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3 years ago

In the engineering design process, testing is the least important step. True False

Engineering

2 answers:

Wittaler [7]3 years ago

7 0

Answer:

false

Explanation:

Romashka-Z-Leto [24]3 years ago

4 0

i think the answer is false

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Give me source code of Simple openGL project. ( without 3D or Animation) simple just.

Ivan

Answer:

Use GitHub or stackoverflow for this answer

Explanation:

It helps with programming a lot

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3 years ago

What is the diffrence between a small block and a big block lets see if yall know

ivann1987 [24]

Answer:

The difference in weight and size?

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It explains itself :P

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2 years ago

Suppose you have two boxes in front of you. One box contains a Thevenin Equivalent (voltage source in series with a resistor) an

fomenos

Answer:

1. Measure the temperature of the boxes and leave them unconnected.

2. Norton reduces his circuit down to a single resistance in parallel with a constant current source. A real-life Norton equivalent circuit would be continuously wasting power (as heat) as the current source dumps energy into the resistor, even when externally unconnected, while a Thevenin equivalent circuit would sit there doing nothing.

3. The Norton equivalent box would get warm and eventually run out of power. The Thevenin equivalent box would stay at ambient temperature.

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3 years ago

Which of the following statements are true concerning AC circuit that contains both resistance and inductance? A. The current an

maw [93]

The current will lag the voltage in AC circuit that contains both resistance and inductance.

Answer: C

Explanation

There is no inductance only circuits in reality.

The circuits containing inductance has also a lower amount of resistance.

The current flows in both resistance and inductance.

There is a drop in the total voltage in resistance and inductance giving rise to the voltage applied in the coil when connected in a series.

An example being inductance coil an AC circuit connected to both resistance and inductance in series.

From the vector diagram, this conclusion can be drawn.

3 0

3 years 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)

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

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

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