|May12-05, 11:24 AM||#1|
One of the waste products of many nuclear power plants is heat. They send it towers in the form of steam and some have dumped heat into nearby rivers.
The basic problem is that when you use steam heated by a nuclear reactor to power a turbine, the steam says fairly hot while no longer being very useful for spinning the turbine and generating electricity.
One solution to this is co-generation, i.e. using steam generated by a reactor first to power a turbine and generate electricity, and then using the waste heat to, for example, heat buildings. At its best, it is highly efficient.
Does anyone know if any nuclear co-generation plants exist on a commercial basis, and if so where and how their track record has been? Are any in the U.S.? Why hasn't this approach been pursued more often? Are there technological barriers?
|May12-05, 12:08 PM||#2|
The steam, which is generated in the core of a BWR, or in a steam-generator in a PWR, is passed through a high pressure turbine and then intermediate and low pressure turbines. The much cooler wet (low quality) steam is sent to the condenser to be condensed to a 'sub-cooled' liquid (single) phase in order to be pumped back to the core (BWR) or steam generator (PWR).
There are safety, as well as economic, issues with regard to additional systems in a nuclear plant. Most nuclear plants are located far from populated areas, such that district heating or process heating would not be practical. Midland was one of the exceptions. Another plant, Shoreham, was considered too close to populated areas to allow operation. There is tremendous political pressure to close the Indian Point plant (2 PWRs, ~1900 MWe), because of it's proximity to heavily populated areas, which were not so populated when they were built.
Let me see if I can dig up some specs on the T and P of a typical turbine train, particularly the exit temperature of the low pressure turbine.
|May12-05, 01:13 PM||#3|
To add something to what Astro has just said, I think that heat exhausted in a common nuclear power plant has a low thermal level (low T).
As far as I know, it has been made attempts to use cogeneration concept in a Rankine Cycle by means of a counter-pressure turbine. I mean, the turbine would expand the steam to higher pressure levels than usual. So the condenser will work at higher pressure and temperature levels ([tex]T\sim 150ºC[/tex]) and therefore the heat exhausted could be recovered for domestic or industrial issues. But you should know that the overall efficiency [tex]\eta=W/Q_h[/tex]of the power plant will decrease. It would produce less electric power.
Storing or transporting heat in form of thermal energy is also a great problem. As Astronuc has said, the transport of heat from a nuclear power plant far away out of the city to inside the city would be a great challenge. Therefore Cogeneration is mostly used in domestic or industrial heat engines, where transporting heat is not so difficult.
Here in Spain there is an special state Law which gives economic incentives to those who employs cogeneration systems. Electrical companies must buy by this Law the electricity produced with small cogeneration systems. I would buy a heat engine and use it as a cogeneration system. If I proof to the state I am producing also available heat, I always would have an electrical company knocking my door to buy my MWatts generated. I don't know how is this in USA.
|May12-05, 05:30 PM||#4|
As the others have said; that's NOT steam coming out of the cooling
towers. Steam isn't visible. It's water vapor which is visible.
The water that is sent to the cooling tower is NOT the water in the
steam cycle that goes through the turbine. After flowing through the
turbine, the steam cycle water goes to the condensor to be condensed
back into water [ which also puts a vacuum on outlet side of the turbine].
The water on the other side of the condensor - the water that absorbs
the latent heat of the steam cycle water is what is sent to the cooling
tower to be cooled. This temperature of this water is not greatly
above ambient temperature. That's why the cooling towers have to
be so big. [ That and there is a LOT of that water passing through the
condensor - after all it has to take away the lion's share of the energy
produced by the reactor. ]
As I recall, there may be a power plant in Sweden, if memory serves, that
uses the waste heat as heating for the commuity.
Michigan's Consumers Power [ now Consumers Energy ] planned a twin
unit reactor plant at Midland Michigan back in the 1970s. The output
of one of the reactors was to supply process steam for the nearby
Dow Chemical plant. Midland is listed on:
Although Unit 1 was to be a twin to Unit 2; it's electrical generating
capacity is shown to be smaller than Unit 2 because process steam for
Dow was to be siphoned off of Unit 1.
Midland was held up by the anti-nukes; Dow needed its steam, pulled out
of the deal and built a fossil plant; and the whole idea was scrapped.
Dr. Gregory Greenman
|May12-05, 07:07 PM||#5|
Thanks all. Yes, I see that I was sloppy in saying that it was the steam from the turbine that would be used, as opposed merely to the excess heat.
