Mechanics of Fluids

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this depth is greater than the critical depth, that is, uniform flow would be tranquil. A gradual transition from rapid to tranquil flow is not possible. As the depth of rapid flow increases, the lower limb of the specific-energy diagram (Fig. 10.18) is followed from right to left, that is, the specific energy decreases. If the increase of depth were to continue as far as the critical value, an increase of specific energy would be required for a further increase of depth to the value corresponding to uniform flow downstream. However, in such circumstances an increase in specific energy is impossible. In uniform flow the specific energy remains constant and the total energy decreases at a rate exactly corresponding to the slope of the bed. For any depth less than that for uniform flow the velocity is higher, and friction therefore consumes energy at a greater rate than the loss of gravitational energy. That is, the energy gradient is greater than the slope of the channel bed, and so, as shown in Fig. 10.22, the specific energy must decrease. Consequently, a hydraulic jump forms before the critical depth is reached so that uniform flow can take place immediately after the jump. It represents a discontinuity in which the simple specific-energy relation is temporarily invalid (because the streamlines are then by no means straight and parallel). The depth of the tranquil flow after the jump is determined by the resistance offered to the flow, either by some obstruction such as a weir or by the friction forces in a long channel. The jump causes much eddy formation and turbulence. There is thus an appreciable loss of mechanical energy, and both the total energy and the specific energy after the jump are less than before. Determining the change in depth that occurs at the hydraulic jump is frequently important. Although a hydraulic jump may be formed in a channel with any shape of cross-section, we shall again consider here only a channel of uniform rectangular cross-section because the analysis for other sections is mathematically complicated. A rearrangement of eqn 10.23 gives h 2 2 + h1h2 − 2u 2 1 h1/g = 0 The hydraulic jump 439 Fig. 10.22
440 Flow with a free surface Putting u1 = q/h1, where q represents the discharge divided by width, we get h1h 2 2 + h2 1 h2 − 2q 2 /g = 0 (10.24) whence h2 =− h1 � �� + � � h2 1 2q2 + =− 2 (−) 4 gh1 h1 2 + � �� h2 1 4 + 2h1u2 � 1 g (10.25) (The negative sign for the radical is rejected because h2 cannot be negative.) Equation 10.24 is symmetrical in respect of h1 and h2 and so a similar solution for h1 in terms of h2 may be obtained by interchanging the suffixes. The depths of flow on both sides of a hydraulic jump are termed the conjugate depths for the jump. We recall that the following assumptions have been made: 1. The bed is horizontal (or so nearly so that the component of weight in the direction of flow may be neglected) and the rectangular cross-section uniform (i.e. the channel is not tapered). 2. The velocity over each of the cross-sections considered is so nearly uniform that mean velocities may be used without significant error. 3. The depth is uniform across the width. 4. Friction at the boundaries is negligible. This assumption is justifiable because the jump occupies only a short length of the channel. 5. Surface tension effects are negligible. The loss of mechanical energy that takes place in the jump may be readily determined from the energy equation. If the head lost in the jump is hj then � � � � hj = h1 + u2 1 2g = h1 − h2 + q2 2g − 2g � 1 h2 − 1 1 h2 � 2 h2 + u2 2 which on substitution from eqn 10.24 yields � h1h hj = h1 − h2 + 2 2 + h2 1h2 �� 1 4 h2 − 1 1 h2 � 2 = � �3/4h1h2 h2 − h1 (10.26) This amount is represented by the distance hj on Fig. 10.23b. This dissipation of energy is a direct result of the considerable turbulence in the wave: friction at the boundaries makes a negligible contribution to it. The frictional forces in the wave are in the form of innumerable pairs of action and reaction and so, by Newton’s Third Law, annul each other in the net force on the control volume considered in deriving eqns 10.13 and 10.23. Fortunately then, the exact form of the jump between sections 1 and 2 is of no consequence in our analysis.
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Mechanics of Fluids
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Mechanics of Fluids Eighth edition
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Contents Preface to the eighth edit
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8 Boundary Layers, Wakes and Other
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Preface to the eighth edition In th
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Fundamental concepts 1 The aim of C
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1.1.1 Molecular structure The diffe
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therefore, to have in place systems
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has been a strong movement in favou
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Notation, dimensions, units and rel
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is nearer 2 m, rather than 1 m. Her
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Properties of fluids 13 The relativ
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that is, p1 − p3 = 1 2BC ϱax (1.
