Air Conditioning Load Calculations PDF

Level 4 Air Conditioning Load Calculations RACM-400 -- -- **Main Sources of Heat Loss:** - **Exterior walls** (exposed to outside air) - **Windows & doors** (conduction and air leakage) - **Roofs & ceilings** (heat rising and escaping) - **Partition panels** (walls separating con...

Level 4 Air Conditioning Load Calculations RACM-400 -- -- **Main Sources of Heat Loss:** - **Exterior walls** (exposed to outside air) - **Windows & doors** (conduction and air leakage) - **Roofs & ceilings** (heat rising and escaping) - **Partition panels** (walls separating conditioned/unconditioned areas) - **Below-grade walls** (basement walls in contact with soil) - **Slab & basement floors** (conductive losses through the ground) - **Floors above unconditioned spaces** (such as garages or crawlspaces) - **Duct losses** (from heating/cooling ducts in unconditioned spaces) - **Air leakage** (cracks, gaps, or construction deficiencies) **Absolute Humidity:** - **Absolute humidity (AH):** is the total amount of water vapor present in a given volume of air, regardless of temperature. It is expressed in grains of moisture or pounds. - **7000 grains of moisture = 1 lb.** **Relative Humidity:** - Relative humidity is a comparison of the amount of water vapor measured in 1 pound of air at the current temperature vs. the maximum amount of water vapor 1 pound of air at this temperature can hold before being **100% saturated**. - Warmer air has the ability to hold **more moisture.** - Ex. You have a 100 ml container with 50 ml of water indicating it is **50% full**, when you add the 50 ml of water to a 200 ml container, the same amount of water is present, however it is now **25% full**. **AHRI Equipment Design Specifications:** - **AHRI** designs their equipment based on these conditions. - **95 F** outdoor ambient temperature. - **80 F** indoor temperature. - **50 %** Relative humidity. - Any condition **outside** of these indicates that the system is not operating to its rated design. - **This means that, the 12,000 btu system is not 12,000 btu if outside of the conditions listed above.** **Design Conditions for Load Calculations:** - The general population is not comfortable at **80 F and 50 %** relative humidity. The below conditions should be applied to load calculations as it is more practical. - Indoor - **75 F** dry bulb for cooling, **70 F** for heating. - Indoor - **50 %** relative humidity. **Design Temperature Difference:** - The design temperature difference is what will be calculated for the TD from the **inside of the structure to the outside**. - TD is the main driving force of heat loss and heat gain into a structure. - In the **summertime**, heat is gained through **infiltration.** - In the **wintertime,** heat is lost through **exfiltration**. - The design TD will be different depending on where you are located. **Calculating TD for Load Calculations**: - You are completed a heat loss, heat gain calculation for a home in **Modesto, California.** - The design TD for cooling would be **TD = 98 F -- 75 F = 23 F**. - The design TD for heating would be **TD = 70 F -- 30 F = 40 F.** **Construction Materials -- R Value:** - Another driving force of heat loss/heat gain is the construction materials used to enclose the structure. - **R Value is used to rate the thermal resistance of insulation materials**. - The **higher** the R value, the harder it is for heat to pass through the material. - If the wall is made up of several different materials, the sum of all of the r values is used. **Calculating R Value:** - Ex. A wall is constructed of **4" brick, 5/8" drywall** on the inside, **¾" plywood**, and **insulating board backed aluminum siding**. - R values -- 4" brick = **0.80**, 5/8" drywall = **0.56**, ¾" plywood = **0.94**, insulating board backed aluminum siding = **1.82.** - Total R value = **4.12.