technical english 1 for engineering students

"Technical English 1 for Engineering Students" là cuốn sách giáo khoa dành cho sinh viên kỹ thuật, cung cấp nền tảng kỹ thuật tiếng Anh vững chắc để giao tiếp hiệu quả trong bối cảnh kỹ thuật. Cuốn sách bao gồm các chủ đề về: * Thuật ngữ kỹ thuật cơ bản và nâng cao * Cấu trúc và ngữ pháp của ngôn ngữ kỹ thuật * Báo cáo kỹ thuật và thuyết trình * Các bài tập đàm thoại và hoạt động thực hành để cải thiện kỹ năng giao tiếp

Ngày tải lên: 28/10/2013, 11:15

... therefore... (1. 10 ) But, from the definition of specific heat at constant volume, Eqn (1. 7), Eqn (1. 10 ) becomes Again from the definition, Eqn (1. 8), Eqn (1. 10 ) gives c ~ = CVT T +p/p (1. 11 ) ... constant (1. 18 ) Further, if the flow is unidirectional and cvT = E, Eqn (1. 18 ) becomes, on cancelling dt, d E +pd(i) =0 (1. 19 ) but differentiating Eqn (1. 10 ) gives (1. 20) Combining Eqns (1. 19 ) ... equation of state, which for perfect gases takes the form p/(pT) = constant = R (1. 12 ) Substituting for p/p in Eqn (1. 1 1) yields the relationship c p - CY = R (1. 13 )... fluids gives +

Ngày tải lên: 08/08/2014, 11:21

... 10 3 10 3 10 7 10 9 11 0 11 0 12 2 12 2 12 5 12 9 13 7 14 1 14 2 14 9 14 9 15 2 15 2 15 6 16 0 16 2 16 5 16 9 17 3 17 4 6.9 Local instability 6 .10 Instability of stiffened panels 6 .11 Failure ... (1. 11 ) and (1. 12 ) we see that (1. 15 ) Equations (1. 14 ) and (1. 15 ) give the maximum shear... aJr - 16 0 N/mm 2 = (iii) and 74,; 16 Basic elasticity Fig 1. 11 Solution of Example 1 ... section beams Problems 10 Stress analysis of aircraft components 10 .1 Tapered beams 10 .2 Fuselages Contents vii 17 4 17 5 i77 18 0 18 8 19 7 19 7 209 211 21 1 220 223 225 232 233

Ngày tải lên: 08/08/2014, 11:21

... luận tiếng Anh Phòng C 2008 -2009 Phòng C 2009 – 2 010 Phòng B 2 010 – 2 011 Phòng B 2 012 - 2 013 Phòng B 2 013 -2 014 Phòng B 2 014 -2 015 Tỉnh C 2 015 -2 016 22 ... discussion 11 4 .1 Results 11 4.2 The discussion 19 III.CONCLUSION 19 Limitations of the study 19 Recommendations for further research 19 Conclusion 20 IV REFERENCE V APPENDICES THANHOF HOÁ NĂM 2 019 TABLE ... Speak and the result as follows: Class Number of the students Percentage Excellent 8B 8C 34 35 10 % Good % 11 ,8 8,6 11 29 31, 4 Avera ge 21 14 % 61, 8 40,0 Bad % 5,9 II The meaning of the research

Ngày tải lên: 08/08/2019, 14:11

... luận tiếng Anh Phòng C 2008 -2009 Phòng C 2009 – 2 010 Phòng B 2 010 – 2 011 Phòng B 2 012 - 2 013 Phòng B 2 013 -2 014 Phòng B 2 014 -2 015 Tỉnh C 2 015 -2 016 22 ... discussion 11 4 .1 Results 11 4.2 The discussion 19 III.CONCLUSION 19 Limitations of the study 19 Recommendations for further research 19 Conclusion 20 IV REFERENCE V APPENDICES THANHOF HOÁ NĂM 2 019 TABLE ... Speak and the result as follows: Class Number of the students Percentage Excellent 8B 8C 34 35 10 % Good % 11 ,8 8,6 11 29 31, 4 Avera ge 21 14 % 61, 8 40,0 Bad % 5,9 II The meaning of the research

