BE Civil Engineering (IOE, TU) Engineering Hydrology (IOE, CE 653) Question Paper 2077 Nepal

(a) Arithmetic-mean method

$$\bar P_{arith} = \frac{112+94+138+76+121}{5} = \frac{541}{5} = \mathbf{108.2\ mm}$$

Thiessen-polygon (weighted) method

$$\bar P_{Th} = \frac{\sum P_i A_i}{\sum A_i}$$ $$\bar P_{Th} = \frac{24151}{215} = \mathbf{112.33\ mm}$$

The Thiessen method is more reliable because it weights each station by the catchment area it represents. The arithmetic mean assumes a uniform gauge distribution and equal representativeness, which is not the case here (areas range from 31 to 55 km$^2$).

(b) Thiessen network construction: plot all stations on the catchment map, join adjacent stations by straight lines to form triangles, draw perpendicular bisectors of each connecting line; the bisectors enclose a polygon around each station. The polygon area lying inside the catchment is the weight for that station.

The isohyetal method should be preferred for orographic / mountainous catchments (e.g. Nepal's hills) where rainfall varies strongly with elevation, because lines of equal rainfall (isohyets) can be drawn to reflect the actual spatial gradient rather than assuming a station controls a fixed area.

(c) Optimum number of gauges

$$N = \left(\frac{C_v}{p}\right)^2 = \left(\frac{21}{8}\right)^2 = (2.625)^2 = 6.89 \approx \mathbf{7\ gauges}$$

Existing gauges $= 5$, so additional gauges required $= 7 - 5 = \mathbf{2\ gauges}$.

(a) $\phi$-index

Total rainfall $P = 15+42+28+11 = 96$ mm. The $\phi$-index is the constant loss rate (mm/hr) such that the rainfall above it equals the runoff.

First assume all 4 hours contribute to runoff:

$$\sum(P_i - \phi) = R \Rightarrow 96 - 4\phi = 56 \Rightarrow \phi = \frac{96-56}{4} = 10\ \text{mm/hr}$$

Check: every hourly depth (15, 42, 28, 11) exceeds $\phi = 10$ mm/hr, so the assumption that all 4 hours contribute is valid.

$$\boxed{\phi = \mathbf{10\ mm/hr}}$$

Verification of effective rainfall: $(15-10)+(42-10)+(28-10)+(11-10)=5+32+18+1=56$ mm ✓

(b) Direct-runoff volume and runoff coefficient

Runoff depth $R = 56$ mm $= 0.056$ m, area $A = 96\ \text{km}^2 = 96\times10^6\ \text{m}^2$.

$$V_{DRO} = R\cdot A = 0.056 \times 96\times10^6 = \mathbf{5.376\times10^{6}\ m^3}$$ $$C = \frac{R}{P} = \frac{56}{96} = \mathbf{0.583}$$

(c)

• The $\phi$-index is the average rainfall intensity above which the rainfall volume equals the runoff volume; it lumps all losses (infiltration, depression storage, interception) into one constant rate and ignores the part of rainfall below $\phi$.
• The $W$-index is the average infiltration rate during the period when rainfall intensity exceeds the infiltration capacity; it is refined by subtracting surface storage and interception, so $W \le \phi$.
• Limitation: a constant $\phi$-index assumes loss rate is uniform in time, whereas actual infiltration capacity decays during the storm (Horton); it therefore over-predicts losses early and under-predicts them late.

(a) Convolution (superposition)

Multiply the UH ordinates by 3 cm (no lag) and by 2 cm (lagged 2 hr), then add.

(b) Total hydrograph = DRO + base flow (5 m$^3$/s)

Peak discharge $= \mathbf{95\ m^3/s}$ occurring at $t = \mathbf{6\ hr}$.

