The Steamboating Forum

I've been doing some calculating on how much water this twin will consume.

The crank driven feedpump is installed, and I must admit, it looks a bit small for the job, but the calculations say it's large enough.

It is .75" bore and .75" stroke.

The engine is a 3+3X4 inch twin, calculations are based on 300 RPM.

Perfect world calcs say that pump will deliver roughly 28 US gallons or 6468 cu inches of water per hour at 300 RPM.

The engine consumes 114 cubic inches of steam per revolution, times 300 RPM = 34,000 times 60 = 2, 034,720 cu inches of steam per hour.

One cubic inch of water will create roughly 1600 cubic inches of steam. BAsed on that, the feedpump will provide enough water per hour to create 10,348,800 cubic inches of steam.

10,348,800(produced) / 2,034,720(needed) = 5.086

One question about the above. I looked around to see what I could find, but it seems the pressure would greatly alter the 1600:1 ratio of steam to water. Is this affected by pressure?

Incidentally, I plan on running around 100 PSI max boiler pressure.

Any input on this is welcome.

-Ron

Are you using a keel condenser?[Or any form of condenser?] This will save water. Is your boiler using a steam atomizing burner?This will use water. Then there's all the drips and leaks that we shouldn't have but do have! I have 20 gal on board and I always carry about 4 gal in jugs.It's always nice to have too much and be able to give some to a fello steamer who didn't bring enough! I've seen other boats with up to 60 gal tanks on board.If you're gone for the weekend it's better to be looking at it than looking for it!! Den

Den,

Thanks for the input. Condensing would only deal with recovering water from expelled steam. It seems anyway, hell I don't know what I'm talking about.

I found this table, and it appears by volume steam is around 1/7 the volume at 100psi that it is at 0 PSI gauge pressure.

http://www.boilerroomservices.com/Facts/SteamTables.pdf

Going back to my calculations which don't factor any losses, this pump has the ability to inject 28 gallons per hour in to the boiler, which is 5 times needed at 0 PSI. However, at 100 it needs 7 times, according to the above table. So the pump looks too small, and now the calculations, say it is. This is running about right..

-Ron

DetroiTug wrote:I've been doing some calculating on how much water this twin will consume.

The crank driven feedpump is installed, and I must admit, it looks a bit small for the job, but the calculations say it's large enough.

It is .75" bore and .75" stroke.

The engine is a 3+3X4 inch twin, calculations are based on 300 RPM.

Perfect world calcs say that pump will deliver roughly 28 US gallons or 6468 cu inches of water per hour at 300 RPM.

The engine consumes 114 cubic inches of steam per revolution, times 300 RPM = 34,000 times 60 = 2, 034,720 cu inches of steam per hour.

One cubic inch of water will create roughly 1600 cubic inches of steam. BAsed on that, the feedpump will provide enough water per hour to create 10,348,800 cubic inches of steam.

10,348,800(produced) / 2,034,720(needed) = 5.086

One question about the above. I looked around to see what I could find, but it seems the pressure would greatly alter the 1600:1 ratio of steam to water. Is this affected by pressure?

Incidentally, I plan on running around 100 PSI max boiler pressure.

Any input on this is welcome.

-Ron

The 1600:1 is incorrect. Steam is measured in pounds per cubic foot. According to Mark's Steam Tables, steam at 115 psi absolute weighs 1 pound per 3.881 cubic feet so:
1. Converting cubic inches to cubic feet, divide 2, 034,720 by 1728 = 1177.45 cubic feet/hr;
2. Coverting cubic feet to pounds, divide 1177.45 by 3.881 = 303.4 pounds water(steam)/hr
Water weighs approximately 8.34 pounds/gallon and 1 gallon = 231 cubic inches then:
3. Coverting pounds to gallons, divide 303.4 by 8.34 = 36.4 gallons/hr;
4. Converting gallons to cubic inches, multiple 36.4 by 231 = 8408.4 cubic inches water/hr
This will theoretically replace water used. But water/steam is tricky stuff and leaks everywhere. The old rule of thumb is: "Design your feedwater pump to deliver 1.5 times as much water as needed" - this also lets you "catch up" if the level is low. So:
5a. Minimum needed quantity of water is 8408.4 times 1.5 = 12,612.6 cubic inches/hour or;
5b. Divide by 60 minutes for quantity water/minute or 12,612.6 divided by 60 = 210.21 cu.in./minute or;
5c. For pump capacity divide quantity/minute by rpm, so 210.21 divided by 300 = 0.7007 cu.in./revolution.
Now you know that you NEED a pump with a minimum swept volume of 0.7007 cubic inches/stroke.

Water hammer can be a problem with piston (and sometimes "ram") feedwater pumps. I (and others) have found that this problem will generally disappear if pump speed is held below 200 rpm. That is 2/3 engine speed and this needs to be taken into account in sizing. Thus, a pump of 1.05105 cubic inches should be more than adequate to supply your engine and allow for whistle blowing and any steam operated auxilliary equipment.

Ron,

Thanks for the reply, and this little pump misses it worse than I had thought.

I'll make another from some bronze and stainless. 1.25" bore X .75" stroke should be a lot closer. I'll use what I have for the short test run.

-Ron

DetroiTug wrote:Ron,

Thanks for the reply, and this little pump misses it worse than I had thought.

I'll make another from some bronze and stainless. 1.25" bore X .75" stroke should be a lot closer. I'll use what I have for the short test run.

-Ron

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