Ngày tải lên: 21/10/2019, 10:59

... luận tiếng Anh Phòng C 2008 -2009 Phòng C 2009 – 2 010 Phòng B 2 010 – 2 011 Phòng B 2 012 - 2 013 Phòng B 2 013 -2 014 Phòng B 2 014 -2 015 Tỉnh C 2 015 -2 016 22 ... discussion 11 4 .1 Results 11 4.2 The discussion 19 III.CONCLUSION 19 Limitations of the study 19 Recommendations for further research 19 Conclusion 20 IV REFERENCE V APPENDICES THANHOF HOÁ NĂM 2 019 TABLE ... Speak and the result as follows: Class Number of the students Percentage Excellent 8B 8C 34 35 10 % Good % 11 ,8 8,6 11 29 31, 4 Avera ge 21 14 % 61, 8 40,0 Bad % 5,9 II The meaning of the research

Ngày tải lên: 31/10/2019, 09:49

... luận tiếng Anh Phòng C 2008 -2009 Phòng C 2009 – 2 010 Phòng B 2 010 – 2 011 Phòng B 2 012 - 2 013 Phòng B 2 013 -2 014 Phòng B 2 014 -2 015 Tỉnh C 2 015 -2 016 22 download by : skknchat@gmail.com download by ... discussion 11 4 .1 Results 11 4.2 The discussion 19 III.CONCLUSION 19 Limitations of the study 19 Recommendations for further research 19 Conclusion 20 IV REFERENCE V APPENDICES THANHOF HOÁ NĂM 2 019 TABLE ... Speak and the result as follows: Class Number of the students Percentage Excellent 8B 8C 34 35 10 % Good % 11 ,8 8,6 11 29 31, 4 Avera ge 21 14 % 61, 8 40,0 Bad % 5,9 II The meaning of the research

Ngày tải lên: 29/03/2022, 19:31

... 7 ( 1 + -M2 + - 1 ( 1 M 2 ) 2 + (I-M2 ) 3 + 75 75 31 P 2 5 7M2 =1 +-+ 7 10 Then 22 2!... Therefore p - 426 70 o + 1 = 1. 4 21 p 10 1 325 Therefore 1 1 + - M 2 = (1. 4 21 )2' ... therefore be calculated by 1 Po - P = AP = 5 PO4 and therefore 1 po -p = - x 1 .22 6 (26 4)' = 426 70NmP2 2 Now In standard conditionsp = 10 1 325 Nm-' Therefore p - 426 70 o + 1 = 1. ... (degrees) 0 10 20 30 40 50 60 70 80 90 10 0 11 0 12 0 p-po (Nm-') +569 +502 +3 01 -57 -392 -597 -7 21 -726 -707 -660 -626 -588 -569 For values of 0 between 12 0" and 18 0",

Ngày tải lên: 08/08/2014, 11:21

... 2 .10 .3 using the axisymmetric form of the NavierStokes equations given in Ex. 5 for axisymmetric stagnation- point flow and show that the equivalent to Eqn (2 .11 8) is 411 ’ + 24 41 ... Now for. .. xo: when xo = rn/2.rrU (3. 25) The local velocity The local velocity q = d m jy =- w dY rn and $ = -tan-' 27r Therefore u =- rn 27r 1 1/x + (y/x) 2 - u X - Uy 12 1 12 2 ... 10 2 Aerodynamics for Engineering Students 3 Transport equation for contaminant in two-dimensional flow field In many engineering applications one is interested

Ngày tải lên: 08/08/2014, 11:21

... using conformal mapping, AZAA J., 12 , 12 81. NACA Report, No. 411 (19 31) . NACA Report, No. 14 2 (19 22). Aeronautical Research Council, Reports and Memoranda No. 910 (19 24). Two-dimensional ... page 16 1. ' N. Zhukovsky 'On the shape of the lifting surfaces of kites' (in German), Z. Flugtech. Motorluftschiffahrt, 1; 2 81 (19 10) and 3, 81 (19 12). 16 8 Aerodynamics ... ~(0 .16 3 2 x 0.009) 0 ~ T T +2 ~ = 0 .45 6 + 6.28320 a CM,,,= - - ( A I - A z ) = (0 .16 3 - 0.0228) = -0 .11 0 4 4 (4. 91) (4. 92) 19 6 Aerodynamics for Engineering Students

