Window Heat Balance Calculation[LINK]

The Glazing Heat Balance Equations[LINK]

The window glass face temperatures are determined by solving the heat balance equations on each face every time step. For a window with N glass layers there are 2N faces and therefore 2N equations to solve. Figure 1 shows the variables used for double glazing (N = 2).

The following assumptions are made in deriving the heat balance equations:

1) The glass layers are thin enough (a few millimeters) that heat storage in the glass can be neglected; therefore, there are no heat capacity terms in the equations.

2) The heat flow is perpendicular to the glass faces and is one dimensional. See “Edge of Glass Corrections,” below, for adjustments to the gap conduction in multi-pane glazing to account for 2-D conduction effects across the pane separators at the boundaries of the glazing.

3) The glass layers are opaque to IR. This is true for most glass products. For thin plastic suspended films this is not a good assumption, so the heat balance equations would have to be modified to handle this case.

4) The glass faces are isothermal. This is generally a good assumption since glass conductivity is very high.

5) The short wave radiation absorbed in a glass layer can be apportioned equally to the two faces of the layer.

The four equations for double-glazing are as follows. The equations for single glazing (N = 1) and for N = 3 and N = 4 are analogous and are not shown.

$$E_o{\epsilon }_1-{\epsilon }_1\sigma {\theta }_1^4+k_1\left({\theta }_2-{\theta }_1\right)+h_o\left(T_o-{\theta }_1\right)+S_1=0$$

$$k_1\left({\theta }_1-{\theta }_2\right)+h_1\left({\theta }_3-{\theta }_2\right)+\sigma \frac{{\epsilon }_2{\epsilon }_3}{1-\left(1-{\epsilon }_2\right)\left(1-{\epsilon }_3\right)}\left({\theta }_3^4-{\theta }_2^4\right)+S_2=0$$

$$h_1({\theta }_2-{\theta }_3)+k_2({\theta }_4-{\theta }_3)+\sigma \frac{{\epsilon }_2{\epsilon }_3}{1-\left(1-{\epsilon }_2\right)\left(1-{\epsilon }_3\right)}\left({\theta }_2^4-{\theta }_3^4\right)+S_3=0$$

$$E_i{\epsilon }_4-{\epsilon }_4\sigma {\theta }_4^4+k_2\left({\theta }_3-{\theta }_4\right)+h_i\left(T_i-{\theta }_4\right)+S_4=0$$

Absorbed Radiation[LINK]

S${}_i$ in Equations [eq:GlazingFace1HeatBalEquation] to [eq:GlazingFace4HeatBalEquation] is the radiation (short-wave and long-wave from zone lights and equipment) absorbed on the i${}^{th}$ face. Short-wave radiation (solar and short-wave from lights) is assumed to be absorbed uniformly along a glass layer, so for the purposes of the heat balance calculation it is split equally between the two faces of a layer. Glass layers are assumed to be opaque to IR so that the thermal radiation from lights and equipment is assigned only to the inside (room-side) face of the inside glass layer. For N glass layers S${}_i$ is given by

$$S_{2j-1}=S_{2j}=\frac{1}{2}\left(I_{bm}^{ext}\cos \varphi A_j^f\left(\varphi \right)+I_{dif}^{ext}{A}_j^{f,dif}+I_{sw}^{\mathrm{int}}{A}_j^{b,dif}\right),j=1toN$$

$$S_{2N}=S_{2N}+{\epsilon }_{2N}{I}_{lw}^{\mathrm{int}}$$

Here

$I_{bm}^{ext}$ = exterior beam normal solar irradiance

$I_{dif}^{ext}$ = exterior diffuse solar incident on glazing from outside

$I_{sw}^{\mathrm{int}}$ = interior short-wave radiation (from lights and from reflected diffuse solar) incident on glazing from inside

$I_{lw}^{\mathrm{int}}$ = long-wave radiation from lights and equipment incident on glazing from inside

${\epsilon }_{2N}$ = emissivity (thermal absorptance) of the room-side face of the inside glass layer

Room-Side Convection[LINK]

The correlation for room-side convection coefficient, $h_i$ , is from ISO 15099 section 8.3.2.2. (Prior to EnergyPlus version 3.1, the value for $h_i$ was modeled using the “Detailed” algorithm for opaque surface heat transfer, e.g. for a vertical surface $h_i=1.31{\left|\Delta T\right|}^{\text{1}/13\text{3}}$ ; see section Detailed Natural Convection Algorithm). The ISO 15099 correlation is for still room air and is determined in terms of the Nusselt number, $Nu$, is:

$$h_i=Nu\left(\frac{\lambda }{H}\right)$$

where

$\lambda$ is the thermal conductivity of air, and

$H$ is the height of the window.

The Rayleigh number based on height, $R{a}_H$ , is calculated using:

$$R{a}_H=\frac{{\rho }^2{H}^{3}g{c}_p\left|T_{surf,i}-T_{air}\right|}{T_{m,f}\mu \lambda }$$

where

$\rho$ is the density of air

$g$ is the acceleration due to gravity,

$c_p$ is the specific heat of air,

$\mu$ is the dynamic viscosity of air, and

$T_{m,f}$ is the mean film temperature in Kelvin given by,

$$T_{m,f}=T_{air}+\frac{1}{4}\left(T_{surf,i}-T_{air}\right)$$

There are four cases for the Nusselt correlation that vary by the tilt angle in degrees, $\gamma$, and are based on heating conditions. For cooling conditions (where $T_{surf,i}>T_{air}$), the tilt angle is complemented so that $\gamma =180-\gamma$.

