Services Science - Heat Energy Through Building Structure

8/3/2019 Services Science - Heat Energy Through Building Structure

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Flow of Heat Energy Through Building StructuresPage 1 of 18

Douglas Buchan

College of North West London

SERVICES SCIENCE

ASSIGNMENT 1

FLOW OF HEAT ENERGY THROUGH BUILDING STRUCTURES

Prepared by

DOUGLAS BUCHAN




Douglas Buchan

Services Science

Flow of Heat Energy Through Building Structures

P.2.3 a) Estimate heat loss through a single leaf 100mm brick wall faced on theinside with 15 mm of plaster when inside air temp is 25°C and external airtemp is 12°C

Calculate “U”

12°C 25°C

R 2, 15mm plasterK = 0.814 W/MK

R 1, 100mm Brick K = 0.45 W/MK

1U =

RT

RT = R 0 + R 1 + R 2 + Ri

1 0.1 0.015 1RT = + + +

16.7 0.45 0.814 8.3

RT = 0.421 m²KW

1

U =RT

1U =

0.421

U = 2.375 W/m²K

Q = U x A x Δt Q = 2.375 x 1 x 13

Q = 30.88 w




Douglas Buchan

P.2.3 b) Estimate the heat transfer through a wall comprising 75 x 100mm studdingat 1m centres faced with 15mm plywood the wall is 10m long by 3m highinside temp is 22°C and outside temp is 4°C.

R 0 R i

Areas:R 1 (Plywood) 10m x 3m = 30m²R 2a (Stud) (75mm x 3m) x 11 = 2.475m²R 2b (Air) (10m x 3m) – 2.475 = 27.525m²R 2 (Plywood) 10m x 3m = 30m²

10m

75 mm

100mm

Wood Sample is 3 m high, 10 m long

K = 0.15 W/MK

Air K = 0.025 W/MK

15mm Plywood K = 0.109 W/MK

R 1

R 2a

R 2b

R 3

Studs at 1m centres

4°C22°C




Douglas Buchan

Calculate heat flow:

Rt = R0 + R1 + R2 + R3 + Ri

d2 R 2 =

((k 2a x A 2a) + K 2b x A 2b))

0.1R 2 =

((0.15 x 2.475) + (0.025 x 27.525))

R 2 = 0.094m²Kw

1 d1 d3 1RT = + + R 2 + +

R 0 x A 1 K 1 x A 1 K 3 x A 3 Ri x A 1

1 0.015 0.015 1RT = + + 0.094 + +

(16.7 x 30) (0.109 x 30) (0.109 x 30) (8.3 x 30)

RT = 0.1092 KW

1U =

RT

1U =

0.1092

U = 9.16 W/m²K

Q = U x A x Δt In this instance “A” is omitted as area is given as 30m.

Q = U x Δt Q = 9.16 x (18)Q = 164.9W




Douglas Buchan

P.2.4 For the example in (a) above select a thickness of polystyrene board that willkeep the wall heat loss below 0.35 W/m²K.

1

U2 = 1 + dU1 K

10.35 = 1 + d

2.375 0.033

10.35 = 0.421 + d

0.033

d 10.421 + 0.033 = 0.35

d 10.033 = 0.35 - 0.421

d = 0.033 x (0.1/35 – 0.421)d = 0.033 x 2.436d = 0.080d = 80mm

80mm board should be used to achieve a heat loss value of 0.35 W/m²K.




Douglas Buchan

P.2.5 For the insulated brick wall in example 2.4 determine whether or not avapour barrier will be required if the external relative humidity is 60% andthe internal relative humidity is 50% (internal temp 25°C and external tempof 12°C).

d

The resistances of each layer is R = K

Rt = R 0 + R 1 + R 2 + R 3 + Ri

1R 0 = = 0.0599m²K/W (outside air film)

16.7

0.1R 1 = = 0.2222m²K/W (brick)

0.45

0.08R 2 = = 2.4242m²K/W (polystyrene board)

0.033

0.015R 3 = = 0.0184m²K/W (plaster)

0.814

1Ri = = 0.1204m²K/W (inside air film)

8.3

Rt = 2.845 m²K/WLayer Δt R

The temperature difference Δt is Wall Δt = RT

Layer R Layer Δt = RT x Wall Δt

(Interface temperature in RED)

