Design and Simulation of Two Stroke Engines

Chapter 8 - Reduction of Noise Emission from Two-Stroke Engines
number of cylinders x rpm
fe = ~ — (8.5.19)
60
and, as the intake silencer is of the low-pass type, the engine speed in question is at the lower
end of the usable speed band. Annand and Roe [1.8] give some further advice on this particular
matter.
There are several other criteria to be satisfied as well in this design process, such as
ensuring that the total box volume and the intake pipe area are sufficiently large so as not to
choke the engine induction process and reduce the delivery ratio. Such a calculation is normally
accomplished using an unsteady gas-dynamic engine model, extended to include the
intake and exhaust silencers [8.21]. However, an approximate guide to such parameters is
given by the following relationship, where Ncy is the number of cylinders, Vsv is the swept
volume of any one cylinder, and d2 is the flow diameter of the intake duct or the carburetor, as
shown in Fig. 5.5:
10 x V s v^7 < Vb < 20 x VSVA/N^ (8.5.20)
0.6d2 < dp < 0.8d2
(8.5.21)
To put this advice in numerical form, revisit the 65 cm 3 chainsaw engine employed as a
design example in Chapters 5, 6 and 7. It is designed to run between 5400 and 10,800 rpm,
cutting wood in a wide-open-throttle condition. The carburetor on the engine has a 22 mm
flow diameter. The lowest engine speed of direct interest is 5400 rpm, where the engine has a
fundamental induction frequency, fe, of 90 Hz. Therefore, for design purposes it will be assumed
that this 65 cm 3 chainsaw engine will not be operated at full throttle below 5400 rpm,
consequently the minimum value for fj is 90 Hz. From Eq. 8.5.20 the minimum box volume,
Vb, should be 650 cm 3 . No chainsaw can ever be so profligate with space, so 500 cm 3 is
probably the maximum which can be afforded for the box. The minimum pipe diameter, dp,
from Eq. 8.5.21, is approximately 15 mm. When the air temperature is at 20°C, so the acoustic
velocity, ao, from Eq. 8.1.1, is 343 m/s. Testing a real pipe length, Lp, of 125 mm, by using
Eq. 8.5.18,
From Eq. 8.5.17:
L eff = L D + —~ = 125 + 7tXl = 136.8 mm
v 4 4
t J5_Y
= 3« UllOOoJ = 87J Hz
LeffVb 2K 136.8 500
1000 10 6
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