Problem Set #3 Additional Solutions

Grismer Ch. 5

2.

Area under the curve should equal 1 cm, so solving for height, we have:

½ base * height = 1 cm

½ (5)* h = 1 cm

4.

1. Area of UHG = 1 cm = 3h, or h = 1/3 cm / hr
2. Rainfall intensity = 1 cm / UHG storm duration = 1/3 cm/hr
3. Runoff duration of 9 hr UHG –Use 3 UHGs and lag second storm by one hour, and third storm by 2 hours. Duration of runoff = 10 hours.
4. 6 –hr UHG max discharge intensity = 1/duration of UHG = 1/6 cm/hr.

Serrano Ch. 3

4. A = 24 km²

Ts = 13 C

Ta = 16 C

W = 23.2 km/hr

Rh=0.72

E = ?

Interpolating:

E_s(13C) =11.35 mm Hg

E_a (16 C)=13.73 mm Hg * 0.72 = 9.89 mm Hg

Eqn. 3.3

E=K * A ^–0.05 * W * (e_s-e_a)

E = 0.054 * 24 ^–0.05 * 23.3 * (11.35-9.89)

6.

• S / D t = 17.5 mm/day

Q = 14.8 mm/day

P = 7 mm/day

K = 0.8

E_p = P + Q + D S / D t = 7 +14.8 – 17.5 = 4.3 mm/day

E = k E_P = 0.8 * 4.0 = 3.4 mm/day

Next day,

D S / D t = -4.8 mm/day

P=Q=0

Therefore, E_p = 4.8 mm/day

E = kE_p = 0.8 * 4.8 = 3.8 mm./day

11.

Ta = 36 C

W = 25 km/hr

PET = kc(a + bf)

f = p(0.46 Ta + 8.13)

soybeans, use 0.35

f = 0.31 * (0.46 * 36 +8.13) = 7.6539

PET = 0.35 * (-2.55 + (1.82 * 7.6539) = 3.98 mm/day