# I Thermodynamics: questions about vacuum engines

1. Nov 20, 2016

### pkv

Hi,
I am trying to understand which thermodynamics process or processes could model a vacuum engine (mostly known by the name "flamelickers") and similar engines..

In a vacuum engine you have a cylinder and piston system that draws hot air from a heat source, cools it and mechanical work is produced from the pressure difference between the cold gas and the atmosphere.

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

These are the numbers I calculated for two simple models.

Data:
Working fluid: 1 mol of N2, 1 bar (101 kpa) at 600K. Density 0.564 kg/m3. Volume 0.0493 m3

Steps:

1.- 1 mol of N2 is heated at ambient pressure to working heat.
Vf = 0.0246 ; From PV = nRT at 300K
Q = Delta Ei + W
For constant pressure:
Q = CV * n * Delta T + P * Delta V
Q = 20.8 * 300 + 101000 * 0.0246 = 8724.6 J heat added

2.- N2 is admitted into the cylinder. Piston stops after drawing 49,3 l (1 mol at 600k).

3.- Valve closes, piston stops, and water spraying cools the working fluid to 300K.
This is a constant volume proccess.

I model this step as an isochoric process, using equation:
Q = (Cv/R) * V * (Pf - Pi)

Result is -6240 J of heat removed from the N2.

I can obtain the same quantity just using temperature delta and CV of N2:
-300 * 20.8 = 6240 J

Pressure now is: PV = nRT -> P = (8.3144598 * 300) / 0.0493 = 50500 pascals

4.-Atmospheric pressure of 1 bar performs compression on the working fluid as piston moves to the bottom. Final pressure of the working fluid is 1 bar.

I am modelling this process for 2 different possibilities: adiabatic and isothermal.

A) Model using simple adiabatic compression.
We know the cylinder will move down until both ends have the same pressure, 1 bar.
Using the adiabatic pressure constant, we can get the final volume.

Pi * (Vi^y) = K const -> 50500 * (0.0493^1.4) = 746.925
Vf^y = K / Pf => 746.925 / 101000 = 0.00739
Vf = 0.00739 ^ (1/1.4) = 0.03 m3

Final temperature will be:
PV = nRT => T = (Pf * Vf) / R => (101000 * 0.03) / 8.3144598 = 365.8K

Work performed on the piston:
W = (K * (Vf^(1-y) - Vi^(1-y))) / (1-y)
W = (746.925 * ((0.03^(1-1.4)) - (0.0493^(1-1.4)))) / (1-1.4)
W = -1368 J

B) Model using isothermal compression.
The water spray continues as the piston compresses the working fluid.

Isothermal work W = n * R * T * log(vf / vi)

Vi = 0.0493
Vf = 0.0246 ; From PV = nRT at 300K
W = 8.3144598 * 300 * log(0.0246 / 0.0493) = -1728.9 J

Heat removed during the process should be equal to work = -1728.9 J

QUESTIONS.

Are my equations well chosen for these processes?
Are the negative works of compression equal to the mechanical working capabilities of the machine? (ignoring looses).

If everything is correct I get thermal efficiencies of:
For adiabatic: 1368 / 8724.6 = 0.156 = 15%
For isothermal: 1728.9 / 8724.6 = 0.198 = 20%

I assume 8724.6 J would be the only heat added during the process, as the isothermal compression is rejecting heat.

2. Nov 21, 2016

### Nidum

Sorry but almost all wrong .

Try just sketching the cycle diagrams and then we'll take a look at the detailed analysis .

3. Nov 21, 2016

### pkv

Oh boy, "almost all wrong".
I hope at least heat addition at constant pressure and heat removal at constant volume are correctly calculated!

I will refrain from drawing the cycle diagrams for now, as I am still grasping basic concepts.
After putting some work in the problem, I think I was totally wrong about my interpretation of the compression work.
Compression work must be done by the piston, until it reaches final 1 bar pressure. The column of air on top of the cylinder can perform as much work as the vacant volume allows. Adding both quantities you get the work available to the machine.

Volume goes from 0.0493 to 0.03. That's 0.0193 m3 for the air column to work.
That would be a cube of sides 0.26m long, and a side area of 0.0718 m2.
For atmospheric pressure, that gives 7284.4 Newtons. Using W = Force * distance we get 1953.1 J.
Total work for the adiabatic model: 1953.1 -1365.0 = 588.2 J
Efficiency: 588.2 / 8734.3 = 0.067 -> 6.7%

Performing the same calculations for isothermal compression I get:
Atmosphere column work = 2502.2 J
Total work for the isothermal model: 2502.2 -1728.9 = 773.4 J
Isothermal model efficiency: 773.4 / 8734.3 -> 8.9%

Is that better or I am still totally lost?

Last edited: Nov 21, 2016
4. Nov 21, 2016

### pkv

For the isothermal model, works seems to balance pretty well: 8724.6 - 6240 - 1728.9 - 773.4 = -17.7
Pretty close to 0...

