Mechanical Engineers Data Handbook Episode 7 pptx

140 MECHANICAL ENGINEER’S DATA HANDBOOK The following are the basic equations normally used for combustion processes.. 148 MECHANICAL ENGINEER’S DATA HANDBOOK 4.2 Flow of liquids in pipe

140 MECHANICAL ENGINEER’S DATA HANDBOOK

The following are the basic equations normally used

for combustion processes A table of elements and

compounds is given

Carbon: C + 0, + CO,; 2C + 0, + 2CO

Hydrogen: 2H, +O, + 2H,O

Sulphur: S + 0, + SO,

Typical hydrocarbon fuels :

C4H8+6O2 + 4 C 0 , +4 H 2 0

C,H60 + 3 0 , + 2C0, +3H,O

Carbon with air (assuming that air is composed of

79% nitrogen and 21% oxygen by volume):

12,32,44 and 28

3.16.3 Molecular weights of elements

and compounds

The molecular weights of elements and compounds

used in combustion processes are listed in the table

Element

Approximate molecular Formula weight

Engine exhaust and frue gas analysis

If the analysis includes the H,O (as steam) produced

by the combustion of hydrogen, it is known as a ‘wet analysis’ Usually the steam condenses out and a ‘dry analysis’ is made

3.16.4 Solid and liquid fuels

THERMODYNAMICS A N D HEAT TRANSFER 141 Combustion products (% volume)

142 MECHANICAL ENGINEER’S DATA HANDBOOK Combustion products (YO volume)

Airfluel ratiofrom the CO, in the exhaust for

fuel consisting of C and H , by weight

+ 0.072 yo H ,

R=2.4- YOC

%CO,

Ratio of carbon to hydrogen by massfiom the

dry exhaust analysis

THERMODYNAMICS AND HEAT TRANSFER 143

For a mixture of gases such as H,, 0,, CO, CH,, etc.,

let V , , V,, V 3 , etc., be the percentage by volume of

gases, 1 , 2 , 3 , etc., containing C, H, and 0, V , and V ,

are the percentage volumes of N, and CO,

Let:

c,, c2, c 3 , etc = the number of atoms of carbon in each gas

h,, h,, h,, etc.=the number of atoms of hydrogen in each gas

o l , o,, 03, etc = the number of atoms of oxygen in each gas

144 MECHANICAL ENGINEER’S DATA HANDBOOK Combustion products (YO volume)

3.16.8 Calorific value of fuels

The calorific value of a fuel is the quantity of heat

obtained per kilogram (solid or liquid) or per cubic

metre (gas) when burnt with an excess of oxygen in a

calorimeter

If H,O is present in the products of combustion as a

liquid then the ‘higher calorific value’ (HCV) is

obtained If the H,O is present as a vapour then the

‘lower calorific value’ (LCV) is obtained

LCV=HCV-207.4%H2 (by mass)

Calorific value of fuels

Higher Lower calorific calorific value value

Light fuel oil Heavy fuel oil Residual fuel oil

42 100 average

17.85

6 .00

32.60 3.37 11.79 10.00

ms = mass flow of steam

&=mass flow of fuel

h, = enthalpy of steam

hw = enthalpy of feed water

THERMODYNAMICS A N D HEAT TRANSFER 145

~ ~

Analysis of solid fuels

Fuel

%mass

Analysis of liquid fuels

Residual fuel oil 88.3 9.5 1.2 1 .o

Analysis of ~pseolis fuels

4.1 Hydrostatics

4 I I Buoyancy

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

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); lo5 N m-2 = lo5 Pa = 1 bar = lo00 milli- bars (mbar)

F Pressure in cylinder p = -

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

A 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:

x =depth of centroid

I = second moment of area of plate about a horizontal

6 =angle of inclined plate to the horizontal

axis through the centroid

Force on plate F=pgxA

I Depth of centre of pressure h=x+-

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

pressure’and is at a greater depth than the centroid A

formula is also given for an angled plate

148 MECHANICAL ENGINEER’S DATA HANDBOOK

4.2 Flow of liquids in pipes and ducts

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’

The ‘continuity equation’ is given as are expressions for the Reynold’s number, a non-dimensional quantity expressing the fluid velocity in terms of the size of pipe, etc., and the fluid density and viscosity

For an incompressible fluid p is constant, also the

energy at 1 is the same as at 2, i.e

If no fluid is gained or lost in a conduit:

