = B + 2hb /sin.

3 A h 2h A 2 1 . Substitute for B P= b + b = + hb . h b tan sin h b sin tan . For a given cross sectional area the minimum value of P occurs when dp/dhb = 0. dP A 2 1 2 2 1 . = 2 + Equate to zero and A = h b and substitute for A. dh b h b sin tan sin tan . h 2 2 1 h 2 1 . B + b h b = h b B + b = hb . tan sin tan tan sin tan . 1 1 1 1 . B = 2h b or B = 2h b K where K = . sin tan sin tan . It can be shown that when this is the case, the bottom and sides are both tangents to a circle of radius hb. When = 90o K = 1 and when = 45o K = 2 -1 = and in fact K is almost a linear function such that K /90.

4 WORKED EXAMPLE Calculate the dimensions of a TRAPEZOIDAL channel with sides at 45o if it must carry m3/s of water with minimum friction given that C = 50 in the Chezy formula and the bed has a gradient of 1 in 1000. SOLUTION. The Chezy formula is uo = C (RhS) or Q = A C (RhS) . 1 1 . B = 2h b = b b = hb/tan 45o = hb sin45 tan45 . A = (B + b) hb = ( hb + hb) hb = hb2. 1 . P = B + 2h b = b + b = b sin45 . Rh = A/P = hb Q = = hb2 x 50( hb/1000)1/2. = hb4 ( hb/1000). = hb5. hb = m B = hb = m 2. SELF ASSESSMENT EXERCISE 1.

5 Calculate the dimensions of a TRAPEZOIDAL channel with sides at 60o to the horizontal if it must carry 4 m3/s of water with minimum friction given that C = 55 in the Chezy formula and the bed has a gradient of 1 in 1200. (hb = m B = m). 2. Calculate the dimensions of a TRAPEZOIDAL channel with sides at 30o to the horizontal if it must carry 2 m3/s of water with minimum friction given that C = 49 in the Chezy formula and the bed has a gradient of 1 in 2000. (hb = m B = m). CRITICAL DEPTH. It requires a lot of Algebra to get to the critical values.

6 Start as before hs = hb + uo2/2g Rearrange to make u the subject u o2 = {2g (h s h b )}. Q = Auo Q2 = A2uo2. A = (B + b)hb Q2 = (B + b)2hb2 uo2. Q2. = (h s h b )(B + b ) h b 2 2. Substitute for uo 2g We cannot differentiate this expression because b is a function of h so we make a substitution first. b = hb/tan . 2 2. Q2 h h2 . = (h s h b ) B + b h b 2 = (h s h b ) Bh b + b Now we need to multiply out. 2g tan tan . Q2 h4 2Bh 3b 2 2 h 4h 2Bh 3b h s 2 3 h5 2Bh 4b . = (h s h b ) B2 h 2b + b2 + = B h b h s + b 2s +.. B h b + b2 +.

7 2g tan tan tan tan tan tan . Now differentiate with respect to hb to find the maximum flow rate for a given specific energy head. 2 QdQ 4h 3 h 6Bh 2b h s 5h 4b 8Bh 3b = 2B2 h b h s + b2 s + 3B2 h 2b . 2gdh b tan tan tan 2 tan . For maximum Flow rate equate dQ/dhb to zero. 4h 3 h 6Bh 2b h s 5h 4b 8Bh 3b 0 = 2B2 h b h s + b2 s + 3B2 h 2b . tan tan tan 2 tan . We can simplify by substituting back hb/tan = b 0 = 2B 2 h b h s + 4b 2 h b h s + 6 Bbh b h s 3B 2 h 2b 5b 2 h 2b 8 Bbh 2b ( ) (. 0 = h b 2B 2 h s + 4b 2 hh s + 6 Bbh s h 2b 3B 2 + 5b 2 + 8Bb ).

8 ( ) (. 0 = 2B 2 h s + 4b 2 hh s + 6 Bbh s h b 3B 2 + 5b 2 + 8Bb ). hb = hc =. (2B 2. + 4b 2 + 6Bb ). h s = Ch s Rearrange to get the critical depth (3B 2. + 5b 2 + 8Bb ). C=. (2B 2. ). + 4b 2 + 6Bb (2B + 4b )(B + b ) (2B + 4b ). = =. (3B 2. ). + 5b 2 + 8Bb (3B + 5b )(B + b) (3B + 5b ). 3. hc =. (2B + 4b ) h or h = (3B + 5b ) h (3B + 5b ) s s (2B + 4b ) c 4h s If B = 0 we have a Vee section h b = h c = as before. 5. 2h If b = 0 we have a rectangular section h b = h c = s as before. 3. There are computer programs for making the calculations such as the one at To find the critical velocity flow rate substitute h s =.

9 (3B + 5b ) h into u 2 = u 2 = {2g (h h )}. (2B + 4b ) c o c s c (3B + 5b ) (3B + 5b ) . u o2 = u c2 = 2g h c h c = 2gh c 1 . (2B + 4b ) (2B + 4b ) . (3B + 5b ) (3B + 5b ) (2B + 4b ) . u c = 2gh c 1 = 2gh c . (2B + 4b ) (2B + 4b ) . B + b . u c = 2gh c . (2B + 4b ) . gh . If B = 0 we have a Vee section u c = c as before. 2 . If b = 0 we have a rectangular section we have u c = {gh c } as before. To find the critical flow rate substitute use Qc = A uc A = (B + b)hc B + b B + b . Q c = (B + b)h c 2gh c Q c = (B + b)h 3/2 2g.

10 (2B + 4b ) . c (2B + 4b ) . g . If B = 0 we have a Vee section Q c = bh 3/2. c as before in a slightly different form 2 . If b = 0 we have a rectangular section we have Q c = Bh 3/2. c g as before. Summary for TRAPEZOIDAL section The critical depth is hc =. (2B + 4b ) h (3B + 5b ) s B + b . The critical velocity is u c = 2gh c . (2B + 4b ) . B + b . The critical flow is Q c = (B + b)h 3/2 2g . (2B + 4b ) . c The major problem exists that solving with these formulae requires a value for b and this depends on the answer itself.