Lifts 200 grams to a height of 500 mm in 18 seconds.
How many Watts of power is it producing?
The jet size was 1.04 mm.
April 22, 2020

Lifted 0.5 m /18 sec = 0.02777 m/s
or 1.666 m/min

Force on the mass due to gravity
F = m.a

Acceleration due to gravity
a = 10 m/s/s gravity of 1g approx.

Mass of object
m = 200 gm = 0.2 Kg

F = 0.2 x 10 = 2
Kg.m/s/s = Newtons

Work done
W = Force x distance travelled

W = 2 x 0.5m = 1 Newton.m

W = 1 Joule

Power P is rate of doing work. It took 18 seconds

P = Work / time

P = 1 / 18 = 0.0555 Joules/sec = 0.0555 Watts = 55.5 milliwatts.
(0.0000733 horsepower)

The string was winding onto a spool with diameter 7.958 mm to give a circumference of 25 mm.
To lift at a rate of 500mm in 18 seconds = 27.7 mm/s the speed of rotation must be turning at a rate of
27.77 / 25 = 1.108 revolutions/sec = 66.48 RPM (revs per minute).
But it is driven by a gear box (worm drive) with 44:1 gear reduction.
Therefore the rotor was spinning under load at 44 x 66.48 = 2,925 RPM (approximately 3000 RPM).
It had been running significantly faster with no load.

The original jet size was 2.25 mm diameter and it took off with great force at 14 psi, but rapidly ran out of steam pressure and the water supply ran out quickly too. Only by reducing to the smallest sized jets 1/32 inch or 0.89 mm was the boiler able to maintain a pressure of 14 pounds per square inch (psi) or one bar.

The difference in jet area is 5.06 mm2 reduced to 0.792 mm2 which is a 6.3 fold reduction. The thrust produced by steam escaping from the jets is equal to the rate of mass ejected through the jets and that is proportional to the area of the jet. Reducing the jet size will reduce the power output at a given RPM by a factor of 6.3. However, it was found in testing that higher boiler pressures can be maintained with smaller jets and this results in higher sustained RPM and this tends to compensate to some degree for the lower thrust. Using a bigger boiler would allow larger jets to be used.

It took 6 minutes to boil 800 ml water. Raising the temperature from about 16C to 100C.
It takes 1 small calorie to heat 1 gram of water by 1 degree C at atmospheric pressure.
1 calorie = 4.186 Joules

calories = 800 ml x (100-16) = 67,200
Joules = 67,200 x 4.186 = 281,299
It took 6 minutes = 360 seconds.
Power consumption = 281,299 / 360 = 781 Joules/sec = 781 Watts


If we assume the boiler keeps absorbing heat at a similar rate during operation as it did to boil the water initially (781 Watts), the efficiency of converting thermal energy in the steam to mechanical energy can be calculated.
Power output ranged from 0.055 to 0.1 Watts.
0.0555 Watts / 781 Watts x 100 = 0.0071 % efficiency !
0.1 Watts /781 Watts x100 = 0.0128 %
And this does not take into account the inefficiency of the steam boiler
which wastes a lot of heat that does not get absorbed. In the purely theoretical calculation of Appendix B the mass of steam per second produced by each jet was calculated and from that we could calculate how much energy was requited to produce that steam. This gave a lower rate of heat absorption of only 114 Watts for two 0.89mm jets instead of 781 Watts. Using this number increases efficiency to about 0.08% which is still extremely low.


Lifts 300 grams to a height of 500 mm in 15 seconds.
How many Watts of power is it producing?
April 25, 2020

Lifted 0.5 m /15 sec = 0.0333 m/s

Force on the mass due to gravity
F = m.a

Acceleration due to gravity
a = 10 m/s/s gravity of 1g approx.

Mass of object
m = 300 gm = 0.3 Kg

F = 0.3 x 10 = 3
Kg.m/s/s = Newtons

Work done
W = Force x distance travelled

W = 3 x 0.5m = 1.5 Newton.m

W = 1 Joule

Power P is rate of doing work. It took 15 seconds

P = Work / time

P = 1.5 / 15 = 0.1 Joules/sec

P= 0.1 Watts = 100 milliwatts.


The string was winding onto a spool with diameter 7.958 mm to give a circumference of 25 mm.
To lift at a rate of 0.0333 m/s or 33.3 mm/s the speed of rotation must be turning at a rate of
33.3 / 25 = 1.332 revolutions/sec = 80 RPM (revs per minute).
But it is driven by a gear box (worm drive) with 44:1 gear reduction.
Therefore the rotor was spinning under load at 44 x 80 = 3520 RPM.
Running free without a load Version 4 with 1mm jets ran at 3,800 RPM by laser pointer.

A larger winch spool with a circumference of 80mm instead of 25mm was used and since it would lift the weight faster the height of the lift was increased from 500mm to 1170mm. On several runs it took an average of 15 seconds. Then the propane ran out and stopped further testing. The results are similar to tests with the smaller spool and slower lifting rate because a smaller weight was used.
The increased tension from the string caused it to pull the worm gear out, preventing it from meshing. An extra pulley was added to the crane to alter the angle of the tension on the pulley. This does not alter the gear ratio.
100 grams to a height of 1170 mm in 15 seconds.
How many Watts of power is it producing?
May 1, 2020

Lifted 1.17 m /15 sec = 0.078 m/s
or 4.68 m/min

Force on the mass due to gravity
F = m.a

Acceleration due to gravity
a = 10 m/s/s gravity of 1g approx.

Mass of object
m = 100 gm = 0.1 Kg

F = 0.1 x 10 = 1
Kg.m/s/s = Newtons

Work done
W = Force x distance travelled

W = 1 x 1.17m = 1.17 Newton.m

W = 1.17 Joule

Power P is rate of doing work. It took 18 seconds

P = Work / time

P = 1.17 / 15 = 0.078 Joules/sec = 0.078 Watts = 78 milliwatts.

