7. The following table relates to the open-circuit curve of a shunt gen-Determine the no-load terminal voltage if the field circuit resistanceIf the speed is now halved, what is the resultant terminal voltage?erator running at 750 rev/min:Generated voltage (V)Field current (A)10 172 300 360 385 3950 1 2 34 5is 125 2.At the reduced speed, what valne of field circuit resistance will give ano-load terminal voltage of 175 V?(N.C.T.E.C., O.2)diti

Answered: Image /qna-images/answer/d0d2213e-6c3e-4c19-98f4-0ca37c6b7793.jpg

7. The following table relates to the open-circuit curve of a shunt erator running at 750 rev/min: Generated voltage (V) Field current (A) 172 300 360 385 395 4 5 10 0 1 2 3 Determine the no-load terminal voltage if the field circuit resistance If the speed is now halved, what is the resultant terminal voltage? is 125 Q. At the reduced speed, what valus of field circuit resistance will give a no-load terminal voltage of 175 V? (N.C.T.E.C., 0.2)

7. The following table relates to the open-circuit curve of a shunt erator running at 750 rev/min: Generated voltage (V) Field current (A) 172 300 360 385 395 4 5 10 0 1 2 3 Determine the no-load terminal voltage if the field circuit resistance If the speed is now halved, what is the resultant terminal voltage? is 125 Q. At the reduced speed, what valus of field circuit resistance will give a no-load terminal voltage of 175 V? (N.C.T.E.C., 0.2)

EBK ELECTRICAL WIRING RESIDENTIAL

19th Edition

ISBN:9781337516549

Author:Simmons

Publisher:Simmons

Chapter29: Service-entrance Calculations

Section: Chapter Questions

Problem 3R: a. What is the ampere rating of the circuits that are provided for the small-appliance loads? _____...

See similar textbooks

Similar questions

Fill in the data for a 16-ampere electric motor, single-phase, no Code letters. a. Branch-circuit protection, non-time-delay fuses: Normal size _____________AMaximum size _____________A Switch size _____________ASwitch size _____________A b. Branch-circuit protection dual-element, time-delay fuses: Normal size _____________AMaximum size _____________A Switch size _____________ASwitch size _____________A c. Branch-circuit protection instant-trip breaker: Normal setting _____________A Maximum setting _____________A d. Branch-circuit protectioninverse time breaker: Normal rating _____________A Maximum rating _____________A e. Branch-circuit conductor size, Type THHN Ampacity _____________A f. Motor overload protection using dual-element time-delay fuses: Maximum size _____________A
PROBLEMS i istic such that rated armay, short.circuit characteristic 5 « ; aa'gﬂ?;ciunﬁ e ion. Rated voltaEo on the sitgap line o ts-unit excitation 1s reduted o, su pf lagging? tor is 80% leading. 51, A synchronous gei current is obtained by obtained by 1.25 per-unit excitat = load at rated voltage, rated VA, and 807 52, Repeat Prob, 5-1 for the case where the foad pover fac 5.3, A synchronous generator has the following rating? 180-KVA, 440-V, 300-rpm, 60-Hz, three-Phase: r~0 and x —02960Q Theuit-sap line it described by the equation £ = 1117, RIS = e b Y-connection, expressed per phase. The short. the field current needed
a) What applications / experiments are necessary to find the Rc parameter in the equivalent circuit given in Figure 1? b) Write the value of the Ir 'current given in Figure 1 with a sentence in the idle running test of the 3-phase Asynchronous motor.
What does a "matched load" mean?What if it is not compatible (matched)? Explain in detail.
The data obtained in the idling test performed on a three-phase star-connected asynchronous motor; It is 1200V, 5500W, 9.5A. The data obtained in the short circuit experiment on the same motor is 250V,It is 14500W and 100A. For the case where the slip of the motor is 5%, create the approximate equivalent circuit of the motor (L type equivalent circuit), draw the equivalent circuit, find the values of the circuit elements and MUST AND MUST write them ON THE EQUIVALENT CIRCUIT. R and X values will be used in circuit elements, not Z.
Choose which answers are correct Reliability evaluation deals with the disturbances arising in the power generation system? a . Adequacy based Reliability Evaluation b . Recursive Technique c . Security based Reliability Evaluation d . Reserve margin Method
405 / 5000 Find the slip, the powers converted,air gap, output and efficiency of a motorthree-phase induction in Y of 138.56 V of voltage ofphase, 60 Hz and 50 hp, which is supplied with 35 A at apower factor of 0.85 lagging. The losses are: copper stator, 2.5 kW rotor, 750 W friction and friction, 650 W and from the core, 1850 W, Negligible miscellaneous losses.
What is load capacity? What is the load capacity of a typical residential area, a city? What precautions are practiced by electric companies in terms of loading capacity? When is distribution box needed in terms of electrical supply?
Q1a. Explain the following terminologies: Binary State Multi State Hard failure Soft failure Q1b. With the aid of a well labelled diagram, explain the stages involved in a bathtub curve #3 The life of an automotive component can be modelled by a Weibull distribution. ∝ = 6.3?105 , ? = 1.3 Calculate F(t) R(t) h(t) at the end of rotating period of 36,000 miles Q1c How should a reliability program be organized in an environment of concurrent engineering?
A 6-phase ∆-connected 1600 V (line-to-line), 150 kW, 50 Hz, six-pole induction driveis used in an electric car. The drive has the following parameter values in Ω/phasereferred to the stator:R1 =0.334 Ω R2 =0.168 Ω Rm1 = 125 ΩX1 = 0.273 Ω X2 = 0.287 Ω Xm1 = 16.78 ΩThe total friction, windage, and core losses may be assumed to be constant at Prot =773 W, independent of load. For a slip value of s=0.015: b ) Compute the speed, output torque, power and stator current d) Calculate the efficiency when the motor is operated at rated voltage andfrequency. answer：b) n= 985 rpm; Torque T= 1.18 x104 Nm; Power= 1.22 x106 W; Stator current I = 175.72 /32.79° (d) Efficiency = 86.4%
Explain the concept of electrical load balancing in power systems. How is load balancing achieved, and why is it important for the stability of the electrical grid?
A Y- connected, 10-KVA, 230- has a field winding resistance of 4.59. At an operation of 0.85 lagging power factor, the excitation current is recorded at 10A. the core loss is 300W, and the mechanical loss is 200W. If the corresponding efficiency is 84.9%. calculate the per- phase armature winding resistance.