22-Agric-B7 Principles of Hydrology · December 2016
Question 1 of 4: Watershed–Lake Water Balance and Storm-Rainfall Intensity
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — November-December 2016 — 04-Agric-B7, Principles of Hydrology. Three-hour, open-book exam; any non-communicating calculator is permitted. Format: any THREE (3) questions constitute a complete exam paper (the first three as they appear in the answer book are marked), each of equal value; most questions require calculations. All four questions are solved here as a complete study resource.
Given. Watershed and lake (Part 1.1) and a 40-minute storm hyetograph read at 5-minute increments (Part 1.2):
Quantity
Value
Watershed drainage area
35 km² = 35,000,000 m²
Lake surface area
70.8 ha = 708,000 m²
Lake inflow, April (rate)
1.5 m³/s
Lake outflow, April (rate)
1.25 m³/s
Lake storage change, April
+1.0 m rise over the lake surface
Watershed rainfall, April
22.5 cm
Base flow fraction of stream flow
40%
Days in April
30 d
Storm hyetograph (incremental depth per 5-minute interval):
Time (min)
0
5
10
15
20
25
30
35
40
Rainfall increment (mm)
—
2
5.1
6.4
5.6
5.3
4.1
3.0
0.8
Find. (1.1a) Lake evaporation for April. (1.1b) The percentage of the watershed's April rainfall that appeared as direct stream flow, given base flow is 40% of the total stream flow. (1.2) The maximum rainfall depth and average intensity in any 10-minute and any 30-minute window of the storm.
Approach. Part 1.1 applies a monthly volumetric water balance, first to the lake alone (solving for evaporation) and then to the watershed as a whole (converting the lake's net stream inflow, less its base-flow share, into a fraction of watershed rainfall); Part 1.2 scans the cumulative hyetograph for the wettest 10-minute and 30-minute windows.
Part 1.1(a) — Lake water balance, solve for evaporation. With seepage neglected, the lake's monthly balance is Inflow + Precipitation on the lake − Outflow − Evaporation = ΔStorage. Over 30 days ($t=30\times86400=2{,}592{,}000\ \text{s}$):
$$V_{in}=1.5\times2{,}592{,}000=3{,}888{,}000\ \text{m}^3$$
$$V_{out}=1.25\times2{,}592{,}000=3{,}240{,}000\ \text{m}^3$$
$$\Delta S=1.0\ \text{m}\times708{,}000\ \text{m}^2=708{,}000\ \text{m}^3$$
$$P_{lake}=0.225\ \text{m}\times708{,}000\ \text{m}^2=159{,}300\ \text{m}^3$$
Solving the balance for evaporation:
$$\begin{aligned}E&=V_{in}+P_{lake}-V_{out}-\Delta S\\&=3{,}888{,}000+159{,}300-3{,}240{,}000-708{,}000\\&=\boxed{99{,}300\ \text{m}^3}\end{aligned}$$
spread over the lake surface this is a depth of $99{,}300/708{,}000=\boxed{0.140\ \text{m}=14.0\ \text{cm}}$ for the month.
Part 1.1(b) — Percentage of watershed rainfall that became direct stream flow. The lake's inflow (3,888,000 m³ for the month) is the watershed's total stream-flow contribution; with base flow 40% of that total, the direct-runoff share attributable to April's rainfall event is the remaining 60%:
$$V_{direct}=0.60\times3{,}888{,}000=\boxed{2{,}332{,}800\ \text{m}^3}$$
The volume of rainfall falling on the 35 km² watershed that month is
$$V_{rain}=0.225\ \text{m}\times35{,}000{,}000\ \text{m}^2=7{,}875{,}000\ \text{m}^3$$
so the percentage of rainfall converted to direct stream flow is
$$\%=\frac{2{,}332{,}800}{7{,}875{,}000}\times100=\boxed{29.6\%}$$
Part 1.2 — Maximum 10-minute and 30-minute rainfall depth and intensity. Sliding a 10-minute (two-increment) window along the hyetograph and summing consecutive increments gives depths of 7.1, 11.5, 12.0, 10.9, 9.4, 7.1 and 3.8 mm for the seven possible 10-minute windows; the maximum is the 10–20 min window:
$$P_{10}=5.6+6.4=\boxed{12.0\ \text{mm}}$$
$$i_{10}=\frac{12.0\ \text{mm}}{10/60\ \text{hr}}=\boxed{72.0\ \text{mm/hr}}$$
A 30-minute (six-increment) window gives depths of 28.5, 29.5 and 25.2 mm for the three possible windows; the maximum is the 5–35 min window:
$$P_{30}=5.1+6.4+5.6+5.3+4.1+3.0=\boxed{29.5\ \text{mm}}$$
$$i_{30}=\frac{29.5\ \text{mm}}{30/60\ \text{hr}}=\boxed{59.0\ \text{mm/hr}}$$
