During the simulation several errors of this type occur: "The area of the divider exceeds the glass opening for the window". Here is an example using one of the windows:
FenestrationSurface: detailed, WAN_C11WAN,! - First name Window,! - Surface type WINDOW_6mm! - Name of the building C11WAN,! - Name of building surface , - Exterior Limit Condition Autocalculate,! - View the Factor to Earth Brises_norte,! - Shading control name WA_6_AAJV125, - Frame and splitter name 1 ,! - Multiplier Autocalculate,! - Number of vertices 0.15, 12.6, 5.72, -X, Y, Z 1 (m) 0.15, 12.6, 6.17, X, Y, Z 2, 9.9, 12.6, 6.17, X-Y, Z3, 9.9, 12.6, 5.72; - X, Y, Z 4 {m}
WindowProperty - FrameAndDivider, WA_6_AAJV125,! - First name 0.125,! - frame width {m} 0.0225,! - Exterior projection of frame {m} 0.0225, - Inner projection of the structure {m} 2.27, - Conductance of the weft {W / m2-K} 1 ,! - Conductance ratio of frame edge glass with glass center conductivity 0.9,! - Solar Frame Absorption 0.9,! - Visible absorption of the frame 0.9,! - Thermal Hemispheric Emissivity of the Frame DividedLite,! - Type of divider 0.4,! - Divider width {m} 0,! - Number of horizontal dividers 6, - Number of vertical dividers 0.178, - Divider outer projection {m} 0.49, - Splitter interior projection {m} 2.27, - Divider Conductivity {W / m2-K} 1 ,! - Conductivity Ratio of Splitter-Edge Glass to Conductance of Glass Center 0.9, - Solar Absorption Separator 0.9,! - Visible Absorption Separator 0.9,! - Thermal Hemispheric Emissivity Divisor 0.9, - Out of Reveal Absorption Solar 0.0625,! - Inside depth of the threshold {m} 0.9,! - Indoor solar absorption 0.0625,! - Inside Reveal Depth {m} 0.9; ! - Within Reveal Solar Absorption
ERROR: * Severe * ProcessSurfaceVertices: Divider area exceeds window opening for window WAN_C11WAN * ~~~ * Window surface = [4.38] m2, dividing area = [23.40] m2.
Proof:
The area of the window is: 9.75 x 0.45 = 4.38 .... Ok The area of the dividers: 0.4 x 0.45 x 6 = 1.08 (the vertical dividers are those parallel to the sides)
To get the 23.4 m2 that the program gives, I must multiply the width of the divider for the width of the window instead of the height, that sounds wrong to me...
Area according to the program: 0.4 x 9.75 x 6 = 23.4 ...
So, I must enter the vertical dividers (parallels to the sides) like horizontal dividers (parallels to the top)? These is a program mistake or I am wrong??
