0 title page - submittal(standard) - wn mech rev2.pdf · handbook—fundamentals and ashrae’s...

Load

HVAC Clinic

Contents

Introduction ..................................................................................................................................... 3

Fundamentals of Heat Transfer ..................................................................................................... 3

Energy Measurement ..................................................................................................................... 4

Load Estimation .............................................................................................................................. 5

Example 1 ........................................................................................................................................ 7

Example 2 ...................................................................................................................................... 14

Weather Design Conditions ......................................................................................................... 18 Design Conditions For Reno, NV (ASHRAE 2005 Fundamentals .......................................................... 18 Rates Of Heat Gain from Occupants (ASHRAE 1997 Fundamentals, Chapter 28, Table 3) ................. 19 Heat Gain for Appliances (ASHRAE 1997 Fundamentals, Chapter 28, Table 5 & 6) ............................. 19 Thermal Properties Of Layers (ASHRAE 1997 Fundamentals, Chapter 28, Table 11) .......................... 20

Roofs .............................................................................................................................................. 21 Roof Group Numbers & CLTD (ASHRAE 1997 Fundamentals, Chapter 28, Table 30 & 31) ................. 21

Walls .............................................................................................................................................. 22 Wall Types Mass Located Inside Wall (ASHRAE 1997 Fundamentals, Chapter 28, Table 33A) ........... 22 Wall Types Mass Evenly Distributed (ASHRAE 1997 Fundamentals, Chapter 28, Table 33B) ............. 23 Wall Types Mass Evenly Distributed (ASHRAE 1997 Fundamentals, Chapter 28, Table 33B) ............. 24 Wall CLTD (ASHRAE 1997 Fundamentals, Chapter 28, Table 33B) ..................................................... 25 Wall Types Mass Located Inside Wall (ASHRAE 1997 Fundamentals, Chapter 28, Table 33A) ........... 28

Glass/Fenestration ....................................................................................................................... 29 U-Factors for Various Fenestration (ASHRAE 2005 Fundamentals, Chapter 29, Table 4) .................... 29 Shading Coefficient (ASHRAE 1997 Fundamentals, Chapter 29, Table 11) ......................................... 30 CLTD for Conduction through Glass (ASHRAE 1997 Fundamentals, Chapter 28, Table 34) ................ 31 Zone Types for Use with SCL (ASHRAE 1997 Fundamentals, Chapter 28, Table 35) ......................... 31 Zone Types for Use with SCL (ASHRAE 1997 Fundamentals, Chapter 28, Table 35) ......................... 32

Infiltration ...................................................................................................................................... 33 Infiltration Models (US Department Of Energy, Table 3, Table 4, Table 5) ............................................ 33

HVAC Clinics Draft - Not For Distribution

Nevada Mechanical Systems Duct Design of 33

Introduction Cooling load estimation is the foundation for which everything is based in HVAC. Without an accurate load, we couldn’t perform the psychrometrics required to determine the cooling airflow or the refrigeration load. An accurate load provides us with the information required to size our plant, determine the fraction of outside air, and most importantly, how to maintain occupant comfort. This particular clinic introduces the reader to cooling load estimation. It is intended to introduce the concepts of estimating building cooling and is limited to introducing the components that make up the load on a building. This includes the variables that affect each of these components, and simple methods used to estimate these load components. It is not intended to teach all the details or latest computerized techniques of how to calculate these loads. In addition, this clinic does not cover the process of determining the heating load. If the principles are understood for cooling load estimation, heating load estimation can easily be inferred.

Fundamentals of Heat Transfer The three primary rules of heat transfer are:

