Which equation states conservation of mass for an incompressible fluid, whether the flow is steady or unsteady?

∂u/∂t + u ∂u/∂x = 0, i.e. the acceleration of every particle is zero

∇ × V = 0, i.e. the flow has no rotation anywhere

∂ρ/∂t = 0, i.e. the density field is steady at every point

For a plane incompressible flow with stream function ψ, a curve along which ψ is constant is

a line along which no fluid crosses, a streamline

a line of constant pressure, along which the speed also stays unchanged

a line of constant vorticity, joining the points of equal fluid rotation

a line along which no fluid crosses, a streamline

a line of constant speed, perpendicular to the direction of flow

Steady inviscid water flows round a horizontal bend, with circular streamlines of radius R and speed V. Moving radially outward across the streamlines, the pressure

increases, because the net inward force needed for the turning is supplied by this pressure gradient

decreases, because the speed is higher near the outer wall of the bend and Bernoulli then lowers the pressure there

is unchanged, as in any horizontal flow, because gravity does not act across the bend

increases, because the net inward force needed for the turning is supplied by this pressure gradient

first increases, then decreases, passing through a maximum at the mid-radius

A steady, incompressible, inviscid flow has non-zero vorticity. Points 1 and 2 lie on different streamlines, with no pump or turbine in the flow. The quantity p + ½ρV² + ρgz at the two points is

generally different, since it is constant only along each streamline unless the flow is irrotational

equal, because the fluid is inviscid and no energy is added or removed anywhere in the flow at all

equal only if the two points are at the same elevation and have the same pressure

generally different, since it is constant only along each streamline unless the flow is irrotational

always larger at the point with the larger speed, whatever the elevations and pressures

A siphon of constant bore draws water from a large open reservoir. Treat the water as ideal. The speed of the water leaving the outlet is set by

the vertical drop from the reservoir surface down to the outlet

the height of the summit of the tube above the reservoir surface, since the water must be lifted over it

the vertical drop from the reservoir surface down to the outlet

the total length of the tube, since a longer tube holds more water

the sum of the summit height and the vertical drop, the total path travelled

Among the venturi meter, the orifice meter and the Pitot-static tube, the largest permanent pressure loss in the line is caused by

the orifice meter, because the jet downstream of the plate mixes violently

the venturi meter, because of its long diffuser and the friction along it

the Pitot-static tube, because it brings the fluid to rest at its nose

all three equally, since the same flow passes through each of them

the orifice meter, because the jet downstream of the plate mixes violently

The equation p + ½ρV² + ρgz = constant is applied between two points on one streamline. Which conditions must hold for it to be valid?

viscous losses between the points are negligible

no pump or turbine does work on the fluid between the points

viscous losses between the points are negligible

no pump or turbine does work on the fluid between the points

the flow is irrotational everywhere in the field

For a steady plane incompressible flow described by a stream function ψ defined by u = ∂ψ/∂y and v = −∂ψ/∂x, which statements are correct?

The continuity equation is satisfied automatically

The volume flow per unit depth between two streamlines equals the difference of their ψ values in magnitude

ψ can be defined for a rotational flow as well as an irrotational one

The continuity equation is satisfied automatically

The volume flow per unit depth between two streamlines equals the difference of their ψ values in magnitude

ψ can be defined for a rotational flow as well as an irrotational one

ψ satisfies Laplace’s equation ∇²ψ = 0 in every such flow

Which statements about flow-measuring devices are correct?

A well-made venturi has a discharge coefficient close to 1, typically 0.95 to 0.99

A sharp-edged orifice has a discharge coefficient well below 1 because the jet contracts to a vena contracta

A Pitot-static tube gives the local velocity at its nose, so a traverse is needed to find the mean velocity in a pipe

A well-made venturi has a discharge coefficient close to 1, typically 0.95 to 0.99

A sharp-edged orifice has a discharge coefficient well below 1 because the jet contracts to a vena contracta

A Pitot-static tube gives the local velocity at its nose, so a traverse is needed to find the mean velocity in a pipe

The flow rate through a venturi is directly proportional to the manometer pressure difference

Differential Analysis: Continuity, Euler and Bernoulli Equations, Venturi, Pitot-Static Tube and Orifice Meter

