• #1
15
10
I am planning to build a Hero's steam engine or Aeolipile. This is the steam engine invented by the Greeks about 2000 years ago. It is attributed to Hero or Heron (10-70 AD) but was also referred to in 15 BC. My grandfather build one for a show. I have inherited it but rather than try to repair it i plan to rebuild it.

http://drevanlewis.com/Hero-Steam-engine.jpg

I have calculated the thrust produced by the jets of steam and the power output compared with the energy required to produce the steam which gives a thermal efficiency of about 1-3%. depending mainly on RPM.

This calculation assumes that the steam nozzles are choked. That occurs when a standing wave forms in the nozzle and prevents the velocity of steam from increasing beyond the speed of sound locally within the steam i.e. mach 1.
This occurs when the pressure difference across the nozzle exceeds a certain critical value.

From EngineeringToolbox.com

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

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

where
pc = critical pressure (Pa)
p1 = inlet pressure (Pa)
n = index of isentropic expansion or compression - or 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 at constant volume. k = 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) = 0.9368 ** 8.407 = 0.577

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
pt = 1 / 0.577 = 1.7331 atm absolute pressure
Subtract 1 atm = 0.7331 atm = 10.77 psi by gauge.

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.

I came across a comment in Physics Forums discussing air escaping from a car tyre, which said that for air the critical pressure is 65 psi. Repeating the above calculations for air gives a similar result of 13 psi. I am surprised that this would occur at such a low pressure of 10-13 psi. Is there something wrong with my calculation?

Any positive or negative comment would be helpful.
 

Answers and Replies

  • #2
256bits
Gold Member
3,383
1,419
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.

You may want to read up some more on choked flow.
https://en.wikipedia.org/wiki/Choked_flow
Choked flow is a limiting condition where the mass flow will not increase with a further decrease in the downstream pressure environment while upstream pressure is fixed. Note that the limited parameter is velocity, and thus mass flow can be increased with increased upstream pressure (increased fluid density).
 
  • #3
15
10
Thanks for the link. It shows a slightly different algebraic rearrangement of the formula I used.

Perhaps I should explain my comment "There is little to be gained by higher pressures. " I had calculated the thrust (which is velocity times mass ejection) produced by the two steam nozzles and found that increasing pressure does increase the density of the steam and therefore increases rate of mass ejection by the nozzle even though the velocity is limited to mach 1. Actually the speed of sound also increases with density and that will increase thrust too.

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 and there is increased risk of mechanical failure (ruptured boiler).

The increased pressure also increases the boiling point and therefore increases the amount of energy required to convert water into steam under pressure. The increased thrust did not seem to be worth the increase in energy required to produce the steam and the increased construction required to build the device to withstand higher pressures. (I need to learn Latex)

Mass Flow through Nozzles:
The mass flow through a nozzle with sonic flow can be expressed as

m = A (n . p . ρ1) ** 1/2 * (2 / (n + 1))**( (n + 1)/2(n - 1) )

where

m = mass flow at sonic flow (kg/s)
A = nozzle area (m2)
ρ = initial density (kg/m3)
p = inlet pressure in Pascals (N.m-2)
 
  • #4
russ_watters
Mentor
20,560
7,209
I came across a comment in Physics Forums discussing air escaping from a car tyre, which said that for air the critical pressure is 65 psi. Repeating the above calculations for air gives a similar result of 13 psi. I am surprised that this would occur at such a low pressure of 10-13 psi. Is there something wrong with my calculation?
I'd be curious to see the actual thread, because that sounds like a misconception I used to hold...
 
  • #5
569
85
Testing in-leakage for vacuum systems (<10mmHg abs) uses calibrated nozzles of various cross-sectional areas. Each nozzle has a stamped flow rate based on sonic flow at the throat. So, the delta-P is basically one atmosphere. Your ~13 psi critical delta-P for air agrees with this.
 
  • #6
15
10
Russ, after a lot of hunting I found where I think you had referred to the critical pressure drop of 65 psi.

https://www.physicsforums.com/threads/velocity-of-compressed-air-blowoff.327909/#post-2291274

Thanks "Insightful". That seems to answer my question about the critical delta-P.

My next problem is to make a boiler that can withstand a gauge pressure of 1 atm, or 4 atm with safety factor of 4. I plan to spin it from sheet copper in a lathe. Probably about 1/32 inches thick. I have little knowledge about strength of material calculations. How should I go about this. Due to the limitation of a 9 inch throw on my lathe I plan to make the tank like a flattened sphere (like an ellipse rotated on its short axis). It would be spun in two pieces shaped like bowls. I also plan to put a cylindrical section between the two bowls to increase volume. These 3 pieces will be welded together with Silphos brazing.
 

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