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The jet propulsion with closed combustion type


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#222 qumf

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Posted 05 July 2006 - 08:25 AM

There are many ways to ignite the mixture gas in the chambers. Each of them has weakness and advantage. Such as, the energy to ignite is small; the flame isn’t stable; the heat load is big…
I always think of if there is a good solution for ignition, which has simple structure, is easy to realize, has enough energy to ignite the fresh gas, has as low as possible heat load to the components. Recently I have a idea: the igniter is in a cavity installed the front door. Inside the resistance wire is heated by electricity continuously or intermittently to rather high temperature. The cavity has two openings, one is open to the enter pipe, when the fresh gas come in the cavity, it will be ignited and with flame. When the position of cavity turn to a chamber with the front door, through another opening of the cavity and by the gas pressure the flame in the cavity will enter the chamber that just finish entering fresh gas, so the fresh gas in chamber begin to burn if they have proper parameters. When the cavity moves away the chamber, the pressure in chamber rise rapidly. By the fresh gas flow through the cavity by two opening, the gas can cool the component around the cavity. When we want ignite the gas in cavity, we just left the opening to enter pipe open, close another opening.

The front door separate the fresh mixture gas in inlet pipes and the fresh gas in chambers, We can take advantage of it. so we made modification inside the front door. the front door is rotating, it can suck the high pressure air from the blow/compressor then add pressure, there are small slots or small holes at the edge of opening of the front door. When the front door close the chambers. the clear air run out through the slots or holes and stop there, These clear air separate the fresh mixture gas to the front door, it protects the front door from burning. It also has function of safety about fresh mixture gas in the inlet pipes.

#223 DFINITLYDISTRUBD

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Posted 07 July 2006 - 07:12 PM

Sorry to go M.I.A. on you again.....soooooooooooooooo tired.....so very very tired..............I desparately need a vacation!...........

Engine scrap!!!!
Starting over!
AAAAAAaaaaaaaaaaaaaaaaaarrrrrrrrrrrrrrrrrrrgggg!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
New design no "borrowed componanents" (salvaged)
no valves , no gates, no f#cking central hub, no f#cking compressor !!!!!!!!!!!!!!!!!!!! IT will be MAGNIFICENT!!!! IT will be BEAUTIFUL!!!!
IT will be the most efficient f#cking turbine engine ever dreamed of!!!!

news soon!

#224 qumf

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Posted 08 July 2006 - 10:42 PM

Sorry to go M.I.A. on you again.....soooooooooooooooo tired.....so very very tired..............I desparately need a vacation!...........

Engine scrap!!!!
Starting over!
AAAAAAaaaaaaaaaaaaaaaaaarrrrrrrrrrrrrrrrrrrgggg!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
New design no "borrowed componanents" (salvaged)
no valves , no gates, no f#cking central hub, no f#cking compressor !!!!!!!!!!!!!!!!!!!! IT will be MAGNIFICENT!!!! IT will be BEAUTIFUL!!!!
IT will be the most efficient f#cking turbine engine ever dreamed of!!!!

news soon!


Your safety is first.
I can't image there is such a good engine.Can you desdribe roughly it to me?draw a draft.

#225 qumf

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Posted 14 July 2006 - 08:53 AM

DFINITLYDISTRUBD

You originally thought the engine too simple, it isn’t appropriate; you just consider the structure of the engine, you simplify the structure further. I guess you don’t understand the work principle at the base. You develop it by your experience and imagine. So you failed. During the course I remind you many times.
I try to analyze the reason you failed: you just use a chamber, and the shape inside chamber is simple. The combustion and the flow of the gas exit at the same time. You didn’t control the flow speed and position of flow well either. I think it is hard to keep the flame steady. The fire is blown down.
On another side, the composition of gas near the flame is also important. You didn't control well either.

So far I still think, depending on your material by great effort to get, you fabricate some necessary parts also, it is compeletely you can succeed.

#226 qumf

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Posted 27 July 2006 - 07:15 AM

The principle is easy to understand, though the theoretical efficiency of the two types of jet engines are close, the compressor in the current engine consume a lot of (over half) energy from the burnt gas, further the compressor's efficiency is low. So the new type engine's actual efficiency is higher than the current .


