Bessler's (4th) Kassel wheel Archimedes screw pump calculations

[eccentrically1]

http://www.youtube.com/watch?v=K8E_zMLCRNg

Should the HP be edited?
2nd wheel ~ .02 hp
3rd ~ .028 hp
4th ~ .033 hp

[jim_mich]

I've still not installed Visual Basic Professional Ver.6 on my computer. I've just been too busy with my work, with getting firewood cut, hauled, and stacked in the basement, and other things that life throws at me.

Once VB is installed then it will probably take 16 to 24 hours or more to write a program to calculate the volume of the turning screw.

The reason I wanted VB Ver.6 is that I'm more familiar with it. I used it for a number of years before I switched computers and lost it. I have Ver.5 installed, but it sucks compared to Ver.6. The biggest slow-down to writing VB code is getting the syntax exactly right. VB Ver.6 has a real good syntax help feature. Also, I have written many sub-program that help animate things like wheels and such, but they will not work with Ver.5 because they were written in Ver.6. Also when I write code the syntax is often wrong because I forget to use the older Ver.5 syntax, and then I cannot remember to older syntax, so I go hunting through my reference books looking for the right older syntax, which is a big waste of time.

The bottom line is, this project will have to wait until I have a free weekend with nothing else planned. Sorry.. it's just not my top priority.

[ovyyus]

No problem Jim, get to it when you can.

FWIW, it's probably not such a difficult problem that a computer program is required to solve it. Archimedes screw pumps are well trodden ground and the current estimates appear close enough. Of course the ramifications of such a low power pump reducing the wheel speed as reported might be worthy of further discussion.

[john]

The facts, as can best be determined by scaling Bessler's drawing:

Screw length = 12 foot (144 inches)
Screw lead = 1/2 foot (6 inches)
Screw ID = 1.5 foot (18 inches)
Screw axle = 1/4 foot (3 inches)
Screw tilt angle = 20º
Wheel square pulley = 1.4 foot across corners (16.8 inches)
Screw square pulley = 2.8 foot across corners (33.6 inches)
Pulley ratio = 2:1 reduction.
Wheel speed = 20 RPM (Ref. not used)
Screw speed = 10 RPM (Ref. not used)
Screw working length = 8 foot (96 inches)
Screw reserve length = 2 foot (24 inches)
Screw maximum working length when H2O level is low = 10 foot
Volume per 'bucket' filled with water = 25% to 35%
Notes:
As used here a 'bucket' is the space available for one 'chunk' of the water to ride.
These bucket volume percentages are low and hi estimates based on research reading.
I wanted to actually calculate the volume but such is very complex requiring a computer program.
The upper percentage volume limit of any screw pump is just shy of 60%.
Most screw pumps have a slope of 30° and an ideal percentage volume around 26%.
A shallower slope allows for a larger volume percentage.
Bessler's pump is drawn as a 20° slope and thus could have had a 35% or greater H2O volume.

==============================

Notes:
I've tried to make my calculation as clear as possible.
I've tried to show all my calculations.
I've done a little bit of rounding of numbers to keep it readable.
My goal was an end result number comparable to lifting a weight from the axle.
If anyone does not understand anything then just ask.


Empty bucket volume = ( Pi × (1.5_ID ÷ 2)^2 - Pi × (1/4_Axle ÷ 2)^2 ) × 1/2_Ft_Lead_Length
Empty bucket volume = 0.859 Cubic_Foot

Bucket volume of H2O = 0.859 × 25% = 0.215 Cubic Foot (Conservative estimate)
Bucket volume of H2O = 0.859 × 35% = 0.301 Cubic Foot (Aggressive estimate)

Ref: Weight of H2O = 62.3 Lbs per cubic foot at 69ºF per machinery's handbook

Bucket H2O weight = 0.215 × 62.3 = 13.378 Lbs (Conservative estimate)
Bucket H2O weight = 0.301 × 62.3 = 18.731 Lbs (Aggressive estimate)

