electrical engineering

Re-dispatch

Re-dispatch

If the constraints of upper or lower power limits are violated for any generator, the power is tied to the corresponding limit as a fixed ...

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Electrical Safety in Medium Voltage

<div dir="ltr" style="text-align: left;" trbidi="on"><div class="wp-caption aligncenter" id="attachment_1507" style="width: 394px;"><a href="http://hayatei.blogspot.com/"><img alt="MV switch disconnector" class="size-full wp-image-1507" height="137" src="http://engineering.electrical-equipment.org/wp-content/uploads/2009/10/switch-disconnector-aid.JPG" title="switch disconnector aid" width="384" /></a><div class="wp-caption-text">MV switch disconnector</div></div><a href="http://www.blogger.com/goog_336343801"><span id="more-1490"></span></a><br />
Last June, I presented a paper at the CIRED conference with a title “Medium Voltage Disconnector s and Safety”… and a main message “MV disconnector s are not electrical safety devices”. The topic could be seen as controversial, but it actually raised some real concern about how the electrical safety could be addressed.<br />
The attendants were a bit surprised by the message, as the confusion is very common. It is probably linked to the fact that, for LV installations, disconnection is set in electrical specifications as a safety function, with relevant requirements and acceptance criteria. However, it is not the case for devices above 1kV, because the earthing of the circuit is always required before any work could be performed on the conductors (except if live working procedures are applied).<br />
When coming to electrical specification, our industry has an “ancient tendency”, a bad habit, which is to forget about the basic analysis of the application. Some speak about the “copy-paste” syndrome…<br />
Such a basic analysis is required for an effective risk assessment, and then the proper definition of the protection measures to be applied. Using the word “safety” associated with a device which is not specified, nor used, with a global understanding of the application leads to an over-confidence which could be dangerous.<br />
Safety is a system concern, and can’t be limited to using any so-called “safety device”.<br />
Some discussion raised around my presentation<br />
<a href="http://hayatei.blogspot.com/"><img alt="http://www7.0zz0.com/2010/12/13/12/570937245.jpg" src="http://www7.0zz0.com/2010/12/13/12/570937245.jpg" /></a><br />
As most situations involving an operator also imply to have one side earthed, the open gap is in such cases not over-stressed, compared to any other insulation in the installation.<br />
However, a special situation has been highlighted by a participant, about the cable test operation; in such a situation, the disconnector is actually used with two live circuits on its sides, and operators are possibly in the vicinity of the cables under test… In this case, the claim is that the disconnector becomes a “safety function”.<br />
The fact is that such situation is not properly taken into consideration when specifying the disconnectors, because dielectic withstand under such situation is not defined nor assessed, and the possible failure mode is unknown.<br />
Field operation has shown that, during such operations, available switchgear is safe, that the risk is low – Safety is defined as a tolerable risk level – but nevertheless, the need is not correctly analysed, and the requirements expressed for the disconnector s have no link with this cable test situation.<br />
Keeping an electrical specification based on wrong assumptions could lead to dangerous situations, for which the electrical specification could be met, but the expected service would be impaired.<br />
And trying to define safety rules on a device, without linking them to the way the device is installed and used, is not enough to achieve a safe operation. So, when the specification use the words “safety device”, the consequences of an over-focus, or over-confidence, could become critical.<br />
Electrical risk does exist. It is better to understand it properly, in order to mitigate it and to prevent as far as possible the possible consequences.<br />
We should train specifiers and operators of any electrical installation for a better understanding of the functions, performances and limitations of the available devices, and also to a better knowledge of the safety analysis ideas.<br />
<br />
<a href="http://engineering.electrical-equipment.org/electrical-distribution/safety-in-mv.html">source here  </a></div>

Electrical Safety in Medium Voltage

MV switch disconnector Last June, I presented a paper at the CIRED conference with a title “Medium Voltage Disconnector s and Safety”… and ...

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High Voltage: circuit breaker is the best!