0.0599R 0 Δt = 2.845 x 13K = 0.27K (12+0.27=12.27°C)

0.2222R 1 Δt = 2.845 x 13K = 1.02K (brick) (12.27+1.02=13.29°C)

2.4242R 2 Δt = 2.845 x 13K =11.08K (polystyrene)(13.29+11.08=24.37°C)

0.0814R 3 Δt = 2.845 x 13K = 0.37K (plaster) (24.37+0.37=24.74°C)

0.1204R i Δt = 2.845 x 13K = 0.55K (24.74+0.55=25.29°C)

[check 13.3K Δt 13K]




Douglas Buchan

Vapour resistance for each layer

Vapour resistivity of brick = 50mn s/gm Vapour resistivity of polystyrene = 250mn s/gm Vapour resistivity of plaster = 55mn s/gm

Rv = vapour resistivity x d Rv = vapour resisitance

Rv Brick = 50mn s/gm x 0.1m = 5gm/mnsRv Polystyrene = 250mn s/gm x 0.08 = 20 gm/mnsRv Plaster = 55mn s/gm x 0.015 = 0.825gm/mns

Rvt = Rv Brick + Rv Polystyrene + Rv PlasterRvt = 5gm/mns + 20gm/mns + 0.825gm/mnsRvt = 25.825

Internal) Vapour pressure @ 25°C 100% relative humidity = 3166.0pa Vapour pressure @ 25°C 50% relative humidity = 3169.3 x 0.5 = 1583.00pa

External) Vapour pressure @ 12°C 100% relative humidity = 1401pa Vapour pressure @ 12°C 60% relative humidity = 1401 x 0.6 = 840.6pa

Δpt = 1583 – 840.6 = 742.4

Layer RvLayer Δp = Rvt x Δpt

5Brick Δp = 25.825 x 742.4 = 143.74pa

20Polystyrene Δp = 25.825 x 742.4 = 574.95pa

0.825Plaster Δp = 25.825 x 742.4 = 23.72pa




Douglas Buchan

25°C

Actual temp

Dew point temp

Brick Polystyrene Plaster

Vapour pressure after plaster layer = 1583 – 23.72 = 1559.28Dew point temp for 1559.28 = 14°C

Vapour pressure after polystyrene layer = 1559.28 – 574.95 = 948.33Dew point temp for 948.33 = 6°C

Vapour pressure after brick layer = 948.33 – 143.74 = 804.59

Dew point temp for 804.59 = 4°C

In conclusion a vapour barrier will not be required as the actual temperature does not fallbelow the dew point temperature.

12°C

13.29°C

12.27°C

24.37°C

24.74°C

14.0° C6.0° C

4.0° C




Douglas Buchan

-25°C30°C

M 2.1 It is proposed to build a coldroom using walls comprising wooden studding(dimensions as in example 2.3 (b) above) faced with 15mm plasterboardskimmed with 5mm plaster on the inside and external grade plywood on theoutside. The coldroom is maintained at -25°C. Choose a suitable material tofill the wall cavity that will keep the heat transfer below 0.6 W/m²K.

Perform a calculation to determine where and if condensation will occur onthe design day of 30 DB 25 WB. Assume 100% RH for air in the coldroom.

1) Calculate thermal transmittance “U” value. d

Find RT. Calculate R for each layer (R= K)

Plaster skim = 0.814w/mK 5mm

Plaster board = 0.814w/mK (15mm)

Pol urathane foam = 0.025w/mK 100mm

Pl wood = 0.109w/mK 15mm

Wood stud = 0.15w/mK 100mm

Plan view of 1m section 3m high.Wall is 10m long

R 0

R 2a

R 1

R 2b

R 3 R 4 R i

Plywood

Stud

Plasterboard Plaster skim

Polyurathane




Douglas Buchan

M.2.1 (a) RT = R 0 + R 1 + R 2 + R 3 + R 4 + Ri

1R 0 = = 0.0599m²K/W

16.7

0.015R 1 = = 0.1376m²K/W

0.109

0.1R 2 = = 2.42857m²K/W

(0.025 x 0.85) + (0.15 x 0.15)

0.015R 3 = = 0.0184m²K/W

0.814

0.005R 4 = = 0.0061m²K/W0.814

1Ri = = 0.1204m²K/W

8.3

RT = 0.0599 + 0.1376 + 2.2857 + 0.0184 + 0.0061 + 0.1204RT = 2.6281 m²K/W

1U = RT

1U = 2.6281

U = 0.38 w/m²K




Douglas Buchan

(Interface temp)

M.2.1 (b) Perform a calculation to determine where and if condensation will occur ondesign day of 30db 25wb. Assume 100% relative humidity for air in thecoldroom.