This chapter answers the first half of Section XE1.5 of XE1 Fluid Mechanics, one of the nine optional sections of the Engineering Sciences (XE) paper, any two of which a candidate chooses in the exam: “Differential equations of mass and momentum for incompressible flows, Euler equation, Bernoulli equation and its application for venturi meter, Pitot-static tube, and Orifice meter.” It teaches the differential continuity equation and the stream function, the differential momentum equation and its inviscid form (the Euler equation), the derivation of the Bernoulli equation along a streamline with every restriction, the pressure terms (static, dynamic, stagnation, piezometric), and the three flow-measuring devices. The numericals are missing velocity components, flow between streamlines, jet and siphon speeds and pressures, and flow rates and velocities from venturi, orifice and Pitot-static readings.

1. Differential continuity equation and the stream function

Apply mass conservation to a fixed box dx dy dz. The net mass leaving per second is [∂(ρu)/∂x + ∂(ρv)/∂y + ∂(ρw)/∂z] dx dy dz and it must equal the rate of fall of the mass inside, so ∂ρ/∂t + ∇ · (ρV) = 0. For an incompressible fluid ρ is constant, and the equation becomes ∂u/∂x + ∂v/∂y + ∂w/∂z = 0, i.e. ∇ · V = 0. It holds for steady and unsteady flow alike: it says that the volume of a fluid element does not change as it moves. In cylindrical coordinates it reads (1/r) ∂(r v_r)/∂r + (1/r) ∂v_θ/∂θ + ∂v_z/∂z = 0.

Using it. To test a field, compute ∇ · V. To find a missing component, integrate: if u = 2xy and v = 0 on y = 0 in a plane incompressible flow, then ∂v/∂y = −∂u/∂x = −2y, so v = −y². In steady compressible flow the condition is ∇ · (ρV) = 0, not ∇ · V = 0, and ∇ · V = 0 says nothing about rotation: a flow can be incompressible and rotational.

Stream function. For a plane incompressible flow define ψ(x, y) by u = ∂ψ/∂y, v = −∂ψ/∂x. Then ∂u/∂x + ∂v/∂y = ∂²ψ/∂x∂y − ∂²ψ/∂y∂x = 0, so continuity is satisfied identically. Along a curve ψ = constant, dψ = −v dx + u dy = 0, so dy/dx = v/u: lines of constant ψ are streamlines. The volume flow per unit depth between two streamlines is |ψ₂ − ψ₁|, and where streamlines crowd together the speed is higher. Worked line: u = 2xy + y², v = −y² gives ψ = xy² + y³/3; between the streamlines through (1, 1) and (2, 1), ψ = 4/3 and 7/3, so the flow is 1.0 m²/s per metre depth.

⚠️ The sign convention of ψ is not universal

Some books write u = −∂ψ/∂y and v = ∂ψ/∂x. Magnitudes of flow between streamlines are unaffected, but the sign of ψ and of a quoted flow is not. Always use the convention in the question; here u = ∂ψ/∂y and v = −∂ψ/∂x.

2. Differential momentum equation and the Euler equation

Newton’s second law for a fluid element of volume dV states: mass × acceleration = net force. The forces per unit volume are the pressure force −∇p, the weight ρg, and the net viscous force. The acceleration of a particle is DV/Dt = ∂V/∂t + (V · ∇)V, the local plus the convective part. So ρ DV/Dt = ρg − ∇p + (viscous term). In the x direction: ρ(∂u/∂t + u ∂u/∂x + v ∂u/∂y + w ∂u/∂z) = ρg_x − ∂p/∂x + (viscous). The viscous term is developed in the next chapter.

Euler equation. Where viscous forces are negligible (outside boundary layers and wakes, or for an ideal fluid), ρ DV/Dt = ρg − ∇p. Setting V = 0 returns the hydrostatic law ∇p = ρg, a useful check. In streamline coordinates, with s along the flow and r measured outward from the centre of curvature (radius R), for steady flow:
along the flow: ρ V ∂V/∂s = −∂(p + ρgz)/∂s
across the flow: ∂(p + ρgz)/∂r = ρV²/R

Read the second relation physically: fluid going round a bend needs a net inward force, and only a pressure that rises outward can supply it. For straight, parallel streamlines R → ∞, so p + ρgz is constant across the stream, the hydrostatic law. Worked line: water at 2 m/s round a horizontal bend of streamline radius 0.5 m has ∂p/∂r = 1000 × 2²/0.5 = 8000 Pa/m, i.e. 800 Pa higher on a streamline 0.1 m further out.

