Relation between thermal resistance and heat storage in building enclosures

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Relation between thermal resistance and heat storage in building

enclosures

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RELATION

BETWEEN

THERMAL RESISTANCE AND HEAT STORAGE IN BUILDING ENCLOSURES

by

G. P. Mitalas

The annual b u i l d i n g heating energy requirement is a function o f t h e

thermal ~esistance of t h e b u i l d i n g enclosure and, t o a lesser e x t e n t , o f t h e mass of the building envelope, In this Note relations t h a t can b e used to estimate the "trade-off" between the mass and the thermal resistance o f t h e b u i l d i n g enclosure* are presented in graphical f o r n . These relations d e f i n e t h e permitted allowances for mass in terms of reduction in thermal resistance a s given in 'Weasures for Energy Conservation in New Buildings 1978'"

**

The changes in t h e annual energy requirement for heating

o f

b u i l d i n g s due t o changes in thermal storage of b u i l d i n g enclosures w e r e d e t e r m i n e d

by computer simulation. l * From t h e r e s u l t s of t h e s e calculations it was

possible to derive a simple relation between changes in the mass o f

building enclesure and corresponding changes

in

thermal r e s i s t a n c e of b u i l d i n g enclosure based

on

the c o n d i t i o n t h a t the annual building heating energy requirement remained constant. F o r example, as t h e mass of an exterior wall is increased, it is p o s s i b l e to decrease t h e wall thermal resistance without changing the b u i l d i n g annual heating energy r e q u i r e m e n t .

The computer simulation considered t h e following f a c t o r s . 1

.

Type

of

b u i l d i n g

a) Sing le - f amily residence. (According

to

"Measu-res for Energy

Conservation in

New Buildings 1978", t h i s type is classified a s

l'Enclosures f o r b u i l d i n g s w i t h low energy requirements f o r lighting, fans and pumps .")

b] Large off ice buif ding. [classified as mEnclosures for b u i l d i n g s

w i t h high energy requirements for l i g h t i n g , f a n s and pumps.")

The 'tthemal r e s i s t a n c e of b u i l d i n g e n c l a s u r e ' ~ ~ the over-all thermal resistance between i n s i d e and outside a i r .

**

This document was prepared f o r t h e NRC A s s o c i a t e Committee on t h e

National Building Code by i t s Stairding Committee on Energy Conservation.

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2 . C l i m a t e

H o u r l y weather data

for

Winnipeg, Ottawa and Vancouver for 1970 were used

for

the c a l c u l a t i o n s

.

3 . k s s of

i n t e r i o r

construction

L i g h t , medium and heavy interior constructions were considered: a) Light

=

150 kg/m2 of f l o o r surface area

b] Medium

=

350 kg/m2 o f f l o o r surface area

c ) Heavy 1=-600 kg/rn2 o f f l o o r surface

4 . Mass of exterior wall

a] L i g h t ,

e .

_g., _frame_wall

b) Medium, e.g., layer of concrete about 1 5 cm thick plus i n s u l a t i n g layer

c) Heavy, e . g . , l a y e r o f concrete about 30 em thick plus i n s u l a t i n g layer

5 . Relative position

of

mass layer with respect to insulating layer

a) m a s s layer

on

t h e i n d o o r side of i n s u l a t i n g layer b) mass Payer on the outdoor side of the insulating layer

6 _Thermostat_{s e t}_-point_s-chedules

a] Single-family residence :

Schedule (i) Constant at 22OC

Schedule

(ii)

Daytime at 22°C Nighttime at 1 8 " ~ b) Large Office b u i l d i n g :

Occupied Unoccupied

Period P e r i o d

Schedule (i) 20,O"C 18.3OC

Schedule (ii) 21.7"C 20 O°C

Schedule (iii) 2 0 . O°C 15.6"C Schedule [ i v ) 22.2"C 2 2 . Z°C

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7 . O r i e n t a t i o n

a] Single-family residence: three orientations used, major glazed facade facing south, west

o r

n o r t h .

b ) Large o f f i c e b u i l d i n g was o r i e n t e d w i t h i t s s i d e s f a c i n g t h e cardinal d i r e c t i o n s

.

8 . Zoning

a ) S i n g l e - f a m i l y residence was considered as a s i n g l e zone.

6) Large o f f i c e building was simulated assuming that each storey of

t h e building is made up of nine zones. Four

of

t h e zones are

t h e four corners; the o t h e r f o u r are t h e s i d e s and the n i n t h is

t h e i n t e r i o r zone. The depth of the exterior zones was taken as 1.6 times t h e h e i g h t of one storey.

