Mechanical Engineers Data Handbook Episode 7 pptx

140 MECHANICAL ENGINEER’S DATA HANDBOOK Rs Mixture strength M,=-x 100% Weak mixture M,< 100% Rich mixture M,> 100% R 3.16.2 Combustion equations 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 + 4C0,+4H20 C,H60 + 30, + 2C0, +3H,O Carbon with air (assuming that air is composed of 79% nitrogen and 21% oxygen by volume): 79 79 21 21 C+O,+-N, + CO,+-N, (by volume) 28 x 79 28 x 79 12C+ 32 0, +- 21 N, +44COz+~ N, (by mass) since the molecular weights of C, 0,, CO, and N, are 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 Benzene C6H6 78 Butane C4H10 58 Carbon C 12 Carbon monoxide co 28 Ethane CZH, 30 Ethanol C,HsOH 46 Ethene C2H4 28 Methane CH4 16 Met h a n o 1 CH,OH 32 Nitrogen N, 28 Octane C8H,* 114 Oxygen 0, 32 Carbon dioxide CO, 44 Hydrogen H 2 Pentane CSHl, 72 Propane C3H8 44 Propene C3H6 42 Sulphur S 32 Sulphur monoxide SO 48 Water (steam) H,O 18 Sulphur dioxide SO, 64 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 Let: c=%C, h=%H,, o=%O,, n=%N,, s=%S, all by mass. (2.67~ + 8h +s-0) Stoichiometric air/fuel ratio R, = 23.3 If x = 0.84~ + 0.3 135s + 0.357n +0.0728ERS + 27.4R y = x + 5h (using E = 0 for a stoichiometric air/fuel ratio) THERMODYNAMICS AND HEAT TRANSFER 141 Combustion products (% volume) Wet analysis Dry analysis C 84 - Y C X 84 - S 7.28 35.7n + 274QR 31.3- - ER, h 500 - Y Y Y Y 0 S 7.28ER, 35.7n + 214OR 31.3- X X X 3.16.5 Hydrocarbon fuels, solid and liquid Weak mixture Let: c = %C, h = %H,, both by mass. (2.67~ + 8h) 23.3 Then: R,= x = 0.84~ + 0.0728ERS + 27.4R y=x+Sh Combustion products (% volume) Wet analysis Dry analysis C 84 - Y C X 84 - h 500 - Y 0 7.28 Y ~ ER, 7.28 PER, X R 2740 - Y R 2740 - X Rich mixture (M, > 100%) 31.3(c+3h) n= Ms (c + 6h) a=0.532n 12 x=a+b+n h y=x+- 2 142 MECHANICAL ENGINEER’S DATA HANDBOOK Combustion products (YO volume) Wet analysis Dry analysis 100: Y a 100; b 100- Y b 100 - X h 50 - Y 0 n 100- Y n 100- X 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 (%CO, + %CO+ %CH,) - %C %H, r= (8.858 -0.422%C02 -0.255%CO + 0.245%CH4 +0.078%H, -0.422%0,) r 100% %C=- 100%; %H,=- (1 +r) (1 +r) 3.16.6 Liquid fuels of the type C#*O, Weak mixture (32p + 8q - 16r) (12p+q+16r) R, = 4.292 n En Ma 100 x = p + 376-+- 4 y=x+- 2 Combustion products (YO volume) CO, H2O 0, NZ l00P Wet analysis Y 100 Dry analysis X 4 50 - Y 0 En Y - En - X 31 600n Ms Y 31 600n THERMODYNAMICS AND HEAT TRANSFER 143 Rich mixture (32p+8q- 16r) (12p+q+ 16r) R, = 4.292 Combustion products (% volume) 4 y=x+- 2 Wet analysis Dry analysis U 100- Y a 100- X b 100- Y b 100 - X Y 0 37 600n M, Y 37 600n Msx 3.16.7 Gaseous fuels For a mixture of gases such as H,, 0,, CO, CH,, etc., let V,, V,, V3, 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, c3, etc. = the number of atoms of carbon in each gas h,, h,, h,, etc.=the number of atoms of hydrogen in each gas ol, o,, 03, etc. = the number of atoms of oxygen in each gas And let: S,=c,V,+c,V,+. . . Sh=h,V,+h,V,+. , . S,=o,V,+o,V,+. . . 'h k = S, + - + - 42 Then: R,= ; k R=R,(i+k) x= 100R+ + V" 21 so Sh 24 144 MECHANICAL ENGINEER’S DATA HANDBOOK Combustion products (YO volume) Wet analysis Dry analysis ’h 100(21R - k) 100( Vn + 79R) 50 - sc + vc 100- Y Y Y Y 100(21R-k) 100( Vn + 79R) X X 0 sc+ vc 100 - X 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 Solid (kJkg-’; 15°C) Anthracite 34 600 33 900 Bituminous coal 33 500 32 450 Coke 30 750 30 500 Lignite 21 650 20 400 Peat 15 900 14500 Liquid (kJ kg-‘; 1YC) Petrol (gasoline) Benzole (crude benzene) Kerosene (paraffin) Diesel Light fuel oil Heavy fuel oil Residual fuel oil 47 OOO 42 OOO 46 250 46 OOO 44 800 44OOo 42 100 average Gas (MJm-’; 15°C; I bar) Coal gas 20.00 Producer gas 6.04 Natural gas 36.20 Blast-furnace gas 3.41 Carbon monoxide 