october 2018 · indore / raipur umesh bhatia 09589344318 [email protected] h.o. and service...
TRANSCRIPT
OCTOBER 2018
Automation
Piyush Patel (Managing Director)Amtech Power Vision
AN ISO 9001 : 2015 COMPANY
Why Lord Krishna is called as Jagadguru?In the last month we celebrated Shri Krishnashtami, the birthday of ourbeloved God Shri Krishna.Why this God in particular is called as Jagadguru, the Guru of wholeuniverse? Well there are many reasons for this, but we will consider oneof them, in this article. It starts with his childhood. Lord Krishna and hisfriends used to catch hold of the Makhan, the white butter, which waskept in a pot, located at a height and was not accessible easily. SoKrishna had found out a technique of reaching the height, by forming akind of pyramid, which is famous as Dahi-handi-pyramid. We all Indians
know about it. But what lessons does it give to the world? It gives very important corporate lessons:1.All cannot be on the top.2.As you rise to the top, the risk also rises.3. Those who are at the bottom, bear the maximum load, so that someone can reach to the top.4. If you want to eat Makhan (want to achieve your goal) you have to reach to the top.5. If you arrange your internal team properly, believe in its strength, then you can achieve your goalwith no dependence on any external resources.No doubt, Shri Krishna who was so visionary in his childhood, later became the Lord, who gaveBhagvad Geeta to the world, which is considered as the ultimate knowledge and universal guidefor day to day life as well as for the ultimate well-being.
Customer’s Voice
Field ProblemCase Study
Motor ControlElectronics Soft-Starter
MODENORM
ESC UP
DOWN ENTER
GROUP
STOP
RESET
FAULT
RUN
OT-iB Series
Electronics Soft-Starter
MODENORM
ESC UP
DOWN ENTER
GROUP
STOP
RESET
FAULT
RUN
OT-iB Series
Time is life.Wasting time
is wastingone’s life.
Customer’s Voice
You question,we answer
Energy saving in Pump
AN ISO 9001 : 2015 COMPANY
With OurNextGen Technology,
Gain Total Control of YourPulp & Paper Mill
CUSTOMER’S VOICE
3
FIELD PROBLEM CASE STUDY
Problem Faced:During Harmonic Study we take the readings at PCC (Point of Common Coupling), for Voltage Distortion and Current Distortion.As perIEEE 519-2014 standard, the I-THD, Total Harmonic Distortion for current is the responsibility of consumer and he needs to keep theTDD – Total Demand Current Distortion within the limits, as defined in Table-1 given below. Now to know this limit, we need to know theI /I ratio (Short circuit current to Demand current ratio) at PCC. Hence it is necessary to know the short circuit capacity of theSC L
transmission line, where PCC is connected. Many a times this information is not available with the consumer. Then how it can be foundout?Table-1: IEEE 519-2014 Standard Current distortion limits
The Solution:Let us see how we can find out short circuit capacity of the transmission line, then we can calculate short circuit current I at PCC.SC
Refer to Figure-1.
PART-1
Transformer secondary impedance including reflectedimpedance from primary
= Z1 x (Vs / Vp) + Zs2
Now Y = Vs / Z22
Hence Z2 = Vs / Y2
Zs = Transformer per unit impedance x Z2
Hence Zs = Z x (Vs / Y)2
Short circuit capacity, at transformer secondary
= Vs / (Transformer secondary impedance)2
= Vs / (Z1 x (Vs / Vp) + Z x (Vs / Y)2 2 2
)
Dividing numerator and denominator by Vs2
= 1 / (Z1 / Vp + Z / Y)2
Now Vp / Z1 = X, hence substituting in above equation,2
= (1 / X + Z / Y)1
We multiply both numerator and denominator, by X (Y / Z)
Hence short circuit capacity, at transformer secondary,Ps = X (Y / Z) / (X + Y / Z) -----------(1)
Now we will express X in terms of Ps, Y and Z.
Ps = X (Y / Z) / (X + Y / Z)
Hence, Ps x (X + Y / Z) = XY / Z
Ps x X - XY / Z = - Ps Y / Z
X( Ps – Y / Z) = -PsY / Z
Hence, X = Ps Y / Z / (Y / Z – Ps)---------(2)
We know Y and Z. Hence if we know Ps, then we can find out X.
