Question 2 of 4: Pond Water Balance, Combination-Method Evaporation and Storm Infiltration

Question 2: Pond Water Balance, Combination-Method Evaporation and Storm Infiltration (33.3 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Three related sub-parts on the same 1,250 ha watershed and its small on-site pond (20 m × 10 m × 1.6 m deep, plan area 200 m²):

Pond depth day 1 = 1.5 m (start of record); pond depth day 10 = 1.2 m (end of record).

2.3 – cumulative storm rainfall (Table 2) and the resulting direct-runoff hydrograph (Figure 1, triangular: 0 at $t=0$, peak 20 cms at $t=5$ hr, back to 0 at $t=12$ hr) over the same 1,250 ha watershed:

[Figure not reproduced: Figure 1 – direct-runoff hydrograph from the 2.3 storm event (reproduced from the source figure: triangular, peak 20 cms at $t=5$ hr, base 0–12 hr). See the official exam paper.]

Find. (2.1) The average daily seepage loss from the pond (mm/day). (2.2) The average daily evaporation rate (mm/day) from the pond by the combination (Penman) method. (2.3) The total infiltration depth (mm) for the 2.3 storm event.

Approach. 2.1 closes a depth water balance on the pond (rainfall in, evaporation out, storage change measured, seepage the unknown residual) over the 10-day record. 2.2 combines an energy-balance (radiation) term and an aerodynamic (mass-transfer) term, weighted by the slope of the saturation-vapour-pressure curve and the psychrometric constant, per the Penman combination method. 2.3 closes a depth water balance on the storm event itself (rainfall in, direct-runoff depth out via the hydrograph volume, evaporation out, infiltration the unknown residual).

1. Part 2.1 — Pond water balance, solve for seepage. The pond has vertical sides (20 m × 10 m rectangular box), so its plan area of 200 m² is constant with depth and a plain DEPTH balance applies over the 10-day record: $$\Delta h = P - E - S_{eep}$$ Summing Table 1: $\sum P = 0+25+0+95+0+0+50+0+0+0=170\ \text{mm}$; $\sum E = 12.5\times7=87.5\ \text{mm}$ (7 non-zero days); observed depth change $\Delta h = 1.2-1.5=-0.300\ \text{m}=-300\ \text{mm}$. Solving for the total 10-day seepage: $$S_{eep,tot}=\sum P-\sum E-\Delta h = 170-87.5-(-300)=\boxed{382.5\ \text{mm}}$$ so the average daily seepage loss is $$S_{eep,avg}=\frac{382.5}{10}=\boxed{38.25\ \text{mm/day}}$$ (equivalently $38.25\ \text{mm}\times200\ \text{m}^2/1000=7.65\ \text{m}^3/\text{day}$).
2. Part 2.2(i) — Slope of the saturation-vapour-pressure curve and psychrometric constant. Using $e_s(T)=0.6108\exp\!\big(17.27T/(T+237.3)\big)$ kPa and $\Delta=4098\,e_s/(T+237.3)^2$ at $T_a=14$°C: $$e_s=0.6108\exp\!\left(\frac{17.27\times14}{14+237.3}\right)=\boxed{1.599\ \text{kPa}},\qquad e_a=RH\cdot e_s=0.65\times1.599=\boxed{1.039\ \text{kPa}}$$ $$\Delta=\frac{4098\times1.599}{(14+237.3)^2}=\boxed{0.1037\ \text{kPa/}^\circ\text{C}}$$ The latent heat of vaporization at 14°C is $\lambda=2.501-0.002361\,T_a=2.468\ \text{MJ/kg}$, and (reading the printed "specific heat of water 1005 J/(kg·K)" as the specific heat of AIR at constant pressure — 1005 J/(kg·K) is the textbook value for $c_p$ of air, and the psychrometric constant always uses $c_p$ of air, not of liquid water — flagged below) the psychrometric constant is $$\gamma=\frac{c_p P}{0.622\,\lambda}=\frac{1.005\times101.3}{0.622\times2467.9}=\boxed{0.0663\ \text{kPa/}^\circ\text{C}}$$
3. Part 2.2(ii) — Radiation (energy-balance) term. Converting net radiation directly to an equivalent evaporation depth rate, $E_{rad}=R_n/(\rho_w\lambda)$: $$E_{rad}=\frac{40\ \text{W/m}^2}{999.2\ \text{kg/m}^3\times2.468\times10^{6}\ \text{J/kg}}=1.622\times10^{-8}\ \text{m/s}=\boxed{1.40\ \text{mm/day}}$$
4. Part 2.2(iii) — Aerodynamic (mass-transfer) term. With the wind run converted to a mean speed, $u_2=172.8\ \text{km/day}=172{,}800/86{,}400=\boxed{2.00\ \text{m/s}}$, the logarithmic wind-profile (Thornthwaite–Holzman) mass-transfer equation gives the aerodynamic evaporation term directly from the roughness height, air/water densities and vapour-pressure deficit (this is exactly the data the question supplies $\rho_a$, $\rho_w$ and $z_0$ for): $$E_{aero}=\frac{0.622\,k^2\rho_a\,u_2\,(e_s-e_a)}{P\,\rho_w\,[\ln(z_2/z_0)]^2}$$ With von Kármán constant $k=0.4$, $z_2=2\ \text{m}$, $z_0=0.0003\ \text{m}$ (so $\ln(z_2/z_0)=\ln(6667)=8.805$), and $e_s-e_a=0.560\ \text{kPa}=560\ \text{Pa}$: $$E_{aero}=\frac{0.622\times0.4^2\times1.23\times2.00\times560}{101{,}300\times999.2\times8.805^2}=1.746\times10^{-8}\ \text{m/s}=\boxed{1.51\ \text{mm/day}}$$
5. Part 2.2(iv) — Combine by the Penman weights. $$E=\frac{\Delta}{\Delta+\gamma}E_{rad}+\frac{\gamma}{\Delta+\gamma}E_{aero}=0.610\times1.40+0.390\times1.51=\boxed{1.44\ \text{mm/day}}$$
6. Part 2.3(i) — Direct-runoff depth from the triangular hydrograph. The hydrograph is a triangle: 0 at $t=0$, peak $Q_p=20$ cms at $t=5$ hr, back to 0 at $t=12$ hr, so its area (runoff volume) is $$V=\tfrac12\times(12\times3600\ \text{s})\times20\ \text{m}^3/\text{s}=\boxed{432{,}000\ \text{m}^3}$$ Spread over the 1,250 ha ($=12{,}500{,}000\ \text{m}^2$) watershed this is a runoff depth of $$Q_{runoff}=\frac{432{,}000}{12{,}500{,}000}\times1000=\boxed{34.56\ \text{mm}}$$
7. Part 2.3(ii) — Event water balance, solve for infiltration. Total storm rainfall from Table 2's cumulative depth is $P=61.0$ mm over a $135/60=2.25$-hr storm duration. Evaporation during the storm, prorated from the stated 5 mm/24 h at a constant rate: $$E_{event}=5\times\frac{2.25}{24}=\boxed{0.469\ \text{mm}}$$ Closing the depth balance $P=Q_{runoff}+E_{event}+F$ for infiltration $F$: $$F=61.0-34.56-0.469=\boxed{26.0\ \text{mm}}$$