The objective of this example is to calculate the cooling load for the office shown in Figure 16 for July 3:00 pm local standard time. This corner office is on the second floor of a two-story office building.
Cooling Loads Using RTS Method
Traditionally, simplified cooling load calculation methods such as the radiant time series (RTS) Method have estimated the total cooling load at a particular design condition by independently calculating each component load (wall, windows, occupants, lighting, etc.) and then summing the component loads. Although the actual heat transfer processes for each component do affect each other, this simplification, known as the principle of superposition, is appropriate for design load calculations and useful to the designer in understanding the relative contribution of each component to the total cooling load.
On the following pages RTS procedures will be demonstrated for calculating (1) load due to internal heat gain, (2) exterior wall load, (3) load for windows with no shading, (4) load for windows with internal shading, (5) load for windows with internal and external shading, and (6) the total room load. All loads will be calculated for July 3:00 pm local standard time. Equations used in these calculations are summarized in Table 26.
Part 1. Cooling load due to internal heat gain.
Objective: Calculate the cooling load due to overhead lighting heat gain at 3:00 pm local standard time.
Solution: Calculation of the lighting load involves the following steps: (a) calculate the 24 h heat gain profile, (b) split those heat gains into convective and radiant components, (c) determine the radiant portion of the load by applying appropriate RTS factors, and (d) sum the convective and radiant load components to determine the total lighting load.
The heat gain profile is calculated using Equation (1) for each hour of the day. Calculations are shown in columns b through e in Table 30. Each heat gain is designated as qi. For example, for 3:00 pm (hour 15):
Next, the lighting heat gain is divided into convective and radiant portions. Table 3 shows that a “Non-in-ceiling fluorescent luminaire” has a radiant factor between 0.5 and 0.57. We will use the higher value of 0.57 or 57%. Therefore 57% of the heat gain is radiant and the remaining 43% is convective. Columns f and g in Table 30 show the results of this calculation: convective heat gain is 161 Btu/h while radiant heat gain is 214 Btu/h.
The convective portion of the heat gain immediately becomes a cooling load. The radiant portion undergoes a conversion process from heat gain to cooling load which is modeled using the appropriate RTS factors. Those factors are obtained from Table 19 using the data for medium weight construction, carpeted floor, and 50% glass. The factors are reproduced in Table 31 in the “non-solar RTS factors, zone type 8” column. The radiant load is calculated using Equation (33). Results are shown in column i of Table 30. For 3:00 PM, for example, the calculation is
Finally, the total lighting cooling load at the designated hour is the sum of convective and radiant portions. Table 30 shows the results in column j. For 3:00 pm the calculation is as follows:
Part 2. Wall cooling load.
Objective: Calculate the cooling load for the spandrel wall section facing 60° west of south for July 3:00 pm local standard time.
Solution: Determining the wall cooling load requires calculation of: (a) the sol-air temperature at the exterior surface (b) the heat input at the exterior surface based on sol-air temperature, (c) the delayed heat gain through the mass of the wall to the interior surface using conduction time series factors, (d) convective and radiant portions of the heat gain, (e) the delayed space cooling load from interior surface radiant heat gain using radiant time series factors, and (f) the total wall load as the sum of convective and radiant load components.
First, calculate the sol-air temperature at 3:00 pm local standard time (LST) on July 21 for a vertical, spandrel glass wall assembly, facing 60° west of south. Key input data for the calculation is shown in Table 32.
Sol-air temperature is calculated using Equation (29). For the dark-colored wall, α/ho = 0.30, and for vertical surfaces, εΔR/ho = 0. The solar irradiance Et on the wall must be determined using the equations in Chapter 14:
Solar Angles:
ψ = surface azimuth = +60°
Σ = surface tilt from horizontal (where horizontal = 0°) = 90° for vertical wall surface
3:00 pm LST = hour 15
Calculate solar altitude, solar azimuth, surface solar azimuth, and incident angle as follows:
From Table 2 in Chapter 14, solar position data and constants for July 21 are
ET = –6.3536 min
δ = 20.44°
Eo = 420 Btu/h·ft2
Local standard meridian (LSM) for Eastern Time Zone = –75°.
