Energy Efficient Buildings Heat Gain/Loss through Walls Introduction To reduce heating and cooling energy use, energy efficient buildings seek to minimize conductive heat gain and loss through a building’s walls, windows, ceiling and floor. To understand how to design and construct more energy efficient walls, it is necessary to understand conduction heat loss through walls. Thus, this chapter begins by presenting the equations for steady- state conduction heat transfer. The conduction equations are used to show to importance of minimizing wall area and thermal breaks and maximizing thermal resistance in order to reduce heat gain and loss. Finally, the use of vapor and moisture barriers to control moisture migration through walls is discussed. Principles of Energy-Efficient Walls Energy Balance Approach The equation for steady-state conduction through walls is: Q=k A ΔT Δx =k A Th - Tc Δx Where Q = rate of heat transfer k = conductivity (Btu ft / hr ft 2 °F) A = surface area (ft 2 ) T = temperature difference between the inside and outside air (F) x = wall thickness (ft) Following this equation, the energy efficiency opportunities are: Reduce wall area Reduce the conductivity of the wall Increase the thickness of the wall 1
Transcript
Energy Efficient BuildingsHeat Gain/Loss through Walls
IntroductionTo reduce heating and cooling energy use, energy efficient buildings seek to minimize conductive heat gain and loss through a building’s walls, windows, ceiling and floor. To understand how to design and construct more energy efficient walls, it is necessary to understand conduction heat loss through walls. Thus, this chapter begins by presenting the equations for steady-state conduction heat transfer. The conduction equations are used to show to importance of minimizing wall area and thermal breaks and maximizing thermal resistance in order to reduce heat gain and loss. Finally, the use of vapor and moisture barriers to control moisture migration through walls is discussed.
Principles of Energy-Efficient Walls
Energy Balance ApproachThe equation for steady-state conduction through walls is:
Q=k A ΔTΔx
=k A Th - TcΔx
Where
Q = rate of heat transferk = conductivity (Btu ft / hr ft2 °F)A = surface area (ft2)T = temperature difference between the inside and outside air (F)x = wall thickness (ft)
Following this equation, the energy efficiency opportunities are:
Reduce wall area Reduce the conductivity of the wall Increase the thickness of the wall
This chapter uses engineering principals for quantifying heat loss through walls as guides to improving the energy efficiency of walls.
Calculating Heat Transfer through Walls
A Conceptual Look and Conduction and ConductivityConduction is the transfer of heat from through a stationary media. On a molecular scale, temperature is a measure of the vibrational energy of atoms. Atoms in high temperature
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molecules vibrate more than the atoms in low temperature molecules. Conduction heat transfer is the transfer of this vibrational energy from high temperature molecules to low temperature molecules. Thus, heat is always transferred from high to low temperature.
In solids, molecules are densely packed and held in lattice structures by molecular bonds. Thus, vibrational energy is easily transferred from high to low temperature molecules. As a consequence, solids have high thermal conductivity. In liquids and gasses, the intermolecular bonds get weaker and the density of molecules get smaller. This makes it harder to transmit vibrational energy from high to low temperature molecules. Thus, liquids and gasses generally have progressively lower conductivities.
It follows that conductivity is generally a function of density. This is the reason people are relatively comfortable in 50 F air, but will become hypothermic in 50 F water. When this thinking is applied to building materials, dense materials like glass and concrete have high thermal conductivity and low thermal resistance. They make poor insulators. Low density materials like foam board or fiberglass batts have low conductivity and make good thermal insulators.
Fourier’s Law of Conduction Fourier’s law of conduction states that the rate of heat flow, Q, through a solid media is proportional to the area, A, and temperature difference T, and inversely proportional to the thickness of the material, x. The constant of proportionality is called the conductivity, k, which is a property of the media.
Q.
∝A ΔTΔx
Q.=k
A (T2−T 1 )x2−x1
where k = constant of proportionality = conductivity
Q=−k A ΔTΔx or
Q=−k A dTdx
In the equation for heat flow, the negative sign can be omitted if ΔT is defined so that heat flows from high temperature, Th, to low temperature, Tc. Thus, conduction heat transfer can be calculated as:
Q=k A ΔTΔx
=k A Th - TcΔx k = conductivity (Btu ft / hr ft2 °F)
Q=U A ΔT=U A (Th - Tc ) U = k / Δx = conductance (Btu / hr ft2 °F)
Q=A ΔTR
=A ( Th - Tc )
R R = 1 / U = Δx / k = thermal resistance (Btu / hr ft2 °F)-1
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Conduction through Composite WallsMost building structures such as walls serve multiple purposes including providing structure, presenting an appealing appearance, retarding vapor and moisture transmission, and providing thermal resistance. To serve these multiple purposes, walls are typically composites of multiple materials. These materials are typically arranged such that heat transfer follows both serial and parallel paths.
