‫`‬

‫‪Subject : HVAc Project‬‬

‫‪Group Names‬‬ ‫‪ID #‬‬
‫غيث مصطفى خالد الشجراوي‬ ‫‪1834666‬‬
‫محمد فؤاد عبد اللطيف صوالحة‬ ‫‪1831570‬‬
‫ريان علي فايز محافظة‬ ‫‪1834666‬‬
‫يوسف عارف نوران الحربي‬ ‫‪1836654‬‬
‫حسن خالد محمد إسماعيل‬ ‫‪1632541‬‬
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External wall Thickness(mm) R

Face brick 70 0.05253
RSI 102 insulation 30 1.44447
Hw concrete block 200 0.19260
Gypsum plaster 20 0.08889

Partitions Thickness(mm) K R
Double face plaster 25 1.2 0.02083
Cement 150 0.95 0.1579
brick(common)

Ceiling Thickness(mm) R
Cement brick with air 180 0.17334
gaps
Lw concrete 60 0.34667
RSI insulation 25 1.20373
Lw concrete 260 0.34667
Built up roofing 20 0.116

Floor Thickness(mm) k R
Cement tiles 50 1.1 0.0454
Aggregates 150 1.05 0.1428
Semi-firm stones 250 1.4 0.1786
`

U W/K.m2
External wall 0.51
Adj 2.275
Ceiling 0.427
Window 3.2
DOOR 2.4
Floor 1.935

Sample of calculation for (bedroom 1)

1-HEATING LOAD CALCULATIONS
From Jordanian code:
𝑇𝑖 = 21℃
𝑇𝑜 = 2℃
𝛥𝑇 = 21 − 2 = 19℃
𝑇𝑖 + 𝑇𝑜 21 + 2
𝑇𝑎𝑑𝑗 = = = 11.5℃
2 2
𝛥𝑇𝑎𝑑𝑗 = (𝑇𝑖 − 𝑇𝑎𝑑𝑗) = 21 − 11.5 = 9.5℃
𝑇𝑔 = 𝑇𝑜 + (5 𝑡𝑜 10) ℃ = 12 ℃
∆𝑇𝑔 = (𝑇𝑖 − 𝑇𝑔) = 21 − 12 = 9℃
1- Heat Due External Walls:-
𝑄̇ 𝑤𝑎𝑙𝑙 = 𝑈𝐴 𝛥𝑇
𝛥𝑇 = 19℃
𝐴 = 28.8𝑚2
𝑄̇ 𝑤𝑎𝑙𝑙 = 279.072 W

2- Heat Due Internal Walls:-

`

𝑄̇ 𝑤𝑎𝑙𝑙, 𝑖𝑛𝑡 = 𝑈𝐴 ∆𝑇𝑎𝑑𝑗

∆𝑇𝑎𝑑𝑗 = 9.5 ℃
𝐴 = 9.92 𝑚2
∑ 𝑄̇𝑖𝑛𝑡𝑒𝑟𝑛𝑎𝑙 = 214.4 𝑊

3- Heat Loss Through The Ceiling:-

𝑄̇ 𝑐𝑒𝑖𝑙 = 𝑈 ∗ 𝐴 ∗ (𝑇𝑜 − 𝑇𝑖)
𝛥𝑇 = 19 ℃
𝐴 = 20 𝑚2
𝑄̇ 𝑐𝑒𝑖𝑙 = 162.26𝑊

4- Through The Windows:-

𝑄̇ 𝑤𝑖𝑛𝑑𝑜𝑤 = UA ΔT
𝛥𝑇 = 19℃
𝐴 = 4 𝑚2
𝑄̇ 𝑤𝑖𝑛𝑑𝑜𝑤 =243.2 W

5- Heat Loss Through The Infiltration:-

1250
𝑄̇ 𝑖𝑛𝑓 = ∗ 𝑁 ∗ 𝑉 ∗ 𝛥𝑇
3600
𝑉 = 3.2 × 4 × 5 = 64 𝑚3
𝑁 = 0.5 for bedrooms
𝛥𝑇 = 19 ℃
𝑄̇ 𝑖𝑛𝑓 = 211.11 W

