MMV211 Fluid MechanicsAnswers to theory questions; only a few figures

Pages refer to 7th Edition in SI Units (F. M. White, Fluid Mechanics).(C. Norberg, April 3, 2013)

Chapter 1 — Introduction

1.1 What is the fundamental difference between a solid and a fluid?

A solid can resist a shear stress by static deformation; a fluid can not. Any shear stress appliedto a fluid, whatever small it may be, will result in motion.

1.2 Give a practical definition of the density of a fluid. Illustrate with a figure.

See Fig. 1.4. Consider a fluid that is known to be non-homogeneous, either a liquid or a gas at notto low a pressure. We want to calculate the local density, the local mass per unit volume. Whenconsidering, for instance, a cubical element surrounding a mathematical point, we can calculatethe ratio between the mass of the cubical element and its volume. For large elemental volumesthere are some variations due to the assumed non-homogeneity. With a successive decrease in thevolume the ratio (density) will become a constant, and remains so down to cube side lengths thatare indeed very small, of the order microns (µm), i.e., a cube volume of the order δV∗ = 1 (µm)3.This small length is called the continuum limit. However, for even smaller cubes the ratio will beill-defined, due to that the volume now contains too few molecules to get a well-defined value.

1.3 (a) For newtonian fluids, shear stresses are proportional to corresponding shear strain rates. Fora simple shear flow, V = u(y)i, show that the shear stress τ is proportional to du/dy.

Consider a small cubical element, at the continuum limit. As a thought experiment, now releasethe fluid element, with its lower side at y, upper side at y+ δy. After a small time δt the elementdeforms in the xy-plane due to the shear action, see Fig. 1.6. If velocity increases with y, thehigher velocity on the upper side (u + δu) means that this side slides further to the right (inx-direction) compared to the lower side, shear angle δθ, sliding length difference δuδt. Since δtis small, δθ ≈ (δu/δy)δt. In the limit δt → 0, dθ/dt = du/dy. Since dθ/dt is the shear strainrate, the shear stress τ is proportional to du/dy (the proportionality constant being the dynamicviscosity µ).

(b) Discuss and illustrate with diagrams some possible characteristics of non-newtonian fluids(liquids).

See Fig. 1.9. When plotting shear stress τ vs. shear strain rate dθ/dt (Fig. 1.9, left), a newtonianfluid appears as a straight line, the gradient (slope) being the viscosity. For a dilatant fluid, alsocalled a shear-thickening fluid, the gradient (the apparent viscosity) increases with dθ/dt. Fora pseudoplastic fluid, also called a shear-thinning fluid, the gradient (the apparent viscosity)decreases with dθ/dt. The so-called Ideal Bingham fluid behaves like a solid up to a certain yieldstress, for higher stresses it behaves like a newtonian fluid. For a newtonian fluid, when exerted toa constant shear rate, there is no time-dependency of the shear stress (Fig. 1.9, right). However,for a rheopectic fluid, the shear stress increases with time; for a thixotropic fluid it decreases (theapparent viscosity decreases).

1.4 State the SI-dimensions of dynamic and kinematic viscosity, respectively. Also provide approx-imate values (with two significant digits) for water and air at standard room conditions (20C,1 atm). At around standard room conditions, what are the separate effects of changing thepressure and temperature, respectively?

[µ] = Pa s; [ν] = m2/s (ν = µ/ρ). At standard room conditions (Table 1.4), µwater = 1.0 ×10−3 Pa s, νwater = 1.0 × 10−6 m2/s; µair = 18 × 10−6 Pa s, νair = 15 × 10−6 m2/s. µwaterdecreases with temperature, µair increases. For both water and air, µ is only dependent of tem-perature (except extreme conditions), as is ν for water since density is (more or less) only de-pendent on temperature. For air, ρ increases with pressure so νair decreases with pressure for agiven temperature; since ρ decreases with temperature, νair increases with temperature for a givenpressure.

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1.5 Explain or define (a) streamline, (b) particle path, (c) streak line.

(a) Everywhere along a streamline the velocity vector is tangent vector, see Fig. 1.16. (b) Aparticle path is the actual path of a fluid element. (c) A streak line is the locus of fluid elements(combined to a line) that previously were released from a certain point in the flow field.

Chapter 2 — Pressure Distribution in a Fluid

2.1 Derive the hydrostatic balance equation ∇p = ρg, valid for a fluid at rest.

In general, the pressure is a scalar point function, p = p(x, y, z, t), and it acts inward on surfacessurrounding a point. Consider a fluid element surrounding a specific point, see Fig. 2.2, but letthe point instead be at the center of the element with sides dx, dy and dz. Since the element isso small the pressure can be treated as constant on each surface, but not necessarily equal to thecenter pressure p if the pressure varies. On the surface at x+dx/2, with area dy dz, the pressure isp+(∂p/∂x)(+dx/2), on surface at x−dx/2 (same area) it is p+(∂p/∂x)(−dx/2). The net pressureforce in the x-direction then is F

(p)x = −(∂p/∂x)dx dy dz. Similarly, F

(p)y = −(∂p/∂y)dx dy dz,

and F(p)z = −(∂p/∂z)dx dy dz, i.e., F(p) = −(dx dy dz)∇p. The gravity force on the element is

ρ(dx dy dz)g, where g is the gravitational acceleration vector. Since the fluid is at rest, the netforce on the element is zero, which means that ∇p = ρg.

2.2 Derive the manometer formula ∆p = (ρm−ρ)g h , where h is the manometer reading (height), ρmthe density of the manometer liquid and ∆p the pressure difference across two horizontal pointsin the fluid of density ρ (U-tube manometer, see Example 2.3).

Connect a U-tube manometer in between two points of a horizontal part of a pipe, across a flowdevice with an assumed pressure difference, see Fig. E2.3. The horizontal points (a) and (b)are at the connections of the U-tube legs, the pressure at (a) is assumed to be highest (pa > pb).The manometer liquid density is ρm, the pipe fluid density is ρ, both assumed to be constant.The height difference of the manometer is h; L is the vertical distance between the highest levelof the manometer liquid and the horizontal at (a) and (b). Within the manometer legs there isa hydrostatic variation, since the fluids within them are at rest, with pressure increasing linearlydownwards. Within the same (connected) fluid at rest, the pressure is the same at a givenelevation (Pascal’s law). Starting from (a), the pressure at the level for the manometer liquid ispa + ρgL+ ρgh, which then must be the same pressure when coming to this horizontal from (b),i.e., pb + ρgL+ ρmgh = pa + ρgL+ ρgh, which means that pa − pb = ∆p = (ρm − ρ)gh.

Chapter 3 — Integral Analysis

3.1 (a) Provide general definitions of volume flow rate and mass flow rate through a surface. Illustratewith a figure.

Consider an elemental area dA, with unit normal vector n, on an arbitrary surface S within aflow field, see Fig. 3.1(a). The volume flow rate through this elemental area is (V · n) dA, where(V · n) = Vn is the area normal velocity. The total volume flow rate is Q =

∫S(V · n) dA. The

mass flow rate through dA is ρVn dA, which means that the total mass flux is m =∫S ρ(V · n) dA.

(b) Show that the mean velocity for laminar fully developed flow in a pipe of circular cross sectionis equal to half the velocity at the centerline. The velocity profile is parabolic.

In general, the mean velocity through an arbitrary surface is V = Q/A, where Q is the volumeflow rate and A is the area of the surface. For a surface normal to the flow within a pipe ofradius R, the cross-sectional area is A = πR2. With a mean velocity profile u(r) across the pipe,r being the local radius, the flow rate is Q =

∫ R0 u(r) 2πr dr. For fully developed laminar flow,

the parabolic velocity profile is u(r) = uCL

[1− (r/R)2

], where uCL is the mean velocity at the

centerline. Introducing η = r/R, the flow rate is Q = 2πR2uCL∫ 10 (1−η2)η dη = πR2uCL/2. This

shows that V = uCL/2.

3.2 Let B be an arbitrary extensive property and b the same property but expressed per unit mass.For a fixed and non-deforming control volume with one inlet (1) and one outlet (2), both assumedone-dimensional, the Reynolds transport theorem, at stationary conditions, reads:

2

d

dtBsys = (b ρ V A)2 − (b ρ V A)1

State without proof the Reynolds transport theorem for an arbitrarily moving but still non-deforming control volume, at non-stationary conditions. Illustrate the convective part in a figure.

For an arbitrarily moving but non-deformable control volume (CV):

d

dtBsys =

∫CV

∂

∂t(b ρ) dV +

∫CSb ρ(V · n)dA

where n is a unit normal vector corresponding to the surface element dA on the control surface(CS), pointing out from the control volume; dV is a volume element within the control volume; Vis the local fluid velocity vector relative to a coordinate system fixed to the control volume. [Whenusing some other (non-moving) coordinate system, V should be replaced with the control-surfacerelative velocity vector Vr = V−Vs, where V is the local fluid velocity vector and Vs is the localsurface velocity vector, see Fig. 3.4].

3.3 Assuming stationary one-dimensional incompressible flow, state without proof the momentumintegral equation applicable to a control volume with one inlet and one outlet. Give a fullaccount for different terms and properties. How can this equation be modified to include velocityvariations across the inlet and outlet? Calculate the correction coefficient (β) for the case of fullydeveloped laminar flow in a straight pipe (circular cross section).

The sum of forces acting on the control volume equals the net outflow of momentum per unittime, i.e.,

∑F = m(Vout −Vin)

where m is the mass flow rate through CV and V is the fluid velocity vector with respect to acoordinate system fixed to CV. When accounting for velocity variations across inlet and outletthe equation can be written as

∑F = m [(βV)out − (βV)in]

where V is the velocity vector based on the mean velocity V = Q/A; β is the momentum fluxcorrection factor, which for a cross-section perpendicular to the flow is

β = A−1∫(u/V )2dA

where u is the local velocity. For fully developed laminar flow in a pipe of radius R, u(r) =uCL

[1− (r/R)2

]. For this flow, uCL/V = 2. The cross-section area is A = πR2; dA = 2πr dr.