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that there is an almost perfect vac
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calorically perfect. (Some writers
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As a liquid is compressed its molec
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or τ = µ ∂u ∂y (1.9) where µ
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shared among the occupants of bb, a
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the origin with slope equal to µ (
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The interplay of these various forc
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non-uniformity is always encountere
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z direction and therefore the same
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Although Reynolds used water in his
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cylinder. For a certain range of Re
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The roles of experimentation and th
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A flow model which assumes steady,
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2.1 INTRODUCTION Fluid statics 2 Fl
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3. dp/dz =−ϱg. (Since the pressu
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that is, dp p =−g dz R T Variatio
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atmosphere it is termed vacuum or s
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at its centreline be p. Then, provi
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possible, and a sloping manometer m
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The amount of liquid B on each side
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In a pressure transducer the pressu
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where the suffixes C and Oy indicat
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surface (where the pressure is atmo
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which has different depths of water
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Hydrostatic thrusts on submerged su
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plane and may then be combined into
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Since all the elemental thrusts are
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neglected when a body is weighed on
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couple acting on the body in its di
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centre of buoyancy on to the transv
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move, but also the centre of gravit
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that is, W(OG) = (W + F)(OB + BM) =
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is given by ∂p ∂x =−ϱax Equi
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Then ∂p/∂ξ = 0 by definition o
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2.5 Two small vessels are connected
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2.18 In the vertical end of an oil
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The principles governing fluids in
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planes B and C (Fig. 3.2), δs bein
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an actual fluid is thus often remar
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density are small. Then eqn 3.9 has
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General energy equation for steady
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− Energy loss to friction/time Ma
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a decrease of pressure (provided th
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in the neighbourhood of B results i
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it is difficult to measure the heig
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The diagram illustrates an orifice
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� depth h below the free surface
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Solution (a) We consider first the
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of the same liquid. A vena contract
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To ensure that the pressure measure
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Since �h = p1 − p2 ϱwg the flo
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For consistency with the mathematic
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showing the essential form of the r
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With the same assumptions used in d
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∴ Velocity of approach = 0.0660 m
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of the base of the notch. Assume th
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Our aim now is to derive a relation
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The velocity, in general, varies fr
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the control volume is � ϱ1u1ux1
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Applications of the momentum equati
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that the resultant force on the flu
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The existence of the reaction may b
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our reference axes attached to the
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of increase of x-momentum is � D
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This is a simplified picture of wha
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since u4 −u1 is the velocity of t
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(1) u 1 (2) (3) (4) Assume a sea le
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and frictional effects to be neglig
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Physical similarity and dimensional
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A well-known example of kinematic s
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polygon. Now a polygon can be compl
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The condition for dynamic similarit
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equality of the Mach numbers is not
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Consider the incompressible flow al
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moving in a straight line with cons
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of the analysis, and its correct im
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Note: a suffix has been attached to
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For example, the parameter Fϱ/µ 2
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The application of dynamic similari
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atmospheric air, the dynamic viscos
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gives rise to waves on the surface.
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To be tested, the model must theref
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it follows that Hence V2 V1 = P2 P1
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5.5 A torpedo-shaped object 900 mm
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Laminar flow between solid boundari
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the place of y: τ = µ ∂u ∂r A
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compensate for the reduction in vel
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The expression for the total discha
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etween the resisting viscous forces
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must be zero so as to meet the requ
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Example 6.2 Two stationary, paralle
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Thus for any value of y the velocit
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Steady laminar flow between moving
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in eqn 6.24 we obtain � D Vp 2
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and the passage of the liquid level
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this book, it is important to give
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6.6.4 Rotary viscometers A simple m
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The torque T on the cylinder of rad
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stationary when the outer cylinder
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Fundamentals of the theory of hydro
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part of the bearing then gives when
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e determined by our analysis.) From
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where a = 6µ�R h3 dh pc dx Funda
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Integrating this between θ = 0 and
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Rearrangement of eqn 6.71 and subst
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6.4 A cylindrical drum of length l
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7.1 INTRODUCTION Flow and losses in
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a maximum value of 2.0. The line CD
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America the friction factor commonl
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course, quite different from the ro
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Variation of friction factor 253 Fi
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straightforward manner without iter
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Distribution of shear stress in a c
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7.5 FRICTION IN NON-CIRCULAR CONDUI
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estimated by simple theory. The pip
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the junction, the curvature of the
