4.14. I.I

HydrostaticsBuoyancyV

p

The apparent weight of a submerged body is less than its weight in air or, more strictly, a vacuum. It can be shown that it appears to weigh the same as an identical volume having a density equal to the difference in densities between the body and the liquid in which it is immersed. For a partially immersed body the weight of the displaced liquid is equal to the weight of the body. 4. I .2

Archimedes principle

Weight of liquid displaced =Weight of body or PLVS= PB V BS Therefore: Vs= VBP B or -=P 2

Submerged bodyLet : W = weight of body V = volume of body = W/pB pB= density of body pL=density of liquid Apparent weight W = W-p,V Then: W = V ( p B - p p , )

PL

VB

PL

4. I.3

Pressure of liquids

The pressure in a liquid under gravity increases uniformly with depth and is proportional to the depth and density of the liquid. The pressure in a cylinder is equal to the force on the piston divided by the area of the piston. The larger piston of a hydraulic jack exerts a force greater than that applied to the small cylinder in the ratio of the areas. An additional increase in force is due to the handleflever ratio.

4.1.4

Pressure in liquids

Gravity pressure p =pgh where: p =fluid density, h =depth. Units are: newtons per square metre (Nm-) or pascals (Pa); lo5N m-2 = lo5 Pa = 1 bar = lo00 millibars (mbar).F Pressure in cylinder p =A

Floating bodyLet :VB=volume of body Vs= volume submerged

where: F=force on piston, A=piston area.

FLUID MECHANICS

147 Symbols used: p =density of liquid A=plate area x =depth of centroid I =second moment of area of plate about a horizontal axis through the centroid 6 =angle of inclined plate to the horizontal

--

-EPressure

FPiston area A

1 1

wHydraulic jackA relatively small force F, on the handle produces a pressure in a small-diameter cylinder which acts on a large-diameter cylinder to lift a large load W:4F 4W a Pressure p =-=-, where F = F nd2 nD2 ,b Load raised W=F-=F,-d2 Force on plate F=pgxA Depth of centre of pressure h=x+h=X+-

I Ax

D2

aD2 bd2

I sin26 (for the inclined plate) Ax

CG = centroid CP=centre of pressure

4. I .5

Pressure on a submerged plate

The force on a submerged plate is equal to the pressure at the depth of its centroid multiplied by its area. The point at which the force acts is called the centre of pressureand is at a greater depth than the centroid. A formula is also given for an angled plate.

148

MECHANICAL ENGINEERS DATA HANDBOOK

4.2

Flow of liquids in pipes and ductsThe continuity equation is given as are expressions for the Reynolds number, a non-dimensional quantity expressing the fluid velocity in terms of the size of pipe, etc., and the fluid density and viscosity.

The Bernoulli equation states that for a fluid flowing in a pipe or duct the total energy, relative to a height datum, is constant if there is no loss due to friction. The formula can be given in terms of energy, pressure or head. 4.2. I

Bernoulli equation

Symbols used: p =pressure p =density h =height above datum V=velocity A = area For an incompressible fluid p is constant, also the energy at 1 is the same as at 2, i.e. E , =E , or p I / p + V:/2+gh,=p,p+ V:/2+ghZ+Energy loss (per kilogram) In terms of pressure: p1 + p v:/2 pgh, = p 2 p ~ : / 2 pgh, Pressure losses

If p1 = p z (incompressible fluid), then: A,V,=A,V, or Q 1 = Q 2 where Q =volume flow rate

4.2.3 Reynolds number (non-dimensional velocity)In the use of models, similarity is obtained, as far as fluid friction is concerned, when: V D V D Reynolds number Re = p -=P is the same for the model and the full scale version. For a circular pipe: D =diameter p =dynamic viscosity v =kinematic viscosity

+

+

+

+

In terms of head: pl/pg v:/2g h , =p,/pg

+

+

+ Vi/2g + h, +Head losses

Velocity pressure p, = p v2/2 Velocity head h, = V2/2g Pressure head h, =p/pg

For a non-circular duct:

-_ D =equivalen?.diameter= 4 x Area - 4A Perimeter P4.2.2Continuity equation

Types o flow fIn a circular pipe the flow is laminarbelow Re N 2000 and turbulent above about Re = 2500. Between these values the flow is termed transitional.