The string was winding onto a spool circumference of 80 mm.
To lift at a rate of 0.078 m/s or 78 mm/s the speed of rotation must be turning at a rate of
78 / 80 = 0.975 revolutions/sec = 58.5 RPM (revs per minute).
But it is driven by a gear box (worm drive) with 44:1 gear reduction.
Therefore the rotor was spinning under load at 44 x 66.48 = 2,574 RPM (approximately 2500 RPM average speed while lifting a load).

The maximum gas flow through a nozzle is determined by the critical pressure when a shock wave forms and prevents further increase in flow. This occurs when the flow equals the speed of sound locally within the gas. The critical pressure ratio defines the pressure gradient which will result in maximum flow. Critical flow nozzles are also called sonic chokes. They establish a fixed flow rate unaffected by the pressure difference and are used for this purpose in engineering applications. (The symbol ** is used to denote 'to the power of' so 2**3 = 8 means 2 to the power of 3.)

The ratio between the critical pressure and the inlet pressure for a nozzle can be expressed as (1)

pc / p1 = (2/(n + 1) ) ** ( n/(n - 1) )      

where
pc = critical pressure (Pa)
p1 = inlet pressure (Pa)
n = index of isentropic expansion or compression - also known as the polytropic constant.

(For a perfect gas undergoing an adiabatic process the index - n - is the ratio of specific heat of the gas at constant pressure divided by the specific heat of the gas at constant volume. cp / cv. )
The values for n in steam where most of the process occurs in the wet region is n = 1.135. (In superheated steam it increases to n = 1.30 and some sources say 1.33 and for air is 1.4. ) We will use:

n=1.135
pc / p1 = ( 2 / (n + 1) ) ** n / (n - 1)  
pc / p1 = ( 2 / 2.135) ** (1.135 / (0.135)
pc / p1 = 0.9368 ** 8.407
pc / p1 = 0.577

pa is atmospheric pressure or 1 atm (outlet pressure)
pt is absolute pressure in the tank (inlet pressure)

At the critical point
pa / pt = 0.577
pt = pa / 0.577

The critical point ratio of 0.577 is similar to that reported elsewhere (1).
pt = 1 / 0.577 = 1.7331 atm absolute pressure
Subtract 1 atm = 0.7331 atm = 10.77 psi = gauge pressure.

Any tank pressure greater than 10.77 psi will produce a sonic choke with velocity limited to mach 1 ! There is little to be gained by higher pressures. The following calculations are for 10-15 psi. Then we can use the equation below for choked flow to calculate mass flow.

The mass flow through a nozzle with sonic flow can be expressed as (2)

mc = Ac (n p1 d1)1/2 (2 / (n + 1))(n + 1)/2(n - 1)    

where
mc = mass flow at sonic flow (kg/s)
Ac = nozzle area (m2)
d1 = initial density (kg/m3)
p1 = inlet pressure in Pascals (N.m-2)

Evaluating the term on the right:
(2 / (n + 1))(n + 1)/2(n - 1)  

( 2 / ( 2.135 ) ** ( 2.135 / (2* 0.135 ) ) = 0.9368 ** 7.9074 = 0.5966
mc = 0.5966 * Ac (n p1 d1)1/2

Thus mass flow is proportional to the area of the nozzle, and proportional to the square root of pressure and density of the gas.

Units N = Kg . m /(sec.sec). Pascals are Newtons per square meter.
mc = A (m.m) ( n(none) . p(N/m.m) . d1(Kg/m.m.m) ) ** 0.5
mc = A (m.m) ( n(none) . p( Kg . m /(sec.sec) /m.m) . d1(Kg/m.m.m) ) ** 0.5
mc = A (m.m) ( Kg .Kg / (sec.sec) / . m / m.m.m.m.m ) ** 0.5
mc = A (m.m) ( Kg/s/m.m)
mc = Kg / s


Calculating mc depends on the jet size
Originally calculated for a jet size of 1.5 mm (1/16 inch is 1.5875) and tank gauge pressure of 14.7 psi (1 atmosphere)
Ac is the area of the jet apperture.

n = 1.135 dimensionless
d1= 1.143 Kg/m3 at a tank gauge pressure of 1 atm (see below)
p1 = absolute tank pressure = 2 atm. We want this in SI units: Pascals
Atmospheric pressure is 101,325 Pascals ( Newtons per square meter )=101.3 kPa
p1 = 101,325 x 2 = 202,650 Pa

mc = 0.5966 * Ac * (1.135 x 1.143 x 202,650 ) ** 0.5
mc = 0.5966 * Ac * 512.7
mc = 305.9 * Ac

Using a single 2.25 mm jet:
Ac = 3.98 sq.mm = 3.98E-6 sq.meters
mc = 305.9 * 3.98E-6
mc = 0.00126 Kg/s
mc = 1.26 g/s ejected
500 ml of water with 2 jets would last 500 / 1.26 /2 / 60 = 3.3 minutes

Using a single 1.5 mm jet:
Ac = 1.5 mm diam. = 2.25 sq.mm = 2.25 E-6 sq.meters
mc = 305.9 * 2.25E-6
mc = 0.000688272023797 Kg/s
mc = 0.688 g/s ejected by a 1.5mm jet
500 ml of water with 2 jets would last 500 / 0.688 / 2 / 60 = 6 minutes

Using a 1mm diameter jet the radius is 0.5mm and
Ac = 0.785 sq mm or 0.785E-6 m.m
mc = 305.9 * 0.785E-6
mc = 0.000240 Kg/s
mc = 0.240 g/s mass ejected per second
500 ml of water with 2 jets would last 500 / 0.240 / 2 / 60 = 17.5 minutes

Using a 0.89 mm diameter jet the radius is 0.445mm and
Ac = 0.445 sq mm or 0.445E-6 m.m
mc = 305.9 * 0.445E-6
mc = 0.000136 Kg/s
mc = 0.136 g/s mass ejected per second
500 ml of water with 2 jets would last 500 / 0.136 / 2 / 60 = 30 minutes


The velocity v is the speed of sound within the steam inside the nozzle.
v = 377 m/s in steam at 100 C or 373 K and 14.7 psi or one atmosphere.
377 m/s = 1357 km/h = 848 MPH.


