In :# Configure Jupyter so figures appear in the notebook %matplotlib inline # Configure Jupyter to display the assigned value after an assignment %config InteractiveShell.ast_node_interactivity='last_expr_or_assign' # import functions from the modsim.py module from modsim import *
This case study is based on Gori, Marincioni, Biddulph, Elwell, "Inferring the thermal resistance and effective thermal mass distribution of a wall from in situ measurements to characterise heat transfer at both the interior and exterior surfaces", Energy and Buildings, Volume 135, 15 January 2017, Pages 398-409, which I downloaded here.
The authors put their paper under a Creative Commons license, and make their data available here. I thank them for their commitment to open, reproducible science, which made this case study possible.
The goal of their paper is to model the thermal behavior of a wall as a step toward understanding the "performance gap between the expected energy use of buildings and their measured energy use". The wall they study is identified as the exterior wall of an office building in central London, not unlike this one.
The following figure shows the scenario and their model:
On the interior and exterior surfaces of the wall, they measure temperature and heat flux over a period of three days. They model the wall using two thermal masses connected to the surfaces, and to each other, by thermal resistors.
The primary methodology of the paper is a Bayesian method for inferring the parameters of the system (two thermal masses and three thermal resistances).
The primary result is a comparison of two models: the one shown here with two thermal masses, and a simpler model with only one thermal mass. They find that the two-mass model is able to reproduce the measured fluxes substantially better.
Tempting as it is, I will not replicate their method for estimating the parameters. Rather, I will
Implement their model and run it with their estimated parameters, to replicate the results, and
leastsq to see if I can find parameters that yield lower errors (root mean square).
leastsq is a wrapper for some venerable FORTRAN code that runs "a modification of the Levenberg-Marquardt algorithm", which is one of my favorites (really).
Implementing their model in the ModSimPy framework turns out to be straightforward. The simulations run fast enough even when we carry units through the computation. And the results are visually similar to the ones in the original paper.
I find that
leastsq is not able to find parameters that yield substantially better results, which suggest that the estimates in the paper are at least locally optimal.
In :m = UNITS.meter K = UNITS.kelvin W = UNITS.watt J = UNITS.joule degC = UNITS.celsius s = UNITS.second
Read the data.
In :data = pd.read_csv('data/DataOWall.csv', parse_dates=, index_col=0, header=0, skiprows=[1,2]) data.head()
The index contains Pandas
Timestamp objects, which is good for dealing with real-world dates and times, but not as good for running the simulations, so I'm going to convert to seconds.
In :timestamp_0 = get_first_label(data)
Subtracting the first
In :time_deltas = data.index - timestamp_0
Then we can convert to seconds and replace the index.
In :data.index = time_deltas.days * 86400 + time_deltas.seconds data.head()
The timesteps are all 5 minutes:
In :np.all(np.diff(data.index) == 300)
Mark the columns of the
Dataframe with units.
In :data.Q_in.units = W / m**2 data.Q_out.units = W / m**2 data.T_int.units = degC data.T_ext.units = degC
Plot the measured fluxes.
In :plot(data.Q_in, color='C2') plot(data.Q_out, color='C0') decorate(xlabel='Time (s)', ylabel='Heat flux (W/$m^2$)')
Plot the measured temperatures.
In :plot(data.T_int, color='C2') plot(data.T_ext, color='C0') decorate(xlabel='Time (s)', ylabel='Temperature (degC)')
In :params = Params( R1 = 0.076 * m**2 * K / W, R2 = 0.272 * m**2 * K / W, R3 = 0.078 * m**2 * K / W, C1 = 212900 * J / m**2 / K, C2 = 113100 * J / m**2 / K)
I'll pass the
make_system, which computes
init, packs the parameters into
Series objects, and computes the interpolation functions.
In :def make_system(params, data): """Makes a System object for the given conditions. params: Params object returns: System object """ R1, R2, R3, C1, C2 = params init = State(T_C1 = Quantity(16.11, degC), T_C2 = Quantity(15.27, degC)) ts = data.index t_end = ts[-1] * s return System(init=init, R=Series([R1, R2, R3]), C=Series([C1, C2]), T_int_func=interpolate(data.T_int), T_ext_func=interpolate(data.T_ext), unit_temp=degC, t_end=t_end, ts=ts)
In :system = make_system(params, data)
Test the interpolation function:
In :system.T_ext_func(0), system.T_ext_func(150), system.T_ext_func(300)
Next we need a slope function that takes instantaneous values of the two internal temperatures and computes their time rates of change.
The slope function gets called two ways.
When we call it directly,
state is a
State object and the values it contains have units.
run_ode_solver calls it,
state is an array and the values it contains don't have units.
In the second case, we have to apply the units before attempting the computation.
require_units applies units if necessary:
The following function computes the fluxes between the four zones.
