Modeling and Simulation in Python

Copyright 2018 Allen Downey

License: Creative Commons Attribution 4.0 International

In [1]:
# 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 module
from modsim import *

Inferring thermal resistance and thermal mass of a wall

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

  1. Implement their model and run it with their estimated parameters, to replicate the results, and

  2. Use SciPy's 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.

Loading the data

First I'll load the units we need from Pint.

In [29]:
m = UNITS.meter
K = UNITS.kelvin
W = UNITS.watt
J = UNITS.joule
degC = UNITS.celsius
s = UNITS.second


Read the data.

In [30]:
data = pd.read_csv('data/DataOWall.csv', parse_dates=[0], index_col=0, header=0, skiprows=[1,2])

Q_in Q_out T_int T_ext
2014-10-05 16:30:00 10.994 6.840 16.92 14.68
2014-10-05 16:35:00 10.952 6.012 16.92 14.69
2014-10-05 16:40:00 10.882 7.040 16.93 14.66
2014-10-05 16:45:00 10.798 8.880 16.93 14.59
2014-10-05 16:50:00 10.756 10.491 16.94 14.50

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 [31]:
timestamp_0 = get_first_label(data)

Timestamp('2014-10-05 16:30:00')

Subtracting the first Timestamp yields Timedelta objects:

In [32]:
time_deltas = data.index - timestamp_0

TimedeltaIndex(['0 days 00:00:00', '0 days 00:05:00', '0 days 00:10:00',
                '0 days 00:15:00', '0 days 00:20:00', '0 days 00:25:00',
                '0 days 00:30:00', '0 days 00:35:00', '0 days 00:40:00',
                '0 days 00:45:00',
                '2 days 23:10:00', '2 days 23:15:00', '2 days 23:20:00',
                '2 days 23:25:00', '2 days 23:30:00', '2 days 23:35:00',
                '2 days 23:40:00', '2 days 23:45:00', '2 days 23:50:00',
                '2 days 23:55:00'],
               dtype='timedelta64[ns]', length=864, freq=None)

Then we can convert to seconds and replace the index.

In [33]:
data.index = time_deltas.days * 86400 + time_deltas.seconds

Q_in Q_out T_int T_ext
0 10.994 6.840 16.92 14.68
300 10.952 6.012 16.92 14.69
600 10.882 7.040 16.93 14.66
900 10.798 8.880 16.93 14.59
1200 10.756 10.491 16.94 14.50

The timesteps are all 5 minutes:

In [34]:
np.all(np.diff(data.index) == 300)


Mark the columns of the Dataframe with units.

In [35]:
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 [36]:
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 [37]:
plot(data.T_int, color='C2')
plot(data.T_ext, color='C0')
decorate(xlabel='Time (s)',
         ylabel='Temperature (degC)')

Params and System objects

Here's a Params object with the estimated parameters from the paper.

In [38]:
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)

R1 0.076 kelvin * meter ** 2 / watt
R2 0.272 kelvin * meter ** 2 / watt
R3 0.078 kelvin * meter ** 2 / watt
C1 212900.0 joule / kelvin / meter ** 2
C2 113100.0 joule / kelvin / meter ** 2

I'll pass the Params object make_system, which computes init, packs the parameters into Series objects, and computes the interpolation functions.

In [39]:
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_end=t_end, ts=ts)

Make a System object

In [40]:
system = make_system(params, data)

init T_C1 16.11 degC T_C2 15.27 degC dtype: o...
R 0 0.076 kelvin * meter ** 2 / watt 1 0.2...
C 0 212900.0 joule / kelvin / meter ** 2 1 ...
T_int_func <function interpolate.<locals>.wrapper at 0x7f...
T_ext_func <function interpolate.<locals>.wrapper at 0x7f...
unit_temp degC
t_end 258900 second
ts Int64Index([ 0, 300, 600, 900, ...

Test the interpolation function:

In [41]:
system.T_ext_func(0), system.T_ext_func(150), system.T_ext_func(300)

(14.68, 14.684999999999999, 14.69)

Implementing the model

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.

  • When 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 [42]:
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 [43]:
compute_flux(system.init, 0, system)

0    -10.657894736842135 delta_degC * watt / kelvin...
1    -3.0882352941176463 delta_degC * watt / kelvin...
2    -7.564102564102562 delta_degC * watt / kelvin ...
dtype: object

Here's a slope function that computes derivatives of T_C1 and T_C2

In [44]:
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 [45]:
slopes = slope_func(system.init, system.ts[1], system)

0     3.555499973097458e-05 delta_degC * watt / joule
1    -3.844086774341106e-05 delta_degC * watt / joule
dtype: object

In [47]:
for y, slope in zip(system.init, slopes):
    print(y, slope*s)

16.11 degC 3.555499973097458e-05 delta_degC * second * watt / joule
15.27 degC -3.844086774341106e-05 delta_degC * second * watt / joule

Now let's run the simulation, generating estimates for the time steps in the data.

