Friday, January 02, 2009

Designing a chamber....

So you want to design a chamber, there are many ways to do this. My method is not real rigorous, but here it is.

First know what you want. For this example:

  • 90% peroxide /kerosene with a solid catalyst.

  • 915 lbs max thrust. (This is 125% of max weight for my notional 180 sec vehicle.)

  • 320 psi feed pressure (This is our hydro pressure /1.25)

  • Optimize for best eff at 475lbs thrust. (This is the mid point on a 180 second burn this number probably needs adjusting upward as more fuel is burned at the beginning of the flight)

  • Operate at FAR or Las Cruces, or the new Mexico space port about 4000 ft altitude. (12.4 psi)

Given a 320 PSI feed pressure we need to allocate some pressure drops, I’ll use 75 lbs in the cat pack and 20 percent in the injector /diffuser.  I got the 75 PSI from a  FMC peroxide catalyst pack design document.  I got the 20% from word of mouth and vague references in Sutton.

So the resultant max chamber pressure is (320–75)*0.8 = 260 PSI.

So at our desired optimum 475 lbs we have a chamber pressure of

Cpopt=(475*260)/915 = 134 PSI.

Next Ill run this through cpropep. I ran lots of cpropep runs to arrive at the 0.17g of fuel for 100gm of 90% peroxide. This is the final input file run:

Propellant X/Y
+466     0.90 g
+976     0.10 g
+797     0.17 g

+chamber_pressure     134 psi
+exit_pressure         12.46 psi

The output is:

Thermo data file: thermo.dat
Propellant data file: propellant.dat
Computing case 1
Frozen equilibrium performance evaluation

Propellant composition
Code  Name                                mol    Mass (g)  Composition
466   HYDROGEN PEROXIDE (100 PC)          0.0265 0.9000   2H  2O
976   WATER                               0.0056 0.1000   2H  1O
797   RP-1                                0.0121 0.1700   2H  1C
Density :  1.228 g/cm^3
3 different elements
H  O  C
Total mass:  1.170000 g
Enthalpy  : -6420.12 kJ/kg

114 possible gazeous species
3 possible condensed species

                       CHAMBER      THROAT        EXIT
Pressure (atm)   :       9.663       5.463       0.848
Temperature (K)  :    2418.562    2205.550    1608.636
H (kJ/kg)        :   -6420.118   -6948.077   -8362.579
U (kJ/kg)        :   -7389.021   -7831.644   -9007.016
G (kJ/kg)        :  -36931.242  -34771.963  -28656.161
S (kJ/(kg)(K)    :      12.615      12.615      12.615
M (g/mol)        :      20.755      20.755      20.755
(dLnV/dLnP)t     :    -1.00000    -1.00000    -1.00000
(dLnV/dLnT)p     :     1.00000     1.00000     1.00000
Cp (kJ/(kg)(K))  :     2.50147     2.45427     2.27083
Cv (kJ/(kg)(K))  :     2.10086     2.05366     1.87022
Cp/Cv            :     1.19069     1.19507     1.21421
Gamma            :     1.19069     1.19507     1.21421
Vson (m/s)       :  1066.56503  1027.58281   875.97064

Ae/At            :                 1.00000     2.44989
A/dotm (m/s/atm) :               157.40662   385.62851
C* (m/s)         :              1520.94574  1520.94574
Cf               :                 0.67562     1.29592
Ivac (m/s)       :              1887.43319  2297.97666
Isp (m/s)        :              1027.58281  1971.02048
Isp/g (s)        :               104.78429   200.98815

Molar fractions

CO                   9.0878e-002 9.0878e-002 9.0878e-002
CO2                  1.2411e-001 1.2411e-001 1.2411e-001
COOH                 3.2035e-007 3.2035e-007 3.2035e-007
H                    1.6249e-003 1.6249e-003 1.6249e-003
HCO                  2.3091e-007 2.3091e-007 2.3091e-007
HO2                  1.8081e-007 1.8081e-007 1.8081e-007
H2                   8.5414e-002 8.5414e-002 8.5414e-002
HCHO,formaldehy      3.0941e-008 3.0941e-008 3.0941e-008
HCOOH                2.3185e-007 2.3185e-007 2.3185e-007
H2O                  6.9535e-001 6.9535e-001 6.9535e-001
H2O2                 1.3262e-007 1.3262e-007 1.3262e-007
O                    3.1412e-005 3.1412e-005 3.1412e-005
OH                   2.4808e-003 2.4808e-003 2.4808e-003
O2                   1.0559e-004 1.0559e-004 1.0559e-004

Lots of data here but I really only want the four items I set in bold.

