BE Civil Engineering (IOE, TU) Engineering Hydrology (IOE, CE 653) Question Paper 2078 Nepal

Hydrologic cycle

The hydrologic (water) cycle is the continuous circulation of water among the atmosphere, land surface and sub-surface. The major processes are evaporation and transpiration (collectively evapotranspiration) from oceans, water bodies and vegetation; condensation forming clouds; precipitation (rain, snow); interception, infiltration and percolation; surface runoff and interflow; groundwater flow; and eventual return to the oceans.

        clouds  <== condensation
   ^               |
   | evaporation   | precipitation
   | / transpir.   v
  ~~~~~ocean~~~~   land ---surface runoff---> streams --> ocean
                    | infiltration
                    v
               groundwater ---base flow---> streams

Water-balance equation

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

where $P$ = precipitation on the surface, $Q_{in}$ = surface inflow, $Q_{out}$ = surface outflow, $E$ = evaporation, $G$ = net seepage to groundwater, and $\Delta S$ = change in storage (all expressed in consistent volume units over the period).

Numerical solution

Area $A = 50\ \text{km}^2 = 50\times10^{6}\ \text{m}^2$. Period $t = 30\ \text{days} = 30\times86400 = 2.592\times10^{6}\ \text{s}$.

Inflow volume: $V_{in} = 12 \times 2.592\times10^{6} = 31.104\times10^{6}\ \text{m}^3 = 31.104\ \text{Mm}^3$

Outflow volume: $V_{out} = 10 \times 2.592\times10^{6} = 25.920\times10^{6}\ \text{m}^3 = 25.920\ \text{Mm}^3$

Precipitation volume: $V_P = 0.180\ \text{m} \times 50\times10^{6}\ \text{m}^2 = 9.000\times10^{6}\ \text{m}^3 = 9.000\ \text{Mm}^3$

Evaporation volume: $V_E = 0.150 \times 50\times10^{6} = 7.500\times10^{6}\ \text{m}^3 = 7.500\ \text{Mm}^3$

Seepage volume: $V_G = 0.040 \times 50\times10^{6} = 2.000\times10^{6}\ \text{m}^3 = 2.000\ \text{Mm}^3$

Applying the balance:

$$\Delta S = V_P + V_{in} - V_{out} - V_E - V_G$$ $$\Delta S = 9.000 + 31.104 - 25.920 - 7.500 - 2.000 = 4.684\ \text{Mm}^3$$

Equivalent depth over the lake:

$$d = \frac{\Delta S}{A} = \frac{4.684\times10^{6}}{50\times10^{6}} = 0.09368\ \text{m} = 93.7\ \text{mm}$$

Change in storage = +4.684 million m³ (a rise), equivalent to +93.7 mm depth over the lake.

Unit hydrograph

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

Assumptions:

1. The rainfall excess is uniformly distributed over the whole catchment.
2. The rainfall excess is at a uniform (constant) rate during the effective duration $D$.
3. Time invariance — the base time of the DRH is constant for a given $D$, irrespective of storm magnitude.
4. Linearity (proportionality and superposition) — DRH ordinates are proportional to the depth of rainfall excess, and the responses of successive blocks add directly.
5. The DRH reflects all fixed catchment characteristics (linear time-invariant system).

Convolution

DRH ordinates from each excess block = (UH ordinate) × (excess depth in cm), shifted by the block start time. Block 1 (2.0 cm) starts at $t=0$; Block 2 (3.0 cm) starts at $t=4$ h.

Check on block 2 shift: at $t=8$ h the block-2 contribution uses UH(4)=25 → $3.0\times25 = 75$; at $t=12$ uses UH(8)=50 → $3.0\times50 = 150$; etc., confirming the table.

Total storm hydrograph ordinates (m³/s) at t = 0,4,8,12,16,20,24,28 h:

$$8,\ 58,\ 183,\ 228,\ 153,\ 88,\ 38,\ 8$$

Peak total discharge = 228 m³/s at t = 12 h.

Muskingum method

In the Muskingum method the storage in a river reach is expressed as a weighted combination of inflow $I$ and outflow $Q$:

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

• $K$ = storage-time constant, approximately the travel time of the flood wave through the reach (units of time).
• $x$ = weighting factor $(0 \le x \le 0.5)$ giving the relative weight of inflow vs. outflow on storage; $x=0$ gives a level-pool (reservoir-type) reach, $x=0.5$ gives pure translation.

The routing equation is:

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

with coefficients

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

and $C_0 + C_1 + C_2 = 1$.

Compute coefficients

$K=12$ h, $x=0.20$, $\Delta t = 6$ h. So $Kx = 2.4$, $0.5\Delta t = 3$.

