# Challenger de Nanchang 2014 (dobles masculino)

ATP Challenger China International – Nanchang :
| Campeones: | Ti Chen Hsien-yin Peng | Ti Chen Hsien-yin Peng | Ti Chen Hsien-yin Peng | Ti Chen Hsien-yin Peng | Ti Chen Hsien-yin Peng | Ti Chen Hsien-yin Peng | Ti Chen Hsien-yin Peng | Ti Chen Hsien-yin Peng | Ti Chen Hsien-yin Peng |
| Subcampeones: | Jordan Kerr Fabrice Martin | Jordan Kerr Fabrice Martin | Jordan Kerr Fabrice Martin | Jordan Kerr Fabrice Martin | Jordan Kerr Fabrice Martin | Jordan Kerr Fabrice Martin | Jordan Kerr Fabrice Martin | Jordan Kerr Fabrice Martin | Jordan Kerr Fabrice Martin |
| Resultado: | 6-2, 3-6, 12-10 | 6-2, 3-6, 12-10 | 6-2, 3-6, 12-10 | 6-2, 3-6, 12-10 | 6-2, 3-6, 12-10 | 6-2, 3-6, 12-10 | 6-2, 3-6, 12-10 | 6-2, 3-6, 12-10 | 6-2, 3-6, 12-10 |
| Modalidades |||||||||||
| Individuales | Masculino | | | |
| Dobles | Masculino | |

## Presentación previa

No hay campeones defensores ya que se trata de la primera edición del torneo.
La pareja de Taiwán formada por Ti Chen y Hsien-yin Peng ganaron el título, derrotando en la final al dúo formado por el australiano Jordan Kerr y el francés Fabrice Martin por 6-2, 3-6, 12-10.

## Cabezas de serie

* Ti Chen / Hsien-yin Peng (CAMPEONES)
* Jordan Kerr / Fabrice Martin (FINAL)
* Hsin-han Lee / Ze Zhang (Semifinales)
* Ryan Agar / Liang-chi Huang (Primera ronda)

## Cuadro

| Primera ronda | Primera ronda | Primera ronda | Primera ronda | Primera ronda | Cuartos de final | Cuartos de final | Cuartos de final | Cuartos de final | Cuartos de final | Semifinales | Semifinales | Semifinales | Semifinales | Final | Final | Final | Final | Final |
| 1 | Ti Chen Hsien-yin Peng | 6 | 6 | |
| WC | Hua Runhao Liu Siyu | 3 | 4 | |
| WC | Hua Runhao Liu Siyu | 3 | 4 | | 1 | T. Chen H-y. Peng | 7 | 6 | |
| | | | | | | Y. Bai D. Wu | 6$^2$ | 4 | |
| | Bai Yan Di Wu | 6 | 6 | | | Y. Bai D. Wu | 6$^2$ | 4 | |
| | Gong Peng-Xiang Gong Xiao | 2 | 2 | | | | | | |
| | Gong Peng-Xiang Gong Xiao | 2 | 2 | | | | | | | 1 | T. Chen H-y. Peng | 6 | 6 |
| | | | | | | | | | | 3 | H-h. Lee Z. Zhang | 4 | 3 |
| 3 | Hsin-han Lee Ze Zhang | 6 | 7 | | | | | | | 3 | H-h. Lee Z. Zhang | 4 | 3 |
| | Nuttanon Kadchapanan Danai Udomchoke | 3 | 5 | | | | | | | | | | |
| | Nuttanon Kadchapanan Danai Udomchoke | 3 | 5 | | 3 | H-h. Lee Z. Zhang | 6 | 4 | 10 | | | | |
| | | | | | | N.S. Balaji S. Sekiguchi | 2 | 6 | 6 | | | | |
| | N.Sriram Balaji Shuichi Sekiguchi | 6 | 6 | | | N.S. Balaji S. Sekiguchi | 2 | 6 | 6 | | | | |
| WC | Chen Long Wu Hao | 3 | 2 | | | | | | | | | | |
| WC | Chen Long Wu Hao | 3 | 2 | | | | | | | | | | | 1 | T. Chen H-y. Peng | 6 | 3 | 12 |
| | | | | | | | | | | | | | | 2 | J. Kerr F. Martin | 2 | 6 | 10 |
| | Maoxin Gong Hiroki Kondo | 6$^5$ | 7$^7$ | [10] | | | | | | | | | | 2 | J. Kerr F. Martin | 2 | 6 | 10 |
| | Nicolas Meister Jeevan Nedunchezhiyan | 7$^7$ | 6$^2$ | [4] | | | | | | | | | |
| | Nicolas Meister Jeevan Nedunchezhiyan | 7$^7$ | 6$^2$ | [4] | | M. Gong H. Kondo | 2 | 7 | 10 | | | | |
| | | | | | | H. Chung J-s. Nam | 6 | 5 | 7 | | | | |
| | Hyeon Chung Ji-sung Nam | 3 | 7$^7$ | [10] | | H. Chung J-s. Nam | 6 | 5 | 7 | | | | |
| 4 | Ryan Agar Liang-chi Huang | 6 | 6$^5$ | [7] | | | | | | | | | |
| 4 | Ryan Agar Liang-chi Huang | 6 | 6$^5$ | [7] | | | | | | | M. Gong H. Kondo | 3 | 4 |
| | | | | | | | | | | 2 | J. Kerr F. Martin | 6 | 6 |
| WC | Ning Yuqing Wang Aoran | 3 | 2 | | | | | | | 2 | J. Kerr F. Martin | 6 | 6 |
| | Denis Matsukevich Valeri Rudnev | 6 | 6 | | | | | | |
| | Denis Matsukevich Valeri Rudnev | 6 | 6 | | | D. Matsukevich V. Rudnev | w/o | | |
| | | | | | 2 | J. Kerr F. Martin | | | |
| | Toshihide Matsui Go Soeda | 6$^5$ | 6 | [6] | 2 | J. Kerr F. Martin | | | |
| 2 | Jordan Kerr Fabrice Martin | 7$^7$ | 3 | [10] |