Giorgio Rizzoni Solutions Manual 6
Electrical engineering by giorgio rizzoni solution manual solutions manual to. Solutions manual for electrical engineering principles applications 6th edition.
Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 17 Chapter 17 Instructor Notes The objective of Chapter 17 is to introduce the foundations for the analysis of rotating electric machines. In Section 17.1, rotating electric machines are classified on the basis of their energy conversion characteristics and of the nature of the electric power they absorb (or generate).
Giorgio Rizzoni Solutions Manual 60
Section 17.2 reviews the physical structure of a DC machine and presents a simple general circuit model that is valid for both motors and generators, including dynamic equations. Section 17.3 contains a brief discussion of DC generators. Section 17.4 describes the characteristics of the various configurations of DC motors, both of the wound stator and permanent magnet types.
The torque speed characteristics of the different configurations are compared, and the dynamic equations are given for each type of motor. The section ends with a brief qualitative discussion of speed control in DC motors. The second half of the chapter is devoted to the analysis of AC machines. In Section 17.5, we introduce the concept of a rotating magnetic field.
The next two sections describe synchronous generators and motors; the discussion is brief, but includes the analysis of circuit models of synchronous machines and a few examples. Circuit models for the induction motor, as well as general performance characteristics of this class of machines are discussed in Section 17.8, including a brief treatment of AC machine speed and torque control. Although the discussion of the AC machines is not particularly detailed, all of the important concepts that a non-electrical engineer would be interested in to evaluate the performance characteristics of these machines are introduced in the chapter, and reinforced in the homework problem set. The homework problems include a mix of traditional electric machinery problems based on circuit models and of more system-oriented problems. Problems 17.24-36 deal with the performance and dynamics of systems including DC motors. These problems are derived from the author’s experience in teaching a Mechanical Engineering System Dynamics course with emphasis on electromechanics, and are somewhat unusual (although relevant and useful for non–electrical engineers) in this type of textbook.
These problems are well suited to a more mature audience that has already been exposed to a first course in system dynamics. Problem 17.39 provides a link to the power electronics topics covered in Chapter 12. All other problems are based on the content of the chapter. Learning Objectives 1. Understand the basic principles of operation of rotating electric machines, their classification, and basic efficiency and performance characteristics.
Understand the operation and basic configurations of separately-excited, permanentmagnet, shunt and series DC machines. Analyze DC generators at steady-state. Analyze DC motors under steady-state and dynamic operation. Understand the operation and basic configuration of AC machines, including the synchronous motor and generator, and the induction machine. Sections 5, 6, 7, 8.
Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 17 Section 17.1: Rotating Electric Machines Problem 17.1 Solution: Known quantities: The relationship of the power rating and the ambient temperature is shown in the table. A motor with Pe = 10 kW is rated up to 85, C.
Find: The actual power for the following conditions. 50, C., b) Ambient temperature is 25 C. A) Ambient temperature is Assumptions: None. Analysis: a) The power at ambient temperature 50, C: Pe' = 10 − 10 × 0.125 = 8.75 kW b) The power at ambient temperature 30, C: Pe' = 10 + 10 × 0.08 = 10.8 kW Problem 17.2 Solution: Known quantities: The speed-torque characteristic of an induction motor is shown in the table. The load requires a starting torque of 4 N ⋅ m and increase linearly with speed to 8 N ⋅ m at 1500 rev min.
Find: a) The steady state operating point of the motor. B) The change in voltage if the load torque increases to Assumptions: None. Analysis: The characteristic is shown below: 17.2 10 N ⋅ m. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 17 Section 17.2: Direct Current Machines Problem 17.3 Solution: Known quantities: Each conductor of the DC motor is −4 6 in. The current is 90 A. The field density is 5.2 × 10 Wb in.
2 Find: The force exerted by each conductor on the armature. Assumptions: None. Analysis: Wb in 2 0.0254 m F = BI × l = 5.2 × 10 × × 90 × 6 in × 2 2 in in (0.0254 m) = 11.06 Nt −4 Problem 17.4 Solution: Known quantities: The air-gap flux density of the DC machine is 4 Wb m 2. The area of the pole face is 2 cm × 4 cm. Find: The flux per pole in the machine. Assumptions: None.
