- #1
OsirisGuy
- 5
- 0
Hey all!
I am thinking of building a box in which the insides can be held a constant temperature. I would like to see if I can determine some parameters for the box.
My first step is to determine heat loss through the box. I'm estimating that the largest difference in heat will be 101°C (Ideally the temperature will be controlled by a PID)
The box will be 1 m3, And will have layers of
| 0.00635 m Plywood | 0.003175 m ABS Plastic | 0.00635 m air space | -
| 0.003175 m ABS Plastic | 0.00635 m Plywood | 0.003175 m ABS Plastic | -
| 0.00635 m air space | 0.003175 m ABS Plastic | 0.00635 m Plywood |
(After these layers, I will assume the surface area is now 0.830103 m2
Thermal Conductivity for Air = 0.024, Plywood = 0.13, ABS = 0.03; and to make things easier I figured Conductance (U) to be 1.0990 and using Fourier's Law [tex]Q = A\frac{DeltaT}{U}[/tex]
I come up with a heat loss of 76.285 Watts. I will just assume that this will occur on every side of the box, so I multiplied by the 6 sides for an estimated heat loss of 457.71 W.
I am now just wondering if my calculations are in the correct direction. Perhaps I am off by a little, but I hope to be heading in the right direction.
The next steps seem to be:
1. Using Peltier Junctions to provide temperature
2. Using a PID to automatically hold temperature at a desired temperature
3. Providing the power and source for the project (hopefully using a "green" method, e.g. solar panels).
Of course, most of this is still theoretical so although Peltiers and solar panels may not be efficient at the moment, I'd still like to incorporate it into the design. Consider this a scrap book or blueprints of sorts in my designing process.
So any takers? Thanks
I am thinking of building a box in which the insides can be held a constant temperature. I would like to see if I can determine some parameters for the box.
My first step is to determine heat loss through the box. I'm estimating that the largest difference in heat will be 101°C (Ideally the temperature will be controlled by a PID)
The box will be 1 m3, And will have layers of
| 0.00635 m Plywood | 0.003175 m ABS Plastic | 0.00635 m air space | -
| 0.003175 m ABS Plastic | 0.00635 m Plywood | 0.003175 m ABS Plastic | -
| 0.00635 m air space | 0.003175 m ABS Plastic | 0.00635 m Plywood |
(After these layers, I will assume the surface area is now 0.830103 m2
Thermal Conductivity for Air = 0.024, Plywood = 0.13, ABS = 0.03; and to make things easier I figured Conductance (U) to be 1.0990 and using Fourier's Law [tex]Q = A\frac{DeltaT}{U}[/tex]
I come up with a heat loss of 76.285 Watts. I will just assume that this will occur on every side of the box, so I multiplied by the 6 sides for an estimated heat loss of 457.71 W.
I am now just wondering if my calculations are in the correct direction. Perhaps I am off by a little, but I hope to be heading in the right direction.
The next steps seem to be:
1. Using Peltier Junctions to provide temperature
2. Using a PID to automatically hold temperature at a desired temperature
3. Providing the power and source for the project (hopefully using a "green" method, e.g. solar panels).
Of course, most of this is still theoretical so although Peltiers and solar panels may not be efficient at the moment, I'd still like to incorporate it into the design. Consider this a scrap book or blueprints of sorts in my designing process.
So any takers? Thanks