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(Ich) has a catchment area of 1250 km2. The Aeron(Iae) has a catchment area of 2.400 * 106 ha. The outflow stream, the Afon Rhondda (Orh) 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?

Question

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(Ich) has a catchment area of 1250 km2. The Aeron(Iae) has a catchment area of 2.400 * 106 ha. The outflow stream, the Afon Rhondda (Orh) 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:

(I(Chwefru)+ I(Nant Aeron) + I(precipitation in lake) ) - (0(Rhonnda) + O(evaporation).
) = SL

where Ii = Inputs, Oi = Outputs and S = change in storage.

Answer 1

i have this same assignment..and i am stuck as well...as far as i understand..u need to get each individual parts in m^3/year.

Answer 2

which process proceds most of the oxygen in earths atmosphere phototsynthesis or transipation

what gas does photosyntheses produce carbon dioxide or oxygen

grasss-grasshopper--snake--hawk
if apart of the food web qouls mORE ARROWS POINT TOWARD OR AWAY FROM 2ND LEVEL

Answer 3

Easha has the right idea. You need to get all the inputs and outputs into the same units.

There's water entering the lake, and there's water leaving the lake. The numbers are just given in different units, which is common in the real world!

There's nothing difficult or hard to understand about it; it's just long and tedious, and requires you to work carefully. Calculate one input at a time, converting each into some standard unit of your choice. m^3/yr will do fine.

It's a nice question in itself, neat and well-constructed, but especially clever because it contains several unusual words, which makes it easy for the teacher to Google where the question has been posted and what answers have been given online. Hi, Teacher! I admire your work. :-)

Answer 4

To show that ΔS (change in storage) is equal to 0, we need to calculate the inputs and outputs of the lake system and check their balance. Let's calculate the inputs and outputs step by step:

1. Input calculations:
- Input from the Afon Chwefru (I(Chwefru)) is the catchment area multiplied by the portion leaving the basin as streamflow. Given that the catchment area is 1250 km2 and 32.0% of the rainfall leaves the basin as streamflow, we can calculate: I(Chwefru) = 1250 km2 * 0.32 = 400 km3.

- Input from the Nant Aeron (I(Aeron)) is given as the annual streamflow amounting to an equivalent water depth. Given that the catchment area of Nant Aeron is 2.400 * 10^6 ha and the streamflow is equivalent to a water depth of 33.87 mm, we need to convert the catchment area to square meters and calculate: I(Aeron) = (2.400 * 10^6 ha * 10^4 m2/ha) * (33.87 mm * 10^-3 m/mm) = 813 m3.

- Input from precipitation in the lake (I(precipitation in lake)) is given as the annual rainfall in the region. Given that the rainfall is 40.0 inches per year, we need to convert it to cubic meters: I(precipitation in lake) = (40.0 inches * 25.4 mm/inch * 10^-3 m/mm) * 10^6 m2 = 1.016 * 10^6 m3.

Now, let's sum up the inputs: I(total) = I(Chwefru) + I(Aeron) + I(precipitation in lake) = 400 km3 + 813 m3 + 1.016 * 10^6 m3.

2. Output calculations:
- Output from the Afon Rhondda (O(Rhonnda)) is given as the average annual flow rate. Given that the flow rate is 1367.6 ft3 s-1, we need to convert it to cubic meters per year: O(Rhonnda) = 1367.6 ft3/s * (0.02832 m3/ft3) * 60 s/min * 60 min/h * 24 h/day * 365 days/year.

- Output from evaporation (O(evaporation)) is given as the annual evaporation from the lake surface. Given that the evaporation is 8.10 * 10^6 m3, we can directly use this value.

Now, let's calculate the outputs: O(total) = O(Rhonnda) + O(evaporation).

3. Change in storage calculation:
Using the continuity equation, ΔS = I(total) - O(total).

Now we have all the necessary values to calculate the change in storage (ΔS) and determine if it equals 0. Substitute the inputs and outputs we calculated into the equation to find ΔS.