A hydraulic jump must be analysed using the momentum equation rather than the energy equation because

the energy loss across the jump is itself the unknown, so the energy equation cannot be closed

momentum is conserved and energy is not, in any flow

the energy loss across the jump is itself the unknown, so the energy equation cannot be closed

the jump is too short for energy to be defined

the flow is unsteady across the jump

In a subcritical flow in an open channel, a disturbance such as a weir placed downstream will

affect the depth upstream, because the flow is slower than the wave celerity

have no effect upstream, since water flows downhill

affect the depth upstream, because the flow is slower than the wave celerity

affect the depth upstream only if the channel slope is steep

cause a hydraulic jump immediately upstream of itself

On a mild slope, the water surface profile upstream of a dam is classified as

M1, a backwater curve with the depth rising in the downstream direction

S1, since a dam creates a steep slope

Which of the following are true of critical flow in an open channel? (One or more)

The specific energy is a minimum for the given discharge

The critical depth depends on the discharge but not on the bed slope or roughness

The specific energy is a minimum for the given discharge

The critical depth depends on the discharge but not on the bed slope or roughness

A surface disturbance can travel upstream

Open-Channel Hydraulics: Specific Energy, the Hydraulic Jump, Uniform and Gradually Varied Flow

A pipe is full and its flow is set by pressure; an open channel has a free surface, and the depth is not given to you — it is part of the answer. That single difference generates the whole subject. Specific energy, E = y + V²/2g measured from the channel bed, is a function of depth with a minimum, and the depth at which it is minimum is the critical depth y_c = (q²/g)^(1/3) for a rectangular channel. Above y_c the flow is deep and slow, subcritical, with Fr < 1; below it the flow is shallow and fast, supercritical, with Fr > 1; and for every specific energy above the minimum there are two depths that carry the same discharge — the alternate depths. Which one a channel is actually at is decided by the slope and by what lies downstream, and that is the second organising idea: a subcritical flow is controlled from downstream and a supercritical flow from upstream, because a disturbance travels at √(gy) and can only move upstream if the flow is slower than that. When a supercritical flow must become subcritical it cannot do so smoothly, because the two depths do not join along the energy curve; it does so through a hydraulic jump, which is violent, short, and destroys energy — and must therefore be analysed by momentum rather than by energy, since the loss is exactly the unknown. Uniform flow is the equilibrium depth a long prismatic channel settles to under Manning’s V = (1/n) R^(2/3) S^(1/2), and comparing that normal depth with the critical depth classifies the slope as mild or steep. Everything between two such states is gradually varied flow, whose profile is fixed by the sign of dy/dx = (S₀ − S_f)/(1 − Fr²) — two signs, and the whole classification of M, S and C curves falls out of them.

Specific Energy, Critical Depth, and the Two Depths That Share It

Specific energy is the energy per unit weight measured from the channel bed, E = y + V²/2g, which for a rectangular channel carrying a discharge q per unit width becomes E = y + q²/(2gy²). Plot it against depth at fixed q and the curve has two limbs. As y gets large the velocity head vanishes and E → y, so the upper limb approaches the 45° line. As y gets small the velocity head blows up as 1/y², so the lower limb runs away to infinity. Between them the curve turns, and at the turning point dE/dy = 0 gives q² = g y³, that is y_c = (q²/g)^(1/3) — the critical depth, at which the specific energy is a minimum. At q = 4 m³/s per m, y_c = 1.1771 m and E_min = 1.5 y_c = 1.7657 m, and two identities are worth checking once: the Froude number at that depth computes to exactly 1.000000, and the velocity head is exactly y_c/2.

🎯 Two depths, one energy, and the reason a channel picks one

Because the specific energy curve has a minimum, every value of E above E_min is reached at two depths, and both carry the same discharge. At E = 2.5 m with q = 4, solving y + 4²/(2gy²) = 2.5 gives y = 2.3527 m with Fr = 0.354, and y = 0.6670 m with Fr = 2.344. These are the alternate depths: the same water, the same energy, the same flow, in two completely different states — one deep and tranquil, one shallow and shooting. Which one a real channel is in is not decided by energy at all, since energy cannot distinguish them. It is decided by the slope, which fixes the normal depth, and by the control, which is where the depth is imposed from outside. That second idea is the one to hold. A disturbance on a free surface travels at the celerity √(gy), so it can propagate upstream only when the flow is slower than that, which is exactly Fr < 1. So a subcritical flow is controlled from downstream — a weir, a gate, a reservoir at the end of the reach sets the depth and the effect reaches back up the channel — while a supercritical flow is controlled from upstream, because nothing downstream can send a signal back against it. That is why a backwater curve extends upstream of a dam for kilometres, and why the flow below a sluice gate is oblivious to what is downstream until it is forced to notice through a jump. A candidate who knows the two alternate depths but not which end the control sits at can compute every number in the problem and still draw the water surface backwards.