In :def compute_flux(state, t, system): """Compute the fluxes between the walls surfaces and the internal masses. state: State with T_C1 and T_C2 t: time in seconds system: System with interpolated measurements and the R Series returns: Series of fluxes """ # unpack the temperatures T_C1, T_C2 = state # compute a series of temperatures from inside out T_int = system.T_int_func(t) T_ext = system.T_ext_func(t) T = [require_units(T_int, system.unit_temp), require_units(T_C1, system.unit_temp), require_units(T_C2, system.unit_temp), require_units(T_ext, system.unit_temp)] # compute differences of adjacent temperatures T_diff = np.diff(T) # compute fluxes between adjacent compartments Q = T_diff / system.R return Q
We can test it like this.
In :compute_flux(system.init, 0, system)
Here's a slope function that computes derivatives of
In :def slope_func(state, t, system): """Compute derivatives of the state. state: position, velocity t: time system: System object returns: derivatives of y and v """ Q = compute_flux(state, t, system) # compute the net flux in each node Q_diff = np.diff(Q) # compute the rate of change of temperature dQdt = Q_diff / system.C return dQdt
Test the slope function with the initial conditions.
In :slopes = slope_func(system.init, system.ts, system)
In :for y, slope in zip(system.init, slopes): print(y, slope*s)
Now let's run the simulation, generating estimates for the time steps in the data.
In :results, details = run_ode_solver(system, slope_func) details
Here's what the results look like.
In :def plot_results(results, data): plot(data.T_int, color='C2') plot(results.T_C1, color='C3') plot(results.T_C2, color='C1') plot(data.T_ext, color='C0') decorate(xlabel='Time (s)', ylabel='Temperature (degC)') plot_results(results, data)
These results are similar to what's in the paper:
To get the estimated fluxes, we have to go through the results and basically do the flux calculation again.
In :def recompute_fluxes(results, system): """Compute fluxes between wall surfaces and internal masses. results: Timeframe with T_C1 and T_C2 system: System object returns: Timeframe with Q_in and Q_out """ Q_frame = TimeFrame(index=results.index, columns=['Q_in', 'Q_out']) for t, row in results.iterrows(): Q = compute_flux(row, t, system) Q_frame.row[t] = (-Q.magnitude, -Q.magnitude) return Q_frame Q_frame = recompute_fluxes(results, system) Q_frame.head()
Let's see how the estimates compare to the data.
In :def plot_Q_in(frame, data): plot(frame.Q_in, color='gray') plot(data.Q_in, color='C2') decorate(xlabel='Time (s)', ylabel='Heat flux (W/$m^2$)') plot_Q_in(Q_frame, data)
In :def plot_Q_out(frame, data): plot(frame.Q_out, color='gray') plot(data.Q_out, color='C0') decorate(xlabel='Time (s)', ylabel='Heat flux (W/$m^2$)') plot_Q_out(Q_frame, data)
These results are also similar to what's in the paper (the bottom row):
In :def compute_error(frame, data): model_Q_in = interpolate(Q_frame.Q_in)(data.index) error_Q_in = model_Q_in - data.Q_in model_Q_out = interpolate(Q_frame.Q_out)(data.index) error_Q_out = model_Q_out - data.Q_out return np.hstack([error_Q_in, error_Q_out])
In :errors = compute_error(Q_frame, data)
Here's the root mean squared error.
Now let's see if we can do any better than the parameters in the paper.
In order to work with
leastsq, we need a version of
params with no units.
In :params = [0.076, 0.272, 0.078, 212900, 113100]
Here's an error function that takes a hypothetical set of parameters, runs the simulation, and returns an array of errors.
In :def error_func(params, data): """Run a simulation and return an array of errors. params: Params object or array data: DataFrame returns: array of float """ print(params) system = make_system(params, data) system = remove_units(system) results, details = run_ode_solver(system, slope_func) Q_frame = recompute_fluxes(results, system) errors = compute_error(Q_frame, data) print('RMSE', np.sqrt(np.mean(errors**2))) return errors
In :errors = error_func(params, data)
leastsq to see if it can do any better.
In :best_params, details = leastsq(error_func, params, data)
The best params are only slightly different from the starting place.
Here's what the results look like with the
In :system = make_system(best_params, data) system = remove_units(system) results, details = run_ode_solver(system, slope_func, t_eval=system.ts) Q_frame = recompute_fluxes(results, system) errors = compute_error(Q_frame, data) print(np.sqrt(np.mean(errors**2)))
The RMS error is only slightly smaller.
And the results are visually similar.
In :plot_Q_in(Q_frame, data)
In :plot_Q_out(Q_frame, data)
Exercise: Try starting the model with a different set of parameters and see if it moves toward the parameters in the paper.
I found that no matter where I start,
leastsq doesn't move far, which suggests that it is not able to optimize the parameters effectively.
In :#16.11 * C
The problem is that it doesn't know whether you want a temperature measurement or a temperature difference.
You can create a temperature measurement like this.
In :T = Quantity(16.11, degC)
If you convert to Kelvin, it does the right thing.
When you subtract temperatures, the results is a temperature difference, indicated by the units.
In :diff = T - 273.15 * K
If you convert a temperature difference to Kelvin, it does the right thing.
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