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UNO [17]
3 years ago
10

A balanced three phase load is supplied over a three-phase , 60 hz, transmission line with each line have a series impedance of

(12.84 j72.76) ohms, the load voltage is 132 kV and the load power is 55 MVA at 0.8 lagging power factor. Determine: a. The ABCD parameters assuming a short line
Engineering
1 answer:
posledela3 years ago
8 0

Answer:

Explanation:

Given a three-phase system

Frequency f=60Hz

Line impedance Z= 12.84 + j72.76 Ω

Then,

The resistance is R=12.84Ω

And reactance is X=72.76Ω

Z=√(12.84²+72.76²)

Z=73.88

Angle = arctan(X/R)

Angle = arctan(72.76/12.84)

Angle=80°

Then, Z=73.88 < 80° ohms

Load voltage is 132 kV

Load power P=55 MWA

Power factor =0.8lagging

the relation between the sending and receiving end specifications are given using ABCD parameters by the equations below.

Vs = AVr + BIr

Is = CVr + DIr

Where

Vs is sending Voltage

Vr is receiving Voltage

Is is sending current

Ir is receiving current

A is ratio of source voltage to received voltage A=Vs/Vr when Ir=0

B is short circuit resistance

B= Vs/Ir when Vr=0

C is ratio of source current to received voltage C=Is/Vr when Ir=0

D is ratio of source current to received current D=Is/Ir when Vr=0

Now,

The load at 55MVA at 132kV (line to line)

Therefore, load current is

Ir= P/V√3

Ir=55×10^6/(132×10^3×√3)

Ir=240.56 Amps

It has a power factor 0.8 lagging

PF=Cosθ

0.8=Cosθ

θ=arcCos(0.8)

θ=36.87°

Therefore, Ir=240.56 <-36.87°

Vr=V/√3

Vr=132/√3

Vr=76.21 kV. Phase voltage

Vr= 76210 < 0° V

For series impedance,

Using short line approximation

Vs = Vr + IrZ

Vs = 76210 < 0° + (240.56 <-36.87° × 73.88 < 80°)

Using calculator

Vs=76210<0° + 17772.5728<(-36.87°+80°)

Vs=76210<0° + 17772.5728<43.13°

Vs=89970.67<7.7°

Also

Is = Ir = 240.56 <-36.87° Amps

Therefore, the ABCD parameters is

A=Vs/Vr

A= 89970.67 <7.7° / 76210 <0°

A=1.181 <7.7-0

A=1.18 <7.7° no unit

B = Vs/Ir

B = 89970.67 < 7.7° / 240.56 <-36.87°

B = 347.01 < 7.7+36.87

B= 347.01 < 44.57° Ω

C= Is/Vr = 240.56 <-36.87° / 76210 < 0°

C= 0.003157 <-36.87-0

C= 3.157 ×10^-3 < -36.87° /Ω

C= 3.157 ×10^-3 < -36.87° Ω~¹

D= Is/Ir

Since Is=Ir

Then, D = 1 no unit.

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Step-by-step explanation:

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Zeq = 0.264 + j0.869 Ω

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Req = 0.264 Ω

Xeq = j0.869 Ω

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Zeq2 = (0.264 + j0.869)/10²

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An incompressible fluid flows between two infinite stationary parallel plates. The velocity profile is given by u=umaxðAy2 + By+
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Answer:

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\frac{Q}{b} = \frac{2}{3} u_{max} h

the ratio is : \frac{V}{u_{max}}=\frac{2}{3}

Explanation:

From the question; the  equations of the velocities profile in the system are:

u = u_{max}(Ay^2+By+C)   ----- equation (1)

The above boundary condition can now be written as :

At y= 0; u =0           ----- (a)

At y = h; u =0            -----(b)

At y = \frac{h}{2} ; u = u_{max}     ------(c)

where ;

A,B and C are constant

h = distance between two plates

u = velocity

u_{max} = maximum velocity

y = measured distance upward from the lower plate

Replacing the boundary condition in (a) into equation (1) ; we have:

u = u_{max}(Ay^2+By+C) \\ \\ 0 = u_{max}(A*0+B*0+C) \\ \\ 0=u_{max}C \\ \\ C= 0

Replacing the boundary condition (b) in equation (1); we have:

u = u_{max}(Ay^2+By+C) \\ \\ 0 = u_{max}(A*h^2+B*h+C) \\ \\ 0 = Ah^2 +Bh + C \\ \\ 0 = Ah^2 +Bh + 0 \\ \\ Bh = - Ah^2 \\ \\ B = - Ah   \ \ \ \ \   --- (d)

Replacing the boundary condition (c) in equation (1); we have:

u = u_{max}(Ay^2+By+C) \\ \\ u_{max}= u_{max}(A*(\frac{h^2}{2})+B*\frac{h}{2}+C) \\ \\ 1 = \frac{Ah^2}{4} +B \frac{h}{2} + 0 \\ \\ 1 =  \frac{Ah^2}{4} + \frac{h}{2}(-Ah)  \\ \\ 1=  \frac{Ah^2}{4}  - \frac{Ah^2}{2}  \\ \\ 1 = \frac{Ah^2 - Ah^2}{4}  \\ \\ A = -\frac{4}{h^2}

replacing A = -\frac{4}{h^2} for A in (d); we get:

B = - ( -\frac{4}{h^2})hB = \frac{4}{h}

replacing the values of A, B and C into the velocity profile expression; we have:

u = u_{max}(Ay^2+By+C) \\ \\ u = u_{max} (-\frac{4}{h^2}y^2+\frac{4}{h}y)

To determine the volume flow rate; we have:

Q = AV \\ \\ Q= \int\limits^h_0 (u.bdy)

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\frac{Q}{b} = \int\limits^h_0 u_{max}(-\frac{4}{h^2} y^2+\frac{4}{h}y)dy \\ \\  \frac{Q}{b} = u_{max}  \int\limits^h_0 (-\frac{4}{h^2} y^2+\frac{4}{h}y)dy \\ \\ \frac{Q}{b} = u_{max} (-\frac{-4}{h^2}\frac{y^3}{3} +\frac{4}{h}\frac{y^2}{y})^ ^ h}}__0  }} \\ \\ \frac{Q}{b} =u_{max} (-\frac{-4}{h^2}\frac{h^3}{3} +\frac{4}{h}\frac{h^2}{y})^ ^ h}}__0  }} \\ \\ \frac{Q}{b} = u_{max}(\frac{-4h}{3}+\frac{4h}2} ) \\ \\ \frac{Q}{b} = u_{max}(\frac{-8h+12h}{6}) \\ \\ \frac{Q}{b} =u_{max}(\frac{4h}{6})

\frac{Q}{b} = u_{max}(\frac{2h}{3}) \\ \\ \frac{Q}{b} = \frac{2}{3} u_{max} h

Thus; the volume flow rate per unit depth is:

\frac{Q}{b} = \frac{2}{3} u_{max} h

Consider the discharge ;

Q = VA

where :

A = bh

Q = Vbh

\frac{Q}{b}= Vh

Also;  \frac{Q}{b} = \frac{2}{3} u_{max} h

Then;

\frac{2}{3} u_{max} h = Vh \\ \\ \frac{V}{u_{max}}=\frac{2}{3}

Thus; the ratio is : \frac{V}{u_{max}}=\frac{2}{3}

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