Questions & explanations
1. What is the difference between confined and unconfined aquifers in terms of groundwater flow?
A confined aquifer is a water-bearing layer sandwiched between two low-permeability layers (aquitards), so water is under pressure and rises in a well above the top of the aquifer. An unconfined aquifer has a water table as its upper boundary, and water flows under atmospheric pressure. In a confined aquifer, pumping causes a drop in pressure (drawdown) that propagates quickly, while in an unconfined aquifer, water is released by draining pores, which is slower. Darcy's law applies to both, but the storage coefficient differs: confined aquifers store water by compressing the aquifer and expanding water, whereas unconfined aquifers store water by draining pore spaces. These differences affect well hydraulics and the analysis of pumping tests.
2. Explain how hydrograph routing (e.g., Muskingum method) is used to predict flood wave movement through a river reach.
Hydrograph routing predicts how a flood wave changes as it moves downstream through a river channel. The Muskingum method uses a storage equation that relates reach storage to inflow and outflow. It has two parameters: K (travel time through the reach) and X (weighting factor for inflow vs. outflow). The method solves the continuity equation (inflow minus outflow equals change in storage) with a linear storage function. Given an inflow hydrograph at the upstream end, you can compute the outflow hydrograph at the downstream end. This is useful for flood forecasting and designing flood control structures. The method assumes a linear relationship between storage and flow, which works well for natural channels with mild slopes.
3. You have two wells: one in a confined aquifer and one in an unconfined aquifer. Both are pumped at the same rate. After 1 hour, which well will have a larger drawdown at the well itself?
The well in the confined aquifer will have a larger drawdown at the well itself after 1 hour. This is because the confined aquifer has a much lower storativity (storage coefficient) than the unconfined aquifer's specific yield. In the confined aquifer, the pressure drop propagates quickly but the water released is small, so the water level in the well drops more. In the unconfined aquifer, water drains from pores, providing a large volume of water, so the water table drops less. For example, typical storativity for confined aquifers is 0.0001 to 0.001, while specific yield for unconfined aquifers is 0.1 to 0.3. Thus, the confined aquifer's drawdown is about 100 times larger for the same pumping rate and time.
4. If a watershed has a 1-hour unit hydrograph peak of 100 m³/s per cm of rainfall, what would be the peak flow from a 3-hour storm that produces 5 cm of excess rainfall uniformly?
First, you need to convert the 1-hour UH to a 3-hour UH using the S-curve or superposition method. For a uniform 3-hour storm, you can sum three 1-hour UHs lagged by 1 hour each, then divide by 3 to get the 3-hour UH. The peak of the 3-hour UH is typically lower than the 1-hour UH peak. Assuming the 3-hour UH peak is, say, 70 m³/s per cm (a typical reduction), then for 5 cm of excess rainfall, the peak flow would be 5 * 70 = 350 m³/s. Alternatively, if you directly use the 1-hour UH with a 3-hour storm, you must apply convolution: divide the storm into 1-hour increments, multiply each by the UH, and sum. The exact peak depends on the UH shape, but the concept is that runoff is proportional to rainfall depth.
5. A pumping test in an unconfined aquifer shows that drawdown is much smaller than in a confined aquifer for the same pumping rate. Explain why.
In an unconfined aquifer, water is released by draining pore spaces as the water table lowers, which provides a large amount of water per unit drop in head (high specific yield). In contrast, a confined aquifer releases water mainly by compression of the aquifer matrix and expansion of water, which yields very little water per unit pressure drop (low storativity). Therefore, for the same pumping rate, the drawdown in an unconfined aquifer is smaller because more water is available from storage. Additionally, the cone of depression in an unconfined aquifer is shallower and spreads more slowly. This difference is important when interpreting pumping test data to estimate aquifer properties.
6. How do you derive a unit hydrograph from a recorded storm event?
To derive a unit hydrograph, you first separate baseflow (groundwater contribution) from the total streamflow hydrograph to get direct runoff. Then, you compute the total volume of direct runoff and divide it by the watershed area to get the depth of excess rainfall. Next, you divide the direct runoff ordinates (flow values at each time step) by that depth to get the unit hydrograph ordinates for that storm duration. For example, if the excess rainfall depth is 2 cm, you divide each direct runoff value by 2 to get the 1-cm UH. The resulting UH has a duration equal to the storm's effective rainfall duration. This UH can then be used to predict runoff for other storms of the same duration.
