Mechanical Engineering Systems 2008 Part 7 pptx

Mechanical Engineering Systems 2008 Part 7 pptx

Mechanical Engineering Systems 2008 Part 7 pptx

... 0.15/(␲0.0025 4 ) = 9 .77 8 × 10 10 Pa s/m 3 ⍀ 1 =8 × 8 × 0.15/(␲0.0015 4 ) = 60.361 × 10 10 Pa s/m 3 Total resistance = 70 .139 × 10 10 Pa s/m 3 The flow rate is then given by: Q = P/⍀ = 120 000 /70 .139 × 10 10 Pa ... laminar flow is necessary. One example from mainstream mechanical engineering is the dashpot. This is a device which is used to damp out any mechanical vibration or to...

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Mechanical Engineering Systems 2008 Part 3 pptx

Mechanical Engineering Systems 2008 Part 3 pptx

... 2 87 × (15 + 273 ) 1 × 10 5 = 7. 52 m 3 /min Swept volume = ␲ × 0.09 2 4 × 0.11 = 7 × 10 –4 m 3 /rev =7 10 –4 × 370 0 2 × 6 = 7. 76 m 3 /min ␩ v = volume induced at reference swept volume = 7. 52 7. 76 = ... the brake power; (c) the mechanical efficiency. ip = P mi A.L.n =3 × 10 5 × ␲ × 0.35 2 4 × 0.5 × 2 .75 × 2 = 79 374 W = 79 . 37 kW bp = T ␻ = (2 .7 × 1.6) × 2 .75 × 2␲ = 7...

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Mechanical Engineering Systems 2008 Part 11 pptx

Mechanical Engineering Systems 2008 Part 11 pptx

... 340 000 79 90 289.0 194 840 .7 292.4 14 .7 21 .7 279 000 6550 2 47. 0 176 834.9 291.6 14.0 18.8 246 000 5890 224.0 76 2 × 2 67 1 97 769.6 268.0 15.6 25.4 240 000 6230 251.0 1 47 753.9 265.3 12.9 17. 5 169 ... 000 4480 188.0 686 × 254 170 692.9 255.8 14.5 23 .7 170 000 4910 2 17. 0 140 683.5 253 .7 12.4 19.0 136 000 3990 179 .0 125 677 .9 253.0 11 .7 16.2 118 000 3480 160.0 Statics...

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Mechanical Engineering Systems 2008 Part 1 pdf

Mechanical Engineering Systems 2008 Part 1 pdf

... cylinder and heated to 75 °C while the volume remains constant. Calculate the heat energy supplied if c v = 70 0 J/kgK. Q = m.c v .␦T Q = 1.5 × 70 0 × (75 – 20) = 57 750 J = 57. 75 kJ Figure 2.1.1 Specific ... 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 5 2 Thermodynamics 7 2.1 Heat energy 7 2.2 Perfect gase...

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Mechanical Engineering Systems 2008 Part 4 potx

Mechanical Engineering Systems 2008 Part 4 potx

... energy equation, Q = W + (U 2 – U 1 ) where: W = 2 67. 7 kJ/kg U 1 = 272 2 kJ/kg U 2 = 26 07. 7 kJ/kg, by interpolation. Q = 2 67. 7 + (26 07. 7 – 272 2) = 153.4 kJ/kg (c) This is +ve, therefore heat energy ... × 2433) = 1814. 37 kJ/kg ␩ T = h 1 – h 2 h 1 – hЈ 2 , 0.85 = 279 8 – h 2 279 8 – 1814. 37 h 2 = 279 8 – 0.85( 279 8 – 1814. 37) = 1962 kJ/kg P = ˙ m(h 1 – h 2 ) = 2( 279 8 – 1962)...

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Mechanical Engineering Systems 2008 Part 5 pps

Mechanical Engineering Systems 2008 Part 5 pps

... be superheated. Figure 2 .7. 7 Example 2 .7. 1 Figure 2 .7. 8 Example 2 .7. 1 96 Thermodynamics Interpolating using entropy values, h 2 = 15 97. 2 + 5 .78 5 – 5.521 5.854 – 5.521 × ( 171 9.3 – 15 97. 2) = 1694 kJ/kg Refrigeration ... h 4 h 2 – h 1 = 124.33 26.6 = 4. 67 (c) Carnot COP ref = T L T L – T H = (–10 + 273 ) (45 + 273 ) – (– 10 + 273 ) = 4 .78 (d) Figure 2 .7. 13 Example 2 .7. 4 F...

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Mechanical Engineering Systems 2008 Part 6 pdf

Mechanical Engineering Systems 2008 Part 6 pdf

... Equation (3.2.1). ␳vD/␮ = 78 0 v 0.3/0.125 = 2000 v = 2000 × 0.125/ (78 0 × 0.3) = 1. 07 m/s Figure 3.2.6 Pressure/flow characteristics for flow along a pipe Figure 3.2 .7 The advantages of ‘streamlining’ h P X AA 1 H Measuring ... exposure record of what the particle does. In streamlined flow we would see that the particle followed exactly the line of flow of the liquid, as in this illust...

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Mechanical Engineering Systems 2008 Part 8 docx

Mechanical Engineering Systems 2008 Part 8 docx

... 2.4 s 3 s 3 = 3.33 m The distance involved in stopping and starting is s 1 + s 3 = 7. 77 m and so s 2 = 350 – 7. 77 = 342.22 m Dynamics 183 (13) A piece of masonry is dislodged from the top of a ... 16 3 37 radians. This takes place in 1 minute, or 60 seconds and so the initial angular velocity of 2600 rev/min becomes 272 .3 rad/s. Hence 0 = 272 .3 – ␣10 ␣ = 27. 23 rad/s 2 For the second...

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Mechanical Engineering Systems 2008 Part 10 doc

Mechanical Engineering Systems 2008 Part 10 doc

... parts of the rod? Each part will experience the same force and thus the stress on the larger diameter part is 20 × 10 3 /( 1 4 ␲ × 0.060 2 ) = 7. 1 MPa and the stress on the smaller diameter part ... the centroid for the uniform thickness sheet shown in Figure 5.1. 27. For the object shown in Figure 5.1. 27, the two constituent parts are different sized, homogeneous, rectangular objec...

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Mechanical Engineering Systems 2008 Part 13 pps

Mechanical Engineering Systems 2008 Part 13 pps

... turbine 57 8, 63 throttle 59 Steady flow processes 17 Steam: flow processes 74 , 76 non-flow processes 84 production 67 8 superheated 67, 68, 71 types 67 Steam flow processes 74 , 76 Steam plant 76 –8 Steam ... 1309 N m 22. 77 2 N m 23. 9.91 s, 370 kN 24. 2 .77 5 kJ 25. 4010 J 26. 259 kJ 27. 12.3 m/s 28. 51 m/s 29. 82% 30. 973 .5 m 31. 17. 83 m/s 32. 15.35 m/s 33. 70 % 34. (a) 6...

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