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marshall27 [118]
1 year ago
11

X cotx expansion using maclaurins theorem.

Engineering
1 answer:
Lemur [1.5K]1 year ago
4 0

It is to be noted that it is impossible to find the Maclaurin Expansion for F(x) = cotx.

<h3>What is Maclaurin Expansion?</h3>

The Maclaurin Expansion is a Taylor series that has been expanded around the reference point zero and has the formula f(x)=f(0)+f′. (0) 1! x+f″ (0) 2! x2+⋯+f[n](0)n!

<h3>What is the explanation for the above?</h3>

as indicated above, the Maclaurin infinite series expansion is given as:

F(x)=f(0)+f′. (0) 1! x+f″ (0) 2! x2+⋯+f[n](0)n!

If F(0) = Cot 0

F(0) = ∝ = 1/0

This is not definitive,

Hence, it is impossible to find the Maclaurin infinite series expansion for F(x) = cotx.

Learn more about Maclaurin Expansion at;
brainly.com/question/7846182
#SPJ1

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(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))

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(b) calculate the resultant supply power factor

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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+11.586j A

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

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

I_{load} =48.115-42.433j 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

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