7. Level Pool Routing

Question 1

What is flood routing?

Answer 1

Calculation of a flood hydrograph at the outlet, given the inflow hydrograph at the inlet,
Used to predict changes in shape of water as it moves through a reservoir.

Question 2

What happens to the flow of water as it moves through a storage (reservoir or river)?

Answer 2

The inflow is translated and attenuated due to friction and gravity.
Translation: delay in time that the peak occurs
Attenuation: reduction in magnitude of flood peak

Question 3

What are the key features of level pool routing?

Answer 3

• the peak of the outflow hydrograph intersects the inflow hydrograph
• there is a unique storage function i.e. S = ƒ(O) as a result of the combination of O = ƒ(h) and S = ƒ(h)

Question 4

Where is the point of maximum storage on a level pool routing hydrograph?

Answer 4

The intersection of the inflow and outflow curves

Question 5

What is the lumped system routing (mid-point) method of level pool routing?

Answer 5

dS/dt = I(t) - O(t)
∆S = I.∆t - O∆t
S(t+∆t) = St + (∆t/2).(I(t)+I(t+∆t) - (∆t/2).(O(t) + O(t+∆t))
Write O in terms of S
Solve for S(t+∆t) using initial conditions

Question 6

What is the convenient version of the lumped system routing (mid-point) method, generally used for tabulated data? What are the steps?
Hint: involves G and Im

Answer 6

G = S/∆t + O/2
Im = [(I(t) + I(t+∆t)]/2
G(t+∆t) = Im + Gt - Ot
1. calculate Im
2. calculate Gt from old time step
3. from the tabulated data of S=ƒ(O), find a pair of values of S(t+∆t) and O(t+∆t) that satisfy the above equation

Question 7

What is the linear reservoir method of lumped system routing?

Answer 7

O = ßS
dS/dt = I(t) - O(t) becomes:
dO/dt = ß.I(t) - ß.O(t)
Can then be solved using mid-point method, solving for O(t+∆t)

Question 8

What is the trapezoid method for natural reservoirs?

Answer 8

S = Dh ( Ao + 2A1 + 2A2 + … + 2An-1 + An ) / 2
S: storage when the water level lies between level (n-1) and n [m3]
Dh: height between measurements [m]
An: area at level ‘n’ [m2]