If no fluid is gained or lost in a conduit: Mass flow m=p,A,V,=p,A,V,

FLUID MECHANICS

149

4.2.4

Friction in pipes

Pressure loss in a pipe ~ ~ = 4L - p - ( N m - ~ ) f v2 D 2Friction factor f This depends on the Reynold's number

The formula is given for the pressure loss in a pipe due to friction on the wall for turbulent flow. The friction factor f depends on both Reynold's number and the surface roughness k, values of which are given for different materials. In the laminar-flow region, the friction factor is given by f = 16/Re, which is derived from the formula for laminar flow in a circular pipe. This is independant of the surface roughness. For non-circular pipes and ducts an equivalent diameter (equal to 4 times the area divided by the perimeter) is used. Let : L=length (m) D 5 diameter (m) V-5 velocity (m s- I ) p=density (kgm-3)

Re=- P V DP

and the relative roughness k/D (for values of k, see table). For non-circular pipes, use the equivalent diameter

D = ,

4xArea -_ -4A Perimeter P

ExampleFor a water velocity of 0.5 m s- ' in a 50 mm bore pipe of roughness k = 0 . 1 mm, find the pressure loss per metre (viscosity=0.001 N - S ~ and p = lo00 kgm-3 - ~ for water).

Laminar region

~~

Critical zone FTurbulent

region

x

t i

0 --

0 .-

C

\

Recr,,

Reynolds number, ReSmooth pipe

150 lo00 x 0.5 x 0.05 a . 5 x 104 0.001 0.1 50

MECHANICAL ENGINEERS DATA HANDBOOK

Reynolds number Re =

Pressure loss pr=pr

+pf2+

Relative roughness k / D =- = 0.002 Friction factor (from chart)f= 0.0073 The mass flow rate is the same in all pipes, i.e. Pressure loss1 1000~0.5~ pf = 4 x 0.0073 x -x = 7 3 N m-2 0.05 2m=m - m 2-etc. - -

where: m l = p A I V l , etc. kgs-

Pipes in parallel Laminar (oiscous) flowThe pressure loss is the same in all pipes: For circular pipes only, the friction factor f= 16/Re. This value is independant of roughness. Pressure loss pr=pr=pf2 =etc.

Typical roughness of pipesRoughness, k (mm) Smooth (k -0)0.05 0.12 0.15 0.25 0.2-1.0 0.3-3.0 1.0-10

The total flow is the sum of the flow in each pipe: Total flow m=hl+m2+. . . where: pf1=4fl-p---,v: pf2=4f2-p-. v: etc. Ll L2 Dl 2 D2 2

Material of pipe (new) Glass, drawn brass, copper, lead, aluminium, etc. Wrought iron, steel Asphalted cast iron Galvanized iron, steel Cast iron Wood stave Concrete Riveted steel

4.2.6 Pressure loss in pipe fittings and pipe section changesIn addition to pipe friction loss, there are losses due to changes in pipe cross-section and also due to fittings such as valves and filters. These losses are given in terms of velocity pressure p(v2/2) and a constant called the K factor.

4.2.5

Pipes in series and parallel

I

I

Sudden enlargement

v: Pressure loss pL=Kp -, where K =2IDI h

Piperoughness

I

Pipes in seriesThe pressure loss is the sum of the individual losses:

FLUID MECHANICS

151

Sudden exitPressure loss p L = p - v: ( K =1) , 2

Losses in valvesGlobe valve wide open K = 1 0 Gate valve wide open K =0.2 Gate valve three-quarters open K = 1.15 Gate valve half open K = 5.6 Gate valve quarter open K =24

I I v,_Sudden contraction

Rounded entryK z 0.05

v: Pressure loss pL= K p 20

0.2

0.4

0.6

0.8

10 .

Re-entrant pipeK0.5 0.45 0.38 0.28 0.14 0

K =0.8-1.o

r l

I ISudden entryPressure l s p , = K p - - ,v: where K z 0 . 5 os 2

BendsThe factor K depends on RID, the angle of bend 0, and the cross-sectional area and the Reynold's number. Data are given for a circular pipe with 90"bend. The loss factor takes into account the loss due to the pipe length.