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particular, the performance of a di
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A pipe bend thus causes a loss of h
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cross-sectional area available in t
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correlating experimental data on th
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to do this. But sub-atmospheric pre
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For systems comprising only short r
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7.8 PIPES IN COMBINATION 7.8.1 Pipe
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2. There can be only one value of h
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So For two pipes in parallel, and s
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7.9 CONDITIONS NEAR THE PIPE ENTRY
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where u denotes the mean velocity i
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Assuming that entry and exit losses
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of one-per-revolution electrical co
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of the piping system, this meter is
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losses, calculate (a) the total pow
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How are the running costs altered i
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main. The depth of water in the tan
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the features of boundary layer flow
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the surface other considerations ar
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and θ = = = � δ u 0 um � δ 0
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The momentum equation applied to th
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1 = y η δ The laminar boundary la
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Table 8.1 Comparison of results for
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and � � ∂f (η) B = ∂η A =
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From equation 8.17, the frictional
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The turbulent boundary layer on a s
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Hence F = 0.037ϱu 2 m l(Re l) −1
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From eqn 8.29 Hence and xt − x0 x
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from it. This breakaway before the
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this expression into eqn 8.31 we ge
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When, for example, a ship moves thr
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separate vortices in an inviscid fl
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8.8.4 Profile drag of two-dimension
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are negligible. With reduction in t
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Fig. 8.14 Drag coefficients of smoo
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sphere is falling through the atmos
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At high Reynolds numbers, the varia
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this does not prevent the continual
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Eddy viscosity and the mixing lengt
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Eddy viscosity and the mixing lengt
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Velocity distribution in turbulent
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then, although at very small values
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as η → 1 and (1 − η) is then
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When f −1/2 − 4 log 10 (d/k) is
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comparable with the velocity of sou
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The x, y and z components of the mo
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y y x i-1, j+1 i, j+1 i+1, j+1 i-1,
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e completely covered by a turbulent
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9.1 INTRODUCTION The flow of an inv
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Now consider another point P ′′
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number of smaller ones of which M a
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distortion, the element is thus not
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Any function φ that satisfies Lapl
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ecome perfect squares - except wher
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fluid, and the flow net indicates t
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9.6.2 Flow from a line source A sou
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higher orders of small magnitudes b
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Combining eqns 9.16 and 9.17 we obt
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9.6.5 Forced (rotational) vortex Th
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For a free vortex qR = constant = K
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source is unable to move to the lef
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Also we note that the source is pos
Page 400 and 401: as shown in Fig. 9.23, encloses all
Page 402 and 403: eqn 9.23 becomes ψ =−U � r −
Page 404 and 405: Thus the magnitude of the velocity
Page 406 and 407: At a stagnation point, qt = 0 and t
Page 408 and 409: function of the combined flow is gi
Page 410 and 411: or, more approximately, that betwee
Page 412 and 413: Then dw dz =−U =−u + iv Hence u
Page 414 and 415: An introduction to elementary aerof
Page 416 and 417: ound the aerofoil, its magnitude be
Page 418 and 419: and lower layers. Since the vortex
Page 420 and 421: where Ɣ0 represents the circulatio
Page 422 and 423: 9.7 An open cylindrical vessel, hav
Page 424 and 425: steadily south-east at 6 m · s −
Page 426 and 427: 10.2 TYPES OF FLOW IN OPEN CHANNELS
Page 428 and 429: The steady-flow energy equation for
Page 430 and 431: where the hs represent vertical dep
Page 432 and 433: to the turbulent rough flow regime
Page 434 and 435: selection of an appropriate value r
Page 436 and 437: 10.6 OPTIMUM SHAPE OF CROSS-SECTION
Page 438 and 439: oughness coefficient n is independe
Page 440 and 441: thrust divided by width of the chan
Page 442 and 443: which the waves follow one another
Page 444 and 445: The conditions for the critical dep
Page 446 and 447: y downstream conditions. In these c
Page 448 and 449: may be obtained by reducing eqn 10.
Page 452 and 453: The dissipation of energy is by no
Page 454 and 455: friction, the steady-flow momentum
Page 456 and 457: Fig. 10.28 Notice that, for a chann
Page 458 and 459: calculated, and then H + u2 1 /2g m
Page 460 and 461: Fig. 10.32 If the depth of flow ove
Page 462 and 463: (Section 10.11.1). Rapid flow can n
Page 464 and 465: where 0 ≤ θ ≤ 90 ◦ and sin
Page 466 and 467: (a) The continuity equation is The
Page 468 and 469: oundaries. It can occur when the fl
Page 470 and 471: into eqn 10.39 we obtain whence dh
Page 472 and 473: Table 10.2 Gradually varied flow 46
Page 474 and 475: The slope of a given channel may, o
Page 476 and 477: direction. The effect of any bounda
Page 478 and 479: This momentum term, however, is pro
Page 480 and 481: With surface tension effects ignore
Page 482 and 483: are for waves of small amplitude) w
Page 484 and 485: when the wave was formed. The lengt
Page 486 and 487: At a particular instant the crests
Page 488 and 489: Hence h = 2.65λ 2π = 2.65 × 225
Page 490 and 491: As the ratio of these velocity comp
Page 492 and 493: water is impulsively displaced and
Page 494 and 495: is relative movement of one layer o
Page 496 and 497: the channel bed a drop of 150 mm in
Page 498 and 499: 11.1 INTRODUCTION Compressible flow
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energy ÷ mass, so that the dimensi
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Energy equation with variable densi
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Solution (a) From the equation of s
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whence (c − u)δρ = (ρ + δρ)
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the sonic velocity as supersonic (M
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velocity into the undisturbed fluid
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whence p2 p1 = 1 + γ M2 1 1 + γ M
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of eqns 11.2 and 11.20 T0 T u2 = 1
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the temperature rises greatly (say
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Fig. 11.10 Oblique shock relations
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however, that for a given upstream
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The deflection through the first wa
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As δθ is very small, cos δθ →
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eqn 11.41. Substituting M 2 − 1 =
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oundaries, both of which are curved
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For supersonic flow eqn 11.45 is no
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Some general relations for one-dime
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of minimum cross-section the veloci
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downstream of the minimum cross-sec
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‘telegraphed’ upstream and a pr
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the external pressure. This compres
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Compressible flow in pipes of const
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Table 11.2 Changes with distance al
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From the Fanno Tables at pc/pB = 0.