6- Heat Loss Through The floor:-

𝑄̇ floor= U*A* ΔTg
`

∆𝑇𝑔 = 9℃
𝐴 = 20 𝑚2
𝑄̇ 𝑓𝑙𝑜𝑜𝑟 = 348.3 𝑊

7- Total Heat:-
𝑄̇𝑡𝑜𝑡𝑎𝑙 = 𝑄̇ 𝑤𝑎𝑙𝑙 + ∑ 𝑄̇𝑖𝑛𝑡𝑒𝑟𝑛𝑎𝑙,𝑡𝑜𝑡𝑎𝑙 + 𝑄̇ 𝑐𝑒𝑖𝑙 + 𝑄̇ 𝑤𝑖𝑛𝑑𝑜𝑤
+ 𝑄̇ 𝑖𝑛𝑓 + 𝑄̇ 𝑓𝑙𝑜𝑜𝑟
𝑄̇𝑡𝑜𝑡𝑎𝑙 = 279.072 + 214.4 + 162.26 + 243.2 + 211.11 + 348.3
= 1458.34𝑊

8- Erorr Percentage {%E} :-

∑𝑄̇𝑡𝑜𝑡𝑎𝑙 = ∑𝑄̇ theo = 1458.34 W

∑𝑄̇𝐻𝐴𝑃 = ∑𝑄̇𝐸𝑋𝑃 = 1292 W

∑ 𝑄̇ 𝑡ℎ𝑒𝑜− 𝑄̇ 𝑒𝑥𝑝
%E= ∑ 𝑄̇ 𝑡ℎ𝑒𝑜
∗ 100%

% E = 11.41%

Error come from the fact that we use the Jordanian code while
the HAP software yses ASHRAE code
`

2-Cooling load calculations :

Solar Time =14:00
July Month
Ti=24 ℃
From table ( A-7)
To=31.9 ℃
Tmax=To=31.9 ℃
Tmin=18.2 ℃
(𝑇𝑚𝑎𝑥+ 𝑇𝑚𝑖𝑛) 31.9+18.2
To,m= = = 25.05 ℃
2 2
2 2
Tadj = Ti+ ( 𝑇𝑜 − 𝑇𝑖) = 24 + ∗ (31.9 − 24) = 29.267 ℃
3 3

1- Heat Gain By The Occupants :

𝑄̇ occ= n[𝑄̇ sen(CLF)occ+𝑄̇ lat]
n :- number of occupants
from tables:-
𝑄̇ lat , 𝑄̇ sen :- from table (4-2)
(CLF)occ :- from table (4-3)
N 2
𝑄̇ lat 30
𝑄̇ sen 65
(CLF)occ 0.61

𝑄̇ occ = 2 × (65 × 0.61 + 30)

𝑄̇ occ= 139.9 W
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2- Heat Gain By External wall :-

𝑄̇ wall = Uwall *A *(CLTD)corr
(CLTD)corr = [(CLTD+LM)*K + (25.5-Ti) + (To,m-29.4)*f ]
Ti=24 ℃
To,m=25.05 ℃
f=1
K = 0.83 (Permanent medium colour wall)
LM >>> from table (9-2)
CLTD >>> from table (9-4) >>> Group D & Group E (by doing
interpolation since the weight =351 kg/𝑚2 )

Surface CLTD LM (CLTD)corr A (𝑚2 )

S 11.1667 -1.6 5.09 15.024
E 19.3 0 13.169 11.824

Uwall = 0.51
𝑄̇ 𝑤𝑎𝑙𝑙, 𝑠𝑜𝑢𝑡ℎ= 39.001 W
𝑄̇ 𝑤𝑎𝑙𝑙, 𝑒𝑎𝑠𝑡 = 79.412 W
𝑄̇ 𝑤𝑎𝑙𝑙, 𝑡𝑜𝑡𝑎𝑙 = 39.001 + 79.412 = 118.4132 𝑊
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3- Heat Gain By Infiltration :-

