The incompressible Navier-Stokes equation ρ DV/Dt = ρg − ∇p + μ∇²V is obtained for a fluid that is

Newtonian, of constant density and constant viscosity

inviscid and of constant density

compressible, with viscosity that varies strongly with temperature

non-Newtonian, with a shear-rate-dependent viscosity

Newtonian, of constant density and constant viscosity

A Newtonian fluid fills the gap between a fixed lower plate and an upper plate moving at constant speed, with zero pressure gradient and laminar flow. Across the gap

the velocity is linear and the shear stress is uniform

the velocity is parabolic and the shear stress is linear, zero at the mid-gap

the velocity is linear and the shear stress rises linearly towards the moving plate

the velocity is uniform at the plate speed and the shear stress is zero

In plane Couette flow the upper plate (gap h) moves at U and the lower plate is fixed, with viscosity μ and a constant pressure gradient dp/dx. Reverse flow first appears at the stationary wall when

dp/dx exceeds 2μU/h², an adverse gradient

dp/dx exceeds μU/h², an adverse gradient

dp/dx is more negative than −2μU/h², a favourable gradient

dp/dx exceeds 4μU/h², an adverse gradient

A mass of fluid rotates like a rigid body with constant angular velocity Ω. The vorticity everywhere in it is

zero, because there is no deformation

A free vortex has v_θ = K/r (K constant) about the origin. Which statement is correct?

The vorticity is zero for r > 0, yet the circulation is 2πK on every circle enclosing the origin

The vorticity is non-zero at every r, and the circulation grows with r

The vorticity is zero for r > 0, yet the circulation is 2πK on every circle enclosing the origin

The vorticity is zero for r > 0, so the circulation is zero on every circle

The vorticity is 2K for r > 0, and the circulation falls as 1/r

A velocity field is u = a y, v = 0 with a > 0, in SI units. Which description is correct?

rotational with ζ_{z} = −a, and with a shear rate of magnitude a

irrotational, because the streamlines are straight and parallel

irrotational, because the flow is incompressible

rotational with ζ_{z} = −a, and with a shear rate of magnitude a

rotational with ζ_{z} = +a, and with no shear deformation

A plane flow has u = x² and v = −2xy. Which statement is correct?

It is compressible and irrotational

It is compressible and rotational

It is incompressible and irrotational

For steady, fully developed laminar flow of a Newtonian fluid in a circular pipe of radius R, which statements are correct?

The velocity is largest on the axis, where it is twice the mean velocity

The shear stress increases linearly from zero on the axis to its maximum at the wall

At a fixed pressure gradient, the flow rate is proportional to R⁴

The velocity is largest on the axis, where it is twice the mean velocity

The shear stress increases linearly from zero on the axis to its maximum at the wall

At a fixed pressure gradient, the flow rate is proportional to R⁴

The wall shear stress equals μ u_{max}/R

Plane Couette flow: a fixed lower plate, an upper plate moving at U at gap h, no pressure gradient. Which statements are correct?

The velocity profile is linear in y

The shear stress is the same at every y, equal to μU/h

The flow rate per unit width is Uh/2

The velocity profile is linear in y

The shear stress is the same at every y, equal to μU/h

The flow rate per unit width is Uh/2

The flow is irrotational, since the streamlines are straight and parallel

Which statements about circulation and vorticity are correct?

By Stokes’ theorem, the circulation round a closed curve equals the integral of the vorticity over a smooth region it encloses

If a flow is irrotational everywhere in a simply connected region, the circulation round every closed curve in it is zero

A loop around the centre of a free vortex has non-zero circulation even though the vorticity is zero at every point of the fluid

By Stokes’ theorem, the circulation round a closed curve equals the integral of the vorticity over a smooth region it encloses

If a flow is irrotational everywhere in a simply connected region, the circulation round every closed curve in it is zero

A loop around the centre of a free vortex has non-zero circulation even though the vorticity is zero at every point of the fluid

