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Llyn y Gweision (a Welsh lake) has a surface area of 10 km2. It is fed by two rivers: the Afon Chwefru and the Nant Aeron. The Chwefru has a catchment area of 1250 km2. The Aeron has a catchment area of 2.400 * 106 ha. The outflow stream, the Afon Rhondda has an average annual flow rate of 1367.6 ft3 s-1 (c.f.s). Rainfall in the region is 40.0 inches y-1, and annual evaporation from the lake surface is 8.10 * 106 m3.The equilibrium amount of water stored in the lake is 0.550 km3. Of the total amount of rainfall over the Chwefru basin, 32.0% leaves the basin as streamflow. For the Aeron basin, the annual streamflow amounts to an equivalent water depth of 33.87 mm over the whole basin. Show that for the above inputs and outputs that ƒ´S = 0. If 10% of the flow from the Chwefru is diverted for municipal water supply, how much will lake level fall in one year; all else staying the same?
A friendly (continuity) equation might be:
(ICh+ IAe + IP(arrow going down) ) - (0Rh + OEv) = (delta)SL
where Ii = Inputs, Oi = Outputs and (delta)S = change in storage.
To show that ƒ´S = 0, we need to calculate the values of the different terms in the equation and see if they balance out to zero.
Let's break down the terms in the equation:
- ICh: The input from the Chwefru basin. It is not explicitly given, but we can calculate it based on the catchment area and the rainfall over the basin. The rainfall over the Chwefru basin is 40.0 inches per year (convert to meters if necessary) and we know that 32.0% of this rainfall leaves the basin as streamflow. So, we can calculate ICh = (0.32 * Rainfall over Chwefru basin).
- IAe: The input from the Aeron basin. We're given that the annual streamflow from the Aeron basin amounts to an equivalent water depth of 33.87 mm over the whole basin. To calculate IAe, we need to convert this depth into a volume by multiplying it with the Aeron catchment area. So, IAe = (33.87 mm * Aeron catchment area * convert to meters if necessary).
- IP(arrow going down): The input from precipitation. We're not given a specific value for this, but it seems to represent additional input from precipitation that is not mentioned in the given information. We can assume that IP(arrow going down) = 0, as there is no other information provided.
- 0Rh: The output to the outflow stream, the Afon Rhondda. We're given the average annual flow rate of the outflow, which is 1367.6 ft3 s-1 (c.f.s). To calculate 0Rh, we need to convert this flow rate into a volume over a year and then convert it into the applicable units. So, 0Rh = (1367.6 ft3 s-1 * convert to cubic meters * convert to km3).
- OEv: The output due to evaporation from the lake surface. We're given that the annual evaporation from the lake surface is 8.10 * 106 m3. So, OEv = 8.10 * 106 m3.
- (delta)SL: The change in storage of the lake. We're given that the equilibrium amount of water stored in the lake is 0.550 km3. So, (delta)SL = 0.550 km3.
Now, let's substitute the calculated values into the equation and see if ƒ´S = 0:
(ICh + IAe + IP(arrow going down)) - (0Rh + OEv) = (delta)SL
(ICh + IAe + 0) - (0Rh + OEv) = 0.550 km3
Now, replace the calculated values and units into the equation and perform the calculations to see if both sides of the equation balance out to zero.
Considering the second part of the question, if 10% of the flow from the Chwefru is diverted for municipal water supply, we can calculate the volume of water diverted using similar steps as above. Then, we can subtract that volume from the equilibrium storage value to find out how much the lake level will fall in one year.