Natural convection question

In summary, the heat transfer from a 60W electric light bulb at 127C to the stagnant air in a room at 27C can be approximated by considering the bulb as a 50mm diameter sphere. Using the equations for Grasshof number and Nusselt number, we can calculate the heat transfer coefficient to be approximately 1.139x10^-5kW/m^2K. This results in a heat transfer of 8.945x10^-3W, which is approximately 0.014% of the power being lost by free convection. However, there may be errors in the calculation of the Grasshof number and the heat transfer coefficient.
  • #1
bill nye scienceguy!
127
0

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[tex]\beta[/tex][tex]\theta[/tex]d^3)/[tex]\nu[/tex]^2
Nu=hd/k

The Attempt at a Solution



I started by working out the Grasshof no. with [tex]\nu[/tex] at 27C= 1.568x10^-5m^2/s, [tex]\beta[/tex]=1/T = 1/27, [tex]\theta[/tex] = 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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  • #2
Check your calculation of [itex]\beta[/itex], [itex]T[/itex] should be in Kelvins. And what value of [itex]k[/itex] are you using? Shouldn't you have [itex]h\approx\mathrm{Nu}[/itex]?
 
  • #3


Your calculations are correct, however, the percentage of power lost by free convection is actually 1.4% instead of 0.014%. This is because the power lost by free convection is calculated by dividing the heat transfer rate by the power input, which in this case is 60W. So, the percentage would be (8.945x10^-3W/60W) x 100% = 1.4%. This means that 1.4% of the power input is lost by free convection.
 

1. What is natural convection?

Natural convection is a type of heat transfer that occurs when a fluid (liquid or gas) is heated, causing it to expand and become less dense. This warmer, less dense fluid rises, while cooler, denser fluid sinks, creating a natural circulation pattern.

2. How is natural convection different from forced convection?

Natural convection is driven by buoyancy forces, while forced convection is driven by external forces such as pumps or fans. In natural convection, there is no external source of energy, while forced convection requires an external source to create the flow.

3. What factors affect natural convection?

The factors that affect natural convection include the temperature difference between the heated and cooled surfaces, the properties of the fluid (such as viscosity and density), the geometry and orientation of the heated surface, and the presence of obstacles or boundaries that may disrupt the flow.

4. How is natural convection used in everyday life?

Natural convection plays a role in many everyday processes, such as boiling water, heating a room with a radiator, and even the circulation of air in a room. It is also important in industrial processes, such as in heat exchangers and cooling systems.

5. Why is natural convection important in thermal engineering?

Natural convection is important in thermal engineering because it is a major mechanism for heat transfer in many systems. Understanding and controlling natural convection is essential for designing efficient and safe thermal systems, such as in power plants, electronics cooling, and building design.

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