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Hydrology Test 2
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Hydrology Test 2
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  • Question 1/10
    1 / -0

    The shape of the recession limb of a hydrograph depends on
    Solutions

    Concept:

    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.

    Factors that influence the hydrograph shape and volume

    i. Meteorological factors

    ii. Physiographic or watershed factors and

    iii. Human factors

    Flood hydrograph has three characteristic regions:

    (i) Rising limb AB: Curve Joining point A, the starting point of the rising curve and point B, the point of inflection. It is concave in shape and the shape depends both upon the rainfall and the catchment parameters.

    (ii) Crest segment BC: Between the two points of inflection B and C with a peak P. For large catchments it generally occurs sometimes after the rainfall has ended.

    (iii) Falling limb/ Recession Limb: Curve CD starting from the second point of inflection C and the shape depends only on catchment parameters as the rainfall has ceased by then.

  • Question 2/10
    1 / -0

    For a catchment with an area of 600 km2 the maximum ordinate of an S-Curve obtained by a unit hydrograph of 4 hour duration is 
    Solutions

    Concept:

    S curve is a hydrograph produced due to continuous effective rainfall at a constant rate for an infinite period .

    The maximum ordinate of S-curve is the equilibrium discharge. It has an initial steep portion and reaches a maximum equilibrium discharge at a time equal to the time base of the first unit hydrograph. The average intensity of ER producing the S-curve is 1/D cm/h and the equilibrium discharge,

    \({Q_e} = 2.778\frac{A}{D}{m^3}/s\) 

    Where,

    A = Area of the catchment in km2

    D = Duration of rainfall in hour.

    Calculation:

    A = 600 km2, D = 4 hour

    \(\therefore {Q_e} = 2.778 \times \frac{{600}}{4} = 416.7\;{m^3}/s\) 

    Hence the maximum ordinate of the S curve is 416.7 m3/s.
  • Question 3/10
    1 / -0

    Match List I (catchment shape) with List II (respective hydrograph) and choose the correct code.

     

    List I

     

    List II

    P

     

    (i)

     

    Q

     

    (ii)

     

    R

     

    (iii)

     

    Solutions

    Concept:

    Direct Runoff Hydrograph: It is a plot of discharge vs time past a specific point due to the direct runoff coming from the catchment without including any contribution from Baseflow.

    Hydrograph from a catchment with a broader end near the outlet will have a higher and early peak with compared to a catchment with narrow end towards the outlet.

    Also, a catchment with two separate broader ends will have two peaks.

    Important Point:

    The fan shaped catchment gives high peak and narrow hydrograph as compared to the fern shaped catchment.

  • Question 4/10
    1 / -0

    Flood discharge for a 100 year return period is calculated using Gumbel’s method.

    The following data is available from the analysis:-

    1) Mean of annual data series = 30000 m3/s

    2) Standard deviation of annual data series = 300 m3/s

    3) Frequency factor = 0.2

    The estimated discharge for a flood of 100 year return period is (in m3/s up to 2 decimal places)

    Solutions

    Concept:

    → Estimation of flood of a particular Return period is done using :

    (a) Empirical Probability Distribution:

    (b) Theoretical Probability distribution:

    (i) Log – Pearson’s type III Method

    (ii) Gumbel’s Method: Flood discharge XT for a return period T years is statistically calculated as below,

    XT = X̅ + KT × σn-1

    Where,

    XT = Flood discharge for T year return period.

    X̅ = Mean of annual data series.

    σn-1 = Standard Deviation of Annual data series.

    KT = Frequency factor.

    Calculation:

    x̅ = 30000 m3/s, kT = 0.2, σ = 300 m3/s

    ∴ XT = X̅ + KT σ

    ⇒ X100 = 30000 + 0.2 × 300 = 30060 m3/s

  • Question 5/10
    1 / -0

    For a catchment area of 180 km2, a until hydrograph when plotted for one hour duration takes a triangular shape with its peal discharge of 50 m3/sec at 10 hours and time base of 20 hours. The phi index is 0.4 cm per hour and base flow is 10 m3/sec. If the there is rainfall of 3.2 cm is hour, the ordinate of flood hydrograph at 15th hour is ______
    Solutions

    Hydrograph for the given catchment:

    Direct until hydrograph ordinate at 15th hour = 15 m3/seconds

    Total rainfall intensity = 3.2 in hours

    Excess rainfall = 3.2 – 0.4 = 2.8 cm

    For 2.8 cm excess:

