# Natural convection question

1. 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?

2. Homework Equations

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

3. 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 im totally wrong. Can anyone assist?

## Answers and Replies

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Mapes
Science Advisor
Homework Helper
Gold Member
Check your calculation of $\beta$, $T$ should be in Kelvins. And what value of $k$ are you using? Shouldn't you have $h\approx\mathrm{Nu}$?