Ngày tải lên: 08/08/2014, 11:21

... coefficients + + + 0.004739 = 0.22079 A1 0.89202 A3 1. 2 51 00 A5 0.66688 A7 0. 011 637 = 0.663 19 A1 f0.98 957 A3 - 1. 3 15 95A5 - 1. 64234 A7 0.0 216 65 = 1. 1 15 73 A1 - 0.679 35 A3 - 0.896 ... Table 5. 1 7~/8 ~ 1 4 3 ~18 71 52 0.382 68 0.707 11 0.923 88 1. ooo 00 0.923 88 0.707 11 -0 .382 68 - 1. ooo 00 0.923 88 -0 .707 11 -0 .38268 1. ooo 00 0.382 68 -0 .707 11 0.923 88 ... and the integral I1 in Eqn (5.23) becomes " 12 To evaluate the integral 12 change variable to x = l/Z so that 1 11 1 1 Finite wing theory 2 31 For large aspect ratios

Ngày tải lên: 08/08/2014, 11:21

... relations: 7 +1 PZ 1 (6. 41) (6.42) Compressible flow 299 Taking y = 1. 4 for air, these equations become: (6.41a) and P2 6-+ 1 P2 - P1 P2 P1 6+- PI (6.42 ~1) Eqns (6.42) ... (6.37) using Eqn (6.36): P1 P1 PI or E- P1 1 =./M: [1- 3 but from continuity u2/u1 = p1/p2, and from the RankineHugoniot relations p2/p1 is a function of (p2/p1). Thus, by substitution: ... directly: =1 p2 xp' = 2YM; - (7- 1) 2YM; - (7 - 1) P1 P2 Y+l Y+l Rearrangement gives for the exit Mach number: (6.49) 302 Aemdynamics for Engineering Students For air

Ngày tải lên: 08/08/2014, 11:21

... + 0 .10 6~ and A/Ae = 1 + 1. 1 78~ where suffix i denotes the value at the entry section Also A, U ,= A U, Au 48 u I e - A - 1 1 .1 78~ + Then -= due dx + -4 8 x 1. 1 78( 1 1 .1 78~ ... 8 x 1. 1 78( 1 1 .1 78~ )-' Finally = 0. 01 78 3.65 + v,= 14 .6 x 0.003 48 x 1. 1 78 x 4 .83 x 0.003 x 0.069 (1 1 .1 78~ )' + 0.0565 m s-' (1 1 1 7 8 ~ ) ~ + + Thus the maximum suction is ... wall. Therefore A" =- 0207 = 0. 011 5m 1. 8 This is the displacement for each wall, so that the total width between side walls at the exit section = 1. 2+2 x 0. 011 5 = 1. 223m.

Ngày tải lên: 08/08/2014, 11:21

... Aero Note 11 0 (N-34736);H.H Pearcey ( 19 6 1) ‘Shock-induced separation and its prevention’, in Boundary Layer & Flow Control, Vol 2 (edited by G.V.Lwhmann), Pergamon, pp 11 7 0 -1 344  514 Aerodynamics ... Hurst (20 01) 'Some aspects of the aerodynamics of Gurney flaps on a double-element wing', Trans of ASME, J of Fluids Engineering, 12 3 ,9 9- 1 04 502 Aerodynamics for Engineering Students ... least, for the aerofoil with the convex distribution. This is a desirable * R.H. Liebeck (19 73) A class of aerofoils designed for high lift in incompressible flow. J. ofdircraft, 10 , 61M17.

Ngày tải lên: 08/08/2014, 11:21

... 0. 013 52 3950 kQ Q(Nm) Pr(kW) 745 19 00 2000 11 13 11 56 31. 67 33.33 10 03 11 15 0.465 0.442 0. 014 36 0. 014 94 4675 5405 930 11 32 210 0 11 89 35.00 12 25 0.4 21 0. 015 38 610 0 13 40 ... 18 00 19 00 2000 210 0 Power (kW) 10 72 11 13 11 56 11 89 The fixed-pitch airscrew of 3.05 m diameter has the following characteristics: J 0.40 0.42 0.44 0.46 0.48 0.50 kT 0 .11 8 0 .11 5 ... 0.42 0.44 0.46 0.48 0.50 kT 0 .11 8 0 .11 5 0 .11 2 0 .10 9 0 .10 6 0 .10 3 kQ 0. 015 7 0. 015 4 0. 015 0 0. 014 5 0. 013 9 0. 013 2 538 Aerodynamics for Engineering Students and is directly coupled to the engine