    Ordinate of direct until hydrograph = 2.8 × 25 = 70 m3/sec

    ∴ Ordinate of flood hydrograph at 15th hour = 70 + base flow = 70 + 10 = 80 m3/sec

  • Question 6/10
    1 / -0

    The effective rainfall hyetograph and a 5-hr unit hydrograph is given below in the table 1 and table 2 respectively

    Time (hr)

    0 - 5

    5 - 10

    Rainfall intensity (cm/hr)

    2

    1

     

    Time (hr)

    0

    5

    10

    15

    20

    25

    Discharge (m3/s)

    0

    25

    50

    75

    25

    0

     

    If the area of catchment is 56700 hectares then, the runoff depth of final direct runoff hydrograph will be

    Solutions

    Time (hr)

    0 – 5

    5 - 10

    Rainfall intensity (cm/hr)

    2

    1

    n

    2 × 5 = 10 cm

    1 × 5 = 5 cm

     

    Time (hr)

    Q(m3/s)

    n, u

    (n = 10 cm)

    n2u(n2 = 5)

    lagged by 5 hrs

    Direct runoff

    Hydrograph (m3/s)

    0

    0

    0

    -

    0

    5

    25

    250

    0

    250

    10

    50

    500

    125

    625

    14

    75

    750

    250

    1000

    20

    25

    250

    375

    625

    25

    0

    0

    125

    125

     

     

     

    0

    0

     

     

     

    Total =

    2625

     

    Runoff depth = \({5\; hr\;\times\;Sum \;of \;DRH}\over{Area\;of\;the\;catchment}\)

     

    \(={{5\times3600\times2625}\over{56700\times10^4}}\) = 0.0833 m = 8.33 cm

  • Question 7/10
    1 / -0

    The ordinates of a flood hydrograph of 2-hour storm is given as follows:

     Time (hour)

    0

    2

    4

    6

    8

    10

    12

     Ordinate (m3/s) 

    3

    7

    12

    15

    11

    5

    3

    If the storm generates 4 cm of excess rainfall and the base flow is 3 m3/s, then the peak ordinate of a 4-hour DRH with excess rainfall of 2 cm is _______.

    Solutions

    Concept:

    A hydrograph is a plot between flood discharge and the duration.

    If the hydrograph contains only direct runoff then it is called direct runoff hydrograph (DRH). But if it also contains the base flow then it is called flood hydrograph.

    Calculation:

    (i)

    (ii)

    (iii)

    (iv)

    (v)

    (vi)

    Time (hr)

    Ordinate of 2 hr flood hydrograph

    Ordinate of 2 hr DRH with 4 cm rainfall excess

    Ordinate of 2 hr DRH lag by 2 hr

    Ordinate of 4 hr DRH of 8 cm rainfall excess

    Ordinate of 4 hr DRH of 2 cm rainfall excess

     

     

    (ii) – Base flow

     

    (iii) + (iv)

    (v) × (2/8)

    0

    3

    0

     

    0

    0

    2

    7

    4

    0

    4

    1

    4

    12

    9

    4

    13

    3.25

    6

    15

    12

    9

    21

    5.25

    8

    11

    8

    12

    20

    5

    10

    5

    2

    8

    10

    2.5

    12

    3

    0

    2

    2

    0.5

    14

     

     

    0

    0

    0

    ∴ Peak ordinate of 4 hour DRH with excess rainfall of 2 cm is 5.25 m3/s.

  • Question 8/10
    1 / -0

    Analysis of data on maximum one-day rainfall depth at Madras indicated that a depth of 280 mm had a return period of 50 years. Determine the probability (in percentage, up to two decimal places) of a one-day rainfall depth equal to or greater than 280 mm at Madras occurring two times in 15 successive years.
    Solutions

    Concept:

    If the probability of an event occurring is P, the probability of the event not occurring in a given year is q = (1 - P).

    The binomial distribution can be used to find the probability of occurrence of the event r times in n successive years.

