BE Civil Engineering (IOE, TU) Engineering Hydrology (IOE, CE 653) Question Paper 2080 Nepal

(a) Normal-ratio method

When the normal annual precipitation of the index stations differs from that of the missing station by more than 10%, the arithmetic-mean method is inadequate and the normal-ratio method must be used. Here the normal annual values vary widely (845–1280 mm), so the normal-ratio method is appropriate.

$$P_S = \frac{N_S}{m}\left(\frac{P_P}{N_P}+\frac{P_Q}{N_Q}+\frac{P_R}{N_R}\right)$$

with $N_S=845$, $m=3$, and the index normals $N_P=1120,\;N_Q=935,\;N_R=1280$.

$$\frac{P_P}{N_P}=\frac{102}{1120}=0.09107$$ $$\frac{P_Q}{N_Q}=\frac{84}{935}=0.08984$$ $$\frac{P_R}{N_R}=\frac{118}{1280}=0.09219$$

Sum $= 0.27310$.

$$P_S=\frac{845}{3}\times0.27310 = 281.667\times0.27310 = 76.92\text{ mm}$$

Missing storm precipitation at S ≈ 76.9 mm.

(b) Double-mass curve

The double-mass curve plots the cumulative precipitation of the test station (Y-axis) against the cumulative mean of the surrounding stations (X-axis). As long as the station record is consistent the plot is a straight line; a change in slope indicates a change in the gauge's environment/exposure. The records after the break are corrected by:

$$P_{corrected}=P_{recorded}\times\frac{S_{before}}{S_{after}}$$

The break occurs at 2070 (gauge obstructed from 2070 onward). Using the period of consistent (pre-2070) data and inconsistent (2073–2076) data:

Original (consistent) regime — years 2067, 2068, 2069: slope $S_{before}=\dfrac{\sum X}{\sum \text{group}}=\dfrac{1130+920+1010}{1110+905+1000}=\dfrac{3060}{3015}=1.0149$

Inconsistent regime — years 2073, 2074, 2075, 2076: slope $S_{after}=\dfrac{644+700+760+880}{920+1000+940+1090}=\dfrac{2984}{3950}=0.7554$

Correction factor $=\dfrac{S_{before}}{S_{after}}=\dfrac{1.0149}{0.7554}=1.3435$

Corrected values:

• 2073: $644\times1.3435 = 865.2$ mm
• 2074: $700\times1.3435 = 940.5$ mm

Corrected precipitation: 2073 ≈ 865 mm, 2074 ≈ 940 mm.

(a) Validity and catchment area

For a unit hydrograph the volume of direct runoff must equal 1 cm of runoff over the catchment.

Volume under UH (using trapezoidal sum with $\Delta t = 4\text{ h}=14400\text{ s}$):

$$\sum Q = 0+25+60+40+20+8+0 = 153\ \text{m}^3/\text{s}$$ $$V = 153 \times 14400 = 2{,}203{,}200\ \text{m}^3$$

This volume corresponds to 1 cm (=0.01 m) depth:

$$A=\frac{V}{d}=\frac{2{,}203{,}200}{0.01}=2.2032\times10^{8}\ \text{m}^2 = 220.32\ \text{km}^2$$

The ordinates rise then recede smoothly to zero and enclose a consistent 1-cm volume, so it is a valid UH.

Catchment area ≈ 220.3 km².

(b) Direct runoff by convolution

Multiply the UH by each excess block and lag the second block by 4 h, then add base flow (10 m³/s).

Check: total DRH ordinates $=0+75+230+240+140+64+16+0=765$. Expected $=153\times(3+2)=765$ ✓.

Peak flood discharge = 250 m³/s at t = 12 h.

Muskingum routing coefficients

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

With $K=12$, $x=0.20$, $\Delta t=6$:

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

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

Check: $C_0+C_1+C_2 = 0.0476+0.4286+0.5238 = 1.000$ ✓

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

Sample for t=6: $0.0476(90)+0.4286(40)+0.5238(40)=4.29+17.14+20.95=42.4$ m³/s. For t=18: $0.0476(110)+0.4286(140)+0.5238(67.46)=5.24+60.00+35.34=100.6$ m³/s.