1. Heat energy cannot be destroyed

2. Heat always flows from a higher temperature source to a lower temperature source

3. Heat can be transferred from one substance to another substance

Heat energy cannot be destroyed. Energy can be transferred from one substance to another substance, but the energy is maintained. This is the first law of thermodynamics and is known as the conservation of energy. Heat always flows from a higher temperature source to a lower temperature source. Heat is a measure of the motion of (i.e. the kinetic energy of) molecules. When an object in motion impacts with an object that is stationary, it imparts some of its energy to the stationary object (via movement and plastic deformation). The same is true of heat. Thus heat always flow from a high temperature source to a low temperature source much the same way a moving object transfer’s energy to a stationary object. Heat energy can be transferred from one substance to another substance by one of three processes; convection, conduction and radiation. Convection is the transfer of energy between two fluids. Convection cannot flow between solids. Conversely, conduction is the transfer of energy between adjacent molecules of a solid. Finally, radiation is the emission of electromagnetic waves from all matter that has a temperature greater than absolute zero. It represents a conversion of thermal energy into electromagnetic energy. Radiation does not require the presence of matter to propagate. An example of a device that exhibits all three methods of heat transfer is a baseboard convector. Heat is transferred from the warm inner tube to the cooler outer fins by conduction. The fins of baseboard then warm the air contained within the casing. The buoyant (less dense) air within the casing then rises and mixes with the cooler room air by the process of convection. Finally, the warm casing surface transfers heat to objects in room air by electromagnetic energy or radiation (figure 1). Radiation does not head the adjacent air; it only heats the objects that the electromagnetic waves impact.

Figure 1. Baseboard Convector



Energy Measurement The IP method of measuring heat is the Btu, or British Thermal Unit. A Btu is defined as the amount of heat energy required to raise one on pound of water 1oF (figure 2). In thermodynamics, we often want to measure the rate of heat transfer, or the amount of heat energy per unit time. This rate of heat transfer is generally measured as a Btu/hr or the amount of heat transfer required to increase one pound of water 1oF in one hour.

Figure 2. Btu

Sensible Vs Latent Heat In environmental building conditioning, we typically deal with two types of energy transfer; sensible and latent heat transfer. Sensible heat transfer is the process of increasing the dry bulb temperature of a substance. As discussed in the psychrometric clinic, latent heat transfer is the process of increasing or decreasing the moisture content of a substance. However, latent energy can also refer to the amount of heat that is released or absorbed during the phase change of a substance. This is analogous to the energy absorbed when boiling water and creating steam (figure 3).

Figure 3. Sensible versus Latent Energy

ASHRAE standard 55, Environmental Conditions for Human Occupancy, defines the psychrometric conditions for human comfort in a conditioned space. These conditions allow for 80% occupant acceptability in terms of comfort (figure 4). Note that as the humidity ratio increases, the acceptable dry bulb temperature decreases. The maximum humidity ratio for comfort is constant and is defined as .012 lb-water/lb-air.



Figure 4. ASHRAE Standard 55

Load Estimation The method for determining space loads used in this clinic will be based on the Load Temperature Difference/Solar Cooling Load/ Cooling Load (CLTD/SCL/CLF) load estimation method developed by ASHRAE. A building typically has two type of cooling load components; space load and system load. Space load components comprise of:

Wall Conduction

Glass Conduction

Roof Conduction

Glass Solar

Partition Conduction

Infiltration

People

Lights

Equipment

Floor

System load components comprise of:

Plenum heat pickup

Fans

Ventilation

As discussed in the psychrometric clinic, space load components are totaled in order to determine the airflow to the zone. Space load plus system load components are totaled in order to determine the total load to the coil.



Figure 5. Load Components

Conduction Conduction through a shaded surface is defined as:

Where: Q = heat gain by conduction, Btu/hr U = overall heat-transfer coefficient of the surface, Btu/hr • ft2 • °F A = area of the surface, ft2 ΔT = dry-bulb temperature difference across the surface, ºF The overall heat transfer coefficient is also called the U-factor. The U-factor is the rate at which heat will be transferred through the structure. A wall is typically composite in nature, consisting of multiple layers of materials. A wall generally consists of multiple layers of materials, each of which has an associated thermal resistance (figure 6).

Figure 6. U-factor



The overall U factor is the inverse of the sum of the thermal resistance of each of these layers.

Where: R = Resistance, hr • ft2 • °F/ Btu For the wall in figure 5. The overall U factor would be

All surfaces also have an indoor and outdoor surface resistance. AHRAE gives the R values of common materials along with indoor and outdoor surface resistance.

Example 1 Find the overall heat transfer coefficient for the wall in figure 7.

