Question

CM&B: 11.3 The principal characteristics of steady flow in a wide rectangular channel are: The mean velocity (V) The frictional resistance (F The density (p) and viscosity (u) of the fluid The depth (y) of flow or the channel length (Z)- as appropriate The bed slope (So) and surface roughness (k) Gravitational acceleration (g) Show that these can be expressed in the form of the following II groups: РVy Reynolds No. Re V Froude No Fr gy ks Relative roughness _ у Bed slope So Bed drag coefficient II II II Il

Answer 1

* Ang : 7 grocefs:- Buckingham I theoleng. V mean velocity (v) → [LT"] frictiona resistance (6 → SMLT) ✓ density (8) » [M23] viscosity (n) [melt] ✓g depth ly [L] of Length (L) (4] (ks) surface Roughness (1 gravitation a acceleration (g) → () geometry term depth (y) Kinematic term - velocity (V) a dynamic term density (8) variableyn) =9, repited variables (m) -3 no of 1-toms - n-m=4 i = moet = (sa b u Iz = Mir = (p2 uby cz) F та = se ® :(+99) 9 ТК = n°РР - (0,49) К * T - [milf'] = (ma+3]" (L777"[234[mu""] [mpleto] = {m ut! ;34, the Ge! Fb-1] .: -61-1=0 / a,+1=0 / -39, +, +4=1 6,- ) a =-11 -3(-1)-1 +4=1 192-1 . 19 = 60' 54 5) the [TF function of Rety poids no.) Puy

* T2: [m 20p] = {ME 3]? [LT] [L] [ML T {mº leto] =(92+1 3392 +62 +6+/ +62-2] I autizo 1-392tb2t =1 -3(-) =2+G=7 [22=-21 : 2 = (81 22 23) .:. 'Ta' is bed drag coefficient. * Tạ - K P3 -[…] (Lil ( [27] = {m23 1393 + b3 + c3+1 g_b3-2] [93=0) |_bz_220 1 -393 + b₃ + c3 =- -2t Q=t z = gorzylg I3 = yg & ² = gy - 1563=-2) I (C =]] ve voy i ara a const) = Froude No & Ty :- {MO E 73 = [mizzau [LT 3 by [L]cu [L} Imour] = frau -304 4 by 4 Cutl (J by] [au=0 s (buzo): -3aut but chat Cu=-1) ... Tu = goryl ks = = Relatie Roughness. Bed slope is (so) is "dimensionless" [m°2 703