Thrust = mc . (v-s)
Where
v= velocity of steam jet, which in a choked jet is the speed of sound locally
s= velocity of “vehicle” through the air or the velocity of the jet in this case.
s= 1319 m/min for a sphere with diameter 12 cm at 3500 RPM.
s= 79 km/h
s is negligible compared with v which is 1357 km/h:

Thrust = mc . v where v is the speed of sound
Thrust = mc . v
For a 2.25 mm jet, Thrust = 377 m/sec . 1.260 g/sec = 0.4750 Newtons
For a 1.50 mm jet, Thrust = 377 m/sec . 0.688 g/sec = 0.2590 Newtons
For a 1.00 mm jet, Thrust = 377 m/sec . 0.240 g/sec = 0.0900 Newtons
For a 0.89 mm jet, Thrust = 377 m/sec . 0.136 g/sec = 0.0512 Newtons

Note: Newtons force is Thrust in m/sec x mass ejected in Kg/sec. I converted to g/sec for clarity.


Speed (s) refers to how fast the nozzle is actually traveling through the air in a circular motion.
(This is to distinguish the velocity of gas ejected from the nozzle v from speed of travel s.)

Assume:
Sphere rotates at 10 revs per second = 600 RPM.
Radius of sphere = 6 cm = .060 m (as in Version 4)
Circumference = 0.377 m

Jet speed s = 0.377 x 10 = 3.77 m/sec at 600 RPM


jet thrust = mc . (v-s)
where v is velocity of sound and s is the velocity or speed of the jet itself. As the jet approaches the speed of sound the thrust reduces to zero. Consequently the jet can never reach the speed of sound but calculating this number will give us some idea of the theoretical upper limits of RPM. If the speed of sound is 377 m/s in steam (about 300 m/s in air), and the circumference=0.377m then the absolute maximum rotation is 377 m/s divided by 0.377 = 1000 revs/s or 60,000 RPM. In practice we achieved 5,400 RPM which is 5,400/60,000 = 0.09 times the speed of sound in steam or 1/11 th. Air resistance, friction and vibration would contribute to the loss of energy. The actual velocity of the jets at 5,400 RPM is 5400x0.377 x 60 = 122 km/h or 73 MPH.
Units:
Work = Force x Distance traveled in Newton.meters = Joules
Power = work per second in Watts = Joules/s = Newton.m/s
Power (P) = speed x Thrust

For one 2.25mm jet P = 3.37 m/s x 0.475 Newtons = 1.600 Watts = 3.200 Watts for 2 jets at 600 RPM = 16.0 Watts at 3000 RPM.
For one 1.5 mm jet P = 3.37 m/s x 0.259 Newtons = 0.873 Watts = 1.746 Watts for 2 jets at 600 RPM = 8.72 Watts at 3000 RPM.
For one 1.0 mm jet P = 3.37 m/s x 0.090 Newtons = 0.303 Watts = 0.606 Watts for 2 jets at 600 RPM = 3.03 Watts at 3000 RPM.
For one 0.89mm jet P = 3.37 m/s x 0.0512 Newtons = 0.172 Watts = 0.345 Watts for 2 jets at 600 RPM = 1.725 Watts at 3000 RPM.


The specific heat of water is 4.186 Joule/gram °C.
The boiling point of water is increased by pressure:

at 0 psi boiling point is 212 F = 100 °C
at 10 psi boiling point is 240 F = 115.5 °C
at 15 psi boiling point is 250 F = 121 °C
at 100 psi boiling point is 338 F = 170 °C

Heating water at 15 psi from room temperature 20°C to its boiling point 121°C is a 101 C increase. The energy required is:
4.186 Joule/gram °C x 101 °C = 422 Joules per gram. (This does not include latent heat of boiling.)

If we are using a 1.5 mm jet it consumes 0.688 grams per second but for two jets we want to boil water at 0.688 gm/s for we need
1.376 gm/s x 422 Joules per gram = 580 J / s = 580 Watts. That is a lot of energy required to produce the steam.


Theoretical Power Output = 8.72 Watts for two 1.5mm jets at 3000 RPM

Efficiency = 8.72 W / 580 W x 100 = 1.5 %

This is for two jets 1.5 mm diameter but the efficiency is the same for any jet size because more power requires more steam in proportion.

The one factor that will alter this result is the speed of rotation it can achieve. This is for 3000 RPM but higher RPM results in proportionately higher power output and proprtionately higher efficiency.

Jet diameter	Thrust	Power produced	Water used	500cc water lasts	Power used*	Efficiency
mm	Newtons	Watts	grams/second	minutes	Watts*	%
2.25	0.475	16.0	2.52	3.3	1063	1.5
1.5	0.259	8.72	1.376	6	580	1.5
1.0	0.090	3.03	0.48	17.5	202.5	1.5
0.89	0.0512	1.725	0.272	30	114.8	1.5

Increased pressure increases the density of ejected gas e.g. at 2 atm tank pressure the density increases from 1.14 to 1.70 but the rate of mass ejection is proportional to the square root of density which changes from 1.067 to 1.303. Meanwhile the speed of sound only increases very slightly so there is relatively little reward for increasing the pressure beyond the critical pressure which is 10 psi if these calculations are accurate. Designing for 15 psi should cover the possibility that some variables cannot be estimated accurately.