Ref: 1 Gallon water weighs 8.328 Lbs.
Gallons H2O per Bucket = 13.378 ÷ 8.328 = 1.61 Gallons (Conservative estimate)
Gallons H2O per Bucket = 18.731 ÷ 8.328 = 2.25 Gallons (Aggressive estimate)

Ref: 1 Liter of water weighs 2.2 Lbs
Liters H2O per Bucket = 13.378 ÷ 2.2 = 6.08 Liters (Conservative estimate)
Liters H2O per Bucket = 18.731 ÷ 2.2 = 8.51 Liters (Aggressive estimate)

Number of buckets being lifted = 8_Ft ÷ 2_per_foot = 16 Buckets (When tank is full)
Number of buckets being lifted = 10_Ft ÷ 2_per_foot = 20 Buckets (When tank is 1/2 full)

Weight H2O lifted = 16 Buckets × 13.378 Lbs_per_Bucket = 214 Lbs (Conservative with full tank)
Weight H2O lifted = 20 Buckets × 18.731 Lbs_per_Bucket = 374 Lbs (Aggressive with 1/2 empty tank)

Load with pulley reduction = 214 Lbs ÷ 2 = 107 Lbs load on wheel (Conservative with full tank)
Load with pulley Reduction = 374 Lbs ÷ 2 = 187 Lbs load on wheel (Aggressive with 1/2 empty tank)

If the bucket volume was 25% and the tank was 1/2 empty then...
Load = 107 Lbs ÷ 8 × 10 = 134 Lbs load on wheel (Conservative with 1/2 empty tank)

If the bucket volume was 35% and the tank was full then...
Load = 187 Lbs ÷ 10 × 8 = 150 Lbs load on wheel (Aggressive with full tank)

----------------------

So... the bottom line...

Assuming a typical screw pump with each bucket filled 25% and with the water tank filled as shown in the drawing, then the load was 107 Lbs, which is slightly less than lifting the 112 Lb weight.

If we assume that the water level in the tank dropped to half full then the load would have increased by about 25% to the equivalent of lifting a load of 134 Lbs, which is about 20% more than lifting the 112 Lbs weight.

Assuming a less typical screw pump with each bucket filled 35% due to the shallow slope angle and with the water tank filled as shown in the drawing, then the load was 150 Lbs, which is about 34% more than when lifting the 112 Lb weight.

If this was the case and the water dropped to half full, then the load would increase to the equivalent of lifting 187 Lbs, which is 67% more than lifting 112 Lbs.

So it is not unreasonable to assume that the wheel might have been able to maintain its 26 RPM speed when lifting the 112 Lb weight (witnesses said that it maintained its speed) but then also slow down when pumping water due to a much greater load that it might have been asked to carry.

At the beginning of this posting are the assumptions. At the end are the possible results. If anyone wants to, they can change the assumption and recalculate the results. Hopefully I've made this clear enough that anyone could do the math.

Also, hopefully I've not made any mistakes.

You guys really have no clue, do you ? A water wheel used buckets.
And if someone built windmills, do you think they would build water mills ?
His clues point to this understanding.

John

[ovyyus]

John, not sure what you're on but I'm tempted to try some :P

What's being discussed here is the water screw which was driven by Bessler's Kassel wheel, depicted in his Kassel wheel illustration and described by witnesses. Capishe?

[wheelrite]

I'm not skilled enough to work out the screw math(s), but I will say that from what we know of the mass/weight of the entire wheel assembly, and what I think is a reasonable amount of weight it could lift at the observed speed (its like my intuitive 'believable ratio of effect'), granted from experience and expectation only not figures, somewere between 1 and 2 hundred pounds would seem 'do-able'. My same 'intuition' tells me this reported work over a period could not be acheived as described by a simple wound/sprung mechanism, or atmospheric, magnetic, or human drive. Only something fairly heavy with weights swinging/dropping/flinging whatever, seems to fit the bill. Not very scientific I know......maybe I rely on it to much and should just learn some more math.
Regards
Jon

[Fletcher]

Basically John, the calcs aren't too bad for the water screw - there will be some margin of error for missing details & assumptions that have to be made - it pays to keep it simple because added complexity does not increase the accuracy of the result with so many unknowns.