Analysis: With B = 4 kG = 0.4 T = 0.4 Wb m 2, we can compute the flux to be: φ = BA = 0.4 × 0.02 × 0.04 = 0.32 mWb 17.4 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 17 Section 17.3: Direct Current Generators Problem 17.5 Solution: Known quantities: A 120V, 10 A shunt motor. The armature resistance is 0.6 Ω. The shunt field current is 2 A. Find: The LVDT equations. Assumptions: None.
Analysis: VL at full load is 120 V and E b = 120 + ( 2 + 10) × 0.6 = 127.2 V Rf = 120 = 60 Ω 2 Eb to be constant, we have: 127.2 = 2.1 A ia = i f = 0.6 + 60 Assuming Therefore: VL = 127.2 − 2.1 × 0.6 = 125.9 V Voltage reg. = 125.9 − 120 = 0.049 = 4.9% 120 Problem 17.6 Solution: Known quantities: A 20 kW, 230V separately excited generator. The armature resistance is 0.2 Ω. The load current is 100 A. Find: a) The generated voltage when the terminal voltage is 230 V. B) The output power. Assumptions: None.
Analysis: If we assume rated output voltage, that is VL = 230V, we have a) The generated voltage is 230 V. B) The output power is 23 kW. If we assume rated output power, that is Pout = 20 kW, we have 17.5 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 17 Problem 17.8 Solution: Known quantities: A 30 kW, 440V shunt generator. The armature resistance is 0.1 Ω and a series field resistance is 200 Ω. Find: a) The power developed at rated load.
B) The load, field, and armature currents. C) The electrical power loss.
Assumptions: None. Analysis: The circuit is shown below: iL if Rf + Ra ia + - Eb Lf RL vL - 30 × 10 3 iL = = 68.2 A 440 440 if = = 2.2 A 200 ia = 70.4 A a) Eb = VL + ia Ra = 440 + 70.4 × 0.1 = 447.04 V P = Ebia = 31.471 kW b) iL = 62.8 A i f = 2.2 A ia = 70.4 A c) Ploss = ia2 Ra + i 2f R f = 1464 W 17.7 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 17 Problem 17.9 Solution: Known quantities: A four-pole 450 kW, 4.6 kV shunt generator. The armature resistance is 2 Ω and a series field resistance is 333 Ω. The generator is operating at the rated speed of 3600 rev min. Find: The no-load voltage of the generator and terminal voltage at half load. Assumptions: None.
Analysis: For n = 3600 rev min, ω m = 377 rad sec: 450 × 10 3 = 97.8 A 4.6 × 10 3 4.6 × 10 3 if = = 13.8 A 333 ia = i f + i L = 111.6 A iL = Using the relation: Eb = VL + ia Ra = 4823.2V At no-load, VL = Eb − ia Ra = 4820.4V At half load, iL = 48.9 A ia = i f + i L = 62.7 A VL = Eb − ia Ra = 4810.7 V Problem 17.10 Solution: Known quantities: A 30 kW, 240V generator is running at half load at 1800 rev Find: The total losses and input power. Assumptions: None. Analysis: 1 rated load = 15 kW 2 At an efficiency of 0.85, the input power can be computed to be: 15 × 10 3 Pin = = 17.647 kW 0.85 Pout = The total loss is: 17.8 min with efficiency of 85 percent. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 17 Section 17.4: Direct Current Motors Problem 17.12 Solution: Known quantities: A 220 V shunt motor. The armature resistance is 0.32 Ω. A field resistance is 110 Ω. At no load the armature current is 6 A and the speed is 1800 rpm.
Find: a) The speed of the motor when the line current is 62 A. B) The speed regulation of the motor.
Assumptions: The flux does not vary with load. Assume a 8 N ⋅ m brush drop.
