GEP 374b Finite Difference Temperature Prediction Lab
Overview
- previously used box but had to wait for steady state
- here we introduce a method to predict the temperature on the way to steady state
- key idea is to step over small time intervals
Safety Concerns
Background
The energy is released by the resistor and must either leave the box through the walls, or heat the air and walls of the box.
\textrm{Energy In} = \textrm{Energy Conducted} + \textrm{Energy to heat box}
We want to predict temperature but unlike earlier examples (loans, batteries) the inputs and outputs aren’t the quantity. That is, we can calculate energy inputs and outputs but we need to relate those to the temperature.
We can use familiar equations to write this energy balance.
P_{elec} \Delta t = mc\Delta T + P_{conducted} \Delta t
\underbrace{P_{elec}\Delta t }_{\text{energy in}} = \underbrace{mc\Delta T}_{\text{converted energy}} + \underbrace{P_{conducted} \Delta t}_{\text{energy lost to conduction}}
The power lost to conduction is related to the temperature inside the box, which we are predicting.
P_{\text{lost to conduction}} = UA (T_{\text{box}} - T_{\text{outside}})
The temperature rise of the material and air of the box is related to the input energy that wasn’t conducted out that converts to internal energy or temperature.
E_{\text{converted}} = mc \Delta T_{box}
Now, we say \Delta T is the difference between the temperature now and the temperature a little bit later.
P_{elec}\ \Delta t = mc \Delta T_{\text{box now later}} + UA \Delta T_{\text{ box in out}}\Delta t
P_{elec}\ \Delta t = mc (T_{next} - T_{now}) + UA (T_{now} - T_{outside}) \Delta t
Solving, we can write
T_{next} = T_{now} + \frac{\Delta t}{mc} \big[P_{elec} - UA(T_{now} - T_{outside})\big]
T_{next} = T_{now} + \frac{\Delta t}{mc} \big[\overbrace{P_{elec}}^{\text{power in}} - \overbrace{UA(\underbrace{T_{now} - T_{outside}}_{\Delta T})}^{\text{power loss}}\big]
Meaning, the next temperature is equal to the temperature now plus the heat capacity temperature rise from the heater energy minus the heat capacity temperature decrease from the energy that left via conduction.
Thermal Conductance Estimate
The section on thermal modeling has these details.
| material | conductivity |
|---|---|
| wood | 0.12 W/m/K |
| glass | 0.8 W/m/K |
| foam | 0.02 W/m/K |
| air film | R-value |
|---|---|
| interior | 0.12 m^2K/W |
| exterior | 0.03 m^2K/W |
Heat Capacity Estimate
| material | property | value |
|---|---|---|
| air | density | 1.2 kg per m^3 |
| MDF | density | 0.75 g per cm^3 |
| air | specific heat capacity | 1.0 J/g/K |
| MDF | specific heat capacity | 1.7 J/g/K |
| foam | specific heat capacity | 1.0 J/g/K |
| foam | density | TBD |
Lab Planning Day
- Review conduction and heat capacity
- Draw diagram of energy balance
- Write formula for energy balance
- Use energy balance formula to solve for next temperature
- Think about spreadsheet strategy to implement formula iteratively
Data Collection Day
Data Analysis Day
We have created a computer model of the temperature of the box. This model has the heating power, box dimensions, and material parameters as inputs.
We will compare the measured data to the modeled data by plotting both on a graph. We will adjust the UA and MC parameters until there is a good fit between the modeled and measured data.
Exercises
1 Euler Method Reflection
- What is clearest about this Euler method?
- What is still unclear about this method?
- What do you predict the temperature curve to look like based on our formula?
2 Box Observations
- Do your observations match your predictions about the shape of the temperature versus time curve?
3 Report Document
Turn in a short report with the following elements
- A typed table of data
- A graph of the data (if you collected two current values)
- A clear explanation of your calculation method
- Reflect on differences between your prediction and your observation