Thunderchild

When I had finished fitting out the workshop, and began construction of the engine in earnest, I decided a name was required. I chose the name "Thunderchild". As literary readers will remember, this was the name of the Royal Navy Ironclad in H.G. Wells' "The War of the Worlds".

The engine as designed will be a 222 N (50lbf) thrust unit. As the engine is relatively small, is my first attempt, and will, I'm sure, be incredibly noisy, "Thunderchild" seemed appropriate.

As you will no doubt have guessed, I decided to build the engine as a series of flanged sub assemblies. Although Kryzcki warns against this in his treatise, it is certainly not without precedent. Whilst researching prior to beginning construction, I saw several examples of one piece chamber/nozzle type units on the internet. I suspected this would involve a great deal of machining to convert a section of round bar into a tapered tube. I decided to try to find a more efficient constructional method. The combustion chamber is essentially a pipe; all that was needed it would seem would be to make a suitable pipe. A unit based on separate sections would make for easier fabrication. It would also allow different injector, nozzle and chamber configurations to be trialled.

The combustion chamber dimensions were calculated using the relations given in Sutton. The parameter used to define the chamber volume is the Chamber Characteristic Length, denoted by L*. This is the ratio of chamber volume to nozzle throat area (Huzel and Huang). It is a substitute for determining the propellant stay time in the chamber.


Values of L* have been predefined for various propellant combinations, and are published in the literature. For a gaseous oxygen/hydrocarbon combination, a value of 1.52 metres (60 inches) is indicated. This figure is quite large, but for a small motor will ensure adequate space for atomisation, mixing and complete combustion of the propellants. The volume of the chamber can be found from:-

Vc = L* At (1)

Where:-

Vc = Chamber volume

At = Throat area

Nozzle theory and thermodynamic relations determined the throat area as 79.40 sq.mm (0.123 sq.inch). I will show these calculations in a later post.

Substituting the knowns into equation (1) produces:-

Vc = 1.52 x 7.94 x 10^-5 (2)

Vc = 0.121 x 10^-3 cubic metres (7.7 cubic inches)

The chamber volume considered in the definition of L* must include the volume of the convergent portion. The theoretical figures resulting from the calculations here give dimensions for a chamber inclusive of a convergent portion. I decided early on that the tubular section of the chamber and the nozzle/convergence would be fabricated as separate units. This has implications for the final exact choice of dimensions for the tubular portion but I will explain these later on. 

The next piece of information required to progress the theoretical chamber sizing calculations is the Contraction Ratio, Ec. This is the ratio of the chamber to throat diameter.

In order to give a usable injector face area, I decided to use an Ec of 5. Figures for Ec greater than 3.5 also enhance stability and energy utilisation efficiency (Sutton). In addition, a value of this order is recommended for thrust chambers smaller than 333 N (75lbf) (Kryzcki).

Hence if Dc = 5Dt,

Dt = 10.06 x 10^-3 m (0.396 inch)

Dc = 50.3 x 10^-3 m (1.98 inch)

Therefore the chamber area Ac is:-

Ac = pi (50.3 x 10^-3/2)^2 (3)

Ac = 1.98 x 10^-3 sq. m

To evaluate the chamber length:-

Lc = Vc/Ac (4)

Substituting the knowns into (4):-

Lc = (0.121 x 10^-3)^3/(1.98 x 10^-3)^2 (5)

Lc = 61 x 10^-3 m (2.4 inch)

The theoretical dimensions for the combustion chamber have now been derived from nozzle theory, thermodynamic relations and published values for Characteristic Length. To summarise, these are:-

Chamber Diameter Dc = 50.3 mm (1.98 inch)

Chamber Length Lc = 61 mm (2.4 inch)

Chamber Area Ac = 1980 sq. mm (3.06 sq. inches)



Next we'll look at the compromises neccesary to convert these theoretical figures into something that can actually be made. I'll tackle this in the next post. Along the way I will explain how the volume required for the convergent portion will be accomodated.