Depth-Area-Duration relationships indicate the areal distribution characteristic of a storm of a given duration. DAD curves involve considerable computational effort and requires meteorological and topographical information of the region.

However, the basic consideration for plotting the relationship is as follows:

For a rainfall of given duration, the average depth decreases with the area in an exponential fashion as given by:

P̅ = P_o × e^{(-k × An)}

Where,

P̅ = average depth in cms over an area A km²,

P_o = highest amount of rainfall in cm at the storm center

K and n are the constants for a given region.

The development of the maximum depth-area-duration relationship is known as DAD analysis.

Concept:

Canal lining and its advantages: It is process of covering or lining the earthen surfaces of a canal with stable, non-erodible lining surface such as concrete, tiles, asphalt etc. Some of the key advantages of lining a canal are the following.

Seepage Control: As the earthen surfaces are lined, so the seepage loss in the lined canal reduces considerably and stops the uncontrolled seepage which takes place in an unlined canal. This in turn saves huge amount of water which can be used for irrigation purposes. ∴ Statement 1 is correct.

Prevention of Water Logging: In case of unlined canals, due to uncontrolled seepage water table often can rise to ground level leading to water logging. By lining canals water logging and subsequent increment in salinity can be prevented. ∴ Statement 2 is correct.

Increased canal capacity: As the surfaces are lined and smooth, so the frictional resistance is considerably less in case of lined canals. Less frictional resistance results in increased flow of water per unit time which leads to more canal capacity. ∴ Statement 3 is correct.

Reduction in maintenance cost: Lining of canal may incur huge capital cost but the maintenance cost reduces substantially. The maintenance cost due to periodic silt removal, periodic removal of weeds and water plants and maintenance cost for minor repairs are substantially reduced by lining canals.  ∴ Statement 4 is incorrect.

Canal drop: It is a structure constructed across a channel to permit lowering down of water level in order to dissipate the surplus energy possessed by the falling water which may otherwise scour the bed and banks of the channel.

Canal Escape: During floods/rains, when canal carries water beyond its full capacity, water is released through Canal Escapes.

Canal Cross regulators: It is provided to admit water into the off taking canal and to regulate the supplies into the canal. It also reulated the discharge to be passed into the canal and controls the silt entry into the canal.

At the same stage, more discharge passes through the river during rising stages than in falling ones.

Note: Stage discharge relation is also known as rating curve.

Flood routing: It is a technique of determining the flood hydrograph at a section of a river by utilizing the data of flood flow at one or more upstream sections.

The two types of routing are as follows:

a) Lumped/Hydrologic flow routing

b) Distributed/Hydraulic routing

Attenuation: It is the technique of reducing the peak value of the outflow. Decreased peak flows contribute less volume to downstream flood events, and peak flows that are delayed high in the watershed may not coincide with peak flows downstream and therefore contribute less to downstream flooding.

The key parameters that define flood attenuation are:

1) decreased peak flows and 2) delayed peak runoff.

Concept:

For calculating the return period from Weibull’s formula, the data is required to be ranked with the highest ordinate having least (first) rank.

For Weibull Formula:

$\mathrm{T}=\frac{\mathrm{N}+1}{\mathrm{m}}$

Where T is the return period, N = Total number of observations and m = rank of the value.

Calculation:

Arranging the data in descending order

Return period for 6.0 cm annual rainfall by using Weibull formula is given by:

$\mathrm{T}=\frac{\mathrm{N}+1}{\mathrm{m}}=\frac{10+1}{7}=\frac{11}{7}$

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.

Steps to derive hydrograph of duration (D) from unit hydrograph is as follows:

1. Analyze the hydrograph and separate base flow.

2. Measure the total volume of DRO under the hydrograph and convert them to inches or mm over the watershed.

3. Convert total rainfall to rainfall excess through infiltration methods, such that rainfall excess = DRO, and evaluate duration D of the rainfall excess that produced the DRO hydrograph.