Case A. $0^{\circ }\le \gamma <{15}^{\circ }$

$$Nu=0.13R{a}_H^{\text{1}/13\text{3}}$$

Case B. ${15}^{\circ }\le \gamma \le {90}^{\circ }$

$$R{a}_{cv}=2.5\times {10}^5{\left(\frac{e^{0.72\gamma }}{\sin \lambda }\right)}^{\text{1}/15\text{5}}$$

$$Nu=0.56{\left(R{a}_H\sin \gamma \right)}^{\text{1}/14\text{4}};forR{a}_H\le R{a}_{CV}$$

$$Nu=0.13\left(R{a}_H^{\text{1}/13\text{3}}-R{a}_{CV}^{\text{1}/13\text{3}}\right)+0.56{\left(R{a}_{CV}\sin \gamma \right)}^{\text{1}/14\text{4}};R{a}_H>R{a}_{CV}$$

Case C. ${90}^{\circ }<\gamma \le {179}^{\circ }$

$$Nu=0.56{\left(R{a}_H\sin \gamma \right)}^{\text{1}/14\text{4}};{10}^5\le R{a}_H\sin \gamma <{10}^{11}$$

Case D. ${179}^{\circ }<\gamma \le {180}^{\circ }$

$$Nu=0.58R{a}_H^{\text{1}/15\text{5}};R{a}_H\le {10}^{11}$$

The material properties are evaluated at the mean film temperature. Standard EnergyPlus pyschrometric functions are used for $\rho$ and $c_p$. Thermal conductivity is calculated using:

$$\lambda =2.873\times {10}^{-3}+7.76\times {10}^{-5}{T}_{m,f}$$

Kinematic viscosity is calculated using:

$$\mu =3.723\times {10}^{-6}+4.94\times {10}^{-8}{T}_{m,f}$$

This correlation depends on the surface temperature of the room-side glazing surface and is therefore included inside the window heat balance interation loop.

Solving the Glazing Heat Balance Equations[LINK]

The equations are solved as follows:

1) Linearize the equations by defining $h_{r,i}={\epsilon }_i\sigma {\theta }_i^3$ . For example, Equation [eq:GlazingFace1HeatBalEquation] becomes

$$E_o{\epsilon }_1-h_{r,1}{\theta }_1+k_1\left({\theta }_2-{\theta }_1\right)+h_o\left(T_o-{\theta }_1\right)+S_1=0$$

2) Write the equations in the matrix form $A\theta =B$

3) Use previous time step’s values of ${\theta }_i$ as initial values for the current time step. For the first time step of a design day or run period the initial values are estimated by treating the layers as a simple RC network.

4) Save the ${\theta }_i$ for use in the next iteration: ${\theta }_{prev,i}={\theta }_i$

5) Using ${\theta }_{2N}$, reevaluate the room-side face surface convection coefficient $h_i$

6) Using the ${\theta }_i$ to evaluate the radiative conductances $h_{r,i}$

7) Find the solution $\theta =A^{-1}B$ by LU decomposition

8) Perform relaxation on the the new ${\theta }_i$ : ${\theta }_i\to \left({\theta }_i+{\theta }_{prev,i}\right)/2$

9) Go to step 4

Repeat steps 4 to 9 until the difference, $\Delta {\theta }_i$, between values of the ${\theta }_i$ in successive iterations is less than some tolerance value. Currently, the test is:

$$\frac{1}{2N}\sum _{i=1}^{2N}|\Delta {\theta }_i|<0.02K$$

If this test does not pass after 100 iterations, the tolerance is increased to 0.2K. If the test still fails the program stops and an error message is issued.

The value of the inside face temperature, ${\theta }_{2N}$, determined in this way participates in the zone heat balance solution (see Outdoor/Exterior Convection) and thermal comfort calculation (see Occupant Thermal Comfort).

Edge-Of-Glass Effects[LINK]

Because of thermal bridging across the spacer separating the glass layers in multi-pane glazing, the conductance of the glazing near the frame and divider, where the spacers are located, is higher than it is in the center of the glass. The area-weighted net conductance (without inside and outside air films) of the glazing in this case can be written

$$\overline{h}=\left(A_{cg}{h}_{cg}+A_{fe}{h}_{fe}+A_{de}{h}_{de}\right)/A_{tot}$$

where

h${}_{cg}$ = conductance of center-of-glass region (without air films)

h${}_{fe}$ = conductance of frame edge region (without air films)

h${}_{de}$ = conductance of divider edge region (without air films)

A${}_{cg}$ = area of center-of-glass region

A${}_{fe}$ = area of frame edge region

A${}_{de}$ = area of divider edge region

A${}_{tot}$ = total glazing area = $A_{cg}+A_{fe}+A_{de}$

The different regions are shown in Figure 2:

Equation [eq:AreaWeightedNetConductance] can be rewritten as:

$$\overline{h}=h_{cg}\left({\eta }_{cg}+{\alpha }_{fe}{\eta }_{fe}+{\alpha }_{de}{\eta }_{de}\right)$$

where

$${\eta }_{cg}=A_{cg}/A_{tot}$$

$${\eta }_{fe}=A_{fe}/A_{tot}$$

$${\eta }_{de}=A_{de}/A_{tot}$$

$${\alpha }_{fe}=h_{fe}/h_{cg}$$

$${\alpha }_{de}=h_{de}/h_{cg}$$

The conductance ratios ${\alpha }_{fe}$ and ${\alpha }_{de}$ are user inputs obtained from Window 5. They depend on the glazing construction as well as the spacer type, gap width, and frame and divider type.

In the EnergyPlus glazing heat balance calculation effective gap convective conductances are used to account for the edge-of-glass effects. These effective conductances are determined as follows for the case with two gaps (triple glazing). The approach for other numbers of gaps is analogous.