Denominator $= K - Kx + 0.5\Delta t = 12 - 2.4 + 3 = 12.6$.

$$C_0 = \frac{-2.4 + 3}{12.6} = \frac{0.6}{12.6} = 0.04762$$ $$C_1 = \frac{2.4 + 3}{12.6} = \frac{5.4}{12.6} = 0.42857$$ $$C_2 = \frac{12 - 2.4 - 3}{12.6} = \frac{6.6}{12.6} = 0.52381$$

Check: $0.04762 + 0.42857 + 0.52381 = 1.000$ ✓

Routing

Start $Q_1 = 10$ at $t=0$.

t = 6 h: $I_2=30,\ I_1=10,\ Q_1=10$

$$Q_2 = 0.04762(30) + 0.42857(10) + 0.52381(10) = 1.429 + 4.286 + 5.238 = 10.95\ \text{m}^3/\text{s}$$

t = 12 h: $I_2=60,\ I_1=30,\ Q_1=10.95$

$$Q_2 = 0.04762(60) + 0.42857(30) + 0.52381(10.95) = 2.857 + 12.857 + 5.736 = 21.45\ \text{m}^3/\text{s}$$

t = 18 h: $I_2=40,\ I_1=60,\ Q_1=21.45$

$$Q_2 = 0.04762(40) + 0.42857(60) + 0.52381(21.45) = 1.905 + 25.714 + 11.236 = 38.86\ \text{m}^3/\text{s}$$

t = 24 h: $I_2=20,\ I_1=40,\ Q_1=38.86$

$$Q_2 = 0.04762(20) + 0.42857(40) + 0.52381(38.86) = 0.952 + 17.143 + 20.354 = 38.45\ \text{m}^3/\text{s}$$

Result

The peak outflow is about 38.9 m³/s, attenuated and lagged relative to the inflow peak of 60 m³/s at $t=12$ h.

Return period and frequency analysis

The return period $T$ of a hydrologic event of a given magnitude is the average interval (in years) between events that equal or exceed that magnitude. It is the reciprocal of the annual exceedance probability $P$: $T = 1/P$. Frequency analysis fits a probability distribution to an observed annual-maximum series so that flood magnitudes corresponding to chosen design return periods (e.g. 50, 100, 1000 yr) can be estimated for design of spillways, bridges, embankments, etc.

Gumbel's method

The Gumbel (Extreme Value Type I) estimate is:

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

where the reduced variate for return period $T$ is:

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

Step 1 — reduced variate for T = 100

$$\frac{T}{T-1} = \frac{100}{99} = 1.010101$$ $$\ln(1.010101) = 0.0100503$$ $$\ln(0.0100503) = -4.60014$$ $$y_{100} = -(-4.60014) = 4.60014$$

Step 2 — frequency factor

$$K = \frac{y_{100} - \bar{y}_n}{S_n} = \frac{4.60014 - 0.5362}{1.1124} = \frac{4.06394}{1.1124} = 3.6533$$

Step 3 — flood magnitude

$$x_{100} = \bar{x} + K\,\sigma_{n-1} = 1200 + 3.6533 \times 350$$ $$x_{100} = 1200 + 1278.7 = 2478.7\ \text{m}^3/\text{s}$$

The 100-year flood is approximately $x_{100} \approx 2479\ \text{m}^3/\text{s}$ (≈ 2480 m³/s).

Thiem's equation (confined aquifer)

For steady, horizontal, radial flow to a fully penetrating well in a confined aquifer of constant thickness $b$, by Darcy's law the discharge across a cylinder of radius $r$ is:

$$Q = A\,K\,\frac{dh}{dr} = (2\pi r b)\,K\,\frac{dh}{dr}$$

Separating variables between two observation wells at radii $r_1, r_2$ with piezometric heads $h_1, h_2$:

$$Q\int_{r_1}^{r_2}\frac{dr}{r} = 2\pi K b \int_{h_1}^{h_2} dh$$ $$Q\ln\!\frac{r_2}{r_1} = 2\pi K b\,(h_2 - h_1)$$

Since drawdown $s = H - h$, we have $h_2 - h_1 = s_1 - s_2$. Hence Thiem's equation:

$$\boxed{\,K = \dfrac{Q\,\ln(r_2/r_1)}{2\pi b\,(s_1 - s_2)}\,} \qquad T = Kb = \dfrac{Q\,\ln(r_2/r_1)}{2\pi(s_1-s_2)}$$

Numerical solution

Given: $Q = 0.05\ \text{m}^3/\text{s}$, $b = 20\ \text{m}$, $r_1 = 25\ \text{m}$, $s_1 = 1.8\ \text{m}$, $r_2 = 80\ \text{m}$, $s_2 = 0.9\ \text{m}$.