3. Bernoulli equation along a streamline and its restrictions

Take the along-the-flow Euler relation for steady flow, multiply by ds and use V dV = d(V²/2): dp + ρ V dV + ρ g dz = 0. With ρ constant, integration gives p + ½ρV² + ρgz = constant along a streamline, or dividing by ρg, p/ρg + V²/2g + z = H, a sum of pressure head, velocity head and elevation. Between two points: p₁ + ½ρV₁² + ρgz₁ = p₂ + ½ρV₂² + ρgz₂. Absolute or gauge pressure may be used if the same one is used at both points.

Restrictions: steady flow; incompressible fluid (for gas, Mach number below about 0.3); inviscid, so no friction losses; along one streamline; no shaft work or heat transfer between the points; gravity the only body force.
The constant changes from streamline to streamline unless the flow is irrotational, when one constant serves the whole field. For unsteady flow a term ρ ∫(∂V/∂t) ds is added.
Where it fails: across a pump or turbine (energy is added or removed), through long pipes, valves and sudden expansions (viscous losses), and between streamlines of a rotational flow. There the energy equation with a pump head and a head-loss term takes over (internal flows, Section XE1.7).

Worked example: Torricelli and siphon. A large open tank with a small orifice at depth h below the free surface: surface and jet are both at atmospheric pressure, the surface speed is negligible, so V = √(2gh) = √(2 × 9.81 × 2) = 6.26 m/s for h = 2 m. If the tank air is at gauge pressure p, V = √(2(p/ρ + gh)). For a siphon with its summit 1.5 m above the reservoir surface and its outlet 2.5 m below it, the outlet speed is V = √(2 × 9.81 × 2.5) = 7.00 m/s. Bernoulli from the surface to the summit (same V as in the tube, constant area) gives p_top = −ρ(V²/2 + g × 1.5) = −1000 × (24.5 + 14.7) = −39.2 kPa gauge. The outlet depth sets the speed, and the summit height sets how low the pressure drops.

4. Static, dynamic, stagnation and piezometric pressure

Pressure terms in the Bernoulli equation
Pressure	Expression	How it appears or is measured
Static	p	Thermodynamic pressure felt by a probe moving with the fluid; read at a wall tap flush with the surface or at the side holes of a static tube
Dynamic	½ρV²	Kinetic energy per unit volume; the difference p₀ − p
Stagnation	p₀ = p + ½ρV²	Pressure where the fluid is brought to rest without loss, at the nose of a Pitot tube (elevation change ignored)
Piezometric	p + ρgz	What a liquid column in an open tube or a manometer connected to a wall tap indicates
Total	p + ½ρV² + ρgz	Constant along a streamline in the Bernoulli conditions

Worked line. Air (ρ = 1.2 kg/m³) at 25 m/s has a dynamic pressure of ½ × 1.2 × 25² = 375 Pa, so a Pitot tube in it senses 375 Pa above the static pressure. Water at 4 m/s has ½ × 1000 × 16 = 8 kPa; if its static pressure is 120 kPa gauge, the stagnation pressure on the nose of an immersed body is 128 kPa. At a stagnation point V = 0, so the pressure there is the highest along that streamline.

🎯 Speed and pressure trade off because of the force balance

Pressure does not drop because speed is high; the fluid speeds up because pressure falls ahead of it. In a converging nozzle the fluid is pushed forward by higher pressure behind it, and Bernoulli records that trade. Gravity enters the same way, as ρgz. This is why a manometer on a horizontal pipe reads static pressure, but one on a vertical pipe also includes ρgz.

