rea 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
- 23 = 2 m3/s over a 4 hour period
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
heyaaa
same concept
same exaple
thanxxxx
