Author Topic: Liquid Fuelled Rocket Engine  (Read 1142 times)

Offline British Reaction Research

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Liquid Fuelled Rocket Engine
« on: September 13, 2017, 01:28:09 PM »
Hello all,

My name is Carl and I'm attempting to build a liquid rocket engine. I'm planning to use nitrous oxide as the oxidiser and isopropanol as the fuel.

I usually post about the project on my blog, British Reaction Research, or on the Jet and Turbines Owners proboard. I've decided to start a thread here too, basing the decision on the fact that the more people who read about what I'm doing then the more there are to spot a mistake if I do something wrong.

This is just an introductory post. The next one I make is going to be about cooling flow rates for the chamber as it was a question that was asked in my post in the Introductions section of this site.

I'm hoping to post later tonight so I'll look forward to hearing any comments you may have on what I've written.

Thanks,

Carl.

Offline billmac

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Re: Liquid Fuelled Rocket Engine
« Reply #1 on: September 14, 2017, 03:01:37 PM »
Carl -

I have just had a quick look at your web site. I think your posts will be very interesting and I look forward to reading them.

Offline British Reaction Research

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Re: Liquid Fuelled Rocket Engine
« Reply #2 on: September 14, 2017, 09:35:57 PM »
Thanks Bill. I'm putting the finishing touches to my cooling post. Want to try to get it right!

Offline British Reaction Research

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Re: Liquid Fuelled Rocket Engine
« Reply #3 on: September 16, 2017, 01:37:12 PM »


This post is going to deal with some tentative calculations on the quantity of heat transferred through the walls of a rocket nozzle I am building. This study is part of my ongoing engine project. To be more specific, I’m interested in finding out how much coolant flow will be required to safely absorb the heat generated by the combustion gases.

It is said that when the engineer who designed the cooling circuit for the Saturn V rocket motor was given his brief, he was told to “make sure it doesn’t melt”. The aims of rocket engine cooling design cannot be put more succinctly than that.
The material I’m using for the nozzle and combustion chamber of my engine is 6082-T6 aluminium. I chose this for a number of reasons, primarily that it has high thermal conductivity, low density combined with high strength and is easily machined.
Strength is one of the properties that I want to conserve since the pressures inside the engine are relatively high. This isn’t easy when you consider that the gases flowing through the chamber and nozzle are somewhere in the region of 2000K (1700 degrees Celsius). Just how exacting a task this is becomes even more apparent when it is realised that the tensile strength of 6082 drops to 90% of the room temperature value at just 420K (about 150 degrees Celsius).

A rocket engine cooling system is really a type of heat exchanger, where the heat from the high temperature gases inside the chamber and nozzle are transferred to the coolant via forced convection and conduction through the chamber and nozzle walls.
The snag is that the coolant is often also the fuel. This means that unlike a normal heat exchanger design in which the coolant flow rate can be made whatever is required to do the job, in a rocket engine we have a flow rate that is already fixed by the mixture ratio of the engine.

I mentioned in my earlier post that I’m using nitrous oxide and isopropanol as propellants in my design. The isopropanol will be the coolant, mixed with about 20% water to improve specific heat capacity.
I’m using an oxidiser to fuel ratio of 2.125. This is very fuel rich but it means that the chamber and nozzle temperatures are much more manageable. To help me in my rocket engine design activities I use a piece of software called Rocket Propulsion Analysis, or RPA. It is produced by Alexander Ponomarenko. This software calculates engine performance parameters based on user inputs from a projected design idea.

So the first thing it would be useful to know in designing the engine cooling system is how much fuel, i.e. coolant flow is there to play with? For an engine design of 8kN (about 1800 lbf) the isopropanol flow rate is 1.4 kg/s. The fuel i.e. coolant flow rate is denoted by the symbol mdot. The highest value of heat flux in the engine is found in the nozzle throat, where the hot gas flow area is smallest.