5. Venturi meter and Pitot-static tube

Venturi meter. A pipe narrows smoothly to a throat and then widens gradually; taps at the inlet (area A₁) and the throat (A₂) feed a differential manometer. Continuity gives V₂ = V₁A₁/A₂, and Bernoulli between the taps gives the ideal flow Q_ideal = A₂ √[ 2Δp* / (ρ(1 − (A₂/A₁)²)) ], with Δp* = (p₁ + ρgz₁) − (p₂ + ρgz₂). The real flow is Q = C_d Q_ideal, where the discharge coefficient C_d (a little below 1, about 0.95–0.99 for a good venturi) allows for friction. A manometer fluid of density ρ_m reads Δp* = (ρ_m − ρ) g h, whatever the inclination of the meter, because the manometer measures the piezometric difference directly.

Worked example. Water in a horizontal venturi, inlet 0.2 m, throat 0.1 m, mercury (ρ_m = 13,600 kg/m³) manometer h = 0.2 m, C_d = 0.98. Δp = 12,600 × 9.81 × 0.2 = 24,721 Pa; A₂ = 7.854 × 10⁻³ m²; A₂/A₁ = 0.25, so (A₂/A₁)² = 0.0625. V₂,ideal = √[2 × 24,721/(1000 × 0.9375)] = 7.262 m/s; Q_ideal = 0.05704 m³/s; Q = 0.98 × 0.05704 = 0.0559 m³/s = 55.9 L/s.

Pitot-static tube. Two concentric tubes: the central one faces the flow and senses p₀, the ring of side holes senses p. Bernoulli along the streamline to the stagnation point gives p₀ − p = ½ρV², so V = √(2Δp/ρ), and with a manometer V = √[2 g h (ρ_m/ρ − 1)]. It gives the local velocity at the nose, so it must be aligned with the flow and traversed across a pipe to find the average. Worked line: a mercury manometer reading 50 mm in water gives V = √(2 × 9.81 × 0.05 × 12.6) = 3.52 m/s.

⚠️ Do not add the elevation head to a manometer reading

In an inclined or vertical venturi the taps are at different heights, but the manometer already registers (p₁ + ρgz₁) − (p₂ + ρgz₂) = (ρ_m − ρ) g h. Use that directly in Q_ideal. Adding ρg Δz again double-counts the elevation. For a Pitot tube in a liquid, use ρ_m/ρ − 1, not ρ_m/ρ, which is only acceptable when the flowing fluid is a gas.

6. Orifice meter and comparison of the three devices

Orifice meter. A thin plate with a sharp-edged hole of diameter d_o in a pipe of diameter D. The flow contracts to a vena contracta whose area is smaller than the hole’s, so C_d is well below 1 (about 0.6 for a sharp-edged orifice). With β = d_o/D the working formula is Q = C_d A_o √(2Δp/ρ) / √(1 − β⁴). The plate is cheap and compact but the jet mixes violently downstream, so the permanent pressure loss is the largest of the three devices.

Worked example. An orifice of 50 mm in a 100 mm pipe (β = 0.5, β⁴ = 0.0625) with C_d = 0.62 and Δp = 20 kPa for water: A_o = 1.9635 × 10⁻³ m², √(2Δp/ρ) = 6.325 m/s, √(1 − β⁴) = 0.9682. Q = 0.62 × 1.9635 × 10⁻³ × 6.325/0.9682 = 7.95 × 10⁻³ m³/s = 7.95 L/s. A larger β makes the denominator smaller, so the same flow gives a smaller pressure difference (a weaker signal), and the orifice then contracts less so the correction gets harder to predict.

Device	Measures	Permanent loss	Remark
Venturi	Flow rate	Small, the diffuser recovers pressure	Costly and long, C_d near 1
Orifice	Flow rate	Largest	Cheap, vena contracta, C_d about 0.6
Pitot-static	Local velocity	Negligible	Point reading, align with flow

