BE Civil Engineering (IOE, TU) Engineering Hydrology (IOE, CE 653) Question Paper 2079 Nepal

(a) Hydrologic cycle

The hydrologic (water) cycle is the continuous, closed circulation of water between the atmosphere, the land surface, and the oceans, driven by solar energy and gravity. Water evaporates from oceans and land, condenses to form clouds, falls as precipitation, and returns to the oceans by surface runoff and groundwater flow.

Major components: evaporation and transpiration (collectively evapotranspiration), condensation, precipitation, interception, infiltration, surface runoff, interflow, and groundwater flow / base flow.

        Clouds (condensation)
          /\        |
   evap. //  \\      | precipitation
        //    \\    v
  [OCEAN] <-- runoff -- [LAND] -- infiltration --> [GROUNDWATER] --> ocean

Water-balance (continuity) equation for a catchment over a time interval:

$$P - R - G - E - T = \Delta S$$

where $P$ = precipitation, $R$ = surface runoff (net outflow), $G$ = net groundwater outflow, $E$ = evaporation, $T$ = transpiration, and $\Delta S$ = change in storage. All terms are volumes (or equivalent depths) over the same period.

(b) Lake evaporation

Work in volumes over the month, with lake area $A = 25\ \text{km}^2 = 25\times10^6\ \text{m}^2$.

Water balance: $\Delta S = P + I_{in} + G - I_{out} - E$, so

$$E = P + I_{in} + G - I_{out} - \Delta S$$

• Precipitation volume: $P = 0.120\ \text{m} \times 25\times10^6\ \text{m}^2 = 3.0\times10^6\ \text{m}^3$
• Surface inflow: $I_{in} = 1.8\times10^6\ \text{m}^3$
• Groundwater inflow: $G = 0.5\times10^6\ \text{m}^3$
• Surface outflow: $I_{out} = 2.6\times10^6\ \text{m}^3$
• Change in storage (rise of 40 mm): $\Delta S = 0.040\ \text{m}\times 25\times10^6\ \text{m}^2 = 1.0\times10^6\ \text{m}^3$

Substitute:

$$E = 3.0 + 1.8 + 0.5 - 2.6 - 1.0 = 1.7\times10^6\ \text{m}^3$$

Convert to equivalent depth over the lake:

$$E_{depth} = \frac{1.7\times10^6\ \text{m}^3}{25\times10^6\ \text{m}^2} = 0.068\ \text{m} = 68\ \text{mm}$$

Daily rate over 30 days:

$$e = \frac{68\ \text{mm}}{30\ \text{d}} = 2.27\ \text{mm/day}$$

Monthly evaporation = 68 mm; daily evaporation rate ≈ 2.27 mm/day.

(a) Normal-ratio method

When the normal annual precipitation at any index (surrounding) station differs from that of the station with the missing record by more than 10%, the arithmetic-mean method is inadequate. The normal-ratio method weights each neighbouring observation by the ratio of normal annual precipitations:

$$P_X = \frac{N_X}{m}\left(\frac{P_A}{N_A}+\frac{P_B}{N_B}+\frac{P_C}{N_C}\right)$$

where $N$ = normal annual precipitation, $P$ = storm rainfall, $m$ = number of index stations. It is preferred whenever the normals vary appreciably (>10%) among stations; otherwise the simple arithmetic mean suffices.

(b) Missing rainfall at X

$$P_X = \frac{1250}{3}\left(\frac{95}{1120}+\frac{78}{980}+\frac{110}{1350}\right)$$

• $95/1120 = 0.08482$
• $78/980 = 0.07959$
• $110/1350 = 0.08148$
• Sum $= 0.24589$

$$P_X = \frac{1250}{3}\times 0.24589 = 416.67\times 0.24589 = 102.5\ \text{mm}$$

Missing storm rainfall at X ≈ 102.5 mm.

(c) Thiessen mean areal rainfall

$$\bar P = \frac{\sum P_i A_i}{\sum A_i}$$ $$\bar P = \frac{4624}{80} = 57.8\ \text{mm}$$

Mean areal rainfall = 57.8 mm.

(a) Unit hydrograph (UH)

A unit hydrograph of a given duration $D$ is the direct-runoff hydrograph resulting from one unit (1 cm) of effective (excess) rainfall generated uniformly over the catchment at a uniform rate during the duration $D$.

