Take the greatest care with your significant digits. If you don't this qusestion is not do able. Llyn Gweision ( A welsh lake) has a surface area of 10km2. It is fed by two rivers: the afron Chwefru and the nant aeron. The Chwefru has a catchment area of 1250km2. The aeron has a catchment area of 2.400* 10 to the power of 6 ha. The outflow stream, the afon Rhondda has average annual flow rate of 1367.6ft3s-1 (c.f.s) Rainfall in the region is 40.0 inches y-1 and annual evaporation from the lake surface is 8.10*10 to the power of 6 m3. The equilibrium amount of water stored in the lake is 0.550km3. Of the total amount of rainfall over the Chwefru basin, 32.0% leaves the basin as streamflow. From the aeron basin, the annual streamflow amounts to an equivalent water depth of 33.87mm over the whole basin.Show that for the above inputs and outputs that change in storage=0. If 10% of the flow from the Chwefru is diverted for municipal water supply, how much will lake level fall in one year; is all else staying the same?

To determine if the change in storage is zero, we need to calculate the input and output of water for the lake and verify if they balance out.

Let's calculate the input and output components:

1. Input Components:
- Rainfall: The region receives 40.0 inches of rainfall per year. To calculate the volume of rainfall, convert inches to meters: 40.0 inches * 0.0254 meters per inch = 1.016 meters. Now, multiply the volume by the surface area of the Chwefru catchment: 1.016 meters * 1250 km^2 = 1.27 km^3. This represents the rainfall over the Chwefru basin.
- Catchment Discharge: To calculate the catchment discharge, multiply the catchment area of the aeron river by the equivalent water depth: 2.400 * 10^6 ha * 33.87 mm = 8.13588 km^3. This represents the annual streamflow from the aeron basin.

2. Output Components:
- Evaporation: The annual evaporation from the lake surface is given as 8.10 * 10^6 m^3. Convert this to km^3: 8.10 * 10^6 m^3 * (1 km^3 / 10^9 m^3) = 0.0081 km^3.
- Outflow: The afon Rhondda has an average annual flow rate of 1367.6 ft^3s^-1. Convert this to m^3s^-1: 1367.6 ft^3s^-1 * 0.0283168466 m^3s^-1 / 1 ft^3s^-1 = 38.688 m^3s^-1. Now, calculate the volume of outflow in a year: 38.688 m^3s^-1 * 60 s/min * 60 min/hr * 24 hr/day * 365 days/yr * (1 km^3 / 10^9 m^3) = 1.220 km^3.

Now, let's add up the inputs and subtract the outputs to calculate the change in storage:

Input = Rainfall over Chwefru basin + Catchment discharge
Input = 1.27 km^3 + 8.13588 km^3
Input = 9.40588 km^3

Output = Evaporation + Outflow
Output = 0.0081 km^3 + 1.22 km^3
Output = 1.2281 km^3

Change in Storage = Input - Output
Change in Storage = 9.40588 km^3 - 1.2281 km^3
Change in Storage = 8.17778 km^3

The change in storage is not equal to zero, indicating that there is a mismatch between the input and output of water. It seems there is an issue with the provided values or calculations.

To answer the second part of the question, we need to calculate the amount of water diverted for municipal water supply from the Chwefru river. We know that 10% of the flow from the Chwefru river is being diverted.

Diverted Flow = 0.1 * Catchment Discharge (Chwefru)
Diverted Flow = 0.1 * 1.27 km^3
Diverted Flow = 0.127 km^3

To determine the amount the lake level will fall, we subtract the diverted flow from the existing amount of water stored in the lake:

Change in Lake Level = Diverted Flow - Change in Storage
Change in Lake Level = 0.127 km^3 - 8.17778 km^3
Change in Lake Level = -8.05078 km^3

The lake level will fall by approximately 8.05078 km^3 in one year, assuming all other factors remain unchanged.