Find temperature difference Δt at each layer.


0.0599R 0 Δt = 2.6281 x 55K = 1.25K (28.75°C)

0.1376R 1 Δt = 2.6281 x 55K = 2.88K (25.87°C)

2.2857R 2 Δt = 2.6281 x 55K =47.88K (-21.96°C)

0.0814R 3 Δt =2.6281 x 55K = 0.39K (-22.35°C)

0.0061R 4 Δt =2.6281 x 55K = 0.13K (-22.48°C)

0.1204R i Δt = 2.6281 x 55K = 2.52K (-25°C)

[check 54.73K wall Δt = 55K]

ii) Find vapour resistance (Rv) for each layer

Vapour resistivity of plywood = 3000mn s/gm Vapour resistivity of polyurethane = 600mn s/gm Vapour resistivity of stud = 70mn s/gm Vapour resistivity of plaster = 55mn s/gm

Rv = vapour resistivity x d

Rv plywood = 3000 x 0.015 = 45gm/MNsRv1 polyurethane = (600x0.1) x 0.85 = 51gm/MNsRv2 stud = (70x0.1) x 0.15 = 1.05gm/MNsRv plaster = 55 x 0.020 = 1.1gm/MNs

Rvt = Rv plywood + Rv polyurethane + Rv stud + Rv plasterRvt = 45 + 51 + 1.05 + 1.1Rvt = 98.15gm/MNs




Douglas Buchan

Internal vapour pressure @ -25°C 100% RH = 63.29paExternal vapour pressure @ 30°C 100% RH = 4246.2pa(from psychrometric chart 30°CDB + 25°CWB = RH 65%)

External vapour pressure @ 30% 65% RH = 4246.2 x 0.65 = 2760.03 pa

Δp for wall = 2760.03 pa – 63.29 pa = 2696.74 pa

RvLayer Δp = Rvt x Δp wall

45

Plywood Δp = 98.15 x 2696.74 = 1236.4 pa

52.05Polyurethane + stud Δp = 98.15 x 2696.74 = 1430.11 pa

1.1Plaster Δp = 98.15 x 2696.74 = 30.22 pa




Douglas Buchan

Dew point temp

Actual temp

Vapour pressure after plywood layer = 2760.03pa – 1236.4pa = 1523.63pa

Dew point temp for 1523.63pa = 14°C

Vapour pressure after stud + polyurethane layer = 1523.63pa – 1430.11pa = 93.52pa

Dew point temp for 93.52pa = -21 °C

Vapour pressure after plaster layer = 1082.65pa – 30.22pa = 63.29pa

Dew point temp for 63.29pa = -25°C

28.75°C

25.87°C

14.0°C

30°C

-21.0°C

Area at risk of condensation

As can be observed from the above diagram there is a risk that condensation will formafter the polyurethane foam layer.

-25.0° C

-22.48° C

-21.96° C

-25° C




Douglas Buchan

D 3.4 For the previous coldroom example, select a suitable vapour barrier to preventmoisture ingress, state where this will be located in the wall and give reasons, performcalculations to determine that the possibility of intersital condensation has beeneliminated.

dFind RT. Calculate R for each layer R= K

RT = R 0 + R 1 + R 2 + R 3 + R 4 + Ri

1R 0 = = 0.0599m²K/W

16.7

0.015

R 1 = = 0.1376m²K/W0.109

0.001R 2 = = 0.0000042m²K/W

236

0.1R 3 = = 2.42857m²K/W

(0.025 x 0.85) + (0.15 x 0.15)

0.015R 4 = = 0.0184m²K/W

0.814

0.005R 5= = 0.0061m²K/W

0.814

1Ri = = 0.1204m²K/W

8.3

RT = 0.0599 + 0.1376 + 0.0000042 + 2.2857 + 0.0184 + 0.0061 + 0.1204RT = 2.6281 m²K/W




Douglas Buchan

(Interface temp)

D 3.4Find temperature difference Δt at each layer.


