Ask question

Ask question

All categories

3 years ago

11

A 500-km, 500-kV, 60-Hz, uncompensated three-phase line has a positivesequence series impedance. z = 5 0.03 1 + j 0.35 V/km and

a positive sequence shunt admittance y = j4.4*10^26 S/km. Calculate:(a) Zc, (b) (gl), (c) the exact ABCD parameters for this line.

1 answer:

Anni [7]3 years ago

7 0

Answer:

A) 282.34 - j 12.08 Ω

B) 0.0266 + j 0.621 / unit

C)

A = 0.812 < 1.09° per unit

B = 164.6 < 85.42°Ω

C = 2.061 * 10^-3 < 90.32° s

D = 0.812 < 1.09° per unit

Explanation:

Given data :

Z ( impedance ) = 0.03 i + j 0.35 Ω/km

positive sequence shunt admittance ( Y ) = j4.4*10^-6 S/km

A) calculate Zc

Zc = $\sqrt{\frac{z}{y} }$ = $\sqrt{\frac{0.03 i + j 0.35}{j4.4*10^-6 } }$

= $\sqrt{79837.128< 4.899^o}$ = 282.6 < -2.45°

hence Zc = 282.34 - j 12.08 Ω

B) Calculate gl

gl = $\sqrt{zy} * d$

d = 500

z = 0.03 i + j 0.35

y = j4.4*10^-6 S/km

gl = $\sqrt{0.03 i + j 0.35* j4.4*10^-6} * 500$

= $\sqrt{1.5456*10^{-6} < 175.1^0} * 500$

= 0.622 < 87.55 °

gl = 0.0266 + j 0.621 / unit

C) exact ABCD parameters for this line

A = cos h (gl) . per unit = 0.812 < 1.09° per unit ( as calculated )

B = Zc sin h (gl) Ω = 164.6 < 85.42°Ω ( as calculated )

C = 1/Zc sin h (gl) s = 2.061 * 10^-3 < 90.32° s ( as calculated )

D = cos h (gl) . per unit = 0.812 < 1.09° per unit ( as calculated )

where : cos h (gl) = $\frac{e^{gl} + e^{-gl} }{2}$

sin h (gl) = $\frac{e^{gl}-e^{-gl} }{2}$

You might be interested in

A square isothermal chip is of width w = 5 mm on a side and is mounted in a substrate such that its side and back surfaces are w

nirvana33 [79]

Answer:

Q(h=200)=0.35W

Q(h=3000)=5.25W

Explanation:

first part h=200W/Km^2

we must use the convection heat transfer equation for the chip

Q=hA(Ts-T∞)

h= convective coefficient=200W/m2 K

A=Base*Leght=5mmx5mm=25mm^2

Ts=temperature of the chip=85C

T∞=temperature of coolant=15C

Q=200x2.5x10^-5(85-15)=0.35W

Second part h=3000W/Km^2

Q=3000x2.5x10^-5(85-15)=5.25W

5 0

3 years ago

In a wheatstone bridge three out of four resistors have of 1K ohm each ,and the fourth resistor equals 1010 ohm. If the battery

Dima020 [189]

Answer:

248.756 mV

49.7265 µA

Explanation:

The Thevenin equivalent source at one terminal of the bridge is ...

voltage: (100 V)(1000/(1000 +1000) = 50 V

impedance: 1000 || 1000 = (1000)(1000)/(1000 +1000) = 500 Ω

The Thevenin equivalent source at the other terminal of the bridge is ...

voltage = (100 V)(1010/(1000 +1010) = 100(101/201) ≈ 50 50/201 V

impedance: 1000 || 1010 = (1000)(1010)/(1000 +1010) = 502 98/201 Ω

__

The open-circuit voltage is the difference between these terminal voltages:

(50 50/201) -(50) = 50/201 V ≈ 0.248756 V . . . . open-circuit voltage

__

The current that would flow is given by the open-circuit voltage divided by the sum of the source resistance and the load resistance:

(50/201 V)/(500 +502 98/201 +4000) = 1/20110 A ≈ 49.7265 µA

8 0

3 years ago

Truckco manufactures two types of trucks: 1 and 2. Each truck must go through the painting shop and assembly shop. If the painti

o-na [289]

He started work at 16 for the North West Company and then the Hudson's Bay Company, becoming a high-ranking officer. From 1851 to 1864, he was Governor of the Colony of Vancouver Island

5 0

3 years ago

Read 2 more answers

Carbon dioxide at 20°C flows in a pipe at a rate of 0.005 kg/s. Determine the minimum diameter required if the flow is laminar (

Vesna [10]

Answer:

the required diameter is 0.344 m

Explanation:

given data:

flow is laminar

flow of carbon dioxide Q = 0.005 Kg/s

for flow to be laminar, Reynold's number must be less than 2300 for pipe flow and it is given as

$\frac{\rho VD}{\mu }$

arrange above equation for diameter

\frac{\rho Q D}{\mu A }<2300

dynamic density of carbon dioxide = 1.47× $10^{-5}$ Pa sec

density of carbon dioxide is 1.83 kg/m³

$\frac{1.83\times 0.0056\times D}{1.47\times 10^{-5}\times \frac{\pi}{4} \times D^{2} }$

$\frac{1.83\times 0.0056}{1.47\times 10^{-5}\times \frac{\pi}{4} \times 2300}= D$

D = 0.344 m

4 0

3 years ago

The top surface of an L = 5mmthick anodized aluminum plate is irradiated with G = 1000 W/m2 while being simultaneously exposed

puteri [66]

Answer:

$J=1963W/m^2$

$q_{rad}=963w/m^2$

$\triangle T= -0.378k/s$

Explanation:

From the question we are told that:

$L=5mm => 5*10^{-3}\\Irradiation G=1000W/m^2\\h=50W/m^2\\T_{infinity} = 25C.\\Plate\ temperature\ T_p= 400 K\\\alpha=0.14\\E=0.76$

at Temp=400K

$E=2702kg/m^2,c=949J/kgk$

Generally the equation for Radiosity is mathematically given by

$J=eG+\in E_p$

$J=(1-\alpha)G+\in \sigma T^4$

$J=(1-0.14)1000+0.76 (5.67*10^_{8}) (400)^4$

$J=1963W/m^2$

Generally the equation for net radiation heat flux $q_{rad}$ is mathematically given by

$q_{rad}=J-G\\q_{rad}=1963-1000$

$q_{rad}=963w/m^2$

Generally the equation for and the rate of plate temp $\triangleT$ is mathematically given by

$\triangle T= -\frac{q_{con} +q_{rad}}{Ecl}$

$\triangle T= \frac{45(400)-(30+273+963)}{(2702*949*0.005)}$

$\triangle T= -0.378k/s$

6 0

3 years ago