Unit Conversions Grams to Watts

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Discussion Overview

The discussion revolves around unit conversions related to calculating power output in watts from thrust measured in grams and velocity in meters per second. Participants are attempting to resolve discrepancies in their calculations and clarify the proper conversion methods for units involved in the power equation.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents a formula for calculating power available based on thrust and velocity, including unit conversion factors.
  • Another participant suggests a correction in the conversion process from lb*ft/sec to watts, proposing a different order of operations for the conversion.
  • A participant reports obtaining a power output of 0.000 watts, expressing concern over the validity of this result in relation to expected efficiency values.
  • A later reply reiterates the 0.000 watts result and emphasizes the need for scientific notation due to the small magnitudes involved, while also noting the lack of efficiency calculations provided by the original poster.
  • There is a suggestion that the scenario described may involve very small power outputs, leading to confusion about the calculations.

Areas of Agreement / Disagreement

Participants express differing views on the correct method for unit conversion and the resulting power output. There is no consensus on the calculations or the interpretation of the results, and the discussion remains unresolved.

Contextual Notes

Participants have not provided complete efficiency calculations, and there are unresolved issues regarding the assumptions made in the unit conversions and the physical context of the problem.

Who May Find This Useful

Students or individuals working on physics problems involving unit conversions, particularly in the context of power calculations related to thrust and velocity.

MD_Programmer
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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
 
Last edited:
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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.
 
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
 
MD_Programmer said:
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.
 

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