 Usefully, RPA also gives us nominal engine dimensions and the nozzle throat diameter of the design mentioned above comes out at about 70mm or 0.07m.
In order to calculate the total amount of heat transferred to the coolant, Q, we must first find the average heat transfer rate, q. To find this we need the nozzle hot gas side heat transfer coefficient, or hg. For other parts of the engine this calculation is slightly complicated by the fact that the hot gas stagnation temperature and the temperature of the gas very close to the wall are not exactly the same, due to boundary layer effects.
What is known as the adiabatic wall temperature or Taw has to be found by multiplying the chamber gas stagnation temperature by a so called recovery factor. Fortunately enough the recovery factor at the throat is 1, so the chamber stagnation temperature can be used as the adiabatic wall temperature at this point.
It should be borne in mind now that this adiabatic wall temperature is not the actual wall temperature; if it was the wall would melt! The point of the cooling system is to prevent this. Therefore, a maximum temperature is defined that the wall can be at without being seriously weakened. This is the hot gas wall temperature or Twg.
 
So, Taw is the temperature of the boundary layer gases adjacent to the wall, whereas Twg is the temperature the wall needs to sit at to retain adequate strength to keep the beast in the cage.
Twg is brought about by the action of the cooling system. Having defined the temperatures inside the nozzle it is time to look at the equations needed to calculate hot gas side heat transfer coefficient hg, average heat transfer rate q, total heat transferred Q and thus the coolant flow rate required, mdot.

At this point the explanation is going to get a bit algebraic. Please bear with me. I know this type of thing looks like gobbledegook and believe me, it is as tedious to type as it is to read. That said an important result should come out of it and one that I hope you, the reader, will be able to help me confirm or deny.
Deep breath, here we go. The equation for the heat transfer coefficient, hg, of a hot gas, in W/m^2K, is:-

hg = 0.026k (p*v/u)^0.8 (1/D)^0.2 (cp*u/k)^0.4  (1)

When:-

k = gas thermal conductivity = 0.2 W/mK
p = gas density at point of interest (throat) = 1.6 kg/m^3
v = gas velocity at point of interest (throat) = 950.7 m/s
u = gas viscosity at point of interest (throat) = 6.9 x 10^-5 kg/m-s
D = throat diameter = 0.07m
cp = gas specific heat at point of interest (throat) = 1.9 x 10^3 J/kg-K

The formula above is taken from Humble, Henry and Larson and is a form of the Colburn equation. All of the above figures were gained from RPA for the engine design in question. Substituting the figures into the formula we get:-

hg = 0.0052 x 749293 x 1.07 x 0.84  (2)

hg = 3502 W/m^2K

Using this heat transfer coefficient we can work out the average heat transfer rate, q, spoken of earlier, and from q can be got the total amount of heat transferred into the coolant, or Q.
Earlier it was mentioned that there are two temperatures of interest in the nozzle flow. These are the adiabatic wall temperature Taw, which can be thought of as the actual gas temperature in the nozzle region. The wall temperature desired to prevent structural failure is Twg.

From RPA:-

Taw = 1940K (1600 degrees Celsius)

And from standard reference works on the high temperature properties of aluminium:-

Twg = 420K (150 degrees Celsius)

This figure for Twg is chosen to give 90% of room temperature UTS for 6082-T6 aluminium, on which the engine chamber and nozzle wall thickness is based.

Given these numbers the average heat transfer rate, q, can be found:-

q = hg x (Taw – Twg)  (3)

Substituting the figures gained so far into the above equation gives:-

q = 3502 x (1940 – 420)

q = 5323040 W/m^2
= 5.32MW/m^2

A lot of rocket engine literature gives these types of figures in Imperial units. I wrestled with these in my early design efforts, but for the sake of my sanity I soon moved over to SI. The figure of 5.32MW/m^2 is about 3.26 BTU/in^2degF in old money. That might sound like a lot, and it is. Remember that the throat has the highest heat flux of any part of the engine and is therefore the most stringent in terms of cooling.

Having discovered q it now becomes possible to calculate what the temperature of the outside wall of the nozzle will be. This is the wall in contact with the coolant and we call this temperature Twc, the temperature of the cooled wall. This temperature is found by evaluating the heat conducted through the nozzle wall.

