Hydrograph MCQ Quiz in తెలుగు - Objective Question with Answer for Hydrograph - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

Let us assume the area of the catchment is A, then the area of triangle (from the shape of unit hydrograph) is equal to the volume of the water accumulated in the catchment area.

\(\frac{1}{2} \times {Q_{Peak}} \times Base\;width = Area\;\left( A \right) \times depth\;of\;water\)

\(\frac{1}{2} \times 20 \times 64 \times 3600 = A \times \frac{1}{{100}}\)

\(A = 230.4 \times {10^6}\;{m^2} = 230.4\;k{m^2}\)

Therefore, the equilibrium discharge of S-curve obtained by using this 8-hour unit hydrograph will be

\({Q_s} = 2.778\frac{A}{{{T_0}}}\)

where A is in km² and T₀ is in hours

\({Q_s} = 2.778 \times \frac{{230.4}}{8}\)

Explanation:

Unit hydrograph: 

A unit hydrograph is a direct runoff hydrograph resulting from one unit (one cm) of constant intensity uniform rainfall occurring over the entire watershed.

Base period of 12 hr unit hydrograph = Base period of 6 hr unit hydrograph + 6

Base period of 12 hr unit hydrograph = 84 + 6 = 90h

Explanation:

Basin lag or Basin lag time:

• It is defined as the measurement of the time between the center of mass of the precipitation (or rainfall excess) to the center of mass of the runoff (on the hydrograph) in the basin. 
• Basin lag is not only a function of basin characteristics but also of storm intensity and speed.
• It also defines as the interval from the center of mass of precipitation to the hydrograph crest.

Hence basin lag in hydrology is the time difference between the centroid of rainfall excess and the centroid of surface runoff.

Explanation:

Flow duration curve:

• Flow-duration curve is a cumulative frequency curve that shows the percent of time specified discharges were equaled or exceeded during a given period.
• It is a graphical method to present the discharge data of a river or a stream.
• it represents flow characteristics of a river for a particular region subjected to recorded flows in the river.

Mass duration curve:

• It is a plot between the stored volume of water in a stream and time in days.
• The curve represents the reservoir capacity.

Hydrograph:

• It is the graphical representation between discharge (rate of flow) and time in a particular river channel.
• it is used in various engineering applications such a estimation flood water and risk analysis of dam etc.

Explanation

Unit hydrograph (UH) is a hydrograph that resulted from 1 cm of rainfall excess depth

We know that

∴ \(rainfall\;excess = \frac{{Volume\;of\;direct\;runoff}}{{Catchment\;area}}\)

The area under triangle = Volume of direct runoff

\(\frac{1}{2} × {q_p} × Base\;period = Area\;of\;catchment × 1\;cm\;runoff\;\;\)

Where qp is peak discharge of unit hydrograph

Given,

Base period = 20 hours, Area of catchment = 500 ha = 50 × 10⁴ m²

\(\frac{1}{2} × {q_p} × Base\;period = Area\;of\;catchment × 1\;cm\;runoff\;\;\)

\(\frac{1}{2} × {q_p} × 20 = 500 \times 10^4m^2 × 1\;cm\;runoff\;\;\)

q_p = 5000 m³/hr

∴ Peak Discharge = 5000 m3/hr

Explanation

Given,

The base period of 4 hr UH is 48 hr 

for 8 hr Unit Hydrograph i.e. lag by 4 Hr 

Hence, the base period of 8 hr storm hydrograph = 48 + 4 = 52 hr

Concept:

Volume of rainfall = Area of hydrograph

Depth of rainfall = Volume of rainfall/Area of catchment

Calculation:

Duration of Storm = 6 hr

Base period = 100 hr = 100 × 3600 sec

Q_P = 40 m³/s

Area of catchment  = 180 km²

The volume of rainfall = Area of Hydrograph \( = \frac{1}{2} \times 100 \times 40 \times 3600 = 72 \times {10^5}{\rm{\;}}{{\rm{m}}^3}\)

Depth of rainfall \( = \frac{{72 \times {{10}^5}}}{{180 \times {{10}^6}}} = 4{\rm{\;cm}}\)

So, from the property of the Hydrograph, peak discharge for unit hydrograph can be calculated.

For runoff of 4 cm, Q_P = 40 m³/s

 As it is asked for 6Hr unit hydrograph, so corresponding to 1 cm runoff depth discharge is calculated

So, for a depth of 1 cm, Q_P = 40/4 = 10 m³/s

Hydrograph: It is a continuous plot of instantaneous discharge (runoff) v/s time.

It results from a combination of physiographic and meteorological conditions in a watershed and represents the integrated effects of climate, hydrologic losses, surface runoff, interflow, and groundwater flow.

Detailed analysis of hydrographs is usually important in flood damage mitigation, flood forecasting, or establishing design flows for structures that convey floodwaters.

Factors that influence the hydrograph shape and volume

i. Meteorological factors

ii. Physiographic or watershed factors and

iii. Human factors

Example of a Hydrograph

Explanation:

Evaporation:

• The process of transformation of liquid water into a gaseous form is called evaporation.

​
Dalton's law:

• It gives the relationship between the rate of evaporation and the difference between the saturation vapour pressure and existing vapour pressure, by the following expression

E_L = C(e_s - e_a)

Where, E_L = Rate of evaporation (mm/day), C = Constant, e_s = Saturation vapour pressure in mm of mercury, e_a = Existing vapour pressure in mm of mercury

Dependency of the rate of evaporation on various factors:

• Increases with an increase in vapour pressure at the water surface and air above it.
• Increases with increases in Air and water temperature.
• Increases when wind speed is above critical Wind speed.
• Decrease with the increase in Atmospheric pressure.

Additional Information

Concept:

Hydrograph:

• A hydrograph is a plot between discharge and time at any given section of a river, channel, etc.
• It is a response of a given catchment to the rainfall input.

Unit hydrograph (UH):

• A unit hydrograph can be defined as the hydrograph of direct runoff resulting from one unit depth (1 cm) of rainfall excess occurring uniformly over the basin for a specified duration.
• A unit hydrograph is based on the principle of linearity and the principle of superposition.
• If a D-hr unit hydrograph is available, a unit hydrograph with other duration such as 2D, 3D, etc. can be obtained easily from the principle of superposition. As the duration of the unit hydrograph increases, the base period also increases, and consequently, the peak ordinate decreases.

Hence, If the peak ordinate of a 4 h unit hydrograph of a basin is 270 m3/s, then the peak ordinate of an 8 h unit hydrograph of the same basin will be Less than 270 m3/s.