Assume the system is in equilibrium.

How much volume of water is discharging [or emptying from the reservoir] every minute for example - from that you can calculate the weight of the water by the height it is being lifted.

You need to estimate the efficiency of the water screw lift e.g. like propeller or turbine efficiency - there are modern day equivalents to get some idea.

This is the Work Done & when factored over time gives you the Joules [N-m] per second which is the Watts [Power].

You'll end up with a likely range of Power & Work Done - Pessimistic, Realistic & Optimistic estimates.

I think ovyyus's calcs seem fairly robust & close enough so I will be interested if Jim's estimates are vastly different.

[nicbordeaux]

I've still not installed Visual Basic Professional Ver.6 on my computer. I've just been too busy with my work, with getting firewood cut, hauled, and stacked in the basement, and other things that life throws at me.

Once VB is installed then it will probably take 16 to 24 hours or more to write a program to calculate the volume of the turning screw.

The reason I wanted VB Ver.6 is that I'm more familiar with it. I used it for a number of years before I switched computers and lost it. I have Ver.5 installed, but it sucks compared to Ver.6. The biggest slow-down to writing VB code is getting the syntax exactly right. VB Ver.6 has a real good syntax help feature. Also, I have written many sub-program that help animate things like wheels and such, but they will not work with Ver.5 because they were written in Ver.6. Also when I write code the syntax is often wrong because I forget to use the older Ver.5 syntax, and then I cannot remember to older syntax, so I go hunting through my reference books looking for the right older syntax, which is a big waste of time.

The bottom line is, this project will have to wait until I have a free weekend with nothing else planned. Sorry.. it's just not my top priority.

Just out of curiosity, Jim, what length of log are you using ? Use a chainsaw or a buzzsaw ? Fell your own trees or buy in ? Open fire or stoves ?

[nicbordeaux]

Sorry to be ignorant, but are we absolutely sure bessler claimed to be lifting water via a screw ? Because with a "in equilibrium" system there are more efficient ways of lifting water...

[jim_mich]

Just out of curiosity, Jim, what length of log are you using ? Use a chainsaw or a buzzsaw ? Fell your own trees or buy in ? Open fire or stoves ?

I cut the logs to 18 inches (0.46 meter).

I have a 16" Poulan chainsaw and an electric Black & Decker alligator saw that will cut up to almost 5" diameter.

I own ten acres (about 4.05 hectare) which is about half wooded. A few years ago Michigan became infested with the Emerald Ash Bore, which is a grub that crawls around between the wood and the bark and kills the trees in one to two years time, then emerges as a green beetle. I had maybe fifty ash trees which are now all dead. So far I've burned maybe ha;f of them.

Last spring I sold about a dozen walnut trees and the buyer left everything smaller than about 14 inches.

I have a 32 year old Rite-way brand wood/gas forced hot air furnace that will take logs up to about 24".

I also have an Ashley wood stove, but have not used it in many years.

Many years ago I bought wood because my trees were too young and small, but now I have too much wood because of the beetle.

Yes, Bessler used a screw to lift water.

[john]

The facts, as can best be determined by scaling Bessler's drawing:

Screw length = 12 foot (144 inches)
Screw lead = 1/2 foot (6 inches)
Screw ID = 1.5 foot (18 inches)
Screw axle = 1/4 foot (3 inches)
Screw tilt angle = 20º
Wheel square pulley = 1.4 foot across corners (16.8 inches)
Screw square pulley = 2.8 foot across corners (33.6 inches)
Pulley ratio = 2:1 reduction.
Wheel speed = 20 RPM (Ref. not used)
Screw speed = 10 RPM (Ref. not used)
Screw working length = 8 foot (96 inches)
Screw reserve length = 2 foot (24 inches)
Screw maximum working length when H2O level is low = 10 foot
Volume per 'bucket' filled with water = 25% to 35%
Notes:
As used here a 'bucket' is the space available for one 'chunk' of the water to ride.
These bucket volume percentages are low and hi estimates based on research reading.
I wanted to actually calculate the volume but such is very complex requiring a computer program.
The upper percentage volume limit of any screw pump is just shy of 60%.
Most screw pumps have a slope of 30° and an ideal percentage volume around 26%.
A shallower slope allows for a larger volume percentage.
Bessler's pump is drawn as a 20° slope and thus could have had a 35% or greater H2O volume.

