BE Civil Engineering (IOE, TU) Engineering Hydrology (IOE, CE 653) Question Paper 2076 Nepal

(a) Hydrologic cycle

The hydrologic cycle is the continuous circulation of water among the atmosphere, land surface, and subsurface, driven by solar energy and gravity.

           ____  clouds  ____
   evaporation /\        |  precipitation
        ^     /  \       v
   [ Ocean / Lake ] <-- runoff <-- [ Land surface ]
        ^                              |  infiltration
   transpiration (plants)              v
                              [ Soil -> Groundwater ] --> baseflow

• Precipitation (P): water reaching the ground as rain, snow, hail.
• Evaporation (E): liquid water converted to vapour from open water/soil.
• Transpiration (T): vapour released by vegetation; combined with E as evapotranspiration.
• Infiltration (F): entry of water into the soil.
• Runoff (R): overland + sub-surface flow reaching streams.

Catchment water-balance equation (over a time interval):

$$P = R + E_T + \Delta S$$

where $P$ = precipitation, $R$ = runoff (surface + groundwater outflow), $E_T$ = evapotranspiration, and $\Delta S$ = change in storage (soil moisture + groundwater + surface). All terms are expressed as equivalent depths or volumes over the catchment.

(b) Lake water balance

$$\Delta S = (\text{Inflow} + \text{P volume}) - (\text{Outflow} + \text{E volume} + \text{Seepage})$$

Area $A = 45\ \text{km}^2 = 45\times10^6\ \text{m}^2$. Period $t = 30\ \text{days} = 30\times86400 = 2.592\times10^6\ \text{s}$.

Stream inflow: $V_i = 3.2 \times 2.592\times10^6 = 8.2944\times10^6\ \text{m}^3 = 8.294\ \text{Mm}^3$

Spillway outflow: $V_o = 2.5 \times 2.592\times10^6 = 6.480\times10^6\ \text{m}^3 = 6.480\ \text{Mm}^3$

Seepage: $V_s = 0.4 \times 2.592\times10^6 = 1.0368\times10^6\ \text{m}^3 = 1.037\ \text{Mm}^3$

Precipitation on lake: $V_P = 0.12\ \text{m} \times 45\times10^6 = 5.40\times10^6\ \text{m}^3 = 5.400\ \text{Mm}^3$

Evaporation from lake: $V_E = 0.09\ \text{m} \times 45\times10^6 = 4.05\times10^6\ \text{m}^3 = 4.050\ \text{Mm}^3$

$$\Delta S = (8.294 + 5.400) - (6.480 + 4.050 + 1.037) = 13.694 - 11.567 = 2.127\ \text{Mm}^3$$

Change in lake level:

$$\Delta h = \frac{\Delta S}{A} = \frac{2.127\times10^6}{45\times10^6} = 0.04727\ \text{m} = 4.73\ \text{cm (rise)}$$

Change in storage $\Delta S \approx +2.13\ \text{Mm}^3$; lake level rises by $\approx 4.73\ \text{cm}$.

(a) Comparison of methods

(b) Computation

(i) Arithmetic mean:

$$\bar P = \frac{125+98+142+110+156}{5} = \frac{631}{5} = 126.2\ \text{mm}$$

(ii) Thiessen method: total area $A = 18+22+15+30+25 = 110\ \text{km}^2$.

$$\bar P = \frac{\sum P_i A_i}{\sum A_i} = \frac{13736}{110} = 124.87\ \text{mm}$$

Arithmetic mean = 126.2 mm; Thiessen mean = 124.87 mm.

Comment: The Thiessen value is slightly lower because the high-reading gauges (C = 142 mm with only 15 km² and E = 156 mm) receive smaller area weights, while the low-reading gauge D (110 mm, large 30 km² area) is given more weight. The arithmetic mean over-states rainfall when high values sit in small influence areas; Thiessen weighting corrects for the uneven gauge distribution.

(a) Unit hydrograph

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

Sherman's assumptions:

1. Time invariance: the DRH for a given excess-rainfall duration is constant (does not change with time).
2. Linearity / proportionality: ordinates of the DRH are proportional to the excess-rainfall depth (e.g. 2 cm gives ordinates twice the UH).
3. Superposition: the DRH from successive rainfall blocks can be added with appropriate time lags.
4. Effective rainfall is uniformly distributed over the whole catchment and constant within $D$.

A 6-hour UH is the DRH from 1 cm of excess rainfall occurring uniformly over 6 hours.

