Given, F in VLD t. 3H
Protection System
A system that protects electrical systems from faults by isolating the problematic part from the remainder of the system, preventing power from being cut from healthy elements, improving system dependability and efficiency is the protection system. Protection devices are the equipment that are utilized to implement the protection system.
Predictive Maintenance System
Predictive maintenance technologies are designed to assist in determining the state of in-service equipment so that maintenance can be scheduled. Predictive maintenance is the application of information; proactive maintenance approaches examine the condition of equipment and anticipate when it should maintain. The purpose of predictive maintenance is to forecast when equipment will fail (depending on a variety of parameters), then prevent the failure through routine and corrective maintenance.Condition monitoring is the continual monitoring of machines during process conditions to maintain optimal machine use, which is necessary for predictive maintenance. There are three types of condition monitoring: online, periodic, and remote. Finally, remote condition monitoring allows the equipment observed from a small place and data supplied for analysis.
Preventive Maintenance System
To maintain the equipment and materials on a regular basis in order to maintain those running conditions and reduce unnecessary shutdowns due to unexpected equipment failure is called Preventive Maintenance (PM).
Instructions: use the handout to answer this in a step by step process use the handout to give a full solution in the given
![Inductor
Сараcitor
dvc
ic = C
Resistor
di̟
VR = iRR
VL = L
dt
dt
Example 1: Find the differential equations that
describe the mesh currents iį and iz in the circuit.
Step 1. Write the mesh equations for the
circuit.
Mesh 1: 8i, + 16 S„(i – iz)dt = v(t)
+ 10iz + 16 SL.(iz – 1)di = 0
8Ω
1H
diz
Mesh 2:
dt
1
v(t)
i2
$102
16
Step 2. Differentiate the mesh equations to Step 3. Then using operators, let D = d/dt.
eliminate the integrals.
Substitution gives:
di
d
(8D + 16) iį – 16i2 = Dv(t)
-16i, + (D² + 10D + 16)i, = 0 (2)
(1)
8
+ 16i
- 16iz =
v(t)
dt
dt
d²iz
+ 10 diz
dt
-16i, +
+ 16iz
= 0
dt2
Step 4. Using the elimination method, multiply We get:
equation (1) by 16 and equation (2) by (8D + 16),
then add the resulting equations. This will eliminate Which simplifies to
(8D3 + 96D² + 288D) iz
= 16Dv(t)
the current variable i1.
(D² + 12D + 36) iz
The differential equation for the current iz is
d²iz
dt2
= 2v(t)
( (8D + 16) i – 16i, = Dv(t)
(8D + 16) l–16i1 + (D² + 10D + 16)iz =
16
diz
+ 12
+ 36 iz = 2v(t)
dt
Step 5. Similarly, if we multiply equation (2) by 16 and equation (1) by (D2 +10D + 16), then add
the resulting equations, we will eliminate the current variable i2.
We get:
(8D3 + 96D² + 288D) i = D(D² + 10D + 16)v(t)
which simplifies to:
(D² + 12D + 36) i =
1
(D² + 10D + 16)v(t)
The differential equation for the current i, is:
d²i,
d²v(t)
dv(t)
+ 2v(t)
di
+ 12
+ 36 i, = 0.125
+ 1.25
dt2
dt
dt2
dt](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6b011a6a-089c-4042-a8e4-44afadcb759f%2F61cab5d6-5d54-45dc-93c1-29b4f4ef2689%2Fxk0z01l_processed.jpeg&w=3840&q=75)
![Given,
F
VLD
t,](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6b011a6a-089c-4042-a8e4-44afadcb759f%2F61cab5d6-5d54-45dc-93c1-29b4f4ef2689%2F037oksc_processed.jpeg&w=3840&q=75)
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