Analysis: a) 1800 = 220 − 2 − 6(0.32) K aφ K aφ = 0.12 ∴n = 220 − 2 − 6(0.32) = 1657 rpm K aφ b)% reg = 1800 − 1657 × 100 = 8.65% 1657 Problem 17.13 Solution: Known quantities: A 50 hp, 550 volt shunt generator. The armature resistance including brushes is 0.36 Ω. Operating at rated load and speed, the armature current is 75 A. Find: What resistance should be inserted in the armature circuit to get a 20 percent speed reduction when the motor is developing 70 percent of rated torque. Assumptions: There is no flux change. Analysis: T = K aφI a I a = 0.7(75) = 52.5 A nR = 550 − 75(0.36) K aφ nR = 523 K aφ 0.8nR = 550 − 52.5 RT 0.8 × 523 = 550 − 52.5RT K aφ ∴RT = 2.51Ω Radd = 2.51 − 0.36 = 2.15 Ω 17.10 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 17 Problem 17.14 Solution: Known quantities: 100 kW, 440V shunt DC motor.
The armature resistance is 0.2 Ω and a series field resistance is 400 Ω. The generator is operating at the rated speed of 1200 rev min. The full-load efficiency is 90 A percent. Find: a) The motor line current. B) The field and armature currents. C) The counter emf at rated speed. D) The output torque.
Assumptions: None. Analysis: At n = 1200 rev min, ω m = 125.7 rad sec, the output power is 100 hp = 74.6 kW. 0.9, we have: 74.6 Pin = = 82.9 kW 0.9 From full-load efficiency of a) From Pin = iSVS = 82.9 kW, we have: 82.9 × 103 iS = = 188.4 A 440 b) 440 = 1.1 A 400 ia = 187.3 A if = c) Eb = VL − ia Ra = 402.5 A d) Tout = Pout = 593.5 N ⋅ m ωm Problem 17.15 Solution: Known quantities: A 240V series motor.
The armature resistance is 0.42 Ω and a series field resistance is 0.18 Ω. The speed is 500 rev min when the current is 36 A.
Find: What is the motor speed when the load reduces the line current to 17.11 21 A. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 17 Analysis: 2π ⋅ n ⋅ T 33,000 2π (820)T 75 = T = 480.4 lb ⋅ ft 33,000 T = K aφI a 480.4 = K aφ (112) K aφ = 4.29 HP = Tn = 4.29(0.85)(84) = 306.2 lb ⋅ ft 550 − 84(0.15) = 973 rpm 0.85(0.65) 2π (973)(306.2) HPn = = 56.7 hp 33,000 nn = Problem 17.18 Solution: Known quantities: 220 VDC shunt motor. The armature resistance is 0.1 Ω and a series field resistance is 100 Ω. The speed is 1100 rev min when the current is 4 A and there is no load. A Find: E and the rotational losses at 1100 rev min. Assumptions: The stray-load losses can be neglected. Analysis: Since n = 1100 rev min corresponds to ω = 115.2 rad sec, we have: iS = 4 A 200 =2A 100 ia = i S − i f = 2 A if = Also, Eb = 200 − 2 × 0.1 = 199.8V The power developed by the motor is: P = Pin − Pcopper loss = 200 × 4 − (22 × 100 + 22 × 0.1) = 399.6W 17.13 G.
Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 17 Problem 17.19 Solution: Known quantities: 230 VDC shunt motor. The armature resistance is 0.5 Ω and a series field resistance is 75 Ω. At 1100 rev min, Prot = 500 W. When loaded, the current is 46 A. A Find: a) The speed Pdev and Tsh. B) ia (t ) and ω m (t ) if L f = 25 H, La = 0.008 H and the terminal voltage has a 115V change. Assumptions: None.
