A new sport car has a drag coefficient of 0.29 and a frontal area of 20ft^2, and is traveling at 100 mi/h.
How much power is required to overcome aerodynamic drag if p = 0,0002378 slugs/ft^3?
Please help me with the steps as well. I really need help
Thank you so much guys.
How much power is required to overcome aerodynamic drag if p = 0,0002378 slugs/ft^3?
Please help me with the steps as well. I really need help
Thank you so much guys.
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Fd = 1/2 * rho * V^2 * A * CD
where
FD is the force of drag, which is by definition the force component in the direction of the flow velocity,
ρ is the mass density of the fluid,
v is the velocity of the object relative to the fluid,
A is the reference area, and
CD is the drag coefficient — a dimensionless constant, e.g. 0.25 to 0.45 for a car.
100 miles * 5280 feet / mile = 5280 * 100 = 528000 feet
v = 528000 feet/3600 second = 146.7 feet/sec
F = 1/2 0.0002378 slugs/ft^3 * (146.7 ft/sec)^2 * 20 ft^2 * 0.29
F = 14.84 slugs ft / sec^2 = 14.84 pounds (believe it or not).
Work = F * d
Work = 14.84 * 528000 = 7835520 foot pounds
Power = W / T = 7835520/3600 = 2176 foot pounds / second
Power = 2176 * 1.36 = 2959 watts
Power = 2959/746 = 3.966 Hp
where
FD is the force of drag, which is by definition the force component in the direction of the flow velocity,
ρ is the mass density of the fluid,
v is the velocity of the object relative to the fluid,
A is the reference area, and
CD is the drag coefficient — a dimensionless constant, e.g. 0.25 to 0.45 for a car.
100 miles * 5280 feet / mile = 5280 * 100 = 528000 feet
v = 528000 feet/3600 second = 146.7 feet/sec
F = 1/2 0.0002378 slugs/ft^3 * (146.7 ft/sec)^2 * 20 ft^2 * 0.29
F = 14.84 slugs ft / sec^2 = 14.84 pounds (believe it or not).
Work = F * d
Work = 14.84 * 528000 = 7835520 foot pounds
Power = W / T = 7835520/3600 = 2176 foot pounds / second
Power = 2176 * 1.36 = 2959 watts
Power = 2959/746 = 3.966 Hp
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[Edit: are you sure that's the right value for the density of air? It looks kinda low..]
Let's start with the formula and see what happens:
C-drag = 2*F-drag/(rho*v^2*A), rho = fluid density, v = velocity relative to fluid, A = reference area.
Do the substitutions:
(100mph = 146.67 ft/s)
0.29 = 2*F-drag / (0.0002378 slugs/ft^3 * (146.7ft/s)^2 * 20ft^2)
Solve for drag force:
F-drag = 0.29 * 2.378x10-4 slugs/ft^3 * 2.15ft^2/s^2 * 20ft^2 / 2
= 14.83 slug-ft/s^2 (pounds)
Work = Force x Distance
Power = Work / Time = Force * Velocity
P = 14.83 slug-ft/s^2 * 146.67 ft/s = 2176 slug-ft^2/s^3
You can convert that to ft-lbs/s then to horsepower or whatever else you like.
[Edit #2: I'm more comfortable with metric units, so I'm going to try it again converted to metric to see if I can get the same answer.]
[Edit #3: Yeah, there's definitely something fishy here. I used 1.2256 kg/m^3 for the density of air and got a value for power = 32.76kW which is about 44 horsepower. Using your value of rho, it came out as only 4 horsepower. I think there's an extra zero in your air-density number..]
Let's start with the formula and see what happens:
C-drag = 2*F-drag/(rho*v^2*A), rho = fluid density, v = velocity relative to fluid, A = reference area.
Do the substitutions:
(100mph = 146.67 ft/s)
0.29 = 2*F-drag / (0.0002378 slugs/ft^3 * (146.7ft/s)^2 * 20ft^2)
Solve for drag force:
F-drag = 0.29 * 2.378x10-4 slugs/ft^3 * 2.15ft^2/s^2 * 20ft^2 / 2
= 14.83 slug-ft/s^2 (pounds)
Work = Force x Distance
Power = Work / Time = Force * Velocity
P = 14.83 slug-ft/s^2 * 146.67 ft/s = 2176 slug-ft^2/s^3
You can convert that to ft-lbs/s then to horsepower or whatever else you like.
[Edit #2: I'm more comfortable with metric units, so I'm going to try it again converted to metric to see if I can get the same answer.]
[Edit #3: Yeah, there's definitely something fishy here. I used 1.2256 kg/m^3 for the density of air and got a value for power = 32.76kW which is about 44 horsepower. Using your value of rho, it came out as only 4 horsepower. I think there's an extra zero in your air-density number..]