The vorticity of a fluid element equals its angular velocity

Navier-Stokes Equation and Exact Solutions; Vorticity, Stream Function and Circulation

This chapter answers the second half of Section XE1.5 of XE1 Fluid Mechanics in the Engineering Sciences (XE) paper: “Navier-Stokes equation and its exact solutions for Couette flow and Poiseuille flow. Concept of fluid rotation, Vorticity, Stream function, and Circulation.” It teaches the incompressible Navier-Stokes equation, its reduction for fully developed parallel flow, plane Couette flow with and without a pressure gradient, plane and pipe Poiseuille flow (profile, flow rate, wall shear, ratio of mean to maximum velocity, the Hagen-Poiseuille law), then rotation, vorticity, the rate of deformation, the stream function and circulation with Stokes’ theorem. The numericals are the force on a moving plate, pressure drop and wall shear in laminar pipes, flow between fixed plates and circulation in a vortex.

1. The Navier-Stokes equation

Put the viscous stresses of a Newtonian fluid into the momentum equation of the previous chapter. For constant density and constant viscosity the viscous force per unit volume simplifies to μ∇²V, and the result is the incompressible Navier-Stokes equation:
ρ (∂V/∂t + (V · ∇)V) = ρg − ∇p + μ∇²V, together with ∇ · V = 0.
The x component is ρ(∂u/∂t + u ∂u/∂x + v ∂u/∂y + w ∂u/∂z) = ρg_x − ∂p/∂x + μ(∂²u/∂x² + ∂²u/∂y² + ∂²u/∂z²). Four unknowns (u, v, w, p) are matched by four equations, so with boundary conditions the problem is closed. With μ = 0 it becomes the Euler equation.

Boundary conditions. At a solid wall the fluid does not slip: its velocity equals the wall velocity (no-slip). The equation is non-linear because of (V · ∇)V, so exact solutions exist only where that term vanishes, which happens in fully developed parallel flow: V = (u(y), 0, 0). Continuity then gives ∂u/∂x = 0, the y momentum equation says p + ρgz does not vary across the flow, and the x momentum equation collapses to
μ d²u/dy² = d(p + ρgz)/dx
with the right-hand side a constant. Every solution below is two integrations of this equation. Gravity is absorbed into the pressure, so for a horizontal flow it is simply dp/dx.

⚠️ Constant viscosity is an assumption, not a law

The form μ∇²V needs constant μ and a Newtonian fluid. If viscosity varies with temperature or position, or the fluid is non-Newtonian, the viscous term has extra pieces and the profiles below no longer hold. Also remember that “fully developed” means the profile no longer changes along the flow, so it needs a long straight channel.

2. Plane Couette flow

A fluid fills the gap h between a stationary lower plate (y = 0) and an upper plate (y = h) moving at speed U in the x direction. With no pressure gradient the equation is d²u/dy² = 0, and the no-slip conditions u(0) = 0, u(h) = U give the linear profile u = U y/h. The shear stress τ = μ du/dy = μU/h is uniform across the gap, which is why the force on a plate of area A is F = μUA/h. The flow rate per unit width is q = Uh/2.

With a pressure gradient G = dp/dx the equation is μu″ = G, and
u = U y/h − (G/2μ) y(h − y)
A favourable gradient (G < 0) adds a parabolic bulge in the direction of the plate; an adverse gradient (G > 0) subtracts it. The flow rate is q = Uh/2 − G h³/(12μ). The wall shear on the stationary plate is τ(0) = μU/h − G h/2, so it reaches zero (the onset of reverse flow at that wall) when G = 2μU/h². Larger adverse gradients make fluid near the stationary wall flow backward. A convenient group is P = −(h²/2μU)(dp/dx): u/U = y/h + P (y/h)(1 − y/h).

Worked example. Oil of viscosity 0.08 Pa·s fills a 2 mm gap and a 0.5 m² plate slides at 1.5 m/s. Shear stress = 0.08 × 1.5/0.002 = 60 Pa, so F = 60 × 0.5 = 30 N. Halving the gap doubles the force; this is the linear-profile result and holds only while the flow is laminar and without pressure gradient.