By introducing η = r/R, β = 8∫ 10 (1− η2)2η dη = 4/3.

3.4 Assuming stationary one-dimensional incompressible flow, state without proof the energy integralequation applicable to a control volume with one inlet and one outlet (extended Bernoulli equa-tion). Give a full account for different terms and properties. How can this equation be modifiedto include velocity variations across the inlet and outlet? Calculate the correction coefficient (α)for the case of fully developed laminar flow in a straight pipe of circular cross section.

The energy equation reads(p

ρg+V 2

2g+ z

)in

=

(p

ρg+V 2

2g+ z

)out

+ hf − hP + hT

p is the static pressure; ρ is the fluid density, g is the gravitational acceleration; V is the velocity; zis the mid-section vertical coordinate (upwards); hf is the head loss; hP and hT are the pump and

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turbine heads, respectively. The terms within the parentheses are called the total head, V 2/(2g)is called the dynamic head. When accounting for cross-sectional velocity variations, the dynamichead can be written as αV 2/(2g), where V is the mean velocity (V = Q/A), and α is the kineticenergy correction factor, which can be calculated from

α = A−1∫(u/V )3dA

where u is the local velocity. For fully developed laminar flow in a pipe of radius R, u(r) =uCL

[1− (r/R)2

]. For this flow, uCL/V = 2. The cross-section area is A = πR2; dA = 2πr dr.

By introducing η = r/R, α = 16∫ 10 (1− η2)3η dη = 2, since (1− η2)3η = η − 3η3 + 3η5 − η7.

Chapter 4 — Differential Analysis

4.1 A velocity field is described in a cartesian coordinate system xyz. Use the rule of chains toexpress the material (substantial) acceleration in local and convective terms. Give a physicaldescription of each term.

The acceleration a of a fluid particle is equal to the time derivative of its velocity, a = dV/dt. Incartesian coordinates, V = (u, v, w), where u = dx/dt, v = dy/dt, w = dz/dt. The coordinatesof the particle are dependent on time, i.e.,

a =d

dtV[ t, x(t), y(t), z(t) ] =

∂V

∂t+∂V

∂x

dx

dt+∂V

∂y

dy

dt+∂V

∂z

dz

dt

This means that

a =∂V

∂t+ (V · ∇)V

where ∇ = (∂/∂x, ∂/∂y, ∂/∂z) is the gradient operator. The first term is acceleration due to localtime variations, it vanishes in steady flow. The second term is called the convective derivativeand it expresses the acceleration of a fluid particle when it moves (convects) in the flow field.

4.2 Derive the differential form of the continuity equation using a cartesian coordinate system. Afterthis, specialize to incompressible flow conditions.

Consider a fixed control volume (CV) surrounding a specific point, see Fig. 4.1. ThroughReynolds transport theorem (with b = 1), mass balance requires∫

CV(∂ρ/∂t) dV +

∫CSρ(V · n) dA =

dmsys

dt= 0

For a differential CV in cartesian coordinates, there is no need for a volume integration in thefirst term, it can be replaced with (∂ρ/∂t)dx dy dz. The second term is the net mass flux out of theCV,

∑mnet,out. Consider first the x-direction. The net mass flux out of CV in this direction is

(∂mx/∂x) dx. Variations over the control surfaces (CS) can be neglected, i.e., mx = (ρ u)dy dz.Other two directions in the same manner gives the total net outflow of mass per unit time:[

∂

∂x(ρ u) +

∂

∂y(ρ v) +

∂

∂z(ρw)

]dx dy dz = [∇·(ρV)] dx dy dz

Dividing with the constant volume (dx dy dz) gives

∂ρ

∂t+∇·(ρV) = 0

Incompressible conditions means that density ρ can be treated as a constant, i.e.,

∇ ·V =∂u

∂x+∂v

∂y+∂w

∂z= 0

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4.3 Show that Ma2 ≪ 1 is a necessary condition for incompressible flow. Provide an engineeringestimate of Ma for which the flow has to be treated as compressible.

Consider, as a special case, one-dimensional steady flow along a horizontal streamline. Thedifferential continuity equation (local mass balance) requires

d

dx(ρV ) = ρ

dV

dx+ V

dρ

dx= 0

A constant fluid density (ρ = const.) implies |V (dρ/dx)| ≪ |ρ dV/dx|, which means |δρ/ρ| ≪|δV/V |. Neglect viscous forces. Local momentum balance then requires dp+ρV dV = 0 (Bernoulliequation). Also neglect heat transfer, i.e., isentropic flow. Then dp = a2dρ, where a is the velocityof sound. When combined with Bernoulli equation, after division with ρa2,

dρ

ρ+ (V/a)2

dV

V= 0 ⇒ |δρ/ρ| = Ma2|δV/V |

Combined with |δρ/ρ| ≪ |δV/V | yields Ma2 ≪ 1, which then is a necessary condition for in-compressible flow. The engineering estimate is that the flow has to be treated as compressible ifMa > 0.3.

4.4 (a) Define the stress tensor σij in a flowing fluid, expressed in terms of the viscous stress tensorτij and the pressure p. What symmetry condition is normally valid for σij?

(a) The stress tensor can be written as σij = −pδij + τij, where p is the pressure, δij is theKronecker delta; δij = 1 for i = j, zero otherwise, and τij is the viscous stress tensor. Unless thefluid exhibits local moments, the stress tensor is symmetric, σji = σij.

(b) Illustrate the components σ11, σ22, σ12 and σ21 in a figure (σ12 = σxy, . . . ).

See Fig. 4.3. In general, σij is the force per unit area acting in the j-direction on a surfacei = const. For instance, τ12 = τxy is the (viscous shear) stress component in the y-directionthat acts in a plane x = const. Stresses with i = j are called normal stresses, others are shearstresses.

(c) Define τxy for a newtonian fluid using cartesian coordinates.

τxy = τyx = µ(∂u/∂y + ∂v/∂x), where µ is the dynamic viscosity.

4.5 Using cartesian coordinates, write out the continuity equation as well as the x-component of themomentum equation for incompressible flow of a newtonian fluid with constant fluid properties.

Continuity equation, incompressible flow: ∇ ·V = 0, which means ∂u/∂x+ ∂v/∂y+ ∂w/∂z = 0.Momentum equation, incompressible flow of a newtonian fluid with constant fluid properties,cartesian coordinates: ρ dV/dt = −∇p+ µ∇2V. Out-written in the x-direction:

ρdu

dt= ρ

(∂u

∂t+ u

∂u

∂x+ v

∂u

∂y+ w

∂u

∂z

)= −∂p

∂x+ µ

(∂2u

∂x2+∂2u

∂y2+∂2u

∂z2

)

4.6 Write out the divergence and rotation (curl), respectively, applied to the velocity field V =(u, v, w) in cartesian coordinates. What do they represent physically?

With ∇ = (∂/∂x, ∂/∂y, ∂/∂z), ∇ ·V = ∂u/∂x + ∂v/∂y + ∂w/∂z, which represents the relativevolume change of a fluid element, per unit time. It equals zero in incompressible flow. The curlof the velocity vector, ∇×V, can be written as a determinant,

i j k∂∂x

∂∂y

∂∂z

u v w=

(∂w

∂y− ∂v

∂z

)i+

(∂u

∂z− ∂w

∂x

)j+

(∂v

∂x− ∂u

∂y

)k

∇×V represents twice the instantaneous anti-clockwise rotational rate of a fluid element.

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4.7 Consider, at zero time, a cubical infinitesimal fluid element in a flow field. Kinematically, theinstantaneous effects on this element from the flow field can be sub-divided into four elementarymotions. Define and describe these elementary motions. Illustrate with a figure.

Decomposition of the general motion (dt→ 0): 1. pure translation (no deformation or rotation);2. linear deformation (normal strain rate); 3. rotation (no deformation); 4. angular deformation(shear strain rate).

4.8 (a) State the general definition of the vorticity vector at a point in a flow field. What does itdescribe physically?

ζ = ∇×V represents twice the instantaneous anti-clockwise rotational rate of a fluid element.

(b) For two-dimensional flow in a plane, show that the vorticity vector only has one component.

If V = (u, v, 0), where u(x, y) and v(x, y), the vorticity vector, according to 4.6, is ζ = ∇×V =(0, 0, ∂v/∂x− ∂u/∂y), i.e., ζx = ζy = 0, ζz = ∂v/∂x− ∂u/∂y.

(c) Show that the only component of the vorticity vector in plane two-dimensional flow canbe interpreted as twice the instantaneous anti-clockwise rotational rate of the diagonal in ainfinitesimal fluid element, initially being quadratic.

See Fig. 4.10. Consider infinitesimal deformations of the fluid element, during a vanishing timeincrement dt→ 0. Due to the differential change in the vertical velocity v in the x-direction, thematerial line BC turns, the anti-clockwise turning angle for dt→ 0 being dα = (∂v/∂x)dx dt/dx.Similarly, due to the change of u in y-direction, the clock-wise turning angle of the material lineBA is dβ = (∂u/∂y)dy dt/dy. For dx = dy, the anti-clockwise turning angle of the diagonal is(dα − dβ)/2. The diagonal turning rate (around the z-axis) is (dα/dt − dβ/dt)/2 = (∂v/∂x −∂u/∂y)/2 = ωz.

(d) How is it possible for an initially irrotational flow to become locally rotational, i.e., to possessvorticity? State at least two mechanisms.

A fluid element might exhibit a tendency to rotate due to: 1. viscous shear forces (Fig. 4.11a);2. density gradients (stratification); 3. non-inertial effects such as the Coriolis acceleration fromthe earth’s rotation; 4. entropy gradients caused by curved shock waves (Fig. 4.11b).

4.9 Consider two-dimensional incompressible flow in a plane, V = (u, v, 0), where u(x, y), v(x, y).

(a) Define the stream function ψ through implicit expressions using u(x, y) and v(x, y). Alsoshow that ψ satisfies the Laplace equation ∇2ψ = 0 when the the flow is irrotational.

