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-Dominant- [34]

3 years ago

6

A three-phase source delivers 4.8 kVA to a wye-connected load with a phase voltage of 214 V and a power factor of 0.9 lagging. C

alculate the source line current and the source line voltage.

1 answer:

BlackZzzverrR [31]3 years ago

5 0

Answer:

a) 7.5A

b) 370V

Explanation:

Note that the Line current is the current through any one line between a three-phase source and load.

From the given question, The source line current =7.5A(approximately)

While the Source line voltage = 370V.

Kindly go through the attached file for a step by step detail of how I arrived at these answers.

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Two metallic terminals protrude from a device. The terminal on the left is the positive reference for a voltage called vx (the o

tresset_1 [31]

Answer:

Vx = 6.242 x 10raised to power 15

Vy = -6.242 x 10raised to power 15

Explanation:

from E = IVt

but V = IR from ohm's law and Q = It from faraday's first law

I = Q/t

E = Q/t x V x t = QV

hence, E =QV

V = E/Q

3 0

3 years ago

Read 2 more answers

An astronomer of 65 kg of mass hikes from the beach to the observatory atop the mountain in Mauna Kea, Hawaii (altitude of 4205

lara [203]

Answer:

$0.845\ \text{N}$

Explanation:

g = Acceleration due to gravity at sea level = $9.81\ \text{m/s}^2$

R = Radius of Earth = 6371000 m

h = Altitude of observatory = 4205 m

Change in acceleration due to gravity due to change in altitude is given by

$g_h=g(1+\dfrac{h}{R})^{-2}\\\Rightarrow g_h=9.81\times(1+\dfrac{4205}{6371000})^{-2}\\\Rightarrow g_h=9.797\ \text{m/s}^2$

Weight at sea level

$W=mg\\\Rightarrow W=65\times 9.81\\\Rightarrow W=637.65\ \text{N}$

Weight at the given height

$W_h=mg_h\\\Rightarrow W_h=65\times 9.797\\\Rightarrow W_h=636.805\ \text{N}$

Change in weight $W_h-W=636.805-637.65=-0.845\ \text{N}$

Her weight reduces by $0.845\ \text{N}$ .

8 0

3 years ago

Only an outer panel is being replaced. Technician A says that removing the spot welds by drilling through both panels allows the

Angelina_Jolie [31]

Answer:

6e66363636633747747363637737373737337374

5 0

3 years ago

Describe with an example how corroded structures can lead to environment pollution?

raketka [301]

According to EonCoat, corrosion is the process of decay on a material caused by a chemical reaction with its environment. Corrosion of metal occurs when an exposed surface comes in contact with a gas or liquid, and the process is accelerated by exposure to warm temperature, acids, and salts.” (1)
Although the word ‘corrosion’ is used to describe the decay of metals, all natural and man-made materials are subject to decay, and the level of pollutants in the air can speed up this process.

5 0

3 years ago

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