The formula required is:-

Twc = Twg – q(t/k)  (4)

Where:-
t = wall thickness = 0.003m (3mm wall thickness based on UTS of 6082 at 420K)
k = wall conductivity = 160 W/m^2K (Average value for 6082)

This equation gives a result in Celsius that will be converted to Kelvin by adding 273. Substituting known values into the equation above gives:-

Twc = 470 – 5.32 x 10^6(0.003/160) = 99.75 deg C

Twc = 372.75K

Temperature figures given in Kelvin can be hard to envisage. Considering that 373K is about 100 deg C, and the desired Twg was 150 deg C, then Twc seems about right given the very high thermal conductivity of aluminium.
Knowing this temperature difference it now becomes possible to calculate the total quantity of heat transferred from the hot gas side to the cooled side of the nozzle.

The formula required is:-

Q = kA(dT/t)  (5)
Where:-

A = heat transfer area = area of the inside wall of the nozzle throat
dT = Temperature difference = 470K-372.75K = 97.25

The inside area of the nozzle throat is simply the circumference multiplied by the length. Since the diameter is 0.07m the circumference is 0.219m. The design length of the throat section is 10mm or 0.01m. Hence the area is about 0.0022m.
Putting these values into the above equation gives:-

Q = 160 x 0.0022 x (97.25/0.003)
Q = 11410 W
= 11.4 kW

The coolant flowrate required to remove this quantity of heat can be found by assuming a coolant inlet temperature, Tinlet, of 293K (about 20 degrees Celsius). As previously stated the throat is the trickiest cooling problem; taking the “cold” coolant (fuel) direct from the tank and putting it through the throat jacket would seem to make sense. Assigning a coolant outlet temperature of, say, 320K (about 50 degrees Celsius), the coolant inlet to outlet temperature difference is 30 degrees.

Remember that the coolant or rather the fuel is being expelled from the tanks under pressure. The pressure drop between the tanks and the injector/chamber will be 0.7MPa (100psi). The boiling point of isopropanol at this temperature is in the region of 413K (140 degrees Celsius). So there is plenty of headroom in terms of the coolant temperature rise.
The formula used to find coolant flow rate is then:-

mdot = Q/(cpdT)  (6)

When:-
cp = coolant specific heat
dT = 30


As the coolant flows through the throat jacket a boundary layer will be set up close to the outside throat wall. The convective heat transfer will take place in this boundary layer, so it seems prudent to use a figure for specific heat capacity at the temperature of the coolant at this point. This is known as a film temperature, Tfilm.

This can be thought of as the average of the cooled wall temperature and the inlet temperature. So far the calculations show the cooled wall will be sitting in the region of Twc = 373K (100 degrees Celsius). The inlet temperature, Tinlet, is 293K (20 degrees Celsius).

Then:-

Tfilm = (Twc – Tinlet)/2  (7)

Tfilm = 313K (40 degrees Celsius)

The specific heat of isopropanol at this temperature is 2.8 x 10^3 J/kg-K. Using this figure it is now possible to use equation (6) to calculate mdot, the flow rate required to cool the nozzle throat.

mdot = Q/cpdT

mdot = 11.4 x 10^3/2.8 x 10^3 x 30

mdot = 0.136 kg/sec

Got there in the end! Recall from the earlier discussion of the engine design that the fuel and hence coolant flowrate is 1.4 kg/s.

To summarise what I think these figures show, a flow rate of about 0.136 kg/sec should be adequate to remove around 11.4kW from the nozzle throat area. This should give a hot gas side temperature of around 150 degrees Celsius and a cooled side temperature of 100 degrees Celsius. The coolant temperature rise will be in the region of 30 degrees Celsius.
If the preceding paragraph sounds like I’m hedging, it is because I am! These figures are very much order of magnitude as opposed to exact; remember, I’m trying to “make sure it doesn’t melt”. What I hope is shown, if I have got it right, is that putting the full fuel flow of 1.4 kg/sec through the throat jacket will be more than ample to cool it.