$$\ln\!\frac{r_2}{r_1} = \ln\!\frac{80}{25} = \ln(3.2) = 1.16315$$ $$s_1 - s_2 = 1.8 - 0.9 = 0.9\ \text{m}$$

Hydraulic conductivity:

$$K = \frac{0.05 \times 1.16315}{2\pi \times 20 \times 0.9} = \frac{0.0581575}{113.097} = 5.142\times10^{-4}\ \text{m/s}$$

Transmissivity:

$$T = Kb = 5.142\times10^{-4} \times 20 = 1.028\times10^{-2}\ \text{m}^2/\text{s}$$

Expressed per day: $T = 1.028\times10^{-2}\times 86400 \approx 888\ \text{m}^2/\text{day}$.

Results: $K \approx 5.14\times10^{-4}\ \text{m/s}$ and $T \approx 1.03\times10^{-2}\ \text{m}^2/\text{s}\ (\approx 888\ \text{m}^2/\text{day})$.

Choice of method

When the normal annual precipitation of the index stations differs from that of the missing station by more than 10%, the normal-ratio method must be used (the simple arithmetic-mean method is valid only if all normals are within 10% of station $A$).

Check: $A$ normal $=920$ mm. Stations $B,C,D$ normals are $1080, 1250, 1400$ — these deviate from $920$ by $17\%, 36\%, 52\%$ respectively, all $>10\%$. Hence the normal-ratio method is required.

Normal-ratio method

$$P_A = \frac{N_A}{m}\left[\frac{P_B}{N_B} + \frac{P_C}{N_C} + \frac{P_D}{N_D}\right]$$

where $N$ = normal annual precipitation, $P$ = storm precipitation, $m = 3$ index stations.

$$P_A = \frac{920}{3}\left[\frac{80}{1080} + \frac{95}{1250} + \frac{110}{1400}\right]$$

Compute the ratios:

$$\frac{80}{1080} = 0.074074,\quad \frac{95}{1250} = 0.076000,\quad \frac{110}{1400} = 0.078571$$

Sum $= 0.074074 + 0.076000 + 0.078571 = 0.228645$

$$P_A = \frac{920}{3}\times 0.228645 = 306.667 \times 0.228645 = 70.12\ \text{mm}$$

The estimated missing storm precipitation at station $A$ is approximately $70.1\ \text{mm}$.

Concept

The $\phi$-index is the constant (average) rate of abstraction such that the rainfall depth in excess of $\phi$ (summed only over intervals where intensity $>\phi$) equals the total rainfall excess (direct runoff).

Total rainfall $P = 1.5 + 3.5 + 5.0 + 2.5 + 1.0 = 13.5\ \text{cm}$. Runoff $R = 7.0\ \text{cm}$, so total losses $= 13.5 - 7.0 = 6.5\ \text{cm}$ over the storm.

Determine which intervals contribute

Each interval is $\Delta t = 1$ h, so intensity (cm/h) equals the depth. We need $\phi$ such that intervals with depth $>\phi$ contribute exactly $7.0$ cm of excess.

Trial — assume all 5 intervals exceed $\phi$: Excess $= (13.5 - 5\phi)$. Setting $= 7.0$: $5\phi = 6.5 \Rightarrow \phi = 1.3$ cm/h. But the smallest interval is $1.0$ cm $< 1.3$, so that interval cannot exceed $\phi$ — assumption invalid.

Trial — assume the $1.0$ cm interval does not contribute (only 4 intervals exceed $\phi$): Contributing rainfall $= 1.5 + 3.5 + 5.0 + 2.5 = 12.5\ \text{cm}$ over $4$ intervals.

$$12.5 - 4\phi = 7.0 \Rightarrow 4\phi = 5.5 \Rightarrow \phi = 1.375\ \text{cm/h}$$

Consistency check: $\phi = 1.375$ cm/h. The $1.5$ cm interval ($>1.375$) contributes, and the $1.0$ cm interval ($<1.375$) does not — consistent with the assumption. ✓

Verify excess: $(1.5-1.375)+(3.5-1.375)+(5.0-1.375)+(2.5-1.375) = 0.125+2.125+3.625+1.125 = 7.0\ \text{cm}$ ✓

The $\phi$-index = 1.375 cm/h (≈ 1.38 cm/h).

Meyer's formula

For a large, deep water body Meyer's empirical formula is:

$$E_L = K_M\,(e_w - e_a)\left(1 + \frac{u_9}{16}\right)$$

where $E_L$ is in mm/day, $e_w, e_a$ in mm of mercury, $u_9$ is the monthly mean wind speed in km/h measured at $9\ \text{m}$ height, and $K_M = 0.36$ for large deep water bodies (0.50 for small shallow ones).

