Mechanics of Fluids

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the control volume is � ϱ1u1ux1 dA1 = cos θ � ϱ1u 2 1 dA1 and so the rate of increase of x-momentum is − cos θ � ϱ1u 2 1 dA1 and this equals the net force on the fluid in the x direction. If the fluid on the solid surface were stationary and at atmospheric pressure there would of course be a force between the fluid and the surface due simply to the static (atmospheric) pressure of the fluid. However, the change of fluid momentum is produced by a fluid-dynamic force additional to this static force. By regarding atmospheric pressure as zero we can determine the fluid-dynamic force directly. Since the pressure is atmospheric both where the fluid enters the control volume and where it leaves, the fluid-dynamic force on the fluid can be provided only by the solid surface (effects of gravity being neglected). The fluid-dynamic force exerted by the fluid on the surface is equal and opposite to this and is thus cos θ � ϱ1u 2 1 dA1 in the x direction. If the jet has uniform density and velocity over its cross-section the integral becomes ϱ1u 2 � 1 cos θ dA1 = ϱ1Q1u1 cos θ (where Q1 is the volume flow rate at inlet). The rate at which y-momentum enters the control volume is equal to sin θ � ϱ1u 2 1 dA1. For this component to undergo a change, a net force in the y direction would have to be applied to the fluid. Such a force, being parallel to the surface, would be a shear force exerted by the surface on the fluid. For an inviscid fluid moving over a smooth surface no shear force is possible, so the component sin θ � ϱ1u 2 1 dA1 would be unchanged and equal to the rate at which y-momentum leaves the control volume. Except when θ = 0, the spreading of the jet over the surface is not symmetrical, and for a real fluid the rate at which y-momentum leaves the control volume differs from the rate at which it enters. In general, the force in the y direction may be calculated if the final velocity of the fluid is known. This, however, requires further experimental data. When the fluid flows over a curved surface, similar techniques of calculation may be used as the following example will show. Applications of the momentum equation 139 Fig. 4.2
140 The momentum equation Example 4.1 A jet of water flows smoothly on to a stationary curved vane which turns it through 60 ◦ . The initial jet is 50 mm in diameter, and the velocity, which is uniform, is 36 m · s −1 . As a result of friction, the velocity of the water leaving the surface is 30 m · s −1 . Neglecting gravity effects, calculate the hydrodynamic force on the vane. Solution Taking the x direction as parallel to the initial velocity (Fig. 4.3) and assuming that the final velocity is uniform, we have Force on fluid in x direction = Rate of increase of x-momentum = ϱQu2 cos 60 ◦ − ϱQu1 = (1000 kg · m −3 � π ) 4 (0.05)2 m 2 × 36 m · s −1� × (30 cos 60 ◦ m · s −1 − 36 m · s −1 ) =−1484 N Similarly, force on fluid in y direction = ϱQu2 sin 60 ◦ − 0 � = 1000 π 4 (0.05)236 kg · s −1� (30 sin 60 ◦ m · s −1 ) = 1836 N Fig. 4.3 The resultant force on the fluid is therefore √ (1484 2 + 1836 2 ) N = 2361 N acting in a direction arctan {1836/(−1484)} = 180 ◦ − 51.05 ◦ to the x direction. Since the pressure is atmospheric both where the fluid enters the control volume and where it leaves, the force on the fluid can be provided only by the vane. The force exerted by the
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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
Page 100 and 101: The principles governing fluids in
Page 102 and 103: planes B and C (Fig. 3.2), δs bein
Page 104 and 105: an actual fluid is thus often remar
Page 106 and 107: density are small. Then eqn 3.9 has
Page 108 and 109: General energy equation for steady
Page 116 and 117: − Energy loss to friction/time Ma
Page 118 and 119: a decrease of pressure (provided th
Page 120 and 121: in the neighbourhood of B results i
Page 122 and 123: it is difficult to measure the heig
Page 124 and 125: The diagram illustrates an orifice
Page 126 and 127: � depth h below the free surface
Page 128 and 129: Solution (a) We consider first the
Page 130 and 131: of the same liquid. A vena contract
Page 132 and 133: To ensure that the pressure measure
Page 134 and 135: Since �h = p1 − p2 ϱwg the flo
Page 136 and 137: For consistency with the mathematic
Page 138 and 139: showing the essential form of the r
Page 140 and 141: With the same assumptions used in d
Page 142 and 143: ∴ Velocity of approach = 0.0660 m
Page 144 and 145: of the base of the notch. Assume th
Page 146 and 147: Our aim now is to derive a relation
Page 148 and 149: The velocity, in general, varies fr
Page 152 and 153: Applications of the momentum equati
Page 154 and 155: that the resultant force on the flu
Page 156 and 157: The existence of the reaction may b
Page 158 and 159: our reference axes attached to the
Page 160 and 161: of increase of x-momentum is � D
Page 162 and 163: This is a simplified picture of wha
Page 164 and 165: since u4 −u1 is the velocity of t
Page 166 and 167: (1) u 1 (2) (3) (4) Assume a sea le
Page 168 and 169: and frictional effects to be neglig
Page 170 and 171: Physical similarity and dimensional
Page 172 and 173: A well-known example of kinematic s
Page 174 and 175: polygon. Now a polygon can be compl
Page 176 and 177: The condition for dynamic similarit
Page 178 and 179: equality of the Mach numbers is not
Page 180 and 181: Consider the incompressible flow al
Page 182 and 183: moving in a straight line with cons
Page 184 and 185: of the analysis, and its correct im
Page 186 and 187: Note: a suffix has been attached to
Page 188 and 189: For example, the parameter Fϱ/µ 2
Page 190 and 191: The application of dynamic similari
Page 192 and 193: atmospheric air, the dynamic viscos
Page 194 and 195: gives rise to waves on the surface.
Page 196 and 197: To be tested, the model must theref
Page 198 and 199: 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
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as shown in Fig. 9.23, encloses all
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eqn 9.23 becomes ψ =−U � r −