Apparent solar time AST
AST = LST + ET/60 + (LSM – LON)/15
= 15 + (–6.3536/60) + [(-75 – (–84.43))/15]
= 14.2388
Hour angle H, degrees
H = 15(AST – 12)
= 15(14.2388 – 12)
= 33.58°
Solar altitude β
sin β = cos L cos δ cos H + sin L sin δ
= cos (33.64) cos (20.44) cos (33.58) + sin (33.64) sin (20.44)
= 0.8434
β = sin−1(0.8434) = 57.5°
Solar azimuth ϕ
cos ϕ = (sin β sin L – sin δ)/(cos β cos L)
= [(sin (57.5)sin (33.64) – sin (20.44)]/[cos (57.5) cos (33.64)]
= 0.2637
ϕ = cos−1(0.2637) = 74.71°
Surface-solar azimuth γ
γ = ϕ – ψ
= 74.71 – 60
= 14.71°
Incident angle θ
cos θ = cos β cos γ sin Σ + sin β cos Σ
= cos (57.5) cos (14.71) sin (90) + sin (57.5) cos (90)
= 0.5197
θ = cos−1(0.5197) = 58.69°
Beam normal irradiance Eb
Eb = Eoexp(–τbmab)
m = relative air mass
= 1/[sinβ +0.50572(6.07995 + β)−1.6364], β expressed in degrees
= 1.1849
ab = beam air mass exponent
= 1.454 – 0.406τb – 0.268τd + 0.021τbτd
= 0.71357
Eb = 420 exp[–0.515(1.18490.71357)]
= 234.9 Btu/h·ft2
= 741 W/m2
Surface beam irradiance Et,b
Et,b = Ebcosθ
= (234.9)cos(58.69)
= 122.1 Btu/h·ft2
Diffuse irradiance Ed − Horizontal surfaces
Ed= Eoexp(–τdmad)
ad = diffuse air mass exponent
= 0.507 + 0.205τb − 0.080τd − 0.190τbτd
= 0.24514
Ed = Eoexp(–τdmad)
= 420 exp[–2.066(1.18490.24514)]
= 48.7 Btu/h·ft2
Diffuse irradiance Ed − Vertical surface with surface azimuth +60°
AI = anisotropy index
= Eb/ Eo
= 234.9 / 420
= 0.5592
Rb = cos θ / sin β
= cos(58.69°) / sin(57.5°)
= 0.6162
Et,d= Ed [AI Rb + (1 – AI)(1 + cos Σ)/2 ]
= 48.7[(0.5592)(0.6162) + (1 – 0.5592)(1 – cos(90°)/2]
= 48.7[0.5683]
= 27.5 Btu/h·ft2
Ground reflected irradiance Et,r
Et,r= (Ebsinβ + Ed)ρg(l – cos Σ)/2
= [234.9 sin(57.5) + 48.7](0.2)[1 – cos(90)]/2
= 24.7 Btu/h·ft2
Total surface irradiance Et
Et= Et,b + Et,d + Et,r
= 122.1 + 27.5 + 24.7
= 174.3 Btu/h·ft2
Sol-air temperature [from Equation (29)]:
Te = to + αEt /ho – εΔR/ho
= 91.6 + (0.30)(174.3) – 0
= 143.9°F
AI = anisotropy index
= Eb/ Eo
= 741 / 1325
= 0.5592
Rb = cos θ / sin β
= cos(58.69°) / sin(57.5°)
= 0.6162
Et,d= Ed [AI Rb + (1 – AI)(1 + cos Σ)/2 ]
= 154 [ (0.5592)(0.6162) + (1 – 0.5592) (1 – cos(90°)) / 2 ]
= 154 [ 0.5683]
= 86.9 W/m2
Ground reflected irradiance Et,r
Et,r= (Ebsinβ + Ed)ρg(l – cos Σ)/2
= [741 sin(57.5) + 154](0.2)[1 – cos(90)]/2
= 77.9 W/m2
Total surface irradiance Et
Et= Et,b + Et,d + Et,r
= 385.1 + 86.9 + 77.9
= 549.8 W/m2
Sol-air temperature [from Equation (29)]:
Te = to + αEt /ho – εΔR/ho
= 33.1 + (0.053)(549.8) – 0
= 62.2°C
This procedure is used to calculate the sol-air temperatures for each hour. Because of the tedious solar angle and intensity calculations, using a simple computer spreadsheet or other computer software can reduce the effort involved. A spreadsheet was used to calculate a 24 h sol-air temperature profile for the data of this example. See Table 33A for the solar angle and intensity calculations and Table 33B for the sol-air temperatures for this wall surface and orientation.