First dimension question how big is the throat?

For 915 lbs thrust at 260 PSI we need the Cf number.

So Throat Area = 915/(260*cf) =2.71 sq inches in area.

The Throat diameter is thus 1.859473 inches.

For thoose who want consistent SI units:

915 lbs is 4080 N  260psi is 179 N per cm^2

So At = 4080/(179*cf) = 17.58 cm^2

converting back to inches for a consistantcy check we have 2.72 sq in.

Now that we have the Area Throat we need Area Exit

Ae=(At*Ae/At) =2.71*2.44989 =6.65 sq in and a diameter of 2.9147 inches.

So now we have the Throat diameter and Exit diameter.

We need the chamber length. This chamber will have Gas/Liquid mixing so we can probably use a shorter chamber than usually, but I’m going to be somewhat pessimistic and use a L* of 50in. I’m also using some fixed catalysts at 7” so my chamber inner diameter is going to be 6.5”.

6.5” diameter is 33 sq in. The At =2.71 so 2.71*50/33 = 4.1 inches.

We now have the major dimensions of the chamber done.

  • Throat 1.859 in diameter

  • Exit 2.915 in diameter

  • Chamber 6.5 in diameter

  • Area before the throat 4.1 in long.

The Now to draw some shapes. One could use the classic 15 degree exit cone that would be easy to machine, but gives up a few percent performance. Or one could try and do a parabolic expansion cone. I used the RAO optimum nozzle approximation. This is covered in Rao’s classic paper or texts like Modern Engineering for design of Liquid propellant Rocket Engines (Huzel/Hwang) . This requires that you pick a couple angles off a graph and fit a parabola to the resultant points. The problem is actually over specified so you have to decide which end of the parabola you want to be exact. The cook book formula gives you 4 equations and 3 unknowns. For the more math inclined you have two specified points and the slope at both points and you want to fit a 2nd order parabola through these. Like I said over specified. (You can see the document in question by doing a google book search look at page 76. Search for Huzel Hwang) So after getting some advice and a code snip-it from arocket I wrote a short program that fits a parabola to the defined points . I assumed a 100% of conical length.

So I drew the chamber in the following steps….

  • Placed a reference  point at 0,0 and one at 0,(1.859/2) (black)

  • Drew an Arc of 1.5 rT to 60 degrees (red)

  • From the chart Fig 4–16 initial angle is ~20 degrees. So I drew an arc blue of 0.382 Rt to 20 degrees.

  • Then I placed a point at the exit position. A 15 degree half angle cone would be 1.97 inches long going from 1.859 to 2.915 inches. So I plot this point referenced from the throat. (Green line)

The result at this point:


Now using my CAD program I read the x,y locations for the end of the blue arc:

Point at (0.239756,0.971775,0). And the end of the green line. Point at (1.97,1.4575,0).

I plug all this into my little program and calculate a script to draw the nozzle. The Program :File Attachment: calcv2.cpp (1 KB). Now I have the basic chamber contour. All I need to do is Add the upper chamber and mirror


Next steps involve giving the chamber some thickness and the design of the catalyst holders and injector/diffuser seals connections etc…. to be continued….












Anonymous said...

I have no knowledge to critic this design but I did want to say thanks for documenting the process.
Very interesting.

Anonymous said...

Thanks for showing how you do this, I'm in the process of designing my chamber at the moment as well.

The only thing is that, I'm not making a traditional combustion chamber so the same numbers wouldn't quite work for me.

Anonymous said...

Thank-you very much. I like that you listed the various tools and texts that you used. Is there any particular reason to use a parabola as the nozzle instead of a third order polynomial? I really have no idea but I suppose, with the rate of expansion you get at the throat, that separation is not a problem at that discontinuity in the derivative you have. It just seems to me that smoother would be better.

Paul Breed said...

Is there any particular reason to use a parabola as the nozzle instead of a third order polynomial?

Weak Math ;-)
I can specify 2 end points, the slope at each end and the fact that I want it to be monotonically decreasing slope.

Anonymous said...

Very nice to read such detailed process rather than just "we made foo". Thanks!

Tangentially, everyone's default is to use JPEG images thinking it is the smallest (including myself) but for technical drawings like this using .PNG would give you way better compression and way better image quality.

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