X can also be expressed in terms of Pt, transformer shortcircuit capacity, as follows:
From (2), X = Ps Y / Z / (Y / Z-Ps)
Y/Z = (Vs / Z2 ) / Z2
= (Vs /( Z x Zs) / Z2
= Pt , as Vs / Zs = Pt, transformer short circuit capacity2
Hence Y / Z = Pt -----------(3)
Hence, X = ---------(4)Pt Ps
Pt – Ps
We can prove this in another way: From figure-1,
Short circuit impedance on transformer secondary side,
Zsc = Zs + Z1(Vs / Vp )2 2
Hence Z1 = (Vp / Vs ) (Zsc - Zs)2 2
Vp / Z1 = Vp x Vs / (Vp ) (Zsc - Zs)2 2 2 2
=Vs / (Zsc – Zs)2
Dividing both numerator and denominator by Vs2
= 1 / (Zsc - Zs) / Vs2
= 1/ (Zsc / Vs – Zs / Vs )2 2
= 1/ (1 / Ps – 1 / Pt)
Hence X = ------- (5)Pt Ps
Pt – Ps
As Vp / Z1 = X, PCC short circuit capacity, in VA; Ps = Vs / Zsc,2 2
Short circuit capacity, at transformer secondary, in VA;
Pt=Vs / Zs, Transformer short circuit capacity .2
Hence once we know Ps, then we can find out X.
From X we can find out Isc = X / Vp
Now how to find out Ps, we will continue in Part-2 of thisField Problem Case Study, in newsletter.December 2018
Z1 = transmission line impedance on primary side oftransformer TR.Zs = transformer secondary winding impedance.Vp = transformer primary voltageZ2 = transformer output impedance as per it's rated capacityZsc = short circuit impedance on transformer secondary sideVs = transformer secondary voltage
Z = transformer per unit impedanceX = PCC short circuit capacity, in VAY = transformer rating, in VAPs = Short circuit capacity, at transformer secondary, in VAPt = Transformer short circuit capacity, in VAPCC = Point of Common Coupling
3≤h<11I /ISC L11≤h<17 17≤h<23 23≤h<35 35≤h≤50 TDD
Maximum harmonic current distortionin percent of IL
Individual harmonic order (odd harmonics) a,b
2.0<20c 1.0 0.75 0.3 0.15 2.5
3.520<50 1.75 1.25 0.5 0.25 4.0
5.050<100 2.25 2.0 0.75 0.35 6.0
6.0100<1000 2.75 2.5 1.0 0.5 7.5
7.5>1000 3.5 3.0 1.25 0.7 10.0
DOC. NO. CS-270TITLE: ENERGY SAVING IN PUMP
Payback Period Approximately = 7-9 Months
Industry : Paper
Application : umpP
Motor rating : AC Induction Motor, 3 Phase-
75kW :
415V :olt
7014RPM :
50Hz :
130A :mp
Previous system : Motor was running on tar delta starter.s
Problem observed : 1) High maintenance cost.
.2) Energy losses due to flow control by valve
Present system : 5 kW7 AXPERT-EAZY Variable Frequency Drive used.
kW (Consumption) Valve Position
Previous system
(Without drive) : 71 3kW 9 % Open
Present system
(With drive ) : 4 kW 100 % Open6
System block diagram:
Merits of new system : 1. Easy & Smooth Operation.
2. Energy Saving due to flow control by VFD.
Economical analysis :
�Energy consumption without VFD - P1 = kW71
�Energy consumption with VFD - P2 = kW64
�Energy aving er ay ((P - P2) 24) = Unitss p d 1 x 168
�Saving er ay @ Rs. 4/- Unit = Rs. /-p d 672
�Saving er ear (Approx. 300 Days) = Rs. /-p y 2,01,600
75 kW, 3-PHASEINDUCTION MOTOR
AXPERT-EAZY
3-PHASE,
415 V,
AC SUPPLY
75 kW
PUMP
FROM AMTECH POWER
5
CUSTOMER’S VOICE
6
PRODUCT OVERVIEW
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Integrated Automation, Drives &Power Quality
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YOU QUESTION, WE ANSWER
Question:
Why the efficiency of an equipment drops when it is not loaded up to its rated capacity?
Any equipment has two types of losses: Fixed losses and variable losses. Fixed losses are same whether theequipment is giving output as per rated capacity or it is loaded partially and its output is a fraction of its ratedcapacity. Variable losses vary in proportion to the loading of the system. For example in Variable FrequencyDrive the power consumed by cooling fan is a fixed loss, whereas the conduction loss in IGBTs is a variable loss.At partial loads the losses are relatively large in relation to the output delivered as compared with those when theoutput is at its rated capacity.
This is clear from the following calculation. If the fixed losses in the system are 5% of its rated capacity and thevariable losses are 0.03% per percent of output delivered, then its efficiency at rated output:
OutputEfficiency (%) = ---------------------
Output + Losses100
= ------------------------ x 100100 + 5 + 0.03 x 100( )
= 92.6%
50Efficiency at 50% output = ----------------------
50 + 5 + 0.03 x 50( )= 88.5%
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