Subject: Re: How many BTU's to raise 4

From: [email-masked] (RAYMOND FRYE)
Date: Wed, 04 Dec 96 12:24:00 -0800

Renewable at [email-masked] writes:

>  350 - 80  delta of 270F    (120PSI)

>  270F * 4.2 Lbs = 1134 BTU's. 

>  but 1200 BTU's needed to vaporize 1 lb H20 to steam, so  4.2 LBS H20
>  would need 5000BTU's??.

>  What I want to know, is how many BTU's would I save if the feedwater
>  was heated to 200F instead of to 80F and then fed to boiler/engine.
----------------------

Starting with your last question:

       approx. (200 - 80) = 120 BTU/lbm (504 BTU)

For your boiler:

       Heating from 80 to 341.3 F everything is liquid.

       approx. (341.3 - 80) = 261.3 BTU/lbm (264.2 is a bit closer)
       has gone into heating.

       At this point you start generating steam. The temperature in the
       boiler will remain at 341.3 F until all the liquid becomes steam.

       796 BTU/lbm goes into making this phase change.

       (To get higher than 341.3 F requires a superheater)

       approx. 3.9 BTU/lbm from the superheater to get to 350F

       264.2 + 796 + 3.9 = 1064 BTU/lbm (~4469 BTU)

   Boiler losses have been ignored, but this is the basic school stuff.
   A good set of steam tables would get you a closer answer and the
   1200 BTU/lbm is likly a rule of thumb that accounts for losses.

   OK, so now if you build something you have to tell us.

Ray

 * RM 1.3 02229 * a somnambulist in the information trafic jam

---

From: Steve Wellcome <[email-masked]>
Date: 3 Dec 1996 15:51:01 GMT

Yep, the phase change takes a *lot* of energy.  But
prior to the phase change, it's a straight 1 btu per
degree (or pretty close), so you'd save 200-80 = 120 btu
per pound by preheating.

The energy sucked up by the phase change was described
by Joseph Black, an 18th-century (I think) Scottish 
clergyman, who used the term "latent heat."

Are you sure that 1200 btu figure is correct?  That
sounds high to me.  I thought it was more like 144 btu,
but perhaps I'm confusing it with the btu in the 
ice/water phase change.

---

From: Jay Hayes <[email-masked]>
Date: Fri, 06 Dec 1996 01:12:30 -0800

renewable wrote:
> 
> BTU=raise one pound of water one degree Farenheit.
> 
> but 1200 BTU's needed to vaporize 1 lb H20 to steam,
> so  4.2 LBS H20 would need 5000BTU's??.
> 
> I don't understand the discrepancy, except that the first
> does not take into account a phase change. Help.
> 
> Real question I am trying to answer is that I know of
> a steam engine/gen  that produces  10 kWe
> consuming  5000BTU's or 1/2 gallon per minute
> with a feedwater of 80F.
> 
> What I want to know, is how many BTU's would
> I save if the feedwater was heated to 200F instead
> of to 80F and then fed to boiler/engine.
> 
> It should take less than 5000BTU with the 200F
> feedwater than the 80F feedwater, but by how
> much less BTU's?
> 
> It involves a phase change so I can't quite grok
> it. Thanks for any help..
> 
> dsg

The first thing to understand this problem is that the water at
120PSI(gage pressure) (is actually 134.7 PSIA (absolute pressure)) boils
at about 350 degrees F. Under this condition the latent heat of
evaperation in order to change the water to steam is really about 871.2
BTUs/lb. The Enthalpy of the saturated vapor(steam) is 1193.2 BTU/lb.

1193.2 BTU/lb x 4.2 lbs = 5011.44 BTU

871.2 BTU/lb x 4.2 lbs = 3659.04 BTU
 
350 deg. - 80 deg. = 270 deg.

270 x 1 BTU/lb x 4.2 lbs = 1134 BTU

3659.04 BTU + 1134 BTU = 4793.04 BTU (that's better than 95 persent
efficient - not bad!)

Heating the feed water will still take BTUs from somewhere. The savings
only enter in to it when you reclaim heat that would otherwise be lost.
An economizer is one way of doing it. An economizer uses excess stack
gas temperature to heat the feedwater just before it enters the bioler. 
A condenser can also be used to reclaim the condensate and some of its
heat from the exhausted steam.