𝑽̇
𝑸̇𝒊𝒏𝒇 = (ho – hi)
𝒗𝒐

from table (6-1) >>> N = 0.5

32𝑚 3 𝑚 3
𝑉̇ = 𝑁 × 𝑉 = 0.5 × 5 × 4 × 3.2 = = 0.0089
ℎ 𝑠

from (A-7) >>> ∅o =40%

comfort Rrelative humidity in summer >>> ∅i=50%
from psychometric chart >>> hi = 48kj/kg , ho = 63 kj/kg, vo = 0.88
𝟎.𝟎𝟎𝟖𝟗
𝑸̇𝒊𝒏𝒇 = ∗ (𝟔𝟑 − 𝟒𝟖) = 151.70 W
𝟎.𝟖𝟖

4- Heat Gain By Glass :-

𝑄̇ 𝑔𝑙𝑎𝑠𝑠 = 𝑄̇ 𝑡𝑟𝑎𝑛𝑠𝑚𝑖𝑡𝑡𝑒𝑑 + 𝑄̇ 𝑐𝑜𝑛𝑣𝑒𝑐𝑡𝑒𝑑
𝑄̇ 𝑡𝑟𝑎𝑛𝑠𝑚𝑖𝑡𝑡𝑒𝑑 = 𝐴 ∗ (𝑆𝐻𝐺 ) ∗ (𝑆𝐶 ) ∗ (𝐶𝐿𝐹 )
(SHG) >>> from table (9-7)
(SC) >>> from table (9-8)>>Double glass (Regular)
(CLF) >>> from table (9-11)
Surface SHG SC CLF Area{m2}
S 227 0.57 0.68 2
E 678 0.57 0.22 2

𝑄̇ 𝑡𝑟𝑎𝑛𝑠𝑚𝑖𝑡𝑡𝑒𝑑, 𝑠𝑜𝑢𝑡ℎ = 2 ∗ 227 ∗ 0.57 ∗ 0.68 = 175.97 𝑊

`

𝑄̇ 𝑡𝑟𝑎𝑛𝑠𝑚𝑖𝑡𝑡𝑒𝑑, 𝑒𝑎𝑠𝑡 = 2 ∗ 678 ∗ 0.57 ∗ 0.22 = 170.0424 W

𝑄̇ 𝑐𝑜𝑛𝑣𝑒𝑐𝑡𝑒𝑑 = 𝑈𝑤𝑖𝑛𝑑𝑜𝑤 ∗ 𝐴 ∗ (𝐶𝐿𝑇𝐷 )𝑐𝑜𝑟𝑟

U = 3.2 from table (5-4)
(CLTD)corr= [(CLTD + LM)*K+(25.5 - Ti) + (To,m - 29.4)f ]
Ti=24 ℃
To,m=25.05 ℃
f=1
K=1 for glass
from table (9-12) >>> CLTD =7
LM >>> from table (9-2)

Surface LM (CLTD)corr Area

S -1.6 2.55 2
E 0 4.15 2

𝑄̇ 𝑐𝑜𝑛𝑣𝑒𝑐𝑡𝑒𝑑, 𝑠𝑜𝑢𝑡ℎ = 16.32 𝑊

𝑄̇ 𝑐𝑜𝑛𝑣𝑒𝑐𝑡𝑒𝑑, 𝑒𝑎𝑠𝑡 = 26.56 𝑊
𝑄̇ 𝑔𝑙𝑎𝑠𝑠 = 𝑄̇ 𝑡𝑟𝑎𝑛𝑠𝑚𝑖𝑡𝑡𝑒𝑑 + 𝑄̇ 𝑐𝑜𝑛𝑣𝑒𝑐𝑡𝑒𝑑
𝑄̇ 𝑔𝑙𝑎𝑠𝑠 = 175.97 + 170.0424 + 16.32 + 26.56 = 388.89 𝑊
`

