The University of Jordan

School of Engineering
Mechanical Engineering Department

Air-conditioning (1) Project

Spring Semester 2024

Instructor: Dr. Osama Ayadi

By:

Name Serial Number

Neamh Mohannd 0202875
Dana Odeh 0207217

May 21,2024
Table of contents:
Design parameter ………………………………………………………………....2
Calculating heating load…………………………………………………………. 4
Walls………………………………………………………………………………5
Doors………………………………………………………………………………6
Windows……………………………………………………………………………6
Sample of calculations………………………………………………………….......7
Using HAP………………………………………………………………………15
Domestic Hot Water………………………………………………………………21
Sizing of boiler……………………………………………………………………21
Radiator selection…………………………………………………………………23
Pipe selection……………………………………………………………………25
Pump selection……………………………………………………………………25
Energy Consumption……………………………………………………………..26
Terminal System…………………………………………………………………..27
Costs Based on the Jordanian Market…………………………………………….28
Sizing of hot water heating system (HWHS)……………………………………..29
Cooling Part……………………………………………………………………30
References ………………………………………………………………………..36

1
 Design parameters:
The design process involves calculating heating loads, selecting
appropriate heating equipment, and designing the distribution system to
meet the specific needs of a building.
Below, I outline the basic steps and considerations for designing a
heating system:
 Understand the Design Conditions:
 Inside Design Conditions: Typically, for winter conditions, an indoor
temperature of 20°C (68°F) is used for residential spaces. However, this
can vary based on local preferences and standards.
The inside temperature that we will work on it is 𝑇𝑖𝑛𝑠𝑖𝑑𝑒 = 21℃
 Outside Design Conditions: Refer to local climate data for the coldest
average temperatures for your area. ASHRAE (American Society of
Heating, Refrigerating and Air-Conditioning Engineers) provides design
temperature guidelines.

Our building is located in Amman:

2
 We will choose to design a heating system at the lowest outside temperature:
(Heating load)
Minimum 𝑇𝑜𝑢𝑡𝑠𝑖𝑑𝑒 = 0℃
We will choose to design a cooling system at the highest outside
temperature: (cooling load)
Maximum 𝑇𝑜𝑢𝑡𝑠𝑖𝑑𝑒 = 38℃

By default, from ASHRAE-Fundamentals tables:

We choose new concrete at angle = 70°, So

Average Ground Reflectance = 0.34
3
 We choose the sands, So soil conductivity =1.5

2. Calculate the Heating Load:

The heating load is the amount of heat energy that needs to be added to
a space to maintain the desired indoor temperature during the coldest
weather.
 This calculation considers various factors:
Building Envelope: Analyze the construction materials and thermal
properties (R-values) of walls, celling, and floors.
Walls: Table (5-6)
Celling & floor: Table (5-7)

4
  Overall heat transfer coefficients for outside wall:

 Overall heat transfer coefficients for inside wall:

Note:
The overall heat transfer coefficient for an outside wall is typically larger
than that of an inside wall. This is because outside walls are exposed to
varying external conditions such as temperature, wind, and sunlight,
which can result in higher heat transfer rates compared to inside walls,
which are often shielded from such extreme conditions.

 Overall heat transfer coefficients for the floor and ceiling:

5
 Windows and Doors: Consider the type, size, and thermal characteristics
(U-values) of all windows and doors.
Table (5-4): Windows
Table (5-5): Doors

1) 35mm-wood 3.1
2) Aluminum 7
3) Steel 5.8

6
Heating Load for the building of the Scientific Schools Lecture Halls

𝑸 = 𝑼𝑨∆𝑻
 Sample of calculation:

Hall_1
Area U (W/
Wall (m²) m².k) T_in(C) T_out(C) Q(W)
N 20.934 1.24 21 0 545.1214
S 26.382 0.86 21 0 476.4589
E 48.07 1.24 21 0 1251.743
W 48.07 0.86 21 0 868.1442
Total 143.456 3141.467
Area U (W/
Windows (m²) m².k) T_in(C) T_out(C) Q(W)
N 11.328 5.6 21 0 1332.173
S 0 0 21 0 0
E 0 0 21 0 0
W 0 0 21 0 0
Total 11.328 1332.173
Ceilling 107.4 0.88 21 0 1984.752
Floor 107.4 0.88 21 0 1984.752
door 5.88 3.1 21 0 382.788
Total 220.68 8825.932

