Tungsten filament-finding length of coil

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SUMMARY

The discussion centers on calculating the length of a tungsten filament coil in an 18W light bulb operating at 2800 K, with a filament diameter of 0.25 mm and an emissivity of 0.35. The Stefan-Boltzmann law is applied using the equation &Q/&t = eσAT^4, where σ is the Stefan-Boltzmann constant (5.67 x 10^-8 W/m²K⁴). The user initially calculated the area (A) as 1.48 x 10^-5 m² and derived the height (h) of the filament as 0.30 cm, later correcting the formula used for the calculation.

PREREQUISITES
  • Understanding of the Stefan-Boltzmann law and its application in thermal radiation.
  • Familiarity with emissivity and its role in heat transfer calculations.
  • Basic knowledge of geometry related to cylinders for calculating surface area.
  • Proficiency in unit conversion, particularly between metric units (e.g., mm to cm).
NEXT STEPS
  • Review the application of the Stefan-Boltzmann law in thermal radiation calculations.
  • Explore emissivity values for different materials and their impact on heat transfer.
  • Learn about the geometry of cylindrical objects and how to calculate their surface area and volume.
  • Investigate the effects of vacuum conditions on thermal radiation in light bulbs.
USEFUL FOR

Students in physics or engineering, electrical engineers designing light bulbs, and anyone interested in thermal radiation principles and calculations.

laural
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Homework Statement


Tungsten wire has an emissivity of 0.35. A tungsten filament in a light bulb can be considered to emit light only from the sides of the filament (consider it a cylinder... they are usually wire "cylinders" wrapped into coils.). If the operating temperature of the filament of a small 18W light bulb is 2800 K and the diameter of the filament is 0.25 mm, then what is the length of the coil (in cm)? Assume the inside of the light bulb is under vaccuum, and that the ends of the filament do not contribute any heat radiation, since they are attached to the circuit...and assume other energy loss in the circuit is negligible.



Homework Equations



I will be using & to represent delta as I am unsure how change in is represented on keyboard.

&Q/&t = ekAT^4 where k is the Stefan-Boltzmann constant and = 5.67 x 10^-8 W/m^2K^4

The Attempt at a Solution



I plugged in 18W as the rate of change of heat per time, and the provided emissivity for e

Plugged 2800 K in for change in temperature.

So:

18W= (0.35)*(5.67 x 10^-8 W/m^2K^4
)*A*(2800^4 K)

Solved A= 1.48x10^-5

Then used A=(pi)r^2h,

And solved h= 0.30 cm

I have been struggling with this problem and am not sure if this is right. I would appreciate it if someone would review my work.

Thank-you.

Edit: I realized I was using the wrong formula. Changed.
 
Last edited:
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Check the calculation.
h = 1.48x10^-5/pix(0.125x10^-3)^2
 

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