CE 4603: Water Resources Engineering — David Wagner 2007/04/09 13:43

Homework #8 Due 4/9

Solve problems 3.2, 3.4, 3.11 (part a), 3.16 (part a), and 3.20 of Water Resources Engineering by David Chin, 2^nd edition, pages 255-257.

3.2

Water flows at 8 m³/s through a rectangular channel 4 m wide and 3 m deep. If the flow velocity is 1 m/s, caculate the depth of flow in the channel. If this channel expands (downstream) to a width of 5 m and the depth decreases by 0.5 m from the upstream depth, then what is the flow velocity in the expanded section?

A=4h ; P=4+2h ; R=

h=2 m

A=5h →

V = 1.07 m/s

3.4

A trapezoidal channel is to be excavated at a site where the permit restrictions require the channel have a bottom-width of 5 m, side slopes of 1.5:1 (H:V), and a depth of flow of 1.8 m. If the soil material erodes when the shear stress on the perimeter of the chnnel exceeds 3.5 Pa, determine the appropriate slope and flow capacity of the channel. Use the Darcy-Weisbach equation and assume that the excavated channel has an equivalent sand roughness of 3 mm. A=(b+my)y = (5+1.5*1.8)*1.8 = 13.86 m² $P = b + 2 sqrt{1+m^2}y = 5+ 2*sqrt{1+1.5^2}*1.8$ = 11.49 m R = {A/P} = 13.86/11.49 = 1.21 tau_0 = gamma R S_f = 3.5 = 9810*1.21*s_f right S_f =

S_f = 0.000295 m/m

${1/sqrt{f}} =-2log_10(k_s/12R) = -2*log_10(3/{12*1.21})= 1.37 right f$ =0.533 $V = sqrt{8g/f} sqrt{R S_f} = sqrt{8*9.81/0.533}*sqrt{1.21*0.000295} = 12.13*0.0189$ =0.229 m/s Q=AV = 13.86*0.229 =

Q = 3.18 m³/s

3.11a

Water flows at 20 m³/s in a trapezoidal channel that has a bottom width of 2.8 m, side slope of 2:1 (H:V), longitudinal slope of 0.01, and a Manning n of 0.015. (a) Use the Manning equation to find the normal depth of flow…

A=(b+my)y = (2.8 + 2*y)*y $P=b+2 sqrt{1+m^2}y = 2.8+sqrt(1+2^2}*y = 2.8 + 2.236*y$

¹⁾

$Q=AV right V=Q/A = {1/n}(A/P)^{2/3}S^{1/2} = 20/{(2.8+2*y)*y} = {1/0.015}*({(2.8 + 2*y)*y}/{2.8 + 2.236*y})^{2/3}*0.01^{1/2}$

y_n = 0.8 m

3.16a

A trapezoidal irrigation channel is to be excavated to supply water to a farm. The design flowrate is 1.8 m³/s, the side slopes are 2:1 (H:V), the longitudinal slope of the channel is 0.1%, Manning's n is 0.025, and the geometry of the channel is to be such that the length of each channel side is equal to the bottom width. (a) Specify the dimensions of the channel required to accomodate the design flow under normal conditions… $b=sqrt{h^2+(2h)^2} = sqrt{5} y$

$A=(b+my)y = (sqrt{5}*y + 2*y)*y = 4.236*y^2$

P=3b=3*sqrt{5}*y = 6.708*y

$R={A/P} = 4.236y^2/6.708y = 0.6315*y$

$Q=AV right V=Q/A = {1/n}R^{2/3}S^{1/2} = 1.8/{4.236*y^2} = {1/0.025}*(0.6315*y)^{2/3}*0.001^{1/2}$

h = y_n = 0.74 m	b = 1.65 m

3.20

Consider the drainage channel and adjacent floodplains showin in Figure 3.58. The Manning roughness coefficients are given by:

Section	n
Left Floodplain	0.040
Main Channel	0.016
Right Floodplain	0.050

and the longitudinal slope is 0.5%. Field tests have shown that the Horton (1933) equation best describes the composite roughness in the channel. Find the capacity of the main channel and the lateral extent of the floodplains for a 100-year flow of 1590 m³/s.

A=30*3+{1/2}*9*3+{1/2}*6*3 = 90 + 13.5+9 =112.5 m² $P=30+sqrt{9^2+3^2}+sqrt{6^2+3^2}=30+srt{90}+sqrt{45}$ =46.2 m R=A/P = 112.5/46.2 =2.435 $Q=AV={112.5/0.016}*2.435^{2/3}*0.005^{1/2}=7031*1.81*.0707$ =

Q_main = 900 m³/s

¹⁾