Conduction through Serial Paths When heat is transferred through materials in a serial arrangement, the total thermal resistance is the sum of the thermal resistances. These thermal resistances should include the thermal resistances of the convection coefficients on each side of the structure. The figure below shows the temperature profile through a wall from a warm inside air temperature, Ti, to a cooler outside air temperature To. The temperature decreases as a result of the thermal resistances of the convection coefficients and wall materials.
In steady-state conduction, the rate of heat transfer between each set of temperature nodes must be equal.
Q=AT i−Twi
Rhi=A
Twi−Tmid
RA=A
Tmid−Two
RB=A
Two−T oRho
Thus, the rate of heat transfer can be written as:
Q=AT i−T oRT where RT = Rhi + RA + RB + Rho
Example
Consider a 60 ft2 wall made of two materials positioned such that heat transfer through the wall occurs in series through both materials. The thermal resistances of the materials are 2.18 hr ft2
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°F /Btu and 0.45 hr ft2 °F /Btu. The thermal resistance of the convection coefficients on interior and exterior surfaces are 0.68 hr ft2 °F /Btu and 0.17 hr ft2 °F /Btu. The inside air and outside air temperatures are 72oF and 50oF. Calculate the heat flow rate through the wall.
RT=Rhi+RA+RB+Rho=0.68+2 .18+0 . 45+0 .17=3 . 48 hrft2 o F/Btu
Q=AT i−ToRT
=6 0 ft2×72 o F−50 o F3 .48 hrft2 o F/Btu
=379 Btu/hr
Conduction through Parallel PathsWhen heat flows through parallel paths, the total thermal resistance can be formulated in terms of the area fractions and thermal resistances of the paths. Consider the following example of parallel heat flow through a wall with a window.
The total rate of heat transfer, Q, is:
Q=
ATRT
(T i−T o)
The total resistance RT can be found in terms of the area fractions, FA, and thermal resistances, R, of the parallel path by noting that:
Parallel paths should be joined at nodes with common temperatures. In most cases this means that parallel paths should be joined at the nodes representing the air temperatures on either side of the structure, since the air is typically well mixed. Combining the nodes at common
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materials in the structure is less accurate because of two-dimensional heat transfer through the common materials.
Example
Consider a wall with gross surface area of 80 ft2 that includes 50 ft2 of wall area, 20 ft2 of door area and a 10 ft2 of window area. The thermal resistance of the wall, door and windows are 15 hr ft2 oF/Btu, 5 hr ft2 oF/Btu and 2 hr ft2 oF/Btu, including convection coefficients. The inside air and outside air temperatures are 72 oF and 50 oF. Calculate the total thermal resistance of the wall including the door and window, and the heat flow rate through the wall.
The area fractions of the wall, window and door are:
Effective Convection Coefficients That Include RadiationHot surfaces lose heat to the surroundings via convection and radiation. The equation for heat loss, Q, to the surroundings at Ta, from a hot surface at Ts, with area A is:
Q = h A (Ts – Ta) + A (Ts4 – Ta
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where h is the convection coefficient, is the Stefan-Boltzman constant (0.1714 · 10-8 Btu/ft2-hr-R4, or 5.67 · 10-8 W/m2-K4), is the emissivity of the surface. Emissivity is the ratio of the radiation emitted from a surface at a given temperature to the radiation emitted by a black surface which emits the maximum radiation possible at the same temperature. Thus, emissivity ranges from 0 to 1.0, and radiation heat loss increases as the emissivity increases. The thermal emissivity of most building materials is between 0.90 and 0.95. To reduce radiation heat loss
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surfaces can be painted with aluminum paint that has as an emissivity of about 0.55 or covered with aluminum foil type material that has as emissivity of about 0.10.
In building applications, the temperature differences between walls and the surrounding environment and air are typically small. In these cases, the heat loss from convection is approximately equal to the heat loss from radiation. To simplify calculations, an effective coefficient, h’, that includes heat loss from both radiation and convection is often used where:
Q = h A (Ts – Ta) + A (Ts4 – Ta
4) = h’ A (Ts – Ta)
Example
Consider a wall with surface temperature of 75 F and emissivity of 0.90. The temperature of the air and surroundings is 50 F. Calculate the effective convection coefficient h’.