ψ is defined so that the continuity equation is fulfilled identically. In cartesian coordinates, thecontinuity equation reads: ∂u/∂x + ∂v/∂y = 0, which is satisfied if u = ∂ψ/∂y, v = −∂ψ/∂x.Irrotational flow means that ζz = ∂v/∂x − ∂u/∂y = 0, which implies −∂2ψ/∂x2 − ∂2ψ/∂y2 =−∇2ψ = 0, i.e., ∇2ψ = 0.

(b) Show that ψ = const. represents streamlines and that the difference in ψ between twostreamlines is equal to the volume rate per unit width.

Along a line ψ = const., dψ = 0; ψ = ψ(x, y) ⇒ dψ = (∂ψ/∂x) dx+(∂ψ/∂y) dy. According to thedefinition of ψ: dψ = −v dx+u dy = 0. An infinitesimal vector along a streamline, (dx, dy, 0), isparallel to the velocity vector (u, v, 0), i.e., v/u = dy/dx or u dy − v dx = 0 = dψ. Now considertwo nearby streamlines, the differential difference in the stream function being dψ. Since therecan not be any velocity component across a streamline the volume flow rate per unit width betweenthese two lines is a constant; and it can be calculated as dQ/b = dq = u dy−v dx = dψ (Fig. 4.8),which means that Q/b = q =

∫ 21 dψ = ψ2 − ψ1.

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Chapter 5 — Dimensional Analysis and Similarity

5.1 (a) What is the principle of dimensional homogeneity, PDH?

All terms in a valid physical-mathematical expression must have the same dimensions. This isthe principle of dimensional homogeneity, PDH.

(b) What are the MLTΘ-dimensions for dynamic (absolute) viscosity µ, heat conductivity k andpower P (work per unit time)?

See Table 5.1. Simple shear flow: τ = µ(du/dy), which from PDH in (a) means that µ =τ/du/dy, where means ”the dimension of”. Primary MLTΘ-dimensions: M for mass,L for length, T for time, and Θ for temperature, respectively. Shear stress is shear force per unitarea, i.e., in SI-units, τ = N/m2. Since force (N) is mass (kg) times acceleration (m/s2),N/m2 = kg/(ms2) = ML−1T−2. Finally, since du/dy = u/y = (m/s)/m = s−1 = T−1,µ = ML−1T−1. Simple heat conduction: Q = −k(dT/dy)A, where Q is heat flux per unittime, k = QdT/dy−1A−1. Heat has the same dimension as energy, i.e., Q = J/s =Nm/s. dT/dy = K/m, and A2 = m2, k = N/(sK) = kgm/(s3K), which means thatk = MLT−3Θ−1, where Θ is the primary dimension for temperature. Power P is work per unittime; work has same dimension as energy (J = Nm), i.e. P = Nm/s, i.e., P = kgm2/s3 =ML2T−3.

(c) State the Π-theorem, in words.

A physical process described as a valid relation between n dimensional variables that satisfiesPDH can be reduced to an equally valid relation between k = n− j non-dimensional variables, so-called Π-groups. The reduction j is the number of dimensional variables, containing all prevailingprimary dimensions, that among themselves cannot form a Π-group, and it is always less thanor equal to the number of primary dimensions in the original relation.

5.2 Define the Strouhal number (St) for periodic vortex shedding downstream of a circular cylinderof diameter d, exposed to a constant flow velocity U perpendicular to the cylinder axis. Assumingincompressible flow, state an approximate value of the Strouhal number over the interval Re =Ud/ν = 300− 105.

Strouhal number, St = fSd/U , where fS is the vortex shedding (or Strouhal) frequency. Forincompressible flow and Re = Ud/ν = 300− 105, St ≃ 0.2; see Fig. 5.2, or Fig. 7 in PM forLab 2b.

5.3 Explain briefly

(a) homologous points and homologous times

Assume that all physical coordinates of a certain geometrical situation are made dimensionlessusing a length scale L. A specific point in this dimensionless coordinate system is then a ho-mologous point to all other coordinate systems, whatever L may be; same relative location, seeFig. 5.7. Assume that the flow within the specific geometry also involves a velocity scale U . Adimensionless time tU/L is then a homologous time for this flow-geometry situation, whatever(the time scale) L/U may be.

(b) geometric, kinematic and dynamic similarity

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For geometric similarity all spatial dimensions in model and prototype conditions should be scaledin the same linear scale ratio, α = Lm/Lp, where L is a (characteristic) length scale. If geometricsimilarity is fulfilled and all fluid velocities at homologous points and times are in the same ratio,in model and prototype scales, then the flow-geometric situation is kinematically similar. Asstated in the text-book (page 331): ”The motions of two systems are kinematically similar ifhomologous particles lie at homologous points at homologous times”. If geometric similarity isfulfilled and all forces acting on fluid particles are in the same ratio at homologous points andtimes, in model and prototype scales, then the flow-geometric situation is dynamically similar,i.e., force polygons look the same (are similar) at homologous points and homologous times, seeFig. 5.7. If the flow is dynamically similar it is also kinematically so, at least for incompressibleflow.

5.4 In practical cases concerning free-surface flows, most model tests have to renounce on the principleof dynamic similarity. Explain why.

For dynamic similarity it is required that both Froude and Reynolds numbers are equal at modeland prototype conditions, Frm = Frp and Rem = Rep. Since Fr = U2/(gL) this requires that

Um/Up =√Lm/Lp =

√α (assuming gm = gp). Since Re = UL/ν, it is also required to have

νm/νp = (Lm/Lp)(Um/Up) = α√α. Most model test are carried out with models that are smaller

than the prototype, α < 1. For easy numbers, assume α = Lm/Lp = 0.09. Equal Froude numberthen requires Um/Up =

√0.09 = 0.3; since velocity now is fixed, an equal Reynolds number then

requires νm/νp = 0.3×0.09 = 0.027. However, since νp is mostly for water the kinematic viscosityin the model tests then should be only about 3% of this value, and it is in fact impossible to findsuch a liquid. Even mercury, the liquid with the lowest kinematic viscosity, only has about one-ninth the kinematic viscosity of water. (Free-surface flow model tests are usually carried out inwater at the correct Froude number, but at a much to low Reynolds number; effects of Reynoldsnumber then need to be estimated from extrapolation.)

5.5 For incompressible flow conditions, the drag D on a smooth spherical object depends only on thediameter d, fluid viscosity µ, fluid density ρ and flow velocity U .

(a) Define the conventional drag coefficient CD, and show, using dimensional analysis, that CDonly depends on the Reynolds number.

By definition:

CD =2D

ρU2A= A = πd2/4 =

8D

πρU2d2

Dimensional relation: D = f(d, µ, ρ, U); five independent variables, n = 5. By inspection ofdimensions it can be seen that the relation only contains three primary dimensions, M, L, andT, e.g., µ = ML−1T−1. ρ, U and d contain all three primary dimensions and they cannotbe combined to a dimensionless quantity (only ρ contains M, only U contains T). Thus, thereduction is j = 3; k = n − j = 2 means that the relation can be reduced to Π1 = g(Π2),where Π1 and Π2 are dimensionless. By inspection it can be seen that ρU2d2 has the samedimension as D; for instance ρV 2 has the same dimension as pressure, which is a force per unitarea; Π1 = D/(ρU2d2). A (well-known) non-dimensional combination of ρ, U , d and µ is theReynolds number, Re = ρUd/µ = Π2. Comparison with the above definition above shows thatCD = CD(Re).

(b) Suggest an alternative drag coefficient that does not include the flow velocity. Explain whythis coefficient could be more appropriate to use, in some cases.

See Example 5.11. If CD is a function of Re only, it means that that also CDRe2 is. Since Re ∝ V ,

CD ∝ U−2, CD = CDRe2 does not contain the velocity. In this case, CD = CDRe

2 = 8ρD/(πµ2).When plotted against Re the graph (or functional relationship) can be used conveniently to getthe velocity directly from a measurement of the drag (for a certain diameter and known fluidproperties); ρ, µ, D → CD → Re (graph, formula) → U using d; see Fig. 5.11.

8

5.6 Show that the Reynolds, Froude and Weber number, respectively, each is a measure of a ratiobetween certain fluid forces in a flow field (mass times acceleration = inertia force).

Definitions: Re = ρUL/µ, Fr = U2/(gL), We = ρU2L/Υ, where U is a characteristic velocity,L is a characteristic length; ρ and µ are the density and dynamic viscosity of the fluid; Υ isthe surface tension coefficient at the liquid (fluid)-gas interface. Let Fi = ma represent inertialforce, Fµ viscous force, Fg = mg gravitational force, and FΥ surface tension force, acting on afluid element in the flow. Since acceleration is change of velocity per unit time, a = dV/dt ∝U/(L/U) = U2/L; mass is proportional to ρL3, which means that Fi ∝ ρL2U2. In simple shearflow, Fµ = µ(du/dy)A ∝ µ(U/L)L2 = µUL. This shows that Fi/Fµ ∝ ρUL/µ = Re. SinceFg ∝ ρL3g it follows that Fi/Fg ∝ U2/(gL) = Fr. Finally, since FΥ ∝ ΥL it follows thatFi/FΥ ∝ ρU2L/Υ = We.

5.7 Ekman (Ek) and Rossby (Ro) numbers are of importance in rotating flows.1 Ekman number is ameasure of the ratio between frictional (viscous) and Coriolis (fictional) forces on a fluid element;Rossby number correspondingly between inertial and Coriolis forces. Define these two numbersif Ω is a characteristic angular rate of rotation (characteristic velocity U ; length L).

Coriolis acceleration: 2Ω × V .