==============================

Notes:
I've tried to make my calculation as clear as possible.
I've tried to show all my calculations.
I've done a little bit of rounding of numbers to keep it readable.
My goal was an end result number comparable to lifting a weight from the axle.
If anyone does not understand anything then just ask.


Empty bucket volume = ( Pi × (1.5_ID ÷ 2)^2 - Pi × (1/4_Axle ÷ 2)^2 ) × 1/2_Ft_Lead_Length
Empty bucket volume = 0.859 Cubic_Foot

Bucket volume of H2O = 0.859 × 25% = 0.215 Cubic Foot (Conservative estimate)
Bucket volume of H2O = 0.859 × 35% = 0.301 Cubic Foot (Aggressive estimate)

Ref: Weight of H2O = 62.3 Lbs per cubic foot at 69ºF per machinery's handbook

Bucket H2O weight = 0.215 × 62.3 = 13.378 Lbs (Conservative estimate)
Bucket H2O weight = 0.301 × 62.3 = 18.731 Lbs (Aggressive estimate)

Ref: 1 Gallon water weighs 8.328 Lbs.
Gallons H2O per Bucket = 13.378 ÷ 8.328 = 1.61 Gallons (Conservative estimate)
Gallons H2O per Bucket = 18.731 ÷ 8.328 = 2.25 Gallons (Aggressive estimate)

Ref: 1 Liter of water weighs 2.2 Lbs
Liters H2O per Bucket = 13.378 ÷ 2.2 = 6.08 Liters (Conservative estimate)
Liters H2O per Bucket = 18.731 ÷ 2.2 = 8.51 Liters (Aggressive estimate)

Number of buckets being lifted = 8_Ft ÷ 2_per_foot = 16 Buckets (When tank is full)
Number of buckets being lifted = 10_Ft ÷ 2_per_foot = 20 Buckets (When tank is 1/2 full)

Weight H2O lifted = 16 Buckets × 13.378 Lbs_per_Bucket = 214 Lbs (Conservative with full tank)
Weight H2O lifted = 20 Buckets × 18.731 Lbs_per_Bucket = 374 Lbs (Aggressive with 1/2 empty tank)

Load with pulley reduction = 214 Lbs ÷ 2 = 107 Lbs load on wheel (Conservative with full tank)
Load with pulley Reduction = 374 Lbs ÷ 2 = 187 Lbs load on wheel (Aggressive with 1/2 empty tank)

If the bucket volume was 25% and the tank was 1/2 empty then...
Load = 107 Lbs ÷ 8 × 10 = 134 Lbs load on wheel (Conservative with 1/2 empty tank)

If the bucket volume was 35% and the tank was full then...
Load = 187 Lbs ÷ 10 × 8 = 150 Lbs load on wheel (Aggressive with full tank)

----------------------

So... the bottom line...

Assuming a typical screw pump with each bucket filled 25% and with the water tank filled as shown in the drawing, then the load was 107 Lbs, which is slightly less than lifting the 112 Lb weight.

If we assume that the water level in the tank dropped to half full then the load would have increased by about 25% to the equivalent of lifting a load of 134 Lbs, which is about 20% more than lifting the 112 Lbs weight.

Assuming a less typical screw pump with each bucket filled 35% due to the shallow slope angle and with the water tank filled as shown in the drawing, then the load was 150 Lbs, which is about 34% more than when lifting the 112 Lb weight.

If this was the case and the water dropped to half full, then the load would increase to the equivalent of lifting 187 Lbs, which is 67% more than lifting 112 Lbs.

So it is not unreasonable to assume that the wheel might have been able to maintain its 26 RPM speed when lifting the 112 Lb weight (witnesses said that it maintained its speed) but then also slow down when pumping water due to a much greater load that it might have been asked to carry.