Figure 7. Example 1

Solution: First, we must sum all of the R values along with the R values of the inside and outside surface resistance. Inside and outdoor surface resistance values as given by the 1997 ASHRAE Handbook—Fundamentals and ASHRAE’s Cooling and Heating Load Calculation Principles manual are .68 and .25 hr • ft2 • °F/ Btu respectively.



Most surfaces are not shaded. They are exposed to direct sunlight during some portion of the day. The sun imparts radiant energy on opaque or non-transparent surfaces. The amount of energy depends on the color of the surface, smoothness of the surface and the angle at which the sun’s rays impact the surface. The radiant energy from the sun is at a maximum when the rays strike at an angle that is 90o from the surface. The angle at which the sun’s rays strike the surface depend on the time of day, the latitude and the month of the year. In addition, a wall has the capacity to store heat. The amount of heat storage is a function of the density and width of the wall. The greater the storage capacity of the wall, the longer the time delay or time lag (figure 8) before the heat that impacts the structure enters the space.

Figure 8. Time Lag

The cooling load temperature difference (CLTD) is used to account for the added heat from radiant heat transfer due to the sun shining on exterior walls, roofs, and windows, and the capacity of the wall and roof to store heat. The CLTD is substituted for ΔT in the equation to estimate heat transfer by conduction.

Tables for various wall and roof types, as well as correction factors can be found in the 1997 ASHRAE Handbook—Fundamentals and ASHRAE’s Cooling and Heating Load Calculation Principles manual. Solar Radiation Through Glass Earlier in this clinic, we estimated the heat transferred through glass windows by the process of conduction. However, a large part of the solar heat energy that shines on a window or skylight is radiated through the glass and transmitted



directly into the space. The amount of solar heat radiated through the glass depends primarily on the reflective characteristics of the glass and the angle at which the sun’s rays strike the surface of the glass. The equation used to predict the amount of solar heat gain through glass is:

Where: Q = heat gain by solar radiation through glass, Btu/hr A = total surface area of the glass, ft2 SC = shading coefficient of the window, dimensionless SCL = solar cooling load factor, Btu/hr • ft2 . Factors which affect the solar cooling load factor are:

Time of day

Month

Direction that the window faces

Latitude

Construction of interior partition walls

Type of floor covering

Internal shading devices (if any)

The solar cooling load factor (SCL) is used to estimate the rate at which solar energy radiates into the space, heats up the internal components (furniture, partitions, etc), and is then released into the space. Much like the CLTD, the SCL factor accounts for the capacity of the space to absorb and store heat. The shading coefficient (SC) is used to define how much of the radiant solar energy which strikes the outer surface of the window is transmitted through the window and into the space. The shading coefficient for a particular window is determined by comparing its reflective properties to a standard reference window. A higher shading coefficient represents the ability of a window to reflect more radiant energy. Internal shading devices can reduce the amount of solar energy passing through a window. The effectiveness of such shading devices depends on their ability to reflect incoming solar radiation back through the window. Lighter colored blinds or drapes are more effective than dark colored shading devices. External shading devices act in much the same manner as internal shading devices. External shading devices reflect the incident solar energy and prevent it from entering the space.

Internal Heat Gains

The next component of a load analysis is to account for the internal load gains. Internal load gains consist of the heat introduced by people, lighting, and equipment. People introduce both sensible and latent heat gains. Those gains are a function of activity level and vary with age, physical size, gender and type of clothing. ASHRAE Handbook—Fundamentals gives both sensible and latent heat gains for people based on activity type based on an average person. The equations for calculating the sensible and latent heat gain produced by people are:

Where: QS = sensible heat gain from people, Btu/hr QL = latent heat gain from people, Btu/hr CLF = cooling load factor, dimensionless



Much like the SCL for solar heat gain, the cooling load factor (CLF) is used to account for ability of the space to absorb and store heat. The CLF depends on the construction of the interior partition walls, the type of floor covering, the total number of hours that the space is occupied and the number of hours since the people entered the space. ASHRAE Handbook—Fundamentals give CLF for various space types. Lighting Often, the single greatest internal load producing component is lighting. The equation to estimate the heat gain from lighting is:

