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Arte-miy333 [17]

2 years ago

9

(a) In a 3-phase, 4-wire system, the currents are in the A, B, and C lines under abnormal conditions of loading were as follows:

IA = 100/30° A, 1b = 50/300° A, and Ic = 30/180° A
(i) Calculate the positive, negative, and zero-phase sequence currents in the A-line.

(ii) Calculate the return current in the neutral conductor.

(b) Check by sketching IA, IB, and Ic as the sum of the appropriate symmetrical components.

1 answer:

Verizon [17]2 years ago

5 0

Answer:

Check the attached image below

Explanation:

Kindly check the attached image below to get the step by step explanation to the question above.

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The wall of drying oven is constructed by sandwiching insulation material of thermal conductivity k = 0.05 W/m°K between thin me

masha68 [24]

Answer:

86 mm

Explanation:

From the attached thermal circuit diagram, equation for i-nodes will be

$\frac {T_ \infty, i-T_{i}}{ R^{"}_{cv, i}} + \frac {T_{o}-T_{i}}{ R^{"}_{cd}} + q_{rad} = 0$ Equation 1

Similarly, the equation for outer node “o” will be

$\frac {T_{ i}-T_{o}}{ R^{"}_{cd}} + \frac {T_{\infty, o} -T_{o}}{ R^{"}_{cv, o}} = 0$ Equation 2

The conventive thermal resistance in i-node will be

$R^{"}_{cv, i}= \frac {1}{h_{i}}= \frac {1}{30}= 0.033 m^{2}K/w$ Equation 3

The conventive hermal resistance per unit area is

$R^{"}_{cv, o}= \frac {1}{h_{o}}= \frac {1}{10}= 0.100 m^{2}K/w$ Equation 4

The conductive thermal resistance per unit area is

$R^{"}_{cd}= \frac {L}{K}= \frac {L}{0.05} m^{2}K/w$ Equation 5

Since $q_{rad}$ is given as 100, $T_{o}$ is 40 $T_ \infty$ is 300 $T_{\infty, o}$ is 25

Substituting the values in equations 3,4 and 5 into equations 1 and 2 we obtain

$\frac {300-T_{i}}{0.033} +\frac {40-T_{i}}{L/0.05} +100=0$ Equation 6

$\frac {T_{ i}-40}{L/0.05}+ \frac {25-40}{0.100}=0$

$T_{i}-40= \frac {L}{0.05}*150$

$T_{i}-40=3000L$

$T_{i}=3000L+40$ Equation 7

From equation 6 we can substitute wherever there’s $T_{i}$ with 3000L+40 as seen in equation 7 hence we obtain

$\frac {300- (3000L+40)}{0.033} + \frac {40- (3000L+40)}{L/0.05}+100=0$

The above can be simplified to be

$\frac {260-3000L}{0.033}+ \frac {(-3000L)}{L/0.05}+100=0$

$\frac {260-3000L}{0.033}=50$

-3000L=1.665-260

$L= \frac {-258.33}{-3000}=0.086*10^{-3}m= 86mm$

Therefore, insulation thickness is 86mm

8 0

3 years ago

Two wastewater treatment plant workers (one male and one female) are exposed to hydrogen sulfide in confined spaces in the treat

Vlada [557]

Answer:

Go to explaination for the details of the answer.

Explanation:

In order to determine the lifetime (75 years) chronic daily exposure for each individual, we have to first state the terms of our equation:

CDI = Chronic Daily Intake

C= Chemical concentration

CR= Contact Rate

EFD= Exposure Frequency and Distribution

BW= Body Weight

AT = Average Time.

Having names our variables lets create the equations that will be used to derive our answers.

Please kindly check attachment for details of the answer.

5 0

2 years ago

An 800-kg drag racer accelerates from rest to 390 km/hr in 5.8 s. What is the net impulse applied to the racer in the first 5.8

marissa [1.9K]

Answer:

Impulse =14937.9 N

tangential force =14937.9 N

Explanation:

Given that

Mass of car m= 800 kg

initial velocity u=0

Final velocity v=390 km/hr

Final velocity v=108.3 m/s

So change in linear momentum P= m x v

P= 800 x 108.3

P=86640 kg.m/s

We know that impulse force F= P/t

So F= 86640/5.8 N

F=14937.9 N

Impulse force F= 14937.9 N

We know that

v=u + at

108.3 = 0 + a x 5.8

$a=18.66\ m/s^2$

So tangential force F= m x a

F=18.66 x 800

F=14937.9 N

6 0

3 years ago

Should the ship breaking business continue why or why not?

