Design Heating &ventilation system for Hospital

1. DESIGN OF HVAC SYSTEM FOR FIRST
FLOOR FOR SCIENCE AND TECHNOLOGY
HOSPITAL
Sana’a University – faculty of engineering-mechanical
engineering
Team:
MOHAMMED HABEB
ADEB HASAN
MOHAMMED AL-GAORI
MOATH RADMAN
Supervised
by:
DR. ABDULJALIL AL-ABIDI

2. Introduction 1
Cooling load calculation 4
Building description 3
Duct design 5
Conclusions
&Recommendations 9
Outline :
Pipe design 6
Terminal units 7
Problem statement 2
Absorption chiller 8

3. Heating, ventilation, and air conditioning (HVAC) is the
technology of indoor environmental comfort. Its goal is
to provide thermal comfort and acceptable indoor air
quality. ... It is one of the most important factors for
maintaining acceptable indoor air quality in buildings
0
20
40
60
80
100
120
44% of a commercial building’s energy consumption is
attributed to HVAC systems
Introduction 1
w w w . b l u e h a t m e c h a n i c a l . c o m

4. Hospital buildings are very important building and one of
the few types of building that run at full scale twenty four
hours a day, so they need to make HVAC system
to maintain the temperature, air quality, airflow and humidity to
create the most comfortable environment possible for the
patients and to control hazards and control of infection, removal
of odors and bacteria
Introduction 1

5. In practice, HVAC of hospital depend on air
conditioning system configuration and design
parameters selection and this based on
knowledge, experience, and skill of system
designer and using of standards
Introduction 1

6. HVAC for hospital is a little different due to
The parameter for each room and department
Introduction 1

7. Problem statement 2
h t t p s : / / w w w. hy d ro l - e a r t h - sy s t - s c i . n e t / 1 7 / 3 9 3 7 / 2 0 1 3 /
Due Hot temperature climate in Yemen
Importance of Air conditioning system for
hospitals

8. • Air-conditioning system in Yemen
depend on air-conditioning using
compressor
• large amount of electrical power
required
• the cost of electrical energy at Yemen
increased
• missing the local grid and the
increasing of fuel cost
• The using of solar cooling is necessary
for energy saving
Problem statement 2

9. Building description 2
- Location (15.35°N) (44.27°E)
- First floor
- Area = 1970 m2
Thickness
(mm)
K
W /m.°C
Layer
150
1.05
Soft stone
20
1.2
Cement concrete
200
1
Block cement
20
1.2
Cement plaster
10
2.9
Ceramic
Components of External Wall

10. Building description 2
` ICU
Sterilization Operation rooms
Cardiac
Operation room department
Sterilization
ICU
Cardiac department
Recovery room
Dressing rooms
Break rooms

11. Cooling loads calculation 3
There are two methods used for loads calculations
B) Technical Method
A) Manual Method

12. DB Temperature 21°C
Relative humidity 50%
Pressure Positive
Exhaust air 200-400 cfm less than
supply
Ventilation 15-20 ACH
Operation room parameters
Cooling loads calculation 3
taking operation room 1 (OP-Z1) with 35 m2
and 2.8m height
as sample, which is located in the SE and NE
of building.

13. Overall heat transfer coefficient for walls U=1.787 w/m2 .C. 𝑅𝑡ℎ =
1
𝑈𝑤
= 𝑅𝑖 + ෍
𝑗=1
𝑛
𝑥𝑗
𝑘𝑗
+ 𝑅𝑜
A) Manual Method
Overall heat transfer coefficient for windows U=5.097w/m2 .C.
Outside design condition: 30.20 C Db.
Inside design condition: 210C Db and 50% RH.
Occupancy: 5m2/person=7 people with sensible heat gain 82
W/person, and latent heat gain 79 W/person
Lights: 17 W/m2 of floor area.
Equipment: 110 W/m2.
Cooling loads calculation 3