Let Fi represent the inertial force on a fluid element, mass times (linear) acceleration. Like-wise,let FΩ represent the (fictional) Coriolis force. With characteristic length L, characteristic velocityU , and fluid density ρ, the mass is proportional to ρL3. (Linear) acceleration is velocity changeper unit time, which means that it is proportional to U/(L/U) = U2/L, i.e., Fi ∝ ρU2L2. Let Fµrepresent viscous force. In simple shear flow, Fµ = µ(du/dy)A ∝ µ(U/L)L2 = µUL. From thegiven definition, the Coriolis acceleration is proportional to ΩU , i.e., FΩ ∝ ρL3ΩU . This showsthat (i) Fµ/FΩ ∝ µ/(ρΩL2) = ν/(ΩL2) = Ek, (ii) Fi/FΩ ∝ U/(ΩL) = Ro.

Chapter 6 — Viscous Flows in Ducts

6.1 Make a sketch on the variation of pressure and the velocity profile development for laminar flowin a horizontal pipe connected to a large reservoir. The submerged inlet is rounded. State anapproximate expression for the entrance length.

See Fig. 6.6, (Le/d)lam ≈ 0.06Re.

6.2 Define the hydraulic diameter Dh in duct flows. Find the hydraulic diameter for a circular and asquare cross-section. In engineering calculations, state the commonly accepted Reynolds numberover which the flow can be regarded as fully turbulent.

Hydraulic diameter, Dh = 4A/P, where A is the cross-sectional area perpendicular to the flow,P is the wetted perimeter. For a circular cross-section, A = πd2/4, P = πd ⇒ Dh = d; square:A = s2, P = 4s ⇒ Dh = s. With Reynolds number based on Dh, ReDh

, it is commonly acceptedin engineering duct flows that the flow is fully turbulent when ReDh

> 4000.

6.3 Derive the Hagen-Poiseuille formula for the volume flow rate in a straight circular pipe (fullydeveloped laminar flow). The suggested starting point is a force balance on a cylindrical element.

Force balance on a horizontal cylindrical element with radius r (pipe radius R) and length L:(p1 − p2)πr

2 − τ2πrL = 0, where τ = −µ(du/dr). Cleaning up gives du/dr = −C1r, whereC1 = (p1 − p2)/(2µL) > 0. Integration gives u = −C1r

2/2 + C2. For r = R the velocity is zero(no-slip), i.e., C2 = C1R

2/2, which gives u = (C1R2/2)[1− (r/R)2]. Maximum velocity for r = 0

(pipe center), C1R2/2 = umax. The volume flow rate is

Q =

∫ R

0u(r) 2πr dr = 2πumaxR

2∫ 1

0(1− η2)η dη = πR2umax/2

This shows that Q = πR4(p1 − p2)/(8µL), which is the Hagen-Poiseuille formula.

1Ekman and Rossby were Swedes; Vagn Walfrid Ekman (1874–1954) professor at Lund University 1910–1939.

9

6.4 Define the Darcy-Weisbach friction factor f for pipe flow. How is the (major) pressure losscalculated? What can be stated generally for f in fully developed laminar duct flow, regardlessof the cross section?

The pressure loss due to wall friction can be written as

∆pf = fL

Dh

ρV 2

2

This defines the Darcy-Weisbach friction factor f ; V is the mean velocity (V = Q/A), L is thepipe length, and Dh is the hydraulic diameter (Dh = 4A/P). For fully developed laminar ductflow, flam = C/ReDh

, where C depends on the type of cross-section (circular: C = 64).

6.5 Make an account for the recommended scheme of obtaining the pressure loss over a certain lengthin a duct of constant but non-circular cross section, assuming fully developed turbulent flow. Theexpression for the corresponding pressure loss in fully developed laminar flow is known.

The recommended scheme is to use the concept of effective diameter to calculate the frictionfactor, f = ϕ(Reeff , ϵ/Deff), e.g. from Haaland’s formula, eq. (6.49); Reeff = ρ V Deff/µ. Bydefinition, the frictional pressure loss is calculated from ∆pf = f(L/Dh)ρV

2/2, where Dh is thehydraulic diameter, and L is the pipe length. The concept is based on that the laminar frictionfactor is forced to be flam = 64/Reeff . Since flam = C/ReDh

, where C is known, it follows that

Deff

Dh=

ReeffReDh

=64

C

6.6 (a) Make a sketch of mean velocity profiles in fully developed laminar and turbulent flow in apipe (circular cross section). In turbulent flow, what is the influence of Reynolds number?

See Fig. 6.11. The laminar profile is a parabolic curve. For turbulent flow, an increasedReynolds number gives a more flat profile in the middle but a stronger variation closer to thewall (steeper velocity gradient).

(b) State an approximate power-law expression for the mean velocity profile in fully developedturbulent pipe flow. Use this expression to determine the ratio between the average velocity andthe mean velocity at the center of the pipe. What is the influence of Reynolds number?

Approximate velocity profile, fully turbulent flow: u/umax ≃ (1 − r/R)m, where the exponent0 < m < 1 decreases with increasing Re. Let η = r/R, i.e., u/umax = (1− η)m. The flow rate is

Q =

∫ R

0u 2πr dr = 2πR2umax

∫ 1

0(1− η)mη dη

After the substitution (1− η = ξ, dη = −dξ) the integral becomes

∫ 1

0ξm(1− ξ) dξ =

∫ 1

0(ξm − ξm+1) dξ =

1

m+ 1− 1

m+ 2=

1

(m+ 1)(m+ 2)

Since Q = πR2V , V/umax = 2(m+ 1)−1(m+ 2)−1. (Re ≈ 5× 104: m = 1/7, V/umax = 0.82.)

6.7 Make a sketch in a semi-logarithmic diagram of the time-mean velocity profile for pipe flow or afully developed turbulent boundary-layer. Use inner scaling for velocity and distance from wall,u+ and y+, which both should be defined. Mark out different regions in the diagram and state,if possible, pertinent expressions for the velocity in these regions. For boundary-layers, what isthe influence of pressure gradient? What is the influence of wall roughness?

See Fig. 6.10, linear vertical axis with u+ = u/u∗, logarithmic horizontal axis with y+ = u∗y/ν,where y is the vertical distance from the wall; u∗ is the friction velocity, u∗ =

√τw/ρ. Very

close to the wall, in the viscous sublayer (y+ ≤ 5), there is a linear velocity variation, u+ =y+. For high enough Reynolds numbers, there is a layer (sometimes called the overlap layer)

10

with a logarithmic velocity variation, u+ = κ−1 ln y+ + B, where κ ≃ 0.41; for a smooth wall(ϵu∗/ν = ϵ+ < 5), B ≃ 5.0. The logarithmic variation is valid within 50ν/u∗ < y < 0.25δ,where δ is the boundary layer thickness; for a pipe δ = R. For y/δ > 0.25, the velocity is higherthan indicated from the logarithmic function. For boundary layers, the deviation increases withincreasing pressure gradient. (For pipe flow, the maximum deviation only is about one unit ofu+.) On roughness effects and if ϵ+ > 5, there is a downward shift of the logarithmic function,i.e., B decreases with increasing ϵ+, see Fig. 6.12(a).

6.8 Explain in detail how a Pitot-static tube (Prandtl tube) works and derive an expression for thevelocity. State an approximate condition for the expression to be valid and discuss advantagesand disadvantages of this method for measuring local velocity.

See Fig. 6.30. Neglecting viscous effects the Bernoulli equation along the (assumed horizontal)stagnation streamline approaching the probe gives: p + ρU2/2 = p0, where p0 is the stagna-tion pressure at the tip, U is the (undisturbed) fluid velocity, p is the (undisturbed) fluid staticpressure. By its’ special design, the pressure at the position of the circumferential holes on thePrandtl tube, about 8 tip diameters downstream, is equal to p. This gives U =

√2(p0 − p)/ρ.

A simple measurement of the pressure difference (p0 − p), and known fluid density ρ gives thefluid velocity. The approximate condition for validity is ReD > 100, based on the tip diameter D(not ReD > 1000 as stated in the book). The probe is robust, reliable, and rather insensitive toyaw misalignment, see figure insert. A disadvantage is that it is only possible to measure slowlyvarying velocities; another is that the probe is intrusive, it might affect the flow development.

6.9 Describe briefly the concept of hot-wire anemometry (HWA). State some advantages and disad-vantages with this method of measuring local velocity, as compared with some other method.

See pages 421/2, and PM for lab 2b. The laser-Doppler anemometer (LDA, see page 422 andFig. 6.29h) is a candidate for comparison.

6.10 Describe two ways of measuring volume flow rate in a pipe. The two methods should be basedon different basic principles.

See pages 423 to 436; turbine meter (Figs. 6.32/3), vortex meter (Fig. 6.34), ultrasonic flow-meter (Fig. 6.35), rotameter (Fig. 6.36), Coriolis mass flowmeter (Fig. 6.37), laminar flowelement (Fig. 6.38), and flowmeters based on the Bernoulli obstruction theory, e.g. the Venturimeter, the thin-plate orifice, or the flow nozzle (Fig. 6.40).

Chapter 7 — Boundary Layers and Flow past Immersed Bodies

7.1 (a) Give a practical definition of the boundary layer thickness δ.

The most common practical definition: δ = δ99, where δ99 is the vertical distance from the wall tothe point where the velocity equals 99% of the velocity outside the boundary layer, u(y = δ99) =0.99U .

(b) How does δ vary along a flat, wide plate when exposed to a tangential oncoming flow?Power-law expressions in different regions should be stated. Illustrate with a figure.

See Fig. 7.1. Close to the leading edge the boundary layer is laminar (laminar boundary layer,LBL). Eventually, for a long enough plate, there will be transition to a turbulent boundary layer(TBL). The transition normally starts within 5×105 < Rex < 3×106, see Fig. 7.6; Rex = Ux/νis based on the distance x from the leading edge. For a LBL and Rex > 500 (approx.), δ ∝ x1/2

(Blasius solution); for a TBL, δ increases more rapidly, approximately as δ ∝ x0.8.

7.2 Define the displacement thickness δ∗, the momentum thickness θ and the shape factor H for aflat-plate boundary layer assuming two-dimensional incompressible flow. Explain why δ∗ is calledthe displacement thickness.

