Singles
1989 Virginia Slims of Nashville
Champion	Leila Meskhi
Runner-up	Helen Kelesi
Score	6–2, 6–3
Draw	32 (4Q/2LL)
Seeds	8

Susan Sloane was the defending champion, but lost in the semifinals to Helen Kelesi.

Leila Meskhi won the title by defeating Kelesi 6–2, 6–3 in the final.^[1] Meskhi became the first soviet woman in 15 years to win a professional tournament at the United States, after Olga Morozova won at Philadephia in 1974.^[2]

Seeds

Draw

Bottom half

	First round							Second round							Quarterfinals							Semifinals
		S Martin	3	7	3
		W White	6	5	6				W White	6	7
		C Porwik	7	6					C Porwik	3	6
		K Kessaris	6	3												W White	2	4
		M McGrath	4	4											4	S Sloane	6	6
		B Bowes	6	6					B Bowes	2	6
	Q	M Mazzotta	2	1				4	S Sloane	6	7
	4	S Sloane	6	6																		4	S Sloane	2	3
	8	J Pospíšilová	6	6																		2	H Kelesi	6	6
		P Louie Harper	3	3				8	J Pospíšilová	6	6
		C Cunningham	6	6					C Cunningham	2	0
	LL	A Farley	2	0											8	J Pospíšilová	2	3
		ML Daniels	6	6	6										2	H Kelesi	6	6
		L Allen	7	3	4				ML Daniels	6	4
		J Hetherington	4	6				2	H Kelesi	7	6
	2	H Kelesi	6	7

References

External links

• Official results archive (ITF)
• Official results archive (WTA)