4. Divide the ordinates of the DRO hydrograph by the volume in inches or mm and plot these results as the UH for the basin. Timebase is assumed constant for storms of equal duration and thus it will not change.

5. Check the volume of the UH to make sure it is 1.0 inch or 1.0 mm, and graphically adjust ordinates as required.

Peak of DRH = Peak of UH × Rainfall excess

Explanation:

The DRH and the UH are two hydrographs of the same rainfall duration.

Thus the ordinates of DRH = Ordinates of UH × Rainfall excess

Hence the ordinates of curve B = ordinates of curve A × 2.5

Concept:

Silt Control Devices: The entry of silt into a canal, which takes off from a diversion head works, can be reduced by constructing certain special works which are called silt control works or silt control devices. These works may be classified into the following two types:

Silt Excluders:

• Silt excluders are those works which are constructed on the bed of the river at the upstream of the head regulator.
• The clearer water enters the head regulator and consequently the silted water enters the silt excluder. In this type of works, the silt is, therefore, removed from the water before it enters the canal.

Silt Ejectors:

• Silt ejectors, also called silt extractors, are those devices which extract the silt from the canal water after the silted water has travelled a certain distance in the off-take canal.
• These works are, therefore, constructed on the bed of the canal, and a little distance downstream from the head regulator.

∴ The correct combination for Silt Excluders is 1-B and 1-C and the correct combination for Silt Ejectors is 2-A and 2-D.

Runoff depth $=\frac{36hec}{120hec}=0.03m$

Runoff depth (R) = 30 mm

Rainfall depth (P) = $\frac{36}{0.45}=80mm$

 ϕindex = $\frac{P-R}{\mathrm{\Delta }t}$

 ϕindex = $\frac{80-30}{12}$

 ϕindex = $4.167mm/hr$

Concept:

Duty: It is the number of hectares of land irrigated for full growth of a given crop by a supply of 1 cumec of water continuously during the entire base period of that crop.

Delta: The total water depth required by a crop to attain its full maturity in its base period.

Base Period: The time period that elapsed for the instant of the showing of crop to the instant of its harvesting is called base period or crop period.

The relation between Duty (D), Delta (Δ) and Base period (B) is given as:

$\mathbf{\Delta }\left(\mathbf{m}\mathbf{e}\mathbf{t}\mathbf{r}\mathbf{e}\mathbf{s}\right)=\frac{8.64B}{D}$

Calculation:

Area to be irrigated = Irrigation intensity × area = 0.5 × 500 = 250 hectares

Duty $\left(D\right)=\frac{8.64\times B}{\mathrm{\Delta }}=\frac{8.64\times 15}{0.6}=216$ hectares/cumec

Discharge, $Q=C_{d}A\sqrt{2gH}$ 

Where, C_d = Coefficient of discharge,

A = Area of cross-section,

H = Head,

g = acceleration due to gravity

Q $=0.58\times \left(4\times 1\right)\times \sqrt{2\times 9.81\times 8}$

= 29.066 m³/s

Total discharge = No. of spillways × Q = 4 × Q = 4 × 29.066 = 116.26 m³/s

Concept:

In order to determine the peak flood value (X_T) for a particular return period, the reduced variate (Y_T) is initially calculated for that particular return period

$Y_T=\ln \left\{\ln \left(\frac{T}{T-1}\right)\right\}$

Calculations:

Returns Period of the flood is given

(T) = 100 years

Therefore, reduced variate (y_T)

$Y_T=-\ln \left\{\ln \left(\frac{100}{100-1}\right)\right\}$

$Y_T=-\ln \left\{\ln \left(\frac{100}{99}\right)\right\}$

Y_T = 4.6

Concept:

The monthly Consumptive use in a field can be obtained by Blaney-Criddle formulae i.e.