L " 2

K

1.0 0.4 0.2 0.18 0.2 0.27

0.33 0.4

152

MECHANICAL ENGINEER'S DATA HANDBOOK

Cascaded bendsPlate : K = 0 2 .

K =0.05 aerofoil vanes, 0.2 circular arc plate vanes

Aerofoil : K

-

0.05

4.3

Flow of liquids through various devicesFlow in channels depends on the cross-section, the slope and the type of surface of the channel.

Formulae are given for the flow through orifices, weirs and channels. Orifices are used for the measurement of flow, weirs being for channel flow.

4.3. I

Orifices

Let: C , = coefficient of discharge C, =coefficient of velocity C , =coefficient of contraction H =head A = orifice area Aj =jet area

TIT

FLUID MECHANICS

153

Values o C, f

oiie type rfcRounded entry Sharp edged Borda reentrant (running full) External mouthpiece

C d Nearly 1.00.61-0.64

About 0.72 About 0.86

+!I 4Arrangement0

0

c-, I-

0

=

L

O

4.3.2

Weirs, vee notch and channels

Unsuppressed weirFlow Q =2.95C,(b-0.2H)H1.

Suppressed weirFlow Q=3.33bH1.

flow Q=2.36C,tan-H2.

e

2

Vee notch

where C,=discharge coefficient

154

MECHANICAL ENGINEERSDATA HANDBOOK

ChannelsSymbols used : m =hydraulic mean radius =A/P i=slope of channel C =constant =87/[ 1 + (K/&)] A=flow area P =wetted perimeter Mean velocity V = Flow rate Q = V A

Values of KSurface Clean smooth wood, brick, stone Dirty wood, brick, stone Natural earth

K0.16

~Jmr

0.28 1.30

Maximum discharge for given excavationChannel Rectangular Trapezoidal Condition Arrangement

d=b/2 Sides tangential to semicircle

4.3.3

Venturi, orifice and pipe nozzle

These are used for measuring the flow of liquids and gases. In all three the restriction of flow creates a pressure difference which is measured to give an indication of the flow rate. The flow is always proportional to the square root of the pressure difference so that these two factors are non-linearly related. The venturi gives the least overall pressure loss (this is often important), but is much more expensive to make than the orifice which has a much greater loss. A good compromise is the pipe nozzle. The pressure difference may be measured by means of a manometer (as shown) or any other differential pressure device. The formula for flow rate is the same for each type.

Let : D =pipe diameter d =throat diameter p =fluid density p, =density of manometer fluid p1=upstream pressure p =throat pressure C, =coefficient of discharge h =manometer reading Flow rate Q = C,E 4

/?

FLUID MECHANICS

155

I

InletThroat

Values of C,Cd

Venturi Orifice plate Nozzle

0.974.99 0.60 0.92 to 0.98

4.44.4. I

Viscosity and laminar flowviscosityV+dV

In fluids there is cohesion and interaction between molecules which results in a shear force between adjacent layers moving at different velocities and between a moving fluid and a fixed wall. This results in friction and loss of energy. The following theory applies to so-called laminar or viscous flow associated with low velocity and high viscosity, i.e. where the Reynolds number is low.

At

Dejnition of viscosityIn laminar flow the shear stress between adjacent layers parallel to the direction of flow is proportional to the velocity gradient. Let : V=velocity y =distance normal to flow p =dynamic viscosity dV Shear stress 7=constant-=pd Y dV d Y

Flat plate moving overjixed plate of area AForce to move plate F=7A=pAA

VY

--Iy

Flua.mkKitypmfile

156

MECHANICAL ENGINEER'S DATA HANDBOOK

Kinematic viscosityKinematic viscosity = Dynamic viscosity DensityP

in a circular pipe is parabolic, being a maximum at the pipe centre.

or v = -

P

Dimensions o viscosity fDynamic viscosity: ML- TKinematic viscosity: L2 T- '

'

Units with conversions from Imperial and other unitsDynamic viscosity Kinematic viscosity

