# Chamber Wall Thickness for Liquid Fuel Rocket

In, How to Design, Build, and Test Small Liquid-Fuel Rocket Engines, by Krzycki, page 18 has the following equation:

$$t_w = \dfrac{PD}{16000}$$

Assuming the following: Pressure ($P$) in combustion chamber is 500 psi. Mean Diameter of Combustion Chamber ($D$) is 2 inches. Working Stress (as given in the book) 8,000 psi for the Copper.

Chamber Wall Temperature of 100°C (I know that's low, help me out here, I'm new).

1. Solve for Combustion Chamber wall thickness.

2. Working Stress is listed as about 8,000 psi in the book, is working stress the same as proportional limit?

While I am a mechanical engineering student, this question isn't homework. I'm a hobbyist with a passion for rocketry. I know this is my first post here, I don't want to get off on the wrong foot. Please substitute different values if that helps! I was hoping someone could walk me through this. Specifically what I don't understand, if I multiply a pressure variable (500 psi) by a diameter (2 inches), what unit of measure is the answer? Obviously $$500*2=1000$$ But 1000 what? $$1000/16000=0.0625$$ .... Is 0.0625 inches? That's 1.5mm and that doesn't seem thick enough at all....

• Welcome to Engineering! Item (1) looks like a homework question. In order for such questions to be answered in this site, we need you to add details describing the precise problem you're having. What have you tried to solve this yourself? Please edit your question to include this information. – Wasabi Apr 19 '16 at 13:36
• it seems you tried to edit your question with a new handle? can you login again as matt? – mart Apr 19 '16 at 13:59
• I would also have a look at this question. – grfrazee Apr 21 '16 at 17:22

1. Solve for Combustion Chamber wall thickness.

Eqn. (22) of the text you reference shows the following:

$$S = \frac{PD}{2t_w},$$

which mechanical engineers will recognize as the equation for stress in a pipe wall given an internal pressure. This equation may be rewritten to solve for wall thickness $t_w$, as

$$t_w = \frac{PD}{2S}.$$

Reading the paragraph above (23), the allowable stress $S$ is limited to 8,000 psi. The author rewrites (23) to solve for the wall thickness $t_w$, as

$$t_w = \frac{PD}{16000\mathrm{psi}},$$

but he drops the units out of the equation. This is generally a thing to avoid since it leads to confusions like what you're experiencing, especially for someone unfamiliar with whatever unit system the author is working in.

Thus, if you have an internal pressure of 500 psi and a diameter of 2 inches,

$$t_w = \frac{(500\mathrm{psi})(2\mathrm{in})}{16000\mathrm{psi}} = 0.0625\mathrm{in} = \frac{1}{16} \mathrm{in}.$$

Given that an internal pressure stresses the vessel in membrane tension only, this is reasonable with the pressure and diameter provided.

1. Working Stress is listed as about 8,000 psi in the book, is working stress the same as proportional limit?

Working stress is given as a fraction of the proportional limit (a.k.a. yield stress). The working stress includes a factor of safety against yield.

If I do a search for copper pipe alloy, it looks like typical copper tube one might buy at a home store is alloy C10200, C10300, C10800, C12000, or C12200 (source here). Referring to the ASME Boiler & Pressure Vessel Code, 2013, Section II, Part D, Table Y-1, these material alloys have a yield strength of 9,000 psi and a tensile strength of 30,000 psi.

So, it seems that your reference material is using 89% of yield or 27% of tensile strength as the working stress. Whether or not this is acceptable is up to you.

• > the allowable stress S is limited to 8,000 psi. Is is correct for copper. What is about stainless steel? – Robotex Feb 19 '17 at 19:00
• @Robotex, The allowable stress of stainless depends on the specification and grade. Consult your code for this value. – grfrazee Feb 19 '17 at 23:15
• steel 420 I'm planning to 3d-print this part from steel. So, it will be steel sand using in printers. – Robotex Mar 21 '17 at 16:34