Mechanical Engineering Systems 2008 Part 4 potx
Mechanical Engineering Systems 2008 Part 4 potx
... h 2 = 947 .4 kJ/kg From the superheat tables at 30 bar, 40 0°C, h 1 = 3231 kJ/kg 3231 – h 2 = 947 .4, h 2 = 3231 – 947 .4 = 2283.6 kJ/kg h 2 = 2283.6 = h f + x.h fg at 0.05 bar, 2283.6 = 138 + x. 242 3, ... u f + xu fg = 561 + 0.763(2 544 – 561) = 20 74 kJ/kg Q = W + (U 2 – U 1 ) = 80.35 + (20 74 – 20 74. 3)0. 346 = 80.25 kJ Example 2.6.17 0.1 m 3 of steam at a pressure of 14 bar is ex...
Ngày tải lên: 11/08/2014, 15:20
Mechanical Engineering Systems 2008 Part 1 pdf
... constant pressure. Q = m.c p .␦T 80 000 = 2 × 900 × ␦T ␦T = 80 000 1800 = 44 .44 °C Final temperature = 10 + 44 .44 = 54. 44 C Latent heat Latent means ‘hidden’, and is used in this connection because, ... 100°C, Q 4 = m × latent heat of vaporization water Q 4 =4 × 2256.7 = 9026.8 kJ Total heat energy = Q 1 + Q 2 + Q 3 + Q 4 = 81.6 + 1 340 + 1680 + 9026.8 = 12 128 .4 kJ. Mechanical...
Ngày tải lên: 11/08/2014, 15:20
Mechanical Engineering Systems 2008 Part 3 pptx
... spring rate = 390 70 × 0.8 = 4. 46 bar = 4. 46 × 10 5 N/m 2 ip = P m A.L.n × number of cylinders = 4. 46 × 10 5 × × 0.15 2 4 × 0.2 × 5.5 2 × 4 = 17 339 W = 17. 34 kW Example 2 .4. 8 During a test, a 2-cylinder, ... – 293 47 3 – 293 = 0.791 W c = c p (T 2 – T 1 ) = 1.005 (47 3 – 293) = 180.9 kJ/kg TЈ 4 T 3 = p 4 p 3 ␥–1 ␥ , TЈ 4 = 953 × 1.01 4. 04 1.33–1 1.33 , TЈ 4...
Ngày tải lên: 11/08/2014, 15:20
Mechanical Engineering Systems 2008 Part 5 pps
... was isentropic, hЈ 2 = 146 9.9 + (5.3 14 – 4. 962) (5.397 – 4. 962) × (1613 – 146 9.9) = 1585.7 kJ/kg 0.85 = hЈ 2 – h 1 h 2 – h 1 = 1585.7 – 144 6.5 h 2 – 144 6.5 , h 2 = 1610.3 kJ/kg Power = m(h 2 – h 1 )= 4. 5 60 (1610.3 ... we want the rate of heat energy transfer, i.e. J/s = W. (5 04 – 2 04) = Q 1 × 1 0.0095 44 Q = (5 04 2 04) × 1 × 1 × 44 0.0095 = 1 389 47 4 W = 1389.5 kW...
Ngày tải lên: 11/08/2014, 15:20
Mechanical Engineering Systems 2008 Part 6 pdf
... × area =(g4/2) × (4 × 5 .4) = 1000 × 9.81 × 2 × 4 × 5 .4 = 42 3.8 kN Therefore F 2 = 2 42 3 800 N We could also express F 2 as (gh/2) × (h × 5 .4) and so 1000 × 9.81 × 5 .4 × h 2 /2 = 2 42 3 800 giving ... 9.81 = 49 .050 kN Hence the resultant tension in the cable with the pillar partly submerged is: 49 .050 – 25.506 = 23. 54 kN Buoyancy This phenomenon of solid objects being a...
Ngày tải lên: 11/08/2014, 15:20
Mechanical Engineering Systems 2008 Part 7 pptx
... pipes 10 3 2 345 68 10 4 10 5 10 6 10 7 22233 344 4555666888 10 8 10 8 2 2 3 3 4 4 5 5 6 6 8 8 – 0.00001 0.00005 Laminar flow Laminar flow Relative roughness 0.0001 0.0002 0.00 04 0.0006 0.0008 0.001 0.002 0.0 04 0.006 0.008 0.01 0.015 0.02 0.03 0. 04 0.05 0.000 ... flow: h f = 4fL d V 2 2 g so the pressure drop is given by P = g · 4fL d · V 2 2 g Using Q = VA = Vd 2 /4, P =...
Ngày tải lên: 11/08/2014, 15:20
Mechanical Engineering Systems 2008 Part 8 docx
... acceleration then become 2 = 1 + ␣t (4. 1.10) = 1 t + ␣t 2 /2 (4. 1.11) = t( 1 + 2 )/2 (4. 1.12) 2 2 = 1 2 + 2␣ (4. 1.13) Velocit y 0 Tim e t u v ( + )/2uv 1 74 Dynamics Adding the two areas ... s 1 Therefore s 1 = 4. 44 m Similarly for 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 = 350 –...
Ngày tải lên: 11/08/2014, 15:20
Mechanical Engineering Systems 2008 Part 10 doc
... diameter? ( 24) A uniform beam of length 5.0 m rests on supports 1.0 m from the left-hand end and 0.5 m from the right-hand end. A Figure 5.1 .43 Problem 14 Figure 5.1 .44 Problem 15 Figure 5.1 .45 Problem ... concrete a modulus of elasticity of 14 GPa. The area of the column that is steel is 4 × 1 4 × 25 2 = 1963 mm 2 . The area of the column that is concrete is 45 0 × 45 0 – 1963...
Ngày tải lên: 11/08/2014, 15:20
Mechanical Engineering Systems 2008 Part 11 pptx
... I xx cm 4 Section modulus Z cm 3 Area of section cm 2 838 × 292 226 850.9 293.8 16.1 26.8 340 000 7990 289.0 1 94 840 .7 292 .4 14. 7 21.7 279 000 6550 247 .0 176 8 34. 9 291.6 14. 0 18.8 246 000 5890 2 24. 0 762 ... 267 197 769.6 268.0 15.6 25 .4 240 000 6230 251.0 147 753.9 265.3 12.9 17.5 169 000 44 80 188.0 686 × 2 54 170 692.9 255.8 14. 5 23.7 170 000 49 10 217.0 140 683.5 253....
Ngày tải lên: 11/08/2014, 15:20
Mechanical Engineering Systems 2008 Part 13 pps
... kJ 36. 41 7.8 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 . 12 892.78 m/s 44 . 6.37 m/s 45 . 0.69 m/s right to left 46 . 4. 42 m/s 47 . –5.125 ... 18% 4. 21% 3 04 Solutions to problems 19. 137.5 mm 20. 165.5 mm 21. 11.7 × 10 6 mm 4 , 2.2 × 10 6 mm 4 22. 158 mm 23. d /4 24. (a) 71 mm, 74. 1 × 10 6 mm 4 , (b)...
Ngày tải lên: 11/08/2014, 15:20