Answer:
a) 4160 V
b) 12 kW and 81 kVAR
c) 54 kW and 477 kVAR
Explanation:
1) The phase voltage is given as:

The complex power S is given as:


The line current I is given as:

The phase voltage at the sending end is:

The magnitude of the line voltage at the source end of the line (
b) The Total real and reactive power loss in the line is:

The real power loss is 12000 W = 12 kW
The reactive power loss is 81000 kVAR = 81 kVAR
c) The sending power is:

The Real power delivered by the supply = 54000 W = 54 kW
The Reactive power delivered by the supply = 477000 VAR = 477 kVAR
Answer:
(a) the percent thermal efficiency is 27.94%
(b) the temperature of the cooling water exiting the condenser is 31.118°C
Explanation:
Answer:
119.35 mm
Explanation:
Given:
Inside diameter, d = 100 mm
Tensile load, P = 400 kN
Stress = 120 MPa
let the outside diameter be 'D'
Now,
Stress is given as:
stress = Load × Area
also,
Area of hollow pipe =
or
Area of hollow pipe =
thus,
400 × 10³ N = 120 ×
or
D² = tex]\frac{400\times10^3+30\pi\times10^4}{30\pi}[/tex]
or
D = 119.35 mm
Answer:
The displacement from t = 0 to t = 10 s, is -880 m
Distance is 912 m
Explanation:
. . . . . . . . . . A
integrate above equation we get

from information given in the question we have
t = 1 s, s = -10 m
so distance s will be
-10 = 12 - 1 + C,
C = -21

we know that acceleration is given as
[FROM EQUATION A]
Acceleration at t = 4 s, a(4) = -24 m/s^2
for the displacement from t = 0 to t = 10 s,

the distance the particle travels during this time period:
let v = 0,

t = 2 s
Distance ![= [s(2) - s(0)] + [s(2) - s(10)] = [1\times 2 - 2^3] + [(12\times 2 - 2^3) - (12\times 10 - 10^3)] = 912 m](https://tex.z-dn.net/?f=%3D%20%5Bs%282%29%20-%20s%280%29%5D%20%2B%20%5Bs%282%29%20-%20s%2810%29%5D%20%3D%20%5B1%5Ctimes%202%20-%202%5E3%5D%20%2B%20%5B%2812%5Ctimes%202%20-%202%5E3%29%20-%20%2812%5Ctimes%2010%20-%2010%5E3%29%5D%20%3D%20912%20m)