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

so far, i have concluded that

input for chwefru would be 1250 km2 and for iput for aeron it is 2.400*10^6 ha. but do i change the units? and i don't understand wat to do for percipitation

Answer 2

hey..i have this same question problem and i also posted this same question here in this site..and as far as i understand u need to get each in m^3/year...!!

Answer 3

m^3/ year as your final units?!

Answer 4

no..i meant that input shud all b in m^3/year...den output also..and the other individual parts as well..so change in storage will also b in m^3/year..and change in storage must b zero..!!

i was wondering...by any chance do u have a question number 1a.b.c.d.e??..if so..can u tell me wat value did u get for each (in u have dis rainfall related question den)..!

Answer 5

no need for 1a b c d e..i think my one is already right!!:)

Answer 6

To show that ΔS = 0, we need to evaluate the inputs and outputs of the lake system and determine if they balance each other out.

Let's begin by calculating the inputs:

1. Input from the Afon Chwefru (I(Chwefru)):
The catchment area of the Afon Chwefru is given as 1250 km^2. If 32% of the rainfall over the Chwefru basin leaves as streamflow, we can calculate the input:
I(Chwefru) = 0.32 * 40.0 inches * 1250 km^2
= 51200 km^3

2. Input from the Nant Aeron (I(Nant Aeron)):
The catchment area of the Nant Aeron is given as 2.400 * 10^6 Ha. To convert to km^2, we divide by 100 (1 Ha = 10,000 m^2 = 0.01 km^2):
I(Nant Aeron) = 33.87 mm * 2.400 * 10^6 Ha * 0.01 km^2/Ha
= 81288 km^3

3. Input from precipitation in the lake (I(precipitation in lake)):
The rainfall in the region is given as 40.0 inches per year. To convert to cubic kilometers, we need to consider the lake's surface area of 10 km^2:
I(precipitation in lake) = 40.0 inches * 0.0254 m/inch * 10 km^2
= 40.64 km^3

Next, let's calculate the outputs:

1. Output through the Afon Rhondda (0(Rhondda)):
The average annual flow rate of the Rhondda is given as 1367.6 ft^3/s. To convert to cubic kilometers, we need to multiply by the conversion factor for ft^3 to m^3 and then divide by 10^9 to get km^3:
O(Rhondda) = 1367.6 ft^3/s * 0.028316847 m^3/ft^3 * 1 km^3/10^9 m^3
= 5.8803808 km^3

2. Output through evaporation (O(evaporation)):
The annual evaporation from the lake surface is given as 8.10 * 10^6 m^3. To convert to kilometers, we divide by 10^9:
O(evaporation) = 8.10 * 10^6 m^3 / 10^9 km^3
= 0.0081 km^3

Now, let's calculate the change in storage (ΔS):

ΔS = (I(Chwefru) + I(Nant Aeron) + I(precipitation in lake)) - (O(Rhondda) + O(evaporation))
= (51200 km^3 + 81288 km^3 + 40.64 km^3) - (5.8803808 km^3 + 0.0081 km^3)
= 136528.632 km^3 - 5.8884808 km^3
≈ 136522.744 km^3

Since the change in storage is not exactly zero, we can conclude that ΔS is not equal to 0. There is a positive change in storage.

Now, let's consider the situation where 10% of the flow from the Chwefru is diverted for municipal water supply. We need to calculate the amount of water diverted and subtract it from the inputs:

Diverted flow from Chwefru = 0.1 * I(Chwefru)

The new continuity equation becomes:
(I(Chwefru) - 0.1 * I(Chwefru) + I(Nant Aeron) + I(precipitation in lake)) - (O(Rhondda) + O(evaporation))
= S_new

Simplifying the equation:

(0.9 * I(Chwefru) + I(Nant Aeron) + I(precipitation in lake)) - (O(Rhondda) + O(evaporation))
= S_new

We can now calculate S_new using the given values and the calculated inputs and outputs.