The Hydraulic Jump: Solve It With Momentum, Not Energy

A supercritical flow that has to become subcritical cannot slide along the specific energy curve from one alternate depth to the other, because doing so would require gaining energy on the way. What happens instead is a hydraulic jump: a short, turbulent, air-entraining transition in which the depth rises abruptly and a large part of the energy is destroyed. The analysis has to be done by momentum, not energy — and the reason is exact rather than stylistic. The energy loss is precisely the quantity you do not know, so an energy equation contains two unknowns and cannot be solved; a momentum equation across the jump contains only the pressure forces and the momentum flux, both computable from the depths, so it closes. Equating the specific force on either side of a rectangular channel gives the sequent depth relation y₂/y₁ = ½(√(1 + 8Fr₁²) − 1).

🧠 A jump that destroys a third of the energy arriving at it

Take q = 4 m³/s per m arriving at a depth of y₁ = 0.5 m. The velocity is 8 m/s and Fr₁ = 8/√(9.81 × 0.5) = 3.612, firmly supercritical. The sequent depth is y₂ = 0.5 × 0.5 × (√(1 + 8 × 13.048) − 1) = 2.3164 m, a rise of a factor of 4.63, and downstream Fr₂ = 0.362, subcritical as required. Now count the energy. E₁ = 0.5 + 8²/19.62 = 3.762 m and E₂ = 2.3164 + 1.7268²/19.62 = 2.468 m, so the loss is 1.294 m — 34.4 per cent of the energy that arrived. Check it against the closed form ΔE = (y₂ − y₁)³/(4y₁y₂) = 1.8164³/(4 × 0.5 × 2.3164) = 1.294 m ✓, and the agreement to six figures is the sign that the momentum solution and the energy accounting are consistent. Three things follow. The loss grows very fast with Fr₁, as the cube of the depth difference, so a jump at Fr₁ = 2 dissipates only a few per cent while one at Fr₁ = 9 dissipates over 70 — which is why a stilling basin below a spillway is designed to force the jump to occur inside it, where the energy can be destroyed harmlessly on concrete rather than on the riverbed. The jump is a one-way device: it converts supercritical to subcritical and never the reverse, because the reverse would require creating energy. And the position matters as much as the existence — a jump that forms too far downstream has already scoured what it was meant to protect, which is why the tailwater depth is checked against y₂ rather than assumed adequate.

Uniform Flow, and What Makes a Slope Mild

In a long prismatic channel of constant slope the depth settles to the value at which the component of gravity down the slope exactly balances the boundary resistance. That is the normal depth y_n, and Manning’s equation gives it: V = (1/n) R^(2/3) S^(1/2), with R = A/P the hydraulic radius. Work an example. A rectangular channel 5 m wide flowing 1.5 m deep at n = 0.015 on a slope of 0.001 has A = 7.5 m², P = 8 m and R = 0.9375 m, so V = 2.019 m/s and Q = 15.15 m³/s. Note that n is not dimensionless and carries units that make the formula work only in SI, and that the exponent on R means a channel’s efficiency depends on its shape, not only its area — which is why the most hydraulically efficient rectangular section is one with b = 2y, maximising R for a given area.

The useful step is the comparison that follows. For this channel q = 15.15/5 = 3.029 m³/s per m, so y_c = (3.029²/9.81)^(1/3) = 0.978 m. Since y_n = 1.5 m > y_c = 0.978 m, the normal flow is subcritical — Fr = 0.526 confirms it — and the slope is classified mild. Check it a second, independent way by computing the slope that would make this same channel critical at this same discharge: S_c = 0.003453, or 1 in 290, against the actual 1 in 1000. The actual slope is milder, and the two tests agree. That classification is not vocabulary; it is what determines which gradually varied profiles can exist on the reach, and therefore which way the water surface curves as it approaches any control.