Where: Q = sensible heat gain from lighting, Btu/hr 3.41 = conversion factor from W to Btu/hr Watts = total energy input to lights CLF = cooling load factor (dimensionless) Ballast factor = 1.2 for fluorescent lights, 1.0 for incandescent lights Equipment Equipment can be a very large contributor to internal heat loads. With the proliferation of the use and increased processing powers of personal computers, computer loads can be as high as 1.5 watts/sq-ft. Additional equipment such as printers, coffee makers, etc can additionally contribute load. ASHRAE Handbook—Fundamentals gives both sensible and latent heat gains for various types of equipment loads. All equipment loads must be summed and accounted for when calculating the load. Infiltration In buildings, air can leak into and out of the building through cracks around windows, door and small cracks in the building exterior. The direction of air leakage (into or out) will depend on space pressurization and wind direction. Air leaking into a space is called infiltration. Air that leaks into the space can introduce both sensible and latent loads. Three Methods Infiltration Load Calculation

Air change method

Crack method

Effective leakage-area method

For the purposes of this discussion, we will focus on the air change method. The air change method is the easiest but least accurate method. The air change methods uses ASHRAE tables to estimate the expected number of air changes per hour based on building construction. The infiltration airflow is then calculated as:

The sensible and latent load introduced by infiltration then be calculated as:

Where:

Infiltration airflow = quantity of air infiltrating into the space, cfm Volume of space = length × width × height of space, ft3



Air change rate = air changes per hour 60 = conversion from hours to minutes ΔT = design outdoor dry-bulb temperature minus the desired indoor dry bulb temperature, oF ΔW = design outdoor humidity ratio minus the desired indoor humidity ratio, grains of water/lb of dry air The crack method is based upon the average quantity of air known to enter through cracks around windows and doors when the wind velocity is constant. The effective leakage area method takes wind speed, shielding, and “stack effect” into account. The effective leak method should be used when calculating infiltration for high rise buildings. System Loads System loads are loads that impact the refrigeration coil but do not directly impact the space. System loads include ventilation loads, plenum/return duct heat pickup and fan heat gain. Outdoor air is used to remove contaminants from the indoor air. Outdoor air introduced through the buildings HVAC system is called ventilation. ASHRAE 62.1 provides equations to determine the quantity of ventilation air based on space type and occupant density. This air must typically be cooled and dehumidified before it can be provided to the space. The cooling and dehumidification of the ventilation air provides an additional system load to the analysis. The

equations for determining the system ventilation load are:

. Where:

Ventialtion airflow = quantity of air infiltrating into the space, cfm ΔT = design outdoor dry-bulb temperature minus the desired indoor drybulb temperature, oF ΔW = design outdoor humidity ratio minus the desired indoor humidity ratio, grains of water/lb of dry air Fans cans be another significant contributor to system heat. The amount of system heat is a function of the fan efficiency, motor efficiency airflow and external static pressure. Supply fans can introduce system load either before the refrigeration coil or after the refrigeration coil depending on the type of system. In blow through systems, fan heat is added before the coil. In draw through systems, fan head is added after the coil. The position of the fan relative to the coil (blow through or draw through) will affect the psychrometrics and thus system airflow determination. Exhaust fans will additionally add heat to the system, whereas return fans reject all of their heat outside the system and thus do not add additional system heat. The equation for determining the amount of heat a fan introduces to a system is:

Where: 745.7 = conversion from watts to bhp 3.41 = conversion from btu’s to watts Fan motor bhp = actual fan motor brake horse power required to drive the fan, bhp Lastly, duct heat gain must be accounted for. Typically, supply duct heat gain is minimal. Supply ductwork is often insulated, minimizing potential heat gain. Return systems may be either ducted or use the return plenum. Open plenum return systems typically experience some degree of heat pickup from lighting and/or roof loads. Supply duct and return system heat pickup should be accounted for during a load analysis.



Load Analysis Finally, a load analysis can be performed. First, we must summarize all of our load components. Those load components are summed as they impact the space load and system loads (figure x).