Thrust is determined by the mass of steam ejected per second and the power required to create the steam is also proportional to the mass of steam ejected per second. Since thrust and power are proportional to the area of the jet, increasing the area of the nozzle has the same effect as increasing the number of nozzles. Doubling the jet nozzle area doubles the power but efficiency remains the same. Doubling the number of nozzles doubles the power, but efficiency remains the same.


Increased RPM will increase power proportionately, without requiring more steam production. It therefore increases efficiency proportionately as well. It is difficult to estimate just how fast the sphere will spin.


The theoretical maximum revolutions would occur when the jets reach the speed of the steam leaving the jet nozzles. This is the speed of sound 377 m/s. Factors such as friction and air resistance are ignored here. Coincidentally the circumference of a 12 cm diameter sphere is 0.377 m. So maximum revolutions is 377 / 0.377 = 1000 revs per second = 60,000 RPM. This is not going to happen as the equation for thrust is mc.(v-s) and as the speed of the jets (s) approaches the speed of sound (v) the thrust reduces towards zero. But gives us an upper limit. If we combine high speed at 12,000 RPM with the the largest nozzle diameter of 2.25mm we could achieve about 50 to 64 W of power but it would require a large boiler.


1 Watt = 1 Newton . m/s
The force exerted by a 1 Kg weight hanging on a string is
F = m.g
W = F . d where d represents distance traveled
P = W / sec
P = F.d/s
P = m.g.d/s
L = d/s where L is speed of lifting

L = P /( m . g ) ie at 1 Watt
L = 1 Watt /( 1 Kg. 10 m. s-1. s-1)
L = 0.1 ( N.m .s-1 ) / N
L = 0.1 m .s-1 = 0.1 m/s
L = 10 cm/s

0.1 Watt could lift 100 gm at approximately 10 cm/sec
1 Watt could lift 1 Kg at approximately 10 cm/s
12 Watt lifts 12 Kg at 0.1 m/s = 10 cm/sec or 6 m / minute.

If we substitute water, one Watt could pump 1 Kg or 1 liter of water at a rate of 10 cm/second or 600 cm/minute.
In other words one Watt could pump water to a height of 600 cm at one liter / min.
or 30 cm (one foot) at 20 liters per minute (1/3 liter per second).

If we have a water wheel turning at one turn per second or 60 RPM (with a 10:1 pulley ratio from the spinning sphere at 600 RPM) and lifting water 30 cm it would need buckets to carry 20 liters/60 RPM = 1/3 liter per turn = 333 ml. If it had 10 buckets then the size would be 33.3 ml per bucket. Similar calculations can be applied to an Archimedes Screw.

REFERENCES for Appendix B are listed with other references.

It took 6 minutes to boil 800 ml water, raising the temperature from about 16C to 100C at atmospheric pressure.
It takes 1 Calorie (kilocalorie) to heat 1 Kg of water by 1 degree C at atmospheric pressure.
It takes 1 calorie to heat 1 g of water by 1 degree C at atmospheric pressure.
(Note: the small calorie, usually written with a small c is the heat required to raise the temperature of 1 gm instead of 1 Kg).
One calorie equals 4.186 Joules according to the SI standards.
One Calorie equals 4186 Joules according to the SI standards.
To heat 800 g of water from 16 to 100C (a difference of 84C) requires 800x84=67,200 calories
67,200 calories x 4.186 = 281,299 Joules.
It took 6 minutes = 360 seconds.
Power consumption = 281,299 / 360 = 781 Joules/sec = 781 Watts.
We assume the boiler continues to absorb this amount of heat during operation.
Note that at a gauge pressure of 15 psi the water will boil at 120C but this test was done at atmospheric pressure.


If the boiler continues to absorb 781 Watts of heat during operation, the efficiency of converting this thermal energy in the steam into mechanical energy is
0.0555 Watts / 781 Watts = 0.000071 = 0.0071% efficiency !
for Hero Engine version 3 running at 3000 RPM with 0.89 mm jets.
And this does not take into account the inefficiency of the steam boiler
which wastes a lot of heat that does not get absorbed.



https://en.wikipedia.org/wiki/Rotation_around_a_fixed_axis
https://en.wikipedia.org/wiki/Moment_of_inertia
https://en.wikipedia.org/wiki/Torque

These calculations were done before the engine was built
and differs from the final modified designs.
This assumes a relatively small sphere 4 cm radius with two 1.5 mm jets extended out to 6 cm radius.

Moment of inertia of a Point Mass:
m = Mass
r = radius

I = m . r**2


Each nozzle can be considered as a point of mass
Density of copper = 8.92 gm/cc
Mass estimated:
2 cm pipe
diameter D = 0.8 cm
radius R = 0.4 cm
thickness T = 0.15 cm
volume V = 2.Pi.R.T
mass = 2.Pi.R.T .d
Distance from axis, radius r=6 cm

I = m.r**2
I = 2.Pi.R.T .d . 6**2
I = 121 gm.cm.cm (for each nozzle)


The original shaft through the center of the sphere is mainly hollow.
Diameter =1.2 cm
Radius r = 0.6 cm
Thickness T=1.5 mm =0.15 cm
Length L= 15 cm

I = m.r**2

m = Volume X density of metal
V = 2 Pi. r . T . L
m = 8.482300 x 8.92 = 75.66 gm
I = 75.66 x 0.6**2

