... học - Giới thiệu tổng quan môn học - Cơ sở lý luận xây dựng mô hình kinh tế lượng phântích hồi quy, thể mối quan hệ biếnđịnh lượng mở rộng biếnđịnh tính - Ước lượng các hệ số đại lượng mô ... Chương 1: NHỮNG VẤN ĐỀ CƠ BẢN CỦA MÔ HÌNH HỒI QUI ĐƠN Phântích hồi qui 1.1 Bản chất phântích hồi qui 1.2 Phântích hồi qui mối quan hệ khác Số liệu phântích hồi qui 2.1 Các loại số liệu 2.2 Nguồn ... qui với biến độc lập biến giả 1.3 Hồi qui với nhiều biến giả Hồi qui với biến lượng biến chất 2.1 Biến chất có hai phạm trù 2.2 Biến chất có nhiều hai phạm trù Hồi qui với nhiều biến lượng biến...
... Để tiếtkiệm tài ngun hoi nhân lực tài có cơng tác thẩm định dự án, áp dụng loạt giai đo n khác qui trình thẩm định Mỗi giai đo n thực với sở liệu mang độ xác cao Vào cuối giai đo n, tới định ... phântích trải khía cạnh quan trọng dự án Giai đo n xác định, thẩm định thiết kế dự án bao hàm loạt khâu thẩm định điểm định, đưa tới kết chấp thuận hay bác bỏ dự án Tồn qui trìnhphân chia cách ... đơi với việc tính tốn lợi ích mức q cao Thơng tin tốt khơng thiên lệch cóquan thẩm định dự án chịu tốn thời gian tiền Những chi phí bù đắp lại nguồn lực tiếtkiệmcó thơng tin tốt nhờ tránh việc...
... claim or cause arises in contract, tort or otherwise DOI: 10.1 036/ 0071417990 For more information about this title, click here Contents Preface v Staff vi How to Use the ... Electrical and magnetic units 635 Dimensional formulas of common quantities .635 Internal energy and generalized work 636 General rules of integration 637 Schematic electronic ... passes in quantities controlled by the choke plate and the throttle plate { er horn } ˙ air horsepower [MECH ENG] The theoretical (minimum) power required to deliver the specified quantity of...
... illustration, information can be either quantitative or nonquantitative Quantitative information is information that is expressed in numbers Examples of nonquantitative information are visual impressions, ... programs, and newspaper stories Accounting is primarily concerned with quantitative information Accounting is one of several types of quantitative information Accounting information is distinguished ... Purpose of Accounting ILLUSTRATION Information 1–1 Types of information Consists of Nonquantitative information Quantitative information Consists of Accounting information Nonaccounting information...
... transformer is IS = 20 kVA ∠ − 36. 87° A = 41.67∠ − 36. 87° A 480 V The secondary current referred to the primary side is IS′ = j16.7 Ω I S 41.67∠ − 36. 87° A = = 2.50∠ − 36. 87° A 16.67 a 25 Therefore, ... = ∠ − 36. 87° A = 5∠ -36. 87° A VS = The secondary voltage referred to the primary side is V ′ = aV = 100∠0° V S S The secondary current referred to the primary side is I I S ′ = S = 10∠ − 36. 87° ... The primary circuit voltage is given by ( VP = VS ′ + I S ′ Req + jX eq ) VP = 100∠0° V + (10∠ − 36. 87° A )( 0.20 Ω + j 0.750 Ω ) = 106.2 ∠2.6° V The excitation current of this transformer is 106.2∠2.6°...
... kV )(0.8) I S = 362 .3∠ − 36. 87° A The voltage on the primary side of the transformer (referred to the secondary side) is VP ′ = VS + I S Z EQ VP ′ = 13,800∠0° V + ( 362 .3∠ − 36. 87° A )( 0.38 ... becomes Iline ′ = Iload = 13.2∠0° kV = 26.37∠ − 36. 89° A 0.60∠53.1° Ω + 500 36. 87° Ω The load voltage is Vload = Iload Z load = ( 26.37∠ − 36. 89° A )(500 36. 87° Ω ) = 13.185∠ − 0.02° kV The ratio of ... PF (15 kV )( 0.8) I S ′ = 16,667∠ − 36. 87° A The voltage on the primary side of the transformer is VP = VS ′ + I S ′ Z EQ,P VP = 15,000∠0° V + (16,667∠ − 36. 87° A )( 0.0135 + j0.0563 Ω ) = 15,755∠2.24°...