7
To calculate the Area:
𝑸 = 𝑼𝑨∆𝑻
 Celling= 12.65*8.49
=107.3985W
We choose (U = 1.08); because the ceiling exposed to outside condition
𝑄 = 1.08 ∗ 107.3985 ∗ 21
=1984.72428
 Floor = 12.65*8.46
=107.3985W
We choose (U = .88) because the ceiling exposed to inside condition
𝑄 = 0.88 ∗ 107.3985 ∗ 21
=1984.72428W
 Door: we have 2 doors
The dimensions were obtained from paper diagrams and the door is wood;
So (U= 3.1)

8
Area of 2 doors =2*1.4*2.1
=5.88
𝑄 = 5.88 ∗ 3.1 ∗ 21
=382.788W

 Windows:

In North = (1.7*1.92) + (4.2*1.92) =11.328

The single glass is U = 5.6
𝑄 = 113.28 ∗ 5.6 ∗ 21=1332.1728 kJ

 Walls:
 The North wall = 8.49*3.8- N windows
=32.262-11.328
=20.934
U outside walls is 1.24
𝑄 = 20.934 ∗ 1.24 ∗ 21 = 545.12136 kJ
 The South wall = 8.49*3.8 – door = 32.262 - 5.88 =26.382W
9
 U inside wall = 0.86
𝑄 = 26.382 ∗ .86 ∗ 21 = 476.45892

 The east and west walls:

=12.65*3.8 =48.07
East : 𝑄 = 48.07 ∗ 21 ∗ 1.24
= 1251.7428W
West :𝑄 = 48.07 ∗ 21 ∗ .86
=868.1442W
 Total Q in walls =3141.46728 W

 Total Q in room = 8825.932 W

10
To calculate the sensible heat loss due to infiltration:
Using table 6.1

V̇f = N * VRoom
Sample of calculation for Hall 6:

We use N (Air change per hours )for classrooms = 2.0

So

V̇f = 2.0 * 107.4*3.8*2 = 816.24 m3/h

Q̇s,f = 𝜌o V̇f CP ∆T

11
By Interpolation we get the density and the specific heat in constant
pressure @ T = 0 ° + 273 = 273 K from table 2.1 :

For CP:

300−273 1.0057−𝑥
= = 1.0055
300−250 1.0057−1.0053

For density:

300 − 273 1.1774 − 𝑥

= = 1.305
300 − 250 1.1774 − 1.4128

Q̇s,f = 𝜌o V̇f CP ∆T = 1.305*816.24*1.0055*21 = 22 492.087 W.

To get the Humidity Ratio for the winter months: from Appendix A-7
(December – January – February – March)

12
 72 + 75 + 70 + 62
= 69.75 %
4

 We assumed the humidity ratio (HR) = 70% so we can get the information needed
from the carrier psychometric
----------------------------------------------------------------------------------------------------

We selected the Sand

and value of 1.5, this
is important for the
heating not cooling.

We selected that the

circumference for the
building is as shown
the angle is 70 degrees
and the surface is new
concrete.

This table shows that the solar reflectance on the ground that circle the building from
every side with the selected material and specific angle, it effects the heating load
and that important to consider in our calculations.

13
Figure (1): Map of the Ground Floor

Figure (2): Map of the First Floor.

14
 Using Hap:

15
 We selected the Sand and
value of 1.5, this is
important for the heating
not cooling

Using hap by
selecting the space
usage it gives us the
same data as in table
4-6 from the HVAC
book [1]

16
 For the activity
level cafeteria
work involves
various physical
activities and
responsibilities
we expect it to be
Medium work.

A recessed,
unvented fixture
radiates energy
only to the walls
and floor in a space.