3. Plane Poiseuille flow between fixed plates

Both plates are stationary (U = 0) at y = 0 and y = h and the flow is driven by a pressure fall, G = dp/dx < 0. Integrating μu″ = G with u(0) = u(h) = 0:
u = −(G/2μ) y(h − y)
The profile is a parabola with its maximum at the mid-plane. Writing Δp/L = −G:
u_max = (h²/8μ)(Δp/L); mean velocity V = (h²/12μ)(Δp/L); flow per unit width q = (h³/12μ)(Δp/L).
Hence V/u_max = 2/3. The shear stress τ = μ du/dy varies linearly, zero at the mid-plane and τ_w = (h/2)(Δp/L) at each wall, a result that follows from a force balance on the fluid between the walls and holds even without the profile.

⚠️ Gap h versus half-gap

Books that place y = 0 on the centre-line use the half-height b = h/2, giving u_max = b²/(2μ)(Δp/L) and q = 2b³/(3μ)(Δp/L). These agree with the formulas above, since b²/2 = h²/8. Use the length the question calls h and do not mix the two.

Worked example. Oil (μ = 0.05 Pa·s) between fixed plates 4 mm apart, with Δp/L = 2000 Pa/m: u_max = (0.004)² × 2000/(8 × 0.05) = 0.08 m/s, V = 0.0533 m/s, q = (0.004)³ × 2000/(12 × 0.05) = 2.13 × 10⁻⁴ m²/s, and τ_w = 0.002 × 2000 = 4 Pa. Doubling the gap at the same gradient multiplies the flow by 8.

4. Pipe Poiseuille flow and the Hagen-Poiseuille law

For steady, fully developed laminar flow in a circular pipe of radius R the x momentum equation in cylindrical form is (μ/r) d(r du/dr)/dr = dp/dx. Integrating twice, with du/dr finite at r = 0 and u(R) = 0:
u = (Δp/4μL)(R² − r²), u_max = Δp R²/(4μL) at the axis.
Integrating over the area gives Q = πR⁴Δp/(8μL) = πD⁴Δp/(128μL) (the Hagen-Poiseuille law), so the mean velocity is V = R²Δp/(8μL) and V/u_max = 1/2. The shear stress is τ = μ(−du/dr) = (Δp/2L) r, linear from zero at the axis to τ_w = RΔp/(2L) = 4μV/R = 8μV/D.

Consequences. At fixed Δp/L the flow goes as D⁴, so at fixed Q the pressure drop goes as D⁻⁴: halving the diameter raises Δp sixteen-fold. Writing the Darcy friction factor as f = 8τ_w/(ρV²) gives f = 64/Re, with Re = ρVD/μ, valid for laminar flow (Re below about 2300). The kinetic-energy flux of this profile is twice that of a uniform flow of the same mean speed, which matters when energy equations are written for laminar pipes.

Worked example. Oil (μ = 0.12 Pa·s, ρ = 900 kg/m³) flows at 0.2 L/s through a 20 mm pipe 10 m long. V = 2 × 10⁻⁴/(3.1416 × 10⁻⁴) = 0.637 m/s and Re = 900 × 0.637 × 0.02/0.12 = 95, so the flow is laminar. Δp = 128 × 0.12 × 10 × 2 × 10⁻⁴/(π × (0.02)⁴) = 61.1 kPa. Check with the wall shear: τ_w = 8 × 0.12 × 0.637/0.02 = 30.56 Pa, and Δp = 4τ_wL/D = 4 × 30.56 × 10/0.02 = 61.1 kPa.

