Using these symbols, and assuming that all quantities are in a consistent system
ID: 285674 • Letter: U
Question
Using these symbols, and assuming that all quantities are in a consistent system of units (no unit conversions required): i. Write the water-balance equation for the drainage basin. ii. Write the water-balance equation for the lake iii. Write the expression for the residence time of the lake a. b. Below are data for Lake Winnipesaukee, New Hampshire's largest lake Dainage basin: An® 940 km 2; Po = 1,240 mm/yr, eu = 500 mm/yr Lake i. Use the equations you derived in 8a to estimate average lake evaporation, e A! = 180 km": PL = 952 mm/y, Q = 15.1 ms/s; z." 18 m Does the answer differ if you use the drainage-basin equation (8.a.i) or the lake equation (8.a.ii)? ii. Compare the estimatets) of e with the given value of p What are possible explanations for the differences? ii. Calculate the residence time for the lake. iv. Compare the magnitudes of the inflows to the lake from drainage-basin runoff and precipitation. What are the water-quality implications of these numbers? Compare the magnitudes of the outflows to the lake via streamflow and evaporation. v.Explanation / Answer
2. i)
Water-balance equation for drainage basin
P=E+Q+S
Where precipitation P, evapotranspiration E, run off Q & soil water storage S
ii) Water-balance equation for lake
Qinput - Qoutput = dv/dt = A dh/dt [where V=lake water volume; dv/dt= rate of volume change
A lake water area; h water depth]
Qinput = (Precipitation + stream inflow + groundwater inflow + diffuse runoff )
Qoutput = (Evapo-transporation + stream outflow + groundwater outflow)
iii) Residence time of the lake = Total volume of lake/ Rate of Qinput = A.h / [d(Qinput)/dt]
b. i)
from water-balance equation for lake, we get
PL * A + Q = eL * A then, PL = Q/A + eL where eL =average lake evaporation
Q= 15.1 m3/s * 365 days = 15.1 m3/s * (365*24*3600 )s
Q/A = 15.1 * (365*24*3600) m3/ (180*106) m2 = 2.64552 m/yr = 2645.52 mm/yr
eL = PL + Q/A= 952 +2645.52 = 3597.52 mm/yr
hence, average lake evaporation is 3597.52 mm/yr.
if it’s used in drainage equation, then evaporation will be negative which is not possible.
ii) Answer: evaporation of drainage basin is only 500 mm/yr which is very less compare to that of lake. Possible reason can be water exposed area which is very high in the case of lake. Water exposed to sun make evaporation rate high.
iii) Answer: for lake, reservoir volume = A* h = 180*106 * 18 m3
input flow per year= PL * A + Q = 647553600 m3
Residence time= reservoir volume/ input flow per year = 5 yrs (round off)
iv) Answer: for drainage basin,
PL =Q + ed
1240 mm/yr * A= Q + 500 mm/yr. *A
Q= 740 mm/yr *A= (740 *10-3 )* 940* 106 = 695,600,000 m3
Inflows to the lake 647,553,600 m3 ,
Precipitation volume over drainage basin = 1240 mm/yr * A = (1240 *10-3 )* 940* 106 = 11656,000,000 m3
Run-off in the drainage basin is higher than the total inflow of lake.
Water-quality implication: as the run-off is high, the residence time in the drainage basin of water will be less. Thus it will improve the quality of water and also involve in a recycling of natural nutrient.
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