Load Component Space Load System Load

Conduction through walls, roof & windows

X X

Solar radiation through windows X X

Conduction through ceiling, interior partitions walls and floor

X X

People X X

Lights X X

Equipment & Appliances X X

Infiltration X X

Fans X

Ventilation X

Supply and Return Duct/Plenum X

However, the load components will fluctuate with the time of day. For example, roof conduction for a flat roof will typically peak at the same time the ambient temperature peaks (generally somewhere between 3pm and 6pm). Solar radiation will peak when the incident angle from the sun is at a maximum angle relative to the exposed window or wall. An eastern facing window for example will experience its maximum solar radiation exposure in the morning. Conversely, a western facing window will experience its maximum solar radiation exposure in the afternoon. So, the question becomes when does the particular zone experience its maximum load? Is it at 9 am, 12 pm, 3 pm or some other time (figure 9)?

Figure 9. Time Of Peak Cooling Load

The answer to the question is determined by summarizing all of the load components at one hour intervals during the occupied hours of the building. This is the method employed by computer load analysis tools. However, due to the time required to perform an hourly analysis, this is impractical for a hand calculated load analysis. If we are going to manually calculate a building load, we will sum the loads at 9am, 12pm, 3pm and 6 pm. After the space and system load peaks have been determined along with its associated peak time of day, a psychrometric analysis can be performed (please see Clinic One for a review of psychrometrics). As discussed in the psychrometric clinic, the intersection of the cooling coil curve with the sensible heat ratio (SHR) line determines the cooling supply air temperature. The SHR is determined from the cooling load analysis. Recall that:



The sensible and latent heat gains used in the SHR equation are determined from the load analysis. Having determined the peak load and SHR, the cooling airflow and cooling coil discharge air temperature can be calculated. Block Load Versus Sum Of The Peaks Often times, a single air handler will serve multiple spaces. The example in figure 10 shows a building with two rooms. Room 101 has a wall with a window with a western exposure and room 102 with a wall and a window with an eastern exposure.

Figure 10

Room 101 with the western exposure will likely peak in the afternoon. Room 102 with the eastern exposure will likely peak in the morning. Assuming the loads shown in table 2:

Room 101 102 Total

9 am 5 tons 8 tons 13 Tons

6 pm 10 Tons 5 Tons 15 Tons

The peak load occurs at 6 pm and totals 15 tons. This is known as the block load. The block load is the single instance of time in which the loads are the greatest. If the air handler was a variable air volume (VAV) unit, it would be sized for the block load. It would only need to deliver the worst case block load, or 15 tons. However, if the air handler was a constant volume unit, it would need to be able to deliver the equivalent of 8 tons of airflow to room 102 during all hours of operation and 10 tons of airflow to room 101 during all hour of operation. This is because the supply fan cannot unload or deliver a reduced quantity of airflow. Thus the fan would need to be sized to deliver 18 tons (10 tons for 101 and 8 tons for 102). The air handler fan would be sized to deliver the sum of the peaks airflow. The sum of the peaks is defined as the peak load delivered to each space regardless of the time in which it occurs. The refrigeration or coil load would still be sized for the block load. Similarly, regardless of system type, ductwork must always be sized for the sum of the peaks. This is why VAV air handler size can typically be reduced when compared to constant volume air handlers. Additionally, the discussion of block load versus sum of the peak loads underlines the importance of placing spaces with similar load profiles together on constant volume systems. Placing spaces with vastly differing load profiles on the same constant volume air handler can waste unnecessary energy and create control and comfort related issues.



Example 2 Single Space Cooling Example

Open-plan office space located in a single-story office building in Reno, Nevada

Floor area = 45’ x 60’

Floor-to-ceiling height = 15’ (3’ Plenum)

Ducted Return

Desired indoor conditions = 75oF dry bulb, 50% relative humidity during cooling season

West-facing wall, 15’ height x 45’ long, constructed of 4” heavy weight concrete block with 1” stucco siding, 3” insulation and ¾” gypsum board on the inside

Eight double glazed 1/8” clear, fixed (e=0.2 on surface 3, ¼” airspace) windows mounted in aluminum frames with thermal break. Each window is 4’ wide x 5’ high.

Flat, 45’ x 60’ roof constructed of steel deck with 3.64” insulation and an acoustic tile drop ceiling (3’ plenum).

Space is occupied from 8:00 a.m. until 5:00 p.m. by 14 people doing moderate active work.