If you can reclaim some BTUs you can use the 1 BTU/lb H2O/deg.F to get a
rough idea of how much savings. Or you can check a set of steam tables
to get a more exact answer. I'm sure there must be some software that
does it too.  

Hope this helps,

Jay

---

From: [email-masked] (Eric W. Dalton Jr)
Date: Fri, 6 Dec 1996 01:27:55 GMT

Check me if I'm wrong, The change from ice/water takes 540
calories/gm
and the change from water to steam takes 80 calories/gm.
Converting
to BTU/lb I get 799 BTU and 118 BTU respectively. 

JD
The above values are reversed.  
The phase change values are 80 cal/g to convert ice to water
and 540 cal/g to convert water to steam.

Eric W. Dalton Jr. Science Department  Fauquier High School 
705 Waterloo Road   Warrenton Va. 22186   540 364 6100

---

From: [email-masked] (JD)
Date: Tue, 03 Dec 96 18:01:37 GMT

In article <[email-masked]>,
>but perhaps I'm confusing it with the btu in the 
Check me if I'm wrong, The change from ice/water takes 540 calories/gm
and the change from water to steam takes 80 calories/gm. Converting
to BTU/lb I get 799 BTU and 118 BTU respectively. 

JD

---

From: [email-masked]
Date: 6 Dec 1996 22:23:49 GMT

Dear Jay:

I have noted your great expertise in manipulating numbers,
now perhaps you would like to help me with this question.

In my native Spain, we have a Steam Ram in our workshop,
this machine was probaly build in the late 1900 and has been
used in the manufacturer of coins. For the last 25 years nobody
has touched the machine, because nobody can still remember
the tonnage of the RAM. Can you help ?

The machine worked according to the following principle.

An upright solid steel cylinder of 12 " diametre and a length of 6 ft, was

elevated with saturated steam by 8 ft, and mechanically locked.

On demand the cylinder was released and then crashed down
with its full weight into the forming  tool, creating the coin image.
A steam cushion was then used to absorb some of the vibration.

My question is:

What is the tonnage of the RAM ?

Your help is very much welcome.

Alberto.

---

From: [email-masked] (Scott Logan)
Date: Sat, 07 Dec 1996 15:24:35 GMT

On 6 Dec 1996 22:23:49 GMT, [email-masked] wrote:

OK, in your e-mail, you said you were looking for the IMPACT FORCE
(not mentioned in your original post).

You mention a steam cushion was used to absorb some of the vibration.
Without knowing what cushioning effect that may have, the impact can
not be calculated.  Also, the friction caused by the sleeve around the
cylinder will have some effect.

If your cylinder were being dropped in a vacuum, with no other forces
acting upon it, and if it were stopped with no deceleration, it would
hit with a force of blow of approx. 18,450 lbs.

Again, all figures and formulae from Machinery's Handbook.

Gee, I love getting homework on weekends <VBG>!

Scott Logan
[email-masked]
www.loganact.com

---

From: Jay Hayes <[email-masked]>
Date: Sat, 07 Dec 1996 22:20:55 -0800

> Your help is very much welcome.

Alberto

I understand that the steel cylinder is a solid ram 12" D x 72"H.
It free falls 8 feet, striking the die.

The ram's weight would be about 2303 lbs. The WORK performed by the
falling ram would be about 18424 ft-lb. 

To find the force exerted you must first find the velocity of the
falling ram. The ram drops 8 feet at the gravitational acceleration 32
ft/second sqrared. Thats a velocity of about 22.63 feet/second. Velocity
equals the square root of two  times the gravitational constant times
the distance. So you have a 2303 lb ram falling at a velocity of 22.36
ft/sec for about .3535 sec (thats 8 ft/ 22.63 ft/sec).

Force = mv/gt = (2303 lb x 22.63 ft/sec)/(32 ft/sec2 x .3535 sec)
Force = 4607.22 lbs

One may rate the machine at 2.3 tons, another might rate it 9.2 tons.
Where have we seen this type of discrepency recently? (Compressor horse
power ratings)

Steam drop hammers were in their prime around the turn of the century
but are not used much today. Some designs allowed steam to power the ram
down there by greatly increasing its tonnage. This was common in steam
forging hammers of that era. 