From definitions, δ∗ =∫ δ0 (1− u/U) dy, θ =

∫ δ0 (u/U)(1− u/U) dy; H = δ∗/θ. δ∗ is the (upward)

vertical displacement of the streamline that comes from the undisturbed flow and reaches the edgeof the boundary layer, at a specific distance x from the leading edge, see Fig. 7.4.

11

7.3 It can be shown from momentum-integral analysis that the skin-friction coefficient for the flat-plate boundary layer is cf = 2τw/(ρU

2) = 2 dθdx . The velocity outside the boundary layer isconstant (= U). For a laminar boundary layer of thickness θ the velocity profile can be roughlyapproximated as a linear function: u/U = A+Bη, where η = y/δ.

Determine the constants A,B and derive an expression for cf as a function of Rex = Ux/ν.Hint: θ = δ/C, where δ is a function of x and C is a constant (pure number).

A = 0 because of the no-slip condition, B = 1 since at y = δ the velocity is U , u/U = y/δ.From definition, θ = δ

∫ 10 η(η − 1) dη = δ/6, which means that τw/(ρU

2) = (1/6)dδ/dx. τwis the wall shear stress, τw = µ(∂u/∂y)y=0, in this case τw = µU/δ. Equality with τw frommomentum analysis gives µU/δ = ρU2(1/6)dδ/dx, or with separated variables δ dδ = (6ν/U)dx,where ν = µ/ρ. Integration from x = 0 where δ = 0, to position x with δ(x) gives δ2/2 = 6νx/U ,or δ =

√12νx/U . Combination with above gives cf = 2µ/(ρUδ) =

√ν/(3Ux) = 1/

√3Rex. (The

exact solution from Blasius gives 0.664 rather than 1/√3 = 0.577.)

7.4 Describe how the shape of the velocity profile in a boundary layer, close to the wall, is dependenton the pressure gradient. Explain why boundary-layer separation only can occur in regions withdecelerating outer velocity.

At the wall, irrespective of whether the boundary layer is laminar or turbulent, µ(∂2u/∂y2)y=0 =dp/dx. This means that the pressure gradient dp/dx determines the curvature of the velocityvariation u(y) at wall. Bernoulli equation applies just outside the boundary layer, p + ρU2/2 =const., and since the pressure variation across a boundary layer is negligible, dp/dx = −ρdU/dx.A decelerating outer velocity, dU/dx < 0, means that the pressure increases in the flow direction.Boundary layer separation occurs for zero wall shear stress, τw = µ(∂u/∂y)y=0, and this situationcan occur only with dp/dx > 0 since it implies a positive curvature at the wall with zero gradient,see Fig. 7.7.

7.5 Define the the drag coefficient CD in flow around a sphere. Describe in detail and in a diagramhow CD for a smooth sphere varies with the Reynolds number Re (in detail means that certainvalues on the axes should be marked; logarithmic axes are preferred). What is the asymptoticexpression for CD when Re ≪ 1? What is the influence of surface roughness?

By definition, D = CDAρU2/2. For flow around a sphere, A = πd2/4, where d is the diameter;

Reynolds number, Re = ρUd/µ = Ud/ν. In the limit Re → 0, CD = 24/Re (Stokes solution).In a log-log diagram, CD vs. Re, see Fig. 7.16, the Stokes solution is a straight line pointingdownwards, and it is valid up to Re ≈ 1. At higher Re, the drag coefficient decreases more slowly.At around Re = 103, it levels out at CD ≈ 0.5, and it stays at around this value up to aboutRe = 2× 105. Within a very limited Re-range at around Re = 3× 105, the drag coefficient dropsrapidly, from CD ≃ 0.5 to CD ≃ 0.2 (drag crisis). With increasing Re, the CD recovers somewhat,the indicated value at Re = 107 being CD ≈ 0.3. With a roughened surface, the drag crisis occursat a lower Re, otherwise the above sequence is similar; see Fig. 5.3 for the comparable case ofa circular cylinder in cross-flow.

7.6 Derive the integral condition D(x) = ρ bU2 θ(x), valid for a constant-pressure flat-plate boundarylayer (θ is the momentum thickness, x the distance from the leading edge, b is the plate widthand U is the velocity outside the boundary layer).

See Example 3.11. Inflow at x = 0, velocity vector V = U i; outflow at x, V = u(y) i. Uppercontrol surface is along the streamline that passes through y = δ (where u = U) at x, it joinsback to y = h at x = 0; lower surface at y = 0 (solid wall, u = 0). Momentum balance inx-direction, stationary conditions, constant pressure:

∫CS ρu(V · n) b dy =

∑Fx = −D, where

D is the (one-sided, frictional) drag on the plate. Along the stream-surface, (V · n) = 0; atinflow, n = −i; outflow, n = i, i.e., −D =

∫ h0 ρU(−U)b dy +

∫ δ0 ρu

2b dy, or since ρ is constant,

D = ρU2bh − ρb∫ δ0 u

2 dy. Mass balance requires Uh =∫ δ0 u dy. Combining with previous result

gives D = ρb∫ δ0 u(U − u) dy = ρ bU2 θ, where θ =

∫ δ0 (u/U)(1− u/U) dy.

7.7 The momentum equations for a two-dimensional boundary layer can be simplified drastically byusing order-of-magnitude arguments. Carry out these steps and show in particular the conse-

12

quences for the pressure field. The equations of motion for stationary, incompressible flow withnegligible effects of gravitation reads

∂u

∂x+∂v

∂y= 0

u∂u

∂x+ v

∂u

∂y= −ρ−1 ∂p

∂x+ ν

(∂2u

∂x2+∂2u

∂y2

)

u∂v

∂x+ v

∂v

∂y= −ρ−1 ∂p

∂y+ ν

(∂2v

∂x2+∂2v

∂y2

)

The boundary layer thickness δ is much smaller than some characteristic dimension of the bodyL and the body surface radius of curvature, respectively. The Reynolds number based on theoncoming constant velocity U and L is very high (Re = UL/ν ≫ 1).

Non-dimensional variables, with estimates of magnitude orders: x = x/L ∼ 1, y = y/L ∼ δ/L =δ ≪ 1, u = u/U ∼ 1, v = v/U , p = (p− p∞)/(ρU2); non-dimensional equations:

∂u

∂x+∂v

∂y= 0

u∂u

∂x+ v

∂u

∂y= −∂p

∂x+Re−1

(∂2u

∂x2+∂2u

∂y2

)

u∂v

∂x+ v

∂v

∂y= −∂p

∂y+Re−1

(∂2v

∂x2+∂2v

∂y2

)

Since at constant y, u =∫ x0 (∂u/∂x) dx = (∂u/∂x)avgx ∼ (∂u/∂x)avg ∼ 1, it follows that

∂nu/∂xn ∼ 1. With the same reasoning using u =∫ y0 (∂u/∂y) dy = (∂u/∂y)avgy ∼ (∂u/∂y)avgδ ∼

1, ∂nu/∂yn ∼ δ−n. ∂u/∂x ∼ 1 in the the continuity equation requires that also ∂v/∂y ∼ 1,since neither term can be neglected. But v = (∂v/∂y)avgy ∼ (∂v/∂y)avgδ, so v ∼ δ, i.e.,v ≪ u. All terms in the x-momentum equation must have the same order of magnitude, in-ertia must balance pressure and viscous forces. However, the first viscous term can be neglectedsince ∂2u/∂x2 ≪ ∂2u/∂y2. Both inertial terms in x-momentum equation are ∼ 1, which showsthat ∂p/∂x ∼ 1, and therefore p ∼ 1. Also, since ∂2u/∂y2 ∼ δ−2, it follows that Re−1 ∼ δ2,i.e., Re ≫ 1. Inertial terms in the y-momentum equation are ∼ δ. Other terms must have thesame order of magnitude, which shows that ∂p/∂y ∼ δ, or |∂p/∂y| ≪ 1; p depends on x only. Indimensional variables, p(x) only, which means that there is a negligible pressure variation acrossthe boundary layer. Finally, since all terms are ∼ δ, the whole y-momentum equation can bedropped! The remains are the so-called boundary-layer equations, eq. (7.19):

∂u

∂x+∂v

∂y= 0

u∂u

∂x+ v

∂u

∂y= −ρ−1 dp

dx+ ν

∂2u

∂y2

Also see lecture notes, or hand-outs from the book by Prandtl & Tietjens (exchange students).Note: If given as an exam question, all steps above are not required; for instance, orders ofmagnitudes for ∂nu/∂xn, ∂nu/∂yn, etc., can be written out directly.

Introduction to Turbulence

T.1 Make a brief account of (at least) four characteristics of turbulence (fully developed turbulentflow).

13

See page T-1, and pages 357–360 in White. (i) irregularity or randomness; (ii) high diffusivityor spreading rate; (iii) large Reynolds numbers, turbulence often originates as an instabilityof laminar flow at high enough Re; the Re is always large if based on large-scale turbulencecharacteristics; (iv) three-dimensional vorticity fluctuations; (v) high dissipation rate, rapid decayif supply energy is lost.

T.2 Introduce the Reynolds decomposition of turbulent fluid motion (velocity and pressure) and carryout the time-averaging procedure for the continuity equation in cartesian coordinates assumingtime-mean stationary, incompressible turbulent flow.

See page T-4, and page 371 in White. At a certain position in a flow field, the local velocitycomponents ui, where i = 1, 2, 3, and the local pressure p can be written as

ui(t) = ui + u′i(t)

p(t) = p+ p′(t)

where the overline means time-averaging. The fluctuating components (with a prime) have a zerotime-average, e.g., p′ = 0. The continuity equation for incompressible flow reads

∂u

∂x+∂v

∂y+∂w

∂z= 0

Now take the time-average of this whole equation. Since A+B = A + B for any quantities Aand B, it is enough to look at one term. Since the procedures of time-averaging and derivation

can be carried out in any order, and A = A, it follows that

∂

∂x(u+ u′) =

∂u

∂x+∂u′

∂x=

∂

∂xu+

∂

∂xu′ =

∂u

∂x

which means that

∂u

∂x+∂v

∂y+∂w

∂z= 0

T.3 Consider a two-dimensional boundary layer of thickness δ over a smooth surface in stationary,incompressible flow. Suppose that this boundary layer can be divided into two partially overlap-ping cross-flow regions: (1) the near-wall region with very high mean velocity gradients and (2)the outer region with small mean velocity gradients.