<div dir="ltr" style="text-align: left;" trbidi="on"><br />
<a href="http://engineering.electrical-equipment.org/wp-content/uploads/2010/05/ringmaster_exterieur_small.jpg" imageanchor="1"><img alt="ringmaster outdoor" border="0" class="alignright size-full wp-image-2363" height="300" src="http://engineering.electrical-equipment.org/wp-content/uploads/2010/05/ringmaster_exterieur_small.jpg" title="ringmaster_exterieur_small" width="225" /></a><br />
<br />
Every time I hear or read something about high voltage fuses, I’m convinced that15 years ago in the UK, we made the right choice by moving away from high voltage fuses to circuit-breakers.<br />
If we take the first argument I read in a previous article, that a fuse-switch is a cost effective solution, this is in fact a myth. In the past it was true due to the physical size of the old circuit-breakers.  Now, compact RMUs with dedicated circuit breakers and self-powered relays have the same life time costs as fuse-switch solutions and have far more useful features.<br />
<span id="more-2340"></span><br />
The main, or should I say only, technical argument for the fuse claiming high fault clearance speed is biased.  It is extremely unlikely that a 3-phase short circuit would occur inside the transformer without forewarning by a low magnitude fault, for example an inter-turn fault.  In this example a fuse would be unable to clear the fault, whereas a circuit-breaker works perfectly on something that is similar to an overload.  This is even truer in other countries, on networks having either isolated or resonant neutral systems.  Also, even on high magnitude faults, circuit-breakers function well, clearing faults quickly enough.<br />
Self-powered relays are simple and easy to set.  Also, settings can be changed as the installation evolves.  This removes doubt over the selection of the proper fuse when replacement is needed either after a fault or for other reasons.<br />
Upstream and downstream coordination is much easier with circuit-breakers.  Network management is better facilitated.  In case of a fault on the network, the overall time, including fault finding up to re-energising is reduced with circuit-breakers compared to the use of fuse-switch solutions.<br />
More arguments exist but, I think it is enough to understand why there is a trend towards the use of circuit-breakers instead of fuse-switches.<br />
Richard Douglas</div>

High Voltage: circuit breaker is the best!

Every time I hear or read something about high voltage fuses, I’m convinced that15 years ago in the UK, we made the right choice by movin...

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Operating cost of thermal plants

<div dir="ltr" style="text-align: left;" trbidi="on"><div style="color: red;"><span style="font-size: small;">According to the input-output curve of thermal power plant,</span></div><div style="color: red;"><br />
</div><div style="color: red;"><span style="font-size: small;"> the cost function is simplified in a quadrate form:</span></div><div style="color: red;"><span style="font-size: small;"><br />
C = α + β*P + γ*P2</span></div><div style="color: red;"><span style="font-size: small;"><br />
This relation gives a good approximation to the actual </span></div><div style="color: red;"><br />
</div><div style="color: red;"><span style="font-size: small;">variation of the cost with the variation of the output power</span></div><div style="color: red;"><br />
</div><br />
<span style="color: red; font-size: small;">Economic Dispatch Neglecting Losses</span><br />
<div style="color: red;"><br />
</div><div style="color: red;"><span style="font-size: small;">It is assumed that all generators are connected at the same bus without losses in transmission </span></div><div style="color: red;"><br />
</div><div style="color: red;"><span style="font-size: small;">lines The economic dispatch problem neglecting all losses in the transmission<br />
<br />
system is defined </span></div><div style="color: red;"><span style="font-size: small;"><br />
</span></div><div style="color: red;"><span style="font-size: small;"><br />
</span></div><div style="color: red;"><span style="font-size: small;">Steps of obtaining the solution:</span></div><div style="color: red;"><span style="font-size: small;"><br />
</span></div><div style="color: red;"><span style="font-size: small;"><br />
</span></div><div style="color: red;"><span style="font-size: small;">Define the coefficients of the cost functions of all<br />
generating units (β i and γg g i, i=1,2,…n) , , )</span></div><div style="color: red;"><span style="font-size: small;"><br />
• For a certain load demand, calculate the value of λ</span></div><div style="color: red;"><span style="font-size: small;"><br />
• Depending on the calculated value of λ, compute the individual power P1, P2, ... Pn</span></div><div style="color: red;"><span style="font-size: small;"><br />
corresponding to incremental fuel cost of production</span></div><div style="color: red;"><span style="font-size: small;"><br />
• Check for max. and min. output power for each plant. In case of violating the generation limits, </span></div><div style="color: red;"><br />
</div><div style="color: red;"><span style="font-size: small;">the output power of that generator is fixed at its extreme limit</span></div><div style="color: red;"><span style="font-size: small;"><br />
• The remaining load demand is calculated and distributed among the remaining generators</span></div><div style="color: red;"><span style="font-size: small;"><br />
</span></div><div style="color: red;"><span style="font-size: small;">Graphical solution</span></div><div style="color: red;"><span style="font-size: small;"><br />
Graphically, the Lagrange multiplier λ can be defined by plotting the value of β + 2γ*Pfor all generators on the same graph</span></div><div style="color: red;"><span style="font-size: small;"><br />
A random value of γ is selected and the corresponding generated power from each unit is obtained</span></div><div style="color: red;"><span style="font-size: small;"><br />
The total power is compared with the total demand If the demand exceeds the generated power, the line representing the Lagrange multiplier λ is shifted up to increase the power and vice versa</span></div><div style="color: red;"><span style="font-size: small;"><br />
</span></div><div style="color: red;"><span style="font-size: small;">Graphically, the Lagrange multiplier λ can be defined by plotting the value of β + 2γ*Pfor all </span></div><div style="color: red;"><br />
</div><div style="color: red;"><span style="font-size: small;">generators on the same graph</span></div><div style="color: red;"><br />
</div><div style="color: red;"><span style="font-size: small;"> A random value of λ is selected and the corresponding generated power from each unit is obtained</span></div><div style="color: red;"><br />
</div><div style="color: red;"><span style="font-size: small;"> Check for max. and min. output power for each plant. In case of violating the generation limits, </span></div><div style="color: red;"><span style="font-size: small;">the output power of that generator is fixed at its extreme limit</span></div><div style="color: red;"><br />
</div><div style="color: red;"><span style="font-size: small;"> The total power is compared with the total demand</span></div><div style="color: red;"><br />
</div><div style="color: red;"><span style="font-size: small;"> If the demand exceeds the generated power, the line representing the Lagrange multiplier λ is<br />
shifted up to increase the power and vice versa</span></div><div style="color: red;"><br />
</div><div style="color: red;"><br />
</div></div>