** **Calculating U value:** - The U-value is the **thermal conductivity** of the material. - The **higher** the U-value, the **easier** it is for heat to pass through the material. - U-Value is the **inverse** of the R-value. - Ex. R value of **4.12**, **U = 1 / 4.12 = 0.24** - The units for U-value are: **btus/hr/f/sq.ft** **Heat Loss, Heat Gain Formula:** **Net Wall Area:** - **NET WALL AREA:** *Net wall area* is a term that is used to describe the area of a wall after the doors and windows are cut into it. The net wall area can be expressed mathematically as: - Net Wall Area = **Gross Wall Area - Door Area - Window Area - Area of Other Components** - Let's take a look at the wall in figure 42.12. This wall has a gross wall area of 300 ft2, which is simply the height of the wall multiplied by the width of the wall. To find the net wall area, we would subtract the area of the window and door. - Net Wall Area = Gross Wall Area - Door Area - Window Area - Net Wall Area = 300 ft2 - Door Area - Window Area - Net Wall Area = 300 ft2 - (3\' x 7\') - (6\' x 3\') - Net Wall Area = 300 ft2 - 21 ft2 -18 ft2 - Net Wall Area = 300 ft2 - 39 ft2 - Net Wall Area = 261 ft2 **Heat Loss Through Doors and Windows:** - Wall: 2 x 6 studs, R-19 cavity insulation, - R-4 board insulation - Window: Double pane with 0.50\" of air space - Door: 2.25\" solid wood - TD across panel: 40F **To determine the heat loss via the construction materials, the** **U-values are needed. These values are obtained as follows:** - Wall U-value: **0.055** (from figure 42.10) - Window U-value: **0.49** (from figure 42.8 and using U = 1/R, 1/2.04 = 0.49) - Door U-value: **0.27** (from figure 42.8 and using U = 1/R, 1/3.70 = 0.27) **The heat transfer rate through the window is found as follows:** - Q = U-value x TD x Area - Q = 0.49 x 40 F x (3 x 6) ft2 - Q = 0.49 x 40 F x 18 ft2 - **Q** = **352.8 Btu/h** **Similarly, the heat transfer rate through the door is found as** **follows:** - Q = U-value x TD x Area - Q = 0.27 x 40F x (3 x 7) ft2 - Q = 0.27 x 40F x 21 ft2 - **Q** = **226.8 Btu/h** **We can then determine the heat loss for the entire wall by** **using the following:** - QEntire Wall = QNet Wall + QWindow + QDoor - QEntire Wall = \[0.055 x 261 ft2 x 40F\] + QWindow + QDoor - QEntire Wall = 574.2 Btu/h + 352.8 Btu/h + 226.8 Btu/h - **QEntire Wall = 1153.8 Btu/h** **Heat Loss Through Roofs:** - See fig. **42.13** below: - **Height of Triangles**: 6 ft (16 ft peak height -- 10 ft wall height) - **Base of Triangles:** 8 ft (half the width of the house) - **Area of a Triangle**: [*a*^2^ + *b*^2^ = *c*^2^]{.math.inline} - Area = [6^2^ + 8^2^ = *c*^2^]{.math.inline} - Area = 36 + 64 = [*c*^2^]{.math.inline} - Area = √100 - Area = 10 - Since the house is 25 ft long, the area is: A=25ft x 10 = **250 sq.ft** - Remember we only calculated half of the roof, so the area will be **500 sq. ft.** - *Let's say the roof has a U value of **0.30** and a TD of **70 F**.* - ***Qroof** = 0.30 x 70 F x 500 sq.ft **= 10,500 btu/hr*** ***Heat Loss through Ceilings to Unconditioned Attics:*** - *Example: The footprint of a house is **400 sq.ft** and a ceiling is separating the occupied space from the attic. The ceiling is insulated with **R-30** insulation, and the TD is **60 F**.* - ***Qceiling** = 0.033 x 60 F x 400 sq.ft = **792 btus.\ *** ***Heat Loss Through Partition Walls**:* *A partition wall is a wall that separates an **interior conditioned** space from an **unconditioned interior** space, like a garage. The formula is the same, however the TD changes. Ex. Outside temperature is **0 F**. The garage temperature is **40 F**, and the inside design temperature is still **70 F**. The TD used in the formula is **30 F**, not 70 F.* ***Heat Loss Through Below and Above Grade Walls:*** - *Heat loss through above and below grade walls need to be calculated separately as the U values will differ from above to below grade. Heat loss through below grade walls decreases the further you go down.