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Section 6: Heat Transfer - Part II ENV-2D02 (2006):Energy Conservation – power point versions of lectures. Will be available on WEB later in Week 1Introduction to organisation of course and Field Course 2Revision of Simple Economic Analysis 3Thermal comfort: physical and physiological aspects. What temperatures do we actually need? 4Energy use by sector: 5Energy GDP relationships: Energy Balance Tables. 6HEAT TRANSFER: U Values 7Heat Losses from Buildings – Effect of Built Form: Dynamic Effects 8 Introduction to Energy Management - order may be swapped with section 9 9 Energy Management Continued: Energy Targets: Building Regulations 10Electricity Conservation 11Thermodynamics 12Combined Heat and Power 13The Heat Pump 14Energy Conservation Measures at UEA 15Energy Analysis: Concluding Remarks

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Main Objective of the Lecture To apply the basic ideas of Heat Transfer from previous lecture to estimating the thermal properties of typical building materials. To provide the tools to allow Heat Loss Estimations from buildings to be made [covered in next lecture ] >> hence to estimate potential savings.

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Summary of Last Lecture – Key Point Heat is lost from a building by: Conduction But heat is also lost by: Convection Radiation from the wall surfaces Resistance to Heat Flow Estimated by where k is conductivity of material d is thickness brick plaster External boundary layer Internal boundary layer

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Surface Resistance Analysis of heat flow by convection and radiation is more complex. Can be approximated in most situations for buildings by additional resistance layers. Cannot be used if surface temperature are substantially different from surrounds – e.g. a hot water pipe/radiator. Consequences of Boundary Layers Surface Temperature of window (on room side) is BELOW room temperature - for single glazing it will be about 7 0 C if outside temperature is 0 o C and internal temperature is 20 o C External surface temperature will be above surrounding air. Internal Surface Temperatures are important as they affect Mean Radiant Temperature and hence Thermal Comfort. External Surface Temperatures can affect weathering properties of bricks.

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6.5 Internal and External Surface Resistances Typical values for surface resistances (m 2 o C W -1 ): Vertical Heat Flow 0.11 for upward flow through floors/roof 0.15 for downward flow through floors Horizontal Heat Flow R int = internal surface R ext = 0.08 sheltered external surface 0.06 normal 0.03 severe Note: - the orientation of windows is important in heat loss calculations as the external resistance is a significant proportion of the total resistance;

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6.6 Resistances of Air-Spaces - thermal conductivity of air-spaces is very small, and heat transfer is mostly by radiation and convection, - values are given in tables, but can be divided generally into two categories:- unventilated air spaces (or low ventilation) – resistance is about Examples:- air-space in modern cavity walls, air-space in double glazing, air-space between ceiling & underside of felt (post-war houses). ventilated air spaces – the resistance is about Examples:- older cavity walls and air-space between ceiling and underside of tiles (pre-war houses).

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6.7 Derivation of 'U'-values for 3 types of wall Many standard constructions have U- values in Tables Non-standard constructions do not – including many new types. Example 1 6 components:- 1) external surface layer 2) outer brick layer 3) cavity 4) inner brick layer 5) plaster 6) internal surface layer Fig. 6.6 Heat flow through wall of 1950's construction conductivity of brick = 1.0 Wm -1 o C -1 conductivity of plaster = 0.7 Wm -1 o C -1 External surface resistance brick cavity plaster Internal surface resistance

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6.7 Derivation of 'U'-values for 3 types of wall Resistance = where k = conductivity d = length of heat flow paths (thickness in this case) resistance of brick = = 0.11 m 2 o C W -1 resistance of plaster = = 0.02 m 2 o C W -1 Effective resistances of air spaces are:- internal boundary m 2 o C W -1 external boundary m 2 o C W -1 air-cavity 0.18 m 2 o C W -1 So total resistance = = m 2 o C W -1 ===========

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6.7 Derivation of 'U'-values for 3 types of wall Total Resistance = m 2 o C W -1 since U = U = 1.67 W m -2 o C -1 Note: that the external resistance is relatively small < 10% of total resistance U value for walls varies little with exposure normally [only a few per cent at most].

