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Two balanced Y-connected loads in parallel, one drawing 15kW at 0.6 power factor lagging and the other drawing 10kVA at 0.8 powe

NemiM [27]

Answer:

(a) attached below

(b) $pf_{C}=0.85$ $lagging$

(c) $I_{C} =32.37 A$

(d) $X_{C} =49.37$ Ω

(e) $I_{cap} =9.72 A$ and $I_{line} =27.66 A$

Explanation:

Given data:

$P_{1}=15 kW$

$S_{2} =10 kVA$

$pf_{1} =0.6$ $lagging$

$pf_{2}=0.8$ $leading$

$V=480$ $Volts$

(a) Draw the power triangle for each load and for the combined load.

$\alpha_{1}=cos^{-1} (0.6)=53.13$ °

$\alpha_{2}=cos^{-1} (0.8)=36.86$ °

$S_{1}=P_{1} /pf_{1} =15/0.6=25 kVA$

$Q_{1}=P_{1} tan(\alpha_{1} )=15*tan(53.13)=19.99$ ≅ $20kVAR$

$P_{2} =S_{2}*pf_{2} =10*0.8=8 kW$

$Q_{2} =P_{2} tan(\alpha_{2} )=8*tan(-36.86)=-5.99$ ≅ $-6 kVAR$

The negative sign means that the load 2 is providing reactive power rather than consuming

Then the combined load will be

$P_{c} =P_{1} +P_{2} =15+8=23 kW$

$Q_{c} =Q_{1} +Q_{2} =20-6=14 kVAR$

(b) Determine the power factor of the combined load and state whether lagging or leading.

$S_{c} =P_{c} +jQ_{c} =23+14j$

or in the polar form

$S_{c} =26.92$ °

$pf_{C}=cos(31.32) =0.85$ $lagging$

The relationship between Apparent power S and Current I is

$S=VI^{*}$

Since there is conjugate of current I therefore, the angle will become negative and hence power factor will be lagging.

(c) Determine the magnitude of the line current from the source.

Current of the combined load can be found by

$I_{C} =S_{C}/\sqrt{3}*V$

$I_{C} =26.92*10^3/\sqrt{3}*480=32.37 A$

(d) Δ-connected capacitors are now installed in parallel with the combined load. What value of capacitive reactance is needed in each leg of the A to make the source power factor unity?Give your answer in Ω

$Q_{C} =3*V^2/X_{C}$

$X_{C} =3*V^2/Q_{C}$

$X_{C} =3*(480)^2/14*10^3$ Ω

(e) Compute the magnitude of the current in each capacitor and the line current from the source.

Current flowing in the capacitor is

$I_{cap} =V/X_{C} =480/49.37=9.72 A$

Line current flowing from the source is

$I_{line} =P_{C} /3*V=23*10^3/3*480=27.66 A$

8 0

3 years ago

the increase of current when 15 V is applied to 10000ohm rheostat which is adjusted to 1000ohm value

Anastasy [175]

Given data:
•) applied voltage = 15 V
•). Resistance = 1000 ohm

Required:
•). The magnitude of current= ?

•••••••••••••SOLUTION•••••••••••••

We can find the relation ship between current, voltage and resistance with the help of Ohms law.

According to ohms law;

V= IR.

Rearranging the above equation;

I= V/ R

Putt the values in the above equation; we get

I= 15V/ 1000ohm

I = 0.015 A( ampere)

••••••••••••••• CONCLUSION•••••••

The value of the current would be 0.15 ampere when Resistance is equal to 1000 and that of Voltage is equal to 15 V.

4 0

3 years ago

A what is marked by two sets of double yellow lines with each set having a broken line on the inside and a solid line on the out

yKpoI14uk [10]

Answer:

(C) Center left turn lanes!

Explanation:

I'm positive this answer is correct! :)

Have a nice day!

5 0

3 years ago

To determine if a product or substance being used is hazardous, consult:__________.

qwelly [4]

Answer:

Option B: An MSDS

Explanation:

A dictionary is used to check up the meaning of general words and not for checking if a substance being used is hazardous. Option A is wrong.

MSDS means "Material Safety Data Sheet" and it contains documents with information that relates to occupational health & safety for checking various substances and products. Thus, option B is correct.

SAE stands for Society of Automotive Engineering and their standards pertain to mainly Automobiles. Thus option C is wrong.

EPA guidelines are mainly for checking facility and environmental health and safety compliance. Thus, option D is wrong.

3 0

4 years ago

Select the correct answer.

andre [41]

Answer:

A. energy transformations

Explanation:

8 0

2 years ago