Key takeaways

Incompressible continuity is ∇ · V = 0, valid for steady and unsteady flow. Steady compressible flow needs ∇ · (ρV) = 0. For plane flow ψ is defined by u = ∂ψ/∂y, v = −∂ψ/∂x; ψ = constant lines are streamlines and |Δψ| is the flow per unit depth.
The momentum equation is ρ DV/Dt = ρg − ∇p + viscous terms, and without viscous terms it is the Euler equation. Across curved streamlines ∂(p + ρgz)/∂r = ρV²/R; across straight ones p + ρgz is constant.
Bernoulli: p + ½ρV² + ρgz is constant along a streamline for steady, incompressible, frictionless flow with no shaft work. It fails across pumps and in lossy flow, and the constant differs between streamlines unless the flow is irrotational.
Stagnation pressure is p₀ = p + ½ρV². Torricelli speed is √(2gh), and a siphon’s speed is set by the outlet depth while its minimum pressure is set by the summit height.
Venturi: Q = C_d A₂ √[2Δp/(ρ(1 − (A₂/A₁)²))] with Δp = (ρ_m − ρ) g h independent of inclination. Orifice: Q = C_d A_o √(2Δp/ρ)/√(1 − β⁴). Pitot-static: V = √(2Δp/ρ) for local velocity.
Of the three devices the orifice has the largest permanent loss and the Pitot-static tube the smallest; C_d is near 1 for a venturi and about 0.6 for a sharp orifice. Flow rate is proportional to √(manometer reading).

Practice questions (16)

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.

