Mechanical Engineering Systems 2008 Part 11 pptx
Mechanical Engineering Systems 2008 Part 11 pptx
... T-section: Total moment = 250 × 30 × 115 + 100 × 50 × 50 = 1 .11 × 10 6 mm 4 Hence the distance of the centroid from the base is (total moment)/(total area): Distance from base = 1 .11 × 10 6 250 × 30 + 100 ... beam, the bending moment, for that part of the beam to the right, is 3 × 4 – 9 × 0.5 = +7.5 kN m. Example 5.4.2 A uniform cantilever of length 3.0 m (Figure 5.4 .11) has a uniform...
Ngày tải lên: 11/08/2014, 15:20
Mechanical Engineering Systems 2008 Part 3 pptx
... ferries, has the following particulars: Power output 8200 kW Operating cycle 4-stroke Number of cylinders 20, in ‘V’ configuration Bore 265 mm Stroke 315 mm Operating speed 115 0 rpm Dimensions 7.4 ... 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 (115 9 – 918) + 1.004(1573 – 115 9), using a mass of 1kg = 173 + 416 = 589 kJ/kg Heat energy rejecte...
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Mechanical Engineering Systems 2008 Part 7 pptx
... 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 cushion an impact. A piston ... level Fluid mechanics 145 Examples of laminar flow in engineering We have already touched on one example of laminar flow in pipes which is highly relevant to engineering but it is worth...
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Mechanical Engineering Systems 2008 Part 1 pdf
... energy = Q 1 + Q 2 + Q 3 + Q 4 = 81.6 + 1340 + 1680 + 9026.8 = 12 128.4 kJ. Mechanical Engineering Systems Thermodynamics 11 Example 2.1.4 A gas with a specific heat at constant pressure, c p = 900 ... split this distance into ten million subdivisions called measures. In French this is Mechanical Engineering Systems Richard Gentle Peter Edwards Bill Bolton OXFORD AUCKLAND BOSTON...
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Mechanical Engineering Systems 2008 Part 4 potx
... definition). u = 114 9 + 0.8(2597 – 114 9) = 2307.4 kJ/kg (d) 5 kg of steam at 40 bar, t = 500°C. From superheat tables, u = 3099 kJ/kg. For 5 kg, u =5 × 3099 = 15 495 kJ. Thermodynamics 77 Figure 2.6 .11 T/s ... x(7.918), x = 0.814 hЈ 2 = h f + x.h fg = 138 + (0.814 × 2423) = 2110 .3 kJ/kg T = h 1 – h 2 h 1 – hЈ 2 = 3231 – 2283.6 3231 – 2110 .3 = 0.845 = 84.5% (b) The h/s chart An h/s c...
Ngày tải lên: 11/08/2014, 15:20
Mechanical Engineering Systems 2008 Part 5 pps
... 13.3 = ˙ m(1420 – 237.2), m = 0. 0112 kg/s (b) Condenser heat rejection = ˙ m(h 2 – h 3 ) = 0. 0112 (1591.7 – 237.2) = 15.2 kW (c) Compressor power = m(h 2 – h 1 ) = 0. 0112 (1597.1 – 1420) = 1.98 kW ... condenser in kW; (d) the compressor power in kW. See Figures 2.7 .11 and 2.7.12. Figure 2.7 .11 Example 2.7.3 Figure 2.7.12 Example 2.7.3 h A 114 Fluid mechanics From this list of differen...
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Mechanical Engineering Systems 2008 Part 6 pdf
... 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 illustration of a particle flowing along in ... the individual particles stream along it. Turbulent flow is when the individual particles do not follow any regular path but their overall motion gives rise to the liquid flow. The particles ar...
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Mechanical Engineering Systems 2008 Part 8 docx
... in engineering terms that is only half the story. Forces in engineering dynamics are generally used to move something from one place to another or to alter the speed of a piece of machinery, particularly ... calculations is to introduce the idea of mechanical efficiency to represent how close a piece of mechanical equipment is to being ideal in energy terms. Mechanical efficiency =...
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Mechanical Engineering Systems 2008 Part 10 doc
... 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 the co-ordinate ... with one part of it having a diameter of 60 mm and the other part a diameter of 30 mm (Figure 5.3.9) and is subject to an axial force of 20 kN. What will be the stresses in the two p...
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Mechanical Engineering Systems 2008 Part 13 pps
... F EF –23.1 kN, F FG +23.1 kN, F GH +11. 5 kN, F HI 11. 5 kN, F IJ +11. 5 kN, F JK 11. 5 kN, F KL +11. 5 kN, F LM 11. 5 kN, F MN + 11. 5 kN, F AN 11. 5 kN, F CD +11. 5 kN, F CF +34.6 kN, F BH +40.4 kN, F BJ +28.9 ... inverted U-tube 119 differential mercury U-tube 119 –20 simple tube 117 U-tube 117 –18 Manometry 116 Marine diesel engines (case study) 51–2 Mass 184–6 Mass momen...
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