building services assignment

Building Services AssignmentSizing Radiators for my Home

Sean O’KeeffeClass 2B

Construction StudiesNational Certificate in Construction

Assignment due: 13 Feb 2007

Lecturer: K Coleman

History of Property:My house is located in Rockforest, Killavullen, just outside Mallow parish.

The house is a three bedroom bungalow of approximately 90 square metres. It was built in 1964. I purchased the house in 2004.

Originally, the house had a suspended timber floor. (This is evident from the vents that were left in the external walls (below floor level)). This was replaced with a ground supported concrete slab at some point. However, no insulation was placed under the slab. This came to my attention when I was putting channels in the slab to facilitate upgrading of the central heating, when I first purchased the property.

I also insulated the attic as there was no insulation in the attic. I used 150mm deep glass wool insulation. The roof is of a cold roof construction.

Upvc double glazed windows and doors had been put in the house in the recent past.

The external walls of the house were constructed with cavity blocks.

The chimneys were sealed by a previous owner.

Assumptions: It is assumed there is no internal heat loss. Therefore I have not

considered heat loss through internal walls or doors. The external wall is of cavity block construction. I have viewed the

wall for the purpose of determining heat loss as a homogenous structure. Plastered on the inside and outside. With two leaves to the wall of 40mm aerated concrete and a cavity in the middle 130mm wide. This is not really correct since the cavity is breached at the ends of the blocks by the blocks.

The roof is of a cold roof construction. I have considered the attic space as external to the house because its air temperature should be the same as the outside air temperature.

I assumed a value of 0.75 air changes/hour for calculating heat loss due to ventilation.

2

Calculations of U-values:

External Wall:

The above diagram shows a section through the external wall of the house.

It is composed of: 70mm sand and cement render on the outside 210mm cavity block with a 130mm cavity in the middle 10mm sand and cement render inside the block 5mm of gypsum skim coat

Kvalues:Aerated concrete – 0.14 W/mKSand + cement render – 0.53 W/mKGypsum plaster – 0.40 W/mK

Rvalue = Thickness of material/ Kvalue

80mm Aerated concrete = 0.08/0.14 = 0.571 mK/W80mm Sand + cement render = 0.08/0.53 = 0.151mK/W

3

5mm Gypsum plaster = 0.005/0.40 = 0.013mK/W

Rvalue for air is 0.18mK/W

Calculations of Uvalue for external wall

Rsi = 0.123 mK/WRso = 0.055 mK/W

Uvalue = 1/ Sum of Rvalues= 1 / 0.123 +0.055 +0.571 +0.151 +0.013 +0.18 = 0.915W/mK

Ground Supported Slab:

The above diagram is of a 150mm deep ground supported slab.

Kvalue: Structural Concrete = 1.40W/mK

Rvalue: 0.150/1.40 = 0.107 mK/W

Uvalue: 1/ 0.106 + 0.150 + 0.107 = 2.755 W/mK

(I choice the Rsi and Rso values from the third page of our handout, which apply to the emissivity of internal surfaces. I considered the bottom of the slab to be internal because it is not exposed to the elements.)

4

Ceiling:

The above diagram shows a section through the ceiling showing the following (The ceiling is closer to level than shown here):

150mm Glass wool insulation 20mm Gypsum plasterboard 5mm Gypsum plaster

Kvalue:Gypsum Plasterboard- 0.15 W/mKGypsum Plaster- 0.40 W/mKGlass wool Insulation -1.02 W/mK

Rvalue:Gypsum Plasterboard- 0.020/0.15 = 0.133 mK/WGypsum Plaster - 0.005/0.40 = 0.0125 mK/WGlass wool insulation – 0.150/0.034 = 4.412 mK/W

Uvalue

1/ 0.106+ 0.150+ 0.133+ 0.0125+ 4.412 = 0.208 W/mK

(I used the same values again for the Rsi and Rso as with the ground slab. I considered the top of the ceiling as internal because it is in the attic.)

Windows and Doors:

I was not able to establish a realistic Uvalue for double glazing using the Kvalues in the handout i.e. 1.02 W/mK for glass .I did obtain a Uvalue for double glazing in the irish-energy.ie website. Uvalue for double glazing- 2.8 W/mK

5

Calculations to determine Heat Loss and Radiator Sizes

Rather than calculating the fabric heat loss and ventilation heat loss for each room individually I used a spreadsheet into which I put the formulas. This allowed me just to add the room, window, door and external wall dimensions for each room individually.

I have shown below how the spreadsheet was setup before dimensions were inserted.

Room:Length C3Width C4Height C5

Window:Width C8Height C9

Door:Width C12Height C13

Component: AreaU

valueTemp.

Diff Qf

FloorD16= c3*c4

E16= 2.755

F16=14.0

G16=D16*E16*F16

WindowD17= c3*c9

E17= 2.800

F17=14.0

G17=D17* E17*F17

DoorD18=

c12*c13E18= 2.800

F18=14.0

G18=D18* E18*F18

External Walls D19

E19= 0.915

F19=14.0

G19=D19* E19*F19

CeilingD20

=D16E20= 0.208

F20=14.0

G20=D20* E20*F20

Volume of room D23= D16*C5

: Sum of Areas: D24=SUM(D16:D20)

Sum of Qf's: D25=SUM(G16:G20)

Number of air changes: 0.75

Inside environmental Temp: D30=D25/(4.8*D24)+21

Heat loss due to Vent: D32=0.33*D28*D23*(D30-7)

Total heat loss: D34=D32+D25

Radiator size: D36=D34/870

6

Bedroom 1:

The room dimensions can be seen in the plan shown on page 1. The dimensions for the two windows are 1.400 x 1.150. I entered these

as one window 2.80 x 1.150

Room:Length 4.250Width 3.550Height 2.650

Window:Width 2.800Height 1.150

Door:WidthHeight

Component: Area Uvalue Temp. Diff QfFloor 15.088 2.755 14.0 581.925Window 3.220 2.800 14.0 126.224Door 0.000 2.800 14.0 0.000External Wall 11.590 0.915 14.0 148.468Ceiling 15.088 0.208 14.0 43.935

Volume of room 39.982

Sum of areas 44.985

Sum of Qf's 900.552 W

Number of air changes 0.750

Inside environmental Temp 25.171 C

Heat Loss due to Vent 179.808 W

Total Heat Loss 1,080.359 W

Radiator size 1.242 metres sq.

A radiator (870W/ metre square) of 1.242 square metres would be required for bedroom 1.

7

Bedroom 2



Door:WidthHeight



Sum of areas 30.712





Total Heat Loss 746.478 W


8

Bedroom 3Room:Length 3.300Width 2.760Height 2.650


Door:WidthHeight



Sum of areas 34.275







9

Bathroom

The bathroom has two windows. (1.150 x 0.480 and 1.150 x 0.920) I have entered these as one window 1.150 x 1.400.



Door:WidthHeight



Sum of areas 17.197







10

Kitchen



Door:WidthHeight



Sum of areas 54.731

Sum of Qf's 1,006.056 W




Total Heat Loss 1,191.728 W


11

Living Room



Door:WidthHeight



Sum of areas 44.284







12

Hall

Door:Width 0.950Height 2.150



Sum of areas 24.028







13

Utility Room



Door:Width 0.950Height 2.120



Sum of areas 14.825







14

building services assignment

Documents

wmk gypsum plaster

mkw uvalue

wmk rvalue

mkw gypsum plaster

wmk sand

internal heat loss

mkw rvalue

wmk glass wool insulation