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Steam Engine Question - Help Understanding Needed

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Sep the Gui
#1
Jan5-13, 02:18 AM
P: 3
Ok, so I have complete faith in the tried and proven laws of thermodynamics, including the part of the second law which states that it is impossible to create a heat engine working in cycle that takes in a certain amount of heat energy, and converts all the heat energy it takes in, into Mechanical work.

But I am counfused, with regards to how a steam engine would not violate that law.

In particular, this is what I am confused about: Couldn't you build a steam engine, where the steam is ran through just enough rotors that the temperature gained from heating the steam is in effect all lost to turning the rotors in the turbine, and thereby is returend to the boiler in the state that it left it in, thereby making no need for a cold sink for the heat to be dumped into?

I know that my reasoning is flawed somewhere, but I'm not sure where . . . . .
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Jobrag
#2
Jan5-13, 05:42 AM
P: 475
There are two problems one is mechanical losses, but even assuming a perfect world with frictionless bearings and perfect seals, for a steam engine to work there has to be delta T somewhere, (between the boiler flue gas and the boiler water, or between the condenser cooling water and the LP steam being condensed) those delta Ts are where you loose efficiency.
russ_watters
#3
Jan5-13, 11:45 AM
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P: 22,244
Welcome to PF.

The problem is that "just enough rotors" (turbine stages) is an infinite number of stages. Each stage only extracts some fraction of the remaining energy, so it might look something like this, if each stage extracts half the remaining energy:

1 stage = 1/2 the energy
2 stages= 1/2 + 1/4 = 3/4
3 stages= 1/2 + 1/4 + 1/8 = 7/8
4 stages= 1/2 + 1/4 + 1/8 + 1/16 = 15/16

Etc. And each stage is dealing with lower pressure steam, which means each successive stage is larger than the last. So you see, there is a quickly diminishing return.

jim hardy
#4
Jan5-13, 04:53 PM
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Steam Engine Question - Help Understanding Needed

I hope i'm not off topic here -

and thereby is returend to the boiler in the state that it left it in,..

Note your steam exits the turbine at quite low pressure.
To get it back into boiler its pressure must be raised back up to that of the boiler.
To that end a phase change is employed - because to reach any given pressure it takes far less work to pump an incompressible fluid than a compressible one,,,

look at your vdp + pdv terms

http://www.engineeringtoolbox.com/sp...nes-d_629.html

The condenser in a steam power plant ideally just discards the heat of vaporization.
They intentionally mix steam with the condensate to minimize subcooling.

http://www.concosystems.com/sites/de...olved-oxyg.pdf

Condensate subcooling occurs frequently due to inadequate
original condenser design, especially in not having allowed
enough vapor to bypass the tube bundles and reheat the
condensate as it cascades down through them. Some remedies
may be found in adding steam sparging to the hotwell, recycling
the condensate through sprays or increasing the laning between
the tube bundles. Clearly, should there be an inherent degree of
subcooling contained within the original design, it is only
subcooling in excess of this value that should be considered as
amenable to correction by some form of backpressure control
action.
old jim
russ_watters
#5
Jan5-13, 06:15 PM
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Quote Quote by jim hardy View Post
I hope i'm not off topic here -

Note your steam exits the turbine at quite low pressure.
Lol, good point -- you need both the condenser and the pump!
Curl
#6
Jan5-13, 11:42 PM
P: 757
What you are asking for is to adiabatically expand steam until it turns back into liquid - it is impossible.
When you expand steam it cools, yes, but it will not condense because you have increased volume. That's why you need the cold sink. Sorry.
Sep the Gui
#7
Jan7-13, 06:27 PM
P: 3
Ok, I think I understand it now. I didn't realize that in order to get the steam back into the boiler, you would have to raise it to the same pressure as the boiler. So I assume, if you didn't use the condenser, you probably would have to use the same amount of work to adiabaticaly compress the steam, as the steam exerted on the rotors while adiabaticaly expanding, thereby resulting in zero net work. Haha, and yes, I guess I was expecting the steam to adiabatically expand into a liquid. Thanks!
flatmaster
#8
Jan8-13, 08:45 PM
P: 504
Quote Quote by jim hardy View Post
I hope i'm not off topic here -

The condenser in a steam power plant ideally just discards the heat of vaporization.
They intentionally mix steam with the condensate to minimize subcooling.

old jim
This makes sense to me. All you want to do is turn the fluid back into a liquid to decrease the volume to make VdP smaller. Why cool the liquid further if you're just going to heat it up again?