14. 1
2
3
4
𝑇𝑜𝑚 = 𝑇𝑜 −
𝐷𝑅
2
= 23.75
𝐶𝐿𝑇𝐷𝑐𝑜𝑟𝑟 = 𝐶𝐿𝑇𝐷 − 𝐿𝑀 𝐾 + 25.5 − 𝑇𝑖 + 𝑇𝑂,𝑚 − 29.4 𝑓
𝑄𝑠𝑜𝑙𝑎𝑟 = 𝐴𝑈ሺ𝐶𝐿𝑇𝐷𝑐𝑜𝑟𝑟) = 705.7 𝑊
𝑄𝑙𝑖𝑔ℎ𝑡 = 𝑄
𝑊
𝑚2 𝐴 = 581 𝑊
𝑄𝑜𝑐𝑐𝑢𝑝. = 𝑄𝑆𝐻 + 𝑄𝐿𝐻 = 956 w
൯
𝑄𝑒𝑞𝑢𝑖𝑝. = 𝑄 Τ
𝑤 𝑚2
∗ 𝐴 𝑚2
= 3850 𝑊 5
𝑄𝑡𝑟𝑎𝑛𝑠 = 𝐴 𝑆𝐻𝐺𝐹 𝐶𝐿𝐹 𝑆𝐶 = 2227.2 W
Cooling loads calculation 3

15. Q inf il =mf(hi
__ho) =27.6 W 6
Qv=(
𝑉𝑓
𝑣𝑜
) *(ho-hi) =6650 W 7
The Total Heat Gain
𝑄𝑡𝑜𝑡𝑎𝑙 = 𝑄𝑠𝑜𝑙𝑎𝑟 𝑒𝑓𝑓𝑒𝑐𝑡𝑠 + 𝑄𝑡𝑟𝑎𝑛𝑠 + 𝑄𝑙𝑖𝑔ℎ𝑡 + 𝑄𝑜𝑐𝑐𝑢𝑝 + 𝑄𝑒𝑞𝑢𝑖𝑝 + 𝑄𝑖𝑛𝑓𝑖𝑙+ =8.34 KW =2.385 TR
The Total coil load
𝑄𝑡𝑜𝑡𝑎𝑙 = 𝑄𝑠𝑜𝑙𝑎𝑟 𝑒𝑓𝑓𝑒𝑐𝑡𝑠 + 𝑄𝑡𝑟𝑎𝑛𝑠. + 𝑄𝑙𝑖𝑔ℎ𝑡 + 𝑄occc𝑢𝑝 + 𝑄𝑒𝑞𝑢𝑖𝑝 + 𝑄𝑖𝑛𝑓𝑖𝑙+ 𝑄𝑣𝑒𝑛𝑡.
= 14.99 KW =4.28 TR
Cooling loads calculation 3

16. B) Technical Method
Using REVIT program to calculate cooling loads
the program performs the following tasks:
• Calculates design cooling loads for zones
• Determines required supply and return airflow flow rates for zones
Cooling loads calculation 3

17. SketchUp
First
SketchUp creates and the
building geometry in REVIT
and gives the defining for
each zone
Second
separate the departments
into spaces and collect spaces
to zones
Such as OP-Z1 and we do
that for all the building
Cooling loads calculation 3

18. Cooling loads calculation 3

19. Cooling loads calculation 3
REVIT input data
2.8m
The modified data done according to ASHRAE

20. Cooling loads calculation 3

21. R e s u l t s
Cooling loads calculation 3

22. Cooling loads calculation 3
Validation
Load (Kw) Manual method Technical method
Total load 15 15.48
Ventilation load 6.657 6.859
Error = 3.2%
Which of these methods is preferred ……..??????
Technical method is better than manual method due to
1- The manual method takes all the loads at the same time
2- other consideration like fan heat

23. Duct design 4
conduits or passages used in ventilation, and air
conditioning (HVAC) to deliver and remove air
Figure shows galvanized iron material that is used in
ducts design for this work because its availability ,
corrosion resistance and lowest cost.

