Doubles
1994 Romanian Open
Champions	Wayne Arthurs Simon Youl
Runners-up	Jordi Arrese José Antonio Conde
Score	6–4, 6–4

Menno Oosting and Libor Pimek were the defending champions, but competed this year with different partners. Oosting teamed up with Federico Mordegan and lost in the first round to Saša Hiršzon and Goran Ivanišević, while Pimek teamed up with Filip Dewulf and also lost in the first round to Jon Ireland and Jack Waite.

Wayne Arthurs and Simon Youl won the title by defeating Jordi Arrese and José Antonio Conde 6–4, 6–4 in the final.^[1][2]

Seeds

Draw

Draw

	First round							Quarterfinals							Semifinals							Final
	1	F Mordegan M Oosting	3	4
	WC	S Hiršzon G Ivanišević	6	6				WC	S Hiršzon G Ivanišević	6	1	4
		V Flégl K Kučera	4	2					W Arthurs S Youl	3	6	6
		W Arthurs S Youl	6	6												W Arthurs S Youl	6	6
	4	J Ireland J Waite	6	6												M Blackman R Furlan	4	2
		F Dewulf L Pimek	3	4				4	J Ireland J Waite	6	6
		M Blackman R Furlan	6	6					M Blackman R Furlan	7	7
	Q	P Gauthier C Gómez-Díaz	3	2																			W Arthurs S Youl	6	6
		G Köves L Markovits	6	6																			J Arrese JA Conde	4	4
	WC	A Pavel R Sabău	7	7				WC	A Pavel R Sabău	6	2	6
		A Gaudenzi B Larkham	6	6					A Gaudenzi B Larkham	4	6	4
	3	N Broad G Van Emburgh	4	4											WC	A Pavel R Sabău	4	6	4
	WC	G Cosac CP Porumb	6	3	6											J Arrese JA Conde	6	4	6
		J Arrese JA Conde	1	6	7				J Arrese JA Conde	6	7
		Ģ Dzelde B Mota	3	3				2	T Carbonell F Roig	4	6
	2	T Carbonell F Roig	6	6

References

External links

• Official results archive (ATP)
• Official results archive (ITF)