6- Heat Gain By The Light :-

𝑸̇𝑳𝒕 = 𝑁 ∗ 𝑃𝐿𝑡 (𝐹𝑢 𝐹𝑏 )(𝐶𝐿𝐹 )𝐿𝑡
𝑃𝐿𝑡 :- the lighting intensity [W]
𝐹𝑢 = 1
𝐹𝑏 = 1.2
(𝐶𝐿𝐹 )𝐿𝑡 (𝐹𝑖𝑥𝑡𝑢𝑟𝑒 𝑌) = 0.775
𝑸̇𝑳𝒕 = 𝟐 ∗ 𝟑𝟔 ∗ 𝟏 ∗ 𝟏. 𝟐 ∗ 𝟎. 𝟕𝟕𝟓 = 𝟔𝟔. 𝟗𝟔 𝑾

7- Heat Adjacent :-
∆𝑇𝑎𝑑𝑗 = (𝑇𝑖 − 𝑇𝑎𝑑𝑗 ) = 24 − 29.267 = −5.267 ℃
𝑼𝒂𝒅𝒋 = 𝟐. 𝟐𝟕𝟓
𝑄̇ 𝑖𝑛𝑡. 𝑤𝑎𝑙𝑙 = 𝑈𝐴 𝛥𝑇𝑎𝑑𝑗
𝐴 = 14 𝑚2
𝑄̇ 𝑖𝑛𝑡. 𝑤𝑎𝑙𝑙 = −167.754 𝑊

𝑼𝒅𝒐𝒐𝒓 = 𝟐. 𝟒
𝑨𝒅𝒐𝒐𝒓 = 𝟐𝒎𝟐
𝑄̇ 𝑖𝑛𝑡. 𝑑𝑜𝑜𝑟 = −25.282 𝑊
∑ 𝑄̇𝑎𝑑𝑗 = −25.282 − 167.754 = −193.0356 𝑊

8-heat due ceiling :

Uceiiing = 0.427
A = 4*5 = 20 𝒎𝟐
`

𝑄̇ 𝑐𝑒𝑙𝑖 = 𝑈 ∗ 𝐴 ∗ (𝑇𝑜 − 𝑇𝑖)

𝑄̇ 𝑐𝑒𝑙𝑖 = 0.427 ∗ 20 ∗ (31.9 − 24)
𝑄̇ 𝑐𝑒𝑙𝑖 = 67.466 𝑊

NOTE : THE HEATT TRANSFER DUE FLOOR IS NEGLECTED

9- total cooling load

∑ 𝑄̇𝑇𝑜𝑡𝑎𝑙 = 𝑄̇𝑐𝑒𝑖𝑙𝑖𝑛𝑔 + 𝑄̇𝑤𝑎𝑙𝑙 + 𝑄̇𝐿𝑡 + 𝑄̇𝑜𝑐𝑐 + 𝑄̇𝑔𝑙𝑎𝑠𝑠 + 𝑄̇𝑖𝑛𝑓 + ∑ 𝑄̇𝑎𝑑𝑗

∑ 𝑄̇𝑇𝑜𝑡𝑎𝑙 = 67.466 + 118.4132 + 66.96 + 139.9 + 388.89 + 151.7

− 193.0356
= ∑ 𝑄̇𝑇𝑜𝑡𝑎𝑙 = 1.1 ∗ (740.29 ) = 814.32 𝑊
We multiply the total cooling load by 10% safety factor since thre is
miscellaneous load an emergency loads.