Fluorescent lighting in space = 2 W/ft2

Computer and office equipment in space = 0.5 W/ft2 plus one coffee maker. Room 101 is surrounded by spaces that are conditioned to the same temperature as this space. Find the space load at 5 PM. Solution: First, we should find the climatic data for Reno. From ASHRAE table 1B (1997 Fundamentals).

0.4% 1% Range of DB

DB MWB DB MWB

95 61 92 60 35.5

All values for conduction must be corrected for actual design conditions. For example, at the bottom of Table 32 (ASHRAE 1997 Fundamentals) Notes 2 reads:

Note that two sets of data are listed. One set of data reads 0.4%. The second reads 1%. These numbers denote the percentage of hours out of the year that the design conditions exceed the listed temperatures. For example, the 0.4% column list a dry bulb of 95oF and a wet bulb of 61oF. Thus, 0.4% of the time (out of a year consisting of 8760 hours per year) the temperature exceeds 95oF dry bulb and 61oF wet bulb. For this example, we will calculate the load at 0.4%. Calculating the design conduction correction factor:

Substituting tm:



Next, we should calculate the conduction through the west facing wall. Recalling that conduction is:

We need to find the heat transfer coefficient for the wall. The resistance of each of the materials is shown in the table below (ASHRAE 1997 Fundamentals Table 11).

Material Resistance

Outside Surface Resistance .33

1” Stucco .21

4” Heavy Weight Concrete (C3) .71

3” Insulation 1.19

¾” Gypsum .15

Inside Surface Resistance .69

Calculating the resistance of the wall:

Calculating the heat transfer coefficient:

Before we can determine the CLTD of the wall, we must determine the wall type. ASHRAE table 33A (1997 Fundamentals) lists a stucco wall with a principle wall material of 4” heavy weight concrete (C3) at a wall type 2. Having found the wall type, we enter table 32 (ASHRAE 1997 Fundamentals) for wall number 2. A west facing wall a 5 PM has a CLTD of 59. Solving for the corrected CLTD for the west wall:

Solving for the conduction through the west wall:

Note that we are assuming that 100% of the wall load is being transferred to the space. This is a conservative assumption based on the knowledge that some of the wall conduction load will be transferred to the plenum (we have a 3’ plenum). If the system was an open return system, we would assume that 3’ of the conduction would be transferred to the plenum load and not the space load. Next, we will move to conduction for glass. ASHRAE Table 5 (2005 Fundamentals) lists the U factor ¼” airspace double coated, e=.2, fixed operation, aluminum with thermal break windows as having a heat transfer coefficient of 0.54. Table 34 (AHSHRAE 1997 Fundamentals) lists the CLTD’s for Conduction through glass. At 5 PM, the CLTD is 13. However, we must correct the CLTD according to the conduction correction factor found in the previous example.



Solving for glass conduction:

Having solved glass conduction, we next find the glass solar load. In order to find glass solar, we must first find the shading coefficient (SC) and solar cooling load (SCL) for the west facing glass. The SC can be found from ASHRAE table 11 (1997 Fundamentals). For double glazed clear 1/8” e=.2 on surface 3,, the SC is 0.81. Like CLTD, SCL accounts for the space’s capacity to absorb heat. In order to determine the zone type based on heat capacity, we examine ASHRAE table 35B. For a space with three interior walls constructed primarily of gypsum, no inside window shades and carpeted floors, the glass solar zone type is A. Entering ASHRAE table 36 (1997 Fundamentals) for zone type A, a west wall at 5 PM has a SCL of 192. Solving for glass solar:

Moving on to roof conduction, we first must find the roof R value. Using a procedure similar to that of the wall, we find the total resistance of each material (ASHRAE 1997 Fundamentals Table 11).