(c) Assumptions of unit-hydrograph theory

1. Constant base length (time invariance): for a given duration of effective rainfall, the base time of the DRH is constant regardless of storm intensity.
2. Linearity / proportionality: DRH ordinates are directly proportional to the depth of effective rainfall of the same duration.
3. Superposition: the DRH from successive blocks of effective rainfall can be added with appropriate time lags; rainfall is uniform in space and time over the unit duration.

(a) Routing coefficients

Denominator $D = 2K(1-x) + \Delta t = 2(12)(0.80) + 6 = 19.2 + 6 = 25.2$

$$C_0 = \frac{\Delta t - 2Kx}{D} = \frac{6 - 2(12)(0.2)}{25.2} = \frac{6-4.8}{25.2} = \frac{1.2}{25.2} = 0.0476$$ $$C_1 = \frac{\Delta t + 2Kx}{D} = \frac{6+4.8}{25.2} = \frac{10.8}{25.2} = 0.4286$$ $$C_2 = \frac{2K(1-x) - \Delta t}{D} = \frac{19.2-6}{25.2} = \frac{13.2}{25.2} = 0.5238$$

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

(b) Routing equation: $O_{n} = C_0 I_{n} + C_1 I_{n-1} + C_2 O_{n-1}$

(c) Peak attenuation and lag

• Inflow peak $= 68\ \text{m}^3/\text{s}$ at $t = 12$ hr.
• Outflow peak $\approx 45.84\ \text{m}^3/\text{s}$ at $t = 24$ hr.
• Attenuation $= 68 - 45.84 = \mathbf{22.16\ m^3/s}$ (about 33% reduction).
• Lag of peak $= 24 - 12 = \mathbf{12\ hr}$.

(a) Gumbel flood estimation

The Gumbel frequency factor is $K_T = \dfrac{y_T - \bar y_n}{S_n}$, where the reduced variate is

$$y_T = -\ln\!\left[\ln\!\left(\frac{T}{T-1}\right)\right]$$

and the estimate is $x_T = \bar x + K_T\, s$.

For $T = 50$ yr:

$$y_{50} = -\ln\!\left[\ln\tfrac{50}{49}\right] = -\ln[\ln 1.02041] = -\ln(0.020203) = 3.9019$$ $$K_{50} = \frac{3.9019 - 0.5296}{1.0864} = \frac{3.3723}{1.0864} = 3.1041$$ $$x_{50} = 2150 + 3.1041\times560 = 2150 + 1738.3 = \mathbf{3888\ m^3/s}$$

For $T = 100$ yr:

$$y_{100} = -\ln\!\left[\ln\tfrac{100}{99}\right] = -\ln[\ln 1.01010] = -\ln(0.010050) = 4.6001$$ $$K_{100} = \frac{4.6001 - 0.5296}{1.0864} = \frac{4.0705}{1.0864} = 3.7468$$ $$x_{100} = 2150 + 3.7468\times560 = 2150 + 2098.2 = \mathbf{4248\ m^3/s}$$

(b) A "100-year flood" is a flood of magnitude $\approx 4248\ \text{m}^3/\text{s}$ that has a probability $P = 1/T = 0.01$ of being equalled or exceeded in any single year (it does not mean it occurs once every 100 years exactly).

Risk of at least one exceedance in $n = 20$ years:

$$R = 1 - (1-P)^n = 1 - (1-0.01)^{20} = 1 - (0.99)^{20} = 1 - 0.8179 = \mathbf{0.182\ (18.2\%)}$$

(c) Limitations of Gumbel's method

1. It assumes the annual maximum series is independent and identically distributed and that the data exactly follow the EV-I distribution; short records (here 24 years) give wide confidence bands for $T = 100$ yr.
2. It assumes stationarity (no trend due to land-use change or climate change) and uses tabulated $\bar y_n,\ S_n$ that depend on sample size, introducing sampling error for large extrapolations.

