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3 years ago

10

PLEASE HELP! I'm in the middle of a test and the teacher didn't go over the material!

1 answer:

stepan [7]3 years ago

3 0

Answer:

1. 4/d and 1

2. Engineering tech with 3, Engineer with 1, and Scientist with 2

Explanation:

I'm pretty sure but tell me if im wrong

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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

What do you think will happen to the demand curve if there is a sudden surplus for a certain commodity?

schepotkina [342]

Answer:

the curve will flatten due to the supply and demand theory.

7 0

2 years ago

When in regular operation, ash and other debris should be removed _______ from the combustion chamber of a wood-burning heater.

Klio2033 [76]

Answer:

Daily from the combustion chamber of a wood-burning heater.

<h3>Explanation:</h3>

As a wood stove heats up, it radiates heat through the walls and top of the stove
This radiant heat warms the immediate area and can be carried into other parts of the home via the home's natural airflow.
Electric or convection-powered fans can help circulate this heat to warm a larger area.

To learn more about it, refer

to brainly.com/question/23275071

#SPJ4

8 0

1 year ago

When a man returns to his well-sealed house on a summer day, he finds that the house is at 35°C. He turns on the air conditioner

ipn [44]

Answer:

$\dot W = 1.667\, kW$

Explanation:

A well-sealed house means that there is no mass interaction between air indoors and outdoors. Hence, cooling process is isochoric. The heat removed by the air conditioner is:

$\dot Q_{L} = \frac{m_{air}}{\Delta t} \cdot c_{v, air} \cdot (T_{o}-T_{f})\\\dot Q_{L} = \frac{800\, kg}{(30\, min)\cdot (\frac{60\, s}{1 \, min} )}\cdot (0.7 \frac{kJ}{kg \cdot K}) \cdot (15\, K)\\\dot Q_{L} = 4.667\, kW$

The power drawn by the air conditioner is:

$\dot W = \frac{\dot Q_{L}}{COP_{R}} \\\dot W = \frac {4.667\, kW}{2.8}\\\dot W = 1.667\, kW$

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3 years ago

4 grams of a saturated liquid are converted to a saturated vapor by being heated in a weighted piston–cylinder device arranged t

antiseptic1488 [7]

Answer:T2= 351.6k

Explanation:

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3 years ago