Effect of flow rate of shielding gas on distribution of particles in coaxial double-tube helium atmospheric pressure plasma jet

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Abstract

In the application of atmospheric pressure plasma jet, the influence of ambient gas cannot be ignored, especially in some specific scenarios which are highly sensitive to ambient particles. Coaxial double-tube plasma jet device is a promising method of controlling the chemical properties of jet effluent by restraining the mutual diffusion between jet effluent and ambient gas. In this work, the discharge characteristics and chemical properties of coaxial double-tube helium atmospheric pressure plasma jet at different flow rates of shielding gas are studied numerically, and the model is validated by experimental optical images. The results illustrate the enhanced discharge at the high flow rate, the weaker discharge at the low flow rate, and discharge behaviors without shielding gas as well. With the increase of shielded gas flow rate, the particle density increases in the discharge space, which can be attributed to the wider main discharge channel caused by the increase of shielding gas flow rate. In addition, the analysis shows the great difference in ion fluxes affected by the flow rate of the SG between the contour lines of different helium mole fractions. This study further reveals that different discharge positions have a great influence on the generation of nitrogen and oxygen particles, thus deepening the understanding of influence of shielding gas flow rate on discharge behavior, and may open up new opportunities for the further application of plasma jet.

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Corresponding author:Dai Dong,ddai@scut.edu.cn

Funds:Project supported by the National Natural Science Foundation of China (Grant No. 51877086).

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边界	表达式	备注
AX	—	对称轴
BC	u_i= 3 slm,c= 1	工作气体入口
DE	u_o,c= 0	屏蔽气体入口
FG	u= 0.1 m/s,c= 0	环境空气入口
GW	p= 1 atm, ${{\boldsymbol{n}}} \cdot {D_{\text{d} } }\nabla c = 0$	—
BPO,CQRD, UW,ESTF	u= 0 m/s, ${{\boldsymbol{n}}} \cdot {D_{\text{d} } }\nabla c = 0$	—

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边界	表达式	备注
IPO	V=V₀, 方程(9)—方程(12)	外施电压
HX	—	对称轴
IJ,KL	$- {\boldsymbol{n}} \cdot {\boldsymbol{D}} = 0$, $- {\boldsymbol{n} } \cdot {\boldsymbol{\varGamma} } {\text{e} } = 0$, $- {\boldsymbol{n}} \cdot {\boldsymbol{\varGamma}} {\varepsilon } = 0$
TV	V= 0, $- {\boldsymbol{n}} \cdot {\boldsymbol{\varGamma}} {\text{e} } = 0$, $- {\boldsymbol{n}} \cdot {\boldsymbol{\varGamma}} {\varepsilon } = 0$	接地
TM,XYV	V= 0	接地
UV,LST,JQRK	方程(9)—方程(12), 方程(14), 方程(15)

