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:
Where
Q = rate of heat transfer
k = conductivity (Btu ft / hr ft2 °F)
A = surface area (ft2)
DT = temperature difference between the inside and outside air (F)
Dx = 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 Conductivity
Conduction 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 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 DT, and inversely proportional to the thickness of the material, Dx. The constant of proportionality is called the conductivity, k, which is a property of the media.
/where k = constant of proportionality = conductivity
or
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:
k = conductivity (Btu ft / hr ft2 °F)
U = k / Δx = conductance (Btu / hr ft2 °F)
R = 1 / U = Δx / k = thermal resistance (Btu / hr ft2 °F)-1
Conduction through Composite Walls
Most 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.
Thus, the rate of heat transfer can be written as:
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 °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.
Conduction through Parallel Paths
When 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:
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:
Q = Qwin + Qwall
= UAwinΔT + UAwallΔT
= (UAwin + UAwall) ΔT
= [[(Awin / Rwin) + (Awall / Rwall)] ΔT = (AT / RT) ΔT
where and
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 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:
FAwall = 50 ft2 / 80 ft2 = 0.625
FAwin = 10 ft2 / 80 ft2 = 0.125
FAdoor = 20 ft2 / 80 ft2 = 0.25
The thermal resistance of the wall is:
The total heat transfer is:
Effective Convection Coefficients That Include Radiation
Hot 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) + s A e (Ts4 – Ta4)
where h is the convection coefficient, s is the Stefan-Boltzman constant (0.1714 · 10-8 Btu/ft2-hr-R4, or 5.67 · 10-8 W/m2-K4), e 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 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) + s A e (Ts4 – Ta4) = 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 [DT]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)
Radiation heat loss from the wall is:
Q/A = s e (Ts4 – Ta4) = 0.1714 · 10-8 (Btu/ft2-hr-R4) 0.9 [(75+460)4 – (50+460)4]
Q/A = 22.02 (Btu/hr-ft2)
The total heat loss from the wall is:
Q/A = 13.74 (Btu/hr-ft2) + 22.02 (Btu/hr-ft2) = 35.76 (Btu/hr-ft2)
The effective convection coefficient including radiation is:
h’ = Q/A / (Ts – Ta) = = 35.76 (Btu/hr-ft2) / (75 F – 50 F) = 1.43 (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:
EXTERIOR / h’ / R’ = (1/h’)Summer / 4.0 (Btu/hr-ft2-F) / 0.25 (hr-ft2-F/Btu)
Winter / 6.0 (Btu/hr-ft2-F) / 0.17 (hr-ft2-F/Btu)
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:
INTERIOR / h’ / R’ = (1/h’)Warm floor or cold ceiling / 1.64 (Btu/hr-ft2-F) / 0.61 (hr-ft2-F/Btu)
Vertical wall / 1.47 (Btu/hr-ft2-F) / 0.68 (hr-ft2-F/Btu)
Warm ceiling or cold floor / 1.09 (Btu/hr-ft2-F) / 0.92 (hr-ft2-F/Btu)
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 Values
A 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 Walls
These values and heat transfer equations can be used to calculate heat loss and gain through walls.
Example
Calculate Rtotal = RT for typical stud wall. Assume winter conditions.
Rhi = .68 (Btu / hr ft2 °F)-1
Rgypsum = .45
Rstud = () 0.833 = 3.02
Rinsulation = () 3.2 = 11.6
Rsheathing = 1.32
Rsiding = .81
Rho = .17
Rstud path = Rhi + Rgypsum + Rstud + Rsheathing + Rsiding + Rho = 6.45 (Btu / hr ft2 °F)-1
Rinsulation path = Rhi + Rgypsum + Rinsulation + Rsheathing + Rsiding + Rho = 15.03 (Btu / hr ft2 °F)-1
then
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 Area
From 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.