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Hunter-Best [27]

3 years ago

9

I don't know what is this

1 answer:

pychu [463]3 years ago

8 0

That means “ if possible then link”

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Yall know what this is called?

aliya0001 [1]

Answer:

oof no bro

Explanation:

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3 years ago

Source water pollution in Madagascar

vodka [1.7K]

Answer:

What is the question?

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3 years ago

What’s the difference between quality and quantity

Lesechka [4]

Quality is how good something is. Like how good the material is. Like the quality of the IPhone X is better than the quality of the iPhone 6s + because it has more features and can sustain better hold
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I dont know exactly if that makes sense but there ya go
YW!

4 0

3 years ago

Read 2 more answers

An industrial plant consists of several 60 Hz single-phase motors with low power factor. The plant absorbs 600 kW with a power f

Gelneren [198K]

Answer:

(a) $Q=332$ $kvar$ and $C=5.66$ $uF$

(b) $pf=0.90$ lagging

Explanation:

Given Data:

$P=600kW$

$V=12.47kV$

$f=60Hz$

$pf_{old} =0.75$

$pf_{new} =0.95$

(a) Find the required kVAR rating of a capacitor

$\alpha _{old}=cos^{-1}(0.75) =41.41$ °

$\alpha _{new}=cos^{-1}(0.95) =18.19$ °

The required compensation reactive power can be found by

$Q=P(tan(\alpha_{old}) - tan(\alpha_{new}))$

$Q=600(tan(41.41) - tan(18.19))$

$Q=332$ $kvar$

The corresponding capacitor value can be found by

$C=Q/2\pi fV^{2}$

$C=332/2*\pi *60*12.47^{2}$

$C=5.66$ $uF$

(b) calculate the resultant supply power factor

First convert the hp into kW

$P_{mech} =250*746=186.5$ $kW$

Find the electrical power (real power) of the motor

$P_{elec} =P_{mech}/n$

where $n$ is the efficiency of the motor

$P_{elec} =186.5/0.80=233.125$ $kW$

The current in the motor is

$I_{m} =(P/\*V*pf)$

The pf of motor is 0.85 Leading

Note that represents the angle in complex notation (polar form)

$I_{m} =(233.125/12.47*0.85)$

$I_{m}=18.694+11.586$ $j$ $A$

Now find the Load current

pf of load is 0.75 lagging (notice the minus sign)

$I_{load} =(600/12.47*0.75)$

$I_{load} =48.115-42.433$ $j$ $A$

Now the supply current is the current flowing in the load plus the current flowing in the motor

$I_{supply} =I_{m} + I_{load}$

$I_{supply}= (18.694+11.586)+(48.115-42.433)$

$I_{supply} =66.809-30.847j$ $A$

or in polar form

$I_{supply} =73.58$ °

Which means that the supply current lags the supply voltage by 24.78

therefore, the supply power factor is

$pf=cos(24.78)=0.90$ lagging

Which makes sense because original power factor was 0.75 then we installed synchronous motor which resulted in improved power factor of 0.90

8 0

3 years ago

A particle moves along a straight line such that its position is defined by s = (t2 - 6t + 5) m. Determine the average velocity,

Dominik [7]

Answer:

0 m/s , 3 m/s , 2 m/s^2

Explanation:

Given : s(t) = ( t^2 - 6t + 5)

v(t) = ds / dt = 2t - 6

s(0) = 5 m

s(6) = (6)^2 - 6*6 + 5 = 5 m

Vavg = ( s(6) - s(0) ) / 2 = 0 m\s

Find the turning point of particle:

ds/dt = 0 = 2t - 6

t = 3 sec

s(3) = 3^2 -6*3 + 5 = - 4

Total distance = 5 - (-4) + (5 - (-4)) = 18 m

Total time = 6s

Average speed = Total distance / Total time = 18 / 6 = 3 m/s

Taking derivative of v(t) to obtain a(t)

a (t) = dv(t) / dt = 2 m/s^2

7 0

3 years ago