Next. use the sol-air temperature to calculate the heat input at the exterior surface of the wall using Equation (30). Table 33B shows the results of this calculation in column f. The calculation for 3:00 pm is
With the heat input at the exterior surface known, the conduction heat gain at the interior surface of the wall can be calculated using Equation (31). This involves applying the conduction time series to calculate the delay as heat flows through the wall. In Table 16, the most similar wall construction is wall number 2. This is a spandrel glass wall that has similar mass and thermal resistance. Conduction time series factors for this wall are also shown in Table 31. Results of this calculation are shown in Table 33B in column g. For example, the calculation for 3:00 pm is:
Next, split the conduction heat gain into convective and radiant portions. Table 14 shows that for conduction heat gain through walls and floors the convective fraction is 54% and the radiant fraction is 46%. Table 33B shows the division of the total heat gain into convective and radiant portions for all hours of the July design day in columns h and i. For 3:00 pm the calculation is:
The convective heat gain will immediately become a cooling load. The radiant portion of the heat gain will undergo a conversion process to cooling load that is modeled using the RTS factors for the room. The same RTS factors used in Part 1 for lighting load calculation will be applied to the wall calculation since the room conditions are the same. Table 31 lists these factors in the “non-solar RTS factors, zone type 8” column. Table 33B shows the result of this calculation in column k. The calculation for 3:00 pm is as follows:
Finally, the total wall cooling load is the sum of the convective and radiant loads. Table 33B shows these results in column l. For July 3:00 pm the calculation is as follows:
Part 3. Window cooling load without internal or external shading
Objective: Calculate the cooling load for the 40 ft2 window facing 60° west of south for July 3:00 pm local standard time, without considering internal or external shading. Internal and external shading will be considered in Parts 4 and 5, respectively.
Solution: Determining the window cooling load requires calculation of: (a) the 24 h window heat gain profile, (b) the convective and radiant portions of the heat gain, (c) the conversion of the radiant heat gain into cooling load using RTS factors, and (d) the total window load as the sum of convective and radiant loads.
First calculate the window heat gain profile using Equations (12) to (15). Respectively, these equations calculate the direct beam solar heat gain qb, the diffuse solar heat gain qd, the conduction heat gain qc, and the total fenestration heat gain. The calculation uses solar irradiance and solar angle values calculated in Part 2. The solar irradiance and solar angle values along with the window heat gain results are shown in Table 34 in columns b through d and h through k. For July 3:00 pm (hour 15), for example, Part 2 calculated:
Et,b =122.1 Btu/h·ft2
Et,d=27.5 Btu/h·ft2
Er=24.7 Btu/h·ft2
θ=58.7°
From Chapter 15, Table 10, for glass type 5d,qb15 = AEt,b
SHGC(θ) = SHGC(58.7) = 0.400 (interpolated)
〈SHGC〉D = 0.41
The indoor attenuation coefficient (IAC) is 1.0 because internal shades are not considered in this calculation.
Then, applying Equations (12) to (15) for July 3:00 pm:
Next, divide the window heat gain into convective and radiant components. Because loads for windows without internal shading (blinds, drapes, etc.) are being calculated, the direct beam solar gain must be treated separately from the diffuse and conduction heat gains. The direct beam heat gain is treated as 100% radiant, and the conversion from heat gain to load uses special solar RTS factors different from those used to convert other types of radiant heat gain to load. Therefore, the total window heat gain must be divided into three parts: (1) the direct beam heat gain, (2) the radiant part of the conduction plus diffuse solar heat gain, and (3) the convective part of the conduction plus diffuse solar heat gain. In Table 35, columns b to d show the direct beam component of heat gain. For conduction and diffuse solar heat gains, Table 14 shows that for conduction heat gains through windows where the SHGC is less than 0.5, the convective fraction of the heat gain is 54% and the radiative fraction is 46%. Columns f to j in Table 35 show how the sum of diffuse solar and conduction heat gain is split into convective and radiant portions.
