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CE 4603: Water Resources Engineering — David Wagner 2007/02/27 15:08

Homework #4 Due 2/26

Solve problems 5.28, 5.30, 5.32, 5.37, and 5.41 of Water Resources Engineering by David Chin, 2nd edition, pages 588-590.

5.28

Pine woods [p.376]: n=1; a=0.05 mm; b=0.20; I = a + bP^n → I = 0.05 + 0.20 P

For a 30-min storm: i=2819/{t+16} = 2819/46 = 61.3 mm/hr - > P = 31 mm →

  • I = 0.05 + 0.20 P = 0.05+.2*31 = 6.2 mm

5.30

[p.386] f0 = 200 mm/h ; fc = 60 mm/h ; k = 4 min-1 ; i≥200 mm/hr ; t = 50 min f_p = f_c + (f_0 - f_c) e^{-kt} = 60 + (200-60)*e^{-4t} = 60 + 140 e^{-4t}

  • f_p50 =60 + 140*e^{-4*50}=60 mm/hr

5.32

[p.386-387]F = [{f_c/k} ln(f_0-f_c)+{f_0/k}] - {f_c/k}ln(f_p-f_c)-{f_p/k} = [{50/180} ln(150-50)+{150/180}] - {50/180}ln(f_p-50)-{f_p/180}

  • F = 1.28-0.833*ln(f_p-50) - {f_p/180}
  • t: 0-20 min, f<fc → no ponding, F= i_1 delta_t + i_2 delta_t = (20+40)*{10/60} = 10 mm
  • t: 20-30 min, f>fc → ponding possible F = 10 = 1.28-0.833*ln(f_p-50) - {f_p/180} → fp=50
  • 80>50 → Ponding occurs at 20 minutes

5.37

[p.396-400]

  • min I=10 → NRCS Soil Group A
  • ½ac lots → Curve Number 54
  • Normal soil → ARC II ( no adjustment)
  • i = 203/{(t+7.24)^0.73} = 203/{67.24^0.73} = 9.4 cm/hr → P=9.4 cm
  • CN=1000/{10+0.0394S} = 54 right S=25.38*({1000/54} - 10) = 68.15 mm
  • Q={(P-0.2S)^2} / {P+0.8S} : P>0.2s 9.4 > 1.36✔
  • Q={(9.4-0.2*6.8)^2} / {9.4+0.8*6.8} = 64.64/14.8 = Q=4.356 cm

5.41

Since the CRCS curve-number model is useful only for 24-hr rainfall amounts, only 24-hr periods should be considered. First, the 1-day event.

  • <m>f = {S^2i} / {(P+0.8S)^2} = {21^2*{31/24}} / {(31+0.8*21)^2} = 570/2285</m>=0.25 cm/hr

S=210 mm is given. By definition, the curve number CN = 1000/{10+0.0394S} = 1000/{10+.0394*210} = 54.7 and does not vary with infiltration capacity fp; the relationship is a horizontal straight line. If S is not constant as given, the following formula could be rewritten to solve for CN in terms of fp, for the precipitation P in each 24-hr period under consideration.

F = [{f_c/k} ln(f_0-f_c)+{f_0/k}] - {f_c/k}ln(f_p-f_c)-{f_p/k} = {(P-0.2S)^2} / {P+0.8S}

But, you will also need the values of f0, fc, and k.

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