Basic assumptions:

1. Time invariance: the effective-rainfall hyetograph–to–DRH relationship does not change with time; the same excess rainfall always produces the same DRH.
2. Linearity (proportionality): DRH ordinates are directly proportional to the volume of effective rainfall.
3. Superposition: the DRH of a complex storm equals the sum of the DRHs of its individual rainfall blocks, suitably lagged.
4. Effective rainfall is uniform in space and constant in intensity over the duration $D$.

(b) DRH by convolution

Multiply UH ordinates by 3 (first block) and by 2 (second block, lagged by 4 h), then add.

Direct-runoff hydrograph ordinates (m³/s): 0, 75, 200, 190, 96, 24, 0 at t = 0, 4, 8, 12, 16, 20, 24 h.

Peak DRH = 200 m³/s at t = 8 h.

(a) Muskingum method

The Muskingum method models channel storage $S$ as a linear function of weighted inflow $I$ and outflow $Q$:

$$S = K\,[\,xI + (1-x)Q\,]$$

where $K$ = storage-time constant (≈ travel time of the flood wave through the reach, units of time) and $x$ = dimensionless weighting factor ($0 \le x \le 0.5$) describing the relative importance of inflow vs outflow on storage (prism + wedge storage). Combining with continuity gives the routing equation:

$$Q_2 = C_0 I_2 + C_1 I_1 + C_2 Q_1$$

with

$$C_0=\frac{-Kx+0.5\Delta t}{K-Kx+0.5\Delta t},\quad C_1=\frac{Kx+0.5\Delta t}{K-Kx+0.5\Delta t},\quad C_2=\frac{K-Kx-0.5\Delta t}{K-Kx+0.5\Delta t}$$

and $C_0+C_1+C_2=1$.

(b) Routing

Compute the denominator $D = K - Kx + 0.5\Delta t$:

• $Kx = 12\times0.2 = 2.4$
• $0.5\Delta t = 0.5\times6 = 3$
• $D = 12 - 2.4 + 3 = 12.6$

Coefficients:

• $C_0 = \dfrac{-2.4+3}{12.6} = \dfrac{0.6}{12.6} = 0.0476$
• $C_1 = \dfrac{2.4+3}{12.6} = \dfrac{5.4}{12.6} = 0.4286$
• $C_2 = \dfrac{12-2.4-3}{12.6} = \dfrac{6.6}{12.6} = 0.5238$

Check: $0.0476+0.4286+0.5238 = 1.000$ ✓

Routing $Q_2 = C_0 I_2 + C_1 I_1 + C_2 Q_1$:

t = 6 h ($I_1=40, I_2=120, Q_1=40$):

$$Q = 0.0476(120)+0.4286(40)+0.5238(40)=5.71+17.14+20.95=43.8\ \text{m}^3/\text{s}$$

t = 12 h ($I_1=120, I_2=180, Q_1=43.8$):

$$Q = 0.0476(180)+0.4286(120)+0.5238(43.8)=8.57+51.43+22.94=82.9\ \text{m}^3/\text{s}$$

t = 18 h ($I_1=180, I_2=120, Q_1=82.9$):

$$Q = 0.0476(120)+0.4286(180)+0.5238(82.9)=5.71+77.14+43.43=126.3\ \text{m}^3/\text{s}$$

Outflow: Q(6 h) ≈ 43.8 m³/s, Q(12 h) ≈ 82.9 m³/s, Q(18 h) ≈ 126.3 m³/s.

(a) Return period and frequency analysis

The return period (recurrence interval) $T$ of a hydrologic event of given magnitude is the average interval, in years, between events that equal or exceed that magnitude. Its exceedance probability is $P = 1/T$. Flood frequency analysis fits a probability distribution to a series of annual maximum floods so that the magnitude of a flood for any chosen $T$ (design flood) can be estimated by extrapolation.

Gumbel (EV-I) formula:

$$x_T = \bar x + K\,\sigma_{n-1}, \qquad K = \frac{y_T - \bar y_n}{S_n}, \qquad y_T = -\ln\!\left[\ln\!\frac{T}{T-1}\right]$$

where $K$ = frequency factor, $y_T$ = reduced variate, and $\bar y_n, S_n$ = reduced mean and reduced standard deviation (functions of sample size $N$).