Operating cost of thermal plants

According to the input-output curve of thermal power plant, the cost function is simplified in a quadrate form: C = α + β*P + γ*P2 This ...

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Economic Dispatch Problem part 2

<div dir="ltr" style="text-align: left;" trbidi="on"><div class="post-body entry-content"> <div dir="ltr" style="text-align: left;"><div style="color: red;"><span style="font-size: medium;"><a href="http://hayatei.blogspot.com/2010/12/economic-dispatch-problem_13.html">Economic Dispatch Problem part 1</a></span></div><div style="color: red;"><span style="font-size: medium;"><br />
</span></div><div style="color: red;"><span style="font-size: medium;">The incremental heat rate “the heat rate” is the ratio of input </span></div><div style="color: red;"><br />
</div><div style="color: red;"><span style="font-size: medium;">fuel to the corresponding output power</span></div><div style="color: red;"><span style="font-size: medium;"><br />
</span></div><div style="color: red;"><span style="font-size: medium;">From the incremental heat rate curve, the incremental fuel</span></div><div style="color: red;"><br />
</div><div style="color: red;"><span style="font-size: medium;"> rate is deduced With F is the input fuel and P is the output </span></div><div style="color: red;"><br />
</div><div style="color: red;"><span style="font-size: medium;">power The reciprocal of the incremental fuel rate is known as </span></div><div style="color: red;"><br />
</div><div style="color: red;"><span style="font-size: medium;">the incremental efficiency</span></div><div style="color: red;"><span style="font-size: medium;"><br />
To operate at the maximum fuel efficiency, the point of the</span></div><div style="color: red;"><br />
</div><div style="color: red;"><span style="font-size: medium;"> minimum heat rate has to be defined </span></div><div style="color: red;"><br />
</div><div style="color: red;"><span style="font-size: medium;">Multiplying the incremental fuel rate by the fuel cost, the</span></div><div style="color: red;"><br />
</div><div style="color: red;"><span style="font-size: medium;"> incremental fuel cost is obtained in ($/MWh)</span></div><div style="color: red;"><span style="font-size: medium;"><br />
</span></div><div style="color: red;"><span style="font-size: medium;"><br />
</span></div><div style="color: red;"><span style="font-size: medium;">Example</span></div><div style="color: red;"><span style="font-size: medium;"><br />
The input fuel in (Btu/h) for a power plant with minimum and maximum power of 10 and 100 MW</span></div><div style="color: red;"><span style="font-size: medium;"><br />
respectively is given by:</span></div><div style="color: red;"><span style="font-size: medium;"><br />
F=(40+4*P + 0.012*P2)*106</span></div><div style="color: red;"><span style="font-size: medium;"><br />
where P is the generated power in (MW). Plot the input </span></div><div style="color: red;"><br />
</div><div style="color: red;"><span style="font-size: medium;">output curve of the plant. Calculate the heat rate and plot its</span></div><div style="color: red;"><br />
</div><div style="color: red;"><span style="font-size: medium;"> curve against the output power.</span></div><div style="color: red;"><span style="font-size: medium;"><br />
Assuming a fuel cost of 0.12*10-6 $/Btu, calculate</span></div><div style="color: red;"><span style="font-size: medium;"><br />
the incremental fuel cost in $/MWh and plot its</span></div><div style="color: red;"><span style="font-size: medium;"><br />
curve against the output power.</span></div><div style="color: red;"><span style="font-size: medium;"><br />
</span></div><div style="color: red;"><span style="font-size: medium;"><br />
</span></div><div style="color: red;"><span style="font-size: medium;">The fuel is given as:</span></div><div style="color: red;"><span style="font-size: medium;"><br />
F=(40+4*P + 0.012*P2)*106</span></div><div style="color: red;"><span style="font-size: medium;"><br />
Also, the heat rate can be calculated from equation</span></div><div style="color: red;"><span style="font-size: medium;"><br />
input fuel (Btu/h)</span></div><div style="color: red;"><span style="font-size: medium;"><br />
output power (MW)</span></div><div style="color: red;"><span style="font-size: medium;"><br />
Heat rate = The fuel cost as a function of the output power is:</span></div><div style="color: red;"><span style="font-size: medium;"><br />
C=0.12*10-6 *F= 4.8 + 0.48*P + 0.0014*P2 ($/h)</span></div><div style="color: red;"><span style="font-size: medium;"><br />
The incremental fuel cost is:</span></div><div style="color: red;"><span style="font-size: medium;"><br />
Ifr = 0.48 + 0.0028*P ($/MWh)<br />
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Economic Dispatch Problem part 2

