4D11 Building Physics LECTURE 1. 1 Introduction 2 HEAT TRANSFER IN BUILDINGS

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1 4D11 Building Physics LECTURE 1 Michaelmas 2008 Allan McRobie 1 Introduction The energy we use in keeping the inside of buildings at comfortable temperatures contributes between 40-50% of the world s carbon. Building physics has thus risen rapidly up both the political agenda and the list of engineering imperatives for the coming century. In this course you will first learn the underlying physical principles that govern a building s behaviour, then in each case move on towards understanding current design strategies. We begin, though, with the basic design strategy for low carbon building design: STEP 1: ENERGY DEMAND REDUCTION STEP 2: RENEWABLES There are a host of available strategies for the first step - energy-efficient appliances, good insulation, an air-tightness and/or natural ventilation strategy, heat pumps, intelligent solar shading, thermal storage, etc. Only after the energy demand has been substantially reduced should one begin to look at supplying the required energy with renewables. Government targets (such as supplying 20 percent of power from renewable sources) are far easier to hit after a concerted effort at energy demand reduction. (Note that this general strategy is somewhat at odds with the Solar Decathlon philosophy, where the dominant focus is on Step 2.) 2 HEAT TRANSFER IN BUILDINGS Within the built environment, it is the heating and cooling of buildings that exert the greatest energy demands. The embodied energy in construction materials is typically only a small fraction of the energy that a building will consume over its lifetime. It is therefore worth concentrating on how a building behaves in operation. Before one can optimise or minimise the energy demand, one needs to understand the fundamental physical processes involved. There are three forms of heat transfer - conduction, convection and radiation. We shall consider each in turn. We begin with the simplest of all models of heat transfer in a room, illustrated in Figure 1. The corresponding equation states 1

2 Figure 1: A schematic of heat-transfer in the simplest model of a room that, at Steady State, the heat input required to maintain an internal temperature of T internal is given by Q input = Q fabric + Q vent (1) where Q fabric is the heat lost by conduction through the building fabric (walls, roof, floor) and Q vent is the ventilation heat loss, this equalling the heat required to warm the incoming air from the external to the internal temperature. Since heat flows are a bit like electrical currents, this equation is analogous to Kirchhoff s First Law for currents entering and leaving a junction - in this case a hypothetical node in the interior of the room. We shall develop this electric analogy further as we proceed. In a nutshell, the fabric heat loss is given by Q fabric = ( i A i U i ) T (2) where U i are the U-values (i.e. the thermal transmittances) of the individual walls, floor and ceiling, each of which have corresponding areas A i and T = T i T o is the difference between the inside and outside temperatures. The ventilation heat loss is given by Q vent = c p ṁ T (3) where c p is the specific heat capacity at constant pressure of air, and ṁ is the mass flow rate of the air. 2

3 2.1 Conduction heat loss through a wall Under steady-state conditions the heat flux q (i.e. the flow of heat per unit area) that conducts through a building element made of a single material (such as a concrete wall) is proportional to the temperature difference (T inside T outside ) across the element; inversely proportional to the thickness L of the element. These lead to the steady-state heat conduction equation q = k L (T i T o ) (4) The proportionality constant k is the thermal conductivity, a property of the material. It is clear from the equation that the units of k must be [W/mK]) The ratio k/l (with units W/m 2 K) is known as the thermal transmittance. This is the U-value. When multiplied by the area, we obtain UA, the thermal conductance. The reciprocal of the U-value, L/k, is called the thermal resistance or R-value. U-values are used in the UK and Europe, R-values are used in the US, often in crazy imperial units of degrees Fahrenheit, square feet hours per British Thermal Unit, (ft 2 F h / Btu ). SI units for the R-value are m 2 K/W. The heat flows through the various different surfaces in a room, (the walls, the floor, etc) are obviously in parallel. Q fabric = Q walls + Q floor + Q ceiling (5) = (A walls U walls + A floor U floor + A ceiling U ceiling ) T (6) and so we ADD THE CONDUCTANCES. However for heat flow through a multi-layer wall of different materials, the thermal resistances are in series and we must ADD THE RESISTANCES. This can be readily deduced since at steady state the heat flow is constant through all layers of the wall, but the individual temperature drops must add up to the full temperature drop. T = T si + T 1 + T (7) = q + q + q +... (8) U si U layer1 U layer2 = (R si + R 1 + R )q (9) whence q = 1 T = U T (10) R 1 + R