SI unit: Nsm-2=47.9 N s m-' llbf-s ft -' 1 lbf-hft-2= 17.24 N s m-2 1 poundal-s ft - = 1.49 N s m-2 llbft-' s - ' = 1.49 kg ms 1 slugft-'s-'= 47.9 kgms-'

SI unit: m2s1 ftz s-

' ''Velocity distribution

' =Om29 mz s1ft2h- = 334 mz s-

'

'

Flow Q=?c (Pi -p2)r4 8PLMean velocity V = (Pi -P2)rZ 8PL Maximum velocity V , =2V

Viscosity o water fApproximate values at room temperature: p=10-3Nsm-z y = 10-6mZs-l Temperature ("C)0.01 20 40 60 80 100

4.4.3

Laminar flow between flat plates

Dynamic viscosity ( x 10-3Nsm-i) 1.755 1.0020.65 1

Flow Q = (PI - p 2 W 12pL Mean velocity V = (Pi-Pz)t2 12pL Maximum velocity V,=$ V

0.4620.350

0.278

4.4.2

Laminar flow in circular pipes

The f o is directly proportional to the pressure drop lw for any shape of pipe or duct. The velocity distribution

FLUID MECHANICS

157

4.4.4

Flow through annulus (small gap)

Mean velocity V=

Qz ( R 2- r 2 )

0

U e formula for flat plates but with B =zD,, where D, s is the mean diameter.

0

Flow through annulus (exact formula)Flow Q =- (PI - p 2 ) ( R 2- r 2 ) (R2 r 2 ) - + 8uL?K

(R2 r 2 ) R In -

1is proportional to the mass flow rate. Examples are of jets striking both fixed and moving plates.

4.5

Fluid jets

If the velocity or direction of a jet of fluid is changed, there is a force on the device causing the change which

Change of momentum of aJIuid streamLet : m=mass flow rate=pAV VI =initial velocity V z=final velocity p =fluid density A=flow area

For flow in one direction, the force on a plate, etc., causing a velocity change isF = h ( VI - V 2 )

4.5. I Jet on stationary plates

It1I' i

Jet on u p a t plateIn this case V2 =0, and if VI = Vl

A

II I

F=mV=

Flat plate, e = Boc

Angled plate, 8c90"

158

MECHANICAL ENGINEER'SDATA HANDBOOK

Jet on angled plateF=pAP(l-cost?) in direction of VIFor

e=90", F = ~ A V , For t?=180,F=2pAV2.

VMoving flat plate

't'

Moving angled plate

ExampleAngled plate.

e= 180'

If r = -0.4,

4.5.2

Jet on moving plates

t?= 170, V = 10 ms- I , A = 4 cm2 ( = 4 x V 10-4m3)and p=lOOOkgm-'. Then P=lOOox 4x x lo3 x 0.4(1-0.4)(1 -COS 170)=190.5 watts

U

Jet on a p a t plateF=pAV(V- U ) where: U =plate velocity. Power P=FU=pAVU(V-U)= p A V3r( 1- r )

Jet on jixed curved vaneIn the x direction: F,= pA V2(cost?, +cos e, ) In the y direction: F, = p A p(sin 8, -sin e, )

Jet on moving curved vanesina cost?, sin 8, where: r = -

U where: r = -

V'

Jet on angfed plate

UV'

F=pAV(V-U)(l--cost?) in direction of V P = p ~ V ~ r ( l - r ) (-cost?)p l

FLUID MECHANICS

159

occurs when the boat speed is half the jet speed and maximum power is attained. When the water enters the front of the boat, maximum efficiency occurs when the boat speed equals the jet speed, that is, when the power is zero. A compromise must therefore be made between power and efficiency. Let: V=jet velocity relative to boat U =boat velocity

m=mass flow rate of jet

Water enters side of boatThrust F =m V(1- r ) Pump power P =m 2 Efficiency q =2r( 1-r); q,,=O.S, at r=0.5.

VZ

U

F,=mV

(

1-- ?i:)sina

sina cos8, sin 8, sina cos8, sin 8, where: V=jet velocity, a=jet angle, 8, =vane inlet angle, O2 =vane outlet angle.