    \({{\rm{P}}_{\left( {{\rm{r}},{\rm{n}}} \right)}} = {}_{}^{\rm{n}}{{\rm{C}}_{\rm{r}}}{{\rm{P}}_{\rm{r}}}{{\rm{q}}^{{\rm{n}} - {\rm{r}}}} = \frac{{{\rm{n}}!}}{{\left( {{\rm{n}} - {\rm{r}}} \right)!{\rm{r}}!}}{{\rm{P}}^{\rm{r}}}{{\rm{q}}^{{\rm{n}} - {\rm{r}}}}\)

    Where

    P(r,n) = Probability of a random hydrologic event (rainfall) of given magnitude and exceedance probability P occurring r times in n successive years.

    E.g.

    a) The probability of an event of exceedance probability P occurring 2 times in n successive years is

    \({{\rm{P}}_{\left( {2,{\rm{n}}} \right)}} = \frac{{{\rm{n}}!}}{{\left( {{\rm{n}} - 2} \right)!2!}}{{\rm{P}}^2}{{\rm{q}}^{{\rm{n}} - 2}}\)

    b) The probability of the event not occurring at all in n successive years is

    P(0,n) = qn = (1 - P)n

    c) The probability of the event occurring at least once in n successive years

    P1 = 1 – qn = 1 – (1 - P)n

    Calculation:

    Given: P = 1/50 = 0.02

    For rainfall occurring two times in 15 successive years.

    n = 15, r = 2

    \({{\rm{P}}_{\left( {2,15} \right)}} = \frac{{15!}}{{13!2!}} \times {\left( {0.02} \right)^2} \times {\left( {0.98} \right)^{13}} = 15 \times \frac{{14}}{2} \times 0.0004 \times 0.769 = 0.0323 = 3.23\% \)

  • Question 9/10
    1 / -0

    The peak of a flood hydrograph due to a 3-h duration isolated storm in a catchment is 180 m3/s. The total depth of 3-h rainfall is 2.7 cm and an average loss due to infiltration is 0.3 cm/hr. The peak discharge of the 3-hour unit hydrograph for a base flow of 30 m3/s is (in m3/s up to 2 decimal places)
    Solutions

    Concept:

    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.The unit hydrograph is the unit pulse response function of a linear hydrologic system.

    The unit hydrograph is a simple linear model that can be used to derive the hydrograph of any duration resulting from any amount of excess rainfall for the given area.

    Peak of UH\( \frac{{Peak\;of\;DRH}}{{Rainfall\;Excess}}\)

    Calculation:

    Peak of flood hydrograph = 180 m3/s

    Base flow = 30 m3/s

    ∴ Peak of DRH = 180 – 30 = 150 m3/s

    Total rainfall = 2.7 cm

    Infiltration loss = 0.3 cm/hour

    ∴ Rainfall excess = 2.7 – 0.3 × 3 = 1.8 cm

    ∴ Peak of 3 – h unit hydrograph \(= \frac{{Peak\;of\;DRH}}{{Rainfall\;Excess}} = \frac{{150}}{{1.8}} = 83.33\;{m^3}/s\)

  • Question 10/10
    1 / -0

    An average rainfall of 16 cm occurs over a catchment during a period of 12 hours with uniform intensity. The unit hydrograph (unit depth 1 cm, unit duration = 6 hr) of the catchment rises linearly from 0 to 30 cumec in six hours the fall linearly to zero in 12 hr. ϕ index of the catchment is 0.5 cm/hr. Base flow in the river is 5 cumec. The peak discharge resulting from direct run-off hydrograph will be ______ cumec.
    Solutions

    Time

    6 hr UH

    S-curve

    Addition

    S-curve

    SA

    12 hr delayed

    S-curve

    (SA - SB)

    12 hr UH

    \(= \left( {\frac{{{S_A} - {S_B} \times 6}}{{12}}} \right)\)

    0

    0

    -

    0

     

    0

    0

    3

    15

    -

    15

     

    15

    7.5

    6

    30

    0

    30

     

    30

    15

    9

    22.5

    15

    37.5

     

    37.5

    18.75

    12

    15

    30

    45

    0

    45

    22.5

    15

    7.5

    37.5

    45

    15

    30

    15

    18

    0

    45

    45

    30

    15

    7.5

    21

     

    45

    45

    37.5

    7.5

    3.75

     

     

     

    45

    45

    0

    0

    Run-off produced by 12 hr storm,

    R = P – ϕt

    = 16 – 0.5 × 12

    = 10 cm

    ∴ Peak flow resulting from direct Run-off hydrograph = Run-off × Ordinate of UH = 10 × 22.5 = 225 cumec

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