The results of the computer simulations i n d i c a t e d t h a t in t h e

relation between thermal resistance and mass f o r building enclosures, the

fol

lowing are s i g n k f i c a n t f a c t o r s : 1 . Type

of

b u i l d i n g .*

2. Position of mass layer with respect to t h e insulating layer.

3. Mass of i n t e r i o r .

4 . Mass of mass layer of t h e b u i l d i n g enclosure.

5. Thermostat s e t - p o i n t schedule (in t h e case o f large o f f i c e buildings)

.

The t h e m s t a t s e t - p o i n t schedule is not treated s e p a r a t e l y as a d i s t i n c t variable in t h e _{presentation of}_{t h e}_relation_between _thermal

r e s i s t a n c e and mass, s i n c e t h e operation of t h e building is n o t controlled

b y the designer. The average of t h e results for a l l the thermostat s e t - point schedules

is

taken a s the representative value. A l s o ,

in

a similar way, t h e average of t h e results was taken f o r the o t h e r weaker variables, such as o r i e n t a t i o n and climate. Consequently, the t h e m 1 resistance/ mass elations presented in F i g s . 1 to 4

are

for:

a) Building types*

b) P o s i t i o n of the mass in the building enclosure r e l a t i v e to insulating layer.

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In lrMeasures f o r Energy Conservation in New Buildings 197811

designations, such as single-family residences and large o f f i c e b u i l d i ~ ~ g s , are n o t used; buildings are c l a s s i f i e d according to t h e r a t e o f . h e a t l o s s relative to internal h e a t gain. Usually, a single-family residence has a high r a t e 05 heat loss r e l a t i v e t o internal h e a t g a i n . Results f o r such

residences can be used, therefore, as examples of

the

c l a s s d e f i n e d a s

'7enclosures "for buildings w i t h low energy requirements for lighting, fans

and pumps." The

thermal

resistanceJmass relations for t h e s e types o f enclosures are given in Figs. 1 and 3 . Examples of the other group of enclosures, i.e., '7enclosures for buildings w i t h h i g h energy requirements

f o r

lighting, fans and pumps," are given by the thermal resistancc/mass relations o f large o f f i c e b u i l d i n g s (Figs. 2 and 43

.

Figures I and 2 define the t h e r m a l resistance/mass r e l a t i o n f o r t h c mass Layer on t h e outside of the insulating layer of the b u i l d i n g

enclosure. Experience i n d i c a t e s t h a t the e f f e c t i v e n e s s of the mass layer depends on the t h e r m a l resistance between room a i r and the mass l a y e r .

Any resistance i n a d d i t i o n t o t h a t o f t h e a i r film reduces t h e effective- ness of t h e _mass_{l a y e r}_as_an_{i n t e r i o r}_heat_{storage component.} _In_{t h i s}

Note anything other than d i r e c t c o n t a c t between mass layer and space a i r [where d i r e c t contact means a thermal r e s i s t a n c e equal or less than

0 . 2

**r n 2 * ~ / w )**

is considered as an i n s u l a t i n g layer. Examples of this layer are false ceilings, sprayed asbestos f i r e p r o t e c t i o n , as well as t h e main

insulating l a y e r in the building enclosure.

Figures 3 and 4 define t h e t h e r m a l resistanceJmass r e l a t i o n f o r the

mass

layer

an

the i n s i d e and

in

direct contact with controlled space a i r . The use of Figures 3 and 4 is not straightforward because t h e addltion of a mass layer augments t h e i n t e r i o r mass and the annual h e a t i n g r e q u i r e - m e n t of t h e building depends on the combined effect of all the interior

mass. I n a d d i t i o n , t h e mass e f f e c t is not linear since _the_{increase of} mass beyond approximately 500 kg/rn2 of f l o o r surface area has l i t t l e

effect on t h e annual building heating requirement. T h i s can b e seen in

F i g s . 3 and 4 where t h e extensions of t h e ''R" curves are parallel to the mass axes beyond 500 kg/m2 o f floor surface area.

Example P ~ o b l e m s t o Determine t h e Allowance

in

Terms of

Thermal Resistance for t h e Mass Layer i n a Building Enclosure Using F i g u r e s I and 2

Figures

1 and 2

can

be used t o estimate the change in t h e r i a l

resistance 05 t h e b u i l d i n g enclosure which is e q u i v a l e n t to t h e layer a f mass f o r the condition where the mass layer does n o t c o n t r i b u t e to the

i n t e r i o r mass, i . c . , a mass layer that i s not

in

direct c o n t a c t w i t h temperature controlled space a i r .