(a) State and define a velocity scale that is characteristic for the turbulent motions across thewhole boundary layer.

See page T-12, and page 372 in White. The friction velocity, u∗ =√τw/ρ, where τw is the local

time-mean wall shear stress, is a characteristic velocity scale of wall-bounded turbulence.

(b) State relevant length scales in the two regions and provide the basic criterion that motivatesthe assumption of a sub-division of the boundary layer.

See page T-12, and page 372 in White. Inner region: ℓ = ℓv = ν/u∗, viscous length scale;outer: ℓ = ℓy = δ (or ℓ = ℓy = R for pipe flow), outer length scale. Since by observationℓy/ℓv = u∗δ/ν = Re∗ ≫ 1, the basic assumption is that for Re∗ → ∞ the dynamically importantturbulent motions within the inner and outer region are loosely independent.

(c) At distances sufficiently far from the wall, viscous effects can be neglected. Together withthe above (+ some further reasoning) it can be inferred that the mean velocity gradient isproportional to the ratio between the characteristic (turbulent) velocity scale and the prevalentlength scale. Show that this leads to that the velocity variation within the overlapping regionmust be logarithmic, i.e., u = C1 ln y + C2.

14

See pages T-15/6. Since u∗ is a relevant velocity scale we have that du/dy ∝ u∗/ℓ, where ℓ is themost characteristic length scale of the turbulent eddies. In the outer region (2) with small meanvelocity gradients and large eddies, ℓ ∼ δ, where δ is the boundary layer thickness; close to thewall with large gradients, in the inner region (1), ℓ ∼ ν/u∗. For high enough ratio between δ andν/u∗, Re∗ = u∗δ/ν ≫ 1, and within a layer that overlaps the two regions, neither one of thesetwo scales can be the most characteristic one. The only remaining possible length scale to usethen is the wall distance y, i.e., du/dy ∝ u∗/y, which by integration gives u = u∗(a ln y + b).

T.4 Define the root-mean-square value (RMS-value) of a quantity that has a non-zero time average.Sketch how the RMS-values of the fluctuating components of the velocity vector vary across aconstant-pressure turbulent boundary layer. The velocities should be scaled using the frictionvelocity u∗.

See page 371 in White and pages T-3/13; uRMS is the standard deviation of the fluctuations

around the mean value u, uRMS = u′ =√T−1

∫ T0 (u− u)2 dt, where T → ∞. Across the constant-

pressure (flat-plate) TBL, the fluctuations are largest in the stream-wise direction (u′), smallestperpendicular to the wall (v′); the levels of span-wise fluctuations are roughly in between, w′ ≈(u′ + v′)/2. These RMS-components are of the same order of magnitude as the friction velocity,the maximum value being u′ ≈ 2.5u∗ (occurs very close to the wall).

Chapter 8 — Incompressible Potential Flow

8.1 For irrotational flow (∇×V = 0) there exists a velocity potential function ϕ.

(a) Define ϕ and derive its’ differential equation for irrotational incompressible flow. What isthis equation called?

From mathematics it’s a fact that if a vector field has no curl it can be expressed as the gradient ofa scalar function. In this case the scalar is called the velocity potential, V = ∇ϕ. Incompressibleflow means that ∇ ·V = 0. Combined with the definition of ϕ, ∇·(∇ϕ) = ∇2ϕ = 0. This iscalled the Laplace equation.

(b) State boundary conditions for ϕ in flows around a body without any effect of free surfaces.

At the (assumed solid) body surface the velocity normal to the boundary is zero, ∂ϕ/∂n = 0,where n is perpendicular to the body surface. On outer boundaries, if sufficiently far away,it can be inferred that the velocity vector is known, i.e., in cartesian coordinates that V =(∂ϕ/∂x, ∂ϕ/∂y, ∂ϕ/∂z) is known.

(c) For a given potential field ϕ, how is the pressure field determined?

For incompressible flow with ∇×V = 0 inserted into the Navier-Stokes equation, the result isthe Bernoulli equation:

ρ∂ϕ

∂t+ p+

ρV 2

2+ ρgz = C(t)

where p is the pressure, V = |∇ϕ|, and z is upwards. If the flow is stationary, ∂ϕ∂t = 0 and C is

a pure constant, same at all locations (not only along a streamline).

8.2 (a) Define circulation Γ along a closed curve C.

Γ =∮C V · ds, where ds is an anti-clockwise arc vector along the closed line C; V is the velocity

vector, see Fig. 8.7.

(b) The following applies for a line vortex placed at the origin: vθ = K/r, vr = vz = 0. Determinethe circulation along a closed curve that encompasses the origin.

Let the closed curve be a circle of radius a, ds = a dθ θ ⇒ V · ds = K dθ; Γ =∫ 2π0 K dθ = 2πK.

8.3 Derive the governing equation for streamlines belonging to a doublet of strength λ = 2 am placedat the origin, along the x-axis. Sketch some streamlines.

Hint: The stream function for a line source with strength m placed at x = −a, y = 0:

15

ψk = m tan−1 y

x+ a

Trigonometric identity: tan(α− β) =tanα− tanβ

1 + tanα tanβ

The sink at x = a has stream function ψs = −m tan−1 yx−a . Superposition gives

ψ/m = tan−1 y

x+ a− tan−1 y

x− a= α− β

Taking the tangent on both sides and using the trigonometric identity gives

tan(ψ/m) = tan(α− β) =y/(x+ a)− y/(x− a)

1 + y2/(x2 − a2)=

−2ay

x2 − a2 + y2

When taking the arc-tangent on both sides, and noting that tan−1 ϵ = ϵ when ϵ → 0 can be usedwhen a→ 0, the final result is (see Fig. 8.12 for streamlines):

ψ =−2may

x2 + y2=

−λyx2 + y2

8.4 The two dimensional potential flow around a circular cylinder is governed by superposition of adoublet at the origin (ψ1 = −λr−1 sin θ) and a parallel flow (ψ2 = U∞r sin θ).

Determine

(a) the velocity field (vr = r−1 ∂ψ∂θ and vθ = −∂ψ

∂r ), (b) the radius a, (c) the pressure distributionalong the surface of the cylinder, expressed as a pressure coefficient Cp =

p−p∞12ρU2

∞, where p∞ is the

pressure at large distances upstream along the stagnation line.

(a) Superposition gives ψ = −λr−1 sin θ + U∞r sin θ + C = U∞r sin θ[1− λ/(r2U∞)

]+ C.

Velocity components:

vr =1

r

∂ψ

∂θ=

(1− λ

U∞r2

)U∞ cos θ

vθ = −∂ψ∂r

= −(1 +

λ

U∞r2

)U∞ sin θ

(b) At r = a (at the surface of the cylinder) we must have vr = 0, i.e., a =√λ/U∞.

(c) Bernoulli equation between a point far upstream at θ = π where p = p∞ and (vr = −U∞,vθ = 0) and a point on the cylinder surface with p = ps, vθ = −2U∞ sin θ:

p∞ +ρU2

∞2

= ps +ρv2θ2

= ps +ρ

2U2∞ 4 sin2 θ

The pressure coefficient is

Cp =p− p∞ρU2

∞/2= 1− 4 sin2 θ

(See Fig. 7.13)

16

8.5 (a) Use the condition of irrotational flow together with vr = 0 to derive the velocity field for aline vortex (stationary, incompressible flow in a plane).

Hint:

ζz =1

r

∂

∂r(rvθ)−

1

r

∂vr∂θ

∇ ·V =1

r

∂

∂r(rvr) +

1

r

∂vθ∂θ

Incompressible flow, ∇ ·V = 0, with vr = 0 gives ∂vθ/∂θ = 0, which means that vθ is onlydependent on r, vθ(r); irrotational flow, ζz = 0, then gives

d

dr(rvθ) = 0 ⇒ vθ =

C

r

where C is a pure constant.

(b) The flow field in the outer parts of a tornado can be approximated as a line vortex, theinner parts like a solid-body rotation. Use these approximations to derive simplified analyticalequations for the velocity and pressure field within a tornado. Sketch the variations of velocityand pressure with the radial distance r from the center of the tornado.

Hint: Momentum balance in radial direction ⇒ ρ v2θ/r =∂p∂r

See Problem P8.14, Fig. P8.14. Solid-body rotation: vr = 0, vθ = Kr, i.e., dp/dr = ρK2r.Integration gives p = ρK2r2/2 + K2. Line vortex: vr = 0, vθ = C/r, i.e., dp/dr = ρC2/r3.Integration gives p = −ρC2/(2r2) + C2. Let p = p∞ at large distance from the origin, C2 = p∞.Let the pressure at the origin be p(r = 0) = K2 = pc. Equal velocities at matching point at r = Rgives vθ,R = C/R = KR, and pc = p∞ − ρv2θ,R. (As a numerical example, let vθ,R = 50 m/s in

air-flow with ρ = 1.2 kg/m3; this gives pc − p∞ = −3 kPa.)

8.6 (a) Consider incompressible potential flow around a circular cylinder with added circulation Γ.The axis of the cylinder is at the origin. The velocity along the cylinder surface (r = a) in polarcoordinates is

vθ = −2U∞ sin θ +Γ

2πa

where U∞ is the undisturbed velocity upstream of the cylinder.

Show, using integrations, that the drag is zero and that the lift per unit width is −ρU∞Γ.

Note,∫ 2π0 (sin θ)2dθ = π.

