Hydrology Test 1

Concept:

Runoff coefficient \(= \frac{{{\rm{Runoff}}}}{{{\rm{Rainfall}}}}\)

Given: Area = 100 hectares, Rain fall coefficient (C) = 0.4, Rainfall depth = 5 cm

Calculation:

Volume of Rainfall in m³ = Area × Rainfall Depth = (5 × 10^-2) × (10000 × 100)

⇒ Total Rainfall volume (m³) = 5 × 10⁴ m³

⇒ Total Runoff = 0.4 × (5 × 10⁴) = 20000 m³

Total duration = 10 hours

∴ Average streamflow \(= \frac{{20000}}{{{\rm{\;}}10}}\) = 2000 m³/hour

Important Point:

Streamflow, or discharge, is the volume of water flowing in a stream channel expressed as per unit time.

1 hectare = 10⁴ m²

Actual Evapo-transpiration is the quantity of water that is actually removed from a surface due to the processes of evaporation and transpiration.

Potential Evapo-transpiration is a measure of the ability of the atmosphere to remove water from the surface through the processes of evaporation and transpiration assuming no control on water supply.

a) If water supply to the plant is adequate, soil moisture will be at field capacity then, the ratio of AET to PET =1

b) If water supply to the plant is inadequate, then the ratio of AET to PET less than 1

c) In clayey soils, AET/PET almost equal to 1.

d) When the soil moisture approaches the permanent wilting point, AET tends to 0.

Hence the range of ratio of actual evapo-transpiration to potential evapo-transpiration is lies between 0 to 1.

Concept:

Average velocity for a stream is calculated as follows:

a) In shallow stream: stream of depth ≤ 3 m.

The velocity is measured at 0.6 times the depth of flow from water surface, the velocity so obtained is taken as average velocity.

V̅  = V_0.6d

b) In moderately deep streams:

\({\rm{\bar V}} = \frac{{{{\rm{V}}_{0.2{\rm{d}}}} + {{\rm{V}}_{0.8{\rm{d}}}}}}{2}\)

c) In river having flood flows and which are wide enough, the average velocity is measured within 0.5 times the depth below the surface.

V̅ = K × V_s

According to IS 4987 : 1968

1. In plains

→ 1 rainguage upto 500 km² shall be sufficient

2. In a region of an average elevation of 1 km from sea level

→ 1 raingauge in 250 km² – 400 km² shall be sufficient

3. In areas predominantly hilly and where heavy rainfall

→ 1 raingauge should be sufficient in not more than 150 km² it economically feasible.

Concept:

Optimum number of Rain gauges:

It is the number of Rain gauges required to calculate the average rainfall value with a certain acceptable percentage of error.It is given as,

\(N = {\left( {\frac{{{C_v}}}{e}} \right)^2}\) 

Where,

e = Allowable error

C_v = Coefficient of variance

Also,

\({C_V} = \frac{{Standard\;Deviation}}{{Mean\;Precipitation}}\)

Calculation:

P_m = 110 cm, σ = 25 cm, e = 0.1

\(\therefore {C_v} = \frac{\sigma }{{{P_m}}} = \frac{{25}}{{110}} = 0.2273\) 

\(\therefore N = {\left( {\frac{{{C_v}}}{e}} \right)^2} = {\left( {\frac{{0.2273}}{{0.1}}} \right)^2} = 5.16\) 

Adopting 6 rain gauges, additional number of rain gauges required = 6 – 4 = 2

Important Point:

Concept:

Return period: It is the average time interval after which flood of a given magnitude is expected to be equalled or exceeded.

\(T = \frac{1}{p}\)

Where, p = probability of occurrence or exceedance of an event in a year.

By using Binomial probability distribution, probability of occurrence of an event ‘r’ times in ‘n’ years is given as:

P = nCr (p)r (q)n-r

where, p = probability of occurrence in a year.

q = probability of non-occurrence in a year.

For n = design life; the probability of non-occurrence of an event in ‘n’ years is called the Reliability of the structure.

For n = design life; the probability of occurrence of an event once in ‘n’ years is called Risk associated with the structure.

Calculation:

Let the design life of the structure be ‘n’ years.