Neglecting the very small resistance of the glass layers, the center-of-glass conductance (without inside and outside air films) can be written as:

$$h_{cg}={\left({\left(h_{r,1}+h_{c,1}\right)}^{-1}+{\left(h_{r,2}+h_{c,2}\right)}^{-1}\right)}^{-1}$$

where

h${}_{c,k}$ = convective conductance of the k${}^h$ gap

h${}_{r,k}$ = radiative conductance of the k${}^h$ gap or:

$$h_{r,k}=\frac{1}{2}\sigma \frac{{\epsilon }_i{\epsilon }_j}{1-\left(1-{\epsilon }_i\right)\left(1-{\epsilon }_j\right)}{\left({\theta }_i+{\theta }_j\right)}^3$$

$\epsilon$${}_i$, $\epsilon$${}_j$ = emissivity of the faces bounding the gap

$\theta$${}_i$, $\theta$${}_j$ = temperature of faces bounding the gap (K)

Equation [eq:HbarConductanceEquation] then becomes:

$$\overline{h}=\left({\eta }_{cg}+{\alpha }_{fe}{\eta }_{fe}+{\alpha }_{de}{\eta }_{de}\right){\left({\left(h_{r,1}+h_{c,1}\right)}^{-1}+{\left(h_{r,2}+h_{c,2}\right)}^{-1}\right)}^{-1}$$

We can also write $\overline{h}$ in terms of effective convective conductances of the gaps as:

$$\overline{h}={\left({\left(h_{r,1}+{\overline{h}}_{c,1}\right)}^{-1}+{\left(h_{r,2}+{\overline{h}}_{c,2}\right)}^{-1}\right)}^{-1}$$

Comparing Equations [eq:HbarwithetaalphaEquation] and [eq:HbarwithhrandhcEquation], we obtain:

$$h_{r,k}+{\overline{h}}_{c,k}=\left({\eta }_{cg}+{\alpha }_{fe}{\eta }_{fe}+{\alpha }_{de}{\eta }_{de}\right)\left(h_{r,k}+h_{c,k}\right)$$

Using ${\eta }_{cg}=1-{\eta }_{fe}-{\eta }_{de}$ gives:

$${\overline{h}}_{c,k}=h_{r,k}\left({\eta }_{fe}\left({\alpha }_{fe}-1\right)+{\eta }_{de}\left({\alpha }_{de}-1\right)\right)+h_{c,k}\left(1+{\eta }_{fe}\left({\alpha }_{fe}-1\right)+{\eta }_{de}\left({\alpha }_{de}-1\right)\right)$$

This is the expression used by EnergyPlus for the gap convective conductance when a frame or divider is present.

Apportioning of Absorbed Short-Wave Radiation in Shading Device Layers[LINK]

If a shading device has a non-zero short-wave transmittance then absorption takes place throughout the shading device layer. The following algorithm is used to apportion the absorbed short-wave radiation to the two faces of the layer. Here f${}_1$ is the fraction assigned to the face closest to the incident radiation and f${}_2$ is the fraction assigned to the face furthest from the incident radiation.

$$f_1=1,f_2=0if{\tau }_{sh}=0$$

Otherwise:

$$\begin{array}{cc}{f}_1=0,f_2=0& \text{if~}{\alpha }_{sh}\le 0.01\\ f_1=1,f_2=0& \text{if~}{\alpha }_{sh}>0.999\\ \begin{array}{c}{f}_1=\frac{1-e^{0.5\ln (1-{\alpha }_{sh})}}{{\alpha }_{sh}}\\ f_2=1-f_1\end{array}\}& \text{if~}0.01<{\alpha }_{sh}\le 0.999\end{array}$$

Window Frame and Divider Calculation[LINK]

For the zone heat balance calculation the inside surface temperature of the frame and that of the divider are needed. These temperatures are determined by solving the heat balance equations on the inside and outside surfaces of the frame and divider.

Frame Temperature Calculation[LINK]

Figure 3 shows a cross section through a window showing frame and divider. The outside and inside frame and divider surfaces are assumed to be isothermal. The frame and divider profiles are approximated as rectangular since this simplifies calculating heat gains and losses (see “Error Due to Assuming a Rectangular Profile,” below).

Frame Outside Surface Heat Balance[LINK]

The outside surface heat balance equation is:

$$Q_{ExtIR,abs}-Q_{IR,emitted}+Q_{conv}+Q_{cond}+Q_{abs}=0$$

where

$Q_{ExtIR,abs}$ = IR from the exterior surround (sky and ground) absorbed by outside frame surfaces

$Q_{IR,emitted}$ = IR emitted by outside frame surfaces

$Q_{conv}$ = convection from outside air to outside frame surfaces

$Q_{cond}$ = conduction through frame from inside frame surfaces to outside frame surfaces

$Q_{abs}$ = solar radiation (from sun, sky and ground) plus IR from outside window surface absorbed by outside frame surfaces (see “Calculation of Absorbed Solar Radiation,” below).

The first term can be written as the sum of the exterior IR absorbed by the outside face of the frame and the exterior IR absorbed by the frame’s outside projection surfaces.

$$Q_{ExtIR,abs}={\epsilon }_1{E}_o{A}_f{F}_f+{\epsilon }_1{E}_o{A}_{p1}{F}_{p1}$$

where $\epsilon$${}_1$ is the outside surface emissivity.

