Mechanical Engineering Systems 2008 Part 6 pdf

Mechanical Engineering Systems 2008 Part 6 pdf

Mechanical Engineering Systems 2008 Part 6 pdf

... volume of a cylinder of radius 0 .6 m and length 2.3 m. Volume of pillar submerged = ␲0 .6 2 × 2.3 = 2 .60 m 3 Weight of water displaced = 2 .60 × 1000 × 9.81 = 25.5 06 kN The tension in the cable with ... once we have converted from x to h in the first part of the calculation then the mercury plays no further part. Pressure difference = 790 × 9.81 × 3. 86 = 29 909 Pa Equal volumes sw...

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Mechanical Engineering Systems 2008 Part 1 pdf

Mechanical Engineering Systems 2008 Part 1 pdf

... of vaporization water Q 4 =4 × 22 56. 7 = 90 26. 8 kJ Total heat energy = Q 1 + Q 2 + Q 3 + Q 4 = 81 .6 + 1340 + 168 0 + 90 26. 8 = 12 128.4 kJ. Mechanical Engineering Systems Thermodynamics 11 Example ... equation 54 2 .6 Steam 66 2.7 Refrigeration 89 2.8 Heat transfer 101 3 Fluid mechanics 112 3.1 Hydrostatics – fluids at rest 113 3.2 Hydrodynamics – fluids in motion 135 4 Dynamic...

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Mechanical Engineering Systems 2008 Part 3 pptx

Mechanical Engineering Systems 2008 Part 3 pptx

... TЈ 4 = 67 6 K ␩ T = 0.84 = T 3 – T 4 T 3 – TЈ 4 ,= 953 – T 4 953 – 67 6 T 4 = 953 – 0.84(953 – 67 6) = 720.3 K W T = c p (T 3 – T 4 ) = 1.15(953 – 720.3) = 267 .6 kJ/kg Net work = W T – W c = 267 .6 – ... 41. 86 MJ/kg. ␩ m = bp ip ip = bp ␩ m = 3200 0.9 = 3555 .6 kW ip = P m A.L.n 3555 .6 8 = P mi × ␲ × 0.4 2 4 × 0.54 × 6. 67 2 P mi = 3555 .6 × 4 × 2 8 × ␲ × 0.4 2 × 0.54 × 6. 67 =...

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Mechanical Engineering Systems 2008 Part 4 potx

Mechanical Engineering Systems 2008 Part 4 potx

... 0.9(2593 – 828) = 24 16. 5 kJ/kg u 2 = u f + x.u fg = 66 9 + 0.4(2 568 – 66 9) = 1428 .6 kJ/kg Change in internal energy = U 2 – U 1 = (1428 .6 – 24 16. 5)0.787 = –778 kJ/kg (c) Heat energy transferred, ... 2798 kJ/kg, s 1 = 6. 029 kJ/kgK At 0.04 bar, for isentropic expansion, sЈ 2 = s 1 = 6. 029 = 0.422 + x(8.051), x = 6. 029 – 0.422 8.051 = 0 .69 6 hЈ 2 = h f + x.h fg = 121 + (0 .69...

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Mechanical Engineering Systems 2008 Part 5 pps

Mechanical Engineering Systems 2008 Part 5 pps

... 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 × 550 + ln 60 50 50 + ln 100 60 0.09 + ln 160 100 0.07 + 1 0. 16 × 15 Q = 1350.89 0.0 36 36 + 0.003 64 6 + 5 .67 58 + 6. 7143 + 0.417 = 105.15 W/m length To find the ... was isentropic, hЈ 2 = 1 469 .9 + (5.314 – 4. 962 ) (5.397 – 4. 962 ) × ( 161 3 – 1 469 .9)...

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Mechanical Engineering Systems 2008 Part 7 pptx

Mechanical Engineering Systems 2008 Part 7 pptx

... Carburettor 0.025 0.020 0.015 0.010 0.009 0.008 0.007 0.0 06 0.005 0.004 0.003 0.002 Friction factor Re y nolds number Turbulent flow Turbulent flow Smooth pipes Smooth pipes 10 3 234 568 10 4 10 5 10 6 10 7 22233344455 566 6888 10 8 10 8 2 2 3 3 4 4 5 5 6 6 8 8 – ... pipes 10 3 234 568 10 4 10 5 10 6 10 7 22233344455 566 6888 10 8 10 8 2 2 3 3 4 4 5 5 6 6 8 8 – 0.00001 0.00005 Lam...

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Mechanical Engineering Systems 2008 Part 8 docx

Mechanical Engineering Systems 2008 Part 8 docx

... revolution is equal to 2␲ radians and so 260 0 revolutions equals 16 337 radians. This takes place in 1 minute, or 60 seconds and so the initial angular velocity of 260 0 rev/min becomes 272.3 rad/s. Hence 0 ... 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 piec...

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Mechanical Engineering Systems 2008 Part 10 doc

Mechanical Engineering Systems 2008 Part 10 doc

... ␴ B A B 1.5 × 10 6 = 14.3␴ c × 1 963 × 10 6 + ␴ c × 200 537 × 10 6 Hence the stress on the concrete is 6. 56 MPa and that on the steel is 93.8 MPa. Example 5.3 .6 A rod is formed with one part of it ... distance of the centroid of the composite from X. Thus: ¯ x = 160 × 100 × 50 + 260 × 60 × 230 160 × 100 + 260 × 60 = 138.9 m Figure 5.1. 26 Circular arc Figure 5.1.27 Example...

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Mechanical Engineering Systems 2008 Part 11 pptx

Mechanical Engineering Systems 2008 Part 11 pptx

... × 292 2 26 850.9 293.8 16. 1 26. 8 340 000 7990 289.0 194 840.7 292.4 14.7 21.7 279 000 65 50 247.0 1 76 834.9 291 .6 14.0 18.8 2 46 000 5890 224.0 762 × 267 197 769 .6 268 .0 15 .6 25.4 240 000 62 30 251.0 147 ... 251.0 147 753.9 265 .3 12.9 17.5 169 000 4480 188.0 68 6 × 254 170 69 2.9 255.8 14.5 23.7 170 000 4910 217.0 140 68 3.5 253.7 12.4 19.0 1 36 000 3990 179.0 125 67 7.9...

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Mechanical Engineering Systems 2008 Part 13 pps

Mechanical Engineering Systems 2008 Part 13 pps

... bar 3. 65 % 4. 46. 87 kJ/kg rejected, 180.1 kJ/kg supplied, 61 6.9 kJ/kg 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 .67 5 kW, 22.4% 5. 6. 64 bar, 32% 6. 6. 455 ... 3. 36 bar 2. 563 .4 K 3. 0.31 m 3 , 572 K 4. 0.2 76 m 3 , 131.3°C 5. 47 .6 bar, 304°C 6. 0.31 m 3 , 108.9°C 7. 1.32 8. 1.24, 3 56. 8 K 9. 60 0 cm 3 , 1. 068 bar, 252 .6...

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