(a) Hydrologic cycle

            (condensation)        clouds
   ^  ----------------------------->  ~~~~~~
   |                                    | precipitation
  evaporation /                        v
  transpiration   <-- interception -- LAND ---> surface runoff --> rivers
   |  (from ocean, lakes, plants)       |                              |
   |                            infiltration                           v
   ^----------------- groundwater flow <-- percolation ----------->  OCEAN

Main processes: evaporation, transpiration (evapotranspiration), condensation, precipitation, interception, infiltration, percolation, surface runoff, groundwater flow. It is a closed cycle driven by solar energy and gravity.

(b) Water budget of the lake

Water-budget equation (as equivalent depth over the lake):

$$\Delta S = P + \frac{(Q_{in}-Q_{out})\,t}{A} - E$$

Convert volumes to depth over area $A = 25\times10^6\ \text{m}^2$, $t = 30\times86400 = 2.592\times10^6$ s.

Net inflow volume $= (6.0-4.5)\times2.592\times10^6 = 1.5\times2.592\times10^6 = 3.888\times10^6\ \text{m}^3$ Equivalent depth $= \dfrac{3.888\times10^6}{25\times10^6} = 0.15552\ \text{m} = 155.52\ \text{mm}$

Known depths: $P = 95$ mm, $\Delta S = +180$ mm.

$$E = P + (\text{net inflow depth}) - \Delta S = 95 + 155.52 - 180 = \mathbf{70.5\ mm}$$

The lake lost about 70.5 mm of water depth to evaporation during the month.

(a) Horton's equation

$$f(t) = f_c + (f_0 - f_c)\,e^{-kt}$$

At $t = 2$ hr:

$$f(2) = 1.5 + (8.0-1.5)e^{-0.45\times2} = 1.5 + 6.5\,e^{-0.9} = 1.5 + 6.5(0.40657) = 1.5 + 2.643 = \mathbf{4.14\ cm/hr}$$

(b) Total infiltration depth over $T = 3$ hr (area under the curve):

$$F = \int_0^T f\,dt = f_c T + \frac{f_0 - f_c}{k}\left(1 - e^{-kT}\right)$$ $$F = 1.5(3) + \frac{6.5}{0.45}\left(1 - e^{-0.45\times3}\right) = 4.5 + 14.444\,(1 - e^{-1.35})$$ $$e^{-1.35} = 0.25924 \Rightarrow F = 4.5 + 14.444(0.74076) = 4.5 + 10.70 = \mathbf{15.20\ cm}$$

(c) Factors that reduce infiltration capacity during a storm

1. Soil-moisture build-up / saturation of the upper soil layers as the storm proceeds (the main cause of the Horton decay).
2. Surface sealing / compaction by raindrop impact (inwash of fine particles clogging pores); also reduced soil temperature, increased surface crusting, and entrapped air.

(a) Rational-method peak discharge

Design intensity at $t_c = 25$ min:

$$i = \frac{6000}{25+35} = \frac{6000}{60} = 100\ \text{mm/hr}$$

Rational formula (SI, with $A$ in km$^2$, $i$ in mm/hr):

$$Q_p = \frac{C\,i\,A}{3.6} = \frac{0.55 \times 100 \times 2.4}{3.6} = \frac{132}{3.6} = \mathbf{36.7\ m^3/s}$$

(b) The rational method assumes the rainfall lasts at least as long as the time of concentration ($t_d \ge t_c$), so that the entire catchment contributes simultaneously to the outlet and the peak occurs at $t = t_c$ with uniform intensity over the whole area.

(c) Weighted runoff coefficient

$$C_w = \frac{\sum C_j A_j}{\sum A_j} = (0.30)(0.90) + (0.70)(0.40) = 0.27 + 0.28 = 0.55$$

Here the area-weighted coefficient is 0.55, identical to part (a), so the revised peak discharge is unchanged:

$$Q_p = \frac{0.55\times100\times2.4}{3.6} = \mathbf{36.7\ m^3/s}$$

(a) Transmissivity (Thiem equation for a confined aquifer)

$$T = \frac{Q\,\ln(r_2/r_1)}{2\pi\,(s_1 - s_2)}$$

with $Q = 30\ \text{L/s} = 0.030\ \text{m}^3/\text{s}$, $r_1 = 15$ m, $r_2 = 60$ m, $s_1 = 2.4$ m, $s_2 = 0.9$ m.

