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2018 Tianjin Open – Doubles

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Doubles
2018 Tianjin Open
Final
ChampionsUnited States Nicole Melichar
Czech Republic Květa Peschke
Runners-upAustralia Monique Adamczak
Australia Jessica Moore
Score6–4, 6–2
Events
Singles Doubles
← 2017 · Tianjin Open · 2019 →

Irina-Camelia Begu and Sara Errani were the defending champions, but Errani is serving a doping suspension. Begu played alongside Barbora Krejčíková, but they lost in the quarterfinals to Han Xinyun and Darija Jurak.

Nicole Melichar and Květa Peschke won the title, defeating Monique Adamczak and Jessica Moore in the final 6–4, 6–2.

Seeds

Draw

Key

Draw

First Round Quarterfinals Semifinals Final
1 Romania I-C Begu
Czech Republic B Krejčíková
6 6
WC China Y Liu
China M Wang
2 0 1 Romania I-C Begu
Czech Republic B Krejčíková
4 6 [8]
  United States S Santamaria
Russia Y Sizikova
2 1 China X Han
Croatia D Jurak
6 4 [10]
  China X Han
Croatia D Jurak
6 6 China X Han
Croatia D Jurak
4 6 [5]
4 United States K Christian
United States D Krawczyk
2 6 [4] Australia M Adamczak
Australia J Moore
6 1 [10]
  Spain L Arruabarrena
Switzerland T Bacsinszky
6 4 [10] Spain L Arruabarrena
Switzerland T Bacsinszky
63 1r
  Australia M Adamczak
Australia J Moore
6 1 [10] Australia M Adamczak
Australia J Moore
77 2
  China Y Duan
China Y Wang
4 6 [8] Australia M Adamczak
Australia J Moore
4 2
  Japan M Doi
Netherlands E Wacanno
6 6 3 United States N Melichar
Czech Republic K Peschke
6 6
  China X Jiang
China Q Tang
4 4 Japan M Doi
Netherlands E Wacanno
3 5
  Russia E Rodina
Australia A Tomljanović
3 6 [6] 3 United States N Melichar
Czech Republic K Peschke
6 7
3 United States N Melichar
Czech Republic K Peschke
6 4 [10] 3 United States N Melichar
Czech Republic K Peschke
4 6 [10]
WC China H Guo
China L Zhu
5 3 Russia N Dzalamidze
Russia V Kudermetova
6 3 [6]
  United States J Cako
India P Thombare
7 6 United States J Cako
India P Thombare
4 4
  Russia N Dzalamidze
Russia V Kudermetova
77 1 [10] Russia N Dzalamidze
Russia V Kudermetova
6 6
Alt India A Raina
United Kingdom E Webley-Smith
64 6 [5]

References