(b) Deriving the 6-h UH

Since 4 cm of excess rainfall produced the given DRH, by proportionality divide each ordinate by 4:

Volume check / catchment area: the volume under the UH equals 1 cm of runoff over the catchment area $A$.

Sum of UH ordinates $= 0+6+20+35+30+20+10+4+0 = 125\ \text{m}^3/\text{s}$.

With uniform time step $\Delta t = 6\ \text{h} = 21600\ \text{s}$:

$$V_{UH} = \left(\sum U_i\right)\Delta t = 125 \times 21600 = 2.70\times10^6\ \text{m}^3$$

This equals $1\ \text{cm} = 0.01\ \text{m}$ depth over $A$:

$$A = \frac{V_{UH}}{0.01} = \frac{2.70\times10^6}{0.01} = 2.70\times10^8\ \text{m}^2$$ $$A = \frac{2.70\times10^8}{10^6} = 270\ \text{km}^2$$

6-hour UH ordinates: 0, 6, 20, 35, 30, 20, 10, 4, 0 m³/s; catchment area ≈ 270 km².

(a) Muskingum method

The Muskingum method treats reach storage as the sum of prism storage (depends on outflow) and wedge storage (depends on inflow−outflow):

$$S = K\,[\,x\,I + (1-x)\,O\,]$$

• $K$ = storage-time constant (≈ travel time of the flood wave through the reach), units of time.
• $x$ = weighting factor ($0 \le x \le 0.5$) measuring the relative importance of inflow; $x=0$ → reservoir (level-pool) routing, $x=0.5$ → pure translation.

Combining the storage equation with the continuity equation gives the routing relation:

$$O_2 = C_0 I_2 + C_1 I_1 + C_2 O_1$$

with

$$C_0 = \frac{\Delta t - 2Kx}{2K(1-x) + \Delta t},\quad C_1 = \frac{\Delta t + 2Kx}{2K(1-x) + \Delta t},\quad C_2 = \frac{2K(1-x) - \Delta t}{2K(1-x) + \Delta t}$$

and $C_0 + C_1 + C_2 = 1$.

(b) Computation of coefficients

Denominator $D = 2K(1-x) + \Delta t = 2(12)(0.8) + 6 = 19.2 + 6 = 25.2$.

$$C_0 = \frac{6 - 2(12)(0.2)}{25.2} = \frac{6 - 4.8}{25.2} = \frac{1.2}{25.2} = 0.0476$$ $$C_1 = \frac{6 + 4.8}{25.2} = \frac{10.8}{25.2} = 0.4286$$ $$C_2 = \frac{19.2 - 6}{25.2} = \frac{13.2}{25.2} = 0.5238$$

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

Routing table

$O_2 = 0.0476\,I_2 + 0.4286\,I_1 + 0.5238\,O_1$

Outflow hydrograph (m³/s): 10.00, 10.95, 21.83, 42.96, 45.84, 42.63, 36.71, 29.80, 22.51.

Peak attenuation: inflow peak = 68 m³/s (at 12 h); outflow peak = 45.84 m³/s (at 24 h).

$$\text{Attenuation} = 68 - 45.84 = 22.16\ \text{m}^3/\text{s};\quad \text{lag} = 24 - 12 = 12\ \text{h}$$

The peak is attenuated by ≈ 22.16 m³/s and delayed by ≈ 12 h.

(a) Return period and Gumbel formula

Return period $T$ is the average interval (in years) between events that equal or exceed a given magnitude. Probability of exceedance in any one year is

$$P(X \ge x_T) = \frac{1}{T}$$

and the non-exceedance probability is $1 - 1/T$.

Gumbel (EV-I) frequency formula:

$$x_T = \bar x + K\,\sigma_{n-1}, \qquad K = \frac{y_T - \bar y_n}{S_n}$$

where the reduced variate is

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

and $\bar y_n$, $S_n$ are the reduced mean and reduced standard deviation depending on sample size $n$.

(b) 50-year flood

Reduced variate for $T = 50$:

$$y_T = -\ln\!\left[\ln\!\left(\tfrac{50}{49}\right)\right] = -\ln\!\left[\ln(1.020408)\right] = -\ln(0.0202027) = 3.9019$$

Frequency factor:

$$K = \frac{y_T - \bar y_n}{S_n} = \frac{3.9019 - 0.5362}{1.1124} = \frac{3.3657}{1.1124} = 3.0256$$

Flood magnitude:

$$x_{50} = \bar x + K\,\sigma_{n-1} = 2500 + 3.0256 \times 650 = 2500 + 1966.6 = 4466.6\ \text{m}^3/\text{s}$$

The 50-year flood ≈ 4467 m³/s.