Drag, D = −∫ 2π0 (ps − p∞) cos θ ba dθ; lift, L = −

∫ 2π0 (ps − p∞) sin θ ba dθ. Bernoulli equation,

p∞ + ρU2∞/2 = ps + ρv2θ/2, gives

Cp =ps − p∞ρU2

∞/2= 1− 4 sin2 θ + 4β sin θ − β2

where β = Γ/(2πaU∞). The integral of cos θ times any power of sin θ over a full cycle 2π is zero(including the power of zero, which is a constant). All terms for the drag are of this kind, whichmeans that the drag is zero. The integral over 2π of any odd power of sin θ is zero. The onlyremaining term contributing to lift then is the term with sin2 θ. The lift per unit width becomes

L/b = −ρU2∞2

a(4β)π = −ρU∞Γ

(b) Describe, using streamlines, how the velocity field around the cylinder looks like for differentvalues of β = K/(U∞a), where K is the vortex strength. How is K related to Γ?

See Fig. 8.14, K = Γ/(2π); note the case with β = 2, for which the two stagnation points jointogether at θ = 90; for higher β, the stagnation point moves upwards along the y-axis.

17

8.7 State the Joukowsky lift theorem for incompressible, potential plane flow. Define quantities andillustrate with a simple sketch.

According to inviscid theory, the lift per unit width of any closed contour (any cylinder of anyshape) in a uniform stream equals ρU∞Γ, where Γ is the net circulation around the contour.The direction of the lift is 90 from the stream direction, rotating opposite to the circulation. Asimple sketch could be for instance a wing profile moving to the left with a resulting clock-wisecirculation. The lift is then upwards.

8.8 A two-dimensional (AR → ∞), symmetric and thin wing is suddenly brought into a constantvelocity. The angle of attack is small but not zero. Describe the flow development with specialemphasis on the flow at the trailing edge and the generation of a circulation and thus a lift forceon the wing. (See Fig. 7.23).

See Fig. 7.23. Very soon after start-up the flow is close to being fully irrotational, negligiblecirculation, negligible change of momentum, negligible lift, Fig. 7.23(a). At this instant, thewall friction is not yet developed and therefore the near-wall flow can in fact turn around thetrailing edge, despite the sharp turn. After the turn, the near-wall flow comes to a stagnationpoint where it joins with the other stream, leaving the surface. Very soon after this, as the wallfriction develops, the increasing adverse pressure gradient towards the stagnation point (on theupper side) will result in flow separation, which quickly moves to the trailing edge, forming avortex, see Fig. 7.23(b). The rotation (circulation) of this vortex is anti-clockwise, since therestill is a tendency for the flow to pass over the trailing edge. The total flow has to be without anyresulting circulation so there is a build-up of an equal amount of clock-wise circulation aroundthe wing. Because of this circulation there is also a net downward change in momentum whencomparing the upstream and downstream parts close to the wing, and accordingly, there is anupward lift on the wing. As the wing moves further to the left the circulation of the vortexincreases, as do the wing-bounded circulation, proportional to the lift, Fig. 7.23(c). Eventuallythere is no more circulation added to the vortex and it will remain essentially at the place wherethe wing started. The trailing edge flow is now very smooth, the flow leaves the wing without anysignificant separation or vortex roll-up, see Fig. 7.23(d). The wing-bounded circulation is nowfully developed, as is the lift. (The circulation that is built up around the wing can be demonstratedfrom an experiment when the wing suddenly stops, a clock-wise vortex, of equal strength as thestarting vortex, will then be released from the wing.)

8.9 (a) State the Kutta condition. How can this condition be motivated physically?

The Kutta condition: the flow leaves the trailing edge of the (working) wing in a smooth fashion,see Fig. 8.22. For an edge with some finite angle, or a somewhat blunt one, there is a stagnationpoint very close to the trailing edge. For a wing with a cusp, very sharp, edge the local velocitieson either side of the wing are more or less equal. The condition is motivated physically from pureobservation, this is how the flow behaves close to the trailing edge of a lift-generating wing. TheKutta condition is used to provide the wing theory with the correct amount of circulation.

(b) Write out the mathematical expression of the lift coefficient CL valid for a slender butcambered two-dimensional wing profile at small angles of attack. The maximum camber ish≪ C, where C is the wing chord. Also state the position of the resultant lift force (the centerof pressure, CP). Illustrate schematically.

CL = 2π sin(α+ β), where α is the angle of attack, β = tan−1(2h/C). [For slender wings, thereis no support from experiments to verify the influence of thickness from eq. (8.71)]. Since thetheory is valid only for small camber angles β, the formula can also be written CL = 2π(α+ β),where β = 2h/C. The center of pressure is at the one-quarter-chord point, (x/C)CP = 1/4.

8.10 (a) Discuss shortly and illustrate with figures how the lift force on a finite-span wing gives riseto an induced drag.

On a lift-generating wing there is a tendency for the flow on the lower (pressure) side to moveto the upper (suction) side. On a finite-span wing this short-cut flow happens at the wing tips.This leakage between the lower and upper side introduces a swirling motion that is swept away

18

downstream, wing tip vortices. When looking from behind (Fig. 8.28), the two wing tip vortexsystems have opposite rotation and each superimposes a downwash flow in the downstream regionbetween the wing tips. In particular, there will be a downwash that is also felt on the plane itself.On the plane, during horizontal cruising conditions, the apparent wind will then be as if the windcame a little from above; the effective angle of attack will be less than the actual.

From finite-span wing theory, there is no drag but there is lift, which is perpendicular to theapparent wind direction. Since this is not the actual wind direction there will be a contributionfrom this lift force in the actual streamwise direction, a lift-induced drag, see figure above.

(b) State the correct formula for the lift coefficient CL under the assumptions of elliptical planformarea, small thickness and camber, small angles of attack, high Re, and for a given wing span-to-chord ratio AR.

CL =2π(α+ β)

1 + 2/AR

where AR = b/C, β = 2h/C (C is the mean chord, h is the maximum camber).

(c) Sketch how AR influences CD and CL as a function of the angle of attack α.

See Fig. 7.27.

8.11 Describe what is meant by the concept of added mass. How can the added mass be estimatedwith a known potential flow around the body? State without derivation the added mass forpotential flow around a sphere.

When a body accelerates in a fluid it also accelerates parts of the fluid. Since this fluid massacceleration must also be accounted in a momentum balance for the body, the body behaves as if itwere heavier by an amount that is called the hydrodynamic or added mass (mh). The added masscan be calculated if the total kinetic energy of the fluid relative motion is set equal to the addedmass times the kinetic energy per unit mass of the body, mhU

2/2, where U is the body velocity.From potential theory and the flow around a sphere, mh is one-half of the displaced fluid mass.

Chapter 9 — Compressible Flow

9.1 Use conservation laws of mass and momentum to derive an expression for the velocity C of a finitepressure wave which propagates in an otherwise still fluid. The thickness of the wave-front canbe assumed to be infinitely small so that so that one-dimensional approximations are applicable.After this, specialize to a vanishing small wave amplitude wave assuming adiabatic conditions(velocity of sound).

Consider a fixed thin control volume surrounding the wave front, see Fig. 9.1. With thisperspective the flow through the control volume (CV) is stationary. The incoming velocity isequal to C, outgoing velocity equal to C − ∆V , where ∆V is the (assumed low) fluid velocitydragged along behind the wave front. The wave front is associated with an increase in fluid den-sity ρ and static pressure p. By mass balance, the mass flux coming in must equal the mass

19

flux coming out. Equal areas A on both sides gives m = ρAC = (ρ + ∆ρ)A(C − ∆V ). Thismeans that ∆V = C(∆ρ/ρ)/(1 + ∆ρ/ρ) > 0. The sum of forces on CV must equal the netoutflow of the momentum. The only forces acting in the shock-normal direction are those due topressure, viscous normal stresses can be neglected. Momentum balance in the normal direction:pA− (p+∆p)A = m(Vout − Vin) = ρAC(−∆V ), i.e., ∆p = ρC∆V . Insertion of the expressionfor ∆V yields

C =

√∆p

∆ρ

(1 +

∆ρ

ρ

)

In the limit of infinitesimal strength ∆ρ/ρ → 0 the wave speed becomes the velocity of sounda. No temperature gradients except within CV means that the flow can be treated as adiabatic.Likewise for vanishing velocity gradients means frictionless conditions, i.e., isentropic flow, flowat constant entropy s,

a =

√(∂p

∂ρ

)s

9.2 (a) Define the concepts of stagnation temperature and stagnation pressure.

Stagnation temperature is the temperature of the fluid when brought to rest adiabatically; stagna-tion pressure is the pressure of the fluid when brought to rest isentropically, that is adiabaticallyand without any losses.

(b) Assuming compressible flow with a perfect gas, derive a relationship between stagnationtemperature T0, static temperature T , k = cp/cv and Mach number Ma.

Apply the energy equation between two points along an expanding stream tube in which the fluid isbrought to rest adiabatically, from velocity V and temperature T to V = 0 and T = T0; stationaryflow. Neglected potential energy variations gives h+ V 2/2 = h0, where h is the enthalpy per unitmass. Since ∆h = cp∆T for a perfect gas, T0/T = 1 + V 2/(2cpT ). Mach number: Ma = V/a,where a =

√kRT for an ideal gas, R = cp − cv is the gas constant; cp = kR/(k − 1). Plug-in

gives

T0T

= 1 +k − 1

2Ma2

(c) Again assuming a perfect gas, derive a relationship between stagnation pressure p0, staticpressure p, k = cp/cv and Mach number Ma.

Isentropic relation: pvk = const. (v is the specific volume). Since pvk = pp−k(pv)k = p1−k(RT )k

for an ideal-gas law (pv = RT ), it follows that T k/(k−1)/p = const., or p0/p = (T0/T )k/(k−1), or

when using the result from above,

p0p

=

(1 +

k − 1

2Ma2

)k/(k−1)

9.3 (a) Assuming one-dimensional adiabatic flow of a perfect gas, write out a complete set of equationswhich governs the thermodynamic state immediately downstream of a stationary normal-shockwave in a nozzle,

Index 1 just upstream of the shock, index 2 just downstream; A1 = A2 since the shock is so verythin. Mass balance: ρ1V1 = ρ2V2; momentum balance: p1 − p2 = ρ1V1(V2 − V1); energy balance(perfect gas): T1 + V 2

1 /(2cp) = T2 + V 22 /(2cp); ideal-gas: p1/(ρ1T1) = p2/(ρ2T2). Assuming

the upstream conditions and cp = kR/(k − 1) to be known, we have four equations and fourunknowns (p2, T2, V2, ρ2). However, because of the quadratic terms from the velocity, there aretwo solutions, but only one is possible since the entropy must increase, s2 > s1.

