Apologize if probably this is duplicate question. But I have tried to search if someone have asked such this question.

I have a Honda engine, GX200, 196cc, with it's gross power is 6.5HP. I would like to make its exhaust to blow something. For that purpose, I need to modify the muffler to become something like shown by the picture below. But before I modify, I need to know how strong will be the blowing force right from the exhaust output from the piston? Is any formula to calculate the force? That modification is intended to utilize it heat and its wind blowing.

Engine exhaust/muffler

  • $\begingroup$ Check out turbo chargers, there are many sources showing the calcs for the power that can be delivered. $\endgroup$
    – Solar Mike
    Nov 10, 2018 at 5:40
  • $\begingroup$ But this engine is not using turbo charger. But however, I will check of it is relevant. $\endgroup$ Nov 10, 2018 at 6:13
  • $\begingroup$ I did not say that engine was using a turbo charger - I can see clearly, from the image, that it does not have one, but the theory used to calculate turbos may help you... $\endgroup$
    – Solar Mike
    Nov 10, 2018 at 6:16
  • $\begingroup$ Is this for a airplane that has 4 engines - which are replacements for the originals and have had new mounts etc made... $\endgroup$
    – Solar Mike
    Nov 10, 2018 at 6:21
  • $\begingroup$ Not actually. But I am trying hkw to use such this small engine to power airplane. This engine weight is 17kg net. Airplane engine sould be around 1kg every 2HP. But this engine is 1kg around 0.4HP. Mean, the power of this engine will be consumed to lift the engine weight itself. Need to calculate and to design very accurately. $\endgroup$ Nov 10, 2018 at 6:45

1 Answer 1


To calculate force at exhaust port specifically is of little use, but nevertheless, it is possible.

Honda GX200 specs - Max power 6.5HP@3600 rpm or 60 rps. We will calculate thrust at this rps. 196 cc engine. The exhaust port diameter from diagrams is estimated to be around 2.8cm. Assuming complete combustion of exhaust gases we will use the density of air at 1000k as a substitute as the density is subject to large variation due to active air-fuel ratio, load conditions etc. Density = .3716kg/m^3 Engine is pumping .196L of exhaust gases every 2 revolutions as it is a 4 stroke engine. Therefore at 60 rps, it is pumping .196L 30 times every second. That is equal to 5.88L/s or .00588m^3/s. This is the volumetric flow rate.

The cross sectional area of the exhaust port is pi*(d/2)^2 = .0006157m^2 Therefore the Exhaust gas velocity is .00588/.0006157 = 9.54m/s We can now calculate thrust or force by using the following equation F=density* Me* Ve =.3716*.00588*9.54 = .020 N of force. This is negligible and not enough to force the engine forward r in any particular direction, but can be used as an effective blower or pump of sorts with appropriate tubing. The only way to increase thrust at this rpm would be to restrict the exhaust manifold by reducing diameter and increasing length. This would cause the exhaust gases to be forced through a smaller space and hence will develop the tendency to escape quicker, ultimately increasing exhaust velocity. Hope this answer helped.

  • 1
    $\begingroup$ Have you considered the effect on the engine if you increase the back pressure by reducing the manifold diameter? $\endgroup$
    – Solar Mike
    Feb 28, 2019 at 9:16
  • $\begingroup$ How did you derive this assumption: "Assuming complete combustion of exhaust gases we will use the density of air at 1000k as", and what is this: "Density = .3716kg/m^3"? $\endgroup$ Mar 1, 2019 at 2:46
  • $\begingroup$ I agree that if we reduce the muffler diameter, then the pressure will be higher, Bernoulli's principle. But, will it not to force back the air to the combustion chamber? $\endgroup$ Mar 1, 2019 at 2:48

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