The exterior IR incident on the plane of the window, E${}_o$, is the sum of the IR from the sky, ground and obstructions. For the purposes of the frame heat balance calculation it is assumed to be isotropic. For isotropic incident IR, F${}_f$ = 1.0 and F${}_{p1}$ = 0.5, which gives:

$$Q_{ExtIR,abs}={\epsilon }_1{E}_o(A_f+\frac{1}{2}{A}_{p1})$$

The IR emitted by the outside frame surfaces is:

$$Q_{ExtIR,emitted}={\epsilon }_1\sigma (A_f+A_{p1}){\theta }_1^4$$

The convective heat flow from the outside air to the outside frame surfaces is:

$$Q_{conv}=h_{o,c}(A_f+A_{p1})(T_o-{\theta }_1)$$

The conduction through the frame from inside to outside is:

$$Q_{cond}=k{A}_f({\theta }_2-{\theta }_1)$$

Note that A${}_f$ is used here since the conductance, k, is, by definition, per unit area of frame projected onto the plane of the window.

Adding these expressions for the Q terms and dividing by A${}_f$ gives:

$$E_0{\epsilon }_1\left(1+\frac{1}{2}{\eta }_1\right)-{\epsilon }_1\left(1+{\eta }_1\right){\theta }_1^4+h_{o,c}\left(1+{\eta }_1\right)\left(T_0-{\theta }_1\right)+k\left({\theta }_2-{\theta }_1\right)+S_1=0$$

where S${}_1$ = Q${}_{abs}$/A${}_f$ and

$${\eta }_1=\frac{A_{p1}}{A_f}=\left(\frac{p_{f,1}}{w_f}\right)\frac{H+W-(N_h+N_v)w_d}{H+W+2w_f}$$

We linearize Equation [eq:E0epsilon1Equation] as follows. Write the first two terms as:

$${\epsilon }_1(1+{\eta }_1)\left[E_o\right(1+\frac{1}{2}{\eta }_1\left)/\right(1+{\eta }_1)-{\theta }_1^4]$$

and define a radiative temperature:

$$T_{o,r}=\left[E_o\right(1+\frac{1}{2}{\eta }_1\left)/\right(1+{\eta }_1){]}^{1/4}$$

This gives:

$${\epsilon }_1(1+{\eta }_1)[T_{o,r}^4-{\theta }_1^4]$$

which, within a few percent, equals:

$${\epsilon }_1(1+{\eta }_1)\frac{{(T_{o,r}+{\theta }_1)}^3}{2}(T_{o,r}-{\theta }_1)$$

Defining an outside surface radiative conductance as follows:

$$h_{o,r}={\epsilon }_1(1+{\eta }_1)\frac{{(T_{o,r}+{\theta }_1)}^3}{2}$$

then gives:

$$h_{o,r}(T_{o,r}-{\theta }_1)$$

The final outside surface heat balance equation in linearized form is then:

$$h_{o,r}(T_{o,r}-{\theta }_1)+h_{o,c}(1+{\eta }_1)(T_o-{\theta }_1)+k({\theta }_2-{\theta }_1)+S_1=0$$

Frame Inside Surface Heat Balance[LINK]

A similar approach can be used to obtain the following linearized inside surface heat balance equation:

$$h_{i,r}(T_{i,r}-{\theta }_2)+h_{i,c}(1+{\eta }_2)(T_i-{\theta }_2)+k({\theta }_1-{\theta }_2)+S_2=0$$

where

$$T_{i,r}=\left[E_i\right(1+\frac{1}{2}{\eta }_2\left)/\right(1+{\eta }_2){]}^{1/4}$$

$${\eta }_2=\frac{A_{p2}}{A_f}=\left(\frac{p_{f,2}}{w_f}\right)\frac{H+W-(N_h+N_v)w_d}{H+W+2w_f}$$

and E${}_i$ is the interior IR irradiance incident on the plane of the window.

Solving Equations [eq:horEquation] and [eq:hirEquation] simultaneously gives:

$${\theta }_2=\frac{D+CA}{1-CB}$$

with

$$A=\frac{h_{o,r}{T}_{o,r}+h_{o,c}{T}_o+S_1}{h_{o,r}+k+h_{o,c}}$$

$$B=\frac{k}{h_{o,r}+k+h_{o,c}}$$

$$C=\frac{k}{h_{i,r}+k+h_{i,c}}$$

$$D=\frac{h_{i,r}{T}_{i,r}+h_{i,c}{T}_i+S_2}{h_{i,r}+k+h_{i,c}}$$

Calculation of Solar Radiation Absorbed by Frame[LINK]

The frame outside face and outside projections and inside projections absorb beam solar radiation (if sunlight is striking the window) and diffuse solar radiation from the sky and ground. For the outside surfaces of the frame, the absorbed diffuse solar per unit frame face area is:

$$Q_{abs,sol}^{dif}=I_{ext}^{dif}{\alpha }_{fr,sol}\left(A_f+F_{p1}{A}_{p1}\right)/A_f=I_{ext}^{dif}{\alpha }_{fr,sol}\left(1+0.5\frac{A_{p1}}{A_f}\right)$$

If there is no exterior window shade, I${}^{dif}$${}_{ext}$ includes the effect of diffuse solar reflecting off of the glazing onto the outside frame projection, i.e.,

$$I_{ext}^{dif}\to I_{ext}^{dif}\left(1+R_{gl}^{f,dif}\right)$$

The beam solar absorbed by the outside face of the frame, per unit frame face area is:

$$Q_{abs,sol}^{bm,face}=I_{ext}^{bm}{\alpha }_{fr,sol}\cos {\beta }_{face}{f}_{sunlit}$$

The beam solar absorbed by the frame outside projection parallel to the window x-axis is:

$$Q_{abs,sol}^{bm,h}=I_{ext}^{bm}{\alpha }_{fr,sol}\cos {\beta }_h{p}_{f1}\left(W-N_v{w}_d\right)f_{sunlit}/A_f$$

Here it is assumed that the sunlit fraction, f${}_{sunlit}$, for the window can be applied to the window frame. Note that at any given time beam solar can strike only one of the two projection surfaces that are parallel to the window x-axis. If there is no exterior window shade, I${}^{bm}$${}_{ext}$ includes the effect of beam solar reflecting off of the glazing onto the outside frame projection, i.e.,

$$I_{ext}^{bm}\to I_{ext}^{bm}\left(1+R_{gl}^{f,bm}\right)$$

The beam solar absorbed by the frame outside projection parallel to the window y-axis is:

$$Q_{abs,sol}^{bm,v}=I_{ext}^{bm}{\alpha }_{fr,sol}\cos {\beta }_v{p}_{f1}\left(H-N_h{w}_d\right)f_{sunlit}/A_f$$

Using a similar approach, the beam and diffuse solar absorbed by the inside frame projections is calculated, taking the transmittance of the glazing into account.