Which equation states conservation of mass for an incompressible fluid, whether the flow is steady or unsteady?
1. ∂u/∂x + ∂v/∂y + ∂w/∂z = 0, i.e. ∇ · V = 0
2. ∂u/∂t + u ∂u/∂x = 0, i.e. the acceleration of every particle is zero
3. ∇ × V = 0, i.e. the flow has no rotation anywhere
4. ∂ρ/∂t = 0, i.e. the density field is steady at every point
Show answer
Answer: A — ∂u/∂x + ∂v/∂y + ∂w/∂z = 0, i.e. ∇ · V = 0
For constant ρ the general form ∂ρ/∂t + ∇ · (ρV) = 0 reduces to ∇ · V = 0, valid for steady and unsteady flow because it only says that a fluid element keeps its volume. ∇ × V = 0 is the irrotationality condition, a different property; incompressible flow can be rotational. The acceleration-free statement and the steady-density statement are not continuity.
In a plane incompressible flow the x-velocity is u = 2x + y (SI units) and v = 0 on the line y = 0. The y-velocity field is
1. v = 2y
2. v = −2y
3. v = −2y − x
4. v = −y/2
Show answer
Answer: B — v = −2y
Continuity gives ∂v/∂y = −∂u/∂x = −2, so v = −2y + f(x). The condition v = 0 on y = 0 for all x forces f(x) = 0. The positive sign would make the flow expand, which an incompressible fluid cannot do; adding −x violates the condition on y = 0.
For a plane incompressible flow with stream function ψ, a curve along which ψ is constant is
1. a line of constant pressure, along which the speed also stays unchanged
2. a line of constant vorticity, joining the points of equal fluid rotation
3. a line along which no fluid crosses, a streamline
4. a line of constant speed, perpendicular to the direction of flow
Show answer
Answer: C — a line along which no fluid crosses, a streamline
On ψ = constant, dψ = −v dx + u dy = 0, so dy/dx = v/u, the slope of the velocity vector: the velocity is tangent to the curve and no fluid crosses it. Pressure, vorticity and speed are in general not constant along a streamline.
Steady inviscid water flows round a horizontal bend, with circular streamlines of radius R and speed V. Moving radially outward across the streamlines, the pressure
1. decreases, because the speed is higher near the outer wall of the bend and Bernoulli then lowers the pressure there
2. is unchanged, as in any horizontal flow, because gravity does not act across the bend
3. increases, because the net inward force needed for the turning is supplied by this pressure gradient
4. first increases, then decreases, passing through a maximum at the mid-radius
Show answer
Answer: C — increases, because the net inward force needed for the turning is supplied by this pressure gradient
The Euler equation across the streamlines gives ∂p/∂r = ρV²/R > 0: the pressure rises outward and supplies the centripetal force. The statement that pressure is unchanged holds only for straight parallel streamlines (R → ∞).
A steady, incompressible, inviscid flow has non-zero vorticity. Points 1 and 2 lie on different streamlines, with no pump or turbine in the flow. The quantity p + ½ρV² + ρgz at the two points is
1. equal, because the fluid is inviscid and no energy is added or removed anywhere in the flow at all
2. equal only if the two points are at the same elevation and have the same pressure
3. generally different, since it is constant only along each streamline unless the flow is irrotational
4. always larger at the point with the larger speed, whatever the elevations and pressures
Show answer
Answer: C — generally different, since it is constant only along each streamline unless the flow is irrotational
Integrating the Euler equation along a streamline gives a constant for that streamline only; between streamlines the constant changes by an amount tied to the vorticity. Only for irrotational flow does one constant hold over the whole field. Being inviscid is not enough, and equal elevation does not remove the dependence on streamline.
A siphon of constant bore draws water from a large open reservoir. Treat the water as ideal. The speed of the water leaving the outlet is set by
1. the height of the summit of the tube above the reservoir surface, since the water must be lifted over it
2. the vertical drop from the reservoir surface down to the outlet
3. the total length of the tube, since a longer tube holds more water
4. the sum of the summit height and the vertical drop, the total path travelled
Show answer
Answer: B — the vertical drop from the reservoir surface down to the outlet
Bernoulli from the free surface (atmospheric pressure, negligible speed) to the outlet (atmospheric pressure) gives V = √(2gΔz), with Δz the drop to the outlet. The summit height does not enter the speed; it only fixes how low the pressure falls at the summit. Length matters only through friction, which the ideal-fluid assumption removes.
A Pitot-static tube in a water stream gives a manometer difference that is doubled when the stream speed is changed. By what factor did the speed change?
1. 2
2. 4
3. √2
4. ½
Show answer
Answer: C — √2
Δp = p₀ − p = ½ρV², so V ∝ √Δp. Doubling Δp raises V by √2 ≈ 1.41. A factor of 2 would apply if Δp were proportional to V, and a factor of 4 if V were proportional to Δp².
Among the venturi meter, the orifice meter and the Pitot-static tube, the largest permanent pressure loss in the line is caused by
1. the venturi meter, because of its long diffuser and the friction along it
2. the Pitot-static tube, because it brings the fluid to rest at its nose
3. all three equally, since the same flow passes through each of them
4. the orifice meter, because the jet downstream of the plate mixes violently
Show answer
Answer: D — the orifice meter, because the jet downstream of the plate mixes violently
The sharp-edged plate forces a vena contracta and then a poorly controlled expansion, so most of the throat pressure drop is lost. The venturi’s diffuser recovers most of it, and the Pitot-static tube is a thin probe that disturbs the flow very little.