Economic Dispatch Problem part 1 The incremental heat rate “the heat rate” is the ratio of input fuel to the corresponding output power ...

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Economic Dispatch Problem

<div dir="ltr" style="text-align: left;" trbidi="on"><span style="color: red; font-size: medium;">Economic load dispatch concerns with the operating cost rather than the fixed cost</span><br />
<span style="color: red; font-size: medium;"><br />
Only fuel cost is considered in the study where all other costs that are depend on</span><br />
<br />
<span style="color: red; font-size: medium;"> the generated power will be included in the expression of the fuel cost</span><br />
<span style="color: red; font-size: medium;"><br />
Obviously, the cost of fuel is concerned since thermal plants are assumed</span><br />
<span style="color: red; font-size: medium;"><br />
An early approach of power dispatch was to supply power from the most efficient </span><br />
<br />
<span style="color: red; font-size: medium;">plant till the point of maximum efficiency and then from the next most efficient plant</span><br />
<br />
<span style="color: red; font-size: medium;"> and so on</span><br />
<br />
<span style="font-size: medium;"><span style="color: blue;">Another approach was to load the machines so that</span></span><br />
<br />
<span style="color: red; font-size: medium;">all units have the same incremental cost of production</span><br />
<span style="color: red; font-size: medium;"><br />
However, the locations of power plants and the trans-mission losses are not</span><br />
<br />
<span style="color: red; font-size: medium;"> considered Generating power in far plants requires supplying the transmission</span><br />
<br />
<span style="color: red; font-size: medium;"> losses in addition to the load demand</span><br />
<span style="color: red; font-size: medium;"><br />
Economic distribution of load demand among generating units necessitates</span><br />
<br />
<span style="color: red; font-size: medium;"> expressing the cost as a function of the generated power</span><br />
<span style="color: red; font-size: medium;"><br />
The performance curve of boiler-turbine-generator set is required, which is called</span><br />
<br />
<span style="color: red; font-size: medium;"> input-output curve</span></div>

Economic Dispatch Problem

Economic load dispatch concerns with the operating cost rather than the fixed cost Only fuel cost is considered in the study where all oth...

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Economic Dispatch Problem

<div dir="ltr" style="text-align: left;" trbidi="on"><span style="color: red; font-size: large;">Economic load dispatch concerns with the operating<br />
cost rather than the fixed cost<br />
Only fuel cost is considered in the study where all<br />
other costs that are depend on the generated power<br />
will be included in the expression of the fuel cost<br />
Obviously, the cost of fuel is concerned since thermal<br />
plants are assumed<br />
An early approach of power dispatch was to supply<br />
power from the most efficient plant till the point of<br />
maximum efficiency and then from the next most<br />
efficient plant and so on</span><br />
<br />
<span style="font-size: large;"><span style="color: blue;">Another approach was to load the machines so that</span></span><br />
<br />
<span style="color: red; font-size: large;">all units have the same incremental cost of<br />
production<br />
However, the locations of power plants and the<br />
trans-mission losses are not considered<br />
Generating power in far plants requires supplying the<br />
transmission losses in addition to the load demand<br />
Economic distribution of load demand among<br />
generating units necessitates expressing the cost as<br />
a function of the generated power<br />
The performance curve of boiler-turbine-generator<br />
set is required, which is called input-output curve</span><div style="color: red;"><br />
</div></div>

Economic Dispatch Problem

Economic load dispatch concerns with the operating cost rather than the fixed cost Only fuel cost is considered in the study where all ot...

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