ECSE_462_HOMEWORK_3%204%20and%205%20solutions%20_2_

ECSE_462_HOMEWORK_3% - ECSE 462 HOMEWORKS 3 4 and 5 PROF F.D GALIANA Sample problems for final exam 3 1 − 2 x2 ⎤ ⎥ Let the currents 2 2−

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Unformatted text preview: ECSE 462 HOMEWORKS 3, 4 and 5 November 22, 2010 PROF. F.D. GALIANA Sample problems for final exam ⎡ 3 1 − 2 x2 ⎤ ⎥ . Let the currents 2 2− x ⎦ ⎣1 − 2 x 1) Consider a 2‐winding device with inductance matrix L ( x ) = ⎢ flowing through the windings be equal to 1 and 2 respectively. Find the voltages induced across the windings if x varies with time according to x(t)=0.1 sin(t). Find the electromechanical force acting on the moving part during this time variation in x. If the mass of the device is 1, is the product of the mass times the acceleration of the moving part equal to the electromechanical force? If not, why? Solution: Solve v = dL( x) dx 1 dL( x) d 2x i . If you find m 2 , it will not be equal to f i . Then, find f = iT d dt dx dx dt 2 , which means that there must be a second external force acting on the moving part. ⎡ 1 1 + x − x2 ⎤ ⎥ . Let the 2 2 − 0.2 x ⎦ ⎣1 + x − x 2) Consider a 2‐winding device with inductance matrix L ( x ) = ⎢ currents flowing through the windings be both equal to 1. Find the electromechanical force acting on the moving part during this change in x. If the mass of the device is 1, what is the equilibrium point? Suppose that the device is also subject to an external force of the form f 0 + f1 x . How would you control the equilibrium point with the parameters f0 and f1 ? For what values of these parameters would the small signal model be stable? Solution: The electromechanical force is given by f ( x) = 1 T dL( x) id i = 2(1 − 2 x) − 0.2 . The dx 2 equilibrium point is given by f ( x) + f 0 + f1 x = 0 , which can be controlled by changing the values of the external force parameters. The governing dynamic equations are dv = f ( x) + f 0 + f1 x dt dx =v dt m Whose Jacobian matrix is, 1⎤ ⎡0 J =⎢ ⎥ ⎣ −4 + f1 0 ⎦ 2 The eigenvalues of this are given by, λ − f1 + 4 = 0 , which will be stable only if f1 − 4 < 0 . 3) The rotating part of an electromechanical device moves at a constant rate of 1 radian per second over a given angle. What is the net torque produced by the device? What is the net mechanical power produced by the device? What is the mechanical energy produced by the device over the change in angle? If the corresponding change in energy stored in the magnetic field of the device is 10 MJ. Find the electrical energy consumed by the device during the change. Solution: Since the device moves at a constant rate its acceleration is zero and therefore its net torque, T, is also zero. Thus, Pmech = ωmT = 0 . If the mechanical power is zero so is the mechanical energy. The electrical energy must therefore be equal to the negative of the change in energy stored in the magnetic field, which is ‐10 MJ. 4) A round rotor machine has a distributed winding in the rotor with a turns density denoted by dN (θ ) = 100sin(2θ ) turns per radian. The angle θ is measured with respect to a reference frame fixed dθ to the rotor. (a) How many poles does this machine have? Ans. 4, the number of ups and downs of the turns density from 0 to 360 degrees. (b) Find the number of turns per pole? Integrate the turns density from θ = 0 to θ = 45 degrees. (c) At what values of θ are the axes of the winding poles? Ans: At θ = 0 and θ = 180 degrees. (d) If the winding is carrying a current i , find the magnetic field in the air gap as a function of θ , H (θ ) , using Ampere’s law. Show all steps, including the location of the contour C. Solution: Apply Ampere’s law for a contour going across the gap centered at θ = 45 degrees. The limits of integration are then θ to 90 degrees minus θ . We then get 2 H (θ ) g = number of turns in that angle interval times the current. (e) If the air gap is 4 cm, what should the winding current be in order for the field density distribution in the air gap to have a maximum value of 1.5 T. At what values