Mechanical Engineering Systems 2008 Part 8 docx

... common shape found in engineering objects such as pulleys, this is given by the formula I = mr 2 /2 S/metres 0 5 10 24 68 t / seconds Slope = d / d s t Velocity / m sV / 0 1 2 24 68 t / second s Calculation ... motion. The resultant force on the smaller mass is therefore T – (200 × 9 .81 × sin 30°) – (0.3 × 200 × 9 .81 × cos 30°) = T – 981 – 509.7 = T – 1490.7 The acceleration is given by...

Ngày tải lên: 11/08/2014, 15:20

... vaporization water Q 4 =4 × 2256.7 = 9026 .8 kJ Total heat energy = Q 1 + Q 2 + Q 3 + Q 4 = 81 .6 + 1340 + 1 680 + 9026 .8 = 12 1 28. 4 kJ. Mechanical Engineering Systems Thermodynamics 11 Example 2.1.4 A ... of water and is at a room temperature of 18 C, and the kettle is 2 kW. Specific heat of water = 4.2 kJ/kgK. Q = m.c. ␦T Q =2 × 4.2 × (100 – 18) = 688 .8 kJ This is the heat ene...

Ngày tải lên: 11/08/2014, 15:20

... ΂ 1.358V 3 16V 3 ΃ 1.4 = 1 .89 9 bar T 5 = T 4 ΂ V 4 V 5 ΃ ␥ – 1 = 1573 ΂ 1.3 58 V c 16 V c ΃ 0.4 = 586 .5 K Heat energy supplied = m.c v (T 3 – T 2 ) + m.c p (T 4 – T 3 ) = 0.717(1159 – 9 18) + 1.004(1573 ... 416 = 589 kJ/kg Heat energy rejected = m.c v (T 5 – T 1 ) = 0.717 ( 586 .5 – 303) = 203.3 kJ/kg Air standard efficiency = 1 – heat rejected heat supplied =1 – 203.3 589 = 0.65 48...

Ngày tải lên: 11/08/2014, 15:20

... 121 + (0.696 × 2433) = 181 4.37 kJ/kg ␩ T = h 1 – h 2 h 1 – hЈ 2 , 0 .85 = 27 98 – h 2 27 98 – 181 4.37 h 2 = 27 98 – 0 .85 (27 98 – 181 4.37) = 1962 kJ/kg P = ˙ m(h 1 – h 2 ) = 2(27 98 – 1962) = 1672 kW Example ... u f + x.u fg = 82 8 + 0.9(2593 – 82 8) = 2416.5 kJ/kg u 2 = u f + x.u fg = 669 + 0.4(25 68 – 669) = 14 28. 6 kJ/kg Change in internal energy = U 2 – U 1 = (14 28. 6 – 2416...

Ngày tải lên: 11/08/2014, 15:20

... 20°C. Figure 2 .8. 8 shows the wall. This problem demonstrates working through part of the wall only. Figure 2 .8. 7 Example 2 .8. 3 Figure 2 .8. 8 Example 2 .8. 4 Thermodynamics 109 t s at 28 bar = 230°C Q ... brick being 0.6 W/mK. See Figure 2 .8. 6. T 1 – T 2 = Q At ΂ s k ΃ (17–5) = Q (6 × 2.9) × (60 × 60) ΂ 0.225 0.6 ΃ Figure 2 .8. 5 Example 2 .8. 1 Figure 2 .8. 6 Example 2 .8. 2 Therm...

Ngày tải lên: 11/08/2014, 15:20

... 1.5/4 = 4.712 m 3 Hence the total volume over the curve is 10. 288 m 3 and the weight pressing down is 4.712 × 0 .8 × 1000 × 9 .81 = 36. 98 kN Now we shall calculate the horizontal force acting on the curve. ... × (4 × 5.4) = 1000 × 9 .81 × 2 × 4 × 5.4 = 423 .8 kN Therefore F 2 = 2 423 80 0 N We could also express F 2 as (␳gh/2) × (h × 5.4) and so 1000 × 9 .81 × 5.4 × h 2 /2 = 2 423 80 0...

Ngày tải lên: 11/08/2014, 15:20

... × 9 .81 ) + 0 .8 2 /(2 × (␲ × 0.6 2 ) 2 × 9 .81 ) – 0 .8 =0 + p 2 /(1000 × 9 .81 ) + 0 .8 2 /(2 × (␲ × 0.3 2 ) 2 × 9 .81 ) 3.3 + 7.645 + 0.0255 – 0 .8 = 0 + p 2 / 981 0 + 0.4 080 10.71 = p 2 / 981 0 + 0.4 080 Therefore: p 2 = ... Carburettor 0.025 0.020 0.015 0.010 0.009 0.0 08 0.007 0.006 0.005 0.004 0.003 0.002 Friction factor Re y nolds number Turbulent flow Turbulent flow Smooth pi...

Ngày tải lên: 11/08/2014, 15:20

... 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 ... a sketch of the body, divide it into a number of composite parts. A hole, i.e. a part having no material, can be considered to be a part having a negative weight or area. (2) Establish t...

Ngày tải lên: 11/08/2014, 15:20

... of section cm 2 83 8 × 292 226 85 0.9 293 .8 16.1 26 .8 340 000 7990 289 .0 194 84 0.7 292.4 14.7 21.7 279 000 6550 247.0 176 83 4.9 291.6 14.0 18. 8 246 000 589 0 224.0 762 × 267 197 769.6 2 68. 0 15.6 25.4 ... 265.3 12.9 17.5 169 000 4 480 188 .0 686 × 254 170 692.9 255 .8 14.5 23.7 170 000 4910 217.0 140 683 .5 253.7 12.4 19.0 136 000 3990 179.0 125 677.9 253.0 11.7 16.2 1 18 000...

Ngày tải lên: 11/08/2014, 15:20

... body 250 Free-body diagram 207 Friction 282 dry 282 fluid 282 kinetic 283 laws 283 on rough inclined plane 285 static 283 , 285 Friction factor, pipe wall 157 8 Friction power 45 Frictional losses: in ... kW, 57.42 kW, 86 3 .8 kJ, 1103 .8 kJ, 1913 .8 kJ 36. 417 .8 m/s 37. 589 kW 38. 7 .85 kW, 20.4 m 39. 14.0 m/s 40. 12.5 m/s 41. (a) 10.5 m (b) 9.6 m/s 42. 5 .87 m/s 43. 12 89 2....

Ngày tải lên: 11/08/2014, 15:20