At the beginning of this posting are the assumptions. At the end are the possible results. If anyone wants to, they can change the assumption and recalculate the results. Hopefully I've made this clear enough that anyone could do the math.

Also, hopefully I've not made any mistakes.

Jim,
I've read this thread some and came up with this as a reply.

Ralph,
Mind if I throw a little math back at you ?
If as it says on the front page, the wheel was
11 feet in diameter and 1 foot wide, then 2r*Pi=
circumference or 11*3.142 = 34.562.
Times 1foot and it is a surface area of 34.562
square feet. If water were used to create the
over balance, then 34.562*1 = x.
(34.562feet * 12)*1 = x, then 414.744*1= 414.7
cubic inches. And 414.7/28=14.8.
This would 14.8 pounds. 14.8/3= 4.9 pounds over
120 degrees of the wheel for every inch of water.
And considering what you've understood about the
water being pumped, it is at a 15 degree inclination,
then it is further reduced by 25% in the energy
needed to pump it or an 8 to 1 reduction.
This would mean that it would consume a low figure
of about 12.5% of the net force of the wheel to
pump it.
As it is, the drawing could also show in a sense
how the wheel works, water gushes from one pump
to another. The easiest place to hide anything is
out in the open.
And with the wheel having a width of 1 foot, then
a tube with a 1 foot diameter full open can hold
1*3.142= 3.142 square inches * 138.25 inches =
434.4. 434.4/28 = 15.5 pounds over about 120 degrees
of arc on the wheel.
And if you consider this 15.5 pounds is at 5 1/2 feet,
then you have 85.25 foot pounds of torque.
How much torque is required to operate the pump ?

I went over this quickly at home. I can clean it up later if you like and it is based on water having a density of 28 cubic inches per pound.
With water covering 90 degrees arc of the wheel, it would be 85/4*3 or still over 63 foot pounds of torque. Want more power, then consider an oil/mercury mix. In Bessler's times, leather was used for canteens to store water. Water would help to keep soft lather hydrated and water tight. Yet, 63 pounds of torque with an 8:1 reduction is the same as saying 63*8=504. If at what was it you mentioned, 125 pounds ? If so, then that based on math would slow the wheel by 25%. Something that might not be very noticeable if the wheel is rotating at les than 20rpm.

John

[rlortie]

john,

For some unknown reason you have addressed the above follow up of Jim's quote to me!

I am not into the mathematics of water nor gaining any applied torque from it. I am a firm objective believer that water seeks its own level.

Therefore you cannot use water to pump water gaining any additional torque without a reduction or increase of ambient pressure on one side of the wheel. The only device I am aware of that utilizes velocity of water for a gain in elevation is a 'Ram Pump'

If I have misinterpreted your meaning, please acknowledge.

Ralph

[john]

Ralph,
You might have misunderstood.
I was going to simplify the math for Jim and anyone else interested.
Going by the dimensions of the wheel on the front page, 1 foot wide and 11 feet tall;
If the base of a column of water is 12 inches wide by 6 inches deep, this is 72 square inches.
For every 1 inch of depth, it is about 2 1/2 pounds. A wheel of 11 feet in diameter has a quarter
section (45 dergees above and below the level of the axle) of about 8 feet. To round it, about 100 inches.
A vertical column of water would weigh about 250 pounds.
The project I am working on would demonstrate mechanics that could utilize this potential.
As such, by having an understanding of about how much power the screw pump would
require would give a starting point in understanding how much water an over balance would
need and if the mechanics I will be demonstrating would be within those required.
One of the unique aspects of what I will be demonstrating is that if needed, vacuum can be used to assist in lifting the water. This would allow weights performing no work to perform work.
In such a scenario, it may be possible to have 4 out of 8 weights working together to achieve an
over balance.

John

[ovyyus]

John, the least you could do is get the Kassel wheel dimensions correct before trying to 'simplify' anything. It was 12 feet diameter x 18 inches thick.

On second thoughts, why not start your own topic to explore your own pet theory rather than cluttering this topic with rambling nonsense.

[eccentrically1]

I think I just threw up a little math