I = 27.23 gm.cm.cm


This was done as an exercise in integration.
Later modified for pipes rather than rods.
Rod Diameter D
Radius of rod R
Length L
density d = 8.92


mass = Volume X density of metal
Consider a small element at distance r along the rod
length of element is delta r =dr
Volume of the element is Pi . R**2 . dr
Mass of the element is d . Pi . R**2 . dr
Moment of inertia of this element at radius r from its axis is
delta i = m . r**2
delta i = d . Pi . R**2 . dr . r**2
delta i = k. r**2. dr
where constant k = d . Pi . R**2
Overall inertia is the sum of all these small elements over the length of the rod.
If the elements are infinitely thin, this is described by an integral:
Integrate k. r**2 .dr over the length of the rod from r1=0 to r2=L

I = k. Integral r**2 .dr
I = k. [1/3 r**3]
I = 1/3 . d. Pi. R**2 . L**3 after substituting r1 and r2
Units of I:
gm.cm**-3 . cm**2 . cm+3 = gm . cm . cm (correct)


Diameter D =0.6 cm
Radius R = 0.3 cm
Thickness T=1.5 mm =0.15 cm
Length of pipe L = 6 cm

Consider a small element at distance r along the rod
length of element is delta r =dr
Volume of the element is 2.Pi . R.T . dr
Mass of the element is d . 2.Pi . R.T . dr
Moment of inertia of this element at radius r from its axis is
delta i = m . r**2
delta i = d . 2.Pi . R.T . dr . r**2
delta i = k. r**2. dr
where constant k = d . 2.Pi . R.T
Overall inertia is the sum of all these small elements over the length of the rod.
If the elements are infinitely thin, this is described by an integral:
Integrate k. r**2 .dr over the length of the rod from r1=0 to r2=L

I = k. Integral r**2 .dr
I = k. [1/3 r**3]
I = 2/3 . d .Pi . R.T. L**3 after substituting r1 and r2

Evaluate numerically:
I = 2/3 . 8.92 Pi * 0.3 * 0.15 x 6**3
I = 181 gm.cm.cm


I = 2/3 . m . r**2

where m = mass of material (metal) in the shell.
m = Area . thickness . density
Sphere has an area of
A = 4 Pi . r**2

Combining equations for area, density and inertia:
I = 2/3 . m . r**2
I = 2/3 r**2 . 4 Pi r**2 * T * d
I = 8/3 Pi r**4 . T . d (units are correct)

Let
radius r = 4 cm
thickness of copper is T = 0.5 mm = .05 cm or 20 thou inches
density of copper d =8.920 gm/cc
I have a sheet of copper 0.075” thick > 1/16”

m = 4 Pi r**2 * t * d
units cm2 * cm * gm/ cm3 = gm
m = 4 Pi * 16 * 0.05 * 8.92 gm/cc = 89.67 gm copper

I = 2/3 . m . r**2
I = 2/3 * 89.67 * 4**2
I = 956.5 gm.cm.cm


I (shaft) = 27.23
I (sphere) = 956
I (radial pipes) = 2 x 181 = 362
I (nozzles) = 2 x 121 = 242
I (TOTAL) = 1587 gm.cm.cm

Units:
I = 1587 E-3 Kg.cm.cm
I = 1587 E-7 Kg.m.m
I = 0.1587 E-3 Kg.m.m


Angular acceleration (alpha) is the applied torque (tau) divided by the moment of inertia ( I ):
alpha = tau / I

Where
alpha = Angular Acceleration ( radians per second per second )
I = moment of inertia ( kg . m.m )
tau = Applied torque (kg.m.m . radians / sec / sec)

Thrust = 0.1832 N for one jet 1.5mm (from the thrust calculations document)
Thrust = 0.3664 N for two jets
Jets positioned r = 6 cm=.060 m from the axis.


Torque Tau = thrust * r
Tau = 0.3664 * .06 m
Tau = 0.0220 N.m

alpha = tau / I

alpha = 0.0220 N.m / (0.1587 E-3 Kg.m.m) = 138.6

units: Kg. m . s-2 .m . Kg-1 . m-2 = s-2 = radians/second/second

alpha = 138.6 radians/second/second

A full circle is 2 Pi radians.

alpha = 138.6 / 2Pi = 22 Revs / sec /sec


Next we can calculate how long it takes to get to a particular angular velocity w

initial velocity wi = 0 therefore if t is time in seconds

w = alpha . t

After 5 seconds

w= 22 revs/s/s . 10 = 110 revs/sec = 6600 RPM

Conversely, how long does it take to reach 1000 RPM ?

w = 1000 RPM = 16.67 revs/sec

t = w / alpha

t = 16.67 / 22 = 0.75 seconds.

Note:
This assumes a relatively small sphere 4 cm radius with two 1.5 mm jets extended out to 6 cm radius.


P = w . Tau

If w = 1000 RPM 16.67 revs/sec

Tau = 0.0220 N.m

P = 16.67 revs/sec . 0.0220 N.m

P = 0.367 N.m/sec

units = N.m/sec = Joules/sec = Watts

For 2 jets = 0.367 W or 367 mW at 1000 RPM.
At 5000 RPM: 1.83 Watts

Still more than one watt which can pump 20 liter/min to 30 cm or one foot.

But we need a worm drive to reduce RPM of output !

Consider making a pump with an archimedes screw 30 cm long with 10 helical cycles and pitch of 3 cm.
Volume of the 30 cm pipe, 3 cm diameter is
V = Pi. r**2 * L
V = Pi. 1.5**2 * 30 = 212 cc
Divided into 10 spiral segments = 21.2 cc /seg
But each segment is only about half full = 10 cc per segment.
The screw could lift 10 cc water 3 cm pitch per turn. It will require 10 turns to lift 10 cc water 30 cm i.e. one turn/cc.
To get 2000 cc / minute we need it to turn 2000 rpm.

If the engine is running at 5000 RPM we only need a 5:2 or 2.5 : 1 reduction which could be done with pulleys. But a certain safety margin would be good. Perhaps 10:1 reduction would be better with a worm drive would be nice. And a clutch to to let it get up to speed too.