10- Erorr Percentage {%E} :-

∑ 𝑄̇𝑇𝑜𝑡𝑎𝑙 =∑ 𝑄̇𝑇ℎ𝑒𝑜 = 814.32 W
∑ 𝑄̇𝐻𝐴𝑃 =∑ 𝑄̇𝑒𝑥𝑝 = 894 W
∑ 𝑄̇𝑇ℎ𝑒𝑜 −∑ 𝑄̇𝑒𝑥𝑝 814.32−894
%𝐸 = ∑ 𝑄̇𝑇ℎ𝑒𝑜
∗ 100% = *100% = 9.78%
814.32

Now we will design the heating system for this residental house
`

Systems
Underfloor heating system
It is a hot water heating method known as floor panel heating.
The underfloor heating using thermoplastic pipes is the most commonly
used heating method .
The purpose that this system is common that it use low hot water
temperature.so that fact results in reducing heat losses from the
system.
Comfort inside temperature = 21 ℃
Outside design temperature =2℃
Inside Relative humidity = 50%
Outside Relative humidity = 40%
Pipe size is (20×2)mm (20mm diameter,2mm thickness)
Type of pipes is PEX -AL-PEX 2mm
1 is saloon(1) , 2 is Saloon(2) , 3 is living room , 4 is library ,5 is bedroom
(1)
6 is bedroom (2) , 7 is bedroom (3) , 8 is kitchen

Space No 1 2 3 4 5 6 7 8

𝑄̇ , 𝑊 2457 1297 1732 1022 684 1292 1131 1465

Floor 26.8 20.1 36.9 23.5 20 20 17.9 19.2

area 𝑚2
`

Now total heating demand

𝑞̇ 𝑡 = ℎ ∗ (𝑇𝑓 − 𝑇𝑖 )
𝑄̇𝑡
𝑞̇ 𝑡 =
𝐴

𝑄̇1 + 𝑄̇2 + 𝑄̇3 + 𝑄̇4 + 𝑄̇5 + 𝑄̇6 + 𝑄̇7 + 𝑄̇8 11080

𝑞̇ 𝑡 = =
𝐴1 + 𝐴2 + 𝐴3 + 𝐴4 + 𝐴5 + 𝐴6 + 𝐴7 + 𝐴8 184.4
𝑊
= 60.086 2
𝑚

So the value is less than 100 as required

𝑄̇1 2457
𝑞̇ 𝑡,1 = = = 91.68 w/𝑚2
𝐴1 26.8

For practical underfloor heating applications

𝑞̇ 𝑐 = 0.48 ∗ 𝑞̇ 𝑡,1
𝑞̇ 𝑐 = 0.48 ∗ 91.68 = 44.006 𝑊
1.12
𝑞̇ 𝑐 = 4.57(𝑇𝑓,1 − 𝑇𝑖 )
𝑇𝑓,1 = 28.555℃
𝑇̅𝑤 = 𝑇𝑓 + ∆𝑇𝑠𝑡 + ∆𝑇𝑐𝑜𝑣.
𝑇̅𝑤 = 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑤𝑎𝑡𝑒𝑟 𝑡𝑒𝑚𝑝𝑟𝑒𝑡𝑢𝑟𝑒
∆𝑇𝑤 = 5℃ so
𝑇𝑤,𝑖𝑛 = 𝑇̅𝑤 + 2.5
∆𝑇𝑠𝑡 = 𝑡𝑒𝑚𝑝𝑟𝑒𝑡𝑢𝑟𝑒 𝑑𝑟𝑜𝑝 𝑡ℎ𝑟𝑜𝑢𝑔ℎ 𝑓𝑙𝑜𝑜𝑟 𝑠𝑡𝑟𝑢𝑐𝑡𝑢𝑟𝑒
∆𝑇𝑠𝑡 = 𝑞̇ 𝑡,1 ∗ 𝑅𝑡ℎ,𝑠
 `

𝑅𝑡ℎ,𝑠 = 𝑡ℎ𝑒𝑟𝑚𝑎𝑙 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝑐𝑜𝑛𝑠𝑡𝑟𝑢𝑐𝑡𝑖𝑜𝑛 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙 𝑎𝑏𝑜𝑣𝑒 𝑡ℎ𝑒 𝑝𝑖𝑝𝑒𝑠

