Mechanical Engineering Systems 2008 Part 1 pdf

... Figure 2.2 .1. p 1 = 2 bar p2= 15 barV 1 = 0. 01 m3V2= 0.004 m3T 1 = 20 + 273 = 293 K T2=?p 1 V 1 T 1 =p2V2T22 × 0. 01 293= 15 × 0.004T2T2= 15 × 0.004 × 2932 × 0. 01 = 879 ... 2.2 .1 p/V diagram.Example 2.2 .1 ContentsSeries Preface vii 1 Introduction: the basis of engineering 1 1 .1 Real engineering 1 1.2 Units 3 1. 3 Units used in this book 52 Thermodynamics 72 .1 Heat ... kJTotal heat energy = Q 1 + Q2+ Q3+ Q4= 81. 6 + 13 40 + 16 80 + 9026.8= 12 12 8.4 kJ. Mechanical Engineering Systems Thermodynamics 11 Example 2 .1. 4A gas with a specific heat at constant pressure,...

Mechanical Engineering Systems 2008 Part 6 pdf

... Density␳kg/m3RelativedensityRDDynamic viscosity␮Pa s or kg/m/sAir 1. 3 1. 3 × 10 –3 1. 85 × 10 –5Water 10 00 1 0.0 01 Mercury 13 600 13 .6 1. 55 × 10 –3Mineral oil 900 0.9 8 × 10 –4Hydrostatic forces on curved surfacesIn ... to metres.Head h = (13 600/850) × (245 /10 00) – 15 0 /10 00= (16 × 0.245) – 0 .15 = 3.92 – 0 .15 = 3.77 mqFhorizontalFverticalFluid mechanics 12 9Total force (Figure 3 .1. 23)Once the two components ... because it has a high density (relativeFigure 3 .1. 10 A piezometertubeFigure 3 .1. 11 A U-tubemanometerAt=0 t= 1 secondV metresQ QV 1 V2A 1 A2 14 0 Fluid mechanicsContinuityHaving studied...

Mechanical Engineering Systems 2008 Part 3 pptx

... ΂ 1. 358V3 16 V3΃ 1. 4= 1. 899 barT5= T4΂V4V5΃␥ – 1 = 15 73 ΂ 1. 358 Vc 16 Vc΃0.4= 586.5 KHeat energy supplied = m.cv(T3– T2) + m.cp(T4– T3)= 0. 717 (11 59 – 918 ) + 1. 004 (15 73 ... cp= 1. 005 kJ/kgK, ␥ = 1. 4For combustion gases, cp= 1. 15 kJ/kgK, ␥ = 1. 33See Figure 2.5.9.TЈ2T 1 =΂p2p 1 ΃␥ 1 ␥, TЈ2= 293 ×΂4.04 1. 01 ΃0.286, TЈ2= 435.4 K␩c=TЈ2– T 1 T2– ... K␩c=TЈ2– T 1 T2– T 1 =435.4 – 293473 – 293= 0.7 91 Wc= cp(T2– T 1 ) = 1. 005(473 – 293) = 18 0.9 kJ/kgTЈ4T3=΂p4p3΃␥ 1 ␥, TЈ4= 953 ×΂ 1. 01 4.04΃ 1. 33 1 1.33, TЈ4= 676...

Mechanical Engineering Systems 2008 Part 4 potx

... x(7. 715 ), x =6.8435 – 0.5597. 715 = 0. 815 h2= hf+ xhfg= 16 3 + 0. 815 (2409) = 212 6 kJ/kgRankine efficiency =h 1 – h2h 1 – h4=3032 – 212 63032 – 16 3= 0. 316 1= 31. 6%Example 2.6 .11 Steam ... superheated.V 1 = 0.394 m3/kgp 1 Vn 1 = p2Vn26 × 0.394 1. 13= 1. 5 × V 1. 132,V2= 1. 13ͱසසසසස6 × 0.394 1. 13 1. 5= 1. 344 m3(a)Examination of the steam tables at 1. 5 bar shows ... 18 14.37 kJ/kg␩T=h 1 – h2h 1 – hЈ2, 0.85 =2798 – h22798 – 18 14.37h2= 2798 – 0.85(2798 – 18 14.37) = 19 62 kJ/kgP =˙m(h 1 – h2) = 2(2798 – 19 62) = 16 72 kWExample 2.6 .13 A steam turbine...

Mechanical Engineering Systems 2008 Part 5 pps

... wasisentropic,hЈ2= 14 69.9 + (5. 314 – 4.962)(5.397 – 4.962)× (16 13 – 14 69.9)= 15 85.7 kJ/kg0.85 =hЈ2– h 1 h2– h 1 = 15 85.7 – 14 46.5h2– 14 46.5, h2= 16 10.3 kJ/kgPower = m(h2– h 1 )=4.560 (16 10.3 ... T6) 1 r 1 h 1 +lnr2r 1 k 1 +lnr3r2k2+lnr4r3k3+ 1 r4h0(length, l =1) Q =2␲.(230 – 15 ) 1 0.05 × 550+ln605050+ln 10 0600.09+ln 16 0 10 00.07+ 1 0 .16 × 15 Q = 13 50.890.036 36 + 0.003 646 + 5.6758 + 6. 714 3 + 0. 417 = 10 5 .15 W/m ... 13 .3=˙m (14 20 – 237.2), m = 0. 011 2 kg/s (b)Condenser heat rejection =˙m(h2– h3)= 0. 011 2 (15 91. 7 – 237.2)= 15 .2 kW (c)Compressor power = m(h2– h 1 )= 0. 011 2 (15 97 .1 – 14 20)= 1. 98 kW (d)Example...

