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MrRa [10]
2 years ago
13

Why the power factor is Low in no load test in induction motor ?​

Engineering
1 answer:
marshall27 [118]2 years ago
5 0

Answer:

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. The reason of the low power factor is high rotor reactance at lower loading. The rotor reactance value gets decreased with increased loading on induction motor..','.

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Write a C program that asks the user to enter two numbers, obtains the two numbers from the user and prints the sum, product, di
Bogdan [553]

Answer:

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Explanation:

Initialize your variable as a float or double since you're going to be using fractions in your answer.

User scanf() to get user input.

Print out the sum, product, quotient, and difference between the two numbers.

8 0
3 years ago
Analyse what effect the building of an airport may have on the decision of how to use an area of land nearby. (6)​
Sonja [21]
An effect might be a customer not wanting to buy it specifically because it’s by an airport, or maybe the customer wants to buy it because it’s right next to the airport, and a lot of people go to the airport so therefore they might go to the building next to the airport.
5 0
3 years ago
what is the transfer function of the loaded filter? express your answer in terms of the variables r , l , rl , and s .
NISA [10]

Loaded, H_{Loaded}(s) = \frac{RR_{L} }{R+R_{L} } /(\frac{RR_{L} }{R+R_{L} }+SL) = \frac{RR_{L}/L }{R+R_{L} } /(\frac{RR_{L} /L}{R+R_{L} }+S) is the loaded filter's transfer function.

A graded filter that, by virtue of its weight and permeability, stabilises the foot of an earth dam or other construction when it is installed at the base of that structure.

Air filters with depth loaded are made to achieve precisely that. They add particles gradually to create air passageways, reducing constriction. You may save time and money by using filters that last longer thanks to them. The bigger particles are caught at the filter's beginning, while the smaller particles are caught as it gets closer. This is intended to avoid rapid surface loading, hence facilitating more airflow. This enables longer-lasting filtration as well.

On the other hand, surface loading filters catch every particle that is on its surface. No matter how big or little the particles are, it doesn't care.

Learn more about Loaded here:

brainly.com/question/20039214

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3 0
11 months ago
A 75 ! coaxial transmission line has a length of 2.0 cm and is terminated with a load impedance of 37.5 j75 !. If the relative p
Cerrena [4.2K]

Answer:

4.26

Explanation:

The wavelength λ is given by:

\lambda=v/f=c/nf\\c=speed\ of\ light=3*10^8m/s,f=frequency=3*10^9Hz,n=permittivity=2.56\\\\\lambda=3*10^8/(2.56*3*10^9)=0.0625\ m\\

Phase constant (β) = 2π/λ

βl = 2π/λ × l

l = 2 cm = 0.02 m

βl = 2π/0.0625 × 0.02=2.01 rad = 115.3°

1 rad = 180/π degrees

Z_L=load\ impedance=37.5+j75\\\\Z_o=characteristic impedance = 75\ ohm\\\\\tilde {Z_L}=Z_L/Z_o=37.5+j75/75=0.5+j

\tilde {Z_{in}}=\frac{\tilde {Z_{L}}+jtan\beta l}{1+j\tilde {Z_{L}}tan\beta l}=\frac{0.5+j+jtan(115.2)}{1+j(0.5+j)tan(115.2)}=0.253-j0.274\\  \\Z_{in}=Z_o\tilde {Z_{in}}=75(0.253-j0.274)=19-j20.5\\\\\Gamma_L=\frac{Z_L-Z_0}{Z_L+Z_o}=\frac{37.5+j75-75}{37.5+j75+75}=0.62\angle 83^o\\\\\Gamma_{in}=\frac{Z_{in}-Z_0}{Z_{in}+Z_o}=\frac{19-j20.5-75}{19-j20.5+75}=0.62\angle -147^o\\\\VSWR=\frac{1+\rho}{1-\rho} =\frac{1+0.62}{1-0.62}=4.26

6 0
3 years ago
Oil with a density of 850 kg/m3 and kinematic viscosity of 0.00062 m2 /s is being discharged by a 5 mm diameter, 40 m long horiz
ra1l [238]

Answer:

Flow rate is 1.82\times 10^{-8} m^{3}/s

Explanation:

Given information

Density of oil, \rho_{oil}= 850 Kg/m^{3}

kinematic viscosity, v= 0.00062 m^{2} /s

Diameter of pipe, D= 5 mm= 0.005 m

Length of pipe, L=40 m

Height of liquid, h= 3 m

Volume flow rate for horizontal pipe will be given by

\bar v=\frac {\triangle P\pi D^{4}}{128\mu L} where \mu is dynamic viscosity and \triangle P is pressure drop

At the bottom of the tank, pressure is given by

P_{bottom}=\rho_{oil} gh=850 Kg/m^{3}\times 9.81 m/s^{2}\times 3 m= 25015.5 N/m^{2}

Since at the top pressure is zero, therefore change in pressure a difference between the pressure at the bottom and the top. It implies that change in pressure is still 25015.5 N/m^{2}

Dynamic viscosity, \mu=\rho_{oil}v= 850 Kg/m^{3}\times 0.00062 m^{2}/s=0.527 Kg/m.s

Now the volume flow rate will be

\bar v=\frac {25015.5 N/m^{2}\times \pi \times 0.005^{4}}{128\times 0.527 Kg/m.s \times 40}=1.82037\times 10^{-8} m^{3}/s\approx 1.82\times 10^{-8} m^{3}/s

Proof of flow being laminar

The velocity of flow is given by

V_{flow}=\frac {\bar v}{A}=\frac {1.82\times 10^{-8} m^{3}/s}{0.25\times \pi\times 0.005^{2}}=0.000927104  m/s

Reynolds number, Re=\frac {\rho_{oil} v_{flow} D}{\mu}=\frac {850 Kg/m^{3}\times 0.000927104 m/s\times 0.005}{0.527 kg/m.s}=0.007476648

Since the Reynolds number is less than 2300, the flow is laminar and the assumption is correct.

5 0
3 years ago
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