* - *Ex. A house with a 40' x 60' footprint, the basement has 8' ceilings, 6' of the wall is below grade. The design TD is 70 F.* ***Area of above grade wall** = height of above grade x perimeter of above grade* - *Area = 2' x (40'+60'+40'+60')* - *Area = 2' x 200' = 400sq.ft* ***Area of Below Grade Wall** = 6' x 200' = 1200 sq.ft.* ***Qbelow** =0.043 x 70 F x 1200 sq. ft = **3612 btu/hr.*** ***Qabove**=0.069 x 70 F x 400 sq. ft = **1932 btu/hr.*** ***Qtotal = 5544 btu/hr.*** ***Heat Loss Through Floors Located Above Unconditioned Spaces:*** - *The heat loss through a floor located above an unconditioned space such as a crawl space or basement needs to be taken into account, similar to the partition walls. Since the unconditioned space is not as cold as the outdoors, the TD will differ.* - *Ex. 40' x 60 ' footprint. R19 insulation. The crawl space is 10 F, so 70 F -- 10 F gives us a TD of **60 F.*** Q = 0.049 x 29.1 F x 2400 sq.ft = **3422.16 btu/hr** **Heat Loss Through Slab Floors:** The formula for calculating heat loss through slab floors differs slightly from what we have been using. The heat loss is calculated using the running feet of exposed slab edge. - **Q = F x P x TD** - **F value =** btu/h per running foot per F. - **P =** Perimeter of exposed edge - **TD =** Temperature difference. - Ex. 40' x 70' footprint. Slab is located on light, dry soil. R-15 vertical insulation. TD = 60 F. Qslab = 0.178 x 60 F x 220' Qslab = **2349.6 btu/h** **Heat Loss Through Cracks, Openings, and Deficiencies:** **Qinfiltration** = 1.08 x cfm x TD The CFM value is an **estimate** of how many cubic feet of air are entering the structure each minute. This can be estimated by determining how many air changes occur each hour **(ACH)** and the **internal volume** of the structure. **Cfm(infiltration)** = ACH x House Volume / 60min Ex. Average construction, 2500 sq.ft, single story house with 8' ceilings, TD is 60 F. **CFM** = 0.50 x (2500 x 8)/60 **CFM** = 0.50 x 20,000/60 = **167 cfm**. **Qinfiltration**= 1.08 x 167 x 60 **Qinfiltration** = **10,821.6 btu/h** **Heat Loss Through Ducts** When warm air is passing through ducts located in attics, ceiling cavities and other unconditioned spaces, heat will be transferred to these areas, causing heat loss. The rate of heat loss depends of a few factors, the **temperature of the air in the ducts**, the **temperature of the surrounding air** in the unconditioned spaces, and the amount of **duct insulation.** The duct losses are calculated as a **percentage of the envelope load**, but the percentage changes based on the **conditions** mentioned above. **Q(ductloss)= envelope load x duct loss factor x duct insulation correction factor** The **envelope load** is simply the sum of the **individual heat losses of the structure**. The heat loss factor is determined by a table similar to the one shown in figure 42.24, which can be found in ASHRAE Manual J. The duct insulation correction value is determined by the duct wall. - R-2, the correction value is **1.48.** - R-4, the correction value is **1.15.** - R-6, the correction value is **1.00.** - R-8, the correction value is **0.93**. Ex. **2500 sq. ft**. envelope with a heat loss calculation of **17,500 btu/hr**. The discharge air temperature is **110 F**, the outside design temperature is **0 F**. The insulation on the duct walls is **R-4**. Q = 17500 x 0.25 x 1.15 = **5,031 btu/hr.