5. Nov 22, 2016

### pkv

Here there are PV diagrams I worked out. Pressure in bars, volume in m3.
Pressure is absolute, as the vacuum engine works under ambient pressure.
To plot the adiabatic I used the P* (V^y) = K condition:
To plot the isothermal I used the P* V = K condition:

Last edited: Nov 22, 2016
6. Nov 22, 2016

### Nidum

I'll have to think some more about the detail of what actually happens during the engine cycle .

Basic problem with your efficiency calculations though is that you don't really know how much energy is being supplied to the engine or how much energy is being wasted externally .

The engine ingests some part of the flame - that is all the information that you have .

You could make an estimate of energy usage based on basic knowledge of engine operation and visual observation or do some proper experiments to get more accurate data .

Last edited: Nov 22, 2016
7. Nov 22, 2016

### pkv

Well, it's true I don't know much about flamelickers. I know these machines were very very simple and extremely inefficient, but there is little info about them. "Less than 5%" is one of the few references I read.
Things I know:

- Known by the names: flamelicker, vacuum engine, atmospheric engine.
- Simple running sequence:
1-Aspiration of hot gas at ambient pressure 1 bar. The admittance valve closes.
2-Gas is cooled inside the cylinder and goes below 1 bar. Atmospheric pressure works on the piston.
3-Gas is expelled.
- They were not optimized for efficiency; most of the heat produced by the flame was lost, as it was not even aspired. Radiance heat was also lost.
- The engine loses power as the cylinder heats, many were water cooled.

I am more interested in modelling the heat engine than examining real flamelickers. These engine were designed for simplicity of operation at small loads, efficiency was not important.

But I have not seen any thermodynamic analysis of this cycle, and that's what I am trying to do.

Thanks for taking a peek at the question anyway.

8. Nov 26, 2016

### pkv

Hi,
after learning a bit more about thermodynamics and studying the Lenoir cycle (simple as a vacuum engine, as it has expansion but no compression), I tried to annalyze again the vacuum engine cycle.
For clarity, I am just describing a cycle which uses adiabatic compression.

Let's describe the gas states.

1.- Ambient temperature and pressure
2.- Hot temperature, ambient pressure
3.- Ambient temperature, low pressure
4.- Medium temperature, ambient pressure

And the steps:

1->2: Isobaric. Air is heated at ambient pressure.
2->3: Isochoric. Gas is admitted in the cylinder and cooled on constant volume. Pressure falls.
3->4: Adiabatic. Gas is compressed until it reaches ambient pressure.
4->1: Isobaric. Gas is freed into the atmosphere and remaining heat is freed.

Example data:

- Cold temp. 300K, hot temp 600K.
- Working fluid: 1 mol of N2
- N2 constants CP = 29.2, CV = 20.8, CP/CV = 1.401

I am using the N2 constants at ambient temperature as approximation; I know there will be an small error.

Step 1->2, isobaric:

Q12 = n. moles * (te - ti) * cp = (600-300) * 29.2 = 8760 J
W12 = press * (vf - vi) = 101000 * (0.049 - 0.025) = 2494.3 J

Step 2->3, isochoric.

Gas will be cooled to 300K; I calculate final pressure using PV = nRT -> 50500 pascals

Q23 = (cv / R) * vol * (pe - pi) = (20.8 / R) * 0.049 * (50500 - 101000) = -6240 J
W23 = 0

The adiabatic expansion should end when the gas reaches ambient pressure.
Calculating the adiabatic condition constant K, I obtain final volume ve = 0.03 m^3

Q34 = 0
W34 = (k * ((ve^(1 - cp/cv)) - (vi ^ (1 - cp/cv))) / (1 - cp/cv) = -1365 J

Step 4->1, isobaric.

Gas is still hot after adiabatic expansion. Temperature is obtained using PV=nRT, T=365.8K
When the gas is freed to atmosphere, heat is released and gas shrinks.

Q41 = n. moles * (te - ti) * cp = (365.8 - 300) * 29.2 = -1922.3 J
W41 = press * (vf - vi) = 101000 * (0.025 - 0.03) = -547.4 J

Work available.

W12 + W34 + W41 = 2494.3 - 1365 - 547.4 = 581.9 J

Efficiency.

Eff. = total work / heat input
Effi. = (W12 + W34 + W41) / Q12 = (2494.3 - 1365 - 547.4) / 8760 = 0.066 or 6.6%

Balance of work and heat.

W = Q
W12 + W34 + W41 = Q12 + Q23 + Q41

2494.3 - 1365 - 547.4 = 8760 - 6240 - 1922.3
2494.3 - 1365 - 547.4 - (8760 - 6240 - 1922.3) = -15.8 J

I guess I am not getting 0 balance due to using approximations cv, cp and cp/cv.