Before I estimated the new type engine's actual efficiency, but I describe too roughly. Now I sort these material and release, I hope somebody can point out the mistake if there is. (Though the case is estimated, I still think it is representative and persuasive)
Firstly I state some premises:
1. The data I chose for the example are not the most advanced, but representative

2. The data is from jet engine, not from turbo-generator. Though their structures are similar, the data are different because they work under different conditions with different requirement.
3. In below analyse I assume that the portion's efficiencies are same if two types engines have the similar function component.
Below I state the example with data to explain.I think it real hard to describe the all main charateristics' relation and variation only with formulations about each course. maybe impossible.

As to the current jet(combustion under constant pressure, there is a theoretical formula. η=1-1/W^[(k-1)/k]
W is in crease rate of compression, in the example I set: W=10,k=1.4, the theoretical efficiency of the current jet is 48.2%.
As to the new type jet engine( close combustion), the efficiency formula: η=1-k*[U^(1/k)-1]/{ W^[(k-1)/k]*(U-1)} You can see the demonstrate course from previous post.
U is the in crease rate of pressure by combustion, here U=8, W is increase rate of gas pressure by blower. W=3 here
I input the data, you can see the theoretical efficiency of this closed combustion jet engine is 50%

By knowledge of Thermodynamics, if without cooling during compression, comparing the energy of compression the gas from 1bar to 10 bar with from 1 bar to 3 bar, the rate is 2.524 times.
By some published material, under common condition, I assume the actual efficiency is 80% for a compressor with compression rate=3. I compare the compression rate 3 and 10. The blower and compressor operate the gas by a stage after a stage. Thus I estimate the actual efficiency of a compressor with compression rate=10 is 65%.(Note: the efficiencies is for whole course, from suck air from atmosphere to completion compression.)
As to current jet engine, by a common case, 55% energy that can be utilizes from burnt gas is given turbine.(the proportion is higher in some jet engines). I also assume the actual efficiency of turbine is 100%
the actual efficiency is: η=(W1-W2)/Q W1:the energy can be utilized from the gas after combustion, W2: the energy the turbine consumes; Q: the chemical energy that the fuel contain.

So η=(W1-W2)/Q=[(W2/0.55)-W2]/Q=0.818W2/Q

The theoretical efficiency: η1=(W1-w2)/Q w2: the consumed energy by turbine theoretically, before I calculated the theoretical efficiency of jet engine, the efficiency of compressor is assumed 100%. the turbine transfer the energy to compressor, the energy consumed by compressor theoretically is less than actual consumed, so the energy consumed by turbine theoretically is less than actual consumed. The rate of the theoretical to the actual is 65% that is the efficiency of the compressor.

So η1=(W1-w2)/Q=[(W2/0.55)-0.65W2]/Q=1.168W2/Q

η/η1=0.70

As to the new type jet engine I proposed, what proportion of whole energy consumed by the turbine that is from the gas after combustion?
55% energy that can be utilizes from burnt gas is given turbine in the current engine, from previous explanation, comparing the energy of compression the gas from 1bar to 10 bar with from 1 bar to 3 bar, the rate is 2.524 times.
Thus after combustion, about 55%/2.524=21.8% energy is consumed by turbine. the course should be thought over . We can understand as below:

We assume that the two types of the engines have the energy on the gas after combustion, by previous analyses, their theoretical efficiency are similar, one is 50%, another is 47.4%. The gas after combustion works, there are two aspects, a portion is to drive turbine, and others is to accelerate the gas itself. 55% is theoretical data, 2.524 times is rather accurate data by calculation, or it has some margin. The last data 21.8% can be actual data.
The actual efficiency for the new type engine is: η=(W1-W2)/Q W1:the energy can be utilized from the gas after combustion that can be utilized; W2: the energy the turbine consumes, Q: the chemical energy that the fuel contain before combustion.
So; η=(W1-W2)/Q=[(W2/0.218)-W2]/Q=3.587W2/Q

The theoretical efficiency is: η1=(W1-w2)/Q w2: the consumed energy by turbine theoretically

When I analyzed theoretical efficiency of the jet engine, I assume the efficiency of blower is 100%, the theoretical energy consumed by the turbine is less than the actual. The rate of the theoretical to actual about the turbine is 80% that is same as the efficiency of blower.
So η1=(W1-w2)/Q=[(W2/0.219)-0.8W2]/Q=3.787W2/Q