Something like the Midland project, using co-generation from a nuclear plant to generate heat for an industrial process (as opposed to residential) would seem to make sense to me.
Also, when one says that efficiency goes down, I would assume that this means only electrical generation efficiency, I would think that a co-generation plant's aggregate efficiency when both the electrical energy and the heat energy put to productive use are combined would be significantly greater than a non-co-generation plant.
|May13-05, 09:14 AM||#6|
It's not really fair to lump the electrical energy and the use of the waste
heat together and define an efficiency that is meaningful.
It's really an "apples and oranges" comparison. Electrical energy has a
high thermodynamic quality - it has zero entropy. The "waste" heat that
is used in co-generation has low thermodynamic quality - it has finite
So lumping these two together as the numerator of some efficiency
ratio doesn't make sense thermodynamically.
The other problem with using the waste heat from a nuclear power
plant is that they tend to be sited far away from either residential
or industrial centers for the most part. [ There are exceptions, like
Consolidated Edison's Indian Point Plant. ]
Therefore, one would have to pipe the waste heat for tens of miles
to a residential or industrial center where it can be used; which is why
it's not practical. You have to take your choice; do you want the
nuclear plants co-located with the population in order to make
effective use of the waste heat, or do you want the nuclear plant
Dr. Gregory Greenman
|May13-05, 04:53 PM||#7|
I'm not sure that it is an apples and oranges comparison. For example, you could have generated the heat with electrical elements instead. If the electricity demand reduced due to the cogeneration plant exceeds the loss of efficiency to the nuclear power generation system, then you have increased the efficiency of the plant.
The idea would be to build an industrial plant contemporaneous with a new nuclear reactor, to be part of a co-generation complex. If you're going to build a new nuclear reactor, why not soak it for all its worth? Ideally, the industrial plant would be highly automated. This would eliminate the long range piping concern.
I agree that it might not make sense for a residential neighborhood.
|May13-05, 05:17 PM||#8|
In spanish it is called F.U.E. (Factor of Utilization of Energy) and is an important coefficient for engineers when they want to evaluate a cogeneration system. As you can see, it sums heat and work, but Morbius is right when saying that is comparing orange and apples. Anyway, it measures how well are you employing the energy exhausted by the fuel.
|May14-05, 12:55 PM||#9|
However, generating heat with electricity is certainly a non-optimum
thing to do.
As above, elecricity is "work" - it is high quality energy with zero entropy.
You pay a price for that high quality - it takes on the order of 2.5 to 3 units
of heat energy to produce a single unit of electricity.
If all you are going to do is turn that single unit of electricity back into
a single unit of heat - then the net effect is that you have expended
2.5 to 3 units of heat energy to recover a single unit of heat energy.
This is also why burning hydrogen produced by electrolosis may be
problematic. In that case, you take about 15 units of heat energy, that
is used in a Rankine steam cycle to produce about 5 units of electric
energy. Those 5 units of electric energy can be used to make hydrogen
with 5 units of chemical energy. When that hydrogen is burned in a car,
with a typical efficiency of ~20%; then the 5 units of chemical energy
will turn into 5 units of heat energy of which 1 unit will be recovered
as useful work to turn the wheels of the car.
So overall - you start out with 15 units of energy in your primary fuel -
and recover a single unit of work.
One may want to do that - if one wants to use non-polluting nuclear
power to substitue for polluting fossil fuels in the transportation
industry - but one can see, there's a steep hill to climb; thermodynamically.
You essentially have to take 2 energy conversion efficiency "hits"
instead of a single one.
Dr. Gregory Greenman
|May14-05, 03:45 PM||#10|
What do you mean by "energy with zero entropy"?
|May16-05, 09:38 AM||#11|
Entropy is a measure of "disorder". It is also a thermodynamic property
of matter. For example, if I tell you the density and specific energy
[ energy per unit mass ] in a material - then I've completely specified the
thermodynamic "state" of that material - including the pressure, the
temperature, and the entropy.
Anytime you transfer an incremental amount of heat energy "dQ", at a
temperature "T", then you also transfer an incremental amount of entropy
"dS" given by dQ = T dS.
So when you transfer heat energy; you transfer entropy. However, work -
like electrical energy or the mechanical energy of a spinning turbine
shaft, carries no such entropy with it. Work is "ordered" energy.
We can see how the concept of entropy affects things by considering the
following derivation of the Carnot efficiency.
