Design Presentation [Compatibility Mode]

7/29/2019 Design Presentation [Compatibility Mode]

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we have it all

n er oor ea ng

ys em es gn


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we have it all

Floor Heating History:

Floor Heating System goes

back more than 2000 years;

ROMAN AGE


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we have it all

2 reasons:


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Ideal Heating Floor Heating Ceiling Heating Radiator HeatingOuter Wall

Radiators HeatingInner Wall


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Design Process:we have it all

Design Process:Heat Load Calculations:

Heat load is ;

=

Where Q = U A (T in - T out)

Q infiltration + Q ventilation

And U = 1/Sum R ; where R is thermal resistance

=

k is the thermal conductivity.

ere s t e area w ere eat s trans erre t roug .

T in; is the room temperature.

T out; is the outside temperature (fixed value for each city).


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Then;

= * *

Q ceiling = U ceiling * Aceiling * (T in T out)

* *oor oor n ou

Qwindow= U window* Awindow* (T in T out)

oor = door door in out

Note that the U-value for windows & doors are obtainedfrom the manufacturer.


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But;

= * * *

QVentilation = f * V * Number of air change per hour * (T in T out)

Where;

Vis the volume

F factor = 12 0 600Rooms in

Resedential No of A-Ch

No External Windows,doors 0.50

As for ventilation the air changevalues are according to required.

One Side External Windows,doors 1.00

Two Sides External Windows,doors 1.50

ree es x erna n ows, oors .

Entrance Halls 2.00


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Note: when calculating the thermal resistance the air films

that are in touch with the wall, ceiling or floor are takennto cons erat on as a convect ona res stance.

Wall Ceiling Floor

Heat transfer Horizantal Upword Downword

Wall Ceiling

Exposed

floor

Wind Speed at 5 m/s at 5 m/s at 0.5 m/s

Ri (m2K/W) 0.12 0.10 0.15Ro (m2K/W) 0.03 0.02 0.09

R = Ri + x1/k1 + x2/k2 + x3/k3 ++ Ro


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The final result is the heat load for each room

Room

No.

Q. Wall

Ext.

Q.

Window

Q.

Door

Q.

Infltrtion

Q.

Infltrtion Q. Floor Q. Ceiling

Q. Total

(kcal/h)

Q. Total/A.

(kcal/h.m^2)

Q.total/A(W/

m^2)

Gr 01 551.43 1324.40 385 1.50 2629.83 1644.62 1644.62 7052.93 107.56 125.1

Gr 02 303.60 385.00 0 1.00 1110.48 1041.70 1041.70 3347.83 80.60 93.7

. . . . . . . . .

Gr 04 76.82 36.96 0 1.00 206.94 194.12 194.12 612.03 180.01 209.3

Gr 05 240.90 481.25 0 1.00 1309.53 1228.42 1228.42 3870.28 161.26 187.5

Gr 06 69.96 77.00 0 1.00 197.31 185.09 185.09 616.77 83.57 97.2

Gr 07 687.72 1108.80 385 1.50 2807.29 1755.60 1755.60 7328.89 348.99 405.8

. . . . . . . . .

Gr 09 646.80 0.00 1732.5 1.50 7700.00 2407.68 2407.68 12841.52 133.77 155.5

Gr 10 366.96 554.40 0 1.00 1366.22 1281.59 1281.59 4182.54 81.85 95.2

Gr 11 125.40 77.00 0 1.00 162.82 152.74 152.74 579.05 26.32 30.6

Gr 12 256.08 770.00 0 1.50 707.04 442.16 442.16 2257.70 128.06 148.9


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Design Process:

28C - 29C --- Live area inside the room.

31C -

32C --- Surrounding areas and bathrooms.


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Design Process:

.

Accordingly, we obtainQ/A(watt/m2 )

From Under Floor Heating Tables wecan then choose the right spacing

Heat Pipe diameter

W/m2 0.16 m 0.20 m 0.25 m

choosing the pipe diameter for the

required system.

0-60 0.30 m 0.30 m 0.40 m

60-70 0.25 m 0.25 m 0.30 m

70-80 0.25 m 0.30m 0.30 m

Note: if the heat is more than UFHsystem can give, then auxiliary heating

80-90 0.20 m 0.25m 0.25 m

90-100 0.20 m 0.20 m 0.25 m

.

is required.

