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CE 4603: Water Resources Engineering — David Wagner 2007/03/04 16:01

Homework #5 Due 3/7

Solve problems 5.51, 5.55, 5.56 of Water Resources Engineering by David Chin, 2nd edition, pages 588-590.

5.51

[pg.415, ex.5.23][p.408,eq.5.94]

i_e = 50 mm/hr ; L = 30 m ; n_asphalt approx 0.012 ; S_0_lot = 0.01

sum{ }{ }{t_c} = t_clot + t_cchannel ; t_clot approx 6.99{{(n L)^0.6}/{i_e^0.4 S_o^0.3}} = 6.99{{(0.012*30)^0.6}/{50^0.4*0.01^0.3}} = 6.99 * 0.5417/{4.782*0.2512} = 3.15 min

Q = 0.02 m^3/s ; n_concrete = 0.013 ; S_0channel=0.006 ; A = 0.20 d ; P = 0.20 + 2 d

Q = {A/n} (A / P)^{2/3} sqrt{S_0} = {{0.20 d}/n} ({0.20 d}/{0.20 + 2 d})^{2/3} sqrt{S_0} = {sqrt{0.006}/0.013} root{3}{{0.008 d^5}/{(0.2+2d)^2}} = 5.96 root{3}{{0.008 d^5}/{(0.2+2d)^2}} = 0.02

right d = 12 cm right A=0.2*d=0.024 m^2 right V_0 = Q/A = 0.02/0.024 = 0.83 m/s
right t_cchannel = L_channel/V_0 = 60/0.83 = 72 s = 1.2 min

tc = 4.35 min

5.55

Assume this problem does not compound 5.52 (which was not assigned) and use ex. 5.24, pg 418-420.

n=0.035 ; A=0.5 ha ; L=30 m ; S_0=0.005 ; C=0.9

t_c = 6.99*{(n L)^0.6} / {{i_e^0.4} S_0^0.3} = 6.99*{(0.035*30)^0.6} / {{i_e^0.4} * 0.005^0.3} = 7.2/{.204*{i_e}^0.4} = 35.3/{{i_e}^0.4}

i_e=C i = 150 C/{(t + 8.96)^0.78}= 150*0.9/{({35.3/{{i_e}^0.4}} + 8.96)^0.78}

ie=12 cm/h

Q_p=C i A =0.9*12*{cm/hr} * {{1m}/{100 cm}} * {{1 hr}/{3600 sec}} * 0.5 ha * {10000*{m^2}/{ha}}=

Qp = 0.15m³/s

5.56

[ex. 5.26, pg.423-424] Find qp for 10 cm of 24-hr Type II precipitation.

A=4.2*km^2 ; A_pond = 0.5% ; CN=79 ; t_c = 3 hr ; P=100 mm

S= {1/0.0394}({1000/CN}-10) = {1/0.0394}({1000/79}-10) = 67.5mm

Q={(P-0.2 S)^2}/{P+0.8 S} = {(100-0.2*67.5)^2}/{100+0.8*67.5} = 7482/154 = 48.6mm

I_a/P = {0.2 S}/P = {0.2*67.5}/100 = 0.135

[Table 5.22]F_p approx 0.93

[Table 5.23]C_0=2.55323 ; C_1=-0.61512 ; C_2 = -0.16403

log(q_u) = C_0 + C_1 log(t_c) + C_2(log(3))^2 - 2.366 =2.55323 -0.61512* log(3) - 0.16403*(log(3))^2 - 2.366 =2.55323 -0.61512* 0.4771 - 0.16403*0.2276 - 2.366 = 2.55323 -0.2935 - 0.03734 - 2.366 = -0.1436 right q_u = 0.7185(m³/s)/cm/km²

q_p = q_u A Q F_p = 0.7185*4.2*4.86*0.93

= qp = 13.7 m³/s

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