Suppose we have a heat engine, working in a cycle [ the working fluid
like the water in a Rankine steam cycle is closed - so the working fluid
comes back to the same thermodynamic state at any point on the cycle].
First, from conservation of energy, we can write the following equation:
Q_in = Q_out + W
That is the sum of the waste heat out, Q_out; plus the useful work W
equals Q_in, the heat provided by the boiler, reactor,....
Now since the working fluid is in a closed cycle that comes back to the
same thermodynamic state at any point in the cycle - then there can
be no "build-up" of entropy. The entropy out has to equal the entropy
put in, plus any entropy created in the cycle. The 2nd Law of Thermodynamics
says that the amount of entropy created is non-negative [ zero or
positive]. The best you can do is to have the created entropy be zero.
But the heat engine can't decrease entropy. Therefore, we have a second
equation concerning the entropy:
S_out = S_in + S_created
The entropy out S_out that is carried by the waste heat Q_out is equal
to the entropy input S_in carried by the heat input Q_in plus any entropy
created. The 2nd law of thermodynamics states the S_created is
greater than or equal to zero - S_created >= 0.
Because of that; I'm going to turn the above equation into an inequality
S_out >= S_in
Now from above, dQ = T dS; and the fact that the output entropy is
exhausted to the environment at a temperature of T_c [ the cold temp ],
Q_out = T_c S_out or S_out = Q_out / T_c
Likewise, the heat input from the boiler or reactor or whatever is at
temperature T_h [ hot temperature ], so we have
Q_in = T_h S_in or S_in = Q_in / T_h
Q_out / T_c >= Q_in / T_h
or turning it around:
Q_in / T_h <= Q_out / T_c
Now from the original conservation of energy equation, we know that
Q_out = Q_in - W; so we substitute that into the above to obtain:
Q_in / T_h <= ( Q_in - W ) / T_c
rearranging the above:
W/T_c <= (Q_in / T_c) - (Q_in / T_h)
Multiply both sides by T_c:
W <= Q_in - Q_in (T_c/T_h) = Q_in [ 1 - (T_c/T_h) ]
Divide both sides by Q_in:
W/Q_in <= [ 1 - (T_c/T_h) ]
Now the work out W divided by the heat input Q_in is the efficiency of
the heat engine. Therefore:
efficiency <= [ 1 - (T_c/T_h) ]
The efficiency of a heat engine is less than or equal to the quantity on
the right hand side. That is the quantity on the right hand side is a limit
on the efficiency of a heat engine - and is known as the Carnot efficiency.
Carnot efficiency = [ 1 - (T_c/T_h) ]
where T_c and T_h are temperatures on an absolute scale - either
degrees Kelvin or degrees Rankine. Since the temperatures are in a
ratio - the "size" of the units doesn't matter - any conversion factor
would divide out. All that is required is that T = 0 at absolute zero, and
not some other temperature like in the Fahrenheit and Celsius scales.
Note that nowhere did we have to specify anything about the particular
heat engine. It doesn't matter how many turbines are in the cycle, or
if reheat is used between turbines, or ..... The above efficiency limit
applies to EVERY heat engine that works in a cycle.
It is because heat carries with it this entropy when ever it is transfered,
according to dQ = T dS ; that gives us this limit. Work - be it mechanical
or electrical doesn't carry entropy with it.
Therefore, a motor that converts electrical energy to mechanical
energy, or a generator that converts mechanical energy to electrical
is not bound by an efficiency limit due to the 2nd Law of Thermodynamics.
Because work doesn't carry this entropy along with it; whereas heat
energy does - work and heat are very different animals thermodynamically.
Work is a higher quality energy than is heat, because it doesn't carry
this entropy "baggage" with it. Therefore, co-mingling work and heat,
and adding them together in some ratio is of questionable use.
When two items are added together to form a sum - one can often
trade off one for the other and preserve the sum. However, with
co-mingled work and heat - you can't do that because you can't slosh
the entropy from the heat to the work component. Work will always
have zero entropy.
Dr. Gregory Greenman
|May16-05, 12:38 PM||#12|
So since dQ/T = dS and electrical energy has no heat energy, dS is zero? Or is it incorrect to even say electrical energy has zero dQ?
|May16-05, 02:34 PM||#13|
Electrical energy, or mechanical energy have no entropy - so dS = 0
when electrical or mechanical energy are transfered.
Heat energy carries entropy. There is a random "disorder" in heat energy -
and entropy is a measure of that disorder.
Dr. Gregory Greenman
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