It does get a little more complicated than this. The type of flow is very important. High heat transfer requires turbulent flow and that means a Reynolds number in excess of 10,000. Fortunately other calculations that I’ve done (don’t worry, I’m not going to show them, I think you’ve suffered enough…) show that Reynolds numbers much higher than this are possible with a flow velocity just over 1 m/sec, based on a flow area of about 0.0013m^2.

If the flow is arranged to be perpendicular to the throat area instead of parallel things get even better, as extremely turbulent flow then results with vortex shedding. This so called cylinder in cross flow arrangement is highly beneficial for heat transfer.
A rough surface also helps to increase heat transfer and induce a higher Reynolds number for a lower velocity. This is why golf balls are dimpled. So if, say, I was to put a knurl on the outside surface of the nozzle area that would very likely increase heat transfer. Another aspect of this is something called nucleate boiling. As the fluid film temperature approaches that of the cooled wall, small bubbles start to appear. These rapidly detach and condense in the cooler surrounding liquid. More bubbles are then produced and the cycle repeats.
 
This process massively increases heat transfer. The surface of a knurl would provide excellent nucleation sites for the bubbles to form. The feed pressure of the fuel could thus be tuned to induce some nucleate boiling.
If you have got this far, thank you for staying with it. Perhaps now you would allow me one last indulgence and tell me if you think I have got it somewhere near right!

Thanks,

Carl.

Offline billmac

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Re: Liquid Fuelled Rocket Engine
« Reply #4 on: September 16, 2017, 04:47:11 PM »
Carl -

That was indeed very interesting. I will need to read through it a few times to get the full picture.

You seem to have quite a bit of headroom in the coolant temperature increase. As you state, for good heat transfer you must have guaranteed turbulent flow - the difference netween that and pure laminar flow is large. Are you using a pump to deliver the fuel or just gas pressure in the fuel tank? If pumped, can you recirculate fuel?

I have recently gone through some heat transfer calculations to design a boiler economiser. In that case we had the luxury of using a more powerful pump and re-circulating the hotwell contents. The end result was a significant increase in boiler performance and a fast recovery rate from heavy steam usage.

Offline British Reaction Research

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Re: Liquid Fuelled Rocket Engine
« Reply #5 on: September 16, 2017, 07:49:00 PM »
Hi Bill,

Thanks for replying and thanks for wading through what I had written. If you were to re-read it with a view to commenting on my approach to the problem then that would be most gracious of you. I would be very grateful indeed.

The boiler economiser you have designed sounds very interesting indeed. I seem to have found the right person to look at my results!

I'm planning on using a gas pressurised feed for the engine, at least to begin with. Some other calcs that I've done to determine the Reynolds number indicate that even with an annulus type parallel flow coolant gap in the jacket of 5mm - easily achievable - the Reynolds number would be in excess of 127000. Anything above 10000 in liquids and you have turbulent flow.

As I mentioned one plan I have is to use cross flow - the coolant flowing perpendicular to the axis of the nozzle. This would induce even greater turbulence and vortex generation, all of which are good for heat transfer. Knurling or milling grooves or slots in the nozzle outside wall would also induce turbulence and eddies, as well as becoming nucleation points for a boiling heat transfer regime.

I think that one thing I didn't mention in my above post is that the numbers would be for steady state operation. As the rocket engine is likely to be run for short durations it will probably never attain steady state conditions.

Another point I forgot to make is that the engine ought to be tested with water as a coolant to begin with. As is well known, water is a formidable thermal fluid and will easily cool the nozzle and chamber in the way described in my post above. The inlet and outlet temperatures can then be measured and from that the performance of the cooling system can be checked. Only when it is proven as safe will I try to regeneratively cool the system with the fuel.

Thanks again for taking the time out to read my post and to comment on it.

Carl.

Offline PK

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Re: Liquid Fuelled Rocket Engine
« Reply #6 on: September 17, 2017, 01:25:47 AM »
Did I miss-read or are those calculations just for the throat section of the motor?
The throat is the hardest part to cool, but it's not the biggest source of heat. Particularly in a development motor where you tend to run conservative (read big) L* chambers.