${\text{C}}_{\text{u}}=\frac{\text{kp}}{40}\times (1.8\text{T}+32)$

Here,

C_u = consumptive use (in cm), k = consumptive use coefficient, p = monthly % sunshine hour, and T = mean monthly temperature (in °C)

Consumptive irrigation requirement (CIR): The consumptive water which needs to be supplied by irrigation is called consumptive irrigation requirement.

CIR = Consumptive use – Effective rainfall

Net Irrigation requirement (NIR): It also contains the leaching requirement with consumptive irrigation requirement.

NIR = CIR + leaching

Solution:

Given: Consumptive use coefficient (k) = 0.75, Effective rainfall = 5.3 cm, Leaching requirement = 3 cm

Total consumptive use (C_u) = Consumptive use in September + consumptive use in October

$\Rightarrow {\text{C}}_{\text{u}}=\frac{0.75\times 6.8}{40}(1.8\times 22+32)+\frac{0.75\times 7.4}{40}(1.8\times 18+32)=18.06\text{ cm}$

CIR = Consumptive use – Effective rainfall

⇒ CIR = 18.06 – 5.3 = 12.76 cm

NIR = CIR + leaching

Concept:

Average rainfall of a catchment is calculated by following three methods:

i) Arithmetical Mean Method

${\mathrm{P}}_{\mathrm{a}\mathrm{v}\mathrm{g}}=\frac{{\mathrm{P}}_1+{\mathrm{P}}_2+\dots {\mathrm{P}}_{\mathrm{n}}}{\mathrm{n}}$

ii) Theisen Polygon Method

${\mathrm{P}}_{\mathrm{a}\mathrm{v}\mathrm{g}}=\frac{{\mathrm{P}}_1{\mathrm{A}}_1+{\mathrm{P}}_2{\mathrm{A}}_2+{\mathrm{P}}_3{\mathrm{A}}_3+\dots {\mathrm{P}}_{\mathrm{n}}{\mathrm{A}}_{\mathrm{n}}}{{\mathrm{A}}_1+{\mathrm{A}}_2+{\mathrm{A}}_3\dots {\mathrm{A}}_{\mathrm{n}}}$

iii) Isohyetal Method

${\mathrm{P}}_{\mathrm{a}\mathrm{v}\mathrm{g}}=\frac{\left(\frac{{\mathrm{P}}_1+{\mathrm{P}}_2}{2}{\mathrm{A}}_1\right)+\left(\frac{{\mathrm{P}}_2+{\mathrm{P}}_2}{2}\right){\mathrm{A}}_2+\left(\frac{{\mathrm{P}}_3+{\mathrm{P}}_4}{2}\right){\mathrm{A}}_3\dots \left(\frac{{\mathrm{P}}_{\mathrm{n}-1}+{\mathrm{P}}_{\mathrm{n}}}{2}\right){\mathrm{A}}_{\mathrm{n}}}{{\mathrm{A}}_1+{\mathrm{A}}_2+{\mathrm{A}}_3+\dots {\mathrm{A}}_{\mathrm{n}}}$

Calculation:

${\mathrm{P}}_{\mathrm{a}\mathrm{v}\mathrm{g}}=\frac{10\times 0.22+12\times 1.06+15\times 2.34+20\times 4.03+25\times 2.03}{0.22+1.06+2.34+4.03+2.03}$

Concept:

The limiting height for an elementary gravity dam profile is given by:

a) Considering uplift condition:

$h_1=\frac{f}{{\gamma }_w\left(G-C+1\right)}$

Where,

f = maximum allowable stress of dam material

γ_w  = density of rotating fluid

G = specific gravity of dam material

C = constant

The value of C is recommended as 1 by U.S.B.R for considering uplifting conditions.

b) Without considering uplift condition:

$h_0=\frac{f}{\left(G+1\right){\gamma }_w}$

Calculation:

$G=\frac{24380N/m^3}{9810N/m^3}=2.485$

$\therefore \frac{h_1}{h_0}=\frac{\frac{f}{{\gamma }_w\left(G-C+1\right)}}{\frac{f}{{\gamma }_w\left(G+1\right)}}=\frac{G+1}{G-C+1}$