Outflow: 42.4, 67.5, 100.6, 103.2, 86.2 m³/s at 6, 12, 18, 24, 30 h. The peak outflow (~103 m³/s) is attenuated below the inflow peak (140 m³/s) and lagged.

(b) Physical meaning of K and x

• $K$ is the storage-time constant (travel time): it has units of time and approximates the time of travel of a flood wave through the reach.
• $x$ is a dimensionless weighting factor that distributes the relative influence of inflow and outflow on the reach storage, $S = K[xI+(1-x)O]$. It reflects the degree of wedge storage / attenuation.
• The practical range is $0 \le x \le 0.5$ (typically 0.2–0.3 for natural streams); $x=0$ gives maximum attenuation (linear reservoir) and $x=0.5$ gives pure translation with no attenuation.

Gumbel's method

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

(a) T = 100 years

Reduced variate:

$$y_T=-\ln\!\left[\ln\frac{100}{99}\right]=-\ln\!\left[\ln(1.010101)\right]=-\ln(0.0100503)=4.6001$$

Frequency factor:

$$K=\frac{4.6001-0.5362}{1.1124}=\frac{4.0639}{1.1124}=3.6532$$

Flood magnitude:

$$x_{100}=620+3.6532\times180=620+657.6=1277.6\ \text{m}^3/\text{s}$$

100-year flood ≈ 1278 m³/s.

(b) Return period of a 1000 m³/s flood

Frequency factor required:

$$K=\frac{x_T-\bar{x}}{\sigma_{n-1}}=\frac{1000-620}{180}=\frac{380}{180}=2.1111$$

Reduced variate:

$$y_T=\bar{y}_n+K\,S_n=0.5362+2.1111\times1.1124=0.5362+2.3484=2.8846$$

From $y_T=-\ln\!\left[\ln\dfrac{T}{T-1}\right]$:

$$\ln\frac{T}{T-1}=e^{-y_T}=e^{-2.8846}=0.05586$$ $$\frac{T}{T-1}=e^{0.05586}=1.05745 \Rightarrow T=\frac{1.05745}{0.05745}=18.4\ \text{years}$$

A flood of 1000 m³/s has a return period of ≈ 18.4 years.

(c) Risk over a 25-year design life

For an event with return period $T=100$ years (annual exceedance probability $P=1/T=0.01$), the risk of at least one exceedance in $n=25$ years is:

$$R = 1-(1-P)^{n} = 1-(1-0.01)^{25}=1-(0.99)^{25}$$ $$(0.99)^{25}=0.77782 \Rightarrow R = 1-0.77782 = 0.2222$$

The risk of the 100-year flood being equalled or exceeded at least once in 25 years ≈ 22.2%.

Given: $Q=2500\ \text{L/min}=\dfrac{2500\times10^{-3}}{60}=0.04167\ \text{m}^3/\text{s}$, aquifer thickness $b=25$ m, $r_1=20$ m ($s_1=3.2$ m), $r_2=80$ m ($s_2=1.1$ m), well radius $r_w=0.30$ m.

(a) Transmissivity (Thiem equation, confined aquifer)

$$Q=\frac{2\pi T (s_1-s_2)}{\ln(r_2/r_1)}\;\Rightarrow\; T=\frac{Q\ln(r_2/r_1)}{2\pi(s_1-s_2)}$$ $$\ln(r_2/r_1)=\ln(80/20)=\ln 4 = 1.3863$$ $$T=\frac{0.04167\times1.3863}{2\pi(3.2-1.1)}=\frac{0.05776}{13.1947}=4.378\times10^{-3}\ \text{m}^2/\text{s}$$

Permeability:

$$K=\frac{T}{b}=\frac{4.378\times10^{-3}}{25}=1.751\times10^{-4}\ \text{m/s}$$

T ≈ 4.38×10⁻³ m²/s (≈ 378 m²/day); K ≈ 1.75×10⁻⁴ m/s (≈ 15.1 m/day).