5. Fluid rotation, vorticity and rate of deformation

A fluid element in a plane flow translates, rotates and deforms. Its angular velocity about the z axis is Ω_z = ½(∂v/∂x − ∂u/∂y), the mean rotation of two initially perpendicular lines. The vorticity is ζ = ∇ × V = 2Ω, so ζ_z = ∂v/∂x − ∂u/∂y. Flow with ζ = 0 everywhere is irrotational. In cylindrical coordinates ζ_z = (1/r) d(r v_θ)/dr for a purely circular flow.

The rate of deformation has a linear part, ε_xx = ∂u/∂x (and ε_yy = ∂v/∂y), whose sum is the volumetric dilation rate, zero for incompressible flow, and a shear part, γ̇_xy = ∂u/∂y + ∂v/∂x, to which the shear stress is proportional, τ_xy = μγ̇_xy. Rotation and shear rate come from the same velocity gradients but are different combinations: the difference for rotation, the sum for deformation.

Vorticity of standard flows
Flow	Velocity field	Vorticity ζ_z	Remark
Simple shear (Couette)	u = a y, v = 0	−a	Rotational although streamlines are straight
Rigid-body rotation	v_θ = Ω r	2Ω	Rotational, no deformation at all
Free vortex	v_θ = K/r	0 for r > 0	Irrotational, singular at the centre

Link with ψ and Bernoulli. For the stream function of the previous chapter, ζ_z = −∇²ψ, so an irrotational plane incompressible flow has ∇²ψ = 0. In an irrotational flow the Bernoulli constant is the same on all streamlines. Worked line: u = x², v = −2xy satisfies ∂u/∂x + ∂v/∂y = 2x − 2x = 0, and ζ_z = −2y − 0 = −2y, so it is incompressible and rotational, with rotation vanishing only on y = 0.

6. Circulation and Stokes’ theorem

The circulation around a closed curve C is Γ = ∮ V · dl, taken counter-clockwise. Stokes’ theorem relates it to the vorticity through the enclosed surface: Γ = ∬ (∇ × V) · dA, which in a plane is Γ = ∬ ζ_z dA. So circulation is the total vorticity enclosed: for a small loop, Γ ≈ ζ_z × area. If the flow is irrotational everywhere inside C, then Γ = 0.

Rigid rotation: Γ = (Ω r)(2πr) = 2Ω × πr², i.e. vorticity 2Ω times area, so a disc of fluid turning at 120 rpm (Ω = 12.57 rad/s) has Γ = 7.11 m²/s on a circle of radius 0.3 m. Free vortex: Γ = (K/r)(2πr) = 2πK on every circle round the centre, independent of r, although ζ_z = 0 for every r > 0: all the vorticity is concentrated at the singular centre. A loop that does not enclose the centre has Γ = 0. Simple shear u = a y: around a counter-clockwise rectangle of area A, Γ = −aA, matching ζ_z = −a.

⚠️ Zero vorticity does not always mean zero circulation

Stokes’ theorem needs a region in which the velocity field is smooth. A loop enclosing the centre of a free vortex has Γ = 2πK ≠ 0 even though ζ = 0 at every point of the fluid, because the region contains a singularity. Likewise, straight streamlines do not imply irrotationality, as simple shear shows.

Key takeaways

The incompressible Navier-Stokes equation is ρ DV/Dt = ρg − ∇p + μ∇²V with ∇ · V = 0, for constant ρ and μ. For fully developed parallel flow it reduces to μ d²u/dy² = d(p + ρgz)/dx, a constant right-hand side, with no-slip at walls.
Plane Couette flow without a pressure gradient has u = Uy/h and uniform shear τ = μU/h. With a gradient, u = Uy/h − (G/2μ) y(h − y), and reverse flow starts at the stationary wall when dp/dx = 2μU/h².
Between fixed plates a distance h apart, u_max = h²Δp/(8μL), V = (2/3) u_max, q = h³Δp/(12μL), τ_w = hΔp/(2L).
In a laminar circular pipe u = (Δp/4μL)(R² − r²), V = u_max/2, Q = πD⁴Δp/(128μL), τ_w = 8μV/D and f = 64/Re. At fixed Q, Δp scales as D⁻⁴.
Vorticity is ζ_z = ∂v/∂x − ∂u/∂y = twice the angular velocity, and ζ_z = −∇²ψ. Rigid rotation has ζ = 2Ω, a free vortex has ζ = 0 away from its centre, and simple shear is rotational although its streamlines are straight.
Circulation Γ = ∮ V · dl = ∬ ζ_z dA (Stokes). A free vortex has Γ = 2πK on any loop round the centre and zero on a loop that excludes it.

