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

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The discussion focuses on calculating the heat transfer from a 60W bulb at 127°C to stagnant air at 27°C, approximating the bulb as a 50mm diameter sphere. The user calculated the Grashof number and Nusselt number, leading to a heat transfer coefficient of approximately 1.139x10^-5 kW/m²K. The final calculation indicated a heat loss of 8.945x10^-3W, which is only 0.014% of the bulb's power, raising concerns about potential errors in the calculations. Participants suggest verifying the temperature in Kelvin and the value of thermal conductivity used, as well as ensuring the heat transfer coefficient aligns with the Nusselt number. The thread emphasizes the importance of accurate unit conversions and parameter values in heat transfer calculations.
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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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