Re strength. If you make the cooling jacket and saddle (I'm assuming you're doing it that way) such that it can take some load, then you can run thinner chamber walls and get better heat flow.

One important point, and I assume you are doing this. I know of no biprop that doesn't rely on film/boundary layer cooling. If you push your rich mixture to the edges of the chamber you can pull 300deg out of the combustion temp near the walls and still get reasonable ISP because the rest is nearer to stoichiometric.

Offline British Reaction Research

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Re: Liquid Fuelled Rocket Engine
« Reply #7 on: September 17, 2017, 10:33:41 AM »
Hello Pk

Thanks for taking the time out to read my post and to comment. You did not misread, these calculations are for the throat section. I wanted to base them on the hardest section to cool to try to get a good feel for the problem.

I take your point that the biggest heat load is not the throat, despite it being the hardest part to cool.

I'm not currently planning on using film cooling as such. That said, the injection system will be putting a lot of fuel against the walls. What I haven't so far mentioned is that I also want to mix some silicon oil in with the fuel. A small percentage only. This will burn with the fuel and the theory is that a glass like compound is formed that deposits on the chamber wall. One issue with heat resistant coatings is that they can Spall and flake off. Using this method the coating is being continuously reapplied throughout the duration of the burn.

This acts as an insulating layer that will reduce heat transfer to the walls. I haven't factored this into my heat transfer calculations as yet.

The L* I'm using is about 0.8 metres, which from what I have read is on the low side for a nitrous/alcohol motor.

Thanks again for your wisdom. I'll definitely be looking at incorporating film cooling.

Carl.

Offline billmac

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Re: Liquid Fuelled Rocket Engine
« Reply #8 on: September 17, 2017, 01:39:31 PM »
I wonder whether the silicon oil might preferentially coat the inside of your cooling ducts. If so that could have bad consequences for the rate of heat transfer that you can acheive. As you probably know, the presence of oil, even in very small quantities is a real no-no for steam boilers for that very reason.

Offline PK

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Re: Liquid Fuelled Rocket Engine
« Reply #9 on: September 17, 2017, 06:22:36 PM »
NOX biprops are (to my understanding) poorly characterised. My own efforts don't really count because the motors were tiny, but I have made a LOT of NOX hybrids over the years.
Based on that I'd suggest a longer combustion chamber. The reason being that it takes time for the N2O to heat up to the 500degC or so that it needs to reach in order for it to let go of it's oxygen.  This (and poor injector design) is what gives that classic bell shaped regression in a hybrid fuel grain.
Maybe build a water cooled, boiler plate, motor first, just to remove some variables and arrive at a correct L*.

For the non rocket nerds: It takes time for a given fuel and oxidiser combination to atomise, mix, and burn and, when you are flowing kilograms of propellant per second, time==distance.

If you make the combustion chamber too short then some of that combustion might not happen inside the motor.  It will look great because there will be an enormous flame emitted from your creation, but the sad reality is that all that heat will contribute nothing to generated thrust. 

You can't (normally) just make the motor a lot longer because the amount of heat you need to remove from the combustion chamber walls is a reasonably linear function of the surface area of those walls and you never have an excess of cooling capacity. 

As such, much work goes into characterising the combustion process of a given propellant combination over a range of pressures, flow rates, and o/f ratios. The end result of this is a set of L* constants which you can poke into some simplified maths to get the length of combustion chamber needed for your motor.

Offline British Reaction Research

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Re: Liquid Fuelled Rocket Engine
« Reply #10 on: September 18, 2017, 08:56:30 AM »
Hello Bill and PK,

Many thanks for taking the time to read and post.

Bill - My plan was to inject the silicon oil into the fuel flow after the cooling circuit, for the reasons you point out. Something similar happens if fuels like kerosine get too hot in cooling channels. The kerosene cracks and produces an insulating substance that can also block flow channels.

PK - Thanks for the wise words. I'm aware that 0.8m is low for an L* for nitrous. I'm just playing with figures at the minute. If I go to, say, 1.25m L* then the chamber length goes to 300mm. No problem, and the diameter is about 140mm so neither a pencil nor a pancake.