17
 Table (1): The detailed Total heating load for the zones in the building.
Ground Floor / Zone 1
Heating Load Heating Load Error
Room HAP excel (KW) percentage
Computer lab 1 13.7 9.35733 31.69832117
Computer lab 2 13.2 10.7137 18.83560606
Office 1 1.6 1.487704 7.0185
Office 2 1.6 2.487704 55.4815
West entrance 0.85 1.507605 77.36529412
East entrance 0.85 2.507605 195.0123529
Corridor east entrance 1.1 1.175368 6.851636364
Corridor west entrance 1.1 2.175368 97.76072727
Corridor 2 3.9 2.813892 27.84892308
Buffet 2.1 1.841497 12.30966667
Cafeteria 9.7 8.128186 16.20426804
Middle entrance outside 4 3.652117 8.697075
Middle entrance inside 5.8 4.739379 18.28656897
Corridor beside Buffet 4.1 2.288802 44.17556098

Total 63.6 54.876257

First Floor / Zone 2

Heating Load Heating Load Error
Room HAP excel (KW) percentage
Hall 1 7.4 6.113125 17.3902027
Hall 2 6.9 5.301048 23.17321739
Hall 3 6.9 5.292542 23.29649275
Hall 4 6.9 5.325979 22.81189855
Hall 5 6.9 5.296414 23.24037681
Hall 6 7.4 6.113125 17.3902027
Storage room 1 1 0.6752768 32.47232
Storage room 2 1 0.7961902 20.38098
corridor 1 2.4 2.944162 22.67341667
corridor 3 5.3 0.7484268 85.87873962
corridor 5 4.1 2.787702 32.00726829
corridor 6+7 3.6 3.929622 9.156166667
extra room 3.8 3.571603 6.010447368

Total 63.6 48.8952158

Second Floor / Zone 3

Heating Load Heating Load Error
Room HAP (KW) excel (KW) percentage
Hall 1 7.4 6.113125 17.3902027
Hall 2 6.9 5.301048 23.17321739

18
 Hall 3 6.9 5.292542 23.29649275
Hall 4 6.9 5.325979 22.81189855
Hall 5 6.9 5.296414 23.24037681
Hall 6 7.4 6.113125 17.3902027
Storage room 1 1 0.6752768 32.47232
Storage room 2 1 0.7961902 20.38098
corridor 1 2.4 2.944162 22.67341667
corridor 3 5.3 0.7484268 85.87873962
corridor 5 4.1 2.787702 32.00726829
corridor 6+7 3.6 3.929622 9.156166667
extra room 3.8 3.571603 6.010447368

Total 63.6 48.8952158

Third Floor / Zone 4

Heating Load Heating Load Error
Room HAP (KW) excel (KW) percentage
Hall 1 7.4 6.113125 17.3902027
Hall 2 6.9 5.301048 23.17321739
Hall 3 6.9 5.292542 23.29649275
Hall 4 6.9 5.325979 22.81189855
Hall 5 6.9 5.296414 23.24037681
Hall 6 7.4 6.113125 17.3902027
Storage room 1 1 0.6752768 32.47232
Storage room 2 1 0.7961902 20.38098
corridor 1 2.4 2.944162 22.67341667
corridor 3 5.3 0.7484268 85.87873962
corridor 5 4.1 2.787702 32.00726829
corridor 6+7 3.6 3.929622 9.156166667
extra room 3.8 3.571603 6.010447368

Total 63.6 48.8952158

Fourth Floor / Zone 5

Heating Load Heating Load Error
Room HAP (KW) excel (KW) percentage
Hall 1 7.4 6.113125 17.3902027
Hall 2 6.9 5.301048 23.17321739
Hall 3 6.9 5.292542 23.29649275
Hall 4 6.9 5.325979 22.81189855
Hall 5 6.9 5.296414 23.24037681
Hall 6 7.4 6.113125 17.3902027
Storage room 1 1 0.6752768 32.47232
Storage room 2 1 0.7961902 20.38098
corridor 1 2.4 2.944162 22.67341667
corridor 3 5.3 0.7484268 85.87873962

19
 corridor 5 4.1 2.787702 32.00726829
corridor 6+7 3.6 3.929622 9.156166667
extra room 3.8 3.571603 6.010447368

Total 63.6 48.8952158

Total 318 250.4571

Since we didn’t calculate the infiltration using Excel we the Error Percentage
values higher than the expected.
And the factor of safety for the total heating load was taking equal to 10%.