Mass flow m=p,A,V,=p,A,V,

A,V,=A,V, or Q 1 = Q 2

where Q = volume flow rate

4.2.3 Reynold’s number (non-dimensional velocity)

In the use of models, similarity is obtained, as far as fluid friction is concerned, when:

VD VD Reynold’s number Re = p - = -

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

Example

For 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

MECHANICAL ENGINEER’S DATA HANDBOOK

150

lo00 x 0.5 x 0.05 0.001 Reynold’s number Re = a 5 x 104

For circular pipes only, the friction factor f= 16/Re

This value is independant of roughness

Typical roughness of pipes

The pressure loss is the same in all pipes:

Pressure loss pr = pr = pf2 = etc

The total flow is the sum of the flow in each pipe:

Total flow m=hl+m2+

Roughness, k Material of pipe (new) (mm)

Glass, drawn brass, copper,

Wrought iron, steel

Asphalted cast iron

Galvanized iron, steel

0.12

0.15 0.25 0.2-1.0

4.2.6 Pressure loss in pipe fittings and

pipe section changes

In 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’

Globe valve wide open K = 10

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

The 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

K 1.0 0.4 0.2 0.18 0.2 0.27 0.33 0.4

152 MECHANICAL ENGINEER'S DATA HANDBOOK

Plate : K = 0.2

Aerofoil : K - 0.05

Cascaded bends

K = 0.05 aerofoil vanes, 0.2 circular arc plate vanes

4.3 Flow of liquids through various devices

Formulae are given for the flow through orifices, weirs

and channels Orifices are used for the measurement of

flow, weirs being for channel flow

Flow in channels depends on the cross-section, the slope and the type of surface of the channel

FLUID MECHANICS 153 Values of C,

orifice type Cd Arrangement

Rounded entry Nearly 1.0

Borda reentrant About 0.72

External mouthpiece About 0.86

154 MECHANICAL ENGINEER’S DATA HANDBOOK

Clean smooth wood, brick, stone 0.16

Dirty wood, brick, stone 0.28

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 propor-

tional 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

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 Reynold’s number is low

Dejnition of viscosity

In laminar flow the shear stress between adjacent

layers parallel to the direction of flow is proportional

to the velocity gradient

156 MECHANICAL ENGINEER'S DATA HANDBOOK

Kinematic viscosity

Dynamic viscosity Density Kinematic viscosity =

Units with conversions from Imperial and other units

Dynamic viscosity Kinematic viscosity

0.462 0.350 0.278

4.4.2 Laminar flow in circular pipes

The flow is directly proportional to the pressure drop

for any shape of pipe or duct The velocity distribution

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

Mean velocity V = (Pi -Pz)t2

12pL

Maximum velocity V,=$ V

FLUID MECHANICS 157

4.4.4 Flow through annulus (small gap)

0

Use formula for flat plates but with B = zD,, where D,

is the mean diameter

Flow through annulus (exact formula)

If the velocity or direction of a jet of fluid is changed,

there is a force on the device causing the change which

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

Change of momentum of aJIuid stream

158 MECHANICAL ENGINEER'S DATA HANDBOOK

Jet on angled plate

F=pAP(l-cost?) in direction of VI

For t?=180°, F=2pAV2

For e=90", F = ~ A V ,

V 't' Moving flat plate

Jet on jixed curved vane

In the x direction: F,= pA V2(cos t?, +cos e,)

In the y direction: F , = p A p(sin 8, -sin e,)

Jet on moving curved vane

sin 8,

where: V=jet velocity, a=jet angle, 8, =vane inlet

angle, O2 =vane outlet angle

4.5.3 Water jet boat

This is an example of change in momentum of a fluid

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

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 boat

2r

(1 + r ) Efficiency q = -

q=0.667, for r=0.5

q = 1.0, for r = 1.0

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

Formulae are given for the compressible flow of a gas

They include isothermal flow with friction in a uniform

pipe and flow through orifices The velocity of sound in

4.6.2 Flow through orifice

Mass flow m=C,A nJ-7-7 29 - pIpln2 1-n

Drag coeilkients for various bodies

162 MECHANICAL ENGINEER'S DATA HANDBOOK Drag coelficients for various bodies (continued)

FLUID MECHANICS 163 Drag codficieats for various bodies (continued)

(b) Cube flow on edge

164 MECHANICAL ENGINEER’S DATA HANDBOOK

Drag coefiients for various bodies (continued)