How Much Heat Does a 60W Bulb Transfer to Air by Natural Convection?

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SUMMARY

The heat transfer from a 60W electric light bulb at 127°C to stagnant air at 27°C was calculated using the Nusselt number (Nu) and Grashof number (Gr). The Grashof number was determined to be 1.8x107, leading to a Nusselt number of 21.702. The convective heat transfer coefficient (h) was calculated as 1.139x10-5 kW/m2K, resulting in a heat loss of approximately 8.945x10-3 W, which is 0.014% of the total power. The calculations were questioned regarding the temperature conversion and the value of thermal conductivity (k).

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  • Understanding of heat transfer principles, specifically natural convection.
  • Familiarity with the Nusselt number (Nu) and Grashof number (Gr) calculations.
  • Knowledge of thermal conductivity (k) and convective heat transfer coefficient (h).
  • Basic proficiency in unit conversions, particularly temperature from Celsius to Kelvin.
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  • Review the derivation and application of the Grashof number in natural convection scenarios.
  • Study the relationship between Nusselt number and convective heat transfer coefficient.
  • Learn about the significance of temperature conversions in thermodynamic calculations.
  • Investigate the impact of different materials on thermal conductivity (k) in heat transfer problems.
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Homework Statement



What is the heat transfer from a 60W electric light bulb at 127C to the stagnant air in a room at 27C. Approximate the bulb to a 50mm diameter sphere. What percentage of the power is lost by free convection?

Homework Equations



Nu=2 + 0.6(Gr^1/4)(Pr^1/3)
Gr=(g\beta\thetad^3)/\nu^2
Nu=hd/k

The Attempt at a Solution



I started by working out the Grasshof no. with \nu at 27C= 1.568x10^-5m^2/s, \beta=1/T = 1/27, \theta = 100 to be Gr = 1.8x10^7.

Pr = 0.707

therefore Nu = 21.702, using the relationship above.

substitute this into Nu = hd/k and h= 1.139x10^-5kW/m^2K

the area of a sphere = 4(pi)r^2= 7.854x10^-3

so finally Q=hAdT

left me with 8.945x10^-3W
which is 0.014% of the power being lost by free convection.

this appears to be out by a factor of 100, but that could just be a coincidence, meaning that I am totally wrong. Can anyone assist?
 
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Check your calculation of \beta, T should be in Kelvins. And what value of k are you using? Shouldn't you have h\approx\mathrm{Nu}?
 

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