CE322 Basic Hydrology
Jorge A. Ramírez
Homework No.5 - Solution
1. Do problems 5.1 and 5.9 from your
textbook.
Problem 5.1: Use the Water Budget method to obtain
an estimate of annual evapotranspiration for the following conditions: Basin
Area A = 8000 mi2; annual
precipitation P = 20 in/yr; average
annual streamflow R = 5000 cfs.
Assume
that the system is in steady state, so that over one year, DS = 0; also, assume that the net groundwater flow out of the
basin is G = 0. Then,
Annual
volume of streamflow in equivalent units of depth:
R = 5000 (ft3/s) 86400 (s/day)
365 (day/yr) 12 (in/ft) /8000 mi2/(5280 (ft/mi))2 = 8.48
in/yr
(E+T) = P - R = 20 (in/yr) - 8.48 (in/yr) = 11.52
in/yr
Problem 5.9: Use the Water Budget method to obtain
an estimate of annual evapotranspiration for the following conditions: Basin
Area A = 2500 mi2; annual
precipitation P = 25 in/yr; average
annual streamflow R = 650 cfs.
Assume
that the system is in steady state, so that over one year, DS = 0; also, assume that the net groundwater flow out of the
basin is G = 0. Then,
Annual
volume of streamflow in equivalent units of depth:
R = 650 (ft3/s) 86400 (s/day) 365
(day/yr) 12 (in/ft) /2500 mi2/(5280 (ft/mi))2 = 3.53
in/yr
(E+T) = P - R = 25 (in/yr) - 3.53 (in/yr) = 21.47
in/yr
2. Use the Penman equation to estimate
evaporation from a (water) surface for the following conditions: temperature of
the surface, To = 290 K;
air temperature, Ta = 300
K; relative humidity, RH = 50%; wind
speed, u2 = 3 mph; surface
pressure, p = 100000 Pa; roughness
length, zo = 0.0001 m; net
available energy, Qn = 180
W/m2.
Penman
Equation:
1) Obtain saturation vapor pressures using
Clausius-Clapeyron equation:
es(Ta) = 611exp((-2.5 106/461.5)(1/300-1/273))
= 3604.95 Pa
es(To) = 611exp((-2.5 106/461.5)(1/290-1/273))
= 1934.12 Pa
2) Obtain D and
g:
D = (3604.95 – 1934.12) Pa / (300 - 290) K = 167.08 (Pa/K)
g = (1004 (J/kg/K) 105 Pa)/0.622/2.5
106 (J/kg) = 64.57 (Pa/K)
3)
Obtain
Drying Power of Air EA (do so by using the Thornthwaite eqn):
ra = (105 Pa)/(287 J/kg/K)/(300 K) = 1.16 kg/m3
EA = (0.622) (1.16 kg/m3) (0.4)2 (3 mi/h)(1609 m/mi)/(3600 s/h) ( 3604.95 Pa)
(1-0.5)/105 Pa/(ln(2/0.0001))2
EA = 2.83 10-5 kg/m2/s
Finally,
using Penman equation:
E = 0.721 (180 W/m2)/(2.5 106 J/kg) + (0.279) (2.83 10-5) (kg/m2/s) = 5.98 10-5 (kg/m2/s) = 5.17 mm/day.
3. For the conditions of Problem 2, use the
energy budget method to obtain an estimate of the evaporation rate. Assume that
the Bowen Ratio is, B = 0.5; and that
the temperature of the evaporated water is Te
= 293 K (Also, remember that Tb
= 273 K). What is the error incurred if one neglects to account for the energy
advected by the evaporated water (e.g.,
the term Qw )?
E
= (180 W/m2)/(2.5 106 J/kg)/(1 + 0.5 + 1004 (J/kg/K) (290
- 273 ) K/2.5 106 J/kg)
E
= 4.7783 10-5 kg/m2/s = 4.128 mm/day
Neglecting
energy advected by evaporated water:
E
= (180 W/m2)/(2.5 106 J/kg)/(1 + 0.5)
E
= 4.8 10-5 kg/m2/s = 4.147 mm/day
Relative
Error % = (4.128 - 4.147)/4.128 (100) = -0.46 %
4. Use the Meyer equation to estimate the
daily evaporation for a lake given that the mean air temperature is 85 F, the
mean surface temperature is 65 F, the wind speed is 5 mph, and the relative
humidity is 25 %.
Meyer
Equation:
1) Obtain saturation vapor pressures using
Clausius-Clapeyron equation:
es(Ta) = 611exp((-2.5 106/461.5)(1/302.6-1/273))
= 4209.98 Pa = 1.23 in Hg
es(To) = 611exp((-2.5 106/461.5)(1/291.5-1/273))
= 2129.26 Pa = 0.62 in Hg
ea = 0.25 es(Ta) = 1052.5 Pa = 0.3075 in Hg
E = 0.36 (0.62 - 0.3075) (1 + 5/10) =
0.16875 in/day = 4.29 mm/day
5. For the conditions of Problem 2, obtain:
a)
an estimate
of evaporation using the Thornthwaite-Holzman equation.
ra = (105 Pa)/(287 J/kg/K)/(300 K) = 1.16 kg/m3
EA = (0.622) (1.16 kg/m3) (0.4)2 (3 mi/h)(1609 m/mi)/(3600 s/h) ( 3604.95 Pa)
(1-0.5)/105 Pa/(ln(2/0.0001))2
EA = 2.83 10-5 kg/m2/s
b)
an estimate
of equilibrium evaporation;
Ee = 0.721 (180 W/m2)/(2.5 106 J/kg) = 5.19 10-5 kg/m2/s
= 4.48 mm/day
c)
an estimate
of the evaporation rate under conditions of minimal advection (partial
equilibrium).
Epe = 1.26 Ee = 5.65 mm/day