The natural convection coefficient for air in contact with a vertical wall can be estimated as (ASHRAE Fundamentals, 1989):
h =0.19 [T]0.33 (Btu/hr-ft2-F) h =0.19 [25]0.33 (Btu/hr-ft2-F) = 0.550 (Btu/hr-ft2-F)
Convection heat loss from the wall is:
Q/A = h (Ts – Ta) = 0.550 (Btu/hr-ft2-F) (75 F – 50 F) = 13.74 (Btu/hr-ft2)
The value of the convection coefficient, h, depends on whether the air movement over the surface is forced or natural due to the buoyancy of air. For exterior walls and structures, wind causes forced convection and the convection coefficient is a function of wind speed. Typical values for effective exterior convection coefficients including radiation are:
For interior surfaces, the natural convection coefficient is a function of the geometry and the temperature difference between the surface and air. For example, consider a warm floor. The air near the wall expands and rises as it is warmed by the floor. Air rising off the floor encounters no friction from the floor and hence rises quickly. The relatively high speed causes the convection coefficient to be high. Next, consider a warm vertical wall. The air near the wall expands and rises as it is warmed by the wall. However, air rising along the wall encounters some friction resistance from the wall and hence rises at a moderate rate with a moderate convection coefficient. Finally, consider air under a warm ceiling. The air near the ceiling expands and rises as it is warmed. However, the ceiling impedes movement and the air is largely stagnant. Thus, the convection coefficient is negligible.
By analogy, cold ceilings also cause adjacent air to cool and drop with relatively high velocities and high convection coefficients. Similarly, cold floors cause adjacent air to pool with low convection coefficients. Typical values for effective interior convection coefficients including radiation are:
Some envelope structures include air spaces. The total thermal resistance of an air space is a function of the convection coefficients and emissivities on the inside walls of the space. As before, convection coefficients depend on the geometry. Typical values for effective thermal resistances, including radiation, across air spaces are:
AIR SPACES R’ = (1/h’)Heat flow up 0.87 (hr-ft2-F/Btu)Horizontal 1.01 (hr-ft2-F/Btu)Heat flow down 1.02 (hr-ft2-F/Btu)
Thermal Resistance ValuesA handy compilation of thermal resistances for building energy applications is shown below. The thermal resistances for surface air films listed below include radiation heat loss; thus radiation heat loss does not need to be explicitly calculated when using these coefficients. The modifications of the surface air film values to include radiation assume that the surroundings are at the same temperature of the air. However, in cases such as heat transfer from a roof to a clear night sky, radiation heat transfer may be much larger than assumed by the modified surface air film coefficients. Thus, in these cases, it may be useful to explicitly calculate convection and radiation heat loss individually rather than by using a modified surface film coefficient. ASHRAE Fundamentals includes an even larger set of thermal properties for most building materials.
Source: Mitchell, J., 1983, “Energy Engineering”, John Wiley and Sons, Inc.
Summary: Calculating Steady State Heat Transfer Through WallsThese values and heat transfer equations can be used to calculate heat loss and gain through walls.
Note that the parallel paths are joined at the air temperature nodes on each side of the wall, and not at the common gypsum or sheathing nodes. This is more accurate, since the temperature of the gypsum and sheathing over the insulation and studs is not identical.
Reducing Heat Loss by Reducing Exterior Wall AreaFrom the conduction equation, one of the ways to reduce heat loss through walls is to reduce exterior wall area. Circular buildings have the largest ratio of enclosed area to wall area. Thus, circular buildings like Mongolian yurts require less wall material and experiencing less heat loss through walls than other buildings with the same floor area.
"
853
Ti To
Side view:
Top view: stud
insu
insu
16" O.C.
"
85
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Nominall (2" x 4") stud = 1 5/8 inch x 3 5/8 inch
stud
stud
wood siding
studs and insulation½ inch gypsum board
½ inch insulated board sheathing
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Universal Public Domain
In rectangular construction, squares have the largest ratio of enclosed area to perimeter. Thus, the long narrow houses in the University of Dayton student neighborhood have much greater heat loss per enclosed area than more square buildings.
Irregularities in buildings add architectural interest, but increase the ratio of wall area to enclosed floor area. To create interesting buildings without compromising energy efficiency, architects have extended porches and garages and used multiple room lines on rectangular building envelopes.