Water enters fiont o boat fThrust F=hV(l -r) Pump power P = m

(vz-U2) vz =m-(I 2 2

-9)

2r Efficiency q =(1+ r )q=0.667, for r=0.5. q = 1.0, for r = 1.0.

4.5.3

Water jet boat

This is an example of change i momentum of a fluid n jet. The highest efficiency is obtained when the water enters the boat in the direction of motion. When the water enters the side of the boat, maximum efficiency

Output power (both cases)P,=mitVlr(I - r )

160

MECHANICAL ENGINEER'S DATA HANDBOOK

4.5.4

Aircraft jet engine

Let : V = jet velocity relative to aircraft U =aircraft velocity m=mass flow rate of air hf=mass flow rate of fuel Thrust T=mU - (m +mf)V Output power P = TU=mU2-(m+mf)UV

r

Side entry

l.OL---

-

-

I/0

I0.5r Front entry

I1.o

Po mar =m -, at r =0.5.4

vz

4.6

Flow of gasespipe and flow through orifices. The velocity ofsound in a gas is defined.

Formulae are given for the compressible flow of a gas. They include isothermal flow with friction in a uniform

Symbols used: p =pressure L =pipe length D =pipe diameter T = temperature C , =discharge coefficient

VI =inlet velocity R =gas constant m =mass flow f = friction coefficient y =ratio of specific heats p =density

FLUID MECHANICS

161

4.6. I

Isothermal flow in pipe

Pressure drop:A,=,,(

1

-/=4

D2 Mass flow m = p , V , n -

where: pr

=(g)

4.6.3

:;cmv s

Velocity of sound in a gas

v,= Jyp/p=J r R T4.6.2 Flow through orifice29 - pIpln2 1-nV Mach number M =-

Mass flow m=C,A

nJ-7-7

4.6.4 Drag cafflcients for various bodiesThe drag coefficient (non-dimensionaldrag) is equal to the drag force divided by the product of velocity pressure and frontal area. The velocity may be that of the object through the air (or any other gas) or the air velocity past a stationary object. Coefficients are given for a number of geometrical shapes and also for cars, airships and struts.

where: n=p21p1; p 1 =pl/RTl Maximum flow when n = -

[A-

'- =0.528 for air.

Drag coeilkients for various bodies Drag D = C,Ap -; p =fluid density; A =frontal area; V = fluid velocity. 2 ShapeV Z

L d

cd

Re -

104

A

Arrangement

162 Drag coelficients for various bodies (continued)ShapeL A

MECHANICAL ENGINEER'S DATA HANDBOOK

d Rectangular flat plate

Cd1.15

Arrangement

I 25

1.161.20

60

Ld

10 30

1.22 1.621.98

co

Long semicircular convex surface

Long circular cylinder

I .oo 0.35

20Ld

3

Long square section f o lw on edge

FLUID MECHANICS

163

Drag codficieats for various bodies (continued)Shape-

L d

Cd

A100

Arrangement

(a) Cube flow on face

1 .os

d2

\(b) Cube flow on edge

Sphere

0.45 0.20

20

nd2 -

4

Long elliptical section

8 4 2

0.24 0.32 0.46

10

164

MECHANICAL ENGINEERSDATA HANDBOOK

D a coefiients for various bodies (continued) rgShapeL dCd

A

Arrangement

Long symmetrical aerofoil

1 6 8 7 5 4

0.005 0.006 0.007 0.008 0.009

800

Ellipsoid

5 25 . 1.25

0.06 0.07 0.13

1004

Streamlined body of circular cross-section

34

0.049 0.051

5 6Solid hemisphere f o on lw convex face

0.060 0.0720.38

500

01 .

nd2

FLUID MECHANICS

165

Drag coefficients for various bodies (continued)

Shape

-

L d

Cd

Re -

104

Alrd4

Arrangement

Hollow hemisphere flow on convex face

0.80

0.1

Hollow hemisphere flow on concave face

1.42

0.1

4

lrd2

(a) High-drag car

>0.55

50

-

(b) Medium-drag car

0.45

50

-

(b)

d-b

(c) Low-drag car