(c) Risk over 10 years

Annual exceedance probability $p = 1/T = 1/50 = 0.02$. The probability of at least one exceedance in $N = 10$ years (the risk):

$$R = 1 - (1-p)^N = 1 - (0.98)^{10}$$ $$(0.98)^{10} = 0.81707 \Rightarrow R = 1 - 0.81707 = 0.18293$$

Risk ≈ 0.183, i.e. about an 18.3 % chance of the 50-year flood being equalled or exceeded at least once in the next 10 years.

Thiem (steady, confined) equation

For steady radial flow to a fully penetrating well in a confined aquifer:

$$Q = \frac{2\pi T (h_2 - h_1)}{\ln(r_2/r_1)} \quad\Rightarrow\quad T = \frac{Q\,\ln(r_2/r_1)}{2\pi (h_2 - h_1)}$$

(i) Transmissivity

$$\ln\!\left(\frac{r_2}{r_1}\right) = \ln\!\left(\frac{50}{10}\right) = \ln 5 = 1.6094$$ $$h_2 - h_1 = 20 - 18 = 2\ \text{m}$$ $$T = \frac{0.025 \times 1.6094}{2\pi \times 2} = \frac{0.040236}{12.566} = 3.2019\times10^{-3}\ \text{m}^2/\text{s}$$

Convert to m²/day ($\times 86400$):

$$T = 3.2019\times10^{-3} \times 86400 = 276.6\ \text{m}^2/\text{day}$$

Transmissivity $T \approx 276.6\ \text{m}^2/\text{day}$.

(ii) Hydraulic conductivity

$$K = \frac{T}{b} = \frac{276.6}{25} = 11.06\ \text{m/day}$$

(Equivalently $K = 3.2019\times10^{-3}/25 = 1.281\times10^{-4}\ \text{m/s}$.)

Hydraulic conductivity $K \approx 11.07\ \text{m/day}$.

(a) Horton's equation

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

• $f(t)$ = infiltration capacity at time $t$ (cm/h)
• $f_0$ = initial infiltration capacity at $t = 0$ (cm/h)
• $f_c$ = final constant capacity as $t \to \infty$ (cm/h)
• $k$ = decay constant (h⁻¹) controlling how fast capacity falls

(b)(i) Capacity at $t = 3\ \text{h}$

$$f(3) = 1.5 + (8 - 1.5)\,e^{-0.4\times3} = 1.5 + 6.5\,e^{-1.2}$$ $$e^{-1.2} = 0.30119 \Rightarrow f(3) = 1.5 + 6.5(0.30119) = 1.5 + 1.9577 = 3.458\ \text{cm/h}$$

$f(3) \approx 3.46\ \text{cm/h}$.

(b)(ii) Cumulative infiltration over 0–3 h

Integrate Horton's equation:

$$F = \int_0^t f\,dt = f_c\,t + \frac{f_0 - f_c}{k}\left(1 - e^{-kt}\right)$$ $$F = 1.5(3) + \frac{6.5}{0.4}\left(1 - e^{-1.2}\right) = 4.5 + 16.25(1 - 0.30119)$$ $$F = 4.5 + 16.25(0.69881) = 4.5 + 11.356 = 15.856\ \text{cm}$$

Total infiltrated depth in 3 h $\approx 15.86\ \text{cm}$ (this is the soil's potential capacity; actual infiltration is limited by available rainfall).

(a) Rational method

The peak runoff is

$$Q_p = \frac{1}{3.6}\,C\,i\,A \;=\; 0.278\,C\,i\,A$$

with $Q_p$ in m³/s, $C$ = dimensionless runoff coefficient, $i$ = rainfall intensity (mm/h) for a duration equal to the time of concentration, and $A$ = catchment area (km²).

Assumptions: rainfall is uniform in space and time; intensity is constant over a duration equal to $t_c$; the return period of $Q_p$ equals that of $i$; $C$ is constant. Limitations: valid only for small catchments (typically < 50 km²); ignores storage and antecedent conditions; gives only the peak, not the full hydrograph.

(b) Time of concentration (Kirpich)

$$t_c = 0.0195\,L^{0.77}\,S^{-0.385}\quad (t_c\ \text{in min},\ L\ \text{in m})$$ $$L^{0.77} = 1500^{0.77} = 309.0,\qquad S^{-0.385} = 0.02^{-0.385} = 4.073$$ $$t_c = 0.0195 \times 309.0 \times 4.073 = 24.5\ \text{min}$$

Time of concentration $t_c \approx 24.5\ \text{min}$.