Practice questions (17)

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.

The incompressible Navier-Stokes equation ρ DV/Dt = ρg − ∇p + μ∇²V is obtained for a fluid that is
1. inviscid and of constant density
2. compressible, with viscosity that varies strongly with temperature
3. non-Newtonian, with a shear-rate-dependent viscosity
4. Newtonian, of constant density and constant viscosity
Show answer
Answer: D — Newtonian, of constant density and constant viscosity
The viscous term simplifies to μ∇²V only when the stress is Newtonian and both ρ and μ are constant. Setting μ = 0 gives the Euler equation instead; variable viscosity and non-Newtonian behaviour add extra terms.
A Newtonian fluid fills the gap between a fixed lower plate and an upper plate moving at constant speed, with zero pressure gradient and laminar flow. Across the gap
1. the velocity is linear and the shear stress is uniform
2. the velocity is parabolic and the shear stress is linear, zero at the mid-gap
3. the velocity is linear and the shear stress rises linearly towards the moving plate
4. the velocity is uniform at the plate speed and the shear stress is zero
Show answer
Answer: A — the velocity is linear and the shear stress is uniform
With d²u/dy² = 0 the profile is linear, so du/dy and hence τ = μ du/dy are the same at every height. The parabolic profile with a vanishing mid-gap stress belongs to a pressure-driven flow between fixed plates; the third description is inconsistent because a linear profile has constant slope.
For steady, fully developed laminar flow of a Newtonian fluid between two fixed parallel plates driven by a pressure gradient, the ratio of mean velocity to maximum velocity is
1. 1/2
2. 2/3
3. 3/4
4. 1
Show answer
Answer: B — 2/3
The profile is u = u_max × 4(y/h)(1 − y/h), whose mean is (2/3) u_max. The value 1/2 belongs to the circular pipe, where the parabola is revolved about the axis, and 1 would be a uniform profile.
Oil flows laminarly through a long circular pipe. The flow rate is held fixed, and the pipe is replaced by one of half the diameter and the same length, with the same fluid. The pressure drop is multiplied by
1. 4
2. 8
3. 16
4. 32
Show answer
Answer: C — 16
The Hagen-Poiseuille law gives Δp = 128μLQ/(πD⁴), so Δp ∝ D⁻⁴ and halving D multiplies Δp by 2⁴ = 16. The factor 4 would follow if only the area effect were counted, and 8 from a D⁻³ law.
In plane Couette flow the upper plate (gap h) moves at U and the lower plate is fixed, with viscosity μ and a constant pressure gradient dp/dx. Reverse flow first appears at the stationary wall when
1. dp/dx exceeds 2μU/h², an adverse gradient
2. dp/dx exceeds μU/h², an adverse gradient
3. dp/dx is more negative than −2μU/h², a favourable gradient
4. dp/dx exceeds 4μU/h², an adverse gradient
Show answer
Answer: A — dp/dx exceeds 2μU/h², an adverse gradient
From u = Uy/h − (G/2μ) y(h − y), the slope at y = 0 is U/h − Gh/(2μ). It becomes zero at G = 2μU/h² and negative beyond, which is flow reversal near the fixed wall. Only an adverse (positive) gradient can do this; a favourable gradient steepens the wall slope.
A mass of fluid rotates like a rigid body with constant angular velocity Ω. The vorticity everywhere in it is
1. zero, because there is no deformation
2. Ω
3. Ω/2
4. 2Ω
Show answer
Answer: D — 2Ω
Vorticity is twice the local angular velocity. For v_θ = Ωr, (1/r) d(r × Ωr)/dr = 2Ω. The absence of deformation means the shear rate is zero, not that the rotation is zero.