Ngày tải lên: 08/08/2014, 11:21

... PB,f13 PB,f/3 2pB,f/3 2PB,f 13 d2pB.f 13 0 FD,f -2 J 213 -2 13 -~ 2 1 3 11 3 11 3 21 3 21 3 ~ 21 3 0 0 0 0 0 @ aF&f /apBqf pDsf PD,f pD,f 0 0 0 (N) @ x 10 6 aFDsf laPD.f FaFB.fIapB,f @ x 10 6 ... FaFB.fIapB,f @ x 10 6 F~FDD,fIm,,f 0 0 0 1 1 1 0 0 320 J2 16 0 640J 213 320 13 320 13 48 013 ... remaining requirement of compatibility Thus, from Eqs (3 .11 ) and (i) = - 2 G de x 2C (-$ +$) ... methods... 48 013 16 013 -1 60J2/3 0 C = 12 6 8 0 0 0 320 320 24 0 0 0 0 C = 880 0 4.6 Application to deflection problems 81 linearly elastic and have cross-sectional areas of 1 8 0 0 ~ ' for the

Ngày tải lên: 08/08/2014, 11:21

... (6 .11 7) For horizontal equilibrium (FT - FB) COS 3 - gttd COS' / =0 (6 .11 8) Taking moments about B w - FTd COS @ + $gttd2Cos2a = 0 z (6 .1 19) Solving Eqs (6 .11 7), (6 .11 8) and (6 .11 9) ... (6 .11 9) for q, FT and FB at = td 2w a (1 -$ tan@) sin 2 FT =-[ z+?( W d cos @ F B = -W z - F ( [ d cos @ (6 .12 0) )] )] 1 - -tan@ 1 tan@ (6 .12 1) (6 .12 2) Equation (6 .10 2) becomes (6 .12 3) ... thin plates 16 9 from which 42 EI EI PCR = - = 2.4 71 - 1 712 12 This value of critical load compares with the exact value (see Table 6 .1) of 7r2EI/ 412 = 2.467EI /12 ; the error,

Ngày tải lên: 08/08/2014, 11:21

... 68 71 72 73 73 75 76 78 83 83 83 84 85 86 86 87 87 89 89 91 91 95 95 96 97 10 1 10 4 10 4 10 4 10 5 10 6 10 7 10 9 11 0 11 0 11 2 Basic concepts and definitions 17 If ... (E) Case (F) span (m) 15 3 3 1. 5 1. 5 3 Relative density 0.533 1 3 1 10 10 Temperature (“C) -24.6 +15 +15 +15 +15 +15 Speed (TAS) (ms-’) 10 0 10 0 10 0 75 54 54 Case (A) ... 716 0 = 0 .11 51 kgm-3 P p=-= RT 287.3 x 216 .5 e=($&) 314 = 1. 19 P 1. 44 x kgm-ls-l 17 1 1. 19 = - 10 -5 = Consider a dimension that, on the aircraft, has a length of 10 ...

Ngày tải lên: 13/02/2014, 09:20

... 68 71 72 73 73 75 76 78 83 83 83 84 85 86 86 87 87 89 89 91 91 95 95 96 97 10 1 10 4 10 4 10 4 10 5 10 6 10 7 10 9 11 0 11 0 11 2 2 Aerodynamics for Engineering Students ... (E) Case (F) span (m) 15 3 3 1. 5 1. 5 3 Relative density 0.533 1 3 1 10 10 Temperature (“C) -24.6 +15 +15 +15 +15 +15 Speed (TAS) (ms-’) 10 0 10 0 10 0 75 54 54 Case (A) ... [l/(~ - l)]p/p in Eqn (1. 1 1) readily gives the enthalpy as (1. 14) YP cpT 7-1P (1. 15) x Contents 7.8 7.9 7 .10 7 .11 7 .12 7 .13 7.7.6 Drag coefficient for a flat plate with wholly...

Ngày tải lên: 05/03/2014, 17:20

... were designed for engineering and sciences graduate students writing for publication in English at Hanyang University in Seoul, Korea. However, most of the material is useful for writing in ... other books? 1. It is research-based The book uses insights from research in the fields of Applied Linguistics and English for Specific Purposes and the emerging field of English for Research ... Research Writing: A guide for English learners to publish in international journals Director, English Writing Lab Center for Teaching and Learning and College of Engineering Hanyang University,...

Ngày tải lên: 19/03/2014, 08:20