η/η1=0.947


I put 48.2% as the theoretical efficiency of the current jet engine into the formula, We can get the actual efficiency 34%
I put 50% as the theoretical efficiency of the new type engine into the formula, I get the actual efficiency 47.35%
To compare the data, the actual efficiency of the new type engine is higher than the current by 39%


It is very pleasant that the efficiency can higher so much, but it is theoretical. I think it is enough if the efficiency can be higher 30% actually.
As to the current engine, the best theoretical efficiency may better than I said. Some parameters may not match well each other, but I think the deviation is small and it can be neglected in this example.


As to the new type engine(closed combustion type), some energy will waste when the gas runs from the branches pipes to general pipe.
I conclude that a little energy be wasted while the gases converge together from the branch pipes because below:
1). The convergence occurs in a small closed space.
2. Before convergence main currents have same speed as well as same direction. During and after convergence there is a proper space and necessary for the course. 3) some small currents join the convergence, but their mass are small and their condition are not harmful to others.


Somebody may think, the actual efficiency of the current jet isn't as high as I state here, It is because the jet must work under a wide range condition, Some parameters aren't operate under best data that can cause the better efficiency. Some waste is ignored.
Really I deal with some course too simply, But I think it won't influenced the conclusion.

#227 DFINITLYDISTRUBD

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Posted 08 August 2006 - 08:01 PM

HI!

Can you say eurika!!!!!!!!1

Preliminary #'s are lookin reeeeeeeeeeeeal good!

113.lbs
97cc total combustion chamber displacement.
300 and change ft/lbs torque @ 17850 rpm.
max. rpm. somewhere in the neighborhood of 75 or 80,000 rpm. (dont want to push it too hard yet.)
fuel consumption - 1 gph @ 45,000 rpm.
Thrust- enough to accellerate my 4,900lb pick-up truck from 0 to 110mph in 11 sec. (if you want actual #'s you'll have to do the math!)

#228 qumf

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Posted 10 August 2006 - 08:15 AM

DFINITLYDISTRUBD :

I have some questions because you don’t state it clearly.
1. Is the “113.lbs” the weight of your engine?
Is the “4,900lb” the weight of truck?
2. Is your engine a jet engine or internal combustion engine, why it has the torque “300 and change ft/lbs” if it is jet engine?
3. How many chambers do you adopt? Do you use only one chamber?
4. How do you control the chamber’s opening and closing? Do you use rotary valves as before.
5. I am not clear the specification of common internal combustion engine. I am sure you have a lot information on this field. You can compare it with others.

I also will calculate “ push force/engine weight

#229 qumf

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Posted 19 August 2006 - 03:16 AM

Annotation:

Regarding the explain below why the actual efficiency of the new type jet engine is higher than the current one, I should correct one sentence:

As to current jet engine, by a common case, 55% energy that can be utilizes from burnt gas is given turbine.(the proportion is higher in some jet engines). I also assume the actual efficiency of turbine is 100%

should be ......55% energy that can be utilized by meanings of the expansion of the gas from burnt gas......I also assume the actual efficiency of turbine is 100% that is based on the difference of the gas's energy before and over the turbine.

#230 qumf

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Posted 19 August 2006 - 03:20 AM

reply#229 is for #226

#231 DFINITLYDISTRUBD

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Posted 19 August 2006 - 08:23 PM

As of yet I'm not sure if it is a jet, a turbine, or a rotary....it's design falls somewhere in between!

Torque refered too above is at the output shaft/axle
thrust is developed by the veins attached to it.
113 is engine weight
4900 is truck weight

#232 qumf

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Posted 21 August 2006 - 06:31 AM

DFINITLYDISTRUBD:

Can you draw a picture of the engine for us, otherwise I can’t understand the structure.

Can you explain the “vein” here?
“300 and change ft/lbs torque @ 17850 rpm.” Is 300 ft/lbs on the shaft with the rotation speed 17850rpm? Is the output shaft’s speed 17850 rpm?

