Differentiate between open loop and close loop control system with suitable example

Differentiate between open loop and close loop control system with suitable example
Basis For
Open Loop System
Closed Loop System
The system whose control
action is free from the
output is known as the
open loop control system.
In closed loop, the output depends on
the control action of the system.
Other Name
Non-feedback System
Feedback System
Controller and Controlled
Amplifier, Controller, Controlled Process,
Depends on calibration
Accurate because of feedback.
Less Stable
Basis For
Open Loop System
Closed Loop System
Traffic light, automatic
washing machine,
immersion rod, TV remote
Air conditioner, temperature control
system, speed and pressure control
system, refrigerator, toaster.
Open Loop Control System
In open loop control system, the output does not affect the control action of the
system. In other words, the system whose working depends on time is known as
the open loop control system. The open loop system is free from the feedback.
Let’s understand this with the help of the few examples.
Closed Loop Control System
The closed-loop control system means the output of the system depends on their
input. The system has one or more feedback loops between its output and input.
The closed-loop system design in such a way that they automatically provide the
desired output by comparing it with the actual input. The closed-loop system
generates the error signal which is the difference between the input and output.
1. If 3% of the electric bulbs manufactured by a company are defective, find the probability that in
a sample of 100 bulbs (a) 0, (b) 1, (c) 2, (d ) 3, (e) 4, and ( f ) 5 bulbs will be defective.
2. Find the probability that (a) more than 5, (b) between 1 and 3, and (c) less than or equal to 2
bulbs will be defective.
3. A bag contains 1 red and 7 white marbles. A marble is drawn from the bag and its color is
observed. Then the marble is put back into the bag and the contents are thoroughly mixed.
Using (a) the binomial distribution and (b) the Poisson approximation to the binomial
distribution, find the probability that in 8 such drawings a red ball is selected exactly 3 times.
1 . (a) Probability of 0 defective bulb is = P( x ; μ) = (e ^ -μ) (μx) / x! where μ = 3; x = 0; e = 2.71828
so P (0 ;3) = 0.0497
(b) P (1 ; 3) = 0.1493
(c) P (2; 3) = 0.2240
(d) P ( 3;3) = 0.2240
(e) P (4;3) = 0.1680
(f) P ( 5,3) = 0.1008
2. (a) P (>5 ; 3) = 0.083
(b) P ( 1< x < 3 ;3) = 0.2241
(c) P ( <= 2 ; 3) = 0.4231
3. (a) by bionomical distribution = 8C3 x (0.125 ) ^ 3 x (0,875) ^ 5 = 0.056
(b) by Poisson distribution = 0.06031...
Bus Admittance Matrix
In a power system, Bus Admittance Matrix represents the nodal admittances of the various buses. With the
help of the transmission line, each bus is connected to the various other buses. Admittance matrix is used to
analyse the data that is needed in the load or a power flow study of the buses. It explains the admittance and
the topology of the network. The following are the advantages of the bus admittance matrix.
1. The data preparation of the bus admittance matrix is very simple.
2. The formation of the bus admittance matrix and their modification is easy.
3. The bus admittance matrix is a sparse matrix thus the computer memory requirement is less.
The amount of current present in the bus can be calculated with the help of formation of the Admittance
matrix. It is expressed as shown above.
In the simplest form, the above matrix can be written as shown below.
I is the current of the bus in the vector form.
Y is the admittance matrix
V is the vector of the bus voltage.
Let us consider the figure given below.
From the above figure, the (3×3) admittance matrix is formed as shown below.
The diagonal elements of the Bus Admittance matrix are known as self-admittances and the off-diagonal
elements are known as mutual admittances.
Steps for Solving Bus Admittance Matrix
The following steps given below are used to solve the Admittance Matrix.
First of all, form the bus Admittance matrix.
Select the reference bus to solve the network.
Define the known variables for all the other types of buses.
Assign the initial values for the voltage and angle for all the buses.
Calculate the power mismatch vector and power injection current.
Apply the various iteration methods like Newton-Raphson, Gauss-Siedel etc.
Check the mismatching vector that whether it is within the prescribed limit of 0.001 per unit. If yes, then stop
the procedure and if no then continues the steps of iteration to obtain the new values.
Recheck the values again, whether the obtain values are within the limit.
Load Flow Study
The flow of electrical power in any interconnected electrical system is termed as Load Flow. The Load Flow
Study is conducted to calculate the voltages at the various buses. It is also used for the study of the shortcircuit conditions or any interconnected power system. The load flow study is essential for the operation of
the power system under an operating condition, its improvement and future expansion.