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Mechanical Engineering Systems 2008 Part 7 pptx

... foreach section of the pipe.⍀ 1 =8 × 10 × 0 .15 /(␲0.00254) = 9.778 × 10 10 Pa s/m3⍀ 1 =8 × 8 × 0 .15 /(␲0.0 015 4) = 60.3 61 × 10 10 Pa s/m3Total resistance = 70 .13 9 × 10 10 Pa s/m3The flow rate ... pipes 10 3234568 10 4 10 5 10 6 10 7222333444555666888 10 8 10 8223344556688– 0.000 01 0.00005Laminar flowLaminar flowRelative roughness0.00 01 0.00020.00040.00060.00080.0 01 0.0020.0040.0060.0080. 01 0. 015 0.020.030.040.050.000 ... 0.32)2× 9. 81) 3.3 + 7.645 + 0.0255 – 0.8 = 0 + p2/9 810 + 0.4080 10 . 71 = p2/9 810 + 0.4080Therefore:p2= 9 810 × (10 . 71 – 0.4080)= 10 1 .1 kPaThe cause of energy lossesEarlier we looked at...

Mechanical Engineering Systems 2008 Part 8 docx

... acceleration thenbecome␻2= ␻ 1 + ␣t (4 .1. 10)␪ = ␻ 1 t + ␣t2/2 (4 .1. 11) ␪ = t(␻ 1 + ␻2)/2 (4 .1. 12)␻22= ␻ 1 2+ 2␣␪ (4 .1. 13)Velocity0Timetuv( + )/2uv 17 4 DynamicsAdding the two ... distance it travels in thenext 10 s.Velocity m/s0Time/secondst 1 t2s 1 4s2t3s3Total area = 350 mSlope = 1. 8 m/s2Slope = – 2.4 m/s2Dynamics 17 5Example 4 .1. 1An electric cart is to ... the journey. (10 ) A car accelerates uniformly from 14 m/s to 32 m/s in a timeof 11 .5 s. Calculate:(a) the average velocity;(b) the distance travelled;(c) the acceleration. (11 ) The same car...

Mechanical Engineering Systems 2008 Part 10 doc

... which reaction R 1 acts gives: 12 × 5 + 10 × 10 + 15 × 15 =20R2Hence R2= 19 .25 kN. Equating the vertical components ofthe forces gives:R 1 + R2= 12 + 10 + 15 and so R 1 = 17 .75 kN.Figure ... gives:¯x =30 × 30 × 15 – 20 × 10 × 2530 × 30 – 20 × 10 = 12 .1 cmTaking moments about the x-axis gives:¯y =30 × 30 × 15 – 20 × 10 × 2530 × 30 – 20 × 10 = 12 .1 cmExample 5 .1. 13Determine the ... 5 .1. 12Figure 5 .1. 29 Example 5 .1. 13Figure 5 .1. 30 Distributed loadof 20 N/m over part of the beamis represented by a block ofarrowsFigure 5 .1. 31 Example 5 .1. 14Example 5 .1. 12Determine the...

Mechanical Engineering Systems 2008 Part 11 pptx

... 753.9 265.3 12 .9 17 .5 16 9 000 4480 18 8.0686 × 254 17 0 692.9 255.8 14 .5 23.7 17 0 000 4 910 217 .0 14 0 683.5 253.7 12 .4 19 .0 13 6 000 3990 17 9.0 12 5 677.9 253.0 11 .7 16 .2 11 8 000 3480 16 0.0Statics ... 16 .1 26.8 340 000 7990 289.0 19 4 840.7 292.4 14 .7 21. 7 279 000 6550 247.0 17 6 834.9 2 91. 6 14 .0 18 .8 246 000 5890 224.0762 × 267 19 7 769.6 268.0 15 .6 25.4 240 000 6230 2 51. 0 14 7 753.9 265.3 12 .9 ... 12 × 10 –6/°C.The force compressing the copper and extending thesteel is:F =(␣A– ␣B)␪΂ 1 EAAA+ 1 EBAB΃=(20 – 12 ) × 10 –6× 10 0 1 100 × 10 9× 1 4␲(0.0602– 0.0352)+ 1 200...

Mechanical Engineering Systems 2008 Part 13 pps

... kN, FDE+23 .1 kN, FEF–23 .1 kN, FFG+23 .1 kN, FGH +11 .5 kN, FHI 11 .5 kN, FIJ +11 .5 kN, FJK 11 .5 kN, FKL +11 .5 kN, FLM 11 .5 kN, FMN+ 11 .5 kN, FAN 11 .5 kN, FCD +11 .5 kN, FCF+34.6 ... kg/s8. 0 .11 m/s9. 0 .15 7 m/s, 1. 22 m/s 10 . 0.637 m/s, 7.07 m/s, 31. 83 m/s 11 . 13 4.7 kPa 12 . 7.62 × 10 –6m3/s 13 . 208 mm 14 . 0. 015 × 10 –6m3/s 15 . 2.64 m, 25.85 kPa 16 . 21. 3 m 17 . 12 .34 m 18 . ... vertically, 10 4 N vertically 10 . 300 N m 11 . 2.732 kN m clockwise 12 . P = 10 3.3 N, Q = 11 5 .1 N, R = 70 .1 N 13 . 36.9°, 15 N m 14 . 216 .3 N, 250 N, 12 5 N 15 . (a) 55.9 mm, (b) 21. 4 mm, (c) 70 mm 16 . 4.7 m 17 ....

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