UnitSolutions - 2 Elevation = 50,000 (ft) x 0.305 (m/ft) =...

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ME 201 Thermodyanmics 1 ME 201 Thermodynamics Practice Problems: Unit Manipulations Perform the following unit manipulations. a. A jet engine provides a thrust (force) of 2,000 lb f with a velocity of 600 km/hr. What is the power produced in horsepower? Solution Power = Force x Velocity Convert to SI Force = 2000 (lb f ) x 4.448 (N/lb f ) = 8896 N Velocity = 600 (km/hr) x 0.278 [(m/s)/(km/hr)] = 166.8 m/s Power = 8896 (N) x 166.8 (m/s) = 1,483,853 (N m/s) = 1484 (kW) Covert to hp Power = 1484 (kW) / 0.7457 (kW/hp) = 1990 hp b. What is the potential energy (in kJ) of a 1.25 ton aircraft at an elevation of 50,000 ft? Solution Potential Energy = Mass x Gravitational Acceleration x Elevation Convert to SI Mass = 1.25 (ton) x 907.18 (kg/ton) = 1134 kg Gravitational Acceleration = 9.8 m/s
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Unformatted text preview: 2 Elevation = 50,000 (ft) x 0.305 (m/ft) = 15,250 m Potential Energy = 1134 (kg) x 9.8 (m/s 2 ) x 15,250 (m) = 1.695 x 10 8 (kg m 2 /s 2 ) Covert to kJ Potential Energy = 1.695 x 10 8 (kg m 2 /s 2 ) = 1.695 x 10 8 (N m) = 1.695 x 10 8 (J) x 0.001 (kJ/J) = 1.695 x 10 5 kJ ME 201 Thermodyanmics 2 c. Determine the kinetic energy (in Btu) of a 0.50g baseball thrown at 97 mph. Solution Kinetic Energy = 1/2 x Mass x Velocity Squared Convert to SI Mass = 0.5 (gm) x 10-3 (kg/gm) = 5 x 10-4 kg Velocity = 97 (mph) x 0.447 [(m/s)/(mph)] = 43.4 m/s Kinetic Energy = 0.5 x 5 x 10-4 (kg) x [43.4] 2 (m 2 /s 2 ) = 0.47 J Covert to Btu Kinetic Energy = 0.478 (J) / 1055 (J/Btu) = 4.45 x 10-4 Btu...
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This note was uploaded on 07/25/2008 for the course ME 201 taught by Professor Somerton during the Spring '06 term at Michigan State University.

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UnitSolutions - 2 Elevation = 50,000 (ft) x 0.305 (m/ft) =...

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