- . . .

110-120 0.10 m 0.10 m 0.15 m

120-130 0.10 m 0.10 m 0.15 m


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Design Process:

L length = Aavailable area / S spacing

The length from the beginning of the room to the cabinet isalso measured and added to the length inside the room.

L length = (Aavailable area / S spacing) + L1

Where L1 is the distance from room to cabinet multipliedy

And since it is recommended that;

The length of a single loop is less than 120m.

The difference between loops lengths are not high.


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Design Process:

Since it is always recommended that the spacing at the,two areas;

1 is for reduced s acin S1 near external walls

A1 = 0.6 * B

A 2 is for normal spacing S2;

=

where A is the total area for the room

en length = 1 1 + 2 2


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Design Processwe have it all

Design Process:The average water temperature Tavg for the system isthen calculated for each zone after according to thelayers above the UFH pipes (the thermal resistances forthese layers)

Chose the designT = TS - TR Normally taken as 8 C

Where:TS

is water supply temperature.

TRis water return temperature.


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10mm Ceramic Tiles

20mm Morter

40mm Sand

20mm Screed

Pipes

Layer X (m) K (w/mK) R (m2K/W) T1 T2

. . . . .

Morter 0.025 1.400 0.018 29.67 31.45

Sand 0.032 0.700 0.046 31.45 36.02

Screed 0.030 1.200 0.025 36.02 38.52. . . . .

Tsup Tret

0.097 0.100 42.99 34.99

0.179

resistivity) 0.100 m2K / W

U (overall heat

transfer coeff.)

10.011 W / m2K


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14 mm Solid wood

3 mm Airgap

20 mm morter

20 mm sand

20 mm Screed

0.002mm Pipes

Layer X (m) K (w/mK) R (m2K/W) T1 T2

Solis wood 0.014 0.150 0.093 29.00 39.73

airgap 0.003 0.100 39.73 51.23

Terrazo 0.025 1.200 0.021 51.23 53.63

Morter 0.020 1.400 0.014 53.63 55.27

Sand 0.020 0.700 0.029 55.27 58.56Screed 0.020 0.870 0.023 58.56 61.20

Pipe 0.002 0.430 0.005 61.20 61.74

Tsup Tret

0.104 65.74 57.74

R (thermal

resistivity) 0.285 m2K / W

U (overall heat

transfer coeff.)

3.513 W / m2K


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es gn rocess:

The flow rate for each zone is calculated and also theflow rate for each branch;

Q = m * Cp * T

Then the pressure drop is calculated for each branchand for the main pipe for each zone.

Where:

Pmain is found according to m main

rom e pressure rop curves.

P branch is found according to m branch


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Design Process:

- = a

-T = 8 C

Then:

TR = 40 - 8/2 = 36 C

S


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In order to calculate the pressure drop

known.

Diameter of main pipe must be known


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Heat load for the zone must be calculated Q (Watts)

Temperature difference for the system must be defined (T)

Mass flow rate is then calculated from equation

M = C * T

Diameter of the main pipe must be defined

Pressure drop per meter length is then found from pipe chart.

s va ue s t en mu t p e y t e engt o t e p pe.


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Heat load for each branch must be defined Q (Watts).

Temperature difference for the branch must be defined (T).

Mass flow rate is then calculated for each branch from theequation: M = Q / (Cp* T)

The critical branch must be defined. Which is the one withhighest pressure drop.

Diameter for each branch must be defined.

ressure rop s t en oun rom p pe c art or eac rancand multiplied by the length of the pipe.

.


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Pressure drop for fittings such as manifolds, elbows,etc. is

Pressure drop for other parts of the system must be calculated.

also calculated (detailed or approximate).

t ree way va ves, stra ners,..

The total pressure drop then is;

The flow rate is known.P = P

(main)

+ P(branch)

+ P(fittings & parts)

The pump is then selected, by finding one that covers therequired flow and pressure drop (head).


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Pump Curve;

The total pressure drop:

P (mbar)

The total flow rate:

m (l/s).

h it ll


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Design Process:

Calculate the total heat load forunder floor heating.

Add other extra loads such as;domestic hot water, other heatingnetworks, swimming pools, etc.

Safety factor should be addeddepending on the boiler efficiency.

h it ll


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Design Process:

balanced UFH system.

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