Voltage magnitude | Vi |
Voltage phase angle δi
Active power Pi
Reactive volt-amperes Qi
The load flow study determining the best location, as well as the optimal capacity of the generating station,
substation and the new lines. The following variables given above are associated with each bus. In the load
flow studies in power systems, three types of buses are identified. In each bus, two variables are known, and
the other two are to be calculated. The buses are classified as follows:Swing Bus
The swing bus is the first one to respond to a changing load condition. The voltage magnitude and phase
angle are specified in this type of bus.
Generator Bus
In this type of bus, voltage magnitude and active power are given, and the phase angle and voltage are to be
Load Bus
In this bus, active power and reactive volt amperes are specified. Buses with neither generator nor load may
be considered as load buses. If any bus in a power system has both load and generator, then the load is
treated as the negative generation.
One of the generator buses is selected as the reference bus for the various reasons such as – The losses in
the system remain unknown until the load flow solution is complete. Hence, one of the generator buses is
made to take the additional real and reactive powers to supply the losses. This bus is, therefore, known as
Swing bus. The voltage throughout the system should be close to 1. Hence the voltage on the slack bus is
assigned to 1 per unit.
The voltage of the slack bus is taken as the reference, and thus its angle is equal to zero. The bus connected
to the largest generating station is selected as the slack bus. This bus is numbered as bus 1. In load flow
problem the single phase representation with positive sequence network is used since power system is
usually balanced under normal condition of operation.
The load flow problem has the following solution.
Consider some values of all the unknown values. It is good to set all the voltage angle to zero value and all
the voltage magnitude to 1 per unit value.
Solve the load flow equations using all the values of voltages, angle and magnitude.
Solve and update the changes of the values of the voltage and magnitude.
Stopping condition is checked, if fulfilled, then the process is stopped otherwise repeat the steps again.
The mathematical calculation of the load flow equation is very tedious and time consuming
Newton Raphson Method
Newton Raphson Method is an iterative technique for solving a set of various nonlinear equations with an
equal number of unknowns. There are two methods of solutions for the load flow using Newton Raphson
Method. The first method uses rectangular coordinates for the variables while the second method uses the
polar coordinate form. Out of these two methods the polar coordinate form is used widely.
Let us understand this method with the help of the equations.
The above equation (3) and (4) can also be written as shown below.
We have Δf = J ΔX
Then I = 1, 2, ….n, I ≠ slack, and if
Then I = 1, 2, ….n,
i ≠ slack,
i ≠ PV bus
Where, the subscripts sp and cal denote the specified and calculated values, respectively, then the equation
(7) can be written as shown below.
The off diagonal and diagonal elements of the sub matrices H, N, M and L are determined by differentiating
equation (3) and (4) with respect to δ and |V|.
Procedure of Newton Raphson Method
The computational procedure for Newton Raphson Method using polar coordinate is given below.
Form Y bus.
Assume the initial value of the bus voltages |Vi|0 and phase angle δi0 for i = 2, 3, …..n for load buses and
phase angles for PV buses. Normally we set the assumed bus voltage magnitude and its phase angle equal
to the slack bus quantities |V1| = 1.0, δ1 = 0⁰.
Compute Pi and Qi for each load bus from the following equation (5) and (6) shown above.
Now, compute the scheduled errors ΔPi and ΔQi for each load bus from the following relations given below.
For PV buses, the exact value of Qi is not specified, but its limits are known. If the calculated value of Qi is
within the limits only ΔPi is calculated. If the calculated value of Qi is beyond the limits, then an appropriate
limit is imposed and ΔQi is also calculated by subtracting the calculated value of Qi from the appropriate limit.
The bus under consideration is now treated as a load bus.
Compute the elements of the Jacobian matrix.
Obtain the value of Δδ and Δ|Vi| from the equation shown below.
Using the values of Δδi and Δ|Vi| calculated in the above step, modify the voltage magnitude and phase angle
at all load buses by the equations shown below.
Start the next iteration cycle following the step 2 with the modified values of |Vi|and δi.
Continue until scheduled errors for all the load buses are within a specified tolerance that is
Where, ε denotes the tolerance level for load buses.
Calculate the line and power flow at the slack bus same as in the Gauss Seidel method.
Advantages of Newton Raphson Method
The various advantages of Newton Raphson Method are as follows:-
It possesses quadratic convergence characteristics. Therefore, the convergence is very fast.
The number of iterations is independent of the size of the system. Solutions to a high accuracy is obtained
nearly always in two to three iterations for both small and large systems.
The Newton Raphson Method convergence is not sensitive to the choice of slack bus.
Overall, there is a saving in computation time since fewer number of iterations are required.
Limitations of Newton Raphson Method
The various limitations are given below.
This solution technique is difficult.
It takes longer time as the elements of the Jacobian are to be computed for each iteration.
The computer memory requirement is large.
Flowchart For Newton Raphson Method
The flowchart is drawn below for Newton Raphson Method using polar coordinates for load flow solutions.
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