4 Thus, for unit area of a multilayer wall, the total resistance R tot is where R tot = R si + R 1 + R 2 + R 3 + R so (11) R si, R so are the thermal resistances of thin internal and external surface air layers alongside the wall surfaces, R 1, R 2, R 3 are the thermal resistances of the various layers of the different wall materials. To recap: The inverse of the (unit area) total resistance R tot is the U-value of the wall construction. Since adjacent areas of the wall are in parallel, the conductance of the whole wall is thus AU (where A is the area of the wall). Similarly, the thermal conductance of the whole room is AU, summed over all walls, the roof and the floor. Obviously, if a wall has different constructions over different areas (e.g. some parts are glazed and others are brickwork, say), the summation is over the different regions of each construction. Obviously one prefers LOW U-values. High U-values (of the order of say 3-5 W/m 2 K, say) are bad. U-values less than 1 W/m 2 Kare good. Simply, the U-value is U = Steady state heat flow per unit area temperature difference = Q/A T i T o (12) Thermal bridges Thermal bridges are localised areas within or at the edge of a panel where there is high heat flow. An example might be metal connectors across the cavity between a double skin masonry wall. In analysis, thermal bridges are usually lumped into the U-value for the panel (although if desired they can be included separately via a separate term). In design, one tries to avoid thermal bridges. Not only do they loose heat but having a cold local internal surface temperature can lead to localised mould growth. 4

5 Typical thermal conductivities: Material Thermal Conductivity W/mK Aluminium 160 Plywood 0.14 Expanded polystyrene 0.03 Brick 0.84 Concrete block 0.22 Dense plaster 0.50 Vermiculite 0.03 Concrete slab 1.00 Roofing felt 0.20 Plasterboard 0.16 Wood wool slab 0.10 Clay soil - water content Clay soil - water content Glass (density 2500kg/m 3 ) 1.05 Rock: Marble (2500kg/m 3 ) 2.0 Rock: Slate (2700kg/m 3 ) 2.0 Sheep s wool Thatch (straw) 0.07 Mineral fibreboard For extensive tables of material thermal properties, see CIBSE Guide A. 3 Surface Resistances You will recall from Part 1 Thermofluids lectures that the heat transfer processes that occur at the interface between a gas and a solid have some rather complex physics, and typically involve the formation of convective boundary layers. The nature of the boundary layer depends on such matters as whether it is alongside a vertical wall, above a floor or below a ceiling. It also depends on the direction of the heatflow - i.e. whether the solid is hot compared to the adjacent air, or vice versa. Such processes can be studied in fine detail using the various dimensionless groups learned in Part 1 - the Prandtl, Stanton, Rayleigh, Grashof and Nusselt numbers, for example. However, for our initial purposes the simple empirical values of the heat transfer coefficients tabulated as resistances in the CIBSE Guide A for common geometrical configurations will suffice. (The heat transfer coefficients h i and h o at inside and outside surfaces are TRANSMITTANCES [W/m 2 K]. It is common to quote their reciprocal, the corresponding RESISTANCES). 5