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序号	反应方程式	速率常数	能量损耗 /eV	参考文献
1	${\rm{e+He\to e+He}}$	f(c, ε) (m³·s^–1)	/	[40]
2	${\rm{e+He\to e+He^{\ast}}}$	f(c, ε) (m³·s^–1)	19.82	[40]
3	${\rm{e+He^{\ast }\to e+He}} $	f(c, ε) (m³·s^–1)	–19.82	[40]
4	${\rm{e+He\to 2e+He^{+}}} $	f(c, ε) (m³·s^–1)	24.587	[40]
5	${\rm{e+N_{2}\to e+N_{2}}} $	f(c, ε) (m³·s^–1)	/	[40]
6	${\rm{e+N_{2}\to e+N_{2}(VIB\, \textit{v}1)}}$	f(c, ε) (m³·s^–1)	0.2889	[40]
7	${\rm{e+N_{2}\to e+N_{2}(VIB\, 3\textit{v}1)} }$	f(c, ε) (m³·s^–1)	0.8559	[40]
8	${\rm{e+N_{2}\to e+N_{2}(VIB\, 4\textit{v}1)} }$	f(c, ε) (m³·s^–1)	1.1342	[40]
9	${\rm{e+N_{2}\to e+N_{2}(VIB \,5\textit{v}1)} }$	f(c, ε) (m³·s^–1)	1.4088	[40]
10	${\rm{e+N_{2}\to 2e+N_{2}^{+}}} $	f(c, ε) (m³·s^–1)	15.6	[40]
11	${\rm{e+O_{2}\to e+O_{2}}} $	f(c, ε) (m³·s^–1)	/	[40]
12	${\rm{e+O_{2}\to O+O^{-}}} $	f(c, ε) (m³·s^–1)	/	[40]
13	${\rm{e+O_{2}\to O_{2}^{-}}} $	f(c, ε) (m³·s^–1)	/	[40]
14	${\rm{e+O_{2}\to e+O_{2}(VIB\, 3\textit{v}1)} }$	f(c, ε) (m³·s^–1)	0.57	[40]
15	${\rm{e+O_{2}\to e+O_{2}(VIB\, 4\textit{v}1)} }$	f(c, ε) (m³·s^–1)	0.75	[40]
16	${\rm{e+O_{2}\to e+O_{2} } }(\rm A1)$	f(c, ε) (m³·s^–1)	0.997	[40]
17	${\rm{e+O_{2}\to e+O_{2}}} $	f(c, ε) (m³·s^–1)	–0.997	[40]
18	${\rm{e+O_{2}\to e+O_{2} } }(\rm B1)$	f(c, ε) (m³·s^–1)	1.627	[40]
19	${\rm{e+O_{2}\to e+O_{2}}} $	f(c, ε) (m³·s^–1)	–1.627	[40]
20	${\rm{e+O_{2}\to e+O_{2}(EXC)}} $	f(c, ε) (m³·s^–1)	4.5	[40]
21	${\rm{e+O_{2}\to e+O+O}} $	f(c, ε) (m³·s^–1)	5.58	[40]
22	${\rm{e+O_{2}\to e+O+O(^{1}D)}} $	f(c, ε) (m³·s^–1)	8.4	[40]
23	${\rm{e+O_{2}\to 2e+O_{2}^{+}}}$	f(c, ε)(m³·s^–1)	12.1	[40]
24	${\rm{e+He^{\ast }\to 2e+He^{+}}} $	$4.661 \times {10^{ - 16} } \times {T_{\text{e} } ^{0.6}} \times { {\rm{e} }^{ - 4.78/T_{\text{e} } } }\,({\rm m}^3{\cdot} {\rm{s} }^{-1})$	4.78	[41]
25	${\rm{e+He_{2}^{\ast }\to 2e+He_{2}^{+}}} $	$1.268 \times {10^{ - 18} } \times {T_{\text{e} }^{0.71} }\times { {\text{e} }^{ - 3.4/T_{\text{e} } } }\, ({\rm m}^3{\cdot} {\rm{s} }^{-1})$	3.4	[41]
26	${\rm{2He^{\ast }\to e+He+He^{+}}} $	4.5 × 10^–16(m³·s^–1)	–15	[41]
27	${\rm{e+He_{2}^{+}\to He^{\ast}+He}} $	$5.386\times10^{-13}\times T_{\rm e}^{-0.5}\rm (m^3{\cdot} s^{-1})$	/	[41]
28	${\rm{e+He^{+}\to He^{\ast}}} $	$6.76\times10^{-19}\times T_{\rm e}^{-0.5}\rm (m^3{\cdot} s^{-1})$	/	[41]
29	${\rm{2e+He^{+}\to e+He^{\ast}}} $	$6.186\times10^{-39}\times T_{\rm e}^{-4.4}\rm (m^3{\cdot} s^{-1})$	/	[31]
30	${\rm{e+He+He^{+}\to He+He^{\ast}}} $	$6.66\times10^{-42}\times T_{\rm e}^{-2}\rm (m^6{\cdot} s^{-1})$	/	[31]
31	${\rm{2e+He_{2}^{+}\to He_{2}^{\ast}+e}} $	1.2 × 10^–33(m⁶·s^–1)	/	[31]
32	${\rm{e+He+He_{2}^{+}\to He_{2}^{\ast }+He}} $	1.5 × 10^–39(m⁶·s^–1)	/	[31]
33	${\rm{e+He+He_{2}^{+}\to He^{\ast }+2He}} $	3.5 × 10^–39(m⁶·s^–1)	/	[31]
34	${\rm{2e+He_{2}^{+}\to He^{\ast }+He+e}} $	2.8 × 10^–32(m⁶·s^–1)	/	[31]
35	${\rm{e+N_{2}\to e+N+N}} $	$1\times10^{-16}\times T_{\rm e}^{-0.5}\times {\rm e}^{{-16}/T_{\rm{e} }}\rm (m^3{\cdot} s^{-1})$	9.757	[42]
