Answer:
the generator induced voltage is 60.59 kV
Explanation:
Given:
S = 150 MVA
Vline = 24 kV = 24000 V

the network voltage phase is

the power transmitted is equal to:

the line induced voltage is

Answer:
The work furnished by the compressor is 
The minimum work required for the state to change is 
Explanation:
The explanation to these solution is on the first, second , third and fourth uploaded image respectively
Answer:
The graph representing the linear inequalities is attached below.
Explanation:
The inequalities given are :
y>x-2 and y<x+1
For tables for values of x and y and get coordinates to plot for both equation.
In the first equation;
y>x-2
y=x-2
y-x = -2
The table will be :
x y
-2 -4
-1 -3
0 -2
1 -1
2 0
The coordinates to plot are : (-2,-4) , (-1,-3), (0,-2), (1,-1) ,(2,0)
Use a dotted line and shade the part right hand side of the line.
Do the same for the second inequality equation and plot then shade the part satisfying the inequality.
The graph attached shows results.
Answer:
The impedance of the circuit depends on the angular frequency of the voltage source.
Explanation:
- In a electric circuit, the magnitude of the impedance, is given by the following expression:

where R = Resistance
Xl = Inductive reactance = ω*L
Xc = Capacitive Reactance = 1/ωC
and ω = angular frequency of the voltage source.
- So, it can be seen that the impedance depends on the value of the constants R,L and C, and on the angular frequency ω.
Given Information:
Output power required = Pout = 2.80 MW
Efficiency = η = 30%
Intensity = I = 1180 W/m²
Required Information:
Effective area = A = ?
Answer:
Effective area = A = 7.907x10³ m²
Step-by-step explanation:
A community plans to build a facility to convert solar power into electrical power and this facility has an efficiency of 30%
As we know efficiency is given by
η = Pout/Pin
Where Pout is the output power and Pin is the input power.
Pin = Pout/η
Pin = 2.80x10⁶/0.30
Pin = 9.33x10⁶ W
The effective area of a perfectly absorbing surface used in such an installation can be found using
A = Pin/I
Where I is the in Intensity of the sunlight in W/m²
A = 9.33x10⁶/1180
A = 7.907x10³ m²
Therefore, the effective area of the absorbing surface would be 7.907x10³ m².