6 3.1 Internal Surface Resistances Surface Direction of heat flow Resistance m 2 K/W Walls Horizontal 0.13 Floors or Ceilings Upward 0.10 Floors or Ceilings Downward 0.17 Roofs (flat or sloping) Upward External Surface Resistances Surface Direction of heat flow Resistance m 2 K/W Sheltered Normal Exposed Walls Horizontal Floors Downward Roofs Upward Ventilation heat loss There is always a dynamic tension in the design of a thermally efficient building between ventilation and heat loss. Fresh air needs to be provided for comfort - not least to keep CO and CO2 levels to acceptably low levels. A minimum flow rate would be 5 litres/sec per person, and recommended levels are higher - around 8 litres/sec per person. Traditionally, builders have relied on the natural leakiness of the building to provide such ventilation - cracks under doors, around window frames and in the building fabric generally. Nowadays, however, in order to minimise heat loss, many energy efficient buildings are adopting a philosophy of build tight - ventilate right. The building envelope is designed to be tightly sealed. Newly-constructed buildings are pressure-tested by means of a large fan attached at the front door. An overpressure of 50 Pa is maintained for 1 hour, and the air-tightness of the building is monitored by seeing how much air passes in through the fan over the hour. The Air Permeability of the building is the hourly air -flow (in m 3 ) per square metre of floor area under these 50 Pa conditions. Results are reported verbally as an Air Permeability of 10, say, meaning that there was a loss of 10m 3 /m 2 over the pressurised hour of the test. An Air Permeability of 10 is a common design standard, but it is still fairly leaky. Sweden has been requiring Air Permeabilities of 3 since Modern eco houses (such as PassivHaus designs) are now aiming for extraordinarily high standards of air-tightness, and Air Permeabilities of 1 and below are beginning to be achieved. To achieve such high air-tightness, one needs specially designed doors. One also needs to eliminate such items as letter-boxes, and cat-flaps are definitely out. The general idea is that with an air-tight building, all ventilation can be 6

7 mechanically controlled to provide ventilation only as and when it is needed by means of a sophisticated MVHR system (Mechanical Ventilation with Heat Recovery). However, there is much current debate about whether such highlysealed boxes may yet prove to be problematic in terms of air quality, dust-mites and mould growth. They may reduce carbon, but they might increase asthma. There is an alternative strategy of having naturally-ventilated buildings. In many circumstances (e.g. retail, various industries, computing, warmer climates) it is cooling rather than heating that is the major contributor to energy demand, and thus allowing external cool air into the building can help limit its carbon/energy demand. Even in cold climates there are low-carbon natural ventilation strategies which stand opposed to the tightly-sealed philosophies. Whatever strategy is adopted, the incoming air on cold days needs to be heated as it enters the building, and since air is comparatively incompressible it follows that an equal quantity of warmed air is simultaneously vented and its hard-won heat content is lost. Various forms of heat-exchangers are possible which extract heat from the exiting air (as in MVHR - Mechanical Ventilation with Heat Recovery). Also, it is possible to pre-heat the incoming air by passing it through underground pipes or labyrinths before it enters the building - the air extracts heat from the large thermal mass of the ground as it enters. Leaving all such advanced concepts aside for the moment, we shall consider the case of un-preheated cold air entering the room and no heat recovery from the warm air on exit. If a mass m of air enters at the external temperature T o the amount of heat required to raise it to the indoor temperature T i is c p m(t i T o ), where c p is the specific heat capacity of air. For a continuous process, the rate at which heat must be supplied is thus c p ṁ(t i T o ), where ṁ is the mass flow rate (in kg/sec). If N ACH is the number of air-changes per hour of the room, the mass flow rate is ṁ = ρn ACH V/3600 (13) where ρ is the density of air (approximately 1.25 kg/m 3 ), V is the room volume (in m 3 ) and 3600 converts the air-changes per hour to air-changes per second. The rate of heat loss through ventilation is thus Q vent = c pρn ACH V (T i T o ) = C v (T i T o ) (14) 3600 where C v = c p ρn ACH V/3600 and is called the ventilation conductance. It has units of W/K, just as does the fabric conductance AU. Taking c p ρ as 1200 J/m 3 K leads to the common expression that C V = (1/3)N ACH V (15) 7

