Answers to Problem Set #1
Serrano Ch. 1 7,11,15
Grismer Ch.1, 2 even
Serrano
7)
Area of watershed is 220 km2
Annual precipitation in the watershed area is 4.2*109
Equivalent depth is
4.2*109/ 220*106 = 19.1 m
This is approximately an order of magnitude too high
11)
(1)
River Storage= 12 * 106 m3
River Inflow = 14 m3/s
River Outflow – 20 m3/s
After 4 hours
Inflow = 18 m3/s
Outflow = 26 m3/s
Average groundwater flow TO river during these fours hours = 9 m3/s
Assuming a linear increase, we average the inflow as:
14 + 18 = 16 m3/s
2
However, this is not the only source of water, since there is also groundwater inflow.
The total flow of water is
16 + 9 = 25 m3/s
The average river outflow can be calculated the same way
20 + 26 = 23 m3/s
2
The change in storage is therefore the difference between inflow and outflow over a 4 hour period:
Inflow- Outflow = Change in Storage
4 hours = 240 minutes = 14,400 seconds
Storage = 2 m3/s * 14,400 = 28,800 m3
(2) Since the river storage at the beginning of the four hour period was 12* 106 m3
and we added 2.22*104 m3, the new storage volume is 1202.8*104 m3.
15)
Surface Area of lake = 14 km2 = 14 * 106 m2
Average inflow rate into lake = 50 m3/s
Average ouflow rate = 25 m3/s
Lake storage increases in 1 day with 14.7 cm
In other units:
Inflow = 50 m3/s * 86400 s/day = 4.32 * 10 6 m3/day
Outflow = 25 m3/s * 86400 s/day = 2.16 * 10 6 m3/day
Storage = 14.7 cm * area of lake = 0.147 m * 14 * 106 m2 = 2.058* 106 m3/day
Writing water balance
Evaporation = Inflow – Outflow – Lake volume change
Evap = 4.32*106 m3/day – 2.16*106 m3/day – 2.058*106 m3/day
Evap = 1.02 * 105 m3/day
Over lake of area 14 * 10 m2, equivalent depth of :
1.02*105 m3/day = 7.3 mm/day
14 * 106 m2
Grismer
Ch. 1
2) Nothing can be concluded about deep percolation losses in the cornfield because there is not
information given about the spatial and temporal distribution of rain during the year. For example, back to back rain storms may result in deep percolation losses, but only account for a fraction of the total rainfall.
4)Water balance calculation
P - ET - Q = D S
760mm – 550 mm – 200mm = 10 mm
6) Water Balance Problem
D S = P- E- R
Evap can be assumed to be negligible due to ½ hour storm
Runoff= 12,500 m3 (given)
Area = 106 m2
P = 10 cm * 1 m * 1 hour * 106 m2 = 50,000 m3
hour cm 2
D S = 50,000 – 12, 500 = 37,500 m3
8)
Area = 1250 * 104 m2
D S = P –E – R –W or D S + E = P – R – W
P = 4.12 m * 1250 * 104 m2= 5.15*107 m3/yr
R = 0.35 m3/s * 86400 s/day * 365 day/yr = 1.1 * 10 7 m3/yr
W = 0.16 m * 1250 * 104 m2 = 2.0 *106 m3/yr
D S + E = 3.85* 107 m3/yr
Municipal use = 50,000 m3/day * 365 days/yr = 1.825*106 m3/yr
E threshold = 3.85* 107 m3/yr-1.825*106 m3/yr = 2.02 * 10 6 m3/yr
10)Monthly figures are totaled for annual numbers
Area = 37 ha
D S = P + I – E –W where I = drain inflow
D S= -18.3 mm * 1m/1000 mm * 37 * 104 m2 = -6771 m3
P = 166 mm * 1m/1000 mm * 37 * 104 m2 = 61420 m3
I = (add columns) = 984,000 m3
E = 1.842 m(average E) * 37 * 104 m2 = 681,540 m3
W = P + I – E -D S
= 61420 + 984,000 – 681540 + 6771 = 370,651 m3
Seepage Rate = 370,651 / 37 * 10 4 m2 = 1 m /yr
% of P + I lose to Evaporation E/(P+I)
= 681,540/1,045,420 = 65.2% leaving
34.8% to seepage
Ch. 2
2) Sprinkling water on the crops increases the energy requirement (loss) required to freeze the
plant. The heat capacity of the water and the energy loss required to freeze it act as a temperature buffer so that the plant, water and ice stay around zero degrees. Plant damage occurs at temperatures appreciably below zero.
clusters of water molecules due to hydrogen bonding. These clusters breakdown as temperature increases due to decreasing H bonding, thus viscocity decreases.
and water. Addition of solute to the water has little effect on the surface layer of water molecules except at relatively high concentrations because the ionic interactions tend to occur in bulk solution. Thus, the height will change little by adding salt to water.