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Act 2 Project, Pacific Gas and Electric
Reducing Wall Heat Loss by Increasing InsulationIn existing buildings, uninsulated wall cavities can be filled with foam or cellulose. The R-value of a 4-inch wall cavity increases from R = 1.01 (hr-ft2-F/Btu) to R = 4 x 3.3 = 13.2 (hr-ft2-F/Btu) after filling with cellulose. The cost of cellulose, which is shredded newspaper treated with a fire retardant is quite moderate. To reduce the likelihood of water condensation on the outside of the insulation after passing through the insulation from the building interior, the interior side of exterior walls can be painted with water-resistant paint.
Photos: NREL http://www.nrel.gov/data/pix
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Foam insulation has a higher R-value (3.6-6.5 hr-ft2-F/Btu per inch) than fiberglass batt insulation (3.2 hr-ft2-F/Btu per inch). All closed-cell polyurethane foam insulation made today is produced with a non-CFC (chlorofluorocarbon) gas as the foaming agent. Polyurethane foam is water-vapor permeable, remains flexible, fire resistant and provides good air sealing. Soy-based, polyurethane liquid spray-foam is also available.
Reducing Wall Heat Loss by Minimizing Thermal BridgingA thermal bridge is a component of low thermal resistance that allows heat to bypass components with high thermal resistances. An example of a thermal bridge is a stud in an insulated frame wall. The studs allow heat to bypass the insulation in the wall cavity. Thermal breaks reduce the overall thermal resistance of an envelope component, and should be eliminated or minimized whenever possible.
To demonstrate the importance of minimizing thermal bridging, consider the example below. Both paths of heat loss have equal heat-loss areas, however path 2 is a thermal bridge with 1/10th the thermal resistance of path 1. Intuition may suggest that the overall thermal resistance would be the average of the two paths, R = 5.5, since the two heat loss areas are equal. However, calculating the thermal resistance shows that the overall thermal resistance is R = 1.82. In other words, the overall thermal resistance is dominated by thermal bridges with low thermal resistances.
An even more graphic way of gaining intuition about the importance of thermal breaks is to consider that heat, like water, follows the path of least resistance. Thus, the thickness of the walls of a dam don’t really matter, if there is a break in the dam wall. The water will flow through the break and not the dam walls. In frame walls, thermal bridging due to studs can be reduced by spacing studs farther apart. For example, 2x4 studs are typically placed 16 inches apart, whereas 2x6 studs are typically placed 24 inches apart. Reducing the “framing fraction” from 1.625”/16” = 0.102 to 1.625”/24” = 0.068 reduces thermal bridging. The calculations below shows that walls with 2x6 studs with 24-inch spacing reduce heat loss by 24% compared to walls with 2x4 studs with 16” spacing. Part of the reduction is due to reduced “framing fraction and part to increased wall cavity insulation. Also note that walls with 2x6 studs with 24-inch spacing use only 3% more wood for studs than walls with 2x4 studs with 16” spacing. Thus, the overall material cost of 2x6 studs about the same as 2x4 studs. Energy efficient builders are also finding ways to reduce window and door headers and other framing to reduce thermal bridging.
Steel studs enable more heat loss that wood studs, since metal is about 300 times more conductive than wood. Despite the poor thermal performance of steel studs, they are
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frequently used in multifamily residences due to fire codes. The following table shows ASHRAE correction factors for metal framing. To use the table, multiply the correction factor and the cavity insulation to get an effective cavity insulation that includes the steel stud. The effective cavity insulation is then added to the other thermal resistances of the wall. To mitigate the effect of the thermal bridge caused by the steel stud, insulated sheathing that covers the entire wall should be used.
2" x 8" 16" R-25 0.31 R-7.82" x 8" 24" R-25 0.38 R-9.6
New construction with metal studs and insulated sheathing.
Example
Compare heat loss between a wall with 2x6 wood studs placed 24-inches apart with R-19 insulation and a wall with 2x6 metal studs placed 24-inches apart with R-19 insulation.