8 Care should be taken with the dimensions - the factor of (1/3) has dimensions such that C v ends up in W/K. 4.1 Summary - The Heat Input Required to Maintain a Steady Temperature Difference To maintain a steady difference between internal and external temperatures, heat must be supplied to the room at the same rate that it is lost through the fabric and through ventilation. Heat must thus be supplied at the rate Q input = Q fabric + Q vent = AU(T i T o ) + C v (T i T o ) (16) Low carbon design aims to minimise the required heat input, thus a typical strategy aims to provide a building with low U-values and low ventilation rates. The above equation essentially allows you to size the boilers and the radiators by considering the conditions on a cold winter day. It tells you little about the overall carbon consumption over the year. For that, you will need to perform some sort of time-stepping transient analysis, probably on a computer, which looks at how often you then have to switch the heating on over the course of a typical year. The equation also tells you little about the behaviour over the course of a summer day - for that you may need to use other approaches, (e.g. the Admittance Method) of which more later. To obtain a first back-of-the-envelope estimate of the average annual energy demand for heating, one could apply the equation assuming say an average temperature difference between inside and outside of say 10 C. (One can only really do this by experience, as it obviously depends on many factors, particularly the local climate). One would also have to estimate and subtract any casual gains -i.e. free heat from other sources such as internal occupants or appliances. 4.2 Example: Philip Johnson s Glass House, Connecticut This is a rather famous rectangular block designed by a famous - but rather unlikeable - architect. The walls are 100 percent glass, and there are no columns supporting the heavy concrete roof. The block is 17m long, 9m wide and 3m high. The walls are single-glazed, and the glazing has a U-value of 5.6W/m 2 K. Although typical for single glazing, this is a high figure for a wall. Energy efficient buildings seek to limit heat wastage by providing low U-values - e.g. significantly less than 1W/m 2 K for a well-insulated wall. Current regulations (e.g. Part L of the UK Building Regulations) typically require U-values of 0.45W/m 2 K and Best Practice pushes for U-values less than 0.2W/m 2 K. 8

9 The roof is 150mm concrete with 12mm plasterboard internally and 12mm of felt externally. The floor is assumed to have a U-value of 0.76 W/m 2 K. The exposures of the external surfaces may be assumed to be Normal. The U-value of the multilayered roof EXAMPLE QUESTION 1. Find and correct the mistake in the following calculation of the roof U-value. The thermal conductivities of the roofing materials are listed in the earlier table. They are 0.16, 1.00 and 0.20 W/mK for plasterboard, concrete slab and roofing felt respectively. For upward heatflow through the roof, surface resistances are R si = 0.10 internally and R so = 0.02 externally. The various layers of the roof are in series, thus resistances are added, viz: Layer (m 2 K/W ) R si Internal Surface Resistance R 1 Plasterboard L/k = 0.012/ R 2 Concrete Slab L/k = 0.150/ R 3 Roofing Felt L/k = 0.012/ R so External Surface Resistance R = The U-value is the reciprocal of the total (unit area) resistance, thus U roof = 1/ R = 1/0.405 = 2.47 W/m 2 K (17) The total fabric conductance for the house The walls, roof and floor are in parallel, and thus conductances UA are added. AU = Uglazing A walls + U roof A roof + U floor A floor (18) = (5.6)( ) + (2.47)(153) + (0.76)(153) (19) = (20) = 1368 W/K (21) Note that almost two-thirds of the fabric heat-loss is through the glass walls. The ventilation conductance Let us assume there is one air-change per hour (1 ACH). The ventilation conductance is (1/3)NV = (1/3)(1)(17 9 3) = 153W/K (22) 9