I take your point about nitrous being reluctant to part with its oxygen. One idea I had was to have a catalyst bed before the chamber. Thus would the Nitrous be decomposed to hot nitrogen and oxygen. This would be hot enough to auto ignite the fuel too.

The snags are three-fold. The first two are less intractable than the last. A part of the cooling problem is moved elsewhere. The catalyst unit needs to be preheated initially to get the reaction going. Suitable catalyst materials are very expensive indeed.

Lastly, there is an error in my heat transfer calcs. I used chamber stagnation temperature for the adiabatic wall temp in the nozzle. It should be about 0.9 times chamber temperature at that location. So not a huge deal, just makes the calcs more conservative.

Carl.
« Last Edit: September 20, 2017, 04:42:15 PM by British Reaction Research »

Offline British Reaction Research

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Re: Liquid Fuelled Rocket Engine
« Reply #11 on: September 19, 2017, 02:58:28 PM »
Hi,

I've been working on some more heat transfer stuff, this time for the chamber. I'll post it once I've finished,

Carl.

Offline British Reaction Research

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Re: Liquid Fuelled Rocket Engine
« Reply #12 on: September 24, 2017, 06:47:30 AM »
Still working on the chamber calculations. Apparently these turbulent flow in pipe type analogies for rocket chamber and nozzle flow tend to overestimate for the nozzle and underestimate for the chamber. This apparently is to do with the high pressure gradient in the nozzle increasing convective transfer.

Offline British Reaction Research

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Re: Liquid Fuelled Rocket Engine
« Reply #13 on: October 27, 2017, 10:04:35 PM »
I'm going to be posting the heat transfer calculations for the chamber in the next few days.

Earlier I said I made an error in the nozzle calcs. I stated that the adiabatic wall temperature should have been 0.9 times the chamber temperature. That is not the case. The recovery factor at the nozzle is 1, so the two temperatures are the same at the nozzle location.

Recovery factor in the chamber is about 0.9.

Offline PK

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Re: Liquid Fuelled Rocket Engine
« Reply #14 on: October 28, 2017, 06:21:44 AM »
It's important to remember that a rocket motor is not like a boiler or a stressed beam.
There have been so many boilers and beams designed that the mathematical models have been carefully honed over the last hundred years with thousands of published documents available describing said models.

In comparison, very few small rocket motors have been designed. There are maybe 50 meaningful papers/books describing aspects of them. 

What I'm trying to say is: "Don't get too hung up on chasing the micron fairies of your design. You'll learn more from the first three motors that you destroy than you will from a spreadsheet, no matter how long you spend staring at it."

Offline AdeV

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Re: Liquid Fuelled Rocket Engine
« Reply #15 on: October 28, 2017, 07:07:39 AM »
Earlier I said I made an error in the nozzle calcs. I stated that the adiabatic wall temperature should have been 0.9 times the chamber temperature. That is not the case. The recovery factor at the nozzle is 1, so the two temperatures are the same at the nozzle location.

I can't believe it's taking this long, I mean, it's not rocket sci....... oh, wait.... it is  :lol:  Only joking. The maths would be totally beyond me (I failed at A-level and have never managed to improve my maths since), so don't mind my feeble attempts at comedy. Looking forward to seeing your designs, prototypes, first firing, and your moon shot :D
Cheers!
Ade.
--
Location: Wallasey, Merseyside. A long way from anywhere.
Or: Zhengzhou, China. An even longer way from anywhere...
Skype: adev73

Offline British Reaction Research

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Re: Liquid Fuelled Rocket Engine
« Reply #16 on: October 28, 2017, 07:57:56 AM »
Hi PK and AdeV,

Thanks for your comments and for taking time to read my post.

I take your point fully PK. Basically I'm trying to do the best I can to produce something reasonably safe that won't melt...or at least I hope it won't. Its interesting to me to try to model the heat transfer (albeit in an extremely rudimentary fashion) and then see if I got it right on the test article.

You are quite right of course that theory is one thing and practice can be downright embarrassing. As you say I will learn more from building it and having it explode than I will from reading books and papers.