11.79 Hydrogen 11.85 43 900 40 200 43 250 43 250 42 100 41 300 4oOOO average 17.85 6 .00 32.60 3.37 11.79 10.00 3.16.9 Boiler emciency This may be based on either the HCV or the LCV. ms(hh -hw) Boiler efficiency E,= m,(HCV or LCV) where: ms = mass flow of steam &=mass flow of fuel h, = enthalpy of steam hw = enthalpy of feed water THERMODYNAMICS AND HEAT TRANSFER 145 ~ ~ Analysis of solid fuels Fuel %mass Moisture Volatile matter (%mass) C H, 0, N, ash (%mass of dry fuel) Anthracite 1 Bituminous coal 2 Lignite 15 Peat 20 90.27 3.00 2.32 1.44 2.97 4 81.93 4.87 5.98 2.32 4.90 25 56.52 5.72 31.89 1.62 4.25 50 43.70 6.48 44.36 1.52 4.00 65 Analysis of liquid fuels %mass Fuel C H, S Ash, etc. Petrol (gasolene) s.g. 0.713 84.3 15.7 0.0 - s.g. 0.739 84.9 14.76 0.08 - Benzole 91.7 8.0 0.3 - Kerosene (paraffin) 86.3 13.6 0.1 - DERV (diesel engine road 86.3 13.4 0.3 - vehicle fuel) Diesel oil 86.3 12.8 0.9 - Light fuel oil 86.2 12.4 1.4 - Heavy fuel oil 86.1 11.8 2.1 - Residual fuel oil 88.3 9.5 1.2 1 .o Analysis of ~pseolis fuels %volume Fuel H, CO CH, C,H, C,H, C,H, 0, CO, N, Coal gas 53.6 9.0 25.0 0.0 0.0 3.0 0.4 3 .O 6.0 Producer gas 12.0 29.0 2.6 0.4 0.0 0.0 0.0 4.0 52.0 Natural gas 0.0 1.0 93.0 0.0 3 .O 0.0 0.0 0.0 3 .O Blast-furnace gas 2.0 27.0 0.0 0.0 0.0 0.0 0.0 11.0 60.0 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. 4. I .2 Archimedes principle Submerged body Let : W= weight of body pB = density of body pL = density of liquid Apparent weight W‘= W-p,V Then: W‘= V(pB-pp,) V= volume of body = W/pB Floating body Let : VB = volume of body Vs = volume submerged V” p” Weight of liquid displaced =Weight of body or PLVS= PB VB PB ‘S P PL VB PL Therefore: Vs = VB - or - = 2 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); 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. FLUID MECHANICS 147 -E Pressure F 11 w Piston area A Hydraulic jack 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: 4F 4W a Pressure p = - = -, where F = F - nd2 nD2 ,b D2 aD2 Load raised W=F-=F, d2 bd2 4. I .5 Pressure on a submerged plate Symbols used: p =density of liquid A=plate area 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+- Ax I sin2 6 Ax h=X+- (for the inclined plate) CG = centroid CP=centre of pressure 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. 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. or pI/p+ V:/2+gh,=p,p+ V:/2+ghZ+Energy loss (per kilogram) In terms of pressure: p1 + p v:/2 + pgh, =p2 + p ~:/2 + pgh, + Pressure losses 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 E, = E, 4.2.2 Continuity equation If no fluid is gained or lost in a conduit: Mass flow m=p,A,V,=p,A,V, If p1 =pz (incompressible fluid), then: A,V,=A,V, or Q1=Q2 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 - = - P” is the same for the model and the full scale version. For a circular pipe: D =diameter p = dynamic viscosity v = kinematic viscosity For a non-circular duct: 4x Area 4A Perimeter P -_ D = equivalen?.diameter = - Types of flow In a circular pipe the flow is ‘laminar’ below Re N 2000 and ‘turbulent’ above about Re = 2500. Between these values the flow is termed ‘transitional’. 149 FLUID MECHANICS L v2 4.2.4 Friction in pipes Pressure loss in a pipe ~~=4f-p-(Nm-~) D2 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) Critical zone Laminar region ~~ FTurbulent region x C 0 0 - ti Friction factor f This depends on the Reynold's number P VD Re=- and the relative roughness k/D (for values of k, see table). For non-circular pipes, use the equivalent diameter P 4xArea 4A Perimeter P -_ D, = - 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 for water). Reynolds number, Re \ Recr,, Smooth pipe [...]... 