Example Problem 1

Determine the allowable reduction of exterior wall thermal

resistance, AR, due to the mass layer in t h e e x t e r i o r wall. The mass l a y e r is not i n d i r e c t contact with i n t e r i o r space air.

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Given Data

Type of Building: B u i l d i n g s with low energy requirements f o r lighting,

fans and pumps (single-family residence)

.

E x t e r i o r wall thermal resistance: R = 3 . 6 _{r n 2 - ~ / w}

Mass of the exterior wall layer: D = 200 kg/m2 of w a l l surface area S o l u t i o n

Use Fig. 1 because building has law energy requirements f o r lighting, fans and pumps and mass l a y e r is n o t in d i r e c t contact w i t h i n t e r i o r

space air.

From curves R = 3 and R = 4 for d = 200 kg/m2 of wall surface area, r e a d

ARg = 0.010 (from curve R = 33 dR4 = 0.018 (from curve R = 4)

Then, by linear interpolation,

AR for R = 3 . 6

~ ' . K / w

is

Thus an e x t e r i o r wall w i t h over-all thermal resistance o f

R = 3.6 r n 2 * ~ / ~ and no mass is c o n s i d e r e d equivalent t o an e x t e r i o r wall

with over-all thermal r e s i s t a n c e R = 3 . 6

-

0.015 = 3.585 m2-K/W

=

3 . 5 9 r n 2 - ~ / h ' and mass layer of 200 kg/m2 of wall surface aFea. The equivalence is on t h e b a s i s of t h e b u i l d i n g annual h e a t i n g requirement ( a l l o t h e r f a c t o r s being equal)

.

Determine t h e allowable r e d u c t i o n of thermal resistance,

AR,

of a

roof-ceiling system due to the mass layer in this system. Mass Layer is n o t in d i r e c t contact w i t h i n t e r i o r space a i r .

Given Data

Type of B u i l d i n g ; Building with h i g h energy r e q u i r e m e n t s of lighting, f a n s and pumps ( l a r g e o f f i c e b u i l d i n g )

Thermal r e s i s t a n c e of

roof-ceiling system: R = 4 . 0

r n 2 - ~ / w

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Solwt ion

Use Fig. 2 because building has high energy requirements

for

l i g h t i n g , f a n s and pumps and mass l a y e r is not

in

d i r e c t c o n t a c t with i n t e r i o r space air.

Using curve R = 4 f o r d = 300 kg/m2 of r o o f surface area, _read

Thus

the use

of a

mass layer of 300

kg/m2

of r o o f surface area allows a reduction of over-all roof-ceiling thermal r e s i s t a n c e o f

R = 4 - 0 r n 2 - ~ / h ' by AR = 0.12 m 2 * ~ / l l .

Using Figures 3 .and 4

F i g u r e s 3 and 4 are used to estimate t h e allowable r e d u c t i o n in thermal resistance of a building enclosure component, i . e . , e x t e r i o r wall

or roof-ceiling system, when the mass layer in this component is in d i r e c t c o n t a c t w i t h t h e i n t e r i o r space air.

The

use of t h e s e f i g u r e s is more complicated than t h e use of F i g s . 1 and 2 therefore the following procedure is recommended.

1) Using g i v e n data c a l c u l a t e the c o n t r i b u t i o n to the interior mass by t h e mass layer of the building enclosure,

a)' Roof -ceiling system :'

where

AL = Surface (plan) area of mass layer t h a t is in d i r e c t contact with i n t e r i o r space a i r , m 2

M

= Mass of layer, kg/m2 of surface area

L

AF = F l o u r

area, m 2

When.computing

contribution to the i n t e r i o r mass by the mass of roof-

ceiling system or floor, allowances should b e made f o r s h a f t areas o r o t h e r discontinuities,

i.e.,

AL should be t h e n e t area of mass layer. In

t h e case of slab-on-grade f l o o r , the mass of t h e concrete f l o o r should be included in the calculations of ME unless i t is n o t in d i r e c t contact w i t h room air ( e - g

.

, if f l o o r is covered by heavy carpet)

.

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b) E x t e r i o r wall:

ME

= (A\y

ML)

/AZ

where

Yf

= Surface (elevation) area of the mass layer in e x t e r i o r wall

A = Area of the floor where t h e i n t e r i o r mass is c m s i d e r e d to be a f f e c t e d by the mass layer in an e x t e r i o r wall. In t h i s Note it i s assumed that

where

\

= t o t a l exterior wall area of the s t o r e y in question.