Gradually Varied Flow: Two Signs Give Every Profile

Between a control and the reach’s normal depth the flow is neither uniform nor abrupt: the depth changes gradually over a long distance, and the governing equation is dy/dx = (S₀ − S_f)/(1 − Fr²). Everything about the shape of the water surface is in the signs of that fraction, and reading them is faster than remembering a table of profile names. The numerator compares the bed slope with the friction slope, so it is positive when the flow is deeper than normal — a deeper flow is slower, so it needs less slope to overcome friction. The denominator is positive when the flow is subcritical and negative when it is supercritical. Combine them: two positives or two negatives give a rising surface, one of each gives a falling one.

The three mild-slope profiles, from the signs alone (y_n > y_c)
Profile	Depth range	`S₀ − S_f`	`1 − Fr²`	`dy/dx`	Where you see it
M1	`y > y_n > y_c`	+	+	+ backwater	Upstream of a dam or weir
M2	`y_n > y > y_c`	−	+	− drawdown	Approaching a free overfall
M3	`y_n > y_c > y`	−	−	+ rising	Below a sluice gate, up to a jump

Two boundary behaviours complete the picture and are worth reasoning out rather than memorising. As the depth approaches the normal depth, S_f → S₀, the numerator goes to zero and dy/dx → 0 — the profile becomes asymptotic to the normal depth line, approaching it but never crossing. As the depth approaches the critical depth, Fr → 1, the denominator goes to zero and dy/dx → ∞ — the water surface becomes vertical, which is the equation announcing that the gradually varied assumption has broken down and something abrupt is about to happen. That is exactly where a hydraulic jump or a free overfall appears, and it is why the theory predicts its own limit. The M3 profile in the table shows both ideas working together: water issuing fast and shallow from under a sluice gate rises along an M3 curve until it can no longer stay supercritical, and the transition to the subcritical M-regime happens through a jump rather than through the curve.

Unsteady flow and mobile channels

Everything above assumed the flow was steady and the bed was rigid. The GATE 2027 revision names both assumptions as things to relax. Unsteady flow — a gate closing, a sluice opening, a flood passing — propagates as a wave, and the quantity that governs it is the celerity of a small disturbance in still water, c = √(gy). At a depth of 2 m that is 4.429 m/s; at 1 m, 3.132; at 4 m, 6.264 — so a deeper channel carries information faster, which is why a flood wave outruns the water in it.

🎯 A surge from a closing gate travels upstream, and the Froude number says whether it can

Close a gate in a channel carrying V = 1.5 m/s at a depth of y = 2 m. A surge forms and travels upstream at V + c relative to the bank — or rather, the disturbance moves at c = 4.429 m/s relative to the water, and the water is moving downstream at 1.5, so the wave moving against the flow travels upstream at c − V = 2.929 m/s and the one moving with it at c + V = 5.929 m/s. Both signs are real, and the sign of c − V is the whole content of the Froude number: here Fr = V/c = 1.5/4.429 = 0.339 < 1, the flow is subcritical, and a disturbance can travel both upstream and downstream. Make the flow supercritical, Fr > 1, and V > c: nothing can travel upstream at all, so a gate closing sends no warning to the water approaching it, which is the physical reason a supercritical reach is controlled from upstream and a subcritical one from downstream — the same rule that decided the direction of the gradually-varied-flow computation earlier in this chapter, arrived at from the wave side rather than the energy side. Mobile channels relax the other assumption. Where the bed is erodible the channel adjusts its own geometry until it is in regime with the sediment load it carries — which is the empirical territory of Lacey and Kennedy in the next chapter — and the governing threshold is the critical shear stress at which grains begin to move. The practical consequence is one every canal engineer meets: a channel designed to a fixed section in erodible material does not stay in that section, so it must be designed either to be non-scouring and non-silting at once, or to be lined and take the question away.

Key takeaways

E = y + q²/2gy² has a minimum at y_c = (q²/g)^(1/3), where Fr = 1 exactly and E_min = 1.5 y_c. Above the minimum every E is shared by two alternate depths carrying the same discharge.
A disturbance travels at √(gy), so subcritical flow is controlled from downstream and supercritical from upstream. Get that wrong and every number can be right while the water surface is drawn backwards.
Solve a jump with momentum, because the energy loss is the unknown. y₂/y₁ = ½(√(1+8Fr₁²) − 1); at Fr₁ = 3.61 the depth rises 4.63× and 34% of the incoming energy is destroyed.
Manning gives the normal depth; comparing it with y_c classifies the slope. The worked channel has y_n = 1.5 > y_c = 0.978, so mild — confirmed independently by S_c = 1 in 290 against an actual 1 in 1000.
Every GVF profile falls out of two signs in dy/dx = (S₀ − S_f)/(1 − Fr²). dy/dx → 0 at normal depth (asymptotic) and → ∞ at critical — where the theory announces its own breakdown and a jump or an overfall appears.