20
 Domestic Hot Water :
Th = 60 , Tc = 0 Cp = 4.18 kJ/kg.K
For schools 2.5 liters/day per day
1 1
Q̇HW = ṀHW CP,W (Th – To) = * 2.5 * 250*4.18 * (60-0)
3600 3600

= 43.60KW
𝑄𝐻𝑊 43.60
ṀHW = = = 1.043062 kg
𝐶𝑃 ∗∆𝑇𝑏𝑖𝑙𝑒𝑟 4.18∗10

 Sizing of boiler :
Qboiler = Q heating load + QDHW = 318 + 43 = 361.6 kW

We select boiler NXR4-411

Figure (3): the boiler chosen

from the company CHAPPEE.

21
Figure (4): the boiler chosen from the company CHAPPEE.

22
 Radiator Selection:
We select the radiators from Fondital company using
its catalog:
But first we should follow some steps:
1- Calculate ∆ 𝑻 and number of sections :

𝐓𝐢𝐧−𝐓𝐨𝐮𝐭 𝟖𝟓+𝟕𝟓
∆𝐓= – T design = − 𝟐𝟏 = 𝟓𝟗
𝟐 𝟐

We take it as its equal to 60

𝑸𝒉𝒆𝒂𝒕𝒊𝒏𝒈 (𝑾)
Sections = 𝑾
𝒉𝒆𝒂𝒕 𝒐𝒖𝒕𝒑𝒖𝒕 (𝒔𝒆𝒄𝒕𝒊𝒐𝒏𝒔)

23
 Table (1): shows the model and number of radiators for the ground floor.
Ground Floor / Zone 1

Heating Load HAP sections per

Room model radiators
(KW) radiator
Computer lab 1 13700 800/100 6 10
Computer lab 2 13200 800/100 6 10
Office 1 1600 500/100 1 10
Office 2 1600 500/100 1 10
West entrance 850
East entrance 850
Corridor east entrance 1100 500/100 1 7
Corridor west entrance 1100 500/100 1 7
Corridor 2 3900 800/100 2 9
Buffet 2100 800/100 1 9
Cafeteria 9700 700/100 4 12
Middle entrance outside 4000 800/100 2 9
Middle entrance inside 5800 800/100 3 9
Corridor beside Buffet 4100 800/100 2 9

Table (2): shows the model and number of radiators for the 1st,2nd,3rd,4th floors.

1st,2nd,3rd,4th Floors / Zone 2,3,4,5

sections per
model radiators
Room Heating Load HAP radiator
Hall 1 7400 700/100 4 9
Hall 2 6900 800/100 3 10
Hall 3 6900 800/100 3 10
Hall 4 6900 800/100 3 10
Hall 5 6900 800/100 3 10
Hall 6 7400 700/100 4 9
Storage room 1 1000 700/100 1 5
Storage room 2 1000 700/100 1 5
corridor 1 2400 700/100 1 12
corridor 3 5300 800/100 3 8
corridor 5 4100 800/100 2 9
corridor 6+7 3600 800/100 2 8
extra room 3800 800/100 2 9

24
 Pipe Selection :
𝑄𝑏𝑜𝑖𝑙𝑒𝑟 361.6
ṁboiler = = 4.18∗10 = 8.6 𝑘𝑔/𝑠 = 8.6 L/s
𝐶𝑝 ∆𝑇

Figure (5): pressure drop per unit pipe length for water flowing in commercial steel
pipes.
We selected the diameter 100 mm because it gives us a better pressure drop that
equal to 100 Pa/m ∆𝑝/𝑙
But if we choose the 80 mm diameter it gives us pressure drop larger than the
required.
L = 1.5*Lfurthest = 1.5*100 = 150 m
∆𝑃 = 100 ∗ 150 = 15 kPa
∆𝑃 15
h= = = 1.529 m
𝜌𝑔 9.81

 Pump Selection:
𝑄𝑏𝑜𝑖𝑙𝑒𝑟 361.6 𝑘𝑔
ṁboiler = = = 8.6 = 30.96 m3/h
𝐶𝑝 ∗∆𝑇 4.18∗10 𝑠

Expansion Tank Selection

expansion tank with volume 750 liters has been selected.

