Take the greatest care with your significant digits. If you don’t, this question is not “do-able”.
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 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(goes down) ) - (0Rh + OEv) = (delta)SL
where Ii = Inputs, Oi = Outputs and S = change in storage
conversions:
1 ha = 104 m2 (a hectare {ha} is an area 100 x 100m);
1 m = 3.2808 feet,
1 km2 = 106 m2 (103 x 103) m2;
1 inch = 25.4 mm;
1 m = 1000 (103) mm,
1” (inch) = 2.54 cm,
1 (Canadian) gallon = 4.55 L,
1 (American) gallon = 3.785L,
1 L = 1000 cm3,
1 mile = 5,280 feet,
1 mile2 = 640 acres
To solve this problem, let's break down the given information and use the provided equation to evaluate the change in storage (∆S) for Llyn y Gweision.
Given information:
- Surface area of Llyn y Gweision (lake) = 10 km^2
- Catchment area of Afon Chwefru (river) = 1250 km^2
- Catchment area of Nant Aeron (river) = 2.400 * 10^6 ha
- Average annual flow rate of Afon Rhondda (outflow stream) = 1367.6 ft^3/s
- Rainfall in the region = 40.0 inches/year
- Annual evaporation from the lake surface = 8.10 * 10^6 m^3
- Equilibrium amount of water stored in the lake = 0.550 km^3
- 32.0% of rainfall over Chwefru basin leaves as streamflow
- Streamflow of Aeron basin = 33.87 mm/year
We'll use the given equation:
(ICh + IAe + IP) - (0Rh + OEv) = ∆SL
Step 1: Convert units
- Convert ha (hectares) to m^2: 1 ha = 10,000 m^2
- Convert ft to m: 1 ft = 0.3048 m
- Convert km^2 to m^2: 1 km^2 = 1,000,000 m^2
- Convert inch to mm: 1 inch = 25.4 mm
- Convert L to cm^3: 1 L = 1000 cm^3
- Convert acre to mile^2: 1 mile^2 = 640 acres
Step 2: Calculate the inputs (I) and outputs (O) for the equation
Inputs (I):
- ICh = Catchment area of Chwefru (in m^2) = 1250 km^2 * 1,000,000 m^2/km^2
- IAe = Catchment area of Aeron (in m^2) = 2.400 * 10^6 ha * 10,000 m^2/ha
- IP = Precipitation (in m^3) = Rainfall in the region (in mm) * Surface area of Llyn y Gweision (in m^2) * (1 inch/25.4 mm) * (1 m/1000 mm)
Outputs (O):
- 0Rh = outflow from Rhondda (in m^3) = Average annual flow rate of Afon Rhondda (c.f.s) * (1 m^3/35.3147 ft^3) * (1 s/year)
- OEv = evaporation from the lake (in m^3) = Annual evaporation from the lake surface (in m^3)
Step 3: Calculate ∆S using the equation
∆S = (ICh + IAe + IP) - (0Rh + OEv)
Step 4: Evaluate ∆S and determine if it is equal to 0. If ∆S = 0, it means that inputs equal outputs and there is no change in storage.
To find out how much the lake level will fall in one year if 10% of the flow from Chwefru is diverted for municipal water supply, we can modify the equation by subtracting the amount of diverted water from the Chwefru input (ICh). You can then use the modified equation to calculate ∆SL. Remember to keep all other variables the same.