Lake Tașaul

Coordinates: 44°21′00″N 28°36′48″E / 44.35000°N 28.61333°E / 44.35000; 28.61333
From Wikipedia, the free encyclopedia
(Redirected from Tașaul Lake)
Lake Tașaul
Lake Tașaul is located in Constanţa County
Lake Tașaul
Lake Tașaul
LocationConstanța County, Northern Dobruja
Coordinates44°21′00″N 28°36′48″E / 44.35000°N 28.61333°E / 44.35000; 28.61333
Primary inflowsCasimcea, Tașaul, Dalufac
Basin countriesRomania
Surface area23.35 km2 (9.02 sq mi)
Max. depth4 m (13 ft)
Surface elevation1 m (3.3 ft)

Lake Tașaul (Romanian: Limanul Tașaul) is a lake in Northern Dobruja, Romania. Formerly an open salt water coastal lagoon, connected with the Black Sea, it was transformed into a freshwater lake in the 1920s. Its area is 23.35 km2 (9.02 sq mi) and its maximum depth is 4 m (13 ft).[1]

References[edit]

  1. ^ M.L. Alexandrov; J. Bloesch (2009). "Eutrophication of Lake Tașaul, Romania-proposals for rehabilitation". Environmental Science and Pollution Research. 16 (Suppl. 1): 42–45. doi:10.1007/s11356-008-0071-7. PMID 19132430. S2CID 39817102.