Risk ≤ 30%

i.e. the probability of occurrence of the event once in n year ≤ 30%

Return period = 40 years

\(∴ p = \frac{1}{{40}} \Rightarrow q = 1 - \frac{1}{{40}} = \frac{{39}}{{40}}\)

Where, p = Probability of occurrence

q = Probability of Non-occurrence

Also,

Probability of occurrence of an event, once in ‘n’ year \(= 1 - {q^n} = 1 - {\left( {\frac{{39}}{{40}}} \right)^n}\) 

∵ Risk ≤ 30%

∴Under limiting conditions;

\(∴ \left( {1 - {{\left( {\frac{{39}}{{40}}} \right)}^n}} \right) = 0.3\)

⇒  n ≈ 14.08 years

Adopting n = 14 years, the maximum design life of the structure is 14 years.

Concept:

There are two types of Infiltration index

1) ϕ - Index

Average rate of infiltration during the period of storm producing runoff

\(\phi - {\rm{Index}} = \frac{{{{\rm{P}}_{\rm{e}}} - {\rm{R}}}}{{{{\rm{t}}_{\rm{e}}}}}\)

Where,

P_e = Rainfall during storm producing runoff

t_e = Duration of storm producing runoff

R = Runoff

2) W - Index

Average rate of infiltration during entire storm duration

\({\rm{W}} - {\rm{Index}} = \frac{{{{\rm{P}}} - {\rm{R-Losses}}}}{{{{\rm{t}}}}}\)

Where,

P = Total rainfall during storm

R = Runoff

t = Duration of total storm

Calculation:

Given,

Runoff volume = 27.45 million m³ = 27.45 × 10⁹ mm³

Basin area = 1830 km² = 1830 × 10⁶ m²

Rainfall intensities at an half and hour interval: 5, 9, 20, 13, 6, 8, 16 and 3 mm/hr 

\({\rm{Runoff\;in\;mm}} = \frac{{27.45 × {{10}^6} × {{10}^3}}}{{1830 × {{10}^6}}} = 15{\rm{\;mm}}\)

P = (5 + 9 + 20 + 13 + 6 + 8 + 16 + 3)/2 = 80/2 = 40 mm

\({\rm{W}} - {\rm{Index}} = \frac{{{\rm{40}} - {\rm{15}}}}{{\rm{4}}} = 6.25\;{\rm{mm/hr}}\)

Elimintating rainfall intensities less than 6.25 mm/hr, i.e 5, 6, 3 mm/hr

∴ P = (9 + 20 + 13 + 8 + 16)/2 = 66/2 = 33 mm

As three intensities of half and hour interval is eliminated

∴ t = 4 - (0.5 × 3) = 2.5 hr

\(\phi - {\rm{Index}} = \frac{{{\rm{33}} - {\rm{15}}}}{{{\rm{2}}.{{\rm{5}}_{}}}} = 7.2\;{\rm{mm/hr}}\)

Concept:

Current Meter:

It is a mechanical device used to measure the velocity of water current. It consists of a rotating element (propeller) which rotates with all angular velocity proportional to stream velocity due to the reaction of the stream. The estimation of ‘a’ and ‘b’ for a current meter is called current meter calibration

The relation between velocity and the number of revolutions completed by the current meter is

V = a × N_s + b

Where,

V = velocity in m/s, and N_s = Number of revolutions done by current meter in 1 second

a & b = current meter constant.

Calculation:

Current meter equation: V = aN_s + b

For point A:

V = 5 m/s, N_s = 50 rev/s

⇒ 5 = 50a + b       ---(1)

For point B:

V = 3 m/s, N_s = 40 rev/s

⇒ 3 = 40a + b       ---(2)

Solving (1) and (2) simultaneously, \(a = \frac{1}{5}\;\;and\;\;b = \; - 5\) 

Thus the general velocity equation for the given current meter is

\(V = \frac{1}{5}{N_s} - 5\)

∴ Velocity at point where N = 60 rev/sec

\(V = a \times 60 + b \Rightarrow V = \frac{1}{5} \times 60 + \left( { - 5} \right)\)  ⇒ V = 7 m/s

Concept:

Inflow volume - Outflow volume = Change in water storage of the lake

Inflow is due to surface runoff and rainfall:

Inflow height \(= \frac{{5 \times 30 \times 24 \times 60 \times 60}}{{5000 \times {{10}^4}}} \times {10^3} + 150\)

∴ Inflow (mm) = 259.2 + 150 = 409.2 mm

Outflow is due to supply of water by municipal body and evaporation:

Outflow \(= \frac{{6 \times 30 \times 24 \times 60 \times 60}}{{5000 \times {{10}^4}}}\times 10^3 + 50\)

Outflow (mm) = 311.04 + 50 = 361.04 mm

∴ Water fluctuation = 409.2 - 361.04