With the heat gains split into three categories, the conversion of heat gain to cooling load can be calculated for each radiant gain. First, for the load due to the direct beam solar heat gain is calculated using Equation (33). The solar RTS factors from Table 20 for medium weight construction, carpeted floor, and 50% window area will be used. These factors are also shown in Table 31 in the column titled “Solar RTS Factors, Room Type 8.” Cooling load results appear in column e of Table 35. For July 3:00 pm the calculation is as follows:
The cooling load due to the radiant portion of the diffuse solar and conduction window heat gain is also calculated with Equation (33). This time the non-solar RTS factors are used. These are the same factors used for conversion of lighting and wall conduction heat gains to load in Parts 1 and 2. The non-solar RTS factors are shown in Table 31 in the column titled “Non-Solar RTS Factors, Room Type 8.” Results appear in column l of Table 35.
The convective portion of the diffuse solar and conduction window heat gains immediately converts to load and is shown Table 35 column k. The total diffuse solar and conduction load is the sum of radiant and convective components and is shown in column m.
Finally, the total window load is the sum of the direct beam solar load, and the diffuse solar plus convective load. These loads are shown in Table 35 column n. For July 3:00 pm the calculation is
Part 4. Window cooling load with internal shading
Objective: Building on the window load calculation in Part 3, calculate the window load considering internal shading due to light-colored mini-blinds. Consider the same 40 ft2 window facing 60° west of south for July 3:00 pm local standard time.
Solution: Calculation of the window cooling load requires the same four steps used in Part 3, but with different application data. Calculate: (a) the 24 h window heat gain profile, (b) the convective and radiant portions of the heat gain, (c) the conversion of the radiant heat gain into cooling load using RTS factors, and (d) the total window load as the sum of convective and radiant loads.
Calculation of the 24 h heat gain profiles requires consideration of the effect of the mini-blinds. This effect is calculated with indoor attenuation coefficients (IAC), and different radiant and convective fractions than an unshaded window.
IAC values depend on several factors: (1) type of shading device, (2) position of shading device relative to window, (3) reflectivity of shading device, (4) angular adjustment of shading device, as well as (5) solar position relative to the shading device. These factors are discussed in detail in Chapter 15. For this example with mini-blinds, the IAC for beam radiation is treated separately from the IAC for diffuse solar gain. The direct beam IAC must be adjusted based on the profile angle of the sun.
The mini-blinds are assumed to be light colored with louver reflectance = 0.8 and louvers positioned at a 45° angle on double-glazed heat absorbing windows. Chapter 15 Table 14B lists IAC factors for window type 5d plus this shading type as IAC(0) = 0.74, IAC(60) = 0.65, IAC(diff) = 0.79, and radiant fraction = 0.54.
At 3:00 pm in July, the profile angle of the sun relative to the window surface is 58.4°. Using interpolation, the beam IAC is 0.652. The diffuse IAC is 0.79. As calculated in Part 3, SHGC(θ) = 0.400 and 〈SHGC〉D = 0.41. Thus, the window heat gains at 3:00 pm are
Calculation of beam solar heat gain is shown in Table 36 columns b through e. Calculation of the diffuse solar heat gain is shown in columns f through i. The conduction heat gain is found in column j.
Because internal shades are used, the direct beam solar heat gain is assumed to be absorbed by the shading device, and a portion immediately becomes a cooling load by convection. The remaining solar heat absorbed by the blind is assumed to be radiated to all surfaces of the room just as the diffuse and conduction heat gains are. As a result, the beam direct solar heat gain is combined with diffuse and conduction heat gains to obtain the total heat gain shown in column k of Table 36.