10 The total heat input required To maintain an internal temperature of 22 C when the external temperature is 1 C requires a steady-state heat input of Q input = Q fabric + Q vent = ( AU)(T i T o ) + C v (T i T o ) (23) = (1368)(23) + (153)(23) W (24) = 31.46kW kW = 35kW (25) For this rather unusual building, 90 percent of the total heat-loss is via conduction through the fabric, and over half the total heat is lost through the glass walls. In one hour, according to this very simple analysis, the house will thus require 35 kwh of energy to keep it warm. Carbon demand For every 1 kwh of energy delivered by a heating system using natural gas around 0.2 kg of CO 2 is produced. For every 1 kwh provided by grid-supplied electricity about 0.4 kg of CO 2 is produced. If heated by gas, the Glass House thus creates 7 kg of CO2 every hour and about twice that if electric heaters are used. This small building can thus easily generate 1 tonne of CO2 in a cold week. Although an architectural icon, the Glass House is hardly an exemplar of low-carbon design - with its high U-value glass walls it is a greenhouse but not a green house. EXAMPLE QUESTION 2. Determine the temperatures at the various interfaces of the roof construction under the steady state conditions assumed above. 4.3 Summary Fabric heat loss is largely determined by the U-values of the walls, floor and roof. An energy-efficient design should aim for low U-values. Ventilation conductance is the other main source of heat loss, and there are competing strategies for approaching the problem of the conflicting demands of fresh air and low ventilation heat loss. Given U-values and a ventilation strategy, the (overly-)simple Steady State heat balance equation allows us to rapidly estimate the size of the heaters required on a cold winter day. However, the equation is based on some rather oversimplified physics and it says little about the carbon consumption over the year. We shall address both these short-comings in the next lecture. 10

11 4D11 Building Physics LECTURE 2 BUILDING HEAT TRANSFER USING LINEAR SYSTEMS THEORY 5 STEADY STATE MODELS Heat transfer in buildings occurs by three physical processes: conduction, convection and radiation. Conduction involves transferring vibrations between adjacent molecules, convection involves heat being carried by moving fluids and radiation involves light particles (photons) being transmitted and absorbed across a space. In all three cases there are equations relating the flow of heat to temperature differences. Conduction is a comparatively simple process governed by a linear law stating that the rate of flow of heat is proportional to the temperature difference, but convection and radiation are rather complicated and governed by nonlinear relations between heat flow and temperature. For example, radiation has the Stefan-Boltzmann Q = kt 4 power law. In order to build easily-solvable models, building engineers therefore LINEARISE the equations. For example, the Stefan-Boltzmann law has temperature in Kelvin, thus a temperature of 15 C = K = 288K, and even a 10 C swing is small in comparison to 288K, thus the change in radiative heat flow caused by a 10 C rise can be readily approximated by the linear equation dq = 4kT 3 dt with T = 288 and dt = 10. Once all the equations have been linearised, complicated heat transfer systems are governed by sets of simultaneous linear equations. These can be written in matrix form Q = HT (26) where Q is a vector of heat inputs, T is a vector of temperatures, and H is a matrix of conductances relating the temperatures to the heat inputs. In this form, we can readily use the basic tools of linear systems theory (such as Matlab) to solve our equations. The other advantage of this form, as we shall see, is that it is directly analogous to any other linear system with which we may be more familiar - structural, electrical, hydraulic, etc. The obvious analogy is to electrical systems. A temperature difference driving a heatflow through a thermal resistance is analogous to a voltage difference driving a current through a resistor. That is: TEMPERATURE = VOLTAGE (the driving potential) HEAT FLOW = CURRENT (the flow) RESISTANCE = RESISTANCE (easy, huh?) 11

12 Figure 2: Two equivalent schematics of the heat-transfer in the room. The fabric resistance = 1/ AU and the ventilation resistance is 1/C v 5.1 The Simplest of All Models Assuming all temperatures and heat flows are steady, the simplest possible model of the heat-flows from a room with inside temperature T i to an outside ambient temperature T o is given by where Q input = Q fabric + Q vent = ( AU)(T i T o ) + C v (T i T o ) (27) the fabric conductance AU ( in W/K )is the sum of the individual surface areas (walls, roof, windows, etc. in m 2 ) multiplied by their respective U- values (W/m 2 K); the ventilation conductance C v = c p ṁ = c p ρn ACH V/3600 = (1/3)N ACH V This equation can be represented by the diagram in Fig. 1 of Lecture 1. It is shown even more schematically in Fig. 2 as a pair of resistances in parallel connecting the internal node to the external node(s). In the first diagram, two external nodes are shown, but as the second diagram makes clear, they are one and the same node (- in the electrical analogue they would be the earth). (Although they are real, resistances have been written as Z since the symbol R tends to mean other things in the CIBSE notation). Clearly, the electrical analysis of this diagram is trivial. We apply Kirchhoff s First Law (current conservation) at the internal node to obtain Q input = Q fabric + Q vent (28) and then applying Ohm s Law across each resistor leads immediately to Equation (2). 12