20

(b) How does the stagnation temperature, Mach number, pressure, temperature, density, entropyand stagnation pressure vary across a normal-shock wave?

Stagnation temperature remains the same since the flow is adiabatic. Out of the rest, they allincreases, except stagnation pressure and Mach number, see Fig. 9.9.

9.4 A converging-diverging nozzle (a Laval nozzle) is connected to a large reservoir with a constantpressure pr. The pressure outside the nozzle, the back pressure, is pb (< pr). The gas is perfect(ideal gas with constant cp and cv). The flow can be considered adiabatic and frictionless.

(a) Sketch the variation of pressure through the nozzle (and just outside of it) for different valuesof the pressure ratio pb/pr. In particular, mark out the positions of shocks and expansion waves.

See Fig. 9.12(a,b).

(b) How does the mass flow rate vary with the pressure ratio?

See Fig. 9.12(c).

9.5 According to the theory of adiabatic frictional flow of a perfect gas in a straight pipe the flowalways strives towards sonic conditions, Ma = 1, regardless of the upstream (inlet) Mach numberMa1. There is a certain pipe length L∗(Ma1) for which the exit Mach number will be exactlyunity. The flow is then choked.

Explain in detail what happens if the pipe length is longer than this critical length, L > L∗. Sub-critical and supercritical inlet flow should be treated separately. Please note that inlet conditionsmight be altered.

See pages 662/3, and Fig. 9.15.

9.6 (a) Describe geometrically the deflection process associated with an oblique shock in a plane. Inparticular, mark out the shock wave angle β and the deflection angle θ.

See Fig. 9.20.

(b) Use control volume analysis to show that the tangential component of the velocity is conservedafter the passage of the oblique shock.

The control volume is so very thin that the closing areas on the sides are negligible. This meansthat the resulting pressure force in the tangential direction is negligible, as is the tangential massflux (and viscous forces),

∑Ft = 0. The mass flux per unit area through the control volume is

ρ1Vn1. Tangential momentum balance requires ρ1Vn1(Vt2 − Vt1) =∑Ft = 0 ⇒ Vt2 = Vt1.

(c) Make a sketch on the relation between the deflection angle and the shock wave angle withMach number Ma1 as parameter. Mark out regions for weak and strong shocks, normal shockwaves, and Mach waves. How does the Mach number affect the maximum deflection angle? Alsomark out the line where Ma2 = 1.

See Fig. 9.23. Mach waves are represented by the β-crossings for θ = 0 (β = µ = sin−1(Ma−11 ));

the normal shock is the point where all curves for θ = 0 join together at β = 90.

9.7 (a) Describe the flow pattern around a thin and pointy wing profile in supersonic two-dimensionalflow at a small angle of attack. The profile has a prismatic cross-section (plane surfaces).

See Fig. 9.27(b), but make it even thinner so that it really is an expansion fan at the lowertrailing edge (as the figure is drawn it ought to be an oblique shock there).

(b) The following formula applies to supersonic two-dimensional flows at small angles θ:

p2 − p1p1

=kMa21√Ma21 − 1

θ

where index 1 refers to the state upstream of the deflection (Ma1 > 1).

Use this formula to derive expressions for CL and CD valid for supersonic flow around a thin flatplate at small angles of attack. The upstream Mach number is Ma∞. The width of the plate (b)is much larger than its chord C.

21

See Fig. 9.27(a). There is an oblique shock at the lower front edge, deflection angle θ = α;expansion fan at the upper frontal edge, deflection angle θ = −α. The upward force perpendicularto the plate is F = (p2−p3)Ap = (p2−p3)bC. Small angle of attack, lift: L = F cosα ≃ F ; drag:D = F sinα ≃ Fα. Since p2 − p3 can be written as p2 − p∞ − (p3 − p∞), it follows that

F =kp∞Ma2∞√Ma2∞ − 1

[α− (−α)] = kp∞Ma2∞√Ma2∞ − 1

2α

By definition of lift coefficient, and using the ideal gas law together with the Mach number:

L = CLρ∞U

2∞

2Ap =

CLAp2

p∞RT∞

(Ma2∞kRT∞) = CLApkp∞Ma2∞/2

This shows that

CL ≃ 4α√Ma2∞ − 1

The drag coefficient is CD ≃ CLα.

Chapter 10 — Open-Channel Flow

10.1 Derive an expression for the wave speed c of a surface wave front with amplitude δy in an openrectangular channel of width b. The undisturbed shallow-water depth is y. After this, specializeto δy/y → 0.

See Fig. 10.4. A wave-front of amplitude δy is moving to the left with a velocity c, enteringstill water at an undisturbed depth of y. The velocity dragged along behind the wave is denotedδV , assumed constant over the depth. Consider a rectangular control volume (CV) of heighty + δy fixed around the wave-front, a steady flow situation. The velocity entering the CV is c,outflow velocity c − δV . Mass balance requires ρcy = ρ(c − δV )(y + δV ), or δV = c δy/(y +δy). Constant density and rectilinear flow means that pressure increases linearly with depth,starting from the constant ambient pressure at the free surfaces. Neglect wall friction, short CV.Momentum balance, inflow at section 1, outflow at section 2 (vertical coordinate h):

∫ y+δy

0(p1 − p2) b dh = bρgy2/2− bρg(y + δy)2/2 = ρcyb(c− δV − c) = −bρcy δV

The L.H.S. is −bρg δy(y + δy/2), i.e., δV = g δy(y + δy/2)/y. Equality for δV gives

c =

[gy

(1 +

δy

y

)(1 +

1

2

δy

y

)]1/2When δy/y → 0, c = c0 =

√gy.

10.2 Consider one-dimensional flow in an open channel of constant cross-section (area A, width atsurface b0, velocity V ).

(a) State the wave speed c0 of an infinitesimal shallow-water surface wave. Define the Froudenumber.

c0 =√gyh, where yh = A/b0 (sometimes called the hydraulic depth); Froude number, Fr = V/c0,

where V is the velocity of the liquid (water).

(b) State c0 and the specific energy E for the rectangular cross-section (b0 = b, A = by).

yh = A/b0 = y gives c0 =√gy. Specific energy, E = y + V 2/(2g).

22

10.3 Water flows in an open channel laid on a certain slope, S0 = tan θ (θ ≪ 1). The water crosssection (and depth) as well as the inner surface condition are constant.

(a) Define hydraulic radius of the cross section, Rh.

Rh = A/P , where A is the cross-section area, P is the wetted perimeter (Fig. 10.2b).

(b) According to Manning (1891),√8g/f ≃ R

1/6h /n, where Rh is in meters and n is a constant.

Describe how the flow rate Q can be determined. Starting point: extended Bernoulli equation.Velocity variations across the cross section can be neglected.

Constant depth (y = yn) means that the velocity is constant, V = V0. When applied to astreamline on the surface with constant pressure, the extended Bernoulli equation gives: z1 =z2 + hf , where z is upward. From geometry (Fig. 10.2a), z1 − z2 = S0L, where L is thehorizontal distance between the sections. Head loss by definition,

hf = fL

Dh

V 20

2g

Using hf = S0L and Dh = 4Rh gives

V0 =

(8g

f

)1/2√RhS0 = Q/A

Using√8g/f ≃ R

1/6h /n, the flow rate, Q = V0A, is

Q = AR2/3h S

1/20 /n

10.4 For open-channel flow in a rectangular channel, derive an expression for the critical depth ycat a given discharge per unit width (q). Show that the Froude number at this depth is unity(Frc = 1).

The critical depth is the depth that gives the minimum specific energy for a constant flow rateper unit width. With q = Q/b = V y the specific energy is E = y + V 2/(2g) = y + q2/(2gy2);dE/dy = 1 − q2/(gy3). When set to zero, y = yc = (q2/g)1/3. When inserted, the minimumvalue is Emin = 3yc/2. At Fr = 1, V = Vc =

√gyc. The specific energy then is Ec = yc + yc/2 =

3yc/2 = Emin. This shows that Fr = 1 when water is running at critical depth.

10.5 Water flows at vanishing slope along a straight open channel. Perpendicular to the flow and onthe bottom of the channel there is a small ridge, a bump, with maximum height (amplitude) ∆h.The cross section is rectangular. The flow over the bump can be treated as one-dimensional andfrictionless. Assume given upstream velocity, upstream depth and ∆h.

(a) Derive an implicit relationship for the depth at the top of the bump.

See Fig. 10.9(a). When neglecting wall friction, E1 = E2 +∆h, where E = y+ V 2/(2g). Sincealso q = V1y1 = V2y2 from mass balance, the velocity V2 can be eliminated, y1 + q2/(2gy21) =y2 + q2/(2gy22) + ∆h. This is a cubic equation for y2, three roots but one is negative. The two(positive) roots are called alternate depths.

(b) Describe the depth variation along the bump, for different ∆h < ∆hmax and upstream Froudenumbers (Fr1 < 1 and Fr1 > 1). Illustrate with a diagram (depth y vs. specific energy E).

A plot of y vs. E for a certain q is shown in Fig. 10.9, also see Fig. 10.8. The upper partof the curve is subcritical flow (Fr < 1), lower is supercritical (Fr > 1), critical at the nose, atEmin. When going up the bump E decreases with ∆h (E2 = E1 −∆h). If Fr1 < 1, upper branch,the surface at the bump crest dips down, since the decrease in depth is larger than ∆h (as shownmore clearly in Fig. 10.8, the upper asymptote is at 45, y = E). If Fr1 > 1, lower branch, thesurface at the bump crest goes up, a bit more than the bump height.

Christoffer Norberg, tel. 046-2228606

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