Material Resistance

Outside Surface Resistance .33

Steel Deck 0

3.64” Insulation 12.0

Air Space 1.0

Acoustic Tile 1.79

Inside Surface Resistance .69

Solving for the resistance of the roof:

Calculating the heat transfer coefficient for the roof:

The cooling load temperature differences for flat roofs is ASHRAE table 30 (1997 Fundamentals). However, like walls, we must account for the roofs ability to absorb heat. Table 31 (ASHRAE 1997 Fundamentals) allows us to account for the heat absorbing capability of the wall. For a roof constructed primarily of steel deck, R=15.81, with the mass evenly placed and a suspended ceiling is a roof type 2. Entering table 30, we get a roof CLTD of 80 at 5 PM. Correcting the CLTD:

Solving for the roof load:

Note that this assumes that the system is a ducted return with no ceiling plenum return air. In this example 100% of the roof load is assumed to be conducted to the space. In most instances, some percentage of the roof load (especially if it



is an open plenum return system) is picked up by the return air. In those situations, we generally estimate the percentage of load that is conveyed to the space and the load that is picked up by the return air. Having solved the wall, roof, glass conduction and glass solar loads, all that remains is to find the internal loads. The first internal load we calculate is the internal load. Recall that the lighting load is:

The ballast factor for fluorescent lights is 1.2. Based on a space that is 2700 sq-ft @ 2 watts/sq-ft, the lighting load is:

Note, we assumed a cooling load factor (CLF) for lighting of 1.0. Generally, the a CLF of 1.0 is a safe assumption for lighting and people based on the fact that the heat absorbed by the space will eventually have to be picked up by the cooling equipment. Next, we should find the load for people. Again, we will assume a CLF of 1.0. ASHRAE table 3 (1997 fundamentals) gives the rates of heat gain from occupants. For moderately active office work, a typical adult male emits 250 btus/hr sensible and 200 btu’s/hr latent. For a space with 14 people, the people load is:

Finally, we must account for miscellaneous office equipment. We have computers and office equipment which consume .5 watts/sq-ft. The miscellaneous computers and office equipment load is:

We have one last load contributing device. That is the coffee maker. ASHRAE table 8 (1997 Fundamentals) gives the recommended heat gain from restaurant equipment. A coffee maker gives off approximately 1500 btu/hr sensible and 790 btu/hr latent. Finally, summarizing and totaling all the space load components gives us:

Load Component Sensible Load (btu/hr) Latent Load (btu/hr)

Wall Conduction 8,382

Window Conduction 713

Window Radiation 24,883

Roof Conduction 12,800

Lights 22,097

People 3,500 2,800

Misc Computers & Equip 4,604

Coffee Maker 1,500 790

Total 78,479 3,590

Note that we did not solve for infiltration. The ACCA gives tables for calculating the air changers per hour based on building construction and wind speeds. For a building with low exhaust CFM and tight construction, there is little to no infiltration and thus no infiltration load. However, if the building is of looser construction, infiltration should be accounted for.



Weather Design Conditions Design Conditions For Reno, NV (ASHRAE 2005 Fundamentals



People, Materials, and Appliance Heat Gains/Properties Rates Of Heat Gain from Occupants (ASHRAE 1997 Fundamentals, Chapter 28, Table 3)

Heat Gain for Appliances (ASHRAE 1997 Fundamentals, Chapter 28, Table 5 & 6)



Thermal Properties Of Layers (ASHRAE 1997 Fundamentals, Chapter 28, Table 11)



Roofs Roof Group Numbers & CLTD (ASHRAE 1997 Fundamentals, Chapter 28, Table 30 & 31)



Walls Wall Types Mass Located Inside Wall (ASHRAE 1997 Fundamentals, Chapter 28, Table 33A)



Wall Types Mass Evenly Distributed (ASHRAE 1997 Fundamentals, Chapter 28, Table 33B)



Wall CLTD (ASHRAE 1997 Fundamentals, Chapter 28, Table 33B)



Wall Types Mass Located Inside Wall (ASHRAE 1997 Fundamentals, Chapter 28, Table 33A)



Glass/Fenestration U-Factors for Various Fenestration (ASHRAE 2005 Fundamentals, Chapter 29, Table 4)



Shading Coefficient (ASHRAE 1997 Fundamentals, Chapter 29, Table 11)



CLTD for Conduction through Glass (ASHRAE 1997 Fundamentals, Chapter 28, Table 34)

Zone Types for Use with SCL (ASHRAE 1997 Fundamentals, Chapter 28, Table 35)



Zone Types for Use with SCL (ASHRAE 1997 Fundamentals, Chapter 28, Table 35)



Infiltration Infiltration Models (US Department Of Energy, Table 3, Table 4, Table 5)

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