24. Duct design 4
Duct work
Types of ducts
Rectangular or square
Round
Flexible
Flexible
Round rectangular

25. Duct design 4
fibrous glass material used to
isolate duct for many purpose
such as thermal insulation and
protection duct from corrosion
Duct insulation

26. Duct Design Procedures
Duct design 4
1. Selection the
dimension of air
outlet(diffusers)
according to flow rate
of air in each room
2. Selection the type of
duct
3. Duct calculation

27. Duct design 4
Selection the diffuser
For OP-Z1 m. air= 406 L/s
Type of duct is rectangular

28. Duct design 4
By knowing the flow rate of room
and recommended velocity using
DUCT SIZER to calculate the
losses and duct size
Velocity 500-700 fpm
Duct size 350 x 350
Friction 0.399 pa/m
Duct sizer

29. Duct design 4
Duct fitting losses calculating by
ASHRAE DUCT FITTIND DATA BASE

30. Duct design 4

31. Duct and pipe design 4 REVIT design

32. Duct design 4
REVIT duct result

33. Duct design 4
Exhaust duct
Exhaust duct criteria
1- Exhaust grill should be
70 mm above floor
according to ASHRAE
application
2- exhaust flow depend
on the pressure of
rooms

34. Duct design 4
Duct accessories
Balancing damper
To control the flow rate reducing or
increasing the airflow depending upon
conditions

35. HEPA filter
Efficiency of HEPA filter reach
to 99.997%
Pressure loss reach to 249 pa
Duct design 4
Filter Design Factors
1. Degree of air cleanliness required.
2. Particulate/contaminate size and form (solid or aerosols).
4. Cost (initial and maintenance).
5. Space requirements.
6. Pressure loss/energy use.
FILTERS
To eliminate particles

36. Duct design 4

37. Duct design 4

38. Duct design 4
After drawing and calculating
duct the main ducts detail
Roof top fan units Actual flow rate
(m3/s)
Actual Pressure
drop (Pa)
Mechanical supply
air (1)
1.056 123.64
Mechanical supply
air (2)
1.08 314.67
Mechanical
supply air (3)
1.9274 334.62
Mechanical supply
air (4)
3.04 772.7
Mechanical supply
air (5)
0.9795 116.3
Mechanical supply
air (6)
1.1859 342
Mechanical exhaust
air (1)
2.8107 349
Mechanical exhaust
air (2)
1.9485 255.5
Mechanical exhaust
air (3)
1.7505 276

39. Duct design 4
Selection of the fans
The following must be considered when selecting a suitable fan:
Desired air volume
Required pressure loss
Roof top fan
units
Actual flow
rate (m3/s)
Actual
Pressure
drop (Pa)
Model standar
d flow
rate
(m3/s)
Standar
d
Pressur
e drop
(Pa)
Mechanical
supply air (1)
1.056 123.64 MUB 042450
EC-A2
1.41 250
Mechanical
supply air (2)
1.08 314.67 MUB 042450
EC-A2
1.75 560
Mechanical
supply air (3)
1.9274 334.62 MUB 062560
EC-A2
3 540
Mechanical
supply air (4)
3.04 772.7 MUB 062630
D4-A2 IE2
3.89 790
Mechanical
supply air (5)
0.9795 116.3 MUB 042450
EC-A2
1.75 600
Mechanical
supply air (6)
1.1859 342 MUB 042450
EC-A2
1.75 430
Mechanical
exhaust air (1)
2.8107 349 MUB 062630
D4-A2 IE2
3.63 660
Mechanical
exhaust air (2)
1.9485 255.5 MUB 062560
EC-A2
3 500
Mechanical
exhaust air (3)
1.7505 276 MUB 062560
EC-A2
3 550

40. Duct design 4
A B

41. Pipe design 5
Piping system used to convey water to
FCU and AHU
In a chilled water system, water is first cooled
in the water chiller the evaporator of
reciprocating, screw, or centrifugal refrigeration
system located in a centralized plant to a
temperature of (6 °C). It is then pumped to the
water cooling coils in AHUs and FCUs.