** **Heat Gain Calculations:** Heat gain calculations are very similar. There are a few different factors to be considered. Lighting, appliances, occupants, and other heat/moisture generating elements. - Through exposed (exterior) walls. - Through doors. - Through roofs, walls, floors, and ceilings. - Through windows. - From people, appliances, lighting, and other elements. - Through cracks, openings, and deficiencies in structure. - Through duct runs through attics, ceiling cavities, and other unconditioned spaces. The indoor design temperature used will be **75 F** for cooling. Determining the design temperature difference varies slightly from heating. There is a separate chart used in order to figure this out. It is a two step process, you will figure out the wall construction, find the 'Group', using the group determined, you will find the ACTUAL TD in the next chart. Ex. 20' x 50' wall, R-19 insulation, 2x6 studs, R-5 board insulation, design TD of 20 F. We can determine from the chart that it is in **Group H**. The next chart we will find **Group H,** wall. **Under our design TD of 20 F**, we will find our new TD of **17.7 F.** This is the TD we will use in our formula for the walls. **Q = 0.052 x 17.7 x 1000 = 920 Btu/hr.** The process is similar for the rest of the calculations in regards to building materials. **Heat Gain Through Doors:** **Heat Gain Through Ceilings:** **\ ** **Heat Gain Through Windows:** Heat gain through windows differs from heat loss because not only do we have heat gain through the windows, we have to take into account the **solar load**. When systems are designed, we cannot control if the occupants leave their blinds or curtains fully open or closed, which can dramatically increase/decrease the heat gain through the windows. There is a chart that will give us a '**multiplier**' depending on the **location** of the windows. The chart is based on curtains/blinds being closed half way. Once the **multiplier** is found, all that is needed is the **area of the window**. Ex. **8' by 6' single pane window**, the design TD is **15 F,** the window is facing **North**. **Qwindow = multiplier x area** Qwindow = 22btu/h/sq.ft x 48 sq. ft = **1056 btu/hr**. Ex. The same window, facing **East.** Qwindow = 63 btu/hr/sq.ft x 48 sq.ft = **3024 btu/hr.** **Heat Gain Occupants, Lighting, Appliances:** Full time **occupants** have a btu load of approximately **430 btu/hr**. This is **230 btus/hr** of sensible heat, and **200 btus/hr** of latent heat through evaporation. **Lighting** has a btu load of **3.412 btus/hr per watt**. **Appliances** in households have a heat load, however when performing the load calculation, we use a default chart to give us these numbers. - **1200 Btu/h:** House with a refrigerator and a vented cooking surface. - **2400 Btu/h:** House with a refrigerator, vented cooking surface, dishwasher, clothes washer, vented clothes dryer, and miscellaneous electronic devices. - **3600 Btu/h:** House with two refrigerators, vented cooking surface, dishwasher, clothes washer, vented clothes dryer, and miscellaneous electronic devices. **Heat Gain Through Cracks, Openings, and Deficiencies:** Heat gain through infiltration is found in a similar way as heat loss. The difference is we need to take into account the latent heat. Formulas: **Qs = 1.08 x cfm x TD** **Ql = 0.68 x cfm x DG** **DG** is the design grain of moisture. On the weather data chart, there is a column for design grains @ 50 % RH. Ex. Single Story, average construction. 10 ft. ceilings, 40' x 60' footprint, design TD 20 F, design grains of moisture @ 50% or 30 grains. **CFM(infiltration) = ACH x house volume / 60 min/hr.** ACH is found on a chart -- similar to the heat loss. CFM = 0.25 x 24,000 / 60 = **100 cfm**. Q = 1.08 x 100 x 20 **= 2160 btu/hr sensible heat.** Q = 0.68 x 100 x 30 **= 2040 btu/hr latent heat.** **Heat Gain Through Duct Runs, Attics, Cavities and Unconditioned Spaces:** Qs = Envelope Load x Duct Load Factor x Duct Insulation Correction **Envelope load** = total sensible heat gain for the structure. **Duct load factor** = found in table. **Duct insulation** = found in table. Ex. **2500 sq.ft** house. Envelope load of **24,000 btu/hr**. Outdoor air design temperature is **90 F**, duct insulation is **R-4** with a correction factor of **1.15**. Design grains of moisture is **20 grains.** Qs = 24,000btu/hr x 0.42 x 1.15 Qs = **11,592 btu/hr.** The latent heat portion can be found directly on the chart. Ql = **1456 btu/hr.**