# Engineering Hydrology Questions and Answers – Flood Routing Equations – Set 2

This set of Engineering Hydrology Multiple Choice Questions & Answers (MCQs) focuses on “Flood Routing Equations – Set 2”.

1. The movement of a flood hydrograph through a reservoir is a type of which of the following?

Explanation: The flow of a flood hydrograph through a channel or reservoir is a case of unsteady flow. In terms of open channel flow, it is categorized as gradually varied unsteady flow.

2. Which of the following represents the basic continuity equation used in hydrologic routing? I = rate of inflow, Q = rate of outflow, S = storage, t = time, y = depth of flow
a) I+Q=$$\frac{dS}{dt}$$
b) I-Q=$$\frac{dS}{dt}$$
c) I+Q=$$\frac{dS}{dy}$$
d) I-Q=$$\frac{dS}{dy}$$

Explanation: The basic continuity equation used in hydrologic routing states that the rate of change of storage with time is equal to the difference between the rate of inflow and the rate of outflow from the reservoir or channel.

3. What is the condition for the selected time interval in the modified continuity equation for hydrologic routing in a reach?
a) It must be shorter than the travel time of flood wave through the reach
b) It must be equal to the travel time of flood wave through the reach
c) It must be longer than the travel time of flood wave through the reach
d) No condition to be satisfied

Explanation: The time interval selected should be shorter than the time of travel of the flood wave through the reach. This is to validate the assumption that the inflow and outflow hydrographs are straight lines in that time interval.

4. Which of the following represents the basic hydrologic routing equation for reach for a given time interval t? I1, Q1 and S1 are the inflow rate, outflow rate and storage at the start of interval t respectively and I2, Q2 and S2 are the inflow rate, outflow rate and storage at the end of interval t respectively.
a) $$(\frac{I_1+Q_1}{2})t-(\frac{I_2+Q_2}{2})t=S_1-S_2$$
b) $$(\frac{I_1+Q_1}{2})t-(\frac{I_2+Q_2}{2})t=S_2-S_1$$
c) $$(\frac{I_1-Q_1}{2})t+(\frac{I_2-Q_2}{2})t=S_1-S_2$$
d) $$(\frac{I_1-Q_1}{2})t+(\frac{I_2-Q_2}{2})t=S_2-S_1$$

Explanation: The basic continuity equation concerning hydrologic routing can be modified for a time interval t as,
I-Q=$$\frac{dS}{dt} ⇒(\frac{I_1+I_2}{2})-(\frac{Q_1+Q_2}{2})=\frac{S_2-S_1}{t}$$
⇒$$(\frac{I_1+I_2}{2})t-(\frac{Q_1+Q_2}{2})t=S_2-S_1$$
⇒$$\frac{I_1}{2} t+\frac{I_2}{2} t-\frac{Q_1}{2} t-\frac{Q_2}{2} t=S_2-S_1$$
∴$$(\frac{I_1-Q_1}{2})t+(\frac{I_2-Q_2}{2})t=S_2-S_1$$.

5. The St. Venant continuity equation for unsteady flow in case of no lateral flow is given by which of the following? Q is outflow rate, T is channel top width, y is depth of flow, x is distance and t is time.
a) $$\frac{∂Q}{∂t}+T \frac{∂y}{∂x}=0$$
b) $$\frac{∂Q}{∂t}+T \frac{∂y}{∂x}+C=0, C≠0$$
c) $$\frac{∂Q}{∂x}+T \frac{∂y}{∂t}=0$$
d) $$\frac{∂Q}{∂x}+T \frac{∂y}{∂t}+C=0, C≠0$$

Explanation: The St. Venant equations in hydraulic routing consists of two equations with partial derivatives. The continuity equation for unsteady open channel flow in differential form is given as,
$$\frac{∂Q}{∂x}+T \frac{∂y}{∂t}=0$$, given that there is no lateral inflow.

6. Which of the following represents the equation of motion of unsteady flow used in hydraulic routing? V is the flow velocity, y is the flow depth, x is the distance, t is time, g is acceleration due to gravity and S is the difference between bed slope and energy line slope.
a) $$\frac{1}{g} (V \frac{∂V}{∂x}+\frac{∂V}{∂t})+\frac{∂y}{∂x}=S$$
b) $$\frac{1}{g} (V \frac{∂V}{∂x}+\frac{∂V}{∂x})+\frac{∂y}{∂x}=S$$
c) $$\frac{1}{g} (V \frac{∂V}{∂t}+\frac{∂V}{∂t})+\frac{∂y}{∂t}=S$$
d) $$\frac{1}{g} (V \frac{∂V}{∂x}+\frac{∂V}{∂x})+\frac{∂y}{∂t}=S$$

Explanation: The second equation in the St. Venant equations of hydraulic routing is the equation of motion of unsteady flow of a flood wave which is derived from the momentum equation and is given as,
$$\frac{∂y}{∂x}+\frac{V}{g} \frac{∂V}{∂x}+\frac{1}{g} \frac{∂V}{∂t}=S_0-S_f$$
Where, S0 is the channel bed slope and Sf is the energy line slope.

7. The average inflow and outflow rates from a reach are 34 m3/s and 27 m3s. Find the change in storage in the reach in a time interval of 11 seconds?
a) Loss of 77 m3
b) Gain of 77 m3
c) Loss of 143 m3
d) Gain of 143 m3

Explanation: Using basic continuity equation,
Inflow rate-Outflow rate=(Change in storage)/time
⇒34-27=(Change in storage)/11 ⇒ Change in storage=(34-27)*11=77 m3
Therefore, there is a gain of 77 m3 of volume in the reach in the given time interval.

8. If the outflow rate in a channel section is twice the inflow rate of 2600 m3/min, how many litres of water is lost from the section per second?
a) 2600
b) 5200
c) 43330
d) 156000

Explanation: Using hydrologic continuity equation,
I-Q=$$\frac{S}{t}⇒2600-(2*2600)=\frac{S}{\frac{1}{60} mins}$$
⇒S=$$\frac{2600-5200}{60}$$=-43.33 m3≅-43330 L
Therefore, 43330 litres of water is lost from the section per second.

Sanfoundry Global Education & Learning Series – Engineering Hydrology.