of θ do these maxima occur? Solution: B(θ ) = μ0 H (θ ) =1.5 then solve for i. (f) This 2‐pole distribution fixed to the rotating rotor turns with respect to a 2 pole, 100 turn concentrated winding in the stator (50 turns per pole pair) so that the two windings’ axes are aligned at t=0. The rotor turns at 1800 rpm. Find the flux linking each stator pole pair. The two should be equal. Find the voltages induced across each stator pole pair and compare the frequency of the induced voltage with the speed of rotation of the rotor. Solution: The axis of the stator winding must be at θ = 0 spanning the range of θ = −45 degrees to θ = 45 degrees. To find the flux linking the stator winding, therefore integrate the flux density B (θ ) from θ = −45 degrees to θ = 45 degrees and multiply the result by 2 (since exactly the same flux links the other pole). To find the voltage across the winding take the derivative wrt time. The frequency of the voltage is twice the speed of rotation thus 3600 rpm or 60 Hz. 5) Consider now a distributed winding on the rotor of a round machine with a turns density dN (θ r ) = 100sin(θ r ) carrying a constant current of ir. The machine has a radius of R, an air gap of g, and dθ r a length of L. The angle θ r is measured with respect to a reference frame fixed to the rotor. The rotor turns at a speed ωm = 120π rad / s . The stator of the same machine has a winding with 100 turns, half concentrated at θ = 0 and the other half at θ = π radians. The angle θ is measured with respect to a 2 reference frame fixed to the stator. Find an expression for the voltage induced across the stator winding. Solution: We repeat the same process as in the previous question to find the field distribution fixed to the rotor which will be of the form Br (θ r ) = K 0ir sin(θ r ) , where K0 depends on the air gap, g. The stator winding is placed in an asymmetric form, with half the turns at 0 degrees (the turns going from 0 to 180 degrees) and the other half at 90 degrees, (the turns going from 90 to 270 degrees). Since θ r = θ − ωmt , the field distribution can be described by Br (θ ) = K 0ir sin(θ − ωm t ) . The flux linking the winding with 50 θ =π turns at θ = 0 is θ =3π /2 degrees is ∫ θ =π /2 ∫ θ =0 K 0ir sin(θ − ωmt ) RLdθ while the flux linking the winding with 50 turns at θ = 90 K 0ir sin(θ − ωmt ) RLdθ . The voltage induced across the entire winding is then, ⎧θ =π ⎫ ⎧θ =3π / 2 ⎫ d ⎨ ∫ K 0ir sin(θ − ωmt ) RLdθ ⎬ d ⎨ ∫ K 0ir sin(θ − ωmt ) RLdθ ⎬ ⎭ ⎭ 50 ⎩ θ =π / 2 50 ⎩θ =0 + = dt dt 100 K 0 RLir {cos ωmt + sin ωm t} 6) Consider a 60 Hz, 6‐pole induction motor whose torque versus slip characteristics are given by T ( s ) = 10 s . The motor is operating in steady‐state. The load torque is given by Tload ( s ) = 3 − 2 s . Find the speed of the motor and the frequency of the rotor currents. Find the rotor winding resistance as a fraction of the rotor winding resistance when the load torque is zero. Find the mechanical power produced by the motor as well as the mechanical energy required over the period of one hour. All units are in SI. Solution: In steady state the two torques are equal, 10 s = 3 − 2 s , so that s = 0.25 . Thus ωm = (1 − s ) ωsynch = 0.75 x 1200 = 800 rpm. The frequency of the rotor currents is ωr = sωs = 0.25x60=15 Hz. The rotor resistance is R2 for any s, however when the load torque is zero we have s=1, s R2 /R = s = 0.25. The mechanical power produced by the motor is s2 120π 0.25 ( 0.75 ) = 7.5π . The mechanical energy is then Pmech = ωmT = (1 − s ) ωsynch10 s = 3 7.5π ( 3600 ) joules. thus 7) Consider an induction motor. The electrical circuit as seen by the rotor is a Thevenin equivalent with voltage source E = 1∠0 and impedance 1+j2. This equivalent circuit is connected in parallel with the rotor, which is represented by two resistances in series, one is the actual rotor resistance, R2 , and the ⎛1− s ⎞ other, ⎜ ⎟ R2 , is the slip‐dependent fictitious resistance whose power consumption represents the ⎝s⎠ mechanical power transferred to the motor shaft. Find the mechanical power produced by the motor in terms of slip and R2. 