How much horsepower does a 2019 F-550 6.7 have?
About 440 horsepower.
The 2019 Ford F-550 relies on the 6.7-liter Power Stroke V8 turbo diesel as its sole engine, paired with a heavy-duty automatic transmission. This power level is a key part of the F-Series Super Duty’s heavy-tow capability, and Ford typically brands the 6.7 Power Stroke with a high torque figure to match its purpose. For most 2019 F-550 configurations, the official rating sits around 440 horsepower, with torque near 925 lb-ft. Exact numbers can vary slightly by emission package and regional specification, but 440 hp is the commonly cited figure for this chassis-year combination.
Engine specifications and what they imply
To give readers a quick reference, here are the core specs tied to the 2019 F-550’s 6.7L Power Stroke powertrain. This set is intended to reflect typical production trucks in this model year.
- Engine: 6.7L Power Stroke V8 Turbo Diesel
- Horsepower: 440 hp (328 kW)
- Torque: 925 lb-ft (1255 Nm)
- Transmission: Heavy-duty automatic (TorqShift, part of the 6R140 family)
In practice, horsepower figures are just part of the story. Torque, gearing, and the vehicle’s weight and upfit all influence real-world performance, especially when towing or operating in demanding conditions like construction sites or rural haul routes.
Notes on variation and context
Horsepower can vary slightly by market, emission-control configuration, and upfit differences. For the most precise figure on a given 2019 F-550, consult the vehicle’s window sticker (Monroney label) or Ford’s official 2019 Super Duty specifications for the exact configuration in question.
Summary
The 2019 Ford F-550 with the 6.7-liter Power Stroke diesel is commonly rated at about 440 horsepower, paired with around 925 lb-ft of torque, delivering robust capability for heavy-duty work. While the core figure is consistent, buyers should verify the exact spec on their specific truck due to regional and upfit variations.
