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3 years ago

14

Considet a three-phase transmission line operating at the sending end voltage of 500kV. Theseries impedance is z= 0.045+j0.4Ωper

phase per km and the shunt admittance is y= j4×10-6Siemens per phase per km. For the following cases, evaluate transmission matrix for shortlength, medium length and long length models, and find the receiving end voltage (per phase)at no load using all three models. What can you conclude from this exercise?
a)When the length of the transmission line is 50 km
b)When the length of the transmission line is 100 km

1 answer:

Roman55 [17]3 years ago

4 0

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An engineer must design a rectangular box that has a volume of 9 m3 and that has a bottom whose length is twice its width. What

Jet001 [13]

Answer:

$Length =3$ $Height = 2$ and $Width = \frac{3}{2}$

Explanation:

Given

$Volume = 9m^3$

Represent the height as h, the length as l and the width as w.

From the question:

$Length = 2 * Width$

$l = 2w$

Volume of a box is calculated as:

$V = l*w*h$

This gives:

$V = 2w *w*h$

$V = 2w^2h$

Substitute 9 for V

$9 = 2w^2h$

Make h the subject:

$h = \frac{9}{2w^2}$

The surface area is calculated as:

$A = 2(lw + lh + hw)$

Recall that: $l = 2w$

$A = 2(2w*w + 2w*h + hw)$

$A = 2(2w^2 + 2wh + hw)$

$A = 2(2w^2 + 3wh)$

$A = 4w^2 + 6wh$

Recall that: $h = \frac{9}{2w^2}$

So:

$A = 4w^2 + 6w * \frac{9}{2w^2}$

$A = 4w^2 + 6* \frac{9}{2w}$

$A = 4w^2 + \frac{6* 9}{2w}$

$A = 4w^2 + \frac{3* 9}{w}$

$A = 4w^2 + \frac{27}{w}$

To minimize the surface area, we have to differentiate with respect to w

$A' = 8w - 27w^{-2}$

Set A' to 0

$0 = 8w - 27w^{-2}$

Add $27w^{-2}$ to both sides

$27w^{-2} = 8w$

Multiply both sides by $w^2$

$27w^{-2}*w^2 = 8w*w^2$

$27 = 8w^3$

Make $w^3$ the subject

$w^3 = \frac{27}{8}$

Solve for w

$w = \sqrt[3]{\frac{27}{8}}$

$w = \frac{3}{2}$

Recall that : $h = \frac{9}{2w^2}$ and $l = 2w$

$h = \frac{9}{2 * \frac{3}{2}^2}$

$h = \frac{9}{2 * \frac{9}{4}}$

$h = \frac{9}{\frac{9}{2}}$

$h = 9/\frac{9}{2}$

$h = 9*\frac{2}{9}$

$h= 2$

$l = 2w$

$l = 2 * \frac{3}{2}$

$l = 3$

Hence, the dimension that minimizes the surface area is:

$Length =3$ $Height = 2$ and $Width = \frac{3}{2}$

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3 years ago

An incompressible viscous fluid flows through a pipe with a flow rate of 1 mL/s. The pipe has a uniform diameter D0 and a length

Nataly [62]

Answer:

$Q_2 = 32$ mL/s

Explanation:

Given :

The flow is incompressible viscous flow.

The initial flow rate, $Q_1$ = 1 mL/s

Initial diameter, $D_1= D_0$

Initial length, $L_1=L_0$

The initial pressure difference to maintain the flow, $P_1=P_0$

We know for a viscous flow,

$\Delta P = \frac{32 \mu V L}{D^2}$

$\Delta P = \frac{32 \mu Q L}{\frac{\pi}{4}D^4}$

$Q \propto \Delta P \times D^4$

$\frac{Q_1}{Q_2}= \frac{P_1}{P_2} \times \left( \frac{D_1}{D_2} \right)^4$

$\frac{1}{Q_2}= \frac{P_0}{2P_0} \times \left( \frac{D_0}{2D_0} \right)^4$

$\frac{1}{Q_2}= \frac{1}{2} \times \left( \frac{1}{2} \right)^4$

$\frac{1}{Q_2}= \frac{1}{32}$

∴ $Q_2 = 32$ mL/s

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3 years ago

Sea water with a density of 1025 kg/m3 flows steadily through a pump at 0.21 m3 /s. The pump inlet is 0.25 m in diameter. At the

myrzilka [38]

Answer:

$\dot W_{pump} = 16264.922\,W\,(16.265\,kW)$

Explanation:

The pump is modelled after applying Principle of Energy Conservation, whose form is:

$\frac{P_{1}}{\rho\cdot g}+ \frac{v_{1}^{2}}{2\cdot g} +z_{1} + h_{pump}=\frac{P_{2}}{\rho\cdot g}+ \frac{v_{2}^{2}}{2\cdot g} +z_{2}$

The head associated with the pump is cleared:

$h_{pump} = \frac{P_{2}-P_{1}}{\rho\cdot g}+\frac{v_{2}^{2}-v_{1}^{2}}{2\cdot g}+(z_{2}-z_{1})$

Inlet and outlet velocities are found:

$v_{1} = \frac{0.21\,\frac{m^{3}}{s} }{\frac{\pi}{4}\cdot (0.25\,m)^{2} }$

$v_{1} \approx 4.278\,\frac{m}{s}$

$v_{2} = \frac{0.21\,\frac{m^{3}}{s} }{\frac{\pi}{4}\cdot (0.152\,m)^{2} }$

$v_{2} \approx 11.573\,\frac{m}{s}$

Now, the head associated with the pump is finally computed:

$h_{pump} = \frac{175\,kPa-81.326\,kPa}{(1025\,\frac{kg}{m^{3}} )\cdot (9.807\,\frac{m}{s^{2}} )} +\frac{(11.573\,\frac{m}{s} )^{2}-(4.278\,\frac{m}{s} )^{2}}{2\cdot (9.807\,\frac{m}{s^{2}} )} + 1.8\,m$

$h_{pump} = 7.705\,m$

The power that pump adds to the fluid is:

$\dot W_{pump} = \dot V \cdot \rho \cdot g \cdot h_{pump}$

$\dot W_{pump} = (0.21\,m^{3})\cdot (1025\,\frac{kg}{m^{3}})\cdot (9.807\,\frac{m}{s^{2}})\cdot(7.705\,m)$

$\dot W_{pump} = 16264.922\,W\,(16.265\,kW)$

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3 years ago

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Answer:

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Explanation:

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Answer:

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