The reason for low power factor in no load test in induction motor is high rotor reactance at lower loading since the rotor reactance value gets decreased with increased loading on induction motor
Usually at no load, the Cosine of the phase angle between the stator current(I1) and the stator applied voltage(V1) is the power factor of an induction motor.
In conclusion, the reason for low power factor in no load test in induction motor is high rotor reactance at lower loading since the rotor reactance value gets decreased with increased loading on induction motor
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why people act like that
Answer:
i dob't know
Explanation:
when the electrical connection to the alternator from the battery is not tight, it can cause?
Answer:
affects the flow of electricity
Explanation:
A loose battery terminal affects the flow of electricity. There is less power going to the electrical systems and the vehicle will not start or start sluggishly. Also, a loose battery terminal causes the car's electrical components like navigation, car lights, and audio among others to dim or fail completely.
Beyond providing a visual representation of an object for the creator, what are the advantages of using CAD software to create 3D digital models of engineering prototypes (compared to physical drawings)? Identify at least two advantages.
Answer:
fasdfsd
Explanation:
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The first three terms of a geometric series are 15,7.5 and 3.75. find the representation of the n th term of tge sequence
Answer:
30 1/2
Explanation:
Consider an FSM with 1024 states and a 2-bit input (00, 01, 10, 11). How many flip-flops will you need to construct the FSM
Answer:
468
Explanation:
Hierarchy of hazard control
Answer:
A system for controlling risk in the workspace.
Describe three differences between liquids and gases in fluid power systems.
Answer:
Liquids have less force of attraction between molecules than solids and more than gases; gases have a very weak force of attraction between the molecules which is the least amongst the three states of matter.
Explanation:
what is the relationship between the voltages of the output and adjust pins of LM1117 when in normal operation
Based on the engineering analysis, the relationship between the voltages of the output and adjust pins of LM1117 when in normal operation is "the LM1117 adjustable version establishes a 1.25V reference voltage, (VREF) between the output and the adjust terminal."
The LM1117 generally has a 1.2V at 800mA of load current.
The LM1117 is known to be a progression of low dropout linear voltage regulators. It is adjustable and be assigned between the output voltage of 1.25 to 13.8 V using just two external resistors.
Hence, in this case, it is concluded that the correct answer is "the LM1117 adjustable version establishes a 1.25V reference voltage, (VREF), between the output and the adjust terminal.
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You work for REV, Incorporated, and check your work email on your mobile phone. According to these documents, if you cannot find your phone, what should you do first?
Answer:
In case whereby you work for rev and check your work email on your mobile phone and You have decided to resign your job and take a position with a different company. Regarding rdm on your mobile phone, you should B. Take your phone to IT Services, and they will wipe it clean.
What is expected to do when leaving the company?
It is expected from you to return anything that belong to the company before leaving the company, but in the case whereby there is still a rdm on your mobile phone, then you can work to the IT services so they can wipe it away.
Find out more on IT Services
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Missing options:
A. Turn your phone to IT Services
B. Take your phone to IT Services, and they will wipe it clean
C. Download RDM to your personal computer
D. Call IT Services for support, and they will assist
E. Nothing. IT Services will automatically remove all data
locate the centroid of the shaded area between the two curves
9514 1404 393
Answer:
(x, y) = (5.76, 1 5/7)
Explanation:
The location of the centroid in the x-direction is the ratio of the first moment of area about the y-axis to the total area. Similarly, the y-coordinate of the centroid is the first moment of area about the x-axis, divided by the area.
For the moment about the y-axis, we can define a differential of area as ...
dA = (y2 -y1)dx
where y2 = √(x/k2) and y1 = k1·x^3
The distance of that area from the y-axis is simply x.
So, the x-coordinate of the centroid is ...
[tex]\displaystyle c_x=\frac{a_x}{a}=\frac{\int{x\cdot dA}}{\int{dA}}\\\\a_x=\int_0^{12}{x(k_2^{-1/2}\cdot x^{1/2}-k_1x^3)}\,dx=\frac{2}{5k_2^{1/2}}\cdot12^{5/2}-\frac{k_1}{5}12^5\\\\a=\int_0^{12}{(k_2^{-1/2}\cdot x^{1/2}-k_1x^3)}\,dx=\frac{2}{3k_2^{1/2}}\cdot12^{3/2}-\frac{k_1}{4}12^4\\[/tex]
For k1 = 4/12^3 and k2=12/4^2, these evaluate to ...
[tex]a_x=115.2\\a=20\\c_x=5.76[/tex]
The y-coordinate of the centroid requires we find the distance of the differential of area from the x-axis. We can use (y2 +y1)/2 for that purpose. Then the y-coordinate is ...
[tex]\displaystyle c_y=\frac{a_y}{a}\\\\a_y=\int_0^{12}{(\frac{y_2+y_1}{2}(y_2-y_1))}\,dx=\frac{1}{2}\int_0^{12}{(\frac{x}{k_2}-(k_1x^3)^2)}\,dx\\\\a_y=\frac{12^2}{4k_2}-\frac{k_1^212^7}{14}=\frac{240}{7}\\\\c_y=\frac{12}{7}\approx1.7143[/tex]
The centroid of the shaded area is ...
(x, y) = (5.76, 1 5/7)
Can I have free money? I want to get a new pc
Answer:
NO!
Explanation:
Consider a 25-mm-diameter and 15-m-long smooth tube that is maintained at a constant surface temperature. Fluids enter the tube at 50°C with a mass flow rate of 0.01 kg/s. Determine the tube surface temperatures necessary to heat water, engine oil, and liquid mercury to the desired outlet temperature of 150°C
Answer: ⚠this is not my answer it from hamzaahmeds⚠
Water: h = 35.53 W/m².k
Engine oil: h = 18.84 W/m².k
Mercury: h = 1.19 W/m².k
Explanation:
Assuming the steady state, one-dimensional heat flow, it is clear that the added to the fluid by tube heat will be equal to the heat transfer through convection outside the tube.
Therefore,
mCΔT = hAΔT
mC = hA
h = mC/A
where,
h = convection coefficient
m = mass flow rate = 0.01 kg/s
C = specific heat capacity of fluid
A = surface area of tube = 2πrL = 2π(0.0125 m)(15 m) = 1.178 m²
FOR WATER:
C = 4186 J/Kg.k
Therefore,
h = (0.01 kg/s)(4186 J/Kg.k)/(1.178 m²)
h = 35.53 W/m².k
FOR ENGINE OIL:
C = 2220 J/Kg.k
Therefore,
h = (0.01 kg/s)(2220 J/Kg.k)/(1.178 m²)
h = 18.84 W/m².k
FOR LIQUID MERCURY:
C = 140 J/Kg.k
Therefore,
h = (0.01 kg/s)(140 J/Kg.k)/(1.178 m²)
h = 1.19 W/m².k