... 0.170 + (1.513 + j1.134 )( − j3 .36) (1.513 + j1.134 ) + ( − j3 .36) Z EQ = 0.010 + j0.040 + 0.00788 + j 0.0525 + 0.040 + j0.170 + (2.358 + j0.109) Z EQ = 2.415 + j0 .367 = 2.443∠8.65° The resulting ... impedance of Load is Z 0.45 36. 87° Ω Z load1,pu = load1 = = 1.513 + j1.134 Z base3 0.238 Ω The per-unit impedance of Load is Z − j 0.8 Ω Z load2,pu = load2 = = − j3 .36 Z base3 0.238 Ω The resulting ... regions by the two transformers If the per-unit base quantities in Region are chosen to be S base1 = 1000 kVA and VL ,base1 = 480 V, then the base quantities in Regions and will be as shown above...
... deg2rad = pi / 180; % Remove phase ambiguities: = 360 ang = ang - 360 ; end while ang < ang = ang + 360 ; end % Simulate the output of the phase angle controller ... Since the frequency of the waveform is 60 Hz, the waveform there are 360 ° in 1/60 s, and the firing angle α is α = ( 4.8 ms ) 360 ° = 103.7° or 1.810 radians 1/ 60 s Note: This problem could also ... (ang >= fire & ang = (fire+180) & ang
... 36. 87° VL ( 2300 V ) The phase voltage of this machine is Vφ = VT / = 1328 V The internal generated voltage of the machine is E A = Vφ + RAI A + jX S I A E A = 1328∠0° + ( 0.15 Ω )( 251∠ − 36. 87° ... is a two-pole 60 Hz machine, to must be turning at 360 0 r/min The required torque is τ APP = (e) PIN = ωm 175.2 kW = 465 N ⋅ m 2π rad (360 0 r/min ) 60 s r The rotor current ... = 60 Hz f e = 400 Hz 3000 r/min 1500 r/min 1000 r/min 750 r/min 600 r/min 500 r/min 428.6 r/min 360 0 r/min 1800 r/min 1200 r/min 900 r/min 720 r/min 600 r/min 514.3 r/min 24000 r/min 12000 r/min...
... rated conditions is E A = Vφ + RAI A + jX S I A E A = 7967∠0° + (1.5 Ω )( 418∠ − 36. 87° A ) + j (12.0 Ω )( 418∠ − 36. 87° A ) E A = 12,040∠17.6° V 120 The magnitude of E A is 12,040 V (b) The torque ... 100-kW, two-pole, three-phase, 60-Hz synchronous generator’s prime mover has a no-load speed of 363 0 r/min and a full-load speed of 3570 r/min It is operating in parallel with a 480-V, 75-kW, ... voltage? SOLUTION The no-load frequency of generator corresponds to a frequency of f nl1 = nm P (363 0 r/min )( ) = = 60.5 Hz 120 120 The full-load frequency of generator corresponds to a frequency...
... − cos−1 ( 0.80) = 36. 87° , so I A = 259 ∠ − 36. 87° A The internal generated voltage E A is E A = Vφ − jX S I A E A = ( 480∠0° V ) − j (1.1 Ω )( 259∠ − 36. 87° A ) = 384 ∠ − 36. 4° V (b) This motor ... so I A = 4619∠ − 36. 87° A Therefore, the internal generated voltage is E A = Vφ + RAI A + jX S I A 139 E A = 7217∠0° + (0.0187 Ω )( 4619∠ − 36. 87° A ) + j (1.716 Ω )( 4619∠ − 36. 87° A ) E A = ... so I A = 5247∠ − 36. 87° A The phase voltage is 11.0 kV / V Therefore, the internal generated voltage is = 6351 E A = Vφ + RAI A + jX S I A E A = 6351∠0° + j ( 0.75 Ω )(5247 ∠ − 36. 87° A ) E A...
