1994 Romanian Open – Doubles

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.

Doubles
1994 Romanian Open
Final
ChampionsAustralia Wayne Arthurs
Australia Simon Youl
Runners-upSpain Jordi Arrese
Spain José Antonio Conde
Score6–4, 6–4
Events
SinglesDoubles
← 1993 ·Romanian Open· 1995 →

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

Key

Draw

First roundQuarterfinalsSemifinalsFinal
1 F Mordegan
M Oosting
34
WC S Hiršzon
G Ivanišević
66WC S Hiršzon
G Ivanišević
614
V Flégl
K Kučera
42 W Arthurs
S Youl
366
W Arthurs
S Youl
66 W Arthurs
S Youl
66
4 J Ireland
J Waite
66 M Blackman
R Furlan
42
F Dewulf
L Pimek
344 J Ireland
J Waite
66
M Blackman
R Furlan
66 M Blackman
R Furlan
77
Q P Gauthier
C Gómez-Díaz
32 W Arthurs
S Youl
66
G Köves
L Markovits
66 J Arrese
JA Conde
44
WC A Pavel
R Sabău
77WC A Pavel
R Sabău
626
A Gaudenzi
B Larkham
66 A Gaudenzi
B Larkham
464
3 N Broad
G Van Emburgh
44WC A Pavel
R Sabău
464
WC G Cosac
CP Porumb
636 J Arrese
JA Conde
646
J Arrese
JA Conde
167 J Arrese
JA Conde
67
Ģ Dzelde
B Mota
332 T Carbonell
F Roig
46
2 T Carbonell
F Roig
66

References