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2 years ago

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A three-phase, 60-Hz, completely transposed 345-kV, 200-km line has two 795,000-cmil (403-mm2) 26/2 ACSR conductors per bundle a

nd the following positive-sequence line constants: z 0.032 + 10.35 /km y j4.2 x 10-6 S/km Full load at the receiving end of the line is 700 MW at 0.99 p.f. leading and at 95% of rated voltage. Assuming a medium-length line, determine the following:
a. ABCD parameters of the nominal π circuit
b. Sending-end voltage Vs, current Is, and real power Ps
c. Percent voltage regulation
d. Transmission-line efficiency at full load

1 answer:

Svetllana [295]2 years ago

7 0

Answer:

B) sending end voltage : Vs-l-l = 345.8 ∠ 26.14⁰ kv

sending end current : Is = 1.241 ∠ 15.5⁰ KA

real power = 730.5 Mw

C) percent voltage regulation = 8.7%

D) Transmission line efficiency = 95.8%

Explanation:

attached is the detailed solution to the problem

Given data:

l = 200 km

z = 0.032 + j0.35 Ω/km

y = j4.2 * 10^-6 S/km

A) find the total series impedance and shunt admittance

B) sending end voltage : Vs-l-l = 345.8 ∠ 26.14⁰ kv

sending end current : Is = 1.241 ∠ 15.5⁰ KA

real power = 730.5 Mw

C) percent voltage regulation = 8.7%

D) Transmission line efficiency = 95.8%

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A converging-diverging nozzle is designed to operate with an exit Mach number of 1.75 . The nozzle is supplied from an air reser

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Answer:

a. 4.279 MPa

b. 3.198 MPa to 4.279 MPa

c. 0.939 MPa

d. Below 3.198 MPa

Explanation:

From the given parameters

M $_{exit}$ = 1.75 MPa

M at 1.6 MPa gives A $_{exit}$ /A* = 1.2502

M at 1.8 MPa gives A $_{exit}$ /A* = 1.4390

Therefore, by interpolation, we have M $_{exit}$ = 1.75 MPa gives A

However, we shall use M $_{exit}$ = 1.75 MPa and A

Similarly,

P $_{exit}$ /P₀ = 0.1878

a) Where the nozzle is choked at the throat there is subsonic flow in the following diverging part of the nozzle. From tables, we have

A $_{exit}$ /A* = 1.387. by interpolation M

Therefore P $_{exit}$ = P₀ × P

Which shows that the nozzle is choked for back pressures lower than 4.279 MPa

b) Where there is a normal shock at the exit of the nozzle, we have;

M₁ = 1.75 MPa, P₁ = 0.1878 × 5 = 0.939 MPa

Where the normal shock is at M₁ = 1.75 MPa, P₂/P₁ = 3.406

Where the normal shock occurs at the nozzle exit, we have

$P_b = 3.406\times 0.939 = 3.198 MPa$

Where the shock occurs t the section prior to the nozzle exit from the throat, the back pressure was derived as $P_b$ = 4.279 MPa

Therefore the back pressure value ranges from 3.198 MPa to 4.279 MPa

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2 years ago

Laminar flow normally persists on a smooth flat plate until a critical Reynolds number value is reached. However, the flow can b

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Answer:

At L = 0.1 m

h⁻_lam = 11.004K W/m^1.5

h⁻_turb = 7.8848K W/m^1.8

At L = 1 m

h⁻_lam = 3.48K W/m^1.5

h⁻_turb = 4.975K W/m^1.8

Explanation:

Given that;

h_lam(x)= 1.74 W/m^1.5. Kx^-0.5

h_turb(x)= 3.98 W/m^1.8 Kx^-0.2

conditions for plates of length L = 0.1 m and 1 m

Now

Average heat transfer coefficient is expressed as;

h⁻ = 1/L ₀∫^L hxdx

so for Laminar flow

h_lam(x)= 1.74 . Kx^-0.5 W/m^1.5

from the expression

h⁻_lam = 1/L ₀∫^L 1.74 . Kx^-0.5 dx

= 1.74k / L { [x^(-0.5+1)] / [-0.5 + 1 ]}₀^L

= 1.74k/L = [ (x^0.5)/0.5)]⁰^L

= 1.74K × L^0.5 / L × 0.5

h⁻_lam= 3.48KL^-0.5

For turbulent flow

h_turb(x)= 3.98. Kx^-0.2 W/m^1.8

form the expression

1/L ₀∫^L 3.98 . Kx^-0.2 dx

= 3.98k / L { [x^(-0.2+1)] / [-0.2 + 1 ]}₀^L

= (3.98K/L) × (L^0.8 / 0.8)