Analysis: 230 = 3.07 A 75 ia = i L − i f = 42.93 A if = ω m = 117.3 rad sec 230 At no load, 117.3 =, therefore, K aφ K aφ = 1.96 At full load, ωm = 230 − 0.5 × 42.93 K aφ The back emf is: Eb = 230 − 0.5 × 42.93 = 208.5V The power developed is: Pdev = Eb I a = 8.952 kW The power available at the shaft is: Po = Pdev − Prot = 8952 − 500 = 8452 W The torque available at the shaft is: Tsh = Po = 72.1 N ⋅ m ωm 17.14 G. Rizzoni, Principles and Applications of Electrical Engineering PSH = 7456 − TSH = Problem solutions, Chapter 17 98.25 × 599.1 = 6867.4 W 100 6867.4 = 69.9 N ⋅ m 98.25 Finally, the efficiency is: eff = PSH = 85.84% Pin Problem 17.21 Solution: Known quantities: 50 hp, 230V shunt motor operates at full load when the line current is 181 A at 1350 rev min. The field resistance is 17.7 Ω. To increase the speed to 1600 rev min, a resistance of 5.3 Ω is cut in via the field rheostat. The line current is increased to 190 A. A Find: a) The power loss in the field and its percentage of the total power input for the 1350 rev b) The power losses in the field and the field rheostat for the 1600 rev min speed.
C) The percent losses in the field and in the field rheostat at min speed. 1600 rev min speed. Assumptions: None. Analysis: a) 230 = 13.0 A 17.7 Pf = ( 230)(13.0) = 2988.7 W If = Pf Pm = 2988.7 = 0.072 = 7.2% ( 230)(181) b) If = 230 = 10 A (17.7 + 5.3) Pf = 102 (17.7) = 1770W PR = 102 (5.3) = 530 W c) Pin = ( 230)(190) = 43,700 W 1770 × 100 = 4.05% 43700 530 × 100 = 1.21%% PR = 43700% Pf = 17.16 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 17 Problem 17.22 Solution: Known quantities: 10 hp, 230 V shunt-wound motor. The armature resistance is 0.26 Ω and a series field resistance is 225 Ω.
The rated speed is 1000 rev min. The full-load efficiency is 86 percent. A Find: The effect on counter emf, armature current and torque when the motor is operating under rated load and the field flux is very quickly reduced to 50 percent of its normal value. The effect on the operation of the motor and its speed when stable operating conditions have been regained. Assumptions: None. Analysis: EC = Kφ n; counter emf will decrease. V − EC; armature current will increase.
Ra T = KφI a; effect on torque is indeterminate. Ia = Operation of a dc motor under weakened field conditions is frequently done when speed control is an important factor and where decreased efficiency and less than rated torque output are lesser considerations. V − I a ra V − I a ra 1000 = Kφ Kφ V − I a ra nnew = 0.5 K φ n= Assume small change in the steady-state value of I a. Then: 1000 0.5 = nnew = 2000 rpm nnew 1 Problem 17.23 Solution: Known quantities: The machine is the same as that in Example 17.7. The circuit is shown in Figure P17.23. The armature resistance is 0.2 Ω and the field resistance is negligible.
N = 120 rev min, I a = 8 A. In the operating region, φ = kI f, k = 200. Find: a) The number of field winding turns necessary for full-load operation. B) The torque output for the following speeds: 1.
N, = 2n n, = n 2, 4. N = 3n c) Plot the speed-torque characteristic for the conditions of part b. Assumptions: None., 17.17 G.
Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 17 TX = 9.54 N ⋅ m ω X = 0.25ω m = 3.14 rad sec, TX = 20.53 N ⋅ m 4. At c) The diagram is shown below: Problem 17.24 Solution: Known quantities: PM DC motor circuit model; mechanical load model. Example 17.9. Find: Voltage-step response of motor. Assumptions: None. Analysis: Applying KVL and equation 17.47 to the electrical circuit we obtain: dI (t ) Eb (t ) 0 VL (t) Ra I a (t ) La a dt or dIa (t) Ra I a (t) K a PM m (t ) VL (t) dt Applying Newton’s Second Law and equation 17.46 to the load inertia, we obtain: d (t) T(t) TLoad (t ) b J dt or La d (t ) b (t) 0 dt since the load torque is assumed to be zero.
To derive the transfer function from voltage to speed, we use the result of Example 17.9 with Tload = 0: KT PM VL (s) m (s) sL a Ra (sJ b) Ka PM KT PM K TPM I a (t) J The step response of the system can be computed by assuming a unit step input in voltage: KT PM 1 m (s) sL a Ra (sJ b) Ka PM KT PM s 17.19.
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