Computation

Vapour-pressure deficit:

$$e_w - e_a = 17.5 - 12.0 = 5.5\ \text{mm Hg}$$

Wind factor:

$$1 + \frac{u_9}{16} = 1 + \frac{16}{16} = 1 + 1 = 2.0$$

Evaporation:

$$E_L = 0.36 \times 5.5 \times 2.0 = 3.96\ \text{mm/day}$$

The estimated daily lake evaporation is $E_L \approx 3.96\ \text{mm/day}$ (about $3.96 \times 30 \approx 119\ \text{mm}$ over a 30-day month).

Thiessen polygon method

The Thiessen polygon method assigns to each rain-gauge a polygon whose every point is nearer to that gauge than to any other. The polygons are formed by drawing perpendicular bisectors of the lines joining adjacent stations. The mean areal rainfall is the area-weighted average of the station rainfalls:

$$\bar{P} = \frac{\sum_{i} P_i\,A_i}{\sum_i A_i}$$

Advantage: Unlike the arithmetic-mean method (which weights all gauges equally), the Thiessen method gives each gauge a weight proportional to the area it represents, so non-uniform gauge spacing is accounted for, giving a more reliable estimate.

Computation

$$\bar{P} = \frac{17750}{160} = 110.94\ \text{mm}$$

The mean areal rainfall over the catchment is approximately $110.9\ \text{mm}$.

Rational method

The peak discharge is:

$$Q_p = C\,i\,A$$

Using SI with $i$ in mm/h and $A$ in km², a convenient form is:

$$Q_p\ (\text{m}^3/\text{s}) = 0.278\,C\,i\,A$$

where the factor $0.278 = 1/3.6$ converts (mm/h)(km²) to m³/s. For a composite catchment, use the area-weighted runoff coefficient.

Step 1 — composite runoff coefficient

$$C_w = \frac{\sum C_j A_j}{\sum A_j} = \frac{0.85(0.5) + 0.55(0.4) + 0.20(0.3)}{1.2}$$ $$= \frac{0.425 + 0.220 + 0.060}{1.2} = \frac{0.705}{1.2} = 0.5875$$

Step 2 — peak discharge

$$Q_p = 0.278 \times C_w \times i \times A = 0.278 \times 0.5875 \times 60 \times 1.2$$ $$Q_p = 0.278 \times 0.5875 \times 72 = 0.278 \times 42.3 = 11.76\ \text{m}^3/\text{s}$$

The estimated peak runoff is $Q_p \approx 11.8\ \text{m}^3/\text{s}$.

Components of a storm hydrograph

A single-peaked storm hydrograph has:

• Rising limb (concentration curve): discharge increases as runoff from progressively distant areas reaches the outlet.
• Crest / peak: maximum discharge, occurring after the centroid of rainfall excess (lag time).
• Recession limb (falling limb): decreasing discharge as storage in the catchment drains; its lower part is fed by groundwater (base flow).
• Base flow: the sustained groundwater contribution underlying the direct runoff.

Base-flow separation (straight-line method): join the point where the discharge starts rising (beginning of direct runoff) to a point $N$ days after the peak on the recession limb by a straight line; ordinates above the line are direct runoff. Here $N\ (\text{days}) = 0.83\,A^{0.2}$ with $A$ in km². (Other methods: fixed-base method and variable-slope method.)

Direct-runoff volume (trapezoidal / simple summation)

With uniform interval $\Delta t = 3\ \text{h} = 3\times3600 = 10800\ \text{s}$, the volume is the area under the DRH. Using the sum of ordinates (the first and last are zero, so simple summation × $\Delta t$ equals the trapezoidal estimate):

Sum of ordinates $= 0 + 40 + 80 + 50 + 20 + 0 = 190\ \text{m}^3/\text{s}$.

$$V_{DRO} = (\Sigma \text{ordinates})\times \Delta t = 190 \times 10800 = 2{,}052{,}000\ \text{m}^3 = 2.052\times10^{6}\ \text{m}^3$$

Depth of rainfall excess

Catchment area $A = 90\ \text{km}^2 = 90\times10^{6}\ \text{m}^2$.

$$d = \frac{V_{DRO}}{A} = \frac{2.052\times10^{6}}{90\times10^{6}} = 0.0228\ \text{m} = 22.8\ \text{mm} = 2.28\ \text{cm}$$

Direct-runoff volume = $2.052\times10^{6}\ \text{m}^3$; depth of rainfall excess = 22.8 mm (2.28 cm).