However a pulley system gives more flexible adjustment of the gear ratio. e.g. ranging from 1:1 to 10:1

And a clutch could simply tighten up the slack in the belt.
Could use a light spring on an over-center cam to tighten the belt over a wide range of pulleys.

Make a worm gear drive with a half inch thread with 12 tpi.
Driving a gear wheel with circumference of 3 inches would have 36 teeth. (radius about 1 inch)

One turn of the worm advances one tooth so it would give 36:1 gear reduction.
3600 RPM input gives 100 RPM output

Multiple start thread would give lower gear ratios but harder to make.
I think the gear wheel could be cut using a tap-yes I did that -see my YouTube chanel.

I actually used a bigger metric thread 20 mm pitch 1.25 mm and a corresponding gear wheel with 44 teeth.

The above calculations show that the largest sphere I can spin is 1.6 liters. The preferred volume of the boiler is one liter for safety reasons.

Imagine a cube enclosing the 1.6 liter sphere which has a radius of 7.25 cm. It would be 14.5 cm along each edge.

If we squash the cube so the the volume is decreased by a
factor of 1.6, the height would reduce to 1/1.6 = 0.625 which happens to be a fraction of 5/8ths of the original height. So the spheroid would be rotated about its short axis which is 5/8th of the long axis.

Short Axis = 7.276 x 0.625 = 4.5475 cm
Long Axis radius = 7.276 cm
If we round the measurements to 7.25 cm short radius and 4.55 cm
V = 1001.8 cc

A better method is to rearrange the equation for an ellipsoid:

Ellipsoid:

x2/a2 + y2/b2 + z2/c2 = 1

the lengths of the principal axes are 2a, 2b, 2c so a, b, c are like the radius of a circle.

Ellipsoid of revolution is a special case where sections across a particular axis are circles.

Volume:

V = 4/3 Pi . a.b.c = 4.1888 x a.b.c

This corresponds with the special case of a sphere

V = 4/3 Pi r**3

And it is the same as 2/3 of an enclosing elliptic cylinder

Half the circumference is the size of copper disc we need to start spinning and it must be less than 9 inches diameter.

The 2D ellipse which we rotate to make the ellipsoid is

x2/r2 + y2/c2 = 1

where
r is the radius
c is the depth of the “dish”

An approximation of the perimeter is defined by an infinite series.
According to Ramanujan:

C = 2 Pi [ ( a**2 + b**2 )/2 ]**0.5

The maximum swing of the lathe I am using is 9 inches or 22.86 cm. I will call the diameter of the flat disk I start with D. The volume restricted to one liter. If I increase the diameter I will have to make it thinner (i.e. c becomes smaller). To calculate the optimum shape of the rotated ellipse made from a 9 inch disc and having a volume of 1000 ml, we have to solve the following two simultaneous equations:

Let D be the diameter of the disc which is half the circumference of the rotated ellipse:
D = 0.5 C

D = Pi [ ( r**2 + c**2 )/2 ]**0.5
Volume of rotated ellipse:
V = 4/3 Pi r**3 . c = 1000 cc

With two equations having two unknowns we should be able to solve:
from V

r**3 = 1000 x 3/4 / (c . Pi) = 750/(c.Pi)

(THERE MAY BE AN ERROR HERE in USING R**2 BY MISTAKE but it worked out OK)
We can substitute r**3 into the equation for D and eliminate it
Pi [ (750 / (c .Pi) + c**2) /2 ] **0.5 = 22.86

Square both sides

Pi **2 [ (750 / (c .Pi) + c**2) /2 ] = D**2

multiply both sides by c.Pi

Pi c**3 - c.Pi * 2 *(D/Pi) **2 +750 = 0

Unfortunately this is a cubic polynomial.
Need a TI 89 ! I don’t have one handy.
Instead we can easily solve this numerically by trial and error
D = 22.86 cm
Start with a guess c = 4 cm
gives c= 2.85.
Lets split the difference c= 3.425.
This gives c= 2.63.
Try 2.63. Gives c=2.42.
Try 2.0 Gives 2.32. So the answer is between 2.32 and 2.42.
Try 2.37 gives 2.379.
OK try 2.375 gives 2.380
SO c = 2.38 is close enough

r**2 = 750/(c.Pi)
r**2 = 750 / ( 2.38 * Pi) = 100.3077
r = 10.01 cm = 3.93 inches
diameter of ellipsoid = 20 cm = 8 in

Diameter of the completed ellipsoid is 10 cm with each dish 2.38 deep. Giving a total depth of 4.76 cm = 1.87 inches
Check:

0.5 C = Pi [ ( r**2 + c**2 )/2 ]**0.5

0.5 C = Pi [ ( 10**2 + 2.38**2 )/2 ]**0.5

0.5 C = 23.43 cm =9.22 in approx 9.25

radius of swing = 4 5/8 which is exactly maximum swing!

We wanted 22.83 cm. This doesn’t allow any overlap!

V = 4/3 Pi r**3 . c
V = 4/3 Pi 10*3 * 2.38
V = 997 cc Good!

What if we reduce r from 10 to 9.5 to give 0.5 cm overlap at the seam?

V = 1000 = 4/3 Pi 9.5**3 * c
c = 3/4 * V / ( Pi 9.5**2 ) = ( 750/Pi ) / 9.5**2 = 238.7 / 9.5**2
c = 2.645 cm


Depth of hemi-ellipse = c = 2.645 cm =26.45 mm = 1.04”
Full thickness ellipsoid = 2.645 x 2 = 5.290 cm = 52.9 mm =2.08”
r = 9.5 cm = 95 mm = 3.74”
diameter = 19 cm = 190 mm = 7.48”
Volume V = 4/3 Pi r**3 . c = 999.9 cc
Size of disk to spin
0.5 C = Pi [ ( r**2 + c**2 )/2 ]**0.5 = 22.68 cm
= 8.929 inches

Make a large disc 9” diameter with a 7.48 cm diameter hole in the center (or a bit smaller) to weld to the ellipsoid to prevent flames from directly reaching the joint, and to keep it clean. The stand can also sit under this 3/4 inch ledge. But the copper is soft and would need to have a shaped edge.