The steam drop hammer was propaply retired because a more efficient,
safer, and easier to use machine was available. A mechanical or
hydraulic press would be a fraction of the size and require much less
energy to do the same amount of work.

The force of a falling body does not have as much punch as a machine
that uses energy with mechanical advantage multiplying its force.

Jay

---

From: [email-masked]
Date: 8 Dec 1996 22:29:22 GMT

In article <[email-masked]>,
[email-masked] (Steve Rayner) writes:

>An acetylene regulator works fine.  You will need an adapter for the
tank.
>I have been using this sort of rig for 20 years.

  Adapter? Are you using the the little MC size tank? Or do you have
regulators for a MC size tank? The standard size regulator made for the
bigger tanks will fit right on a propane tank.

   Wayne Cook

Quality is like buying oats,
        if you want nice, clean, fresh oats you must pay a fair price.

However, 
        if you can be satisfied with oats that have already been through
the horse...
                          
That comes a little cheaper...

---

From: [email-masked]
Date: 8 Dec 1996 23:43:27 GMT

Jay:

I am so delighted that you have been able to respond to my
request for assistance, and provided me with your comprehensive
comments.

Next time, when I go home I have to discuss your findings with
our w/s supervisor and see what we can do next.

In the meantime, I shall remain with great thanks

Alberto
in London.

---

From: [email-masked] (Steve Rayner)
Date: Tue, 10 Dec 1996 11:51:55 GMT

In a previous article, [email-masked] () says:

>news.aol.com!not-for-mail
>From: [email-masked]
>Newsgroups: rec.crafts.metalworking
>Subject: Re: Propane
>Date: 8 Dec 1996 22:29:22 GMT
>Organization: AOL http://www.aol.com
>Lines: 22
>Message-ID: <[email-masked]>
>References: <[email-masked]>
>NNTP-Posting-Host: ladder01.news.aol.com
>X-Admin: [email-masked]
>X-Newsreader: AOL Offline Reader
>

I am using one of the tanks sold as replacements for barbecues.  I will 
get a new one when it's empty.  The propane seems to last forever.  The 
tanks certification has run out.  Testing is 1/4 of the price of a new 
tank, so I might as well replace it.  There are two standards for tank 
fittings in use here now, but only one when I bought the tank.

-- 
Steve. 

******************Knowledge is wasted if it is not shared.*******************

---

From: [email-masked] (Michael Horgan [Lughaid])
Date: Tue, 10 Dec 1996 20:20:51 GMT

Scott Logan wrote:
> Well, I'm not Jay, but carbon steel has a mass of 0.2833 lbs per cubic
> inch.  Your ram is 8143.008 cubic inches (pi*r^2*l) so the weight is
> 2306.9 lbs or 1046.4 kg.
> 
> All figures from Machinery's Handbook.
> 
> Scott Logan
> [email-masked]
> www.loganact.com
Very true, Scott, but misleading as well.  Impact velocity has a large
effect on the pressure of the hammer, as does the striking area. I
used a 125 pound hammer, falling about 5 feet, to strike 1 inch bronze
medallions.  reducing the drop by as little as an inch or two made a
great deal of difference in the designs I could strike, as does the
surface area of the die "at the instant of impact!"  My 125 pound
hammer routinely did  work that required a 45 ton hydraulic press, and
faster too! Without knowing the die size, and impact area, as well as
the free drop distance, you can't determine the "Tonnage,"  IMHO.
Michael Horgan ([email-masked])
http://users.aol.com/Lughaid

---

From: [email-masked]
Date: 10 Dec 1996 22:30:39 GMT

Lugheid:

What prompted me to write this article in the first place had to
do with the estimated tonnage of the RAM, its actual past
performance and a comparative performance of more modern
hydraulic or mechanical knuckle presses.

In fact in our workshop in Spain we found samples of work,
which had been carried out on the steam RAM, which would
have required at least four times the available tonnage by modern
machines.

I will be going back during the summer vacation to Spain,
and we will try to fire up the monster, and take some material
samples with us und see what kind of results we are getting.

When this is doen, I will report back my findings to the Forum.
In the mean time, its a great thank you to all these willing people
who have so kindly contributed, and doen all the number 
crunching work for me.