36	${\rm{e+N_{2}^{+}\to N+N}} $	$4.8\times10^{-13}\times T_{\rm e}^{-0.5}\rm (m^3{\cdot} s^{-1})$	/	[42]
37	${\rm{e+N_{2}^{+}\to N_{2}}} $	$7.72\times10^{-14}\times T_{\rm e}^{-0.5}\rm (m^3{\cdot} s^{-1})$	/	[43]
38	${\rm{e+N_{4}^{+}\to 2N_{2}}} $	$3.22\times10^{-13}\times T_{\rm e}^{-0.5}\rm (m^3{\cdot} s^{-1})$	/	[44]
39	${\rm{2e+N_{2}^{+}\to N_{2}+e}} $	$3.165\times10^{-42}\times T_{\rm e}^{-0.8}\rm (m^6 \cdot s^{-1})$	/	[44]
40	${\rm{e+2O_{2}\to O_{2}+O_{2}^{-}}} $	$5.17\times10^{-43}\times T_{\rm e}^{-1}\rm (m^6{\cdot} s^{-1})$	–0.43	[44]
41	${\rm{e+O_{2}^{+}\to O+O}} $	$6\times10^{-11}\times T_{\rm e}^{-1}\rm (m^3{\cdot} s^{-1})$	–6.91	[44]
42	${\rm{e+O_{2}^{+}\to O_{2}}} $	4 × 10^–18(m³·s^–1)	/	[43]
43	${\rm{e+O_{4}^{+}\to 2O_{2}}} $	$2.25\times10^{-13}\times T_{\rm e}^{-0.5}\rm (m^3{\cdot} s^{-1})$	/	[44]
44	${\rm{He^{\ast}+ 2He \to He_{2}^{\ast }+He}} $	1.3 × 10^–45(m⁶·s^–1)	/	[41]
45	${\rm{He^{+}+2He\to He_{2}^{+}+He}} $	1 × 10^–43(m⁶·s^–1)	/	[41]
46	${\rm{N_{2}+N_{2}+N_{2}^{+}\to N_{2}+N_{4}^{+}}} $	5 × 10^–41(m⁶·s^–1)	/	[44]
47	${\rm{O^{-}+O_{2}^{+}\to O+O_{2}}} $	2 × 10^–13(m³·s^–1)	/	[41]
48	${\rm{O_{2}^{-}+O_{2}^{+}\to O_{2}+O_{2}}} $	2 × 10^–13(m³·s^–1)	/	[41]
49	${\rm{O_{2}^{-}+O_{2}^{+}+O_{2}\to 3O_{2}}} $	2 × 10^–37(m⁶·s^–1)	/	[44]
50	${\rm{O_{2}^{-}+O_{4}^{+}+O_{2}\to 4O_{2}}} $	2 × 10^–37(m⁶·s^–1)	/	[44]
51	${\rm{O_{2}+O_{2}+O_{2}^{+}\to O_{2}+O_{4}^{+}}} $	2.4 × 10^–42(m⁶·s^–1)	/	[44]
52	${\rm{He^{\ast }+N_{2}\to e+He+N_{2}^{+}}} $	7 × 10^–17(m³·s^–1)	/	[41]
53	${\rm{He_{2}^{\ast }+N_{2}\to e+2He+N_{2}^{+}}} $	7 × 10^–17(m³·s^–1)	/	[41]
54	${\rm{He_{2}^{\ast }+O_{2}\to e+2He+O_{2}^{+}}} $	3.6 × 10^–16(m³·s^–1)	/	[43]
55	${\rm{He^{\ast }+O_{2}\to e+He+O_{2}^{+}}} $	2.6 × 10^–16(m³·s^–1)	/	[43]
56	${\rm{He_{2}^{+}+N_{2}\to N_{2}^{+}+2He}} $	5 × 10^–16(m³·s^–1)	/	[41]
57	${\rm{He^{+}+N_{2}\to N_{2}^{+}+He}} $	5 × 10^–16(m³·s^–1)	/	[41]
58	${\rm{He+N_{2}+N_{2}^{+}\to He+N_{4}^{+}}} $	8.9 × 10^–42(m⁶·s^–1)	/	[42]
59	${\rm{He+O_{2}+O_{2}^{+}\to He+O_{4}^{+}}} $	5.8 × 10^–43(m⁶·s^–1)	/	[42]
60	${\rm{He+O_{2}^{-}+O_{2}^{+}\to He+2O_{2}}} $	2 × 10^–37(m⁶·s^–1)	/	[43]
61	${\rm{O_{2}^{-}+O_{2}^{+}+N_{2}\to 2O_{2}+N_{2}}} $	2 × 10^–37(m⁶·s^–1)	/	[43]
62	${\rm{O_{2}^{-}+O_{4}^{+}+N_{2}\to 3O_{2}+N_{2}}} $	2 × 10^–37(m⁶·s^–1)	/	[44]
63	${\rm{N_{2}+O_{2}+N_{2}^{+}\to O_{2}+N_{4}^{+}}} $	5 × 10^–41(m⁶·s^–1)	/	[44]
64	${\rm{O_{2}+N_{4}^{+}\to 2N_{2}+O_{2}^{+}}} $	2.5 × 10^–16(m³·s^–1)	/	[44]
65	${\rm{O_{2}+N+N\to O_{2}+N_{2}}} $	3.9 × 10^–45(m⁶·s^–1)	/	[43]
66	${\rm{O+O+N\to O_{2}+N}} $	3.2 × 10^–45(m⁶·s^–1)	/	[42]
注:f(c,ε)表示速率系数是通过电子能量分布函数(EEDF)使用相关文献中的横截面获得的.c表示He摩尔分数,ε表示平均电子能量(eV),n_e和T_e表示电子密度(m^–3) 和电子温度(eV). 他代表He(2³S)和He(2¹S). He₂^*代表He₂(a³∑u⁺). N₂(VIBv1), N₂(VIB 3v1), N₂(VIB 4v1)和N₂(VIB 5v1)被视为N₂, O₂(VIB 3v1), O₂(VIB 4v1), O₂(A1), O₂(B1)和O₂(EXC)被视为O₂; O(¹D)和O(¹S)被视为O.