13 6 The CIBSE Simple Model The previous model of steady-state heat transfer in a room is the simplest of all models. It is a very useful model for quick first calculations. However, it is perhaps overly simple, and sheds little light on the complicated interactions that are occurring between radiation, conduction and convection inside the room. An attempt to better model these interactions is contained in the CIBSE Guide A Simple Model for steady-state heat transfer. Although an improvement, the model still has numerous shortcomings. The first enhancement in the CIBSE Simple Model is to break up the internal node into several different nodes inside the room. These include: the air node; the operative node; the environmental node; the surface node. The surface node (representative of the internal surface of an exterior wall) and the air node (representative of the air inside the room, of course) are the easiest to understand. As well as heat conducting through and convecting from the wall surface, the surface node absorbs and emits radiation. In reality there are many wall surfaces in a room, each perhaps at a different surface temperature, each radiating to and receiving radiation from all the other surfaces. The use of a single surface node is thus one simplification of this complex interaction. The temperature at the operative and environmental nodes have quite subtle and complicated arguments behind their definitions. However, for now, the environmental node can be thought of as being a point just off the surface of the wall, where convection and radiation heat transfer are occurring. The operative node can be loosely associated with a room occupant, and the operative temperature will be used as a measure of comfort. The key thing to recognise is that thus far we have said little about radiation. How warm a person feels does not depend purely on the air temperature. To observe this, you need merely stand outside on a sunny but cold winter day. Stand in the shadow, and you feel freezing. Take one step into the sunshine and you feel decidedly warmer. The air temperature is identical in both places, but when standing in the sun you are receiving radiation and you feel warmer. Inside a room, similar effects occur.each surface of the room has a temperature and it emits radiation that is characteristic of that temperature. All surfaces of a room talk to each other via this exchange of radiation. The human skin similarly converses in this general radiation exchange and skin facing warm radiant surfaces feels warmer. 13

14 Figure 3: Two equivalent schematics of the CIBSE Simple Model The operative node is intended to capture this concept that how warm you feel is dependent not only on air temperature but also on radiative exchange. Despite the subtleties behind these definitions, they remain crude approximations to the more complex reality. As far as our analysis is concerned, the advantage of these nodes is that they can be readily represented in a heat-flow resistance diagram, as shown in Fig 3. Now that there are four internal nodes, we need to revisit our heat loss equation (2) and see how it has changed. The first thing to note is that the ventilation heat loss still uses the air temperature as the internal temperature. This seems quite natural as that is the temperature we need to heat the incoming air to. However the fabric heat loss now uses - quite naturally also - the environmental temperature, the temperature just inside the wall. Our revised equation (2) is thus Q input = Q fabric + Q vent = ( AU)(T env T o ) + C v (T ai T o ) (29) Now that we have more unknowns, some additional facts are required to solve this model. 14

15 These additional facts are: the operative temperature is defined as T c = (1/2)T ai + (1/2)T s the environmental temperature is defined as T e = (1/3)T ai + (2/3)T s the heat transfer coefficient between a surface S and the air A may be taken as 3 W/m 2 K, thus the total resistance from S to A is 1/(3 A) K/W. Each of these assumptions requires some comment. The operative temperature is clearly a simple average of the air temperature and the average temperature of the surfaces of the room with which the skin is communicating radiatively. The environmental temperature (just inside the wall) is a similar average, but biased more more heavily towards the wall surface temperature. CIBSE Guide A and its explanatory companion CIBSE Volume A Design Data take this model and proceed to construct hundreds of complicated formulae for various cases of interest (see for example Vol A Design Data, pages 5-12 and 5-13). The idea seems to be that an engineer can plug values into these without actually having to think. However, there is really no need for such explicit formulation - the simple resistance model shown in the diagram is all that is needed to solve any particular question. By writing down Kirchhoff s Law at each node, and Ohm s Law across each resistor, the resulting equations in matrix form can be readily solved either in Matlab or by hand. For example, thinking of temperatures as voltages, the voltage drop between the air A and the surface S can be depicted as shown in Fig 4, falling linearly as it would across a long rheostat. The operative temperature is at the mid-point and the environmental temperature is at the two-thirds point (to match the first two of our additional facts above). Figure 4: Using our additional facts, the temperature (or voltage ) drop between the air A and the room surface S allows the resistance between any of the four internal nodes to be determined. 15

