Design and Simulation of Two Stroke Engines

Chapter 2 - Gas Flow through Two-Stroke Engines
Ps side a — Ps side b
c s side a — c s side b
(2.5.3)
(2.5.4)
The solution divides into two different cases, one simple and the other more complex,
depending on whether or not the gas composition is identical on both sides of the boundary.
(i) The simple case of common gas composition
The following is the solution of the simple case where the gas is identical in composition
on both sides of the boundary, i.e., ya and Ra are identical to Yb and Rb- Eq. 2.5.4 reduces to:
G 5a a 0a( X l ~ X 2d) ~ G 5b a 0b( X ld ~ X 2)
(2.5.5)
As the gas composition is common (and although the G5 terms are actually equal they are
retained for completeness), this reduces to:
Eq. 2.5.3 reduces to:
a 0a u 5a
In this simple case, the values of G-ja
The solution becomes straightforward:
X 2d -
^ a 0b G 5b; (X l _ X 2d) - ( X ld ~ X 2) (2.5.6)
G,
(Xl+x2d -1)^.=(xld+x2 - ir b
2Xo -X
and G7b are identical and are simply G7:
x l + x 2d - 1 = x ld + x 2 ~ 1
1-
V
I I a 0a G a G
0b 5b J u VG7
5a
a 0b G 5b
a 0a G 5a
(2.5.7)
(2.5.8)
—± hence p2d = p0X2d 7 (2.5.9)
G,
X ld = X l + X 2d - X 2 hence Pld ~ P0 X ld 7 (2.5.10)
(ii) The more complex case of variable gas composition
The simplicity of reduction of Eqs. 2.5.6 to 2.5.7, and also Eqs. 2.5.7 to 2.5.8, is no longer
possible as Eq. 2.5.7 remains as a polynomial function. The method of solution is to eliminate
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