A free vortex has v_θ = K/r (K constant) about the origin. Which statement is correct?
1. The vorticity is non-zero at every r, and the circulation grows with r
2. The vorticity is zero for r > 0, yet the circulation is 2πK on every circle enclosing the origin
3. The vorticity is zero for r > 0, so the circulation is zero on every circle
4. The vorticity is 2K for r > 0, and the circulation falls as 1/r
Show answer
Answer: B — The vorticity is zero for r > 0, yet the circulation is 2πK on every circle enclosing the origin
(1/r) d(r × K/r)/dr = 0 for r > 0, so the flow is irrotational there. But Γ = (K/r)(2πr) = 2πK, constant for every circle round the origin, because all the vorticity sits at the singular centre. Stokes’ theorem cannot be applied across a singularity, which is why zero vorticity does not give zero circulation here.
A velocity field is u = a y, v = 0 with a > 0, in SI units. Which description is correct?
1. irrotational, because the streamlines are straight and parallel
2. irrotational, because the flow is incompressible
3. rotational with ζ_z = −a, and with a shear rate of magnitude a
4. rotational with ζ_z = +a, and with no shear deformation
Show answer
Answer: C — rotational with ζ_{z} = −a, and with a shear rate of magnitude a
ζ_z = ∂v/∂x − ∂u/∂y = −a and the shear rate ∂u/∂y + ∂v/∂x = a. Straight streamlines do not imply irrotational flow, and incompressibility (∇ · V = 0) is a separate condition from ∇ × V = 0.
A plane flow has u = x² and v = −2xy. Which statement is correct?
1. It is compressible and irrotational
2. It is compressible and rotational
3. It is incompressible and irrotational
4. It is incompressible and rotational
Show answer
Answer: D — It is incompressible and rotational
∂u/∂x + ∂v/∂y = 2x − 2x = 0, so it is incompressible. ζ_z = ∂v/∂x − ∂u/∂y = −2y − 0 = −2y, non-zero except on y = 0, so it is rotational.
For steady, fully developed laminar flow of a Newtonian fluid in a circular pipe of radius R, which statements are correct?
1. The velocity is largest on the axis, where it is twice the mean velocity
2. The shear stress increases linearly from zero on the axis to its maximum at the wall
3. At a fixed pressure gradient, the flow rate is proportional to R⁴
4. The wall shear stress equals μ u_max/R
Show answer
Answer: A — The velocity is largest on the axis, where it is twice the mean velocity; B — The shear stress increases linearly from zero on the axis to its maximum at the wall; C — At a fixed pressure gradient, the flow rate is proportional to R⁴
The profile u_max(1 − r²/R²) has mean u_max/2, shear stress (Δp/2L) r, and Q ∝ R⁴. The wall slope is |du/dr| = 2u_max/R, so τ_w = 2μ u_max/R; the version without the factor 2 is wrong.
Plane Couette flow: a fixed lower plate, an upper plate moving at U at gap h, no pressure gradient. Which statements are correct?
1. The velocity profile is linear in y
2. The shear stress is the same at every y, equal to μU/h
3. The flow rate per unit width is Uh/2
4. The flow is irrotational, since the streamlines are straight and parallel
Show answer
Answer: A — The velocity profile is linear in y; B — The shear stress is the same at every y, equal to μU/h; C — The flow rate per unit width is Uh/2
The first three follow from u = Uy/h. The vorticity is −U/h, uniform and non-zero, so the flow is rotational; straight streamlines never guarantee irrotationality.
Which statements about circulation and vorticity are correct?