In [58]:
results, details = run_ode_solver(system, slope_func)

success True
message The solver successfully reached the end of the...

Here's what the results look like.

In [59]:

T_C1 T_C2
0 16.11 degC 15.27 degC
2589 16.19153931824483 degC 15.119152775835897 degC
5178 16.2528855986394 degC 14.860249399364765 degC
7767 16.294553283278788 degC 14.58496288165028 degC
10356 16.317098239731408 degC 14.348725465924332 degC

In [60]:
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 [74]:
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[0].magnitude, 
    return Q_frame
Q_frame = recompute_fluxes(results, system)

Q_in Q_out
0 10.6579 7.5641
2589 9.93106 12.6725
5178 9.30414 17.6391
7767 8.90193 17.1162
10356 8.79081 16.4555

Let's see how the estimates compare to the data.

In [76]:
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 [77]:
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 [91]:
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 [92]:
errors = compute_error(Q_frame, data)

array([-0.33610526, -0.37832695, -0.39254863, ...,  2.26392988,
       -0.7752054 ,  1.99865932])

Here's the root mean squared error.

In [93]:


Estimating parameters

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 [94]:
params = [0.076, 0.272, 0.078, 212900, 113100]

[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 [101]:
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
    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

Testing error_func.

In [102]:
errors = error_func(params, data)

[0.076, 0.272, 0.078, 212900, 113100]
RMSE 2.17995317290043
array([-0.33610526, -0.37832695, -0.39254863, ...,  2.26392988,
       -0.7752054 ,  1.99865932])

Pass error_func to leastsq to see if it can do any better.

In [103]:
best_params, details = leastsq(error_func, params, data)

[7.600e-02 2.720e-01 7.800e-02 2.129e+05 1.131e+05]
RMSE 2.17995317290043
[7.600e-02 2.720e-01 7.800e-02 2.129e+05 1.131e+05]
RMSE 2.17995317290043
[7.600e-02 2.720e-01 7.800e-02 2.129e+05 1.131e+05]
RMSE 2.17995317290043
[7.60000011e-02 2.72000000e-01 7.80000000e-02 2.12900000e+05
RMSE 2.17995317290043
[7.60000000e-02 2.72000004e-01 7.80000000e-02 2.12900000e+05
RMSE 2.17995317290043
[7.60000000e-02 2.72000000e-01 7.80000012e-02 2.12900000e+05
RMSE 2.17995317290043
[7.60000000e-02 2.72000000e-01 7.80000000e-02 2.12900003e+05
RMSE 2.17995317290043
[7.60000000e-02 2.72000000e-01 7.80000000e-02 2.12900000e+05
RMSE 2.17995317290043

In [104]:

fvec [-0.33610526315786693, -0.378326947375756, -0....
nfev 6
fjac [[-0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0...
ipvt [1, 2, 3, 4, 5]
qtf [-0.33610526315786693, -0.378326947375756, -0....
cov_x None
mesg The cosine of the angle between func(x) and an...
ier 4

In [105]:

'The cosine of the angle between func(x) and any column of the\n  Jacobian is at most 0.000000 in absolute value'

The best params are only slightly different from the starting place.

In [117]:

array([7.600e-02, 2.720e-01, 7.800e-02, 2.129e+05, 1.131e+05])

Here's what the results look like with the best_params.

In [108]:
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)


The RMS error is only slightly smaller.

And the results are visually similar.

In [109]:
plot_Q_in(Q_frame, data)

In [110]:
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.


Notes on working with degC.

Usually I construct a Quantity object by multiplying a number and a unit. With degC, that doesn't work; you get

OffsetUnitCalculusError: Ambiguous operation with offset unit (degC).

In [111]:
#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 [112]:
T = Quantity(16.11, degC)

16.11 degC

If you convert to Kelvin, it does the right thing.

In [113]:

289.26 kelvin

When you subtract temperatures, the results is a temperature difference, indicated by the units.

In [114]:
diff = T - 273.15 * K

16.11 delta_degC

If you convert a temperature difference to Kelvin, it does the right thing.

In [115]:

16.11 kelvin

In [ ]: