EBK MINDTAP FOR HERMAN'S UNDERSTANDING
4th Edition
ISBN: 9781337798761
Author: Herman
Publisher: VST
expand_more
expand_more
format_list_bulleted
Concept explainers
Textbook Question
Chapter 43, Problem 1RQ
What are the disadvantages of low-speed, directly coupled induction motors?
Expert Solution & Answer
To determine
The disadvantages of low-speed, directly coupled induction motor.
Explanation of Solution
The directly coupled induction motors are rarely used in the applications where the required speed is below 500rpm. The induction motor take very less current from the supply main.
The vector sum of load current and magnetizing current lags voltage by around 75-80 degrees and hence, the power factor is low. Due to high magnetizing current, the copper losses of the motor increase. This in turn leads to decrease in the efficiency of the motor.
Thus Disadvantage of low-speed, directly coupled induction motors are low power factor and low efficiency.
Want to see more full solutions like this?
Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
4. Give the dimensions completely of the following figure?
2. Consider the picture shown in figure
and draw the orthographic projection of
the object in first angle projection (in
direction of arrow X).
✗
10
90°
60
60
10
15
40
40
5. How are bearings classified? And draw the half sectional view of a ball bearing and indicate the
various parts of it.
(marks 15)
Chapter 43 Solutions
EBK MINDTAP FOR HERMAN'S UNDERSTANDING
Ch. 43 - What are the disadvantages of low-speed, directly...Ch. 43 - Prob. 2RQCh. 43 - What three alignment checks should be made to...Ch. 43 - Prob. 4RQCh. 43 - Induction motors should not be used below a...Ch. 43 - What is the primary reason for using a pulley...Ch. 43 - How tightly should V-belts be adjusted?Ch. 43 - Refer to Figure 436 and assume the motor is to...Ch. 43 - A machine delivered for installation has a 2-inch...
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.Similar questions
- 3. The Figure shows the isometric view of a machine component along with the view from above and the view from the left, draw the full sectional view from the front. 56 Draw here the full sectional view 90 40 (a) 32 56 B 32arrow_forwardHow do I solve this task?A spring scale should have a capacity for 10kg. The tension spring is mounted with a weight that is a preload in the spring. Choose a spring with a maximum load of 50mm. What distance should the kg weight be and what the preload weight should weigh.F=Fo+c·fFo= preload weight (N)F=10·9.81 → F=98.1 N Trying with a spring 3823f=F- Fo /c → 98.1-9/1.090=81.74 mmarrow_forwardA gearbox consists of four gears. Dimension the gearbox for a gear ratio of 18:1. The gear ratio should be as even as possible. In the first stage, module 4, in the second stage module 5. All shafts are in the same plane. Calculate the distance between Da. Da=45mm Db=40mmu=ωin/ ωout= nin/ nout= dout/ din= Mout/ Min= Zout/ Zind=m·z)da=m·(z+2) (top diameters)df=m·(z-2.5) (bottom diameters)tooth limits z1 z2 13 13-16 14 14-26 15 15-45 16 16-101 17 17-1314 18 18- .......arrow_forward
- Question 2 a) Construct the signal flow graph (SFG) for the block diagram shown in Fig. Q2 (a) and C($) obtain the gain using Mason's formula. R(s) 04 -R() 01 0₂ 0 Hi h Sinded States Text Predictions On Accessibility Unavailable Fig. Q2 (a) H₂ CAarrow_forwardHow do you solve for the force acting on member BC?arrow_forwardA brake jaw, A is pressed against the drum, B. Calculate the brake arm, X(m₂).F= 250NBraking torque = 30Nmµ=0.35Around point A:Fm₂-Nm₁-µm₃=0N=Fm₂/m1+ µm₃MBrake =µ·D/2= µ·D/2MBrake =Fµm₂D/2(m₁- µm₃)(X)m₂=FµD/Mbrake·2(m1- µm₃)(X)m₂=250·0.35.0.3/30·2(0.250-0.35·0.06)=?I don’t get some likely value?arrow_forward
- Q7 (12 Marks) For the system shown in Fig.3: 1- Draw the overall block diagram. 2- Determine the transfer function (Pc(s)/E(s)). Orifice→ Ps Actuating error signs) Flapper Pb+Pb. Nozzle. A X+X+ Ri A I R2 ㅍ think +y Pc+PCarrow_forwardFigures 4: show a pneumatic controller. The pneumatic relay has the characteristic that pc=K pb , where K>0. What kind of control action does this controller produce? a. Derive the mathematical model for the system b. Derive the transfer function Pc(s)/E(s) -Solve step by step Orifice F+Ph R₁ Actuating error signal Flapper Nozzle. x+x F+Pe thinkarrow_forwardThe equation of the turning moment diagram for the three crank engine and the equation of the moment required by a machine connected to this engine are given below: Engine Torque Machine Torque T=10000-500 sin (40) T=10000+2000 sin (20) N.m N.m where radians is the crank angle from inner dead center and the mean engine speed is 300 rpm. It is required to select a proper flywheel (find the moment of inertia of the flywheel in kgm2) and then calculate the power of the engine if the total percentage fluctuation of speed of the flywheel is ±1% of the mean speed. Calculate the angular acceleration of the flywheel when angle is 45°.arrow_forward
- Design a cotter joint to support a axial load of 100kN . Carbon steel material selected whichhas Tensile stress = 100MPa Compressive stress =150MPa; Shear stress =60MPaarrow_forwardDesign a cotter joint to support a axial load of 100kN . Carbon steel material selected whichhas Tensile stress = 100MPa Compressive stress =150MPa; Shear stress =60MPaarrow_forwardI need all the derivations from Bohr's postulates in handwritten formarrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
- Understanding Motor ControlsMechanical EngineeringISBN:9781337798686Author:Stephen L. HermanPublisher:Delmar Cengage LearningElectrical Transformers and Rotating MachinesMechanical EngineeringISBN:9781305494817Author:Stephen L. HermanPublisher:Cengage LearningRefrigeration and Air Conditioning Technology (Mi...Mechanical EngineeringISBN:9781305578296Author:John Tomczyk, Eugene Silberstein, Bill Whitman, Bill JohnsonPublisher:Cengage Learning
- Welding: Principles and Applications (MindTap Cou...Mechanical EngineeringISBN:9781305494695Author:Larry JeffusPublisher:Cengage Learning
Understanding Motor Controls
Mechanical Engineering
ISBN:9781337798686
Author:Stephen L. Herman
Publisher:Delmar Cengage Learning
Electrical Transformers and Rotating Machines
Mechanical Engineering
ISBN:9781305494817
Author:Stephen L. Herman
Publisher:Cengage Learning
Refrigeration and Air Conditioning Technology (Mi...
Mechanical Engineering
ISBN:9781305578296
Author:John Tomczyk, Eugene Silberstein, Bill Whitman, Bill Johnson
Publisher:Cengage Learning
Welding: Principles and Applications (MindTap Cou...
Mechanical Engineering
ISBN:9781305494695
Author:Larry Jeffus
Publisher:Cengage Learning
Fluid Mechanics - Viscosity and Shear Strain Rate in 9 Minutes!; Author: Less Boring Lectures;https://www.youtube.com/watch?v=_0aaRDAdPTY;License: Standard youtube license