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Example 2 As example 1 except that the inner brick leaf is replaced by an aerated block wall i.e. construction used from mid-1960's. conductivity for aerated block = 0.14 Wm -1 o C -1 and resistance of such a block = 0.76 m 2 o C W -1 replaces the inner brick of original wall, new resistance = = m 2 o C W -1 so U-value = = 0.80 Wm -2 o C -1 i.e. a 50% saving in the heat lost through the walls of a house. blockbrick cavity plaster brick

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Example 3 As example 2 except that cavity is filled with insulation conductivity of insulation = 0.04 Wm -1 o C -1 resistance of cavity fill = 0.05 /0.04 = 1.25 m 2 o C W -1 replaces the resistance of 0.18 from the air-cavity New resistance = = m 2 o C W -1 and U-value = = 0.43 Wm -2 o C -1 i.e. approximately half of the value in example 2 and one quarter of the value in example 1. [the U-value for a wall with two brick leaves and cavity insulation is 0.60 Wm -2 o C -1 ]. cavity blockbrick plaster Filled cavity

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Example 4: Single Glazing conductivity of glass = 1 Wm -2 o C -1 i.e. resistance = m 2 o C W -1 (for 3mm glass) –internal surface resistance = –external surface resistance = Thus total resistance = = m 2 o C W -1 and U-value = 5.5 Wm -2 o C -1 Note resistance of glass makes very little contribution to the overall resistance if the external resistance changes (from exposure) then the U-value will also be affected significantly. [Compare this with the situation for the walls (see note to example 1)].

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Temperature Profile through a wall Example 2 with polystyrene layer on inside Assumes internal temperature is 20 o C and external temperature is 0 o C Temperature gradient is highest in insulating materials Greatest in polystyrene layer External Surface Temperature = 1.17 If this falls below 0 o C – danger of ice forming and causing bricks to crumble. Without polystyrene, surface temperature would be around 1 0 C lower And PMV would be approx lower

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Example 6: Pitched Roof Heat Flow: Internal surface resistance >plasterboard > Loft space > felt > Felt – tile airspace > tiles >External surface resistance >

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Example 6: Pitched Roof Resistance to heat flow Vertical m 2 o C W -1 Internal surface resistance = 0.11 Plasterboard = 0.06 Loft space = 0.18 Total vertical = 0.35 Inclined Felt = 0.11 Air space felt – tiles = 0.12 Tiles = 0.04 External surface = 0.04 Total Inclined = 0.31 Total Resistance (if A = 45 o ) = cos 45 = 0.57 A U – value = 1 / R = 1.75 W m -2 o C -1 Pre-war houses do not have felt Some houses in extreme weather areas have boards instead of felt

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Example 6: Pitched Roof A Doubling insulation does not half heat loss Simple Way to examine effects of insulation U value without insulation = 1.75 W m -2 o C -1 Resistance = 1 / 1.75 = 0.57 m 2 o C -1 Add 50 mm of insulation conductivity 0.04 Additional resistance = 0.05 / = 1.43 New total resistance = = 2.00 New U-Value = 1 / 2.0 = 0.5 W m -2 o C -1 With 100 mm New total resistance = = 3.43 New U-Value = 1 / 3.43 = 0.29 W m -2 o C -1 With 150 mm New U-Value = 0.21 W m -2 o C -1

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Example 7 Double Glazing U values 3mm glass = 4mm glass = little difference irrespective of what thickness of glass is used. Double glazing: U value (3mm glass) = Note the U value depends on the thickness of the air-space, which is optimum at about 18-20mm. m 2 0 C W -1 3mm single pane - resistance mm single pane - resistance0.004 internal surface resistance0.123 external surface resistance0.055 air-space resistance0.18

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6.9 Problems associated with thermal bridging Thermal bridging leads to: cold spots on the internal surfaces condensation discolouration where the presence of bridges can be seen. A thermal imaging camera can be used to identify such bridges, but these are often expensive. Insulating the loft in UK houses Place fibre glass between the joists. As the thickness of insulation increases problems of thermal bridging appear. timber joists create a thermal bridge

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Thermal bridging – an example In example, insulation occupies 8/9 th of space (400/450) Heat flows are in parallel so use formula 400 mm 100 mm 50 mm Insulation (150 mm thick) joists i.e. R = 2.45 cf 3.75 if bridging is ignored What is effective resistance of joists and insulation?

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Next Lecture Heat Losses from a House Need to work out U-values and area for: –Walls –Windows –Roof –Floor Other sources of heat Loss –Ventilation Remember: you cannot eliminate heat losses – you can only reduce them. Heat lost must be replaced by heating device

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