(b) 100-year flood

Reduced variate for $T = 100$:

$$y_T = -\ln\!\left[\ln\frac{100}{99}\right]$$

• $100/99 = 1.010101$
• $\ln(1.010101) = 0.0100503$
• $\ln(0.0100503) = -4.60015$
• $y_T = 4.60015$

Frequency factor:

$$K = \frac{y_T-\bar y_n}{S_n}=\frac{4.60015-0.5436}{1.1413}=\frac{4.05655}{1.1413}=3.5544$$

Flood magnitude:

$$x_{100}=\bar x + K\,\sigma_{n-1}=1200+3.5544\times350$$ $$=1200+1244.0=2444.0\ \text{m}^3/\text{s}$$

The 100-year flood ≈ 2444 m³/s.

The $\phi$-index is the constant loss rate (cm/h) such that rainfall in excess of $\phi$ equals the observed runoff.

Total rainfall $= 1.2+2.8+2.0+0.6 = 6.6\ \text{cm}$. Total runoff $= 3.2\ \text{cm}$, so total loss $= 6.6-3.2 = 3.4\ \text{cm}$.

Trial 1 — assume all 4 hours contribute (loss per hour $=\phi$):

$$\phi = \frac{6.6-3.2}{4}=\frac{3.4}{4}=0.85\ \text{cm/h}$$

Check: hour 4 rainfall = 0.6 cm < 0.85 cm, so hour 4 produces no runoff. Assumption invalid — discard hour 4.

Trial 2 — only hours 1, 2, 3 contribute ($M=3$): The rainfall during the non-contributing hour 4 (0.6 cm) is entirely loss. For the 3 contributing hours:

$$\phi = \frac{(1.2+2.8+2.0)-3.2}{3}=\frac{6.0-3.2}{3}=\frac{2.8}{3}=0.933\ \text{cm/h}$$

Check: hour 1 rainfall = 1.2 cm > 0.933 cm ✓; hour 4 = 0.6 cm < 0.933 cm (correctly excluded) ✓. Consistent.

Verify runoff: $(1.2-0.933)+(2.8-0.933)+(2.0-0.933)=0.267+1.867+1.067=3.20\ \text{cm}$ ✓

$\phi$-index = 0.93 cm/h (≈ 0.933 cm/h).

Rational formula (SI form):

$$Q_p = \frac{1}{3.6}\,C\,i\,A$$

where $Q_p$ in m³/s, $C$ = dimensionless runoff coefficient, $i$ in mm/h, $A$ in km², and $1/3.6$ is the unit-conversion constant.

Substitute $C=0.65$, $i=75\ \text{mm/h}$, $A=1.5\ \text{km}^2$:

$$Q_p = \frac{1}{3.6}\times0.65\times75\times1.5 = \frac{73.125}{3.6} = 20.31\ \text{m}^3/\text{s}$$

Peak discharge $Q_p \approx 20.3\ \text{m}^3/\text{s}$.

Assumptions of the rational method:

1. Rainfall intensity is uniform over the entire catchment and constant for a duration at least equal to the time of concentration.
2. The peak runoff occurs when the whole catchment is contributing, i.e. when storm duration equals the time of concentration; the runoff coefficient is constant for a given storm and surface.

Thiem equilibrium equation for a confined aquifer:

$$Q = \frac{2\pi K b\,(h_2 - h_1)}{\ln(r_2/r_1)}$$

Given $K = 30\ \text{m/day}$, $b = 25\ \text{m}$, $h_2 - h_1 = 22-18 = 4\ \text{m}$, $r_2/r_1 = 80/20 = 4$.

Transmissivity $T = Kb = 30\times25 = 750\ \text{m}^2/\text{day}$.

$$Q = \frac{2\pi (750)(4)}{\ln 4}$$

• Numerator: $2\pi\times750\times4 = 6000\pi = 18849.6$
• $\ln 4 = 1.38629$

$$Q = \frac{18849.6}{1.38629} = 13597\ \text{m}^3/\text{day}$$

Convert to L/s:

$$Q = \frac{13597\times1000\ \text{L}}{86400\ \text{s}} = 157.4\ \text{L/s}$$

Steady-state discharge $Q \approx 13{,}597\ \text{m}^3/\text{day} \approx 157\ \text{L/s}$.

