Southwest Airlines WN6808 Flight Status Today

Domestic flight Southwest Airlines WN6808 takes off from Atlanta (ATL) United States to Cleveland (CLE) United States. It's operated by Southwest Airlines. The plane leaves Hartsfield-Jackson Atlanta International Airport at 13:00 America/New_York. The flight is expected to land at Cleveland Hopkins International Airport at 14:40 America/New_York. The flight will last about 1 hours 40 minutes.

WN6808 - SWA6808 Southwest Airlines (WN)
ATL Atlanta United States
Flight ATL to CLE
CLE Cleveland United States
Contact Airlines It may not operate on the date requested
Departure Hartsfield-Jackson Atlanta International Airport
IATA: ATLICAO: KATL
Scheduled 13:00
N None
Arrival Cleveland Hopkins International Airport
IATA: CLEICAO: KCLE
Scheduled 14:40
None B11
WN6808 Detail
  • Type: Domestic Flight
  • Flight Duration: 1 hours 40 minutes
  • Flight Distance: 893 kms / 555 miles
Airline
  • Southwest Airlines
  • IATA: WN
  • ICAO: SWA
  • Operating Days: Monday
Passenger Services
  • Service type: Normal passenger
  • Seats: 175
  • Freight capacity: 8.7 tons
  • Passenger classes: Economy, Premium Economy
More Detail
  • Aircraft: Boeing 737-800 (Scimitar Winglets) Passenger
  • Callsign: F-KICT2
  • Departure Timezone: America/New_York
  • Arrival Timezone: America/New_York
  • Current Time in Atlanta: Monday 2026-01-19 03:47 AM
  • Current Time in Cleveland: Monday 2026-01-19 03:47 AM

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Southwest Airlines Flight WN6808 FAQs

What is the scheduled flight duration for Southwest Airlines WN6808 flight?

On average, nonstop flight takes 1 hour(s) 40 minutes, with the flight distance of 893 km (555 miles).

What type of aircraft is used for the Southwest Airlines WN6808 flight?

All Southwest Airlines WN6808 flights are operated using Boeing 737-800 (Scimitar Winglets) Passenger aircraft.

How many Southwest Airlines WN6808 flights are operated a week?

1 flights per week. The Flight Southwest Airlines WN6808 is operated on Monday.