As concluded in a paragraph 6.2 the better tightness of the gasket with the external groove was achieved thanks to an more even contact pressure distribution on the gasket surface which directly corresponds to larger effective gasket’s width. In order to explain what is an influence of the gasket’s effective width on leakage level, the mechanism of the gas flow via porous ring layer was taken to analysis. The radial flow through the porous layer in the form of ring can be described in accordance with modified Darcy’s expression presented among others in [16, 17]:

$$Q=\frac{2\pi Kh\left({{p}^{2}}_{02}-{{p}^{2}}_{01}\right)}{\eta RT\mathrm{ln}\left(\frac{{r}_{02}}{{r}_{01}}\right)}$$(1)where:

*K–* permeability,

*h–* gasket thickness,

*η –* dynamic viscosity,

R– universal gas constant,

T– gas temperature,

*r*_{01}*, r*_{02} – internal and external radius of the ring respectively,

*p*_{01}, *p*_{02} – internal and external pressure of the ring respectively.

Based on the formula (1) it can be seen that the leakage is proportional to gasket thickness, material permeability and pressure difference, and inversely proportional to thermodynamic gas parameters as well as gasket’s radii which determine the gasket width. To investigate the influence of the gasket’s width on a leakage level only the radii of the ring will be considered. Taking above parameters to consideration, the leakage formula (1) can be reduced to the form:

$$Q\approx \frac{1}{\mathrm{ln}\left(\frac{{r}_{02}}{{r}_{01}}\right)}$$(2)Accordance in the gasket’s model presented in Fig. 15 it can be assumed that the increment x_{01} will be caused enlarging of the radius r_{01}, whereas increment x_{02} will be caused decreasing of radius r_{02}. Depending on the gasket’s radii location, three basic cases will be taken to consideration:

Figure 15 The model of the ring gasket as a porous structure

**Case 0**: Where the radius r_{x}_{1} and radius r_{x}_{2} are meets in one common point located in among of gasket width area. Additionally radius r_{xo}_{1} is greater than radius r_{o}_{1} and radius r_{x}_{2} is smaller than radius r_{o}_{2}. In such conditions the below relationships are:

*r*_{X}_{2}<*r*_{02}; *r*_{X}_{1}>r_{01} and *r*_{X}_{2}/*r*_{X}_{1}=1; so ln(1)=0 and it causes that the leakage tends to infinity *Q*=∞, because the gasket width *b*=0.

**Case 1**: Where the radius *r*_{01} is constant and radius *r*_{X}_{2} decreasing but is greater than *r*_{01}. In such conditions the below relationships are:

*r*_{01}=const. *r*_{X}_{2}>*r*_{01} so *r*_{X}_{2}/*r*_{01}>1 it means that ln(*r*_{02}/*r*_{X}_{1})>1 it means that leakage reaches the finite value greater than zero.

$${Q}^{1}=\frac{1}{\mathrm{ln}\left(\frac{{r}_{02}-{x}_{02}}{{r}_{01}}\right)}$$(3)**Case 2**: Where the radius *r*_{02} is constant and radius *r*_{X}_{1} increasing but is smaller than *r*_{02}. In such conditions the below relationships are:

*r*_{02}=const. *r*_{X}_{1}<*r*_{02} so *r*_{02}/*r*_{X}_{1}>1 hence ln(*r*_{02}/*r*_{X}_{1})>1. In this case the leakage also reaches the finite value greater than zero.

$${Q}^{2}=\frac{1}{\mathrm{ln}\left(\frac{{r}_{02}}{{r}_{01}+{x}_{01}}\right)}$$(4)Compeering the equations (3) and (4), the basic question is which situation is better? case_1- where external radius *r*_{X}_{2} decrease or case_2 - where internal radius *r*_{X}_{1} increase?

In order to respond to above question the investigation of the both cases Q^{1} and Q^{2} in relation to nominal leakage (where the internal and external radius was not changing) was carried out:

$$\begin{array}{r}{Q}_{1}/{Q}_{n}=\frac{\mathrm{ln}\left(\frac{{r}_{02}}{{r}_{01}}\right)}{\mathrm{ln}\left(\frac{{r}_{02}-{x}_{02}}{{r}_{01}}\right)}\end{array}$$(5)and

$${Q}_{2}/{Q}_{n}=\frac{\mathrm{ln}\left(\frac{{r}_{02}}{{r}_{01}}\right)}{\mathrm{ln}\left(\frac{{r}_{02}}{{r}_{01}+{x}_{01}}\right)}$$(6)Putting to equation (5) and (6) the gasket’s nominal dimensions: *r*_{01}=26 mm, *r*_{02}=34.5 mm and considering the radius changing in range: 0≤ *x*_{01}<b for a case 1 and 0≤ *x*_{02}<b for a case 2, the course of relative leakages can be comparing - see Fig. 16.

Figure 16 The course of the relative leakage in case 1 and case 2

It can be seen that the smaller relative leakage was reached at constant internal radius *r*_{01} (means in case 1) than in situation where external radius *r*_{02} was constant. It concluded that the better leakage level was obtained if the radius *r*_{02} is changing instead situation where the radius r_{01} is changing.

On the Fig. 17 the effective internal and external radius of gaskets were determined. Putting those values to the equation (7) the relative leakage values were determined and collected in .

Figure 17 Determination of the gasket’s effective radius, a) standard gasket, b) gasket with symmetric grooves, c) gasket wit external groove

Table 1 Relative leakage based of effective (width) radii of the particular gasket solution

$$Q/{Q}_{n}=\frac{\text{ln}\left(\frac{{r}_{02}}{{r}_{01}}\right)}{\text{ln}\left(\frac{{r}_{EF2}}{{r}_{EF1}}\right)}$$(7)Based on it was proved that gasket with external groove characterizes one order of magnitude smallest relative leakage against to standard gasket which was experimentally proven. The gasket with groove on both sides revealed three times greater leakage than gasket with external groove which was also confirmed by experimental study.

In this paper only the geometrical aspect as an effective gasket width was taken to the consideration. In order to model a real leakage behavior, such parameter as material permeability (which strongly depends on the internal and surface structures as well as applied gasket stress) should be taken into consideration.

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