16 Figure 5: The simple representation of the CIBSE Simple Model The total resistance from A to S is 1/(3 A), thus the resistance between any of the four nodes can be simply proportioned from the diagram. The important values are thus from A to C: resistance = (1/2) of 1/(3 A), conductance = 6 A from C to E: resistance = (1/6) of 1/(3 A), conductance = 18 A from A to E: resistance = (2/3) of 1/(3 A), conductance = 4.5 A This latter value (that the heat transfer coefficient between A and E is 4.5 W/m 2 K) is really all that is needed to get the calculations started. Our resistance diagram is now shown in Figure 5. Nodes S and C - although still there - have been subsumed in this simpler schematic by using equivalent total resistances on branches AE and EO. (Remember that the definition of a U-value includes the surface resistances - i.e. on the inside, from the environmental node E to the surface node S, and something similar on the outside.) Only two more pieces of information are required before we can put this into action: the heat inputs and the control sensors. Heat inputs. In the Simple Model the heat inputs are to the air and environmental nodes. If we insert a radiator into the room it will add convective heat and radiative heat. The total heat input Q inp could thus be split into a radiative fraction rq inp and a convective fraction (1 r)q inp. However, it is argued (see Q4) that 1.5 times the radiative heat should be added at the environmental node and the rest (i.e. (1 1.5r)Q inp ) should be added at the air node. 16

17 Control sensors. In typical applications, one imagines that a thermostat will be set to maintain some temperature at a fixed value. The sensor on this thermostat may measure just the ordinary dry bulb temperature of the air, or perhaps may also sense some of the radiant heat. A common possibility is that the sensor (and the heating system) maintain the air temperature T ai at a fixed value. In this case T ai is thus no longer an unknown in our equations. In the program IESVE, the system can be set (via the Apache/Settings/Building dialog box) to sense and maintain any one of T ai, T c or T e. The fact that any of these three can be the known unknown leads the CIBSE Guide A to provide detailed analytical solutions to all three cases. However each solution follows immediately from our resistance diagrams, and there is no need for us to give the long-winded formula for all three cases. We shall proceed assuming the air temperature is the known fixed value. 6.1 Example: Calculating the heat input required to maintain air temperature As we have said, a general solution can now be obtained via Kirchhoff s Law at every node, and Ohm s Law across every resistor. This only needs to be done once in a lifetime, because it will then be seen that there is a quicker way. Trivially, from Fig 5, Kirchhoff and Ohm give A : Q ai = Q v Q ae = C v (T ai T o ) (h ae A)(Te T ai ) (30) E : Q e = Q ae + Q f = (h ae A)(Te T ai ) + ( AU)(T e T o ) (31) O : Q o = Q v + Q f = C v (T ai T o ) + ( AU)(T e T o ) (32) Trivially, the last equation (which is our Equation (4)) is just the sum of the first two, and so is not independent and did not need to be included, although there is no harm in having done so. The three equations can thus be written in matrix form i.e. Q = HT (33) Q ai C v + H ae H ae C v Q e = H ae H ae + AU T ai AU Q o C v AU C v + T e (34) AU T o It is clear that, like stiffness matrices in structural engineering, these matrix equations can either be derived from first principles as we have just done, or they can just be constructed using some simple rules. First we shall get rid of the third equation by working with temperatures above ambient (e.g. dt ai = T ai T o ). 17