$$\ln\frac{r_2}{r_1} = \ln\frac{60}{15} = \ln 4 = 1.3863$$ $$T = \frac{0.030 \times 1.3863}{2\pi\,(2.4-0.9)} = \frac{0.041589}{2\pi\times1.5} = \frac{0.041589}{9.4248} = 4.413\times10^{-3}\ \text{m}^2/\text{s}$$

Converting: $T = 4.413\times10^{-3}\times86400 = \mathbf{381.3\ m^2/day}$.

(b) Coefficient of permeability

$$K = \frac{T}{b} = \frac{4.413\times10^{-3}}{20} = 2.206\times10^{-4}\ \text{m/s}$$ $$K = 2.206\times10^{-4}\times86400 = \mathbf{19.06\ m/day}$$

(c) In a confined aquifer the water is held under pressure between two impermeable layers; the piezometric (pressure) surface lies above the top of the aquifer and pumping produces a drawdown of the piezometric head without dewatering the aquifer (transmissivity stays constant). In an unconfined (water-table) aquifer the upper boundary is the free water table, pumping actually dewaters part of the aquifer, the saturated thickness decreases, and the flow is governed by Dupuit's equation.

(a) Meteorological factors affecting evaporation

1. Vapour-pressure difference between water surface and overlying air ($e_s - e_a$).
2. Wind speed over the surface (removes saturated air).
3. Air/water temperature (controls $e_s$).
4. Atmospheric (barometric) pressure and relative humidity / solar radiation (any of these accepted).

(b) Meyer's formula

$$E_L = K_M\,(e_s - e_a)\left(1 + \frac{u_9}{16}\right)$$ $$E_L = 0.36\,(25.0 - 14.0)\left(1 + \frac{16}{16}\right) = 0.36 \times 11.0 \times (1+1) = 0.36 \times 11.0 \times 2 = 7.92\ \text{mm/day}$$

Monthly evaporation (30 days):

$$E_{month} = 7.92 \times 30 = \mathbf{237.6\ mm}$$

(c) Other methods to estimate lake evaporation: the water-budget method, the energy-budget method, the Penman combination (mass-transfer + energy) method, or measurement with a Class A evaporation pan multiplied by a pan coefficient (typically $\approx 0.7$).

(a) Storm hydrograph

  Q ^                 * crest segment (contains peak Qp)
    |               *   *
    |   rising    *       * <- point of inflection (end of inflow)
    |   limb     *          *
    |          *               *  recession limb (groundwater depletion)
    |        *                      * * *
    |____*___________________________________> time
        start of    time-to-peak
        rise

• Rising (concentration) limb: steep rise as runoff reaches outlet.
• Crest segment: region around the peak discharge $Q_p$.
• Recession limb: falling part fed mainly by depletion of storage; its starting point of inflection marks the end of direct surface inflow.

(b) Base-flow separation is the process of dividing the total stream hydrograph into direct runoff (surface + quick interflow) and base flow (delayed groundwater contribution).

• Straight-line method: join the point where runoff begins (start of rising limb) to the point on the recession limb where direct runoff ends (often located at $N = 0.83\,A^{0.2}$ days after the peak, $A$ in km$^2$) by a horizontal/straight line; the area below is base flow.
• Master-depletion-curve (normal depletion) method: extend the pre-storm groundwater recession curve forward under the hydrograph until it meets a point vertically below the peak, then connect it smoothly to the inflection point on the recession limb; this follows the natural groundwater depletion.

(c) Recession discharge after 5 days

$$Q_t = Q_0 K_r^{\,t} = 60\times(0.92)^5$$ $$(0.92)^5 = 0.65908 \Rightarrow Q_5 = 60\times0.65908 = \mathbf{39.5\ m^3/s}$$