Unit Conversions Grams to Watts

[MD_Programmer]

Homework Statement

The power available (output) is measured by the velocity at which you are moving and the thrust
required to create that movement. NOTE: Conversions for proper units are provided below
equation.

Power Available: Pout = Tc * v

where: Pout = power available (watts)
Tc = Calibrated thrust (grams)
v = wind tunnel velocity (m/s)

In order to determine the efficiency, the units must cancel. To convert the units so that the output
power is in watts, use the following information:

can't get the power right help?

Homework Equations

1 gram = 0.002205 lbs.
1 m/s = 3.28 ft/s
1 horsepower = 550 lb*ft/sec
1 horsepower = 745.7 watts

v = 3 m/s
Tc = 0.0040 g

The Attempt at a Solution

(0.0040*0.002205) = 8.82 E-6

(3*3.28) = 9.84 ft/s

((8.82 E-6 * 9.84) *550) / 745.7 = 6.401212284 E-5 watts

 

[jackarms]

I think you're right in everything except going from lb*ft/sec to watts. Check your unit conversions. I think you should divide by 550 and then multiply by 745.7.

 

[MD_Programmer]

I got 0.000 watts using that method but that doesn't sound right because my efficiency calculations after are barely 1% and this data should keep to where they don't get above 80% according to the professor

 

[SteamKing]

I got 0.000 watts using that method but that doesn't sound right because my efficiency calculations after are barely 1% and this data should keep to where they don't get above 80% according to the professor

These are extremely small magnitudes you are dealing with, so you should be using scientific notation in your calculations.

You haven't provided any efficiency calculations, so it is not clear what your problem is.

It's hard to envision what sort of mechanism you are dealing with (a gnat on afterburners, perhaps), but whatever it is, only microscopic amounts of power are being generated.

 

[DeeNos]

To convert grams to watts, we need to use the following conversions:
1 gram = 0.002205 lbs
1 m/s = 3.28 ft/s
1 horsepower = 550 lb*ft/sec
1 horsepower = 745.7 watts

We are given that the calibrated thrust is 0.0040 grams and the wind tunnel velocity is 3 m/s. Therefore, we can calculate the power available (Pout) using the equation Pout = Tc * v.

Pout = (0.0040 * 0.002205 * 3 * 3.28 * 550) / 745.7 = 0.006 watts

Therefore, the power available is 0.006 watts. This means that for every 0.006 watts of power available, we need to use 0.0040 grams of calibrated thrust and have a wind tunnel velocity of 3 m/s.