Error Due to Assuming a Rectangular Profile[LINK]

Assuming that the inside and outside frame profile is rectangular introduces an error in the surface heat transfer calculation if the profile is non-rectangular. The percent error in the calculation of convection and emitted IR is approximately 100$\left|L_{profile,rect}-L_{profile,actual}\right|/L_{profile,rect}$, where L${}_{profile,rect}$ is the profile length for a rectangular profile (w${}_f$ + p${}_{f1}$ for outside of frame or w${}_f$ + p${}_{f2}$ for inside of frame) and L${}_{profile,actual}$ is the actual profile length. For example, for a circular profile vs a square profile the error is about 22%. The error in the calculation of absorbed beam radiation is close to zero since the beam radiation intercepted by the profile is insensitive to the shape of the profile. The error in the absorbed diffuse radiation and absorbed IR depends on details of the shape of the profile. For example, for a circular profile vs. a square profile the error is about 15%.

Divider Temperature Calculation[LINK]

The divider inside and outside surface temperatures are determined by a heat balance calculation that is analogous to the frame heat balance calculation described above.

Beam Solar Reflection from Window Reveal Surfaces[LINK]

This section describes how beam solar radiation that is reflected from window reveal surfaces is calculated. Reflection from outside reveal surfaces—which are associated with the setback of the glazing from the outside surface of the window’s parent wall—increases the solar gain through the glazing. Reflection from inside reveal surfaces—which are associated with the setback of the glazing from the inside surface of the window’s parent wall—decreases the solar gain to the zone because some of this radiation is reflected back out of the window.

The amount of beam solar reflected from reveal surfaces depends, among other things, on the extent to which reveal surfaces are shadowed by other reveal surfaces. An example of this shadowing is shown in Figure 4. In this case the sun is positioned such that the top reveal surfaces shadow the left and bottom reveal surfaces. And the right reveal surfaces shadow the bottom reveal surfaces. The result is that the left/outside, bottom/outside, left/inside and bottom/inside reveal surfaces each have sunlit areas. Note that the top and right reveal surfaces are facing away from the sun in this example so their sunlit areas are zero.

The size of the shadowed areas, and the size of the corresponding illuminated areas, depends on the following factors:

• The sun position relative to the window

• The height and width of the window

• The depth of the outside and inside reveal surfaces

We will assume that the reveal surfaces are perpendicular to the window plane and that the window is rectangular. Then the above factors determine a unique shadow pattern. From the geometry of the pattern the shadowed areas and corresponding illuminated areas can be determined. This calculation is done in subroutine CalcBeamSolarReflectedFromWinRevealSurface in the SolarShading module. The window reveal input data is specified in the WindowProperty:FrameAndDivider object expect for the depth of the outside reveal, which is determined from the vertex locations of the window and its parent wall.

If an exterior shading device (shade, screen or blind) is in place it is assumed that it blocks beam solar before it reaches outside or inside reveal surfaces. Correspondingly, it is assumed that an interior or between-glass shading device blocks beam solar before it reaches inside reveal surfaces.

Representative shadow patterns are shown in Figures 5 and 6 for a window with no shading device, and without and with a frame. The case with a frame has to be considered separately because the frame can cast an additional shadow on the inside reveal surfaces.

The patterns shown apply to both vertical and horizontal reveal surfaces. It is important to keep in mind that, for a window of arbitrary tilt, if the left reveal surfaces are illuminated the right surfaces will not be, and vice versa. And if the bottom reveal surfaces are illuminated the top surfaces will not be, and vice versa. (Of course, for a vertical window, the top reveal surfaces will never be illuminated by beam solar if the reveal surfaces are perpendicular to the glazing, as is being assumed.

For each shadow pattern in Figures 5 and 6, equations are given for the shadowed areas $A_{1,sh}$ and $A_{2,sh}$ of the outside and inside reveal surfaces, respectively. The variables in these equations are the following (see also Figure 7):

$d_1$ = depth of outside reveal, measured from the outside plane of the glazing to the edge of the reveal, plus one half of the glazing thickness.

$d_2$ = depth of inside reveal (or, for illumination on bottom reveal surfaces, inside sill depth), measured from the inside plane of the glazing to the edge of the reveal or the sill, plus one half of the glazing thickness.

$L$ = window height for vertical reveal surfaces or window width for horizontal reveal surfaces

$\alpha$ = vertical solar profile angle for shadowing on vertical reveal surfaces or horizontal solar profile angle for shadowing on horizontal reveal surfaces.

$p_1\left(p_2\right)$ = distance from outside (inside) surface of frame to glazing midplane.

${d_2}^{\prime }$ = depth of shadow cast by top reveal on bottom reveal, or by left reveal on right reveal, or by right reveal on left reveal.

${d_2}^{\prime \prime }$ = depth of shadow cast by frame.