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Therefore and M1 = u1 a 24.5 m · s
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drag. The abrupt pressure rise thro
Page 558 and 559:
Analogy between compressible flow a
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is, on ∂2n ∂y2 + ∂2 � n ∂
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and the mass flow rate in the duct.
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11.18 Air flows isothermally at 15
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12.2 INERTIA PRESSURE Any volume of
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Equation 12.3 indicates that u beco
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would come to rest later. Although
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a pressure wave in a gas; here we r
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value in eqn 12.7 gives that is, wh
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the moment disregarded. An increase
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at this point remains constant for
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In addition to the reflection of wa
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is closed so that the area coeffici
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The pressure diagrams show that for
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12.3.4 The method of characteristic
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Pressure transients 577 Multiplying
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and velocities are known, then for
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For pipe 2, and hence K ′ = c2 =
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(instantaneously, we will assume) a
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The one-dimensional continuity rela
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to accommodate instantaneous comple
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12.2 A turbine which normally opera
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13.1 INTRODUCTION Fluid machines13
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flow. By way of illustration we sha
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would obviate the changes of pressu
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demand for electric power - for exa
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Turbines 599 Fig. 13.7 Runner of Pe
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determined, multiplication by the r
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Substituting for �vw from eqn 13.
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ut the larger its value the more bu
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ate of flow possible is achieved wh
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the turbine. For the high heads nor
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The integrals in eqns 13.4 and 13.5
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diagram of Fig. 13.16 we have Simil
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would be sufficient to describe the
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Table 13.1 Type of turbine Approxim
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∴ Hydraulic efficiency = Overall
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If either z (the height of the turb
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∴ Hydraulic efficiency = u1vw1/gH
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shows the general effect of change
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an unmixed blessing and, except for
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orbiting space-craft, for example,
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Losses are also possible at the inl
Page 644 and 645:
Those particles next to the forward
Page 646 and 647:
It should be noted that the values
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∴ (β1)A = arctan(va/u) = arctan(
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so much that the aerofoil stalls (a
Page 652 and 653:
Use these data to deduce the pump c
Page 654 and 655:
Manometric efficiency = 0.75 = gH/u
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and CP = P/ϱω 3 D 5 in place of Q
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yields h 1 = h f = 4fl d ≈ (100Q)
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(iii) the power dissipated by pipe
Page 662 and 663:
Large numbers of test data for pump
Page 664 and 665:
educed by rounding the inlet edges
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further: the turbine part then remo
Page 668 and 669:
for reaction turbines: 1 − η = 1
Page 670 and 671:
75 mm wide at outlet. The blades oc
Page 672 and 673:
Losses at valves, etc. are estimate
Page 674 and 675:
APPENDIX Units and conversion 1 fac
Page 676 and 677:
Table A1.4 (contd.) Force 32.17 pdl
Page 678 and 679:
APPENDIX Physical constants and 2 p
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Kinematic viscosity mm 2 ·s -1 App
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Range II: 11 000 m ≤ z ≤ 20 000
Page 684 and 685:
Table A3.1 (contd.) M 1 M 2 p 2/p 1
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Table A3.2 Isentropic flow M p/po
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Table A3.3 Adiabatic flow with fric
Page 690 and 691:
APPENDIX 4 Algebraic symbols Table
Page 692 and 693:
Table A4.1 (contd.) Symbol Definiti
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Table A4.1 (contd.) Symbol Definiti
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Answers to problems 1.1 56.2 m 3 1.
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8.14 0.002955, 2.866 m · s −1 ,
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Absolute pressure 46 Absolute visco
Page 702 and 703:
Eccentricity 230 Eccentricity ratio
Page 704 and 705:
Mach angle 498 intersection 510-1 r
Page 706 and 707:
Slip in fluid couplings 652 in reci
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