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IrinaK [193]
2 years ago
8

Scheduling can best be defined as the process used to determine:​

Engineering
1 answer:
shutvik [7]2 years ago
8 0

Answer:

Overall project duration

Explanation:

Scheduling can best be defined as the process used to determine a overall project duration.

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technician A says that voltage is electric pressure .Technician B says the amperage is electrical flow who is right
Dmitrij [34]

Answer:

I would say technician A

Explanation:

Definition of voltage- An electromotive force or potential difference expressed in volts.

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Read 2 more answers
1. A loss of braking effectiveness due to too much heat in the brakes is called
motikmotik

Answer:

brake fade. loss of brake effectiveness due to overheating.

Explanation:

8 0
1 year ago
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+11.586j 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} =48.115-42.433j 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

8 0
3 years ago
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