An existing building is suffering from cracks in the exterior walls. The investigating engineer wants to ensure that the foundat
1 answer:
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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 = =
= = 282.6 < -2.45°
hence Zc = 282.34 - j 12.08 Ω
B) Calculate gl
gl =
d = 500
z = 0.03 i + j 0.35
y = j4.4*10^-6 S/km
gl =
=
= 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) =
sin h (gl) =
Answer:
Option (d) 2 min/veh
Explanation:
Data provided in the question:
Average time required = 60 seconds
Therefore,
The maximum capacity that can be accommodated on the system, μ = 60 veh/hr
Average Arrival rate, λ = 30 vehicles per hour
Now,
The average time spent by the vehicle is given as
⇒
thus,
on substituting the respective values, we get
Average time spent by the vehicle =
or
Average time spent by the vehicle =
or
Average time spent by the vehicle =
or
Average time spent by the vehicle = hr/veh
or
Average time spent by the vehicle = min/veh
[ 1 hour = 60 minutes]
thus,
Average time spent by the vehicle = 2 min/veh
Hence,
Option (d) 2 min/veh
Answer:
Maximum number that can be represented by 13 bits is 8192 Instructions
Explanation:
number of instructions = 1000
number of bits = log(1000) x number of register
= 6 bits
Since the complete instruction must have 32 bits, then
remaining number of bits = 32 - 6 = 236
number of registers in instruction = 2
number of bits per register = 26/2 = 13
Maximum number that can be represented by 13 bits =
= 2¹³ = 8192