This should be able to pump about 20 liters per minute. See above.

As above a 1000 cc sphere is 6.2 cm radius 12.4 cm diameter = 4.88”.
The ellipsoid is 19 cm diameter. 6.6 cm diameter difference or 3.3 cm either side for the vertical tubes. That is sufficient I think.

This is determined by
mc = 0.486 gm / sec for a single1.5 mm jet.
mc = 0.486 x 3600 = 1,750 gm / hour for a single 1.5 mm jet.
mc = 3,500 gm / hour for two 1.5 mm jets

So a 1000 cc tank containing 750 cc would last 750/3500 = 12.85 minutes!

It may take a while to build up steam too. How long will it take to pick up steam? Will it run out of steam before it is up to speed. What if it is connected to a load?

In the section 'Size and Design of the Boiler' it was mentioned that copper tubes were passed vertically through the boiler to increase the surface area available for heat transfer resulting in an increase in steam production. The calulation below shows that simply increasing the diameter of a flat bottomed boiler from 20cm to 35cm would provide the same increase in area but 20cm was the largest size that my lathe could handle. It has a "throw" of 9 inches or 22.8 cm and after folding the edge over for a joint flange decreases the endplates to about 8 inches or 20cm. The tubes were 16mm diameter by 15cm long but less than 7.5cm inside the tank, increasing the surface area inside the tank by 640 square centimeters which is equivalent to a square 25cm along each side. This boiler is 20cm or 8 inches in diameter and the area of the circlular base is 314 sq cm. Adding 640 sq cm would increase the area to 954 sq cm. Divide by Pi and take the square root to get the radius and multiply by two for diameter. That would require a tank 35cm or 13.7 inches in diameter. This is still a modest size and similar to my grandfather's kettle boiler.

The above calculations are for a tank with a flat bottom which would have an area of Pi.r squared. It a hemispherical bottom was used, like the cauldron in the drawings which accompany Hero's manuscripts, the area of a sphere is 4 .Pi. r squared. Half a sphere would be 2.Pi.r squared. So we only need half the radius to get the same area as a flat bottom with area Pi.r squared. So the requirement for 350mm diameter reduces to 175mm or 6.9 inches diameter!

Equations: According to Wikipedia(10)

https://en.wikipedia.org/wiki/Pressure_vessel#Scaling_of_stress_in_walls_of_vessel

It is unfortunate that Greek letters tend to cause difficulties in the standard fonts used on web pages so I have changed them to non-Greek letters. eg Pi is 3.1415926 and ** means ‘to the power of’, as used in computer programming.

Volume of a sphere V=4/3 Pi.r**3

Area of a sphere A = 4 Pi . r2

Area of a circle Pi r2

Pressure vessels are held together against the gas pressure due to tensile forces within the walls of the container. The normal (tensile) stress in the walls of the container is proportional to the pressure and radius of the vessel and inversely proportional to the thickness of the walls.

Therefore, pressure vessels are designed to have a thickness proportional to the radius of the tank and the pressure of the tank and inversely proportional to the maximum allowed normal stress of the particular material used in the walls of the container.

Because (for a given pressure) the thickness of the walls scales with the radius of the tank, the mass of a tank (which scales as the length times radius times thickness of the wall for a cylindrical tank) scales with the volume of the gas held (which scales as length times radius squared).

The exact formula varies with the tank shape but depends on the density, d, and maximum allowable stress s of the material in addition to the pressure P and volume V of the vessel. (See below for the exact equations for the stress in the walls.)

Stress in a thin-walled pressure vessel in the shape of a sphere is

sh = sl = p.r / 2t

where
sh is hoop stress, or stress in the circumferential direction,
sl is stress in the longitudinal direction,
p is internal p gauge pressure,
r is the inner radius of the sphere, and
t is the thickness of the sphere wall.

A vessel can be considered "shallow-walled" if the diameter is at least 10 times (sometimes cited as 20 times) greater than the wall depth.

Yield Strength and Ultimate tensile strength of copper (11)

https://en.wikipedia.org/wiki/Ultimate_tensile_strength

Note: Yield strength is used for design. Divide by 4 for a safety factor of 4.

For 99.9% copper, Yeild Strength is 70 MPa when the metal stretches.
But for a 4:1 safety factor assume it is 4 times weaker requiring 17.5 MPa.
One pascal is a newton force per square meter.

d = 8.92 gm/cc

The safety factor generally used for pressure vessels is 3.5 to 4.0.

We plan to use a gauge pressure of one atmosphere 14.7 psi. In SI metric units pressure is measured in Newtons per square meter or Pascals and one atmosphere, often referred to as one bar is 101,342 Pascals. To get absolute pressure (p1) as used in Boyles Laws, we have to add atmospheric pressure to the gauge pressure.

The limit for NZ pressure vessel regulations is a gauge pressure of 200 kPa or about 30 psi. Our design is within these limits (12) (see the pdf “DESIGN, SAFE OPERATION, MAINTENANCE AND SERVICING OF BOILERS NEW ZEALAND”
https://worksafe.govt.nz/topic-and-industry/machinery/working-safely-with-boilers/).

p1 = absolute tank pressure = 2 atm

p1 = 101,342 Pascals x2 = 202 kPa

I have a Boxford engineer's lathe with 9 inches (22.86 cm) diameter swing or throw. When I spin copper I can start with a flat sheet up to 9 inches diameter and spin it into a hemispherical shape, two hemipheres being later joined together to make a sphere:

Circumference of the sphere is C = 45.72 cm

C = 2 Pi . r
r = C / 2Pi
r = 45.72 / 2Pi
r = 7.276 cm radius. This will be the radius of the sphere.