Practice questions (8)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

A rectangular channel carries a discharge of 4 m³/s per metre width. The critical depth, in m, is ______ (round to three decimals)
Numerical answer — type the value.
Show answer
Answer: 1.177
1.177 m. For a rectangular channel y_c = (q²/g)^(1/3) = (16/9.81)^(1/3) = (1.6310)^(1/3) = 1.17711 m. Three checks confirm it, and each is worth running once because they are the identities the topic is built on. First, the Froude number at that depth: V_c = q/y_c = 3.398 m/s and Fr = V/√(gy) = 3.398/√(9.81 × 1.177) = 1.000000 — critical depth is defined by Fr = 1, and the cube-root formula is just that condition rearranged. Second, the velocity head at critical is V_c²/2g = 0.5886 m, which is exactly y_c/2 — a relation specific to rectangular channels and a fast way to spot an arithmetic slip. Third, the minimum specific energy is E_min = y_c + y_c/2 = 1.5 y_c = 1.766 m, so no rectangular channel can pass 4 m³/s per metre on less than 1.766 m of specific energy however its depth is arranged. What the critical depth is not is the depth the channel actually flows at. That is the normal depth, fixed by the slope and roughness through Manning, and the comparison between the two is what classifies the slope: y_n > y_c is mild and subcritical, y_n < y_c is steep and supercritical. Note also that y_c depends on the discharge alone for a given channel shape — not on the slope, not on the roughness, not on n. Change the bed slope and y_n moves while y_c stays exactly where it was.
A hydraulic jump must be analysed using the momentum equation rather than the energy equation because
1. momentum is conserved and energy is not, in any flow
2. the energy loss across the jump is itself the unknown, so the energy equation cannot be closed
3. the jump is too short for energy to be defined
4. the flow is unsteady across the jump
Show answer
Answer: B — the energy loss across the jump is itself the unknown, so the energy equation cannot be closed
The energy loss is itself the unknown. Write the energy equation across the jump and it contains y₁, y₂ and h_L, three quantities of which two are unknown — it cannot be solved. Write the momentum equation and it contains only the pressure forces on the two faces and the momentum flux through them, all computable from the depths and the discharge, so it closes and yields the sequent depth y₂/y₁ = ½(√(1 + 8Fr₁²) − 1). Once y₂ is known, the energy equation becomes useful again — not to find the depths but to compute the loss as a result: for y₁ = 0.5 m and q = 4, E₁ = 3.762 m and E₂ = 2.468 m, so h_L = 1.294 m, which is 34.4 per cent of the incoming energy and matches the closed form (y₂ − y₁)³/(4y₁y₂) to six figures. Option A is loosely worded and wrong as stated: momentum is not conserved in general either — it is conserved here only because the external forces over the short length of the jump are small enough to neglect, and on a sloping or a stepped floor that assumption fails and the analysis changes. Option C is false, since specific energy is perfectly well defined on both sides. Option D misdescribes the flow, which is steady in the mean; the turbulence is intense but the jump does not move once it has positioned itself. The general lesson is worth carrying beyond this topic: use the conservation law whose unknown terms you can evaluate, and keep the other one to compute what you were not able to assume.
Water at 4 m³/s per metre width enters a hydraulic jump at a depth of 0.5 m. The sequent (conjugate) depth, in m, is ______ (round to three decimals)
Numerical answer — type the value.
Show answer
Answer: 2.316
2.316 m. First the incoming Froude number: V₁ = q/y₁ = 4/0.5 = 8 m/s and Fr₁ = 8/√(9.81 × 0.5) = 8/2.2147 = 3.612, comfortably supercritical. Then the sequent depth relation, which comes from equating specific force across the jump: y₂ = ½ y₁ (√(1 + 8Fr₁²) − 1) = 0.25 × (√(1 + 104.38) − 1) = 0.25 × 9.266 = 2.3164 m. The depth ratio is 4.63, and downstream Fr₂ = 0.362, subcritical as it must be. The energy accounting is the interesting part and confirms the answer independently: E₁ = 0.5 + 8²/19.62 = 3.762 m, E₂ = 2.3164 + 1.727²/19.62 = 2.468 m, so 1.294 m — 34.4 per cent of the incoming energy — has been destroyed, and the closed form (y₂ − y₁)³/(4y₁y₂) returns the same 1.294 m. That agreement between a momentum-derived depth and an energy-derived loss is the check to run. Two cautions. Sequent depths are not alternate depths: alternate depths share the same specific energy and are what a smooth transition would connect, while sequent depths share the same specific force and are what a jump connects — the jump exists precisely because the flow cannot move between alternate depths without gaining energy. And the loss grows as the cube of (y₂ − y₁), so it climbs steeply with Fr₁: a jump at Fr₁ = 2 dissipates only a few per cent while one at Fr₁ = 9 dissipates over 70, which is why a stilling basin below a spillway is sized to hold the jump where the concrete can take it.