(b) Drawdown at well face and radius of influence

Using the observed point ($r_2=80$ m, $s_2=1.1$ m) and extrapolating with the same drawdown-vs-ln(r) gradient. The drawdown gradient per unit ln(r):

$$\frac{s_1-s_2}{\ln(r_2/r_1)}=\frac{2.1}{1.3863}=1.5148\ \text{m per ln unit}$$

Drawdown at well face $r_w=0.30$ m (extending from $r_1=20$ m, $s_1=3.2$ m):

$$s_w = s_1 + 1.5148\,\ln\frac{r_1}{r_w}=3.2+1.5148\ln\frac{20}{0.30}=3.2+1.5148\times4.1997=3.2+6.362=9.56\ \text{m}$$

Drawdown at the well face ≈ 9.56 m.

Radius of influence $R$ (where $s=0$), extending from $r_2=80$ m, $s_2=1.1$ m:

$$0 = 1.1 - 1.5148\,\ln\frac{R}{80}\;\Rightarrow\;\ln\frac{R}{80}=\frac{1.1}{1.5148}=0.7262$$ $$R = 80\,e^{0.7262}=80\times2.0672=165.4\ \text{m}$$

Radius of influence ≈ 165 m.

(a) Hydrologic cycle

        Clouds  <-- condensation --
          | precipitation         |
          v                       | evaporation/
  ====== land / vegetation ===== transpiration
   |        |          |  ^         ^
 infiltration  surface   |  |         |
   |       runoff -----> streams/lakes ----> Ocean
   v                                        ^
 groundwater flow ------------------------->

It is the continuous circulation of water: evaporation and transpiration from oceans/land/plants → condensation to clouds → precipitation → interception, infiltration, surface runoff, and depression storage → overland & channel flow and groundwater flow → back to the oceans. It is a closed system driven by solar energy and gravity.

(b) Water balance

$$P + I_{in} - O_{out} - E = \Delta S \;\Rightarrow\; E = P + I_{in} - O_{out} - \Delta S$$

(all expressed as equivalent depth over the lake area $A=5.0\times10^6\ \text{m}^2$, time $t=30\times86400=2.592\times10^6$ s).

Inflow volume: $V_{in}=2.0\times2.592\times10^6 = 5.184\times10^6\ \text{m}^3$ → depth $=5.184\times10^6/5\times10^6 = 1.0368$ m Outflow volume: $V_{out}=1.6\times2.592\times10^6 = 4.1472\times10^6\ \text{m}^3$ → depth $=0.8294$ m Rainfall depth $P=0.090$ m; storage change $\Delta S=0.25$ m.

$$E = 0.090 + 1.0368 - 0.8294 - 0.250 = 0.0474\ \text{m}=47.4\ \text{mm}$$

Evaporation ≈ 47.4 mm for the month.

(a) Horton's infiltration equation

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

where $f(t)$ = infiltration capacity at time $t$, $f_0$ = initial infiltration capacity, $f_c$ = final (steady) infiltration capacity, $k$ = decay constant (1/time), $t$ = time from start of rainfall.

(b) Total infiltration in first 2 hours

Integrate $f(t)$ from 0 to $t$:

$$F(t)=f_c\,t+\frac{(f_0-f_c)}{k}\left(1-e^{-kt}\right)$$

With $f_0=9$, $f_c=1.5$ cm/h, $k=1.2$ h⁻¹, $t=2$ h:

$$f_0-f_c = 7.5\ \text{cm/h},\quad \frac{f_0-f_c}{k}=\frac{7.5}{1.2}=6.25\ \text{cm}$$ $$e^{-kt}=e^{-2.4}=0.090718$$ $$F(2)=1.5\times2 + 6.25\,(1-0.090718)=3.0+6.25\times0.909282=3.0+5.683=8.683\ \text{cm}$$

Total infiltration depth in the first 2 hours ≈ 8.68 cm.