For simplicity it is assumed that, for the case without a frame, the shadowed and illuminated areas extend into the glazing region. For this reason, $d_1$ and $d_2$ are measured from the midplane of the glazing. For the case with a frame, the beam solar absorbed by the surfaces formed by the frame outside and inside projections perpendicular to the glazing is calculated as described in “Window Frame and Divider Calculation: Calculation of Solar Radiation Absorbed by Frame."

The following logic gives expressions for the shadowed areas for all possible shadow patterns. Here:

d1 = $d_1$

d2 = $d_2$

P1 = $p_1$

P2 = $p_2$

f1 = $d_1-p_1$

f2 = $d_2-p_2$

d2prime = ${d_2}^{\prime }$

d2prime2 = ${d_2}^{\prime \prime }$

d12 = $d_1+d_2-{d_2}^{\prime }$

TanAlpha = $\tan \alpha$

A1sh = $A_{1,sh}$

A2sh = $A_{2,sh}$

L = $L$

L1 = average distance to frame of illuminated area of outside reveal (used to calculate view factor to frame).

L2 = average distance to frame of illuminated area of inside reveal (used to calculate view factor to frame).

IF(window does not have a frame) THEN
     IF(d2prime < = d2) THEN
    IF(d12*TanAlpha < = L) THEN
      A1sh = 0.5*TanAlpha*d1**2
      A2sh = d2prime*L + 0.5*TanAlpha*d12**2 - A1sh
    ELSE  ! d12*TanAlpha > L
      IF(d1*TanAlpha < = L) THEN
        A1sh = 0.5*TanAlpha*d1**2
        A2sh = d2*L - 0.5*TanAlpha*(L/TanAlpha - d1)**2
      ELSE  ! d1*TanAlpha > L
        A1sh = d1*L - (0.5/TanAlpha)*L**2
        A2sh = d2*L
      END IF
    END IF
  ELSE  ! d2prime > d2
    A2sh = d2*L
    IF(d2prime < d1+d2) THEN
      IF(d12*TanAlpha < = L) THEN
        A1sh = L*(d2prime-d2) + 0.5*TanAlpha*d12**2
      ELSE  ! d12*TanAlpha > L
        A1sh = d1*L - 0.5*L**2/TanAlpha
      END IF
    ELSE  ! d2prime > = d1+d2
      A1sh = d1*L
    END IF
  END IF
ELSE  ! Window has a frame
 f1 = d1-P1
 f2 = d2-P2
 d2prime2 = FrameWidth/TanGamma
 IF(vertical reveal) THEN  ! Vertical reveal
   IF(InsReveal+0.5*GlazingThickness < = P2) d2 = P2 + 0.001
 ELSE                      ! Horizontal
   IF(bottom reveal surfaces may be illuminated) THEN
     ! Bottom reveal surfaces may be illuminated
     IF(InsSillDepth+0.5*GlazingThickness< = P2) d2 = P2 + 0.001
   ELSE
     ! Top reveal surfaces may be illuminated
     IF(InsReveal+0.5*GlazingThickness < = P2) d2 = P2 + 0.001
   END IF
 END IF
 IF(d2prime < = f2) THEN
   ! Shadow from opposing reveal does not go beyond inside
    !  surface of frame
   IF(d12*TanAlpha < = L) THEN
     A1sh = 0.5*TanAlpha*f1**2
     L1   = f1*(f1*TanAlpha/(6*L)+0.5)
     IF(d2-(d2prime+d2prime2+P2) > = 0.) THEN
       A2sh = (d2prime+d2prime2)*L + &
         0.5*TanAlpha*((d1+d2-d2prime)**2-d1+p2+d2prime2)**2)
       L2   = d2prime2 + 0.5*(d2-(d2prime+d2prime2+P2))
     ELSE  ! d2-(d2prime+d2prime2+P2) < 0.
       ! Inside reveal is fully shadowed by frame and/or
        !opposing reveal
       A2sh = f2*L
       L2   = f2
     END IF
   ELSE  ! d12*TanAlpha > = L
     IF((d1+P2)*TanAlpha < = L) THEN
       A1sh = 0.5*TanAlpha*f1**2
       L1 = f1*((f1*TanAlpha)/(6*L) + 0.5)
       IF((d1+P2+d2prime2)*TanAlpha > = L) THEN
         A2sh = f2*L
         L2   = f2
       ELSE  ! (d1+P2+d2prime2)*TanAlpha < L
         A2sh = f2*L - 0.5*(L-(d1+P2)*TanAlpha)**2/TanAlpha &
            + d2prime2*(L-(d1+P2+d2prime2/2)*TanAlpha)
         L2 = d2prime2 + (L/TanAlpha - (d1+P2+d2prime2))/3
       END IF
     ELSE  ! (d1+P2)*TanAlpha > L
       L2 = f2
       A2sh = f2*L
       IF(f1*TanAlpha < = L) THEN
         A1sh = 0.5*TanAlpha*f1**2
         L1 = f1*((f1*TanAlpha)/(6*L) + 0.5)
       ELSE  ! f1*TanAlpha > L
         A1sh = f1*L - 0.5*L**2/TanAlpha
         L1 = f1-(L/TanAlpha)/3
       END IF
     END IF
   END IF
 ELSE
    ! d2prime > f2 -- Shadow from opposing reveal goes beyond
    ! inside of frame
   A2sh = f2*L
   L2 = f2
   IF(d2prime > = d1+d2) THEN
     A1sh = 0.0
     L1   = f1
   ELSE  ! d2prime < d1+d2
     IF(d2prime < = d2+P1) THEN
       IF(f1*TanAlpha < = L) THEN
         A1sh = 0.5*TanAlpha*f1**2
         L1 = f1*((f1*TanAlpha)/(6*L) + 0.5)
       ELSE  ! f1*TanAlpha > L
         A1sh = f1*L - 0.5*L**2/TanAlpha
             L1   = f1 - (L/TanAlpha)/3
           END IF
         ELSE  ! d2prime > d2+P1
           IF(d12*TanAlpha < = L) THEN
             A1sh = L*(d2prime-(d2+P1)) + 0.5*TanAlpha*d12**2
             L1   = (L*(f1-d12/2)-d12*TanAlpha* &
                (f1/2-d12/3))/(L-d12*TanAlpha/2)
           ELSE  ! d12*TanAlpha > L
             A1sh = f1*L - 0.5*L**2/TanAlpha
             L1   = f1 - (L/TanAlpha)/3
           END IF
         END IF
       END IF
     END IF
     FracToGlassOuts = 0.5*(1.0 - ATAN(FrameWidth/L1)/PiOvr2)
     FracToGlassIns  = 0.5*(1.0 - ATAN(FrameWidth/L2)/PiOvr2)
   END IF  ! End of check if window has frame

