Answer:
The three phase full load secondary amperage is 2775.7 A
Explanation:
Following data is given,
S = Apparent Power = 1000 kVA
No. of phases = 3
Secondary Voltage: 208 V/120 V <em>(Here 208 V is three phase voltage and 120 V is single phase voltage) </em>
<em>Since,</em>
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The formula for apparent power in three phase system is given as:
Where:
S = Apparent Power
V = Line Voltage
I = Line Current
In order to calculate the Current on Secondary Side, substituting values in above formula,
Answer:
The MATLAB Code for this PI Controller will be:
Kp = 350;
Ki = 300;
Kd = 50;
C = pid(Kp,Ki,Kd)
T = feedback(C*P,1);
t = 0:0.01:2;
step(T,t)
Explanation:
When you are designing a PID controller for a given system, follow the steps shown below to obtain a desired response.
Obtain an open-loop response and determine what needs to be improved
Add a proportional control to improve the rise time
Add a derivative control to reduce the overshoot
Add an integral control to reduce the steady-state error
Adjust each of the gains $K_p$, $K_i$, and $K_d$ until you obtain a desired overall response.
The further explanation is attached in the Word File.
Answer:
It studies the process of technological change. Under the field of Technology Dynamics the process of technological change is explained by taking into account influences from "internal factors" as well as from "external factors
Explanation:
Answer:
I'm completely sure that the answer is: The most important rating for batteries is the ampere-hour rating. Ampere-hour is the battery discharge rating. It's used as a measure of charge in your device. It indicates how long your device will work without charging.
Explanation:
Hope this helped!
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) = 