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Thus the magnitude of the velocity
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At a stagnation point, qt = 0 and t
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function of the combined flow is gi
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or, more approximately, that betwee
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Then dw dz =−U =−u + iv Hence u
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An introduction to elementary aerof
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ound the aerofoil, its magnitude be
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and lower layers. Since the vortex
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where Ɣ0 represents the circulatio
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9.7 An open cylindrical vessel, hav
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steadily south-east at 6 m · s −
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10.2 TYPES OF FLOW IN OPEN CHANNELS
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The steady-flow energy equation for
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where the hs represent vertical dep
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to the turbulent rough flow regime
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selection of an appropriate value r
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10.6 OPTIMUM SHAPE OF CROSS-SECTION
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oughness coefficient n is independe
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thrust divided by width of the chan
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which the waves follow one another
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The conditions for the critical dep
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y downstream conditions. In these c
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may be obtained by reducing eqn 10.
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this depth is greater than the crit
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The dissipation of energy is by no
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friction, the steady-flow momentum
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Fig. 10.28 Notice that, for a chann
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calculated, and then H + u2 1 /2g m
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Fig. 10.32 If the depth of flow ove
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(Section 10.11.1). Rapid flow can n
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where 0 ≤ θ ≤ 90 ◦ and sin
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(a) The continuity equation is The
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oundaries. It can occur when the fl
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into eqn 10.39 we obtain whence dh
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Table 10.2 Gradually varied flow 46
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The slope of a given channel may, o
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direction. The effect of any bounda
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This momentum term, however, is pro
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With surface tension effects ignore
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are for waves of small amplitude) w
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when the wave was formed. The lengt
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At a particular instant the crests
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Hence h = 2.65λ 2π = 2.65 × 225
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As the ratio of these velocity comp
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water is impulsively displaced and
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is relative movement of one layer o
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the channel bed a drop of 150 mm in
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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
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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
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Those particles next to the forward
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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
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Use these data to deduce the pump c
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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
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Large numbers of test data for pump
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educed by rounding the inlet edges
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further: the turbine part then remo
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for reaction turbines: 1 − η = 1
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75 mm wide at outlet. The blades oc
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Losses at valves, etc. are estimate
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APPENDIX Units and conversion 1 fac
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Table A1.4 (contd.) Force 32.17 pdl
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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
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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
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APPENDIX 4 Algebraic symbols Table
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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
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Eccentricity 230 Eccentricity ratio
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Mach angle 498 intersection 510-1 r
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Slip in fluid couplings 652 in reci
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