= 0.1428
∆𝑇𝑠𝑡 = 91.68 ∗ 0.1428 = 13.092 ℃
∆𝑇𝑐𝑜𝑣. = 𝑡𝑒𝑚𝑝𝑟𝑒𝑡𝑢𝑟𝑒 𝑑𝑟𝑜𝑝 𝑡ℎ𝑟𝑜𝑢𝑔ℎ 𝑓𝑙𝑜𝑜𝑟 𝑐𝑜𝑣𝑒𝑟𝑖𝑛𝑔 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙
∆𝑇𝑐𝑜𝑣 = 𝑞̇ 𝑡,1 ∗ 𝑅𝑐𝑜𝑣
We used cement tiles as a coverage with R=0.0454
∆𝑇𝑐𝑜𝑣 = 91.68 ∗ 0.0454 = 4.16℃ so
𝑇̅𝑤,1 = 13.092 + 4.16 + 28.555 = 45.807 ℃
𝑇𝑤1,𝑖𝑛 = 45.807 + 2.5 = 48.31℃
𝑚̇𝑤 𝑄̇1 2457 𝑘𝑔 𝐿
= = = 0.1174 (𝑠 )
𝑙𝑜𝑜𝑝 20.93 20.93(1000) 𝑠

Using figure 7-13

𝑚̇𝑤 𝑘𝑔 𝐿 ∆𝑝
for = 0.1174 (𝑠 )and pipe diameter =20mm, 𝐸𝐿 = 0. 225
𝑙𝑜𝑜𝑝 𝑠

Similar calculations are performed for the other seven Spaces

Space 1 2 3 4 5 6 7 8
#NO
𝑞̇ 𝑡 , 91.86 64.53 46.94 43.489 34.5 64.6 63.184 76.3
W/𝑚2
𝑇𝑓 , ℃ 28.555 26.52 25.16 24.88 24.16 26.53 26.65 27.412

𝑇𝑤,𝑖𝑛 ℃ 48.31 41.17 36.5 35.56 33.153 41.188 41.04 44.272

𝑚̇𝑤 0.1174 0.062 0.083 0.049 0.033 0.062 0.054 0.07

𝐿
(𝑠 )/loop
∆𝑃 0.225 0.08 0.12 0.055 0.025 0.08 0.06 0.1
𝐸𝐿
`

𝑚̇𝑤,𝑡 = 𝑚̇𝑤,1 + 𝑚̇𝑤,2 + ⋯ + 𝑚̇𝑤,8

𝑙
𝑚̇𝑤,𝑡 = 0.53
𝑠

Now we should pick up a boiler that satisfied our requirement with its
chimmey and expansion tank
From Jordanian code the domestic hot water rate for one family is 280
L/day
Or (50-70) L/day for the first two person in the family and (30-50) L/day
for every single person increased
In Our design the family consists of 8 person and the average number
of people being in the home
So 𝑀̇ℎ𝑜𝑡,𝑤𝑎𝑡𝑒𝑟 = 315
1
𝑄̇𝐻𝑊 = ∗ 𝐶𝑝,𝑤 ∗ (𝑇ℎ − 𝑇𝑐 )
3600∗∆𝑡

∆𝑡 =
𝑡𝑖𝑚𝑒 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑡𝑜 ℎ𝑒𝑎𝑡 𝑑𝑜𝑚𝑒𝑠𝑡𝑖𝑐 ℎ𝑜𝑡 𝑤𝑎𝑡𝑒𝑟 𝑎𝑛𝑑 𝑖𝑡𝑠 𝑢𝑠𝑢𝑎𝑙𝑙𝑦 1 ℎ𝑜𝑢𝑟
𝑇ℎ : 𝑠𝑢𝑝𝑝𝑙𝑖𝑒𝑑 ℎ𝑜𝑡 𝑤𝑎𝑡𝑒𝑟 𝑡𝑒𝑚𝑝𝑟𝑒𝑡𝑢𝑟𝑒
𝑇𝑐 ∶ 𝑡ℎ𝑒 𝑡𝑒𝑚𝑝𝑟𝑒𝑡𝑢𝑟𝑒 𝑜𝑓 𝑐𝑜𝑙𝑑 𝑤𝑎𝑡𝑒𝑟 𝑡ℎ𝑎𝑡 𝑚𝑢𝑠𝑡 𝑏𝑒 ℎ𝑒𝑎𝑡𝑒𝑑
The temperature difference (𝑇ℎ − 𝑇𝑐 ) is usually taken as 50 to 60 ℃
𝑘𝐽
So (𝑇ℎ − 𝑇𝑐 ) = 55℃ , 𝐶𝑝,𝑤 = 4.186 .𝑘
𝑘𝑔
1∗4.186∗55∗305
𝑄̇𝐻𝑊 = = 20145.125 𝑊
3600