h⁻_turb = 4.975KL^-0.2

Now at L = 0.1 m

h⁻_lam = 3.48KL^-0.5 = 3.48K(0.1)^-0.5 W/m^1.5

h⁻_lam = 11.004K W/m^1.5

h⁻_turb = 4.975KL^-0.2 = 4.975K(0.1)^-0.2

h⁻_turb = 7.8848K W/m^1.8

At L = 1 m

h⁻_lam = 3.48KL^-0.5 = 3.48K(1)^-0.5 W/m^1.5

h⁻_lam = 3.48K W/m^1.5

h⁻_turb = 4.975KL^-0.2 = 4.975K(1)^-0.2

h⁻_turb = 4.975K W/m^1.8

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At L = 0.1 m

h⁻_lam = 11.004K W/m^1.5

h⁻_turb = 7.8848K W/m^1.8

At L = 1 m

h⁻_lam = 3.48K W/m^1.5

h⁻_turb = 4.975K W/m^1.8

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Answer:

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3. The software lifetime and delivery schedule.

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A thermal energy storage unit consists of a large rectangular channel, which is well insulated on its outer surface and encloses

yaroslaw [1]

Answer:

the temperature of the aluminum at this time is 456.25° C

Explanation:

Given that:

width w of the aluminium slab = 0.05 m

the initial temperature $T_1$ = 25° C

$T{\infty} =600^0C$

h = 100 W/m²

The properties of Aluminium at temperature of 600° C by considering the conditions for which the storage unit is charged; we have ;

density ρ = 2702 kg/m³

thermal conductivity k = 231 W/m.K

Specific heat c = 1033 J/Kg.K

Let's first find the Biot Number Bi which can be expressed by the equation:

$Bi = \dfrac{hL_c}{k} \\ \\ Bi = \dfrac{h \dfrac{w}{2}}{k}$

$Bi = \dfrac{hL_c}{k} \\ \\ Bi = \dfrac{100 \times \dfrac{0.05}{2}}{231}$

$Bi = \dfrac{2.5}{231}$

Bi = 0.0108

The time constant value $\tau_t$ is :

$\tau_t = \dfrac{pL_cc}{h} \\ \\ \tau_t = \dfrac{p \dfrac{w}{2}c}{h}$

$\tau_t = \dfrac{2702* \dfrac{0.05}{2}*1033}{100}$

$\tau_t = \dfrac{2702* 0.025*1033}{100}$

$\tau_t = 697.79$

Considering Lumped capacitance analysis since value for Bi is less than 1

Then;

$Q= (pVc)\theta_1 [1-e^{\dfrac {-t}{ \tau_1}}]$

where;

$Q = -\Delta E _{st}$ which correlates with the change in the internal energy of the solid.

So;

$Q= (pVc)\theta_1 [1-e^{\dfrac {-t}{ \tau_1}}]= -\Delta E _{st}$

The maximum value for the change in the internal energy of the solid is :

$(pVc)\theta_1 = -\Delta E _{st}max$

By equating the two previous equation together ; we have:

$\dfrac{-\Delta E _{st}}{\Delta E _{st}{max}}= \dfrac{ (pVc)\theta_1 [1-e^{\dfrac {-t}{ \tau_1}}]} { (pVc)\theta_1}$

Similarly; we need to understand that the ratio of the energy storage to the maximum possible energy storage = 0.75

Thus;

$0.75= [1-e^{\dfrac {-t}{ \tau_1}}]}$

So;

$0.75= [1-e^{\dfrac {-t}{ 697.79}}]}$

$1-0.75= [e^{\dfrac {-t}{ 697.79}}]}$

$0.25 = e^{\dfrac {-t}{ 697.79}}$

$In(0.25) = {\dfrac {-t}{ 697.79}}$

$-1.386294361= \dfrac{-t}{697.79}$

t = 1.386294361 × 697.79

t = 967.34 s

Finally; the temperature of Aluminium is determined as follows;

$\dfrac{T - T _{\infty}}{T_1-T_{\infty}}= e ^ {\dfrac{-t}{\tau_t}}$

$\dfrac{T - 600}{25-600}= e ^ {\dfrac{-967.34}{697.79}$

$\dfrac{T - 600}{25-600}= 0.25$

$\dfrac{T - 600}{-575}= 0.25$

T - 600 = -575 × 0.25

T - 600 = -143.75

T = -143.75 + 600

T = 456.25° C

Hence; the temperature of the aluminum at this time is 456.25° C

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