A rectangular channel 5 m wide flows 1.5 m deep with n = 0.015 on a bed slope of 0.001. The discharge, in m³/s, is ______ (round to two decimals)
Numerical answer — type the value.
Show answer
Answer: 15.15
15.15 m³/s. A = 5 × 1.5 = 7.5 m², the wetted perimeter is P = 5 + 2(1.5) = 8 m — the free surface is not wetted and is never included — so R = A/P = 0.9375 m. Manning gives V = (1/n)R^(2/3)S^(1/2) = (1/0.015)(0.9375)^(2/3)(0.001)^(1/2) = 66.667 × 0.9579 × 0.03162 = 2.019 m/s, and Q = AV = 15.15 m³/s. The wetted perimeter is where most marks are lost: including the water surface gives P = 11, R = 0.68 and an answer 12 per cent low. The step worth taking next is the classification. Here q = 15.15/5 = 3.029 m³/s per m, so y_c = (3.029²/9.81)^(1/3) = 0.978 m. Since y_n = 1.5 m > y_c = 0.978 m, the flow is subcritical — Fr = 2.019/√(9.81 × 1.5) = 0.526 agrees — and the slope is mild. Confirm it a second way by asking what slope would make this same channel critical at this same discharge: S_c = 0.003453, or 1 in 290, against the actual 1 in 1000, so the actual bed is milder than critical and the two tests agree. That classification is not terminology; it fixes which gradually varied profiles can exist on the reach and therefore which way the surface curves near any control. Two properties of Manning worth carrying: n is not dimensionless, so the formula as written holds only in SI units, and the R^(2/3) means efficiency depends on shape — the most efficient rectangular section has b = 2y, maximising R for a given area.
In a subcritical flow in an open channel, a disturbance such as a weir placed downstream will
1. have no effect upstream, since water flows downhill
2. affect the depth upstream, because the flow is slower than the wave celerity
3. affect the depth upstream only if the channel slope is steep
4. cause a hydraulic jump immediately upstream of itself
Show answer
Answer: B — affect the depth upstream, because the flow is slower than the wave celerity
It affects the depth upstream, because the flow is slower than the wave celerity. A small surface disturbance travels relative to the water at the celerity c = √(gy). In a subcritical flow V < c, that is Fr < 1, so the disturbance can propagate upstream against the current, and the influence of a downstream weir, gate or reservoir reaches back up the channel — sometimes for kilometres, as the M1 backwater curve above a dam. That is the whole meaning of the statement that subcritical flow is controlled from downstream. In supercritical flow V > c, no signal can travel upstream, and the flow is controlled from upstream instead — which is why the water leaving a sluice gate is oblivious to what lies ahead of it until it is forced to notice through a jump. Option A appeals to an intuition that is simply wrong: what travels upstream is not water but information, a pressure wave on the free surface, and it can travel against a current slower than itself. Option C reverses the situation, since a steep slope means supercritical flow and therefore no upstream influence. Option D describes what happens in the opposite case — a jump forms where a supercritical flow is forced to become subcritical, not where a subcritical flow meets an obstruction. The practical value of getting this right is that it tells you which end to start the calculation from: a gradually varied flow profile is integrated upstream from the control in subcritical flow and downstream from it in supercritical, and starting at the wrong end produces a profile that diverges instead of settling to the normal depth.
On a mild slope, the water surface profile upstream of a dam is classified as
1. M1, a backwater curve with the depth rising in the downstream direction
2. M2, a drawdown curve
3. M3, rising from below critical
4. S1, since a dam creates a steep slope
Show answer
Answer: A — M1, a backwater curve with the depth rising in the downstream direction