𝑄̇𝑏𝑜𝑖𝑙𝑒𝑟 = 1.1 (𝑄̇𝐻𝑊 + 𝑄̇𝑡 ) = 1.1 ∗ (20145.125 + 11080) =

34.348 𝐾𝑤
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So the suitable boiler is selected from the manufacturing selection

tables for the boilers.
Boiler model is (AR5)
It has 4 sections , 𝑄̇𝑜𝑢𝑡𝑝𝑢𝑡 = 40 𝑘𝑊 ,𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 = 88.1%
Boiler chimney
𝑚̇𝑔
𝐴𝑐 =
𝜌𝑔 𝑣

𝑚̇𝑔 = 𝑚𝑎𝑠𝑠 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑐𝑜𝑚𝑏𝑢𝑠𝑡𝑖𝑜𝑛 𝑔𝑎𝑠𝑒𝑠 𝑙𝑒𝑎𝑣𝑖𝑛𝑔 𝑡ℎ𝑒 𝑐ℎ𝑖𝑚𝑛𝑒𝑦

V= the fuel gases average velocity ,Its value ranges from 3 to 5 m/s
𝑄̇𝐻 34.348 𝑘𝑔
𝑚̇𝐹𝑢𝑒𝑙 = = = 0.0009747
𝑒𝑓𝑓𝑖𝑒𝑐𝑖𝑒𝑛𝑐𝑦 ∗ 𝐶. 𝑉 0.881 ∗ 40000 𝑠
Since 1 kg/s of the fuel produce 25.2 kg/s of combustion gases so
𝑘𝑔
𝑚̇𝑔 = 25.2 ∗ 0.0009747 = 0.02456
𝑠
𝑘𝑔
Assuming that 𝜌𝑔 = 1.1 ,v=4 m/s
𝑠
0.02456
So 𝐴𝑐 = = 5.5823 ∗ 10−3 𝑚2
1.1∗4

Diameter of the chimney is = 84.32 mm

Since the minimum diameter recommended is 126 mm then the
chimney diameter is taken as 126 mm.
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AR5 Boiler
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Annual fuel consumption >>

̇
𝑄 ×3600×24×𝐷𝐷𝐻 ×𝐶𝑑
𝑀̇𝐹𝑢𝑒𝑙 = 𝐻 (𝑇 )×𝐶.𝑉×𝜗
𝑖 −𝑇0

C.V is the calorific value and for diesel fuel =40000

Density=825 kg/𝑚3
𝜗 = 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑜𝑖𝑙𝑒𝑟 = 88.1%
𝑄̇𝐻 = 34.348 𝑘𝑤
𝐷𝐷𝐻 𝑓𝑜𝑟 𝑎𝑚𝑚𝑎𝑛 = 1228
𝐶𝑑 = 0.77
So
34.348∗3600∗24∗1228∗0.77 𝑘𝑔
𝑀̇𝐹𝑢𝑒𝑙 = (21−2)∗40000∗0.881 = 4190.972
𝑠𝑒𝑎𝑠𝑜𝑛