#233 qumf

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Posted 29 September 2006 - 08:15 AM

As to the components bear the heat load in the type jet, I have some thought or solution:
1. Because each chamber has a branch pipe behind it, when the gas spout out of the chamber, it can expand inside branch pipe, thus the temperature decrease. Though the temperature in chamber of the new type engine may be higher than in current engine, the temperature at the tubine can be lower than the current jet’s.( before the gas spouts from the chamber,the pressure in its branch pipe is low because of inertia and the effect of gases' velocity form other branch pipes)
2. Sometimes the chambers touch the burning gas in some position, but the rather higher temperature last a very short time. When the fresh gas enters the chambers, the chambers are cooled.
3. At some position in some components, such as inner side of the front door and back door of the chambers, they touch very high temperature gas all the time. I think out a solution: When the doors (round plates) is closing, from their inner side, near the edge, discharge some fresh air to separate the burning gas to the inner side of the doors. I think it can be realized by the rotating plates and the gas compressed by the blower. The rotating plates (the front door and back door) can increase the gas’ pressure while the gas flow inside by centrifugal force if the inside structure is proper.

#234 qumf

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Posted 12 October 2006 - 05:46 PM

Before I analyzed the pressure increase proportion by combustion with the example of methane gas because its combustion process and outcome is simple. If gasoline or kerosene burn in air, the pressure increase proportion will be different. One cause is they have different heat output. Another cause is the quantity of gas by mol change through combustion and change rates are different in different fuel oil or other something.(the quantity of gas by mol is relative to the volume under standard condition. The combustion in my jet engine is confined in an unvaried volume, thus the proportion have a great effect on pressure.) It may be complicated to analyze precisely just by calculation, it will be best to do experiment.We also can refer some document about the biggest explosion pressure of some fuel with air. i think they have similar course.

#235 qumf

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Posted 07 November 2006 - 08:08 AM

Nobody comments on my post about estimating the actual efficiency of jet engines. I have to always self-examine my thought whether there are some mistakes in these conclusions.
Now I found a improper date I chose in the calculation, I set the efficiency of the compressor 65% , that is really not appropriate, many compressors’ are higher than the date. I recalculated and choose 85% as the efficiency of the compressor, which is almost the highest efficiency for compressor so far. Meanwhile I set the increase ratio of gas pressure by compressor in current jet W=11; set the increase ratio in the new type jet engine W=4.5; the efficiency of compressor/blower always is 85% in any working range. Other premise conditions needn’t change. I recalculate and estimate the actual efficiency of the two types of jet by the solution that you can see in reply#226.
Use formula of theoretical efficiency, for current Jet: η=1-1/W^[(k-1)/k]
For Closed combustion type Jet: η=1-k*[U^(1/k)-1]/{ W^[(k-1)/k]*(U-1)}
We can get: pls see the solution in reply#226
Current Jet: theoretical efficiency: 49.6% Actual efficiency: 41.9%
Closed combustion type Jet: theoretical efficiency:55.54% Actual efficiency: 52.18%
I don’t choose the better parameters for new type jet for higher efficiency because that may cause error if we use the method to estimate.
We can find the rate between the two types of jets’ efficiency smaller than previous example, but still very good. The actual efficiency of the new type jet(closed combustion) is 25% higher than the current jet.
(The compressor has the highest efficiency; it is advantageous for the current jet). I think we can omit at least some stages of the compressor (at the high pressure portion) if we adopt closed combustion type, the new type jet right has the advantage in this filed.
It is no doubt if the two types jet engines adopt the same gas pressure increasing rate by the compressors W, the closed combustion type has much higher efficiency than the current type that is consistent pressure combustion type, on both theoretical and actual efficiency.

#236 GAHD

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Posted 07 November 2006 - 07:27 PM

Pritty cool man, you got any bench-top models worked-out yet?

#237 qumf

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Posted 16 November 2006 - 08:20 AM

Pritty cool man, you got any bench-top models worked-out yet?


I just do what I can do.
Be over urgent, don't obtain.

#238 HydrogenBond

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Posted 18 November 2006 - 12:21 PM

One way to lose the blower or compressor is to make the oxygen available by being part of the fuel blend. For example alcohol and hydrogen peroxide will burn in the absense of air. But this changes jet design into rocket design. It also requires carrying extra oxygen doner material, which may or may not add up to the weight advantage gained by the loss of the blower/compression.