7. Compare the use of a unit hydrograph for flood prediction in a small urban watershed vs. a large rural watershed.
A unit hydrograph for a small urban watershed has a shorter time to peak and higher peak flow because impervious surfaces (roads, roofs) cause rapid runoff. The UH shape is steep and narrow. For a large rural watershed, the UH has a longer time to peak and lower peak flow due to greater storage (soils, vegetation) and longer flow paths. The UH is broader and flatter. Urbanization increases runoff volume and reduces infiltration, so the same rainfall produces a larger and faster flood. Therefore, separate UHs must be derived for pre- and post-development conditions. The unit hydrograph method assumes linearity, which may be less accurate for large watersheds with non-uniform rainfall.
8. If you pump a well in a confined aquifer at a constant rate, how does drawdown change over time according to Theis equation?
The Theis equation describes drawdown (lowering of water level) in a confined aquifer due to constant-rate pumping. Drawdown increases with time and with distance from the well. Initially, drawdown is small, but it grows as the cone of depression expands. The equation uses the aquifer's transmissivity (how easily water flows horizontally) and storativity (how much water is released from storage). For a given distance, drawdown is proportional to the pumping rate and inversely related to transmissivity. The Theis solution assumes the aquifer is homogeneous, isotropic, and infinite in extent. In practice, you can use the Theis equation to predict drawdown at any time and location.
9. How does Darcy's law relate groundwater flow velocity to hydraulic gradient?
Darcy's law states that the groundwater flow rate (discharge) through a porous medium is proportional to the hydraulic gradient (the change in hydraulic head over distance) and the cross-sectional area. The constant of proportionality is the hydraulic conductivity (K), which depends on the medium and fluid. The specific discharge (Darcy velocity) is q = -K * (dh/dl), where dh/dl is the hydraulic gradient. The actual average velocity of water particles is higher because water only flows through pores; it equals q divided by porosity. Darcy's law applies to laminar flow in saturated porous media. It is fundamental for calculating flow rates in aquifers and designing wells.
10. How does the Revised Universal Soil Loss Equation (RUSLE) improve upon USLE?
RUSLE (Revised USLE) improves USLE by updating the factor values with more data and better algorithms. For example, the R factor now accounts for more rainfall patterns, the K factor includes seasonal soil variability, the LS factor uses a more accurate formula for slope length and steepness, the C factor incorporates crop rotations and residue management, and the P factor includes new practices like strip-cropping. RUSLE also allows for computer calculation and is more applicable to a wider range of conditions, including rangeland and disturbed sites. However, like USLE, RUSLE still estimates long-term average soil loss and does not predict deposition or gully erosion.
11. A rainwater tank receives 100 m³ of water with a contaminant concentration of 10 mg/L. If the tank loses 2 m³/day by outflow and the contaminant decays at a rate of 0.1 per day, what is the mass of contaminant after 1 day? Assume complete mixing.
First, find the initial mass: 100 m³ × 10 mg/L = 1,000,000 mg (since 1 m³ = 1000 L, so 100 m³ = 100,000 L; 100,000 L × 10 mg/L = 1,000,000 mg). After 1 day, decay reduces mass by 0.1 per day, so remaining mass from decay = 1,000,000 × e^(-0.1×1) ≈ 904,837 mg. Outflow removes 2 m³ of water, which contains contaminant at the current concentration. But since concentration changes, we need a mass balance. Actually, a simpler approach: use the mass balance equation dM/dt = -kM - (Q/V)M, where k=0.1, Q=2 m³/day, V=100 m³. Solve: M(t) = M0 e^(-(k+Q/V)t) = 1,000,000 e^(-(0.1+0.02)×1) = 1,000,000 e^(-0.12) ≈ 886,920 mg. So after 1 day, about 887,000 mg of contaminant remains.
12. How does stream power change along a river from headwaters to mouth, and what effect does that have on sediment size?
Stream power typically increases downstream as discharge (Q) increases, but slope (S) decreases. In headwaters, slope is steep but discharge is small, so stream power may be moderate. In middle reaches, both discharge and slope contribute to high stream power, enabling transport of coarse sediment. Near the mouth, slope is very gentle, so despite high discharge, stream power may be lower, and only fine sediment is transported. As a result, sediment size decreases downstream: boulders and gravel in headwaters, sand in middle reaches, and silt/clay in deltas. This pattern is called downstream fining and reflects the sorting by stream power.