25
 Figure (5): selected pump.
 Energy Consumption :
The following section shows the energy consumption calculations for the system
using the Degree
Day Method, DD. The table () below shows a summary of the results.
𝑄̇ 𝑋 3600 𝑋 24 𝑋 𝐷𝐷𝐻
𝑄𝐻 𝑡ℎ𝑒𝑜𝑟 = (𝐶𝑑 )
𝑇𝑖 − 𝑇𝑜
We took the value of DDH from table 6-10

Where TO = 0 ℃
318∗3600∗24∗1478
Q̇total = ∗ 0.77 = 1488.972 *106 kJ/season
(21−0)

1489∗10^−6
M fuel = = 47 724.35 kg per season
0.80∗39000

26
Vtank = 47 724.35/850 = 56.146 m3 per season
L = 0.735 JD/Liter
Cost = 56 146 * 0.735 = 41 267.31 JD

 Terminal System:
A terminal HVAC system with a 4-pipe fan coil unit and ductwork is ideal for educational
buildings due to its ability to provide simultaneous heating and cooling, ensuring precise
temperature control across diverse zones. This system enhances comfort, energy efficiency, and
indoor air quality, crucial for a conducive learning environment. It operates quietly, minimizing
disruptions, and offers flexibility for future expansions or layout changes. Though installation
costs may be higher, the long-term benefits in efficiency and durability justify the investment.
Regular maintenance and integration with a building management system ensure optimal
performance and reliability.

27
 COSTS BASED ON THE JORDANIAN MARKET
 Valves of each radiator = 15 JD
We have 158 radiators
Cost =158*15 = 2 370 JD
 Boiler =1
Cost=5500 JD
 Radiator = 35.5 JD/Section
Number of section=565
Cost= 20 057.5 JD
 Pipe= 0.25 JD/meter
Meter= 6500 m
 Cost = 1625JD
DMH cylinder =750 L
 Cost =300 JD
 Pump = 3500 JD

28
 Sizing of hot water heating system (HWHS) :

Figure (6): HWHS Ground floor.

Figure (7): CASSETTE Places Ground floor.

29
 Cooling Part

Figure (8): The sun path from website andrew marsh.

30
31
Sample of calculation :
 West wall :
In the west we do not have any part except the wall to calculate so, assuming
the 14:00 clock and the 12 June and assuming it
-Group A of 101.6 mm Face Brick ( insulation or air space +203.2 mm
common brick)
U=0.9W/m2. K

32
- K=0.83 for Light color

Area= 46.56 m

33
The latitude month (LM)
LM= 0 , at lat.angle = 24°

The inside design temperature = 21°C

From table (9-4) the CLTD=11°C

Using equation (9-3) (CLTD)corr

(𝑇𝑚𝑎𝑥+𝑇𝑚𝑖𝑛)
𝑇𝑂,𝑀 =
2
(27.9+15)
𝑇𝑂,𝑀 =
2

𝑇𝑂,𝑀 =21.45°C
(CLTD)_corr=(CLTD+LM)*k+(25.5-Ti) +(𝑇𝑂,𝑀 -29.4) *ƒ
(CLTD)_corr=(11+0)*0.83+(25.5-23)+(21.45-29.4) *1

34
(CLTD)_corr=3.68C

So now we have the equation: Q _̇ conv=UA〖(CLTD)〗_corr=

0.86*46.56*3.68=147.353W
2)North wall , Area = 29.412 m

Q ̇_conv=UA〖(CLTD)〗_corr= 1.24* 29.412 *3.68=134.213W

3)East wall, Area = 48.45 m

Q ̇_conv=UA〖(CLTD)〗_corr= 1.24* 48.45 *3.68=221.09W

3)South wall, Area = 30.596m

Q ̇_conv=UA〖(CLTD)〗_corr= 0.86* 30.596*3.68=96.83W

Total Q_conv= 147.353+134.213+221.09+96.83=599.67W

35
References
1. EXCHANGE_8_2.pdf
2. 4480_CTC_GAMMA_RADIATORI_FN_EN.pdf
3. NXR4 anglais CI389C2.pdf
4. heating and air conditioning (1).pdf
5. NXR4 Brochure.pdf
6. york_fan_coil_units_catalogue_2019-3988.pdf
7. MULTI_V_Catalog[20210528_123125635].pdf
8. ‫( أسعار المحروقات في األردن اليوم | توتال األردن‬totalenergies.jo)
9. Grundfos UMS 65-30 PN6 3~ - (thepumpexchange.co.uk)
10. YORK HVAC Specialised Contractor Price List 2021.pdf (bulclima.com)

36