The beam solar reflected from a sunlit region of area $A$ is given by:

$$R=I_{B}A\cos \beta (1-a)$$

where

$R$ = reflected solar radiation [W]

$I_B$ = beam normal irradiance [W/m${}^2$]

$A$ = sunlit area [m${}^2$]

$\beta$ = beam solar angle of incidence on reveal surface

$a$ = solar absorptance of reveal surface

All reflected radiation is assumed to be isotropic diffuse. For outside reveal surfaces it is assumed that $R/2$ goes toward the window and $R/2$ goes to the exterior environment. Of the portion that goes toward the window a fraction $F_1$ goes toward the frame, if present, and $1-F_1$ goes toward the glazing.

The view factor $F_1$ to the frame calculated by assuming that the illuminated area can be considered to be a line source. Then the area-weighted average distance, $L_1$, of the source to the frame is calculated from the shape of the illuminated area (see above psuedo-code). Then $F_1$ is related as follows to the average angle subtended by the frame of width $w_f$:

$$F_1=\frac{{\tan }^{-1}\left(w_f/L_1\right)}{\pi /2}$$

For the portion going towards the frame, $\left(R/2\right)F_1{a}_f$ is absorbed by the frame (where $a_f$ is the solar absorptance of the frame) and contributes to the frame heat conduction calculation. The rest, $\left(R/2\right)F_1(1-a_f)$ , is assumed to be reflected to the exterior environment.

If the glazing has diffuse transmittance ${\tau }_{diff}$, diffuse front reflectance ${\rho }_{diff}^f$, and layer front absorptance ${\alpha }_{l,diff}^f$, then, of the portion, $\left(R/2\right)(1-F_1)$, that goes toward the glazing, $\left(R/2\right)(1-F_1){\tau }_{diff}$ is transmitted to the zone, $\left(R/2\right)(1-F_1){\alpha }_{l,diff}^f$ is absorbed in glass layer $l$ and contributes to the glazing heat balance calculation, and $\left(R/2\right)(1-F_1){\rho }_{dif{f}^{}}^f$ is reflected to the exterior environment.

The beam solar absorbed by an outside reveal surface is added to the other solar radiation absorbed by the outside of the window’s parent wall.

For inside reveal surfaces it is assumed that $R/2$ goes towards the window and $R/2$ goes into the zone. Of the portion that goes toward the window a fraction $\left(R/2\right)F_2$ goes toward the frame, if present, and $\left(R/2\right)(1-F_2)$ goes toward the glazing ($F_2$ is calculated using a method analogous to that used for $F_1$). For the portion going towards the frame, $\left(R/2\right)F_2{a}_f$ is absorbed by the frame and contributes to the frame heat conduction calculation. The rest, $\left(R/2\right)F_2(1-a_f)$, is assumed to be reflected back into the zone.

If the glazing has diffuse back reflectance ${\rho }_{diff}^b$, and layer back absorptance ${\alpha }_{l,diff}^b$, then, of the portion $\left(R/2\right)(1-F_2)$ that goes toward the glazing, $\left(R/2\right)(1-F_2){\tau }_{diff}$ is transmitted back out the glazing, $\left(R/2\right)(1-F_2){\alpha }_{l,diff}^b$ is absorbed in glass layer $l$ and contributes to the glazing heat balance calculation, and $\left(R/2\right)(1-F_2){\rho }_{dif{f}^{}}^b$ is reflected into the zone.

The beam solar absorbed by an inside reveal surface is added to the other solar radiation absorbed by the inside of the window’s parent wall.

Shading Device Thermal Model[LINK]

Shading devices in EnergyPlus can be on the exterior or interior sides of the window or between glass layers. The window shading device thermal model accounts for the thermal interactions between the shading layer (shade, screen or blind) and the adjacent glass, and between the shading layer and the room (for interior shading) or the shading layer and the outside surround (for exterior shading).

An important feature of the shading device thermal model is calculating the natural convection airflow between the shading device and glass. This flow affects the temperature of the shading device and glazing and, for interior shading, is a determinant of the convective heat gain from the shading layer and glazing to the zone air. The airflow model is based on one described in the ISO Standard 15099, “Thermal Performance of Windows, Doors and Shading Devices—Detailed Calculations” [ISO15099, 2001]. (Between-glass forced airflow is also modeled; see “Airflow Windows.”)

The following effects are considered by the shading device thermal model:

• For interior and exterior shading device: Long-wave radiation (IR) from the surround absorbed by shading device, or transmitted by the shading device and absorbed by the adjacent glass. For interior shading the surround consists of the other zone surfaces. For exterior shading the surround is the sky and ground plus exterior shadowing surfaces and exterior building surfaces “seen” by the window.

• Inter-reflection of IR between the shading device and adjacent glass.

• Direct and diffuse solar radiation absorbed by the shading device.