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反应	c_He= 98%轮廓线上化学反应速率 /(mol·m^–2·s^–1)	c_He= 95%轮廓线上化学反应速率 /(mol·m^–2·s^–1)	c_He= 90%轮廓线上化学反应速率 /(mol·m^–2·s^–1)
R41: e + $\rm O_2^+$ → O + O	2.98 × 10^–3	1.27 × 10^–3	3.81 × 10^–4
R46: N₂+ N₂+ $\rm N_2^+ $ → N₂+ $\rm N_4^+$	1.67 × 10^–4	1.61 × 10^–5	2.67 × 10^–7
R51: O₂+ O₂+ $\rm O_2^+$ → O₂+ $\rm O_4^+$	8.86 × 10^–7	3.83 × 10^–6	6.70 × 10^–6
R52: He^*+ N₂→ e + He + $\rm N_2^+ $	1.29 × 10^–3	4.48 × 10^–5	4.96 × 10^–7
R55: He^*+ O₂→ e + He + $\rm O_2^+$	1.28 × 10^–3	4.42 × 10^–5	4.90 × 10^–7
R58: He + N₂+ $\rm N_2^+ $ → He + $\rm N_4^+ $	1.86 × 10^–3	6.92 × 10^–5	5.41 × 10^–7
R63: N₂+ O₂+ $\rm N_2^+ $ → O₂+ $\rm N_4^+ $	4.45 × 10^–5	4.29 × 10^–6	7.09 × 10^–8
R64: O₂+ $\rm N_4^+ $ → 2N₂+ $\rm O_2^+$	1.59 × 10^–3	9.67 × 10^–4	1.10 × 10^–4

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Vol. 71, Issue 16 - 2022-08-20

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