AdeV thanks for your interest and encouragement. You'd have no problem with the maths, it's all just simple arithmetic really. It must be if I can do it!


Offline British Reaction Research

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Re: Liquid Fuelled Rocket Engine
« Reply #17 on: October 28, 2017, 10:39:00 AM »
Here I will attempt to calculate the heat transfer from the combustion gases to the coolant of the cylindrical combustion chamber of the proposed rocket engine. This will allow further calculation of the minimum coolant flow rate required to safely remove this heat.
I am not going to give exhaustive details of the calculation method, as this was comprehensively covered in my previous post dealing with the heat transfer in the nozzle section.

In order to proceed with the heat transfer investigation, it is necessary to calculate the chamber heat transfer coefficient, or hg. RPA gives all of the relevant variables to do this apart from one, this being the velocity of the combustion gases in the chamber.
This velocity is generally defined as “low” in the literature. How low is low though, bearing in mind we are comparing it to sonic and supersonic velocities in the nozzle and divergence? RPA gives the total mass flow rate through the engine as 4.4 kg/sec. Given that the chamber area is 0.0167 metres square and the density of the gases at this point is 2.5 kg per cubic metre, this gives a velocity of about 103.5 m/sec.

So:-

Vchamber = 103.5 m/sec

Putting this and the other pertinent variables into the modified Colburn equation gives a value for hg in the chamber of:-

hg = 1220 W/m^2 K

Now we can work out what the heat transfer rate, or q is. Recall that to do this we specify a temperature that we want the hot gas wall to be at, given the materials thermal and mechanical properties. We will go with 420K as before. We now subtract this from the adiabatic wall temperature and then multiply by hg.

The adiabatic wall temperature in the chamber is derived by multiplying the stagnation temperature by a so called recovery factor, or r. Experiment shows that for practical purposes r = 0.9. RPA gives the stagnation temperature in the chamber as 2170K so the adiabatic wall is at 1953K.

Substituting these values into the relevant equation gives a figure for q:-

q = 1.87MW/m^2 (1.14 BTU/sec/in^2 deg F in old money)

Having gained a figure for q we can go on and calculate the temperature of the cooled wall, or Twc, by finding the heat conducted through the aluminium. To do this we require the thickness of the wall and the material conductivity. Recalling the wall is 3mm thick and the aluminium conductivity is 160 W/m^2 K, we arrive at a figure for Twc of:-

Twc = 385K (112 deg C)

Now we can go on to find the maximum heat transferred into the coolant or Q. This is done by multiplying q by the heat transfer area, this being the outside area of the cylindrical chamber portion. This area will be 0.005 m^2. Hence the total heat transferred, Q, becomes:-

Q = 9.35 kW

Now we can work out how much coolant flow is required to dissipate this much heat. To do this we need to know the specific heat of the coolant at it’s mean temperature within the chamber cooling jacket. This so called film temperature, or Tfilm, is evaluated by halving the sum of the coolant input temperature and Twc. Remember of course that the coolant will have already done a job of work cooling the nozzle before it enters the chamber. We know from the previous calculation that the film temperature in the nozzle was 306K or 33 deg C. If we assume a temperature rise of 30 degrees in the nozzle, which is perfectly within the range of coolant flow that we have, then we have a chamber inlet temperature of 63 deg C or 336K. This gives a Tfilm in the chamber of:-

Tfilm = 360K (85 deg C)

The specific heat of the fuel at this temperature is 3360 J/kgK.

Let us again allow a temperature rise of 30 degrees for the coolant in the chamber. This gives a coolant i.e. fuel temperature of 120 deg C on entering the injector. We already know that the boiling point of our fuel, which is at elevated pressure, is above this.
To calculate the minimum amount of coolant flow, or mdot, required under the above conditions we divide Q by the product of the specific heat of the coolant and the temperature increase desired.

This gives a coolant flow rate of:-

mdot = 0.093 kg/sec

As was previously mentioned, the fuel/coolant flow rate will be in excess of this figure and so it seems highly likely that the chamber will be able to survive the heat load imposed upon it by the combustion gases, given the design fuel flow rate.