The maximum floor surface temperature is 28.555℃ which is in the

comfortable range also the max water inlet temperature is 48.31℃
Now in order to draw the layout for the underfloor heating system : the
recommanded distance between every two pipes is (20-30) cm .but
that’s not a good choice because in that way we may put pipes more
than we need so increasing cost and increasing in heat supplied to the
space more than our requirement.so the only determination for the
distance between every two pipes is the heating load of the space.so to
avoide that as much as we can we will use a computer programe to
draw the pipes layout along every single area of the heated space and
hence estimating the length of the pipes . the peogramme we will use
called (loop cad) which is specialized in designing the underfloor
heating systems.
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Since the location of the manifolds should be in the centre of the house
in order to make the pipe lengths as minimum as we can and every
single manifolds is recommended to assign (6 to 8) loops .so we see
that the best thing to do is to assign two manifolds as shown below.
Note :since the max supply temperature is 48.555 ℃ so we will input
the water temperature to be either 48.55 or 50 ℃
The suggested layout is :

We see here that the manifolds placed at location where the

maintaince if needed is easy and and every one manifolds supply water
for 4 rooms .so nearly is located at the centre of that 4 rooms.
Note: we drop all heating system report in the project file .
Here we will just drop the summary such as the length of the pipes in the spaces

Note : the pipe arrangement is single serpentine.

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The depth of the circuits is 50 mm

The next layout isn’t proffesionial.it is just to show the location of the
manifolds selected from the boiler.and so to show how well the
location of the manifolds is.

From “loop cad “ reports the loop lengths with supply and return loop
lengths is :

Space No
1 2 3 4 5 6 7 8

length loop 49.6/44.2 39.8 49.8 35.8 25 37.5 38 49.2

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Note :the programme is taking the supply and return length as a part
of the total length of the circuit so (loop length +supply pipe length
+return pipe length = the circuit length that reported in the
programme.)

Note: mass flow rate that estimated before have some error and the
loop cad programme have found mass flow rate so we will use the loop
cad reports results.
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So the pump that would be used for manifold ( 1 ) should be satisfy

that 𝑚̇𝑤1,𝑡 = 0.29 , ∆𝑃𝐿𝑜𝑜𝑝1,𝑚𝑎𝑥 + ∆𝑃𝑀𝑎𝑛𝑖𝑓,1 = 9.8 + 6.8 = 16.6 𝑘𝑝𝑎
∆𝑃𝑡 = 16.6 𝐾𝑝𝑎
The pump that would be used for manifold ( 2) should be satisfy that
𝑚̇𝑤2,𝑡 = 0.175 , ∆𝑃𝐿𝑜𝑜𝑝2,𝑚𝑎𝑥 + ∆𝑃𝑀𝑎𝑛𝑖𝑓,1 = 9.5 + 6.5 = 16 𝑘𝑝𝑎
∆𝑃𝑡 = 16𝐾𝑝𝑎

Pex Al-Pex pipes

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3D VIEWS FOR THE PROJECT AND LAYOUTS

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Cooling system design:

Cooling by split unit:

A wall-mounted ductless air conditioner also referred to as a mini split

system or ductless mini split system is an easy, convenient and efficient
way to gain refrigerated air without the expense or hassle of adding
ductwork to a home or business looking for air conditioning.

In our project, we will use PETRA CO. Products.

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Zone Cooling Load W Cooling Load Petra Co. Product

TR selection
Bedroom1 1325 0.38 CPW 9 / PTR MSG 9 C

Bedroom2 1714 0.5 CPW 9 / PTR MSG 9 C

Bedroom3 1382 0.39 CPW 9 / PTR MSG 9 C

Kitchen 8308 2.35 CPW 30 / PTR MSG 30 C

Library 3302 0.93 CPW 12 / PTR MSG 12 C

Living Room 4484 1.28 2 x CPW 9 / PTR MSG 9 C

Saloon1 2056 0.6 CPW 9 / PTR MSG 9 C

Saloon2 1272 0.36 CPW 9 / PTR MSG 9 C

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Sketch for the split unit air conditioning.