According to the table above, the correction factor for a wall with 2x6 metal studs placed 24-inches apart with R-19 insulation is 0.37. The correction factor should be multiplied by the thermal resistance of the cavity insulation to find an effective resistance of the metal studs and cavity insulation. The total thermal resistance of the wall is then calculated as the sum of the all the thermal resistances, including the effective resistance of the metal studs and cavity insulation. 2x6 metal stud w/ fiberglassInputRho (hr-ft2-F/Btu) 0.17Rsid (hr-ft2-F/Btu) 0.81Rpoly (hr-ft2-F/Btu) 7.2Ins thickness (in) 5.625Rins/in (hr-ft2-F/Btu/in) 3.378Rgyp (hr-ft2-F/Btu) 0.45Rhi (hr-ft2-F/Btu) 0.68Correction Factor, CF 0.37CalcsRins (hr-ft2-F/Btu) 19.001Rins,eff (hr-ft2-F/Btu) = Rins x CF 7.030463Rt (hr-ft2-F/Btu) = Rho+Rsid+Rpoly+Rins,eff+Rgyp+Rhi 16.030Ut (Btu/hr-ft2-F) = 1 / Rt 0.062
Thus, the fraction increase in heat loss through a wall with steel studs compared to a with wood studs and identical insulation is:
% Increase = (Ut,metal - Ut,wood) / Ut,wood 65%
For comparison, the correction factor applied to the wood stud wall that would give the same thermal resistance is 0.89.
Structural Insulated Panels (SIPs) completely eliminate framing and the associated thermal breaks. SIPs are constructed by gluing ‘oriented strand board’ (OSB) panels to foam cores. SIPs are constructed in factories and cut to shape before shipping, which speeds onsite assembly. The solid cores also reduce air infiltration.
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(Source: NREL public domain photos)
Typical thermal resistances of SIP walls depend on the type of foam insulation. Expanded polystyrene (EPS) has a lower thermal resistance than extruded polystyrene (XPS).
Reducing Wall Heat Loss by Increasing Wall ThicknessTruss and beam houses reduce framing fraction and allow for thick insulation. The Canadian house below employed truss and beam construction with 11.8-inch thick wall cavities filled with wet-blown cellulose, and won an ASHRAE 1996 Technology award.
Strawbale houses use stacked straw bales for both structural and thermal integrity. The design virtually eliminates thermal breaks and results in thick walls with high thermal resistance and minimal air leakage. .
Reducing Heat Loss by Field Inspection and MeasurementsInfrared cameras can identify under-insulated walls and thermal breaks. The picture below shows an uninsulated wall panel on the left side of the building. It also shows the effectivensess of window shades at reducing heat loss; the upstairs front windows have R=4 window shades while the downstairs windows are uncovered.
The picture below shows where wall cavities have insulated (yellow) and the additional heat loss through the studs (red). The center wall panel is completely uninsulated. In addition, heat is escaping through the attic vents as a result of warm air leaking from the house into the attic.
The picture below shows an insulated wood-frame addition (green) on an un-insulated brick (red) house. The un-insulated brick is much warmer and is losing heat much more rapidly.
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Field Measurement of Thermal ResistanceThe thermal resistance of existing walls, Rw, can be estimated from measurements of outside air temperature, Toa, inside air temperature, Tia, and the temperature of the inside surface, Tis. Consider the diagram shown below.
At steady-state conditions heat flow through the wall equals heat flow from the surface of the wall to the interior air. Solving this energy balance, and substituting an estimate of the thermal resistance of the interior convection coefficient, Rhi, gives a good estimate of the thermal resistance of a wall excluding the interior convection coefficient.
Calculate the thermal resistance of a wall if the outside air temperature is 30 F, the inside air temperature is 70 F, and the temperature of the inside surface if 65 F.
From the table of property data, the thermal resistance of the interior convection coefficient is 0.68 hr-ft2-F/Btu. The thermal resistance of the wall, excluding the interior convection coefficient is:
Wall Construction: Vapor and Moisture BarriersVapor barriers prevent moisture from migrating from the warm humid inside air to dry cold outside air and condensing on exterior side of insulation. Vapor barriers are usually a thin plastic sheet with negligible R-value. In moderate and cold climates, it is important that the vapor barrier be placed inside the insulation, so that water vapor does not diffuse through the insulation and then condense on the exterior side of the insulation.
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In addition to vapor barriers on the inside of the walls, most exterior walls include a moisture barrier just inside of the exterior sheathing. The moisture vapor prevents water from being blown through the sheathing into the insulation cavity during stormy conditions. However, it is important to allow any moisture that does enter the wall to escape. Thus, moisture barriers are typically fabricated to transmit water vapor, while being impervious to liquid water. Moisture barriers also impede infiltration of air through the wall when all joints are taped and sealed.
Vapor barrier on inside of wall insulation. Moisture barrier on outside of wall insulation.
