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Usimov [2.4K]
3 years ago
15

. A roadway is being designed capable of allowing 70 mph vehicle speed. The superelevation around one curve is 0.05 inches per i

nch and the coefficient of side friction is 0.09. (a) Determine the minimum radius of the curve to the vehicle to provide safe travel. (b) Suppose the radius of the curve to the center of the vehicle is only 1500 feet. What speed limit will need to be posted for safe travel through the curve
Engineering
1 answer:
adelina 88 [10]3 years ago
6 0

The minimum radius of the curve to the vehicle to provide safe travel is 2339.2 ft

<u>Explanation:</u>

Given -

Speed allowed, v = 70mph

1 mph = 1.467 fps (feet per second)

Therefore, v = 70 X 1.467 fps

Superelevation, e = 0.05 inches/inch

Coefficient of side friction, fs = 0.09

Minimum radius of the curve, r = ?

We know,

r = v² / g ( fs + e/100 )

where,

g = acceleration due to gravity in feet

g = 32.2 lb.ft / lbf.s²

r = ( 70 X 1.467 )² / 32.2 ( 0.09 + 0.05 )

r = 2339.2 ft

Therefore, the minimum radius of the curve to the vehicle to provide safe travel is 2339.2 ft

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3 years ago
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
Anni [7]

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 } }    

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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

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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}

     

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2 years ago
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G_{B}^{S S D}=\frac{B}{B-C}

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