M1, a backwater curve. Work it from the signs rather than from a remembered picture. A dam raises the depth above the normal depth, so the reach upstream of it has y > y_n > y_c — that is the M1 zone. Now read dy/dx = (S₀ − S_f)/(1 − Fr²). The flow is deeper than normal, so it is slower than normal, so the friction slope S_f is less than the bed slope S₀ and the numerator is positive. The flow is subcritical, so Fr < 1 and the denominator is positive too. Hence dy/dx > 0: the depth increases in the downstream direction, which is exactly what a backwater curve looks like — the surface rises as it approaches the dam. Option B is the M2 zone, y_n > y > y_c, which occurs where a mild channel approaches a free overfall and the depth is drawn down towards critical. Option C is the M3 zone, below critical, which is what happens under a sluice gate on a mild slope before a jump. Option D confuses the structure with the bed: a dam does not change the slope of the channel it sits in, and the S-profiles belong to channels whose bed slope is steep enough that the normal depth is supercritical. Two limiting behaviours make the M1 curve easy to sketch. Far upstream y → y_n, so S_f → S₀, the numerator vanishes and the profile becomes asymptotic to the normal depth line. At the dam the surface approaches the horizontal, since the velocity there is nearly zero. Between the two it is a smooth rising curve, and its length — often several kilometres — is why reservoir backwater is computed rather than guessed.
For a rectangular channel, the minimum specific energy expressed as a multiple of the critical depth is ______
Numerical answer — type the value.
Show answer
Answer: 1.5
1.5. At critical depth Fr = 1, so V_c = √(g y_c) and the velocity head is V_c²/2g = g y_c/2g = y_c/2. Then E_min = y_c + y_c/2 = 1.5 y_c. The relation is specific to a rectangular channel — for a general section the critical condition is Q²T/(gA³) = 1 with T the top width, and the multiple is not 1.5. Check it on the worked example: q = 4 m³/s per m gives y_c = 1.177 m and E_min = 1.766 m, and 1.766/1.177 = 1.500 ✓. Two things this result is good for. It gives an instant lower bound: no rectangular channel can pass a given q on less specific energy than 1.5 y_c, whatever depth it flows at, so if a transition offers less than that, the flow must choke and back up until it has enough. That is exactly what happens at a broad-crested weir or a raised bed — the flow passes through critical at the control and the upstream depth rises to supply 1.5 y_c plus the step. And it makes the specific energy curve easy to sketch: the minimum sits at (1.5 y_c, y_c), the upper limb approaches the 45° line asymptotically as the velocity head vanishes, and the lower limb runs to infinity as y → 0 because the velocity head blows up as 1/y². The one-third/two-thirds split is worth noting too — at critical depth the flow carries two-thirds of its specific energy as depth and one-third as velocity head, which is the most efficient partition available and the reason the condition is called critical.
Which of the following are true of critical flow in an open channel? (One or more)
1. The specific energy is a minimum for the given discharge
2. The Froude number equals 1
3. The critical depth depends on the discharge but not on the bed slope or roughness
4. A surface disturbance can travel upstream
Show answer
Answer: A — The specific energy is a minimum for the given discharge; B — The Froude number equals 1; C — The critical depth depends on the discharge but not on the bed slope or roughness
A, B and C. These are three descriptions of the same condition. A is the definition by which y_c is usually derived: setting dE/dy = 0 on E = y + q²/2gy² gives q² = gy³, hence y_c = (q²/g)^(1/3). B is the same condition rearranged — q² = gy³ is V²y² = gy³, that is V² = gy and Fr = 1 — and the two are equivalent statements, not separate facts. C is the one that catches people, and it matters. The critical depth follows from the discharge and the channel shape alone: no n and no S₀ appear anywhere in y_c = (q²/g)^(1/3). So changing the bed slope moves the normal depth while leaving y_c exactly where it was, and that is precisely why comparing the two classifies the slope as mild or steep. It also means y_c is the right reference against which to judge any change in a channel, since it stays put while everything else moves. D is false, and it is the trap. At critical flow V = c = √(gy), so a disturbance travelling upstream at the celerity is exactly held stationary by the current — it neither advances nor is swept away. Upstream propagation requires V < c, that is subcritical flow. This is why critical flow is a control: a section forced to be critical, at a weir crest or a raised bed, imposes a unique relation between depth and discharge and separates the upstream reach from the downstream one, which is what makes such a section a reliable place to measure flow.