0.0599R 0 Δt = 2.628 x 55K = 1.25K (28.75°C)

0.1376R 1 Δt = 2.628 x 55K = 2.88K (25.87°C)

0.0000042R 2 Δt = 2.628 x 55K = 0.000088K (25.87°C)

2.2857

R 3 Δt = 2.628 x 55K =47.84K (-21.97°C)

0.0814R 4 Δt =2.628 x 55K = 1.7K (-23.67°C)

0.0061R 5 Δt =2.628 x 55K = 0.13K (-23.8°C)

0.1204R i Δt = 2.628 x 55K = 2.52K (-26.32°C)

(Check: wall Δt = 55°C Total of layer Δt= 56.32) (Small inaccuracy due to rounding up.)




Douglas Buchan

D 3.4 Find vapour resistance (Rv) for each layer

Vapour resistivity of plywood = 3000mn s/gm Vapour resistivity of Aluminium Foil = 4000mn s/gm Vapour resistivity of polyurethane = 600mn s/gm

Vapour resistivity of stud = 70mn s/gm Vapour resistivity of plaster = 55mn s/gm

Rv = vapour resistivity x d

Rv plywood = 3000 x 0.015 = 45gm/MNsRv Aluminium Foil = 4000 = 4000gm/MNs ( “d” omitted as has no effect)Rv1 polyurethane = (600x0.1) x 0.85 = 51gm/MNsRv2 stud = (70x0.1) x 0.15 = 1.05gm/MNsRv plaster = 55 x 0.020 = 1.1gm/MNs

Rvt = Rv plywood + Rv Aluminium Foil + Rv polyurethane + Rv stud+ Rv plasterRvt = 45 +4000+ 51 + 1.05 + 1.1Rvt = 4098.15gm/MNs

Internal vapour pressure @ -25°C 100% RH = 63.29paExternal vapour pressure @ 30°C 100% RH = 4246.2pa(from psychrometric chart 30°CDB + 25°CWB = RH 65%)

External vapour pressure @ 30% 65% RH = 4246.2 x 0.65 = 2760.03 pa

Δp for wall = 2760.03 pa – 63.29 pa = 2696.74 pa

RvLayer Δp = Rvt x Δp wall

45Plywood Δp = 4098.15 x 2696.74 = 29.6 pa

4000 Aluminium Foil Δp = 4098.15 x 2696.74 = 2632.15 pa

52.05Polyurethane + stud Δp = 4098.15 x 2696.74 = 34.25 pa

1.1

Plaster Δp = 102.15 x 2696.74 = 0.724 pa(Plasterboard and skim added for ease of calculation)




Douglas Buchan

Dew point temp

Interface temp

D3.4

Vapour pressure after plywood layer = 2760.03pa – 29.6pa =1572.92 pa

Dew point temp for 2730.43pa = 22°C

Vapour pressure after Aluminium Foil layer = 2730.43 pa – 2632.15 pa = 98.28pa


Vapour pressure after stud + polyurethane layer = 98.28pa – 34.25 pa = 64.03pa

Dew point temp for 64.03pa = -25 °C

Vapour pressure after plaster layer = 64.03pa – 0.72pa = pa


28.75°C

25.87°C

14.0°C

30°C

-20.0°C

-25.0°C

As can be observed from the above diagram the risk that condensation will form has nowbeen eliminated by introducing an aluminium foil layer to the outside of the polyurethanefoam. The dew point temperature has dropped yet the interface temp remains similar tothe previous example. It is important to place the barrier on the “warm side” of theinsulation. With a vapour barrier on the cold side of the wall, warm, moist outside air willpenetrate the insulation layer and the moisture vapour will condense as it nears the colder

inner wall and remain there. This can cause rot and diminishes the efficiency of theinsulation.

-25.0° C

-23.67° C

-21.97° C

-25° C




List of Sources for this Assignment are as follows:

Class Handouts (Heat transfer)Thermodynamic properties of moist air tables (handout)

Documents

Services Science - Heat Energy Through Building Structure