My question is given two incompressible fluids with comparable boiling points, will the fluid with the smaller heat of vaporization produce a more efficient cycle? If this is the case, could we possibly use a liquid other than water to avoid it's relatively large heat of vaporization?

http://en.wikipedia.org/wiki/Latent_heat
jim hardy
#9
Jan8-13, 09:23 PM
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I would think so, but that's just my intuitive answer. I do better at math in the mornings
and i'm an electrical type who learned enough thermo to not appear totally ignorant in presence of mechanical types.

Probably there's somebody here who can give you a good answer with numbers quicker than i.
But i'll take a look.
Here's wher i'll start: http://en.wikipedia.org/wiki/Rankine_cycle

Here is an interesting anecdote - well to me any way -

in the 1930's when they were still trying to figure out what water chemistry would allow high temperature and pressure in steel boilers,
somebody built a two stage plant in downtown Boston.
The high temperature turbine used mercury vapor
and its exhaust heated water for a steam turbine.

The Electrical Year Book, 1937,[2] contained the following description of a mercury vapour turbine operating in commercial use:

The advantage of operating a mercury-vapour turbine in conjunction with a steam power plant lies in the fact that the complete cycle can be worked over a very wide range of temperature without employing any abnormal pressure. The exhaust from the mercury turbine is used to raise steam for the steam turbine. The Hartford Electric Light Co. (U.S.A.) has a 10,000kW turbo-generator driven by mercury vapour, which reaches the turbine at 70 lb. per sq. in. (gauge), 880F. The mercury vapour is condensed at 445F and raises 129,000 lb. steam per hr. at 280 lb. per sq. in. pressure. The latter is superheated to 735F and passed to the steam turbines. During 4 months continuous operation, this plant averaged about 0.715 lb. of coal per kWh of net output, about 43% of the output being from the mercury turbine generator and 57% from the steam plant. On maintained full-load the heat output averages 9800 B.Th.U. per net kWh. It is believed that maintenance costs will be lower than in ordinary steam plant. The back-pressure on the mercury turbine is fixed by the steam boiler pressure; only a small vacuum pump is needed, as there is no air or other gas in the mercury system.
http://en.wikipedia.org/wiki/Mercury_vapour_turbine


9800 was still a respectable heat rate thirty years later after they'd solved the chemisty problems and had steam pressures 10X higher...

Can you imagine permitting that thing today ?
jim hardy
#10
Jan9-13, 11:45 AM
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Well i did take a further look

and as so often happens, my "off the cuff" answer was dead wrong.

Here's a scholarly article i found with Google:
http://www.seas.upenn.edu/~meam502/c...ers/thermo.pdf

Around page 3063 is the answer to your question, see section 3.1.2

And it's obvious in hindsight:
Latent heat defines the length of the horizontal lines on T-S diagram.
That length determines the width of the polygon representing the cycle.
Width X height is area, and area enclosed by the polygon is the work output.
Obviously the wider polygon delivers more work.

So i apologize for the mis-information. I shoulda had more faith in my limited appreciation of basic thermo.

Eq. (6) shows that fluids with higher latent heat give higher unit
work output when the temperatures and other parameters are
defined. The influence of the latent heat can also be explained by
observing the Ts diagram in Fig. 6. Under defined temperatures,
the length of the horizontal line segment is proportional to the latent heat. Long line segment (i.e. latent heat) is expected to
obtain large work output because the area formed by the process of
the cycle is the work output from the turbine
. This result agrees
with the conclusion from Eq. (6). Meanwhile, as it has been
mentioned, Eq. (6) gives the unit mass work output from the
turbine, it can be inferred that fluids with higher density need
smaller equipment setup for same power production. In brief,
working fluids with high density, low liquid specific heat and high
latent heat are expected to give high turbine work output.
It's a pretty good looking article - check it out!

And thanks for the question - obviously i needed a refresher.

old jim
flatmaster
#11
Jan9-13, 03:34 PM
P: 504
Interesting. I haven't read it that closely, but I'm guessing that the "length of the horizontal lines on T-S diagram" is directly related to the expansion ratio of the working fluid from it's liquid state volume to it's largest gas volume.

There is another thing I'm finding confusing when trying to figure this out. Most of the graphs are temp vs. entropy. Why don't they look at the good old Pressure vs. Volume graph or "specific volume" and "specific pressure". Pressure and volume are much more intuitive that entropy.


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