Water flows upward through a vertical venturi meter. Pressure tap 2 (throat) is 0.5 m above tap 1 (inlet). A mercury manometer (density 13,600 kg/m³) connected to the two taps, with the connecting lines full of water (1000 kg/m³), reads h = 0.1 m. Take g = 9.81 m/s². The static pressure difference p₁ − p₂ is closest to
1. 7.5 kPa
2. 13.3 kPa
3. 12.4 kPa
4. 17.3 kPa
Show answer
Answer: D — 17.3 kPa
The manometer reads the piezometric difference (p₁ + ρgz₁) − (p₂ + ρgz₂) = (ρ_m − ρ) g h = 12,600 × 9.81 × 0.1 = 12.36 kPa. Then p₁ − p₂ = 12.36 kPa + ρg(z₂ − z₁) = 12.36 + 4.91 = 17.27 kPa. The value 12.4 kPa is the piezometric difference mistaken for the static difference, 7.5 kPa subtracts the elevation term, and 13.3 kPa uses ρ_m g h and forgets the water in the lines.
The equation p + ½ρV² + ρgz = constant is applied between two points on one streamline. Which conditions must hold for it to be valid?
1. the flow is steady
2. viscous losses between the points are negligible
3. no pump or turbine does work on the fluid between the points
4. the flow is irrotational everywhere in the field
Show answer
Answer: A — the flow is steady; B — viscous losses between the points are negligible; C — no pump or turbine does work on the fluid between the points
Steadiness, absence of friction and absence of shaft work are all needed (together with constant density). Irrotationality is not required along a single streamline; it is only needed to use one constant across different streamlines.
For a steady plane incompressible flow described by a stream function ψ defined by u = ∂ψ/∂y and v = −∂ψ/∂x, which statements are correct?
1. The continuity equation is satisfied automatically
2. The volume flow per unit depth between two streamlines equals the difference of their ψ values in magnitude
3. ψ can be defined for a rotational flow as well as an irrotational one
4. ψ satisfies Laplace’s equation ∇²ψ = 0 in every such flow
Show answer
Answer: A — The continuity equation is satisfied automatically; B — The volume flow per unit depth between two streamlines equals the difference of their ψ values in magnitude; C — ψ can be defined for a rotational flow as well as an irrotational one
Substituting the definition into ∂u/∂x + ∂v/∂y gives zero identically, |Δψ| is the flow between streamlines, and nothing in the definition needs irrotationality. ∇²ψ = −ω_z, so Laplace’s equation holds only when the vorticity is zero.
Which statements about flow-measuring devices are correct?
1. A well-made venturi has a discharge coefficient close to 1, typically 0.95 to 0.99
2. A sharp-edged orifice has a discharge coefficient well below 1 because the jet contracts to a vena contracta
3. A Pitot-static tube gives the local velocity at its nose, so a traverse is needed to find the mean velocity in a pipe
4. The flow rate through a venturi is directly proportional to the manometer pressure difference
Show answer
Answer: A — A well-made venturi has a discharge coefficient close to 1, typically 0.95 to 0.99; B — A sharp-edged orifice has a discharge coefficient well below 1 because the jet contracts to a vena contracta; C — A Pitot-static tube gives the local velocity at its nose, so a traverse is needed to find the mean velocity in a pipe
The first three are standard. The flow rate varies as √Δp, so doubling the flow needs four times the pressure difference; direct proportionality is wrong.
A plane incompressible flow has the stream function ψ = 3x²y − y³ (ψ in m²/s, x and y in m), with u = ∂ψ/∂y and v = −∂ψ/∂x. The volume flow per metre depth between the streamlines through (1, 1) and (2, 1), in m²/s, is ____.
Numerical answer — type the value.
Show answer
Answer: 9
ψ(1, 1) = 3 − 1 = 2 and ψ(2, 1) = 3 × 4 × 1 − 1 = 11. The flow is |11 − 2| = 9 m²/s per metre depth. A check on the field: u = 3x² − 3y², v = −6xy, and ∂u/∂x + ∂v/∂y = 6x − 6x = 0.
A closed tank holds water to a depth of 3 m above a small sharp opening in its side. The air above the water is at a gauge pressure of 50 kPa, and the jet leaves into the atmosphere. Take ρ = 1000 kg/m³ and g = 9.81 m/s², treat the flow as ideal and the surface speed as negligible. The jet speed, in m/s, to two decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 12.60
Bernoulli from the surface (gauge p = 50 kPa, V ≈ 0, height 3 m above the opening) to the jet (gauge p = 0): V = √(2(p/ρ + gh)) = √(2 × (50 + 29.43)) = √158.86 = 12.60 m/s. Ignoring the air pressure gives √(2gh) = 7.67 m/s.
Water (1000 kg/m³) flows through a horizontal venturi meter of inlet diameter 150 mm and throat diameter 75 mm. The pressure difference between the inlet and the throat is 30 kPa and the discharge coefficient is 0.97. The volume flow rate, in litres per second, to one decimal place, is ____.
Numerical answer — type the value.
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Answer: 34.3
A₂ = π(0.075)²/4 = 4.418 × 10⁻³ m²; (A₂/A₁)² = (0.5)⁴ = 0.0625. Q_ideal = A₂ √[2Δp/(ρ(1 − 0.0625))] = 4.418 × 10⁻³ × √(60,000/937.5) = 4.418 × 10⁻³ × 8 = 0.03534 m³/s. Q = 0.97 × 0.03534 = 0.03428 m³/s = 34.3 L/s. Dropping the (1 − (A₂/A₁)²) factor gives 33.2 L/s.
A Pitot-static tube in an air duct (air density 1.2 kg/m³, treated as incompressible) is connected to a water manometer (1000 kg/m³) that reads 25 mm. Take g = 9.81 m/s². The air speed at the tube, in m/s, to one decimal place, is ____.
Numerical answer — type the value.
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Answer: 20.2
Δp = (ρ_w − ρ_air) g h = 998.8 × 9.81 × 0.025 = 244.97 Pa (using ρ_w g h = 245.25 Pa changes nothing at this precision). V = √(2Δp/ρ_air) = √(2 × 244.97/1.2) = √408.3 = 20.2 m/s. Using the water density as if it were the flowing fluid would give about 0.7 m/s.