A lake, drawn schematically below, has a an inflow river with a volume discharge
ID: 285919 • Letter: A
Question
A lake, drawn schematically below, has a an inflow river with a volume discharge rate of
2 x 107 m3 day (about 1/10 the flow rate of the Rhine River, for context), the same flow rate in the river flowing out, and a precipitation rate matching evaporation rate so that steady state is maintained. The lake’s depth is, on average, 5 meters. The lake is roughly circular, about 20-km across, as shown below.
What is the residence time for water in this lake? In other words, how much time, on average, does a molecule of water spend traversing this lake after it enters in the upstream river before flowing out on the other end?
River flow in River flow out Well mixed lake 20 km View of lake fromm aboveExplanation / Answer
Residence time for water in this lake( Whose steady state is maintained) can be calulated by dividing the volume of the lake by the inflow or outflow rate (as both are equal).
Volume of the lake can be calculated by integrating the circle to its height, which is equal to = phi x radius2x depth
radius = 10 kilometer = 10,000 meters
Volume of the lake = 22/7 x (10,000)2 x 5
Volume of the lake = 1,571,428,571 m3
Residence time for water in this lake = 1,571,428,571 m3/2 x 107 m3 day
Residence time for water in this lake = 78.57 days
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