Peak design discharge

With $i = 60\ \text{mm/h}$ (duration $= t_c$):

$$Q_p = 0.278 \times 0.45 \times 60 \times 2.5 = 0.278 \times 67.5 = 18.77\ \text{m}^3/\text{s}$$

Peak design discharge $Q_p \approx 18.77\ \text{m}^3/\text{s}$.

(a) Definitions

• $\phi$-index: the constant average rate of abstraction (cm/h) such that the rainfall depth in excess of it equals the observed direct runoff. It lumps all losses (interception, depression storage, infiltration) into one constant rate.
• W-index: the average infiltration rate during the period when rainfall intensity exceeds the infiltration capacity, computed as $W = (P - R - S)/t_e$, where $S$ is surface storage/interception and $t_e$ the time of rain exceeding capacity.
• Difference: the W-index refines the $\phi$-index by separating out initial losses (interception + depression storage), so $W \le \phi$; for large floods the two nearly coincide.

(b) Determining the $\phi$-index

Total rainfall $P = 1.0 + 3.5 + 5.0 + 2.5 + 1.5 = 13.5\ \text{cm}$. Runoff $R = 8.5\ \text{cm}$; total loss $= 13.5 - 8.5 = 5.0\ \text{cm}$ over the contributing intervals.

Trial – assume all 5 intervals contribute ($\Delta t = 1\ \text{h}$). Then $5\phi = 5.0 \Rightarrow \phi = 1.0\ \text{cm/h}$. Check that every interval depth exceeds $\phi \times \Delta t = 1.0\ \text{cm}$: depths are 1.0, 3.5, 5.0, 2.5, 1.5 — all $\ge 1.0$, so the assumption is consistent (the first interval contributes zero excess but is not below $\phi$).

Excess rainfall per interval ($P_i - \phi$):

$$0,\ 2.5,\ 4.0,\ 1.5,\ 0.5\ \text{cm} \;\Rightarrow\; \text{sum} = 8.5\ \text{cm} = R\ \checkmark$$

$\phi$-index = 1.0 cm/h.

(a) Factors affecting evaporation

• Solar radiation / net energy supply (dominant driver)
• Air and water temperature
• Wind speed (removes the saturated boundary layer)
• Relative humidity / vapour-pressure deficit of the air
• Atmospheric pressure
• Quality of water (salinity/dissolved solids lower evaporation)
• Surface area exposed and depth of the water body

(b) Pan and lake evaporation

Pan evaporation = level drop minus rainfall caught in the pan:

$$E_{pan} = 6.5 - 1.5 = 5.0\ \text{cm/week}$$

Lake evaporation = pan coefficient × pan evaporation:

$$E_{lake} = 0.70 \times 5.0 = 3.5\ \text{cm/week} = 0.035\ \text{m/week}$$

Volume lost from reservoir of area $A = 3.0\ \text{km}^2 = 3.0\times10^6\ \text{m}^2$:

$$V = E_{lake} \times A = 0.035 \times 3.0\times10^6 = 1.05\times10^5\ \text{m}^3$$

Lake evaporation = 3.5 cm/week; volume lost = 1.05 × 10⁵ m³ (= 0.105 Mm³).

(a) Storm hydrograph components

 Q ^                 crest (peak)
   |                * *
   |   rising      *   *  recession
   |    limb      *     *  limb
   |            *        *  * *
   |          *             *   * * *
   |________*___________________*________> t
   | base flow ----------------- base flow
   A (start of rise)    B (end of direct runoff)

• Rising limb: flow increases as the catchment responds to excess rainfall.
• Crest / peak: maximum discharge.
• Recession limb: flow declines as storage drains; reaches the depletion (base-flow) curve at B.
• Base flow: sustained groundwater contribution beneath the direct-runoff hydrograph.

(b) Base-flow separation

• Straight-line method: join the point of rise (A) to the point on the recession (B) where direct runoff ends with a single straight horizontal/sloping line; everything above is direct runoff, below is base flow. Simplest but approximate.
• Constant-discharge method: assume base flow equals the constant discharge prevailing just before the rising limb begins; a horizontal line is drawn from A across to the recession. Suitable where pre-storm base flow is stable.

End of direct runoff (point B): commonly fixed using the empirical relation for the time from the peak to the end of direct runoff:

$$N = 0.83\,A^{0.2}$$

where $N$ is in days and $A$ is the catchment area in km² (i.e. the recession reaches base flow about $N$ days after the peak).