... 266∠0° V − j ( 0.53 Ω )( 251 36. 87° A ) = 362 ∠ − 17.1° V (d) The maximum power that the motor could supply at these conditions is PMAX = 6-15 3Vφ E A XS = ( 266 V )( 362 V ) = 545 kW 0.53 Ω A 100-hp ... 79.4 A ) sin 36. 87° = 62.9 kVAR (d) The internal generated voltage at rated conditions is E A = Vφ − RAI A − jX S I A E A = 440∠0° V − ( 0.22 Ω )( 79.4 36. 87° A ) − j (3.0 Ω )( 79.4 36. 87° A ) ... by the motor is constant, the quantity I A cos θ , which is directly proportional to power, must be constant Therefore, I A2 (0.8) = ( 40 A )(1.00) 159 I A2 = 50 36. 87° A The internal generated...
... 2200 W = 36. 82° ( 24.6 V )(64.5 A ) Therefore, RLR = Z LR cos θ LR = ( 0.2202 Ω ) cos ( 36. 82° ) = 0.176 Ω ′ ⇒ ⇒ R1 + R2 = 0.176 Ω R2 = 0.071 Ω X LR = Z LR sinθ LR = ( 0.2202 Ω ) sin ( 36. 82°) ... 0.204 Ω ) 2 = 0.159 The synchronous speed of this motor is 120 f e 120 ( 60 Hz ) = = 360 0 r/min P 2π rad = ( 360 0 r/min ) = 377 rad/s 1r 60 s nsync = ω sync Therefore the pullout torque of the ... The equivalent impedance of the rotor circuit in parallel with jX M is: 1 ZF = = = 1.539 + j 0 .364 = 1.58∠13.2° Ω 1 1 + + jX M Z j 7.2 Ω 1.625 + j0.17 The phase voltage is 440/ = 254 V, so line...
... VT I L ,rated = ( 460 V )( 274 A ) = 218 kVA and the kVA per horsepower is kVA/hp = 218 kVA = 4 .36 kVA/hp 50 hp This motor would have starting code letter D, since letter D covers the range 4.00-4.50 ... 8-46: τ ind = ZrlBI A (12 cond )( 0.08 m )( 0.3 m )(1.0 T )( 2.54 A ) = a current paths τ ind = 0 .366 N ⋅ m, CW (opposite to the direction of rotation) 8-3 Prove that the equation for the induced ... paths (b) Duplex wave-wound: a = 2m = (2)(2) = paths (c) Triplex lap-wound: a = mP = (3)(12) = 36 paths (d) Quadruplex wave-wound: a = 2m = (2)(4) = paths 8-6 The power converted from one form...
... motor at these conditions is n= EA 236. 2 V no = (900 r/min ) = 1586 r/min E Ao 134 V The power converted from electrical to mechanical form is Pconv = E A I A = ( 236. 2 V )( 25.3 A ) = 5976 W Since ... 101.3 A If I A = 23.3 A, then E A = VT − I A ( RA + RS ) = 240 V − ( 25.3 A )(0.09 Ω + 0.06 Ω ) = 236. 2 V The magnetomotive force is F = NI A = ( 33 turns)( 25.3 A ) = 835 A ⋅ turns , which produces ... % First, initialize the values needed in this program v_t = 120; % Terminal voltage (V) r_a = 0 .36; % Armature + field resistance (ohms) i_a = 9:1:58; % Armature (line) currents (A) % Calculate...
... Plot the torque-speed characteristic for this motor (c) What is its speed regulation? SOLUTION 236 (a) At full load, I A = I L − I F = 100 A − 0.75 A = 99.25 A , and E A = VT − I A ( RA + RS )...
... 480 V ) I A = 60.1∠ − 36. 87° A Therefore, the internal generated voltage E A of the ac machine is E A = Vφ + jX S I A E A = 277∠0° V + j ( 2.0 Ω )( 60.1∠ − 36. 87° A ) = 362 ∠15.4° V (b) When the ... terminal voltage and power of the ac machine are constant, the quantity E A sinδ must be constant 267 E A sin δ = E A sin δ ′ ′ δ ′ = sin −1 EA 362 V sin δ = sin −1 sin15.4° = 14.7° EA 380 V ′ Therefore, ... is Pconv,F = (1 − s ) PAG,F = (1 − 0.05)(386 W ) = 367 W Pconv,B = (1 − s ) PAG,B = (1 − 0.05)(16.2 W ) = 15.4 W Pconv = Pconv,F − Pconv,B = 367 W − 15.4 W = 352 W (d) The output power is POUT...