Rational method

$$Q_p = \frac{C\,i\,A}{360}\quad(\text{Q in m}^3/\text{s, }i\text{ in mm/h, }A\text{ in hectares})$$

Step 1 — design intensity at $t=t_c=25$ min:

$$i=\frac{750}{25+15}=\frac{750}{40}=18.75\ \text{mm/h}$$

Step 2 — peak discharge:

$$Q_p=\frac{0.55\times18.75\times60}{360}=\frac{618.75}{360}=1.719\ \text{m}^3/\text{s}$$

Peak design runoff ≈ 1.72 m³/s.

Assumptions of the rational method:

1. Rainfall intensity is uniform over the entire catchment and constant throughout the storm duration.
2. The storm duration equals the time of concentration (so the whole catchment contributes to the peak).
3. The runoff coefficient $C$ is constant for the storm and representative of the catchment.
4. The return period of the computed peak equals that of the design rainfall.
5. Best suited to small catchments (typically < 50 km²).

(a) Direct runoff volume

Direct runoff ordinates = total − base flow (12 m³/s):

Sum of DRO ordinates $=0+18+63+83+50+28+13+0=255\ \text{m}^3/\text{s}$.

Volume (each ordinate represents $\Delta t=3\text{ h}=10800$ s):

$$V=255\times10800 = 2{,}754{,}000\ \text{m}^3 \approx 2.754\times10^6\ \text{m}^3$$

Runoff depth over $A=150\ \text{km}^2=150\times10^6\ \text{m}^2$:

$$d=\frac{V}{A}=\frac{2{,}754{,}000}{150\times10^6}=0.01836\ \text{m}=18.36\ \text{mm}$$

Direct runoff volume ≈ 2.75×10⁶ m³, equivalent to a runoff depth of ≈ 18.4 mm.

(b) Three components of a single-peaked hydrograph

1. Rising limb (concentration curve): the ascending part from the start of direct runoff to the peak; its shape depends on storm and basin characteristics.
2. Crest segment (peak): the region around the maximum discharge, occurring after rainfall has spread runoff from the entire catchment.
3. Recession limb (falling limb): the descending part from the peak to the resumption of base flow, governed mainly by basin storage characteristics (independent of storm).

Meyer's formula

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

where $E_L$ is in mm/day, vapour pressures in mm Hg, and $u_9$ is the wind speed in km/h at 9 m height.

$$E_L = 0.36\,(17.5-9.0)\left(1+\frac{16}{16}\right)=0.36\times8.5\times(1+1)=0.36\times8.5\times2$$ $$E_L = 6.12\ \text{mm/day}$$

Daily evaporation ≈ 6.12 mm/day.

Two other methods of estimating reservoir evaporation:

1. Pan evaporation method (Class A pan): measured pan evaporation is multiplied by a pan coefficient (≈ 0.70 for a Class A land pan) to obtain lake/reservoir evaporation. Simple and widely used.
2. Water-budget (water-balance) method: evaporation is found from the storage continuity of the reservoir, $E = P + I - O - \Delta S -$ seepage. Conceptually exact but sensitive to errors in the other terms.

(Energy-budget and Penman combination methods are other acceptable answers.)

(a) Arithmetic-mean method

$$\bar{P}=\frac{P_A+P_B+P_C+P_D}{4}=\frac{65+48+80+56}{4}=\frac{249}{4}=62.25\ \text{mm}$$

Arithmetic mean ≈ 62.25 mm.

(b) Thiessen-polygon method

$$\bar{P}=\frac{\sum P_i A_i}{\sum A_i}$$ $$\bar{P}=\frac{4586}{70}=65.51\ \text{mm}$$

Thiessen-weighted mean ≈ 65.5 mm.

Comment: The Thiessen-polygon method is more appropriate because it weights each gauge by the area it represents, accounting for non-uniform gauge distribution. The arithmetic mean is acceptable only when gauges are uniformly spaced and rainfall is fairly uniform over the basin. Here gauge C (high rainfall) also represents the largest area, so the Thiessen value (65.5 mm) is higher than the simple mean (62.25 mm).