(a) Storm hydrograph components

Q
^                crest (peak)
|                  /\
|        rising   /  \  recession
|         limb   /    \   limb
|              _/      \___
|  base flow _/            \____ base flow
+------------------------------------> time

The rising limb reflects increasing direct runoff as the catchment contributes; the crest/peak is the maximum discharge; the recession (falling) limb is the depletion of channel/storage; base flow is the groundwater contribution underlying the direct runoff.

(b) Direct-runoff volume

Use the trapezoidal rule with $\Delta t = 3\ \text{h} = 3\times3600 = 10800\ \text{s}$.

Sum of ordinates (trapezoidal, end ordinates are zero):

$$\sum = \frac{0+0}{2}+40+70+30 = 140\ \text{m}^3/\text{s}$$

Volume:

$$V = \Delta t \times \sum = 10800 \times 140 = 1.512\times10^6\ \text{m}^3$$

Direct-runoff volume $V = 1.512\times10^6\ \text{m}^3$.

Equivalent depth over $A = 90\ \text{km}^2 = 90\times10^6\ \text{m}^2$:

$$d = \frac{V}{A}=\frac{1.512\times10^6}{90\times10^6}=0.0168\ \text{m}=1.68\ \text{cm}$$

Equivalent runoff depth = 1.68 cm.

(a) Factors affecting evaporation

1. Solar radiation / net energy available (the dominant driver).
2. Air (and water) temperature.
3. Wind speed over the surface.
4. Relative humidity / vapour-pressure deficit of the overlying air. (Also atmospheric pressure.)

(b) Pan and lake evaporation

Pan water balance for the week. The drop in level (lowering) is caused by evaporation; rainfall adds water (raises level) and refill adds water:

$$E_{pan} = (\text{measured drop}) + (\text{rainfall added}) + (\text{refill added})$$

Wait — apply continuity carefully. Change in pan level $= P + \text{refill} - E_{pan}$. A net lowering of 52 mm means level fell by 52 mm:

$$-52 = 18 + 10 - E_{pan}\;\Rightarrow\; E_{pan} = 18 + 10 + 52 = 80\ \text{mm}$$

Lake (reservoir) evaporation using pan coefficient $C_p = 0.7$:

$$E_{lake} = C_p \times E_{pan} = 0.7 \times 80 = 56\ \text{mm}$$

Weekly lake evaporation = 56 mm.

(a) Darcy's law and definitions

Darcy's law: the flow rate through a porous medium is proportional to the cross-sectional area and the hydraulic gradient:

$$Q = -K A \frac{dh}{dl}, \qquad v = \frac{Q}{A} = -K\,i$$

where $v$ is the Darcy (specific discharge) velocity and $i = -dh/dl$ is the hydraulic gradient.

• Hydraulic conductivity $K$ (m/day, m/s): the volume of water that flows per unit time through a unit cross-sectional area of aquifer under a unit hydraulic gradient; it depends on both the medium and the fluid.
• Transmissivity $T = Kb$ (m²/day): the rate of flow through the full saturated thickness $b$ of an aquifer per unit width under a unit hydraulic gradient.
• Storage coefficient (storativity) $S$ (dimensionless): the volume of water released from or taken into storage per unit horizontal area of aquifer per unit change in piezometric head.

(b) Calculations

(i) Hydraulic gradient and Darcy velocity

$$i = \frac{\Delta h}{L} = \frac{1.5}{1200} = 1.25\times10^{-3}$$ $$v = K\,i = 25 \times 1.25\times10^{-3} = 0.03125\ \text{m/day} \approx 0.0313\ \text{m/day}$$

(ii) Seepage (actual) velocity

$$v_s = \frac{v}{n} = \frac{0.03125}{0.25} = 0.125\ \text{m/day}$$

(iii) Discharge per unit width ($b = 18\ \text{m}$, width $= 1\ \text{m}$, so area $= 18\ \text{m}^2$)

$$q = T\,i = (Kb)\,i = (25\times18)\times1.25\times10^{-3}=450\times1.25\times10^{-3}=0.5625\ \text{m}^3/\text{day per m width}$$

Results: $i = 1.25\times10^{-3}$; Darcy velocity $v \approx 0.0313\ \text{m/day}$; seepage velocity $v_s = 0.125\ \text{m/day}$; discharge $q \approx 0.563\ \text{m}^3/\text{day per m}$.