Similarly, large concrete columns, beams and f i x e d p a r t i t i o n s t h a t are in d i r e c t c o n t a c t with room air and are located w i t h i n t h e area A2

should be included in the calculation:

M

= ( T o t a l mass of beams, columns and p a r t i t i o n s j / A

C

z

2) Using given d a t a , c a l c u l a t e total interior mass MT, af the space

with floor area A=. The EIT v a l u e should include t h e mass of furnishings, partitions and floor s l a b a s well as the mass of t h e layer in question. As t h i s information is n o t usually a v a i l a b l e , because the position and type of p a r t i t i o n s as w e l l as the type and amount of f u r n i s h i n g s are usually not fixed at t h e time the o u t s i d e envelope is designed, it is

recommended that a fixed valve of 100 kg/m2 of f l o o r area be used as t h e mass of furniture and movable p a r t i t i o n s . Thus

MT = 100

+ M~ + MC c ME kg/m2 of f l o o r area wh ere

9

₌_mass_of_{f l o o r ,} _kg/m2 of floor surface area.

31 Read A R 1 and

AR2 values

against b+- a n d [MT

-

ME) values respectively

using appropriate IrR" curves. Use linear interpolation a s shown in

Example Problem 1

in

case t h e thermal resistance of the building enclosure component is not an i n t e g e r .

4) The difference AR = AR1

-

AR2 i s the allowable r e d u c t i o n of t h e thermal resistance of the b u i l d i n g e n c l o s u r e component due t o the mass

l a y e r in this component when t h i s mass layer i s in direct contact with

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The use o f t h i s procedure is illustrated by Example Problem 3 . E x a m ~ l e Problem 3

Determine the allowable r e d u c t i o n of exterior wall t h e r m a l

resistance,

AR,

based on t h e mass layer of an exterior wall t h a t is in direct contact with i n t e r i o r space air.

Given data:

Type of b u i l d i n g : Building w i t h high energy requirements f ~r

l i g h t i n g , fans and pumps ( l a r g e o f f i c e building) E x t e r i o r wall thermal

resistance : R = 3 . 0 r n 2 - ~ / w

Mass of exterior wall

mass layer:

ML

= 250 kg/m2 of wall surface area Mass of floor slab: MF = 300 kg/m2

of

f l o o r surface area

Surface area o f mass

layer in e x t e r i o r wall: A = 200

_m2

W

Surface area of

exterior wall: = 300 m2

Sa l ut ion

Use Fig. 4 because building has high energy requirements for

lighting, f a n s and pumps and the mass layer is in d i r e c t c o n t a c t with t h e i n t e r i o r space

a i r .

Contribution of mass layer to i n t e r i o r mass:

= 111 kg/m2 of floor area Total interior mass:

= 100 + 300 + 111 = 511 kg/m2 of floor surface area

Using curve

R

= 3 (Fig. 4). f o r

MT

= 511 and far (MT

-

ME) = 400 kg/m2 of floor surface area

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read, re spec t i v e l y ,

then the reduction is

A d d i t i o n a l Notes

13 Consider an exterior wall or roof which is made up of t h r e e layers

such t h a t the i n s u l a t i o n l a y e r is placed between two mass layers and the interior mass layer is

in

direct c o n t a c t w i t h room a i r . In this

case the allowable reduction of exterior wall thermal resistance, AR,

is the sum of two allowable reductions: a) one determined using Figs. I or 2 and b] one determined using F i g s . 3 or 4 .

Consider an

enclosure

which is made up of exterior wall and r o o f such that the mass layers of both are in direct contact w i t h room air. In

this case the t o t a l contribution to the interior mass is t h e sum of the masses of t h e roof-ceiling s y s t e m and exterior wall, i . e . ,

ME = [AL

x-ML)/AF

+ (A\v + ML]/Az. Using F i g s . 3 or 4 and the above value of ME the allowable seduction of the thermal resistance for

exterior wall or roof-ceiling system can be Getermined. Note that only one allowable reduction can be used because both are based on

the t o t a l mass of the room, i .e

. ,

bR calculated for exterior wall or

AR

calculated

f ~ r

roof-ceiling system.

REFERENCES

1. A . Konrad and B.T. Larsen,

ENCORE-CANADA

Computer Program f o r the

Study of Energy Consumption of Residential Buildings in Canada

(preliminary version). Presented at t h e Third International Symposium

on

t h e Use of Computers for Environmental E n g i n e e r i n g Related to Buildings, Banff, May 1978. I

2. G . P . MitaXas, A special compute^ program to simulate large o f f i c e

buildings based on the calculation methods outlined

in

"Procedures

for Determining Heating and Coo l i n g Loads f o r Computerizing Energy Calculations

-

Algorithms for Building Heat Transfer Subroutines", compiled and published by ASHRAE Task Group on Energy Requirements