2 20 0.24 0.32 0.46 10 nd2 - 4 164 MECHANICAL ENGINEER’SDATA HANDBOOK D a coefiients for various bodies (continued) rg Shape d Ellipsoid Streamlined body of circular cross-section Solid hemisphere f o on lw convex face Cd 1 6 8 7 5 4 0.005 0.006 0.0 07 0.008 0.009 800 5 25 1.25 0.06 0. 07 0.13 100 3 4 0.049 0.051 5 6 Long symmetrical aerofoil L - 0.060 0. 072 0.38 A 4 500 01 nd2 Arrangement ...150 MECHANICAL ENGINEER’S DATA HANDBOOK Reynold’s number Re = lo00 x 0.5 x 0.05 a 5 x 104 0.001 Pressure loss pr=pr +pf2+ 0.1 50 Relative roughness k / D =- = 0.002 Friction factor (from chart)f= 0.0 073 The mass flow rate is the same in all pipes, i.e Pressure loss 1 1000~0.5~ pf = 4 x 0.0 073 x -x = 7 3 N m-2 0.05 2 m=m - m 2-etc - - where: m l = p A... 1.0 0.61-0.64 About 0 .72 +!I 4 Arrangement 0 About 0.86 0 c-, I = 0 4.3.2 - Weirs, vee notch and channels Unsuppressed weir Flow Q =2.95C,(b-0.2H)H1.’ Suppressed weir Flow Q=3.33bH1.’ Vee notch e flow Q=2.36C,tan-H2.’ 2 where C,=discharge coefficient L O 154 MECHANICAL ENGINEER’SDATA HANDBOOK Channels Symbols used : m =hydraulic mean radius =A/P i=slope of channel C =constant = 87/ [ 1 + (K/&)] A=flow... 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 V+dV dV d Y Force to move plate F=7A=pAA Iy Flua.mkKitypmfile V Y 156 MECHANICAL ENGINEER'S DATA HANDBOOK in a circular pipe is parabolic, being a maximum at the pipe centre Kinematic viscosity Kinematic viscosity = Dynamic viscosity Density... Angled plate, 8c90" 158 MECHANICAL ENGINEER'SDATA HANDBOOK Jet on angled plate F=pAP(l-cost?) in direction of VI e=90", F = ~ A V , For t?=180°,F=2pAV2 For 't' V Moving flat plate Moving angled plate Example U Angled plate 4.5.2 e= 180' Jet on moving plates If r = -0.4, 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 Jet... L - 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 " 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 bends Plate : K = 0 2 Aerofoil : K K =0.05 aerofoil vanes, 0.2 circular arc plate vanes... the boat, maximum efficiency Water enters fiont o boat f Thrust F=hV(l -r) Pump power P = m (vz-U2) vz =m-(I 2 2 2r Efficiency q =(1+ r ) q=0.6 67, for r=0.5 q = 1.0, for r = 1.0 Output power (both cases) P,=mitVlr(I - r ) -9) 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... m2s- = 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-' 1 ftz s- ' ' =Om29 mz s1ft2h- = 334 mz s- ' ' ' Velocity distribution ' Flow Q=?c (Pi -p2)r4 8PL Viscosity o water f Approximate values at room temperature: p=10-3Nsm-z y = 10-6mZs-l Temperature ("C) Dynamic viscosity ( x 10-3Nsm-i) 0.01 20 40 60 80 100 1 .75 5... flow when n = - '- =0.528 for air Drag coeilkients for various bodies V Z Drag D = C,Ap -; p =fluid density; A =frontal area; V = fluid velocity 2 Shape L d cd Re - 104 A Arrangement 162 MECHANICAL ENGINEER'S DATA HANDBOOK Drag coelficients for various bodies (continued) Shape L - d I 2 1.20 10 30 1.22 1.62 co Arrangement 1.15 5 Rectangular flat plate A Cd 1.98 1.16 60 Ld Long semicircular convex surface... 1 .75 5 1.002 4.4.2 0.65 1 0.462 0.350 0. 278 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 Mean velocity V = (Pi -P2)rZ 8PL Maximum velocity V , =2V 4.4.3 Laminar flow between flat plates Flow Q = (PI - p 2 W 12pL Mean velocity V = (Pi-Pz)t2 12pL Maximum velocity V,=$ V 1 57 FLUID MECHANICS 4.4.4 Flow through . ER, 7. 28 PER, X R 274 0 - Y R 274 0 - X Rich mixture (M, > 100%) 31.3(c+3h) n= Ms (c + 6h) a=0.532n 12 x=a+b+n h y=x+- 2 142 MECHANICAL ENGINEER’S DATA HANDBOOK. Wet analysis Dry analysis C 84 - Y C X 84 - S 7. 28 35.7n + 274 QR 31.3- - ER, h 500 - Y Y Y Y 0 S 7. 28ER, 35.7n + 214OR 31.3- X X X 3.16.5 Hydrocarbon fuels,. with air (assuming that air is composed of 79 % nitrogen and 21% oxygen by volume): 79 79 21 21 C+O,+-N, + CO,+-N, (by volume) 28 x 79 28 x 79 12C+ 32 0, +- 21 N, +44COz+~ N, (by