1 ⎛ 1− s ⎞ 2 . ⎟ R2 , where I = 2 ⎝s⎠ ⎛ R2 ⎞ ⎜1 + ⎟ + 4 s⎠ ⎝ Solution: P = I ⎜ 2 Find the torque. Solution:Since P = (1 − s ) ωsynchT = 1 2 ⎛ R2 ⎞ ⎜1 + ⎟ + 4 s⎠ ⎝ ⎛ 1− s ⎞ ⎜ ⎟ R2 , it follows that, ⎝s⎠ ⎛ ⎞ ⎜ ⎟ 1 ⎟ R2 = I 2 R2 ; in other words, the torque times ω is the power T =⎜ synch 2 ⎜ ⎛ R2 ⎞ ⎟ sωsynch sωsynch ⎜ ⎜1 + ⎟ + 4 ⎟ s⎠ ⎝⎝ ⎠ R2 . dissipated in the fictitious resistor s Use the maximum power transfer theorem to find the maximum torque. Solution: For maximum torque, solve, ⎛ 1 T =⎜ ⎜ ⎜ 1+ 5 ⎝ ( ) R2 = 1 + 2 j = 5 . The maximum torque is then s ⎞ ⎟5 2 ⎟ + 4 ⎟ ωsynch ⎠ If we can control R2 , find an expression for the sensitivity of the maximum motor torque with respect to this resistance (sensitivity is a partial derivative). Solution: ∂T = 0 ; the maximum torque is insensitive to R2. ∂R2 What is the stator current when the motor turns at synchronous speed? Find an expression for the torque at start up. 2 Solution: At start up s=1. This means that I = 1 (1 + R2 ) 2 +4 . The torque at start up is, ⎛ ⎞R 1 T =⎜ ⎟2 2 ⎜ (1 + R ) + 4 ⎟ ωsynch . Since at start up 2 ⎝ ⎠ 8) Consider a DC motor with field winding resistance and current, Rf and If, armature winding resistance and current, Ra and Ia, and terminal voltage, Vt. The internal armature voltage source is proportional to the speed and to the field current. At a speed of 200 rpm and a field current of 10 A, the internal armature voltage is equal to 100 volts. If the field and armature winding resistances are respectively 0.2 and 0.02 ohms find the torque versus speed characteristics when the applied terminal voltage is 100 volts and the DC machine is operated in a self‐excited parallel mode. Solution: E = K ωm I f , that is, E = K ωm I f = 100 = K (200)(10) , so that K=0.05 or E = 0.05 I f ωm . Then, the power consumed by the source is P = ωmT = EI a = E But since I f = (V − E ) = 0.5I ω 100 − 0.5I f ωm . fm Ra 0.02 V 100 ⎛ 100 − 250ωm ⎞ = = 500 , then, T = 250 ⎜ ⎟ . 0.02 R f 0.2 ⎝ ⎠ Find an expression for the losses in the armature and rotor versus speed. Find the motor efficiency in terms of the ratio of power out over power in versus speed. Solution: The field winding loss is I 2 R f = 50, 000 . The armature winding loss is f ⎛ 100 − 250ωm ⎞ ⎛ 100 − 250ωm I Ra = ⎜ ⎟ 0.02 . The power into the motor is 250ωm ⎜ 0.02 0.02 ⎝ ⎝ ⎠ 2 2 a ⎞ ⎟ ⎠ 9) Consider a 60 Hz, 6 pole synchronous motor whose circuit equivalent is represented by a Thevenin equivalent. In per unit, the internal emf of the machine is 1, while the internal reactance is 0.2. If the terminal voltage magnitude is 1, what is the maximum power that this motor can produce? Solution: Since P = EV sin δ = 5sin δ , the maximum power is 5 pu. X What is the maximum torque? Solution: Since a SM turns at only one speed, the maximum torque is 5 / ωsynch = 5 / 120π / 3 pu. At what speed does this motor operate if it is operated at half the maximum torque? Answer: Still at synchronous speed. If we operate at half the maximum torque, how much reactive power does the motor consume? Solution: Half torque means half power which means an angle of 30 degrees. The reactive power ( ) 2 consumed by the motor is Q = 5 V − EV cos δ =5(1‐0.866)=0.67 pu. If the internal voltage of the motor E can be controlled through the field current and is no longer fixed at 1 pu, show that the motor can then be used as a source or a sink of reactive power by adjusting E. For this part, assume that the motor consumes zero real power (this device is called a synchronous condenser and is used to regulate voltage through its reactive power production). Solution: Since P=0, the phase angle is zero, thus, the reactive power consumed by the motor is, Q = 5 (V 2 − EV ) . Thus, if E<V, the motor consumes vars. If E>V the motor generates vars. ...
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This note was uploaded on 09/16/2011 for the course ECSE 361 taught by Professor Franciscodgaliana during the Winter '09 term at McGill.

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