Dmitry [639]

Answer:

Ship-breaking or ship demolition is a type of ship disposal involving the breaking up of ships for either a source of parts, which can be sold for re-use, or for the extraction of raw materials, chiefly scrap. It may also be known as ship dismantling, ship cracking, or ship recycling. Modern ships have a lifespan of 25 to 30 years before corrosion, metal fatigue and a lack of parts render them uneconomical to operate.[1] Ship-breaking allows the materials from the ship, especially steel, to be recycled and made into new products. This lowers the demand for mined iron ore and reduces energy use in the steel making process. Fixtures and other equipment on board the vessels can also be reused. While ship-breaking is sustainable, there are concerns about the use of poorer countries without stringent environmental legislation. It is also labor-intensive, and considered one of the world's most dangerous industries.[2]

In 2012, roughly 1,250 ocean ships were broken down, and their average age was 26 years.[3][4] In 2013, the world total of demolished ships amounted to 29,052,000 tonnes, 92% of which were demolished in Asia. As of January 2020, India has the largest global share at 30%;[5] followed by Bangladesh, China and Pakistan.[6] Alang, India currently has the world's largest ship graveyard,[5] followed by Chittagong Ship Breaking Yard in Bangladesh and Gadani in Pakistan.[6]

The largest sources of ships are states of China, Greece and Germany respectively, although there is a greater variation in the source of carriers versus their disposal.[7] The ship-breaking yards of India, Bangladesh, China and Pakistan employ 225,000 workers as well as providing many indirect jobs. In Bangladesh, the recycled steel covers 20% of the country's needs and in India it is almost 10%.[8]

As an alternative to ship-breaking, ships may be sunk to create artificial reefs after legally-mandated removal of hazardous materials, or sunk in deep ocean waters. Storage is a viable temporary option, whether on land or afloat, though all ships will be eventually scrapped, sunk, or preserved for museums.

6 0

3 years ago

P10.12. A certain amplifier has an open-circuit voltage gain of unity, an input resistance of and an output resistance of The si

klio [65]

complete question

A certain amplifier has an open-circuit voltage gain of unity, an input resistance of 1 \mathrm{M} \Omega1MΩ and an output resistance of 100 \Omega100Ω The signal source has an internal voltage of 5 V rms and an internal resistance of 100 \mathrm{k} \Omega.100kΩ. The load resistance is 50 \Omega.50Ω. If the signal source is connected to the amplifier input terminals and the load is connected to the output terminals, find the voltage across the load and the power delivered to the load. Next, consider connecting the load directly across the signal source without the amplifier, and again find the load voltage and power. Compare the results. What do you conclude about the usefulness of a unity-gain amplifier in delivering signal power to a load?

Answer:

3.03 V 0.184 W

2.499 mV 125*10^-9 W

Explanation:

First, apply voltage-divider principle to the input circuit: 1

$V_{i}= (R_i/R_i+R_s) *V_s = 10^6/10^6+(0.1*10^6)\\$ *5

= 4.545 V

The voltage produced by the voltage-controlled source is:

A_voc*V_i = 4.545 V

We can find voltage across the load, again by using voltage-divider principle:

V_o = A_voc*V_i*(R_o/R_l+R_o)

= 4.545*(100/100+50)

= 3.03 V

Now we can determine delivered power:

P_L = V_o^2/R_L

= 0.184 W

Apply voltage-divider principle to the circuit:

V_o = (R_o/R_o+R_s)*V_s

= 50/50+100*10^3*5

= 2.499 mV

Now we can determine delivered power:

P_l = V_o^2/R_l

= 125*10^-9 W

Delivered power to the load is significantly higher in case when we used amplifier, so a unity gain amplifier can be useful in situation when we want to deliver more power to the load. It is the same case with the voltage, no matter that we used amplifier with voltage open-circuit gain of unity.

4 0

3 years ago