18 As in structural engineering stiffness matrix construction, the three rules are: Rule 1: there is one equation for each free (non-earth, non-support) node. So if there are N free nodes then create an N by N matrix; Rule 2: if a resistor with conductance C connects free node i to earth then add C to the ith diagonal position (i.e. to H ii ). Rule 3: if a resistor with conductance C connects free node i to free node j then add the submatrix [ ] C C (35) C C to the i-th, j-th row/column submatrix (i.e. add C to H ii and H jj and add C to H ij and H ji ). Using these three rules we can trivially write down the two-by-two matrix H relating [Q ai, Q e ] to [dt ai, dt e ] : [ ] Cv + H Q = H dt with H = ae H ae (36) AU + Hae H ae Now, if the heat inputs are known we can readily calculate the excess temperatures above ambient via dt = H 1 Q. In the CIBSE Volume A Design Data the formula for the solution to this is written out fully (Vol A, Eqn A5.158, pa5-14), but with our new understanding of what is being solved, there is now no need for such long-winded detail. If, instead, we have a thermostat and radiator fixing T ai and the radiator has a known radiative fraction r, then we set [ ] [ ] Qai (1 1.5r)Qinp = (37) Q e 1.5rQ inp and solve the system equations for the two unknowns Q inp and dt e. 6.2 EXAMPLE QUESTION 3 Philip Johnson s Glass House, Connecticut Assume all heat input is from radiators with radiative fraction is 0.2. Calculate, by hand, the heat input required to maintain the original air temperature, and calculate the associated environmental temperature. (Write down the heat transfer matrix H and solve to find Q inp and T e ). 6.3 EXAMPLE QUESTION 4 Show that a heat input Q inp with radiative fraction r which inputs rq inp at the surface node S and the remainder (1 r)q inp at the air node A gives the same temperatures and heat flows as the model above (with its funny 1.5r term), at 18

19 least as far as nodes A and E are concerned (i.e. show that the factor 1.5 is a fudge - the radiant fraction rq inp is really input at S, but to make calculations simpler the CIBSE Simple Model inputs it (times 1.5) at the environmental node E. This gives the same answers at A and E but gets the temperature slightly wrong at S). 6.4 A Structural Analogy Without wishing to confuse you, there is another obvious analogy that can be made. Since our heatflow equations are linear and of the form Q = HT, they are directly analogous not only to the electrical equations of the form I = (1/R)V but also to the structural equations of the form F = KX. Heat inputs become applied forces, and temperatures become displacements. Note though that thermal (and electrical) conductances are then analogous to structural stiffnesses. (Perhaps one may have expected resistances to be analogous to stiffnesses). Fig. 6 shows a linear structure equivalent to the thermal model of the Philip Johnson Guest House. It has one degree of statical indeterminacy. Figure 6: A structural analogue of the CIBSE Simple Model, with radiant fraction 0.2. Note that the springs (which look like resistors) have their stiffnesses set equal to the thermal conductances (rather than to the thermal resistances) 19

20 7 SUMMARY The simplest of all models is represented by Figs 1 and 2, and equation (2). The CIBSE Simple Model is represented by Figs 3, 4 and 5, and is represented by the matrix equation Q = H dt (38) which is, more fully [ Qai Q e ] [ ] [ ] Cv + H = ae H ae dtai H ae AU + Hae dt e (39) with H ae = 4.5 A. 7.1 EXAMPLE QUESTION 4 Given steady heat-flux inputs Q to the air and environmental nodes, the steadystate excess temperatures there are given by dt = H 1 Q (40) From Figure 4, the excess operative temperature is dt c = 1 4 dt ai dt e = [ ] [ dt ai dt e ] (41) thus dt c = [ ] H 1 Q (42) The above equation is almost trivially obvious from Figures 4 and 5. Show that it is equivalent to the long-winded CIBSE Guide A formulae 5.29 and 5.30 which state that (in our notation) the operative temperature is dt c = where the factor F cu is given by Q ai + F cu Q e C v + F cu AU (5.29) (43) F cu = 3(C v + 6 A) AU + 18 A (5.30) (44) 20