In this model, the gap between the blind and the window is modelled 
by a separate thermal zone. Details below.

The basic model is as for Experiment 3 that was sent to EMPA.

This, in turn, is the based on the model used in test cell calibration.
 
Therefore it has the same thermal bridge with the extended fin as Energyplus etc.
The calibration model needed a 9 days start-up, but only 1 day is set for this
model as there is more contact with outside conditions.

Note: the orientation was corrected from the calibration model to the correct
orientation (wasn't important in calibration model).

This model has the insulation broken into multiple layers so
there is only 5mm insulation against the steel.
------------
Addition of blind zone
Assume same size as window as it covers all window. Assume all direct solar into "blind"
zone hits the window.
Assume curtain material properties - probably not important; assume given
normal incidence optical properties are OK at all angles of incidence.
Air flow network  - assume openings to side and top to allow buoyancy and
wind induced flow - assuem "sheltered" pressure coeffs, but large uncertainty
here - should do a sensitivity analysis.

----------------------
Internal convection:
This model has buoyancy conv coeffs. This may be more appropriate for
conditions where we have solar radiation entering and greater differences
between surface and air temperatures.

------------
Optical properties:
From WIS, inputting solar properties of the two panes of glass (as used in Expt 3)
Angle	0	40	55	70	80
Tx	0.419	0.399	0.357	0.255	0.119
abs1	0.270	0.279	0.287	0.296	0.283
abs2	0.060	0.064	0.065	0.061	0.046

The transmittances are close to those listed for Glad.
The reflectances were different between WIS and Glad:

e.g. for reflectance from front: 0deg: WIS 0.250, Glad 0.285
e.g. for reflectance from back:  0deg: WIS 0.229, Glad 0.296

but better agreement at other angles (within 0.02 except up to 0.04 at high angles).

In this model simply reduce the absorption and transmission by the external blind shading.
This is equivalent to the blind only influencing the solar transmission.
(An earlier model which manually added shading factors into a shading file did
not work because this only shades the direct transmission - it should also shade
the diffuse). The tranmission through the blind is 21.5%. Therefore reduce the
transmission and absorption factors by multiplying them by 0.215. Assume reflectivity
has increased to take remaining solar.
New data:
Angle	0	40	55	70	80
Tx	0.090	0.086	0.077	0.055	0.026
abs1	0.058	0.060	0.062	0.064	0.061
abs2	0.013	0.014	0.014	0.013	0.010
---------------------------------------------
Window thermal properties:
Centre U-value = 1.144 with hi=7.7 and ho=25.0
Therefore AGR = 1/1.144 - 1/7.7 - 1/25 - 2(0.006/1.0) = 0.692 m2K/W

Note WIS gives central U-value as 1.34 (but surface coeffs not clear - conv int 3 and ext 15?)
This would give AGR = 1/1.34 - 1/7.7 - 1/25 - 2(0.006/1.0) = 0.564 m2K/W
Therefore in this model use the lower AGR. results show some small differences - it does
not seem to make a large difference.

***Note 3.11.05: Newer version of WIS (3.0.1) gives same optical properties but U-value =1.29.
Use this if simulations are repeated.


Linear thermal tranmittance = 0.089 W/mK
Create an equivalent surface with the same heat loss
Say an area under window of length 1.17m and height 0.2m = 0.234m2 and thickness 0.05m
Heat loss = 0.089 * 5.18 (perimeter) = 0.461 W/K
For wall area being replaced, R=1/7.7 + 1/25 + 0.02/0.1394 + .13/0.03578 = 3.947
  - i.e. heat loss is 1/3.949 * 0.234  = 0.059 W/K
Therefore replacement construction should have a heat loss of 0.461 + 0.059 = 0.520 W/K
 - required U- value is .520/0.234 = 2.22; required R = 1/2.22 = 0.45
 - required layer resistance = 0.45 - 1/7.7 - 1/25 = 0.28
 - required layer conductivity = 0.05/0.28 = 0.179 W/mK
  (note the layer resistance >> surf resistances, so won't vary too much.
-------------------------------------------
All insulation thermophysical properties updated with values at expt mean temperature
---------------------------------
Shading and insolation:

No shading 
Assume insolation distribution is diffuse in cell.

Assume no shading onto opaque wall
-----------------------------
Climate:
Climate data - make up 6 minute climate data using 6 minutely data.
  - DN and diffuse data used
  - checked results in res against original data - all looks OK.

Add in 6 minutely data for surrounding guard zone air temperatures (with high convective
coeffs so that the air and surface temperatures are the same)
Derive this data from the hourly data - assume constant over the hour (the temperatures
and fan power changes slowly so this should be OK).
Data is from 23 March to 16 April
Copy data for 23 March to 22 March and do a 1-day start-up for simulation.
In file for reading in from Exp4_temps+intgains.csv to tdf (together with control loop numbers
specified in tdf):
    Column
    ------
	1	Cell air temp	Control loop 1
	2	Ext surf temp - floor	Control loop 4
	3	Ext surf temp - ceiling	Control loop 3	
	4	Ext surf temp - west	Control loop 6	
	5	Ext surf temp - north	Control loop 5	
	6	Ext surf temp - east	Control loop 2	
	7	Int fan power in cell		
----------------------------
Results recovery
All straightforward except conv htc.
Extract convective flux, surface and air temperatures. Then calc htc as:
hc = Q / (Area *(Ts-Tair)) - and take absolute value
Areas used:
 southwall 4.831 m2
 glazing   1.661
 westwall 10.917
 northwall 6.608
 eastwall 10.917
 top      13.452
 base     13.452
----------------------------------------