$\therefore \frac{h_1}{h_0}=\frac{2.485+1}{2.485-1+1}=\frac{3.485}{2.485}$

$\therefore \frac{h_1}{h_0}=1.40$

Piezometric Head = Datum head + Velocity head

∴ Head at the floor bottom = 15 – 3 = 12 m

For the critical failure condition:

Upward pressure due to water head (σ1) should be equal to the pressure exerted by the combined weight of the concrete floor and standing depth of water (σ2)

σ1 = γw × hw = 9.81 × 12 = 117.72 kN/m2

σ2 = γc × tc + γw × hw2 = 2.5 × 9.81 × tc + 9.81 × 2

∴ 117.72 = 24.525 × tc + 19.62

∴ tc = 4 metres

When the given 3-hr unit hydrograph is added to the same 3-hr unit hydrograph placed at a lag of 3 hr, then we obtain the ordinates of a 6-hr hydrograph containing 2 cm runoff. Hence, when this hydrograph ordinate is divided by 2, we can obtain the hydrograph of 6 hr duration containing 1 cm runoff, which will be nothing but 6 hr unit hydrograph.

Maximum ordinate of 6 hr UH = 28.5 m³/s

Delay in the peak = 8 hr – 5 hr = 3 hr.

Concept:

Estimation of missing rainfall data:

i) Arithmetic Mean Method:

It is used if the normal annual precipitation of the nearby stations is 10% of within the normal annual precipitation at station X.

${\mathrm{P}}_{\mathrm{x}}=\frac{{\mathrm{P}}_1+{\mathrm{P}}_2+{\mathrm{P}}_2+\dots {\mathrm{P}}_{\mathrm{n}}}{\mathrm{n}}$

ii) Normal Ratio Method:

It is used if normal precipitation at any of the selected stations differ by more than 10% of selected station.

${\mathrm{P}}_{\mathrm{x}}=\frac{{\mathrm{N}}_{\mathrm{x}}}{\mathrm{N}}\left[\frac{{\mathrm{P}}_1}{{\mathrm{N}}_1}+\frac{{\mathrm{P}}_2}{{\mathrm{N}}_2}+\dots \frac{{\mathrm{P}}_{\mathrm{n}}}{{\mathrm{N}}_{\mathrm{n}}}\right]$

Where,

P₁, P₂, …P_n are respective annual precipitation for different stations.

N₁, N₂, …N_n are respective average annual/normal precipitation for different stations.

Calculation:

Data is missing for station C

N_C = 80 cm

∴ N_C + 10% of N_C = 88 cm and N_C – 10% of N_C = 72 cm

As N_A and N_D lies beyond the range of 72 – 88 cm.

∴ Normal ratio method is to be used.

${\mathrm{P}}_{\mathrm{C}}=\frac{{\mathrm{N}}_{\mathrm{c}}}{\mathrm{n}}\left[\frac{{\mathrm{P}}_{\mathrm{A}}}{{\mathrm{N}}_{\mathrm{A}}}+\frac{{\mathrm{P}}_{\mathrm{B}}}{{\mathrm{N}}_{\mathrm{B}}}+\frac{{\mathrm{P}}_{\mathrm{D}}}{{\mathrm{N}}_{\mathrm{D}}}\right]$

$\therefore {\mathrm{P}}_{\mathrm{C}}=\frac{80}{3}\left[\frac{90}{60}+\frac{60}{75}+\frac{70}{100}\right]$

Concept:

For an irrigation land

SC = Saturation capacity, FC (F_c) = Field capacity, OMC = Optimum moisture content, PWP = Permanent welting point and UWP = Ultimate welting point

1. Equivalent depth of water held at field capacity (x) = S × d × F_c

2. Equivalent depth of water held at PWP (x’) = S × d × (PWP)

3. Available moisture/storage capacity of soil (y) = S × d × (F_c - PWP)

4. Readily available moisture content (d_w) = S × d × (F_c - OMC)

Where,

S = Specific gravity of the soil and d = soil depth

Volume of water = Depth of water × Area of irrigated land

Calculation:

Given: Porosity (n) = 0.35, d = 80 cm = 0.8 m, F_c = 0.40, PWP = 0.20, and Area (A) = 500 m²

Field capacity is defined as the ratio of the weight of the water contained in certain volume of the soil to the weight of the same volume of dry soil.

$\therefore {\mathrm{F}}_{\mathrm{c}}=\frac{{\gamma }_{\mathrm{w}}\times {\mathrm{V}}_{\mathrm{v}}}{{\gamma }_{\mathrm{d}}\times \mathrm{V}}=\frac{{\gamma }_{\mathrm{w}}}{{\gamma }_{\mathrm{d}}}\times \mathrm{n}=\mathrm{S}\times \mathrm{n}$

∴ S = F_c/n = 0.40/0.35 = 1.1428

Now, Depth of water required (y) is given by

y = 1.1428 × 0.8 × 0.7 × (0.4 – 0.2) = 0.128 m

∴ V = A × y = 0.128 × 5000 = 640 m³

Concept:

The specific storage is defined as the volume of water removed or added to storage per unit volume, per unit change in head.

Specific storage can be expressed as follows,

$S_s=\frac{1}{V_s}\times \frac{d{V}_w}{dh}=\left({\alpha }_s+{\beta }_{w}n\right)\rho g$

where

V_S = Total volume of soil

V_w = Volume of water in pores of the soil

h = Hydraulic head

α_s = Compressibility of Soil

β_w = Compressibility of water

Also, the coefficient of storage or storage coefficient (S) can be expressed as follows:

$S=\frac{1}{A}\frac{d{V}_{\omega }}{dh}=S_{s}b$

where

A = Surface area of the aquifer

b = Thickness of aquifer

Calculations:

Given,

η = 0.25, b = 7.8 m, S = 6.5 × 10^-4, A = 1 km², dV_w = 650 m³

$\therefore S=\frac{1}{A}\frac{d{V}_w}{dh}=S_S\times b$

$\Rightarrow dh=\frac{d{V}_{\omega }}{A\times S}$

$\Rightarrow dh=\frac{650m^3}{1k{m}^2\times 6.5\times {10}^{-4}}$

$\Rightarrow dh=\frac{650}{{\left(10\right)}^6\times 6.5\times {10}^{-4}}$

Concept:

Uplift head (h) = Total head (H) – gradient (i) × length travelled (L)

Gradient loss (i) $=\frac{Totalheadloss}{Totallength}$

Calculation:

a) For Bligh’s theory:

L = 2 × 6 + 20 + 2 × 8 = 48 m

Head = 3

Gradient, $i=\frac{3}{48}=0.0625$

Seepage length up to A = 6 × 2 + 10 = 22 m

∴ Uplift head at A = 3 – 0.0625 × 22 = 1.625 m

b) For Lane’s theory:

$L=2\times 6+\frac{20}{3}+2\times 8=34.67m$

Head = 3

∴ Gradient, $i=\frac{3}{34.67}=0.0865$

Seepage length up to A $=6\times 2+\frac{10}{3}=15.33$

∴ Uplift head at A = 3 – 0.0565 × 15.33 = 1.673 m

∴ Difference between two = 1.625 – 1.673 = - 0.0484 m

∴ Absolute difference = 0.0484 m

Concept:

Runoff (R) = Precipitation (P) – Infiltration (F_p) – Evaporation (E)

The infiltration is computed from Horton’s equation.

Horton’s equation: Horton expressed the decay of infiltration capacity with time as an exponential decay and the infiltration depth at any time ‘t’ is given by:

f_p = f_c + (f_o - f_c) e^-kt

where,

f_p = infiltration capacity at any time ‘t’ from the start of rainfall.

f_o = initial infiltration capacity at t = o.

f_c = ultimate infiltration capacity occurring at t = t_c

k = Horton’s decay coefficient

Note: The equation is applicable only when rainfall intensity is greater than or equal to f_p.

Calculation:

For 24-hour storm t_o = 0, t_f = 24 hour

Infiltration depth at any time ‘t’ is given by:

f_t = f_o + (f_o – f_c) × e^{-k x t}

∴ f_t = 0.3 + (1 – 0.3) e^{– 5 × t} = 0.3 + 0.7 e^-5t

The infiltration will be equal to rainfall intensity until infiltration capacity is equal to lesser than rainfall intensity.

∴  0.3 + 0.7 e-5t = 10/24 ⇒ t = 0.353 hr

Runoff up to this time = 0.353 × 10/24 = 0.147 cm

Total infiltration depth after 0.353 hr (f)

$\therefore \mathrm{f}={\int }_{.353}^{24}\left(0.3+0.7{\mathrm{e}}^{-5\mathrm{t}}\right)\mathrm{d}\mathrm{t}={\left(0.3\times \mathrm{t}-\frac{0.7\times {\mathrm{e}}^{-5\times \mathrm{t}}}{-5}\right)}_0^{24}$

∴ f = (0.3t + 0.14e^-5×t)²⁴_0.353

∴f = 7.118 cm

Now,

R = 10 – (7.118 + 0.147) – 0.75 × 0.6 = 2.285 cm

∴ Runoff Volume = 2.285 × 10^-2 × 2 × 10⁶ = 45700 m³.

To derive ‘T’ hour unit hydrograph from S-curve, two S-hydrograph are superimposed on each other with the in-between lag of ‘T’ hour between them.

For the given question, 3 hours unit hydrograph is required and the ordinate is computed at t = 5 hours.

The ordinate at ‘t’ period for ‘T’ hour unit hydrograph obtained from ‘0’ hour S-hydrograph is given by, $\frac{S_1-S_2}{T/D}$

S₁ is the flow at t = 5 hour for S₁ curve

∴ S₁ = 3 – e^-5 (5 - 1) = 2.9731 m³/s

S₂ would be the S-curve value at t = 2 hours

S₂ = 3 – e^-2(2 - 1) = 2.864 m³/s

∴ Ordinate at 5 hour $=\frac{2.9731-2.864}{3/1}=0.036m^3/s$

Concept:

Leaching Requirement (L.R) $=\frac{{\left(E.C\right)}_i}{{\left(E.C\right)}_d}$

Where (E.C)_i = electrical Conductivity value of irrigation water

(E.C)_d = electrical Conductivity value of leaching water

$L.R=1-\frac{C_u}{D_i}$

D_d = -C_u + D_i

Calculations:

Given: D_i = 5.94 mm

(E.C)_i = 0.9 m.mho/cm

(E.C)_d = 2 × 13

= 26 m

L.R = 75% = 0.075

$0.075=\frac{{\left(E.C\right)}_i}{26}$

(E.C)_i = 1.95 m.mho/cm

LR = 1 - $\frac{C_u}{D_i}$

0.075 = 1 - $\frac{C_u}{D_i}$

$\begin{array}{l}0.075=1-\frac{C_u}{5.94}\\ \frac{C_u}{5.94}=1-0.075\end{array}$

C_u = 5.4945 mm

(Precipitation + Inflow) – (Outflow + Evaporation + Seepage loss) = Change in Storage

⇒ Total Precipitation = 0.03 × 10⁶ m³

Total Evaporation = 0.7 × 0.12 × 10⁶ m³

⇒ (0.03 × 10⁶ + 10 × 10⁴) – (20 × 10⁴ + 0.7 × 0.12 × 10⁶) – Seepage loss = -0.20 × 10⁶

[As the storage decrease change is storage is taken as –ve]

⇒ Seepage loss = 0.046 × 10⁶ m³