Drainage basin hydrology and geomorphology practice questions.

During an investigation of river channel change a student uses her own velocity readings, her own channel-width measurements and a set of discharge records downloaded from the national gauging service. How are these data classified?

• The first two are primary data and the gauging records are secondary
• All three count as primary data because the student assembled them all
• All three count as secondary data because instruments produced the readings
• The field measurements are secondary and the downloaded records are primary

Answer: The first two are primary data and the gauging records are secondary

Primary data are collected first-hand by the investigator for this particular investigation, so the origin of the data decides the category, not whether an instrument was involved.

Which statement correctly describes the pattern of erosion and deposition around a meander bend?

• Erosion on the inner bank, deposition on the outer bank
• Erosion on both banks, deposition only on the river bed
• Deposition on both banks, erosion only on the river bed
• Erosion on the outer bank, deposition on the inner bank

Answer: Erosion on the outer bank, deposition on the inner bank

The fastest flow (the thalweg) swings to the outer bank, driving hydraulic action and abrasion there, while the slack water on the inner bank loses energy and deposits a point bar. Reversing the two banks is the classic error.

A student writes: "The river is steepest and most turbulent in its upper course, so its mean velocity must be highest there and must fall towards the mouth." Which statement identifies the error in this reasoning?

• Mean velocity is identical at every point along a river, because discharge is conserved down the channel.
• Mean velocity does fall downstream, but only in reaches where the bed is cut in resistant rock.
• Mean velocity usually rises downstream: the larger channel has a greater hydraulic radius, so less energy is lost to friction.
• Mean velocity cannot be compared between reaches, because velocity depends only on the roughness of the bed.

Answer: Mean velocity usually rises downstream: the larger channel has a greater hydraulic radius, so less energy is lost to friction.

Turbulence is not the same as mean velocity: white water in a boulder-strewn upland reach loses most of its energy to friction against a rough bed and a small hydraulic radius, while the smooth, deep lower channel moves the water faster overall.

Which sequence correctly describes the formation of an oxbow lake on a lowland floodplain?

• Deposition on the outer bends narrows the neck, the river breaks through on the inner bank, the new channel silts up, the old bend carries the main flow.
• Erosion on the outer bends narrows the neck of the meander, the river cuts through the neck at high flow, deposition seals both ends, the abandoned loop becomes a lake.
• The meander migrates downstream until it meets a tributary, the tributary is captured, the old channel is left dry, a lake forms in the dry bed.
• The river deepens its bed by vertical erosion, the meander is stranded above the new channel, the old loop fills with rainwater, a lake forms on the terrace.

Answer: Erosion on the outer bends narrows the neck of the meander, the river cuts through the neck at high flow, deposition seals both ends, the abandoned loop becomes a lake.

The key is lateral erosion on the outer bank of two adjacent bends, which thins the neck between them; the cut-through happens during a flood, and it is deposition at the two ends of the loop that isolates it from the channel.

Bedload pebbles collected 30 km downstream of a river's source are smaller and much rounder than pebbles collected close to the source. Which process is chiefly responsible for that change?

• Abrasion, as the load scrapes and scours the bed and banks of the channel
• Hydraulic action, as water and compressed air force the joints of the bed apart
• Solution, as slightly acidic water dissolves minerals out of the rock in transit
• Attrition, as load particles knock against one another and are worn smooth

Answer: Attrition, as load particles knock against one another and are worn smooth

Abrasion wears away the channel itself, attrition wears away the load; only attrition explains why the transported pebbles become smaller and rounder with distance from the source.

A river channel has a cross-sectional area of 12.5 m² and a mean velocity of 0.84 m s⁻¹. Using Q = A x v, calculate the discharge (to 1 d.p., in m³ s⁻¹). Answer: ___

Answer: 10.5

Q = A x v = 12.5 m² x 0.84 m s⁻¹ = 10.5 m³ s⁻¹. Discharge is a volume per second, so area and velocity are multiplied; adding them is the classic slip in this calculation.