Next, the total heat gain is divided into convective and radiant portions. For window type 5d with mini-blinds, Chapter 15 Table 14B lists a radiant fraction as 0.54. Thus 54% of the heat gain is radiant and 46% is convective. These convective and radiant heat gains are shown in columns l and m in Table 36.
The radiant portion of the heat gain is converted to cooling load using Equation (33). In this equation the “Non-Solar RTS Factors, Zone Type 8” factors from Table 31 are used. Results appear in Table 36 in column o. The convective part of the heat gain immediately becomes a load and is shown in column n.
The total window load is then calculated combining the radiant and convective loads. Results are shown in Table 36 column p. For July 3:00 pm the calculation is
Part 5. Window cooling load with internal and external shading
Objective: Calculate the cooling load for the window in Part 4 with the addition of a 5 ft overhang shading the window.
Solution: As in Parts 3 and 4, determining the window cooling load requires calculation of (a) the 24 h window heat gain profile, (b) the convective and radiant portions of the heat gain, (c) the conversion of the radiant heat gain into cooling load using RTS factors, and (d) the total window load as the sum of convective and radiant loads.
The diffuse solar and conduction heat gain profiles are the same as in Part 4 and are shown in columns h and i of Table 37. The direct beam solar heat gain profile must be recalculated accounting for the shading effect of the overhang. In Chapter 15, methods are described and examples provided for calculating the area of a window shaded by attached vertical or horizontal projections. For July 3:00 pm, the solar position calculated in previous parts of this example is:
From Chapter 15, Equation (33), profile angle Ω is calculated by
From Chapter 15, Equation (35), shadow height SH is
Because the window is 6.4 ft tall, at July 3:00 pm the window is completely shaded by the 5 ft deep overhang. Thus, the shaded window heat gain includes only diffuse solar and conduction gains for this hour. When the shadow height is less than 6.4 ft then part of the window is exposed to direct beam solar. The sunlit area is calculated as A = 6.25(6.4 – Sh) where 6.25 is the window width in feet. The direct beam heat gain is then Sunlit Area × Et,b. Table 37 shows the profile angle, shadow height, sunlit area and direct beam heat gain in columns d through g. Note that the overhang shades the window completely for all hours except hours 16 through 19 when the solar altitude angle is low enough that part of the window is sunlit.
As in Part 4, due to the use of internal shading, direct beam solar heat gain is combined with diffuse solar and conduction heat gains to obtain the total heat gain shown in Table 37, column j. The total heat gain is divided using the 46% convective fraction and 54% radiative fraction used in Part 4 for a window with internal shades. These heat gain components are shown in Table 37, columns k and l.
The radiant heat gain is converted to cooling load using the “Non-Solar RTS Factors, Zone Type 8” from Table 31 as before. The radiant cooling load is shown in Table 37, column n. Combining with the convective load yields the total cooling load listed in Table 37, column o. For July 3:00 pm the total window cooling load is 958 Btu/h.
Part 6. Total room sensible cooling load.
Objective: Calculate the total sensible cooling load for the example office for July 3:00 pm local standard time.
Solution: To calculate the total sensible cooling load for the office, cooling loads for all ten heat gain components must be calculated and then summed.
Part 1 demonstrated the calculation of the lighting load. The same general procedure can be applied to calculating the occupant and equipment cooling loads using those component heat gains.
Part 2 demonstrated the calculation of the cooling load for the spandrel glass wall facing 60° west of true south. The same procedures can be used to calculate wall cooling loads for the other spandrel wall section and the two sections of brick wall. In addition, the same procedures can be used to calculate the roof load. Note that because the ceiling space is used as a return air plenum, we are assuming 30% of the roof load is directly absorbed by the return air and only the remaining 70% reaches the room.
Part 5 demonstrated the calculation of the cooling load for the window facing west of true south, with internal shading and external overhang shading. The same procedures can be used for the window facing east of true south.
Finally the 10 load components are summed to obtain the total sensible load for the room. Table 38 shows the 24 h component and total room load profiles for July. The total room load is in column l. The 3:00 pm load is 3158 Btu/h.