V= 4/3 Pi . r3 = 1613 cc
V in m3 = 1613 /100 /100 /100 = 0.001613 m3


V = 1 liter = 1000 cu cm = 0.001 m3 (There are 1000 liters in a cubic meter)
V = 4/3 Pi . r3

r3 = 3/4 V / Pi

r3 = 0.75 x 0.001 / 3.1425926 = 0.00023873241464

r = 0.0620 m = 6.2 cm (diameter 12.4 cm = 5 inches approx.)

List of parameters used in stress equations:

V = 0.001 m3
Pressure = 1 atm gauge pressure = 101,342 Pascals (N/m2) = 101.3 kPa
P= 101.3 kPa
density
d = 8.92 gm/cc or Kg/liter (or thousands of Kg Tonns/ m3 )
d = 8,920 Kg/ m3
ds = 1.143 Kg/m3 density of steam at a tank gauge pressure of 1 atm

s = 70 MPa Yeild Strength of copper.
Divide this by 4 for a safety factor of 4
s = 17.5 MPa
s = 17,500 kPa


From the equation given above in ‘Stress in thin-walled spherical pressure vessels’

t = p.r / 2 

as above
s Sigma is stress in the material
p is internal gauge pressure,
r is the inner radius of the sphere, and
t is thickness of the sphere wall.

I our example
p= 101 kPa
r = .062
s = 17,500 kPa

t= 101 x 0.062 / (2 x 17,500) = 0.00017954715637
t = 0.00018 m

Minimum required boiler wall thickness with 4:1 safety factor is

t = 0.18 mm
t = 7 thousandths of an inches


Since t is proportional to pressure we could increase the pressure to 8 atm
and t = 56 thousandths of an inch.

If we used copper 1/32 inch thick
= 0.03125 inch
= 0.79375 mm
we could hold a pressure of

p = 2 t . s / r

p = 2 x 0.00079375 m x 17,500 kPa / 0.062 m
p = 448 kPa
p = 4.42 atm
p = 65 psi with 4:1 safety factor.

If we used 1 mm thick copper

p = 2 x 0.001 x 17500 / 0.062

p = 564.5 kPa
p = 5.57 atm
p = 819 psi


This gives similar results.
For a spherical tank, the minimum mass of a pressure vessel is

M = 3/2. P.V. d / s

where:
M is mass of material required for the pressure vessel,
P is the gauge pressure,
V is volume of the vessel,
d is the density of the pressure vessel material,
s is the maximum working stress material can tolerate.

The largest sphere I can spin on the lathe starting with copper discs 9 inches (228 mm) in diameter.

V = 0.001613 m3 = 1.613 liters

Other shapes besides a sphere have constants larger than 3/2 (infinite cylinders take 2), although some tanks, such as non-spherical wound composite tanks can approach this.

M = 3/2 ( 101.3 x 0.001613 ) 8920 / 17,500

M = 0.12499573064733 Kg

M = 125 gm of copper required

Check the units used:
kPa x m3 x Kg/m3 /kPa = Kg this is correct

If the mass is 125 gm and density of copper is 8.92 gm/cc
the volume of copper required is

V = 14 cc.

The area of the sphere with r = 7.276 cm
A = 4 Pi . r2

A= 665 sq cm

Thickness if T = 14 cu cm / 665 sq cm

T = 0.021044 cm = 0.21 mm = 0.00828 inches = 8 thou inches.

This is in the same ball-park as the above direct calculation of thickness.

CONCLUSION For the strength of a copper sphere:

It would be safe to construct a one liter spherical tank using copper 1/32 inches thick (0.8 mm) as only 0.21 mm is required for a gauge pressure of one atmosphere (bar) with a 4:1 safety factor.

I actually used of copper 30 thou inches thick (0.762mm) which is 3.6 times thicker than required and that should increase the safety factor to 4 x 3.6 = 14.5.



1. Critical Pressure
www.physicsforums.com/threads/at-what-pressure-does-steam-exiting-a-nozzle-reach-mach-1.843944/#post-5293979

2. Choked Flow
https://en.wikipedia.org/wiki/Choked_flow

3. Equations for mass flow through nozzles
http://www.engineeringtoolbox.com/nozzles-d_1041.html

4. Properties of steam
http://help.syscad.net/index.php/Water_and_Steam_Properties#Introduction

5. Velocity of air coming out of a nozzle and choked flow
https://www.physicsforums.com/showthread.php?t=694656

6. Calculation of Thrust from a jet in an aeolipile
https://www.physicsforums.com/threads/need-help-with-project-similar-to-a-sprinkler.744967/

7. Pipe flow calculations
http://www.pipeflowcalculations.com/airflow/

8. Thrust Calculations by NASA
https://www.grc.nasa.gov/www/k-12/airplane/thrsteq.html

9. Physical properties of gases
https://browkinnari.files.wordpress.com/2015/07/handbook-of-physical-properties-of-liquids-and-gases-pure-substances-and-mixtures.pdf

10. Calculation of material strength required for a Pressure vessel (Boiler)
https://en.wikipedia.org/wiki/Pressure_vessel#Scaling_of_stress_in_walls_of_vessel

11. Yield Strength and Ultimate tensile strength of copper
https://en.wikipedia.org/wiki/Ultimate_tensile_strength

12. Design, safe operation, maintenance and servicing of boilers in New Zealand
https://worksafe.govt.nz/topic-and-industry/machinery/working-safely-with-boilers/).