1. By Stokes’ theorem, the circulation round a closed curve equals the integral of the vorticity over a smooth region it encloses
2. If a flow is irrotational everywhere in a simply connected region, the circulation round every closed curve in it is zero
3. A loop around the centre of a free vortex has non-zero circulation even though the vorticity is zero at every point of the fluid
4. The vorticity of a fluid element equals its angular velocity
Show answer
Answer: A — By Stokes’ theorem, the circulation round a closed curve equals the integral of the vorticity over a smooth region it encloses; B — If a flow is irrotational everywhere in a simply connected region, the circulation round every closed curve in it is zero; C — A loop around the centre of a free vortex has non-zero circulation even though the vorticity is zero at every point of the fluid
The first three are correct statements of Stokes’ theorem and of the free-vortex singularity. Vorticity is twice the angular velocity, so equating the two is wrong by a factor of 2.
Oil of viscosity 0.08 Pa·s fills a 2 mm gap between a fixed surface and a flat plate of area 0.5 m² sliding parallel to it at 1.5 m/s. Assume a linear velocity profile and no pressure gradient. The force needed to move the plate, in N, is ____.
Numerical answer — type the value.
Show answer
Answer: 30
τ = μU/h = 0.08 × 1.5/0.002 = 60 Pa; F = τA = 60 × 0.5 = 30 N. Using h in mm without conversion would give a result 1000 times too small.
Oil (viscosity 0.12 Pa·s, density 900 kg/m³) flows steadily at 0.2 L/s through a horizontal circular pipe of diameter 20 mm and length 10 m; the flow is fully developed and laminar. The pressure drop over the pipe length, in kPa, to one decimal place, is ____.
Numerical answer — type the value.
Show answer
Answer: 61.1
Q = 2 × 10⁻⁴ m³/s. Δp = 128μLQ/(πD⁴) = 128 × 0.12 × 10 × 2 × 10⁻⁴/(π × 1.6 × 10⁻⁷) = 0.03072/5.0265 × 10⁻⁷ = 61,115 Pa = 61.1 kPa. Check: V = 0.637 m/s, Re = 900 × 0.637 × 0.02/0.12 = 95, so the laminar law applies. The density is not needed for Δp.
Oil of viscosity 0.2 Pa·s flows in fully developed laminar flow in a circular pipe of diameter 50 mm. The centre-line velocity is 0.8 m/s. The shear stress at the pipe wall, in Pa, to one decimal place, is ____.
Numerical answer — type the value.
Show answer
Answer: 12.8
u = u_max(1 − r²/R²), so |du/dr| at the wall is 2u_max/R = 2 × 0.8/0.025 = 64 s⁻¹. τ_w = 0.2 × 64 = 12.8 Pa. Check: V = u_max/2 = 0.4 m/s and 8μV/D = 8 × 0.2 × 0.4/0.05 = 12.8 Pa. Using μu_max/R gives 6.4 Pa, a factor 2 short.
Oil of viscosity 0.05 Pa·s flows between two fixed parallel plates 4 mm apart under a pressure gradient of −2000 Pa/m; the flow is laminar and fully developed. The volume flow rate per metre of plate width, in units of 10⁻⁴ m²/s, to two decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 2.13
q = h³/(12μ) × |dp/dx| = (0.004)³ × 2000/(12 × 0.05) = 6.4 × 10⁻⁸ × 2000/0.6 = 2.133 × 10⁻⁴ m²/s, i.e. 2.13 in the units asked. Using the mid-plane to wall half-gap as h would give one eighth of this.
A circular tank of water turns about its vertical axis at 120 revolutions per minute, with the water rotating as a rigid body. Take π = 3.14159. The circulation round a horizontal circle of radius 0.3 m centred on the axis, in m²/s, to two decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 7.11
Ω = 120 × 2π/60 = 12.566 rad/s. Γ = ∮ V · dl = (Ωr)(2πr) = 2π Ω r² = 2π × 12.566 × 0.09 = 7.106 m²/s. By Stokes: vorticity 2Ω = 25.13 s⁻¹ times area π(0.3)² = 0.2827 m² gives the same 7.106, i.e. 7.11.