• Inter-reflection of solar radiation between shading layer and glass layers.

• Convection from shading layer and glass to the air in the gap (or, for between-glass shading, gaps) between the shading layer and adjacent glass, and convection from interior shading layer to zone air or from exterior shading layer to outside air.

• Natural convection airflow in the gap (or, for between-glass shading, gaps) between shading layer and adjacent glass induced by buoyancy effects, and the effect of this flow on the shading-to-gap and glass-to-gap convection coefficients.

• For interior shading, convective gain (or loss) to zone air from gap airflow.

In the following it is assumed that the shading device, when in place, covers the glazed part of the window (and dividers, if present) and is parallel to the glazing. For interior and exterior shading devices it is assumed that the shading layer is separated from the glazing by an air gap. A between-glass shading layer is assumed to be centered between two glass layers and separated from the adjacent glass layers by gaps that is filled with the same gas. If the window has a frame, it is assumed that the shading device does not cover the frame.

Heat Balance Equations for Shading Device and Adjacent Glass[LINK]

If a window shading device is deployed the heat balance equations for the glass surfaces facing the shading layer are modified, and two new equations, one for each face of the shading layer, are added. Figure 8 illustrates the case of double glazing with an interior shading device.

The heat balance equation for the glass surface facing the gap between glass and shading layer (called in the following, “gap”) is:

$$\frac{E_i{\epsilon }_4{\tau }_{sh}}{1-{\rho }_4{\rho }_{sh}}+\frac{\sigma {\epsilon }_4}{1-{\rho }_4{\rho }_{sh}}\left[{\theta }_5^4{\epsilon }_{sh}-{\theta }_4^4(1-{\rho }_{sh})\right]+k_2\left({\theta }_3-{\theta }_4\right)+h_{cv}\left(T_{gap}-{\theta }_4\right)+S_4=0$$

where

$\tau$${}_{sh}$ = IR diffuse transmittance of shading device

$\epsilon$${}_{sh}$ = diffuse emissivity of shading device

$\rho$${}_{sh}$ = IR diffuse reflectance of shading device ( = 1 - ( $\tau$${}_{sh}$ + $\epsilon$${}_{sh}$))

$\theta$${}_5$ = temperature of the surface of the shading layer that faces the gap (K).

The term 1 - $\rho$${}_4$$\rho$${}_{sh}$ accounts for the inter-reflection of IR radiation between glass and shading layer.

The convective heat transfer from glass layer #2 to the air in the gap is:

$$q_{c,gl}=h_{cv}({\theta }_4-T_{gap})$$

where

T${}_{gap}$ = effective mean temperature of the gap air (K).

h${}_{cv}$ = convective heat transfer coefficient from glass or shading layer to gap air (W/m${}^2$K).

The corresponding heat transfer from shading layer to gap air is:

$$q_{c,sh}=h_{cv}({\theta }_5-T_{gap})$$

The convective heat transfer coefficient is given by:

$$h_{cv}=2h_c+4v$$

where

h${}_c$ = surface-to-surface heat transfer coefficient for non-vented (closed) cavities (W/m${}^2$K)

v = mean air velocity in the gap (m/s).

The quantities h${}_{cv}$ and T${}_{gap}$ depend on the airflow velocity in the gap, which in turn depends on several factors, including height of shading layer, glass/shading layer separation (gap depth), zone air temperature for interior shading or outside air temperature for exterior shading, and shading layer and glass face temperatures. The calculation of h${}_{cv}$ and T${}_{gap}$ is described in the following sections.

The heat balance equation for the shading layer surface facing the gap is:

$$\frac{E_i{\tau }_{sh}{\rho }_4{\epsilon }_{sh}}{1-{\rho }_4{\rho }_{sh}}+\frac{\sigma {\epsilon }_{sh}}{1-{\rho }_4{\rho }_{sh}}[{\epsilon }_4{\theta }_4^4-{\theta }_5^4(1-{\rho }_4({\epsilon }_{sh}+{\rho }_{sh})\left)\right]+k_{sh}({\theta }_6-{\theta }_5)+h_{cv}(T_{gap}-{\theta }_5)+S_{sh,1}=0$$

where

k${}_{sh}$ = shading layer conductance (W/m${}^2$K).

$\theta$${}_6$ = temperature of shading layer surface facing the zone air (K).

S${}_{sh,l}$ = solar radiation plus short-wave radiation from lights plus IR radiation from lights and zone equipment absorbed by the gap-side face of the shading layer (W/m${}^2$K).

The heat balance equation for the shading layer surface facing the zone air is:

$$E_i{\epsilon }_{sh}-{\epsilon }_{sh}\sigma {\theta }_6^4+k_{sh}\left({\theta }_5-{\theta }_6\right)+h_i\left(T_i-{\theta }_6\right)+S_{sh,2}=0$$

where

S${}_{sh,2}$ = solar radiation plus short-wave radiation from lights plus IR radiation from lights and zone equipment absorbed by the zone-side face of the shading layer (W/m${}^2$K).

Solving for Gap Airflow and Temperature[LINK]

For interior and exterior shading devices a pressure-balance equation is used to determine gap air velocity, gap air mean equivalent temperature and gap outlet air temperature given values of zone air temperature (or outside temperature for exterior shading), shading layer face temperatures and gap geometry. The pressure balance equates the buoyancy pressure acting on the gap air to the pressure losses associated with gap airflow between gap inlet and outlet [ISO15099, 2001]. The variables used in the following analysis of the interior shading case are shown in Figure 9.

Pressure Balance Equation[LINK]

The pressure balance equation for airflow through the gap is:

$$\Delta p_T=\Delta p_B+\Delta p_{HP}+\Delta p_Z$$

Here, Δp${}_T$