Ben İngiliz oldum düzelte bilirmiyim

The line voltage of a balanced three-phase transmission line is 4200 V rms. The transmission line has an impedance of 4 6 Z j l

balu736 [363]

Answer:

1.23MVA, 226.74kW, 5.16kV

Explanation:

Parameters Given

line impedance, Zl = (4 + j6) ohms per phase

load voltage, Vl = 4200V

received complex power, S = 1 × 10⁶VA

power factor, cosФ = 0.75

Ф = 41.41°

sinФ = 0.66

Solution:

S = √3 * Vl * I (that is √3 × line voltage × line current)

1 × 10⁶ = √3 × 4200 × I

I = 137.46A

Vl = 4200∠0

I = 137.46∠- 41.41° lagging

source voltage, Vs = load voltage, Vl + voltage drop along the line, Vd

Vd = Zl * I where ( Zl = 4 + j6, = √(4² + 6²)∠tan⁻¹(6/4), = 7.21∠56.31° )

Vd = 7.21∠56.31° × 137.46∠- 41.41°

= 991.22∠14.9°

Vs = Vl + Vd

= 4200∠0° + 991.22∠14.9°

= 4200(cos 0° + j sin 0°) + 991.22(cos 14.9° + jsin 14.9°)

= 4200 + 957.68 + j254.88

= 5157.68 + j254.88

or

= 5163.97∠2.83° V (line voltage at the sending end of the transmission line)

Sending end current, I = 137.46∠-41.41 A

(a) Complex power = √3 × Vs × I

= √3 × 5163.97∠2.83° ×137.46∠-41.41

= 1229477.76∠-38.58°VA

= 1.23∠-38.58MVA

complex power = 1.23MVA

(b) power loss in the three phase line, Pl = 3 × square of line current, I × line impedance, Rl

Pl =3 × I² × Rl where Zl = R + j X = 4 + j6 hence R = 4

= 3 × 137.46² × 4

= 226743.02W

= 226.74kW

(c) from the above, the line voltage at the sending end of the transmission line is = 5163.97V

= 5.16kW

4 0

3 years ago

A student lives in an apartment with a floor area of 60 m2 and ceiling height of 1.8 m. The apartment has a fresh (outdoor) air

USPshnik [31]

Answer:

4

Explanation:

5 0

2 years ago

A fatigue test was conducted in which the mean stress was 46.2 MPa and the stress amplitude was 219 MPa.

sleet_krkn [62]

Answer:

a)σ₁ = 265.2 MPa

b)σ₂ = -172.8 MPa

c) $Stress\ ratio =-0.65$

d)Range = 438 MPa

Explanation:

Given that

Mean stress ,σm= 46.2 MPa

Stress amplitude ,σa= 219 MPa

Lets take

Maximum stress level = σ₁

Minimum stress level =σ₂

The mean stress given as

$\sigma_m=\dfrac{\sigma_1+\sigma_2}{2}$

$2\sigma_m={\sigma_1+\sigma_2}$

2 x 46.2 = σ₁ + σ₂

σ₁ + σ₂ = 92.4 MPa --------1

The amplitude stress given as

$\sigma_a=\dfrac{\sigma_1-\sigma_2}{2}$

$2\sigma_a={\sigma_1-\sigma_2}$

2 x 219 = σ₁ - σ₂

σ₁ - σ₂ = 438 MPa --------2

By adding the above equation

2 σ₁ = 530.4

σ₁ = 265.2 MPa

-σ₂ = 438 -265.2 MPa

σ₂ = -172.8 MPa

Stress ratio

$Stress\ ratio =\dfrac{\sigma_{min}}{\sigma_{max}}$

$Stress\ ratio =\dfrac{-172.8}{265.2}$

$Stress\ ratio =-0.65$

Range = 265.2 MPa - ( -172.8 MPa)

Range = 438 MPa

8 0

3 years ago

Air flows through a 0.25-m-diameter duct. At the inlet the velocity is 300 m/s, and the stagnation temperature is 90°C. If the M

Naddika [18.5K]

Answer:

a. 318.2k

b. 45.2kj

Explanation:

Heat transfer rate to an object is equal to the thermal conductivity of the material the object is made from, multiplied by the surface area in contact, multiplied by the difference in temperature between the two objects, divided by the thickness of the material.

See attachment for detailed analysis

7 0

3 years ago

Hello I need some help with this please. Pick a problem in your school or community that you think could be solved with technolo

bezimeni [28]

Answer:

An AI operated automatic garbage collection system

Explanation:

There is always an issue in my neighbourhood with the garbagemen coming on time so having an automatic system will help in the overall efficiency in the task

7 0

3 years ago

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