Mechanical Engineering Systems 2008 Part 5 pps
Mechanical Engineering Systems 2008 Part 5 pps
... values, h 2 = 159 7.2 + 5. 7 85 – 5. 521 5. 854 – 5. 521 × (1719.3 – 159 7.2) = 1694 kJ/kg Refrigeration effect = h 1 – h 4 = 14 05. 6 – 2 75. 1 = 1130 .5 kJ/kg (a) COP = h 1 – h 4 h 2 – h 1 = 1130 .5 1694 – 14 05. 6 = ... T 6 ) 1 r 1 h 1 + ln r 2 r 1 k 1 + ln r 3 r 2 k 2 + ln r 4 r 3 k 3 + 1 r 4 h 0 (length, l =1) Q = 2.(230 – 15) 1 0. 05 × 55 0 + ln 60 50 50 + ln 100 60 0.09 +...
Ngày tải lên: 11/08/2014, 15:20
Mechanical Engineering Systems 2008 Part 13 pps
... rejected 5. 61%, 5. 67 bar 2.4.2 1. 1841 kW 2. 28.63 kW, 24.3 kW 3. 9. 65 kW 4. 10.9 kW, 8.6 75 kW, 22.4% 5. 6.64 bar, 32% 6. 6. 455 bar, 30.08%, 83.4% 7. 65% 8. 35% 2 .5. 1 1. 1330 kW 2. 1 25 kJ/kg 3. 5 kJ 4. ... m/s, 264 .5 m, 1 .56 5 m/s 2 11. –4.27 m/s 2 , 7 .5 s 12. 430 m 13. 59 . 05 m/s 14. 397 m, 392 m, 88.3 m/s 15. 57 .8 s, 80. 45 s 4.2.1 1. 3.36 kN 2. 2.93 m/s 2 3. 3...
Ngày tải lên: 11/08/2014, 15:20
Mechanical Engineering Systems 2008 Part 1 pdf
... motion 1 35 4 Dynamics 169 4.1 Introduction to kinematics 170 4.2 Dynamics – analysis of motion due to forces 183 5 Statics 204 5. 1 Equilibrium 2 05 5.2 Structures 222 5. 3 Stress and strain 2 35 5.4 ... 222 5. 3 Stress and strain 2 35 5.4 Beams 249 5. 5 Cables 2 75 5.6 Friction 282 5. 7 Virtual work 287 5. 8 Case study: bridging gaps 292 Solutions to problems 2 95 Index 307 Th...
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Mechanical Engineering Systems 2008 Part 3 pptx
... Thermodynamics p 4 V 4 ␥ = p 5 V 5 ␥ p 5 = p 4 V 4 V 5 ␥ = 60 1. 358 V 3 16V 3 1.4 = 1.899 bar T 5 = T 4 V 4 V 5 ␥ – 1 = 157 3 1. 358 V c 16 V c 0.4 = 58 6 .5 K Heat energy supplied = ... MJ/kg. m = bp ip ip = bp m = 3200 0.9 = 355 5.6 kW ip = P m A.L.n 355 5.6 8 = P mi × × 0.4 2 4 × 0 .54 × 6.67 2 P mi = 355 5.6 × 4 × 2 8 × × 0.4 2 × 0 .54 × 6.67 = 1...
Ngày tải lên: 11/08/2014, 15:20
Mechanical Engineering Systems 2008 Part 4 potx
... 85 8 × 0.2613 1. 25 =4 × V 1. 25 2 V 2 = 1.2 5 ͱ සසසසස 8 × 0.2613 1. 25 4 = 0. 455 m 3 /kg V = x.V g at 4 bar 0. 455 = x × 0.4623 x = 0. 455 0.4623 = 0.984 Example 2.6. 15 1 kg of steam at 6 bar, 250 °C ... 17 .5 bar, 300°C, h 1 = 3032 kJ/kg, s 1 = 6.84 35 kJ/kgK. s 1 = s 2 because the expansion is isentropic. 6.84 35 = s f + x.s fg at 0.07 bar 6.84 35 = 0 .55 9 + x(7.7 15) , x =...
Ngày tải lên: 11/08/2014, 15:20
Mechanical Engineering Systems 2008 Part 6 pdf
... 9.81 = 25. 506 kN The tension in the cable with the pillar in air = 5 × 1000 × 9.81 = 49. 050 kN Hence the resultant tension in the cable with the pillar partly submerged is: 49. 050 – 25. 506 = 23 .54 ... second from cubic metres per minute. Q = A 1 v 1 = A 2 v 2 15/ 60 = 0. 05 v 1 = 0.02 v 2 v 1 = 15/ (60 × 0. 05) = 5 m/s v 2 = 15/ (60 × 0.02) = 12 .5 m/s Laminar flow Viscosity Befor...
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Mechanical Engineering Systems 2008 Part 7 pptx
... pipes 10 3 23 456 8 10 4 10 5 10 6 10 7 22233344 455 5666888 10 8 10 8 2 2 3 3 4 4 5 5 6 6 8 8 – 0.00001 0.000 05 Laminar flow Laminar flow Relative roughness 0.0001 0.0002 0.0004 0.0006 0.0008 0.001 0.002 0.004 0.006 0.008 0.01 0.0 15 0.02 0.03 0.04 0. 05 0.000 ... Carburettor 0.0 25 0.020 0.0 15 0.010 0.009 0.008 0.007 0.006 0.0 05 0.004 0.003 0.002 Friction factor Re y nolds numb...
Ngày tải lên: 11/08/2014, 15:20
Mechanical Engineering Systems 2008 Part 8 docx
... this portion of the motion is simply t 2 = 342.22 m/4 m/s = 85. 56 s Hence the total time of travel is ( 85. 56 s + 2.22 s + 1.67 s) = 89. 45 s Note that, within reason, the route does not have to be ... velocity by the time of flight. Distance travelled = 25 × 0 .55 3 = 13.8 m One of the questions that arises often in the subject of trajectories, particularly in its application to sport, i...
Ngày tải lên: 11/08/2014, 15:20
Mechanical Engineering Systems 2008 Part 10 doc
... 1 sin 60° = R 2 × 3 Figure 5. 2.13 Example 5. 2.4 Figure 5. 2.14 Example 5. 2 .5 Figure 5. 2. 15 Example 5. 2 .5 228 Statics Example 5. 2.3 For the plane truss shown in Figure 5. 2.10, determine the reactions ... 5. 2. 25. Figure 5. 2.23 Problem 8 Figure 5. 2.24 Problem 9 Figure 5. 2. 25 Problem 10 Figure 5. 1.28 Example 5. 1.12 Figure 5. 1.29 Example 5. 1.13 Figure 5. 1...
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Mechanical Engineering Systems 2008 Part 11 pptx
... 2 65. 3 12.9 17 .5 169 000 4480 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 1 25 677.9 253 .0 11.7 16.2 118 000 3480 160.0 Statics 257 The ... with a central load F (Figure 5. 4.13), the reactions at each end will be F/2. Figure 5. 4.10 Example 5. 4.1 Figure 5. 4.11 Example 5. 4.2 Figure 5. 4.12 Example 5. 4.2 Figur...
Ngày tải lên: 11/08/2014, 15:20