I would be willing to buy an Air Hammer, say around the
75 kg hammer weight, 300 mm stroke,  would you know 
who sells this equipment from the 1930 to 1950's ?

God save America,

Alberto

---

From: "MARK " <[email-masked]>
Date: 6 Dec 1996 11:53:52 GMT

Eric W. Dalton Jr <[email-masked]> wrote in article
<[email-masked]>...
> 705 Waterloo Road   Warrenton Va. 22186   540 364 6100
I don't know if anyone has brought this up yet, and perhaps you know, but
you need to have the water under pressure to get it to 350F. As the water
boils around 260f (I know its around there) the heat is carried off with
the steem (evapration=cooling). How much pressure I'm not sure but would
think around 10 to 15 psi.
-- 
Mark Worstall
[email-masked]
http://www.bright.net/~ccoiadl
*********************************************************
It stands to reason that when someone speaks
to you of sacrifice, there is someone collecting
the sacrificial offerings. Where there is service,
there is someone being served. The one that speaks
to you of sacrifice speaks of slaves and masters,
and intends to be the master.

---

From: [email-masked] (Scott Logan)
Date: Sat, 07 Dec 1996 12:55:24 GMT

Scott Logan
[email-masked]
www.loganact.com

---

From: [email-masked] (Jens)
Date: Thu, 12 Dec 1996 00:18:20 -0800

[email-masked] (James Kirkpatrick) wrote:

>Actual striking force (not pressure) depends largely on how compressible
>the striking masses are.  If I drop an incompressible object onto another
>incompressible object (this needs to include the planet it's attached to)
>the actual striking fource would be infinite.  The obvious flaw here is that
>nothing is incompressible.

Thank you  .... this is exactly what I thought but I wasn't going to
say anything on the possibility of making a fool out of myself. When
nobody else mentioned this action/reaction business, I thought 'well,
what do I know' ... .guess I still know a few things..

Jens

---

From: [email-masked] (James Kirkpatrick)
Date: 12 Dec 96 12:29:08 MST

In article <[email-masked]>, tony <[email-masked]> writes:
|>
|>O.K., but what about the force when the ram accelerates due to gravity at 32 ft.
|>per second^2 during the distance of its stroke?

I was afraid somebody would ask this.  It's like asking "how long is a string?"

The original question (and Scott's answer) was simply how much did the
ram weigh.

Let's take an extreme opposite example.  Drop a 15-pound bowling ball onto
a pile of foam rubber, and when it first hits, it will start decelerating
(actually, it will first continue to accelerate but at a decreasing rate
until the response force equals the weight of the ball, then it will
decelerate) and slowly come to a stop.  Ignore the springback for now.
Throughout the impact time, the force might not exceed, say, 25 pounds.

Now remove the foam rubber and substitute a thick copper plate.  The same
thing happens but over a significantly shorter distance, perhaps a few
thousandths of an inch.  Decelerating a 15-pound mass traveling a dozen or
so miles per hour in that distance requires significantly higher forces,
perhaps in the many-tons range (I haven't bothered to calcluate, I' trying
to describe it qualitatively here).

So the same thing happens in a falling-ram press.  The actual force will
depend on the compressibility of the thing being pressed.  For soft items the
force will be smaller over a greater distance, and opposite for "hard" items
such as steel.

And pressure is force per unit area, so then we need to take into account
the area of the object.  It also gets difficult to analyze if the object
responds non-linearly to deformation, or if the ram itself absorbs enery
by its own deformation (temporary, hopefully, but unavoidable).

Jim

---

From: [email-masked] (Fitch R. Williams)
Date: Fri, 13 Dec 1996 07:31:09 GMT

>Jens

Wow.  All this ft-lbs stuff.  Inelastic collisions!  I love
it when you write technical!  <G>

Fitch
Lets see, E = IR, I = C dv/dt, V = L di/dt, a+jb, Ciss,
Coss, gate charge, ... ahhhhh, its helping, feeling better
already, millisecond, microsecond, joule, watt-sec,
Millihenry, Farad, ... kilowatt, ,..... OK, ready for next
post.

---

From: tony <[email-masked]>
Date: Mon, 09 Dec 1996 23:11:44 -0500

> All figures from Machinery's Handbook.

O.K., but what about the force when the ram accelerates due to gravity at 32 ft.
per second^2 during the distance of its stroke?