Mechanical Engineering Systems 2008 Part 3 pptx

Mechanical Engineering Systems 2008 Part 3 pptx

Mechanical Engineering Systems 2008 Part 3 pptx

... 1.005(4 73 – 2 93) = 180.9 kJ/kg TЈ 4 T 3 = ΂ p 4 p 3 ΃ ␥–1 ␥ , TЈ 4 = 9 53 × ΂ 1.01 4.04 ΃ 1 .33 –1 1 .33 , TЈ 4 = 676 K ␩ T = 0.84 = T 3 – T 4 T 3 – TЈ 4 ,= 9 53 – T 4 9 53 – 676 T 4 = 9 53 – 0.84(9 53 – ... c p = 1.15 kJ/kgK, ␥ = 1 .33 See Figure 2.5.9. TЈ 2 T 1 = ΂ p 2 p 1 ΃ ␥–1 ␥ , TЈ 2 = 2 93 × ΂ 4.04 1.01 ΃ 0.286 , TЈ 2 = 435 .4 K ␩ c = TЈ 2 – T 1 T 2 – T 1 = 435 .4 –...
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Mechanical Engineering Systems 2008 Part 7 pptx

Mechanical Engineering Systems 2008 Part 7 pptx

... 10 10 Pa s/m 3 ⍀ 1 =8 × 8 × 0.15/(␲0.0015 4 ) = 60 .36 1 × 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 s/m 3 = 1.71 ... number Turbulent flow Turbulent flow Smooth pipes Smooth pipes 10 3 234 568 10 4 10 5 10 6 10 7 22 233 3444555666888 10 8 10 8 2 2 3 3 4 4 5 5 6 6 8 8 – 0.00001 0.00005 Laminar fl...
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Mechanical Engineering Systems 2008 Part 11 pptx

Mechanical Engineering Systems 2008 Part 11 pptx

... 6 230 251.0 147 7 53. 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 217.0 140 6 83. 5 2 53. 7 12.4 19.0 136 000 39 90 179.0 125 677.9 2 53. 0 11.7 16.2 118 000 34 80 ... 10 9 × 1 4 ␲(0.060 2 – 0. 035 2 ) + 1 200 × 10 9 × 1 4 ␲0. 03 = 64 .3 kN The compressive stress acting on the copper is thus: ␴ A = 64 .3 × 10 3 1 4 ␲(0.060 2 – 0. 035 2 ) =...
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Mechanical Engineering Systems 2008 Part 1 pdf

Mechanical Engineering Systems 2008 Part 1 pdf

... 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 gases, gas laws, gas processes 13 2 .3 Work done and ... combustion engines 33 2.5 The steady flow energy 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 1 13 3....
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Mechanical Engineering Systems 2008 Part 4 potx

Mechanical Engineering Systems 2008 Part 4 potx

... to condenser cooling = ˙ m(h 2 – h 3 ) = 1.5(2420 .36 – 230 .2) = 32 85.24 kW Thermal efficiency = h 1 – h 2 h 1 – h 4 = 34 33 – 2420 .36 34 33 – 230 .2 = 0 .31 6 = 31 .6% Note: ᭹ Condenser undercooling ... superheat tables at 30 bar, 400°C, h 1 = 32 31 kJ/kg 32 31 – h 2 = 947.4, h 2 = 32 31 – 947.4 = 22 83. 6 kJ/kg h 2 = 22 83. 6 = h f + x.h fg at 0.05 bar, 22 83. 6 = 138 + x.24...
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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 = 135 0.89 0. 036 36 + 0.0 03 646 + 5.6758 + 6.71 43 + 0.417 = ... work = 210. 63 – 184. 03 = 26.6 kJ/kg (b) COP = h 1 – h 4 h 2 – h 1 = 124 .33 26.6 = 4.67 (c) Carnot COP ref = T L T L – T H...
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Mechanical Engineering Systems 2008 Part 6 pdf

Mechanical Engineering Systems 2008 Part 6 pdf

... × 10 6 × d =(F 2 × 9.57 /3) – (F 1 × 4 /3) = (2 4 23 800 × 9.57 /3) – (4 23 800 × 4 /3) which finally gives d = 3. 59 m Properties of some common fluids Fluid Density ␳ kg/m 3 Relative density RD Dynamic ... to its surroundings. Figure 3. 1 .30 Swimming pool with window Figure 3. 1 .31 Sluice gate Figure 3. 1 .32 A dolphin pool with a viewing tunnel Dye D V Fluid mechanics 137 Sinc...
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Mechanical Engineering Systems 2008 Part 8 docx

Mechanical Engineering Systems 2008 Part 8 docx

... deceleration 0 = 16 – 2 × 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 = 35 0 – 7.77 = 34 2.22 m Dynamics 1 83 ( 13) A piece of masonry is dislodged ... 16 33 7 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 r...
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Mechanical Engineering Systems 2008 Part 10 doc

Mechanical Engineering Systems 2008 Part 10 doc

... × 30 × 15 – 20 × 10 × 25 30 × 30 – 20 × 10 = 12.1 cm Taking moments about the x-axis gives: ¯ y = 30 × 30 × 15 – 20 × 10 × 25 30 × 30 – 20 × 10 = 12.1 cm Example 5.1. 13 Determine the position ... MPa and that on the steel is 93. 8 MPa. Example 5 .3. 6 A rod is formed 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 sub...
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Mechanical Engineering Systems 2008 Part 13 pps

Mechanical Engineering Systems 2008 Part 13 pps

... kJ 2. 51 .3 C 3. 3. 19 kJ, 4.5 kJ 4. 712 kJ 5. 38 0 .3 kJ 2.2.1 1. 24.2 kg 2. 1.05 m 3 3. 45 .37 kg, 23. 9 bar 2.2.2 1. 3. 36 bar 2. 5 63. 4 K 3. 0 .31 m 3 , 572 K 4. 0.276 m 3 , 131 .3 C 5. 47.6 bar, 30 4°C 6. ... 31 0.8 kPa 30 . 3. 775 m, 91 kPa 31 . 4.42 m 32 . 0.000 136 33 . 151 tonnes/hour 34 . 1 in 1060 35 . (a) 1 × 107 (b) 0.0058 (c) 1.95 kPa 36 . 16.8 m 37 . 200...
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