Radiation lamp to heat pizza

[gfd43tg]

Homework Statement

Cheese Board Pizza just hired me to design a radiant pizza warming system. The pizza must be maintained at 250 F until the order is picked up by the delivery driver. The one square foot pizza sits on a ½ inch thick ceramic pan (k = 0.50 BTU /hr ft F) on a counter that is at 85 F. A fan blows air at 70 F over the top of the pizza (the convective film coefficient is estimated to be 2.0 BTU / hr ft2 F). the emissivity (ε) of the pizza is 0.60.

How hot must the radiant heat lamp be to keep the pizza at the desired temperature ?

Homework Equations

The Attempt at a Solution

Hello, I am having some confusion over if the blackbody radiates energy, as well as the pizza radiating energy back at the lamp, which can be approximated as a blackbody.

I originally had 4 terms, a convection and conduction term leaving the pizza, a radiation term leaving the pizza, and a radiation term from the blackbody going to the pizza. However, if I consider the blackbody radiation, I have no equation to find the Q because I don't know the view factor F12 since I don't have a ratio of the smaller side/distance between planes.

 

[gfd43tg]

Solved disregard

 

[berkeman]

Yeah, but now I'm hungry...

 

[collinsmark]

I like pizza.

 

[rezeew]

Hello, as a scientist, I would like to provide a response to the question of how hot the radiant heat lamp must be to keep the pizza at the desired temperature.

Based on the given information, we can use the equation Q = hAΔT to calculate the heat transfer from the pizza to the surroundings (counter and air). Here, Q represents the heat transfer rate, h is the convective film coefficient, A is the surface area of the pizza, and ΔT is the difference between the pizza temperature and the surroundings temperature.

Using this equation, we can calculate the heat transfer rate from the pizza to the surroundings as follows:

Q = (2.0 BTU/hr ft2 F)(1 ft2)(250 F - 85 F) = 330 BTU/hr

Next, we can use the Stefan-Boltzmann Law, which states that the radiant heat emitted by an object is proportional to its emissivity (ε), surface area (A), and temperature (T), to calculate the radiant heat emitted by the pizza.

Q = εσAT^4

Where σ is the Stefan-Boltzmann constant.

Substituting the given values, we get:

330 BTU/hr = (0.60)(5.67 x 10^-8 W/m^2 K^4)(1 ft2)(T^4)

Solving for T, we get T = 369 F.

Therefore, the pizza must be at a temperature of 369 F to emit enough radiant heat to balance out the heat transfer from the surroundings and maintain a temperature of 250 F.

However, since the pizza is on a ceramic pan, there will also be heat transfer through conduction from the pan to the pizza. This can be calculated using the equation Q = kAΔT/d, where k is the thermal conductivity of the ceramic pan, A is the surface area of the pizza, ΔT is the temperature difference between the pan and the pizza, and d is the thickness of the pan.

Assuming a temperature difference of 119 F (369 F - 250 F) and a pan thickness of 0.5 inches, we get:

Q = (0.50 BTU/hr ft F)(1 ft2)(119 F)/(0.5 in) = 142.8 BTU/hr

Therefore, the total heat transfer from the surroundings and the pan to the pizza is 330 BTU/hr + 142