42. After flowing through the coils, the chilled water
increases in temperature up to (12 C ) and then returns to
the chiller
Pipe design 5

43. Pipe design 5
Type of Piping system
a closed system is one in which the flow of water is not exposed
to the atmosphere at any point.
Two-Pipe Direct Return System
Closed system
Piping Material
Steel- Schedule 40

44. Manual calculation
Pipe design 5

45. Section Component Diameter
(mm)
Flow
rate
(kg/s)
Velocity
)m/s)
Equivalent
length
(mm)
NO. of
component
Total
length
(mm
)
P
∆
(pa/m)
P
∆
)pa)
Sect. 1 Pipe 32 2.37 2.46 ‫ــــــــــــ‬ ‫ــــــــــــ‬ 1.1 2000 10200
Elbow 1 2 2
Tee 2 1 2
Total ‫ــــــــــــ‬ ‫ــــــــــــ‬ 5.1
Sect. 2 Pipe 32 1.79 1.85 ‫ــــــــــــ‬ ‫ــــــــــــ‬ 3.2 1200 6240
Tee 2 1 2
Total ‫ــــــــــــ‬ ‫ــــــــــــ‬ 5.2
Sect. 3 Pipe 20 .74 2.15 ‫ــــــــــــ‬ ‫ــــــــــــ‬ 14 3000 87300
Tee 1.2 1 1.2
Elbow 0.6 12 7.2
Pall valve 6.7 1 6.7
Total ‫ــــــــــــ‬ ‫ــــــــــــ‬ 29.1
A.H.U.01 6880
Sect. 4 Pipe 32 1.79 1.85 ‫ــــــــــــ‬ ‫ــــــــــــ‬ 3.2 1200 6240
Tee 2 1 2
Total ‫ــــــــــــ‬ ‫ــــــــــــ‬ 5.2
TOTAL 116860
Pipe design 5
Result of manual
calculation

46. Pipe design 5
Pipe flow wizard program
Input Data
Output data

47. Pipe design 5
REVIT drawing and calculating
First draw pipe and indicate
if it supply or return
Second insert the flow water
for AHU or FCU and the
pressure drop for each AHU
or FCU

48. Pipe design 5
Inserting data to FCU
or AHU based on
catalogs
ሶ
𝒎 =
𝑸𝒄𝒄
𝑪𝒑ሺ𝑻𝒊𝒏 −𝑻𝒐𝒖𝒕 )
(kg/s)
Water flow calculated
by this equation
𝐶𝑝= 4.186 k J/ kg °C
𝑇𝑜𝑢𝑡= 12 0 C 𝑇𝑖𝑛 = 7 oC
For FCU01 the zone load 𝑸𝒄𝒄= 8.688kw …. ሶ
𝒎 = 0.415 L/s

49. F.C.U. Spaces Q(W) V(L/s) mw(L/s)
F.C.U. 01 • Break 3
• Break 4
• Break 5
8688 416.8 0.4151
F.C.U. 02 • Break 1
• Break 2
• Store NR break
11651 557.6 0.55667
F.C.U. 03 • Monitoring 9967 309.6 0..47621
F.C.U. 04 • Dressing NR STR 4659 244.3 0.2226
F.C.U. 05 • Fixing NR STR
• Sub NR STR
6904 241 0.32986
F.C.U. 06 • Sub way 01 9621 470.1 0.45968
F.C.U. 07 • Sub way 02 4089 159.1 0.19537
F.C.U. 08 • Sub way 03 10204 400.7 0.48753
F.C.U. 09 • Replacing dressing zone 6831 321 0.3264
F.C.U. 10 • Replacing NR CATH 7284 347.3 0.348
F.C.U. 11 • ICU services rooms 6242 313.15 0.29823
F.C.U. 12 • CATH services rooms 3984 342.17 0.19
F.C.U. 13 • Services room
• Electric room
• Hall with stair
12556 464 0.59991
F.C.U. 14 • Replacing zone 11052 396.5 0.528
Pipe design 5

50. A.H.U. Spaces Q(W) V(L/s) mw(L/s)
A.H.U. 01 • OP-Z1 15485 405.4 0.73985
A.H.U. 02 • OP-Z2 12230 423.5 0.58433
A.H.U. 03 • OP-Z3 13847 414.1 0.661586
A.H.U. 04 • OP-Z4 7689 344.9 0.36737
A.H.U. 05 • OP-Z5 18195 616.7 0.86933
A.H.U. 06 • I.C.U. 31777 876.75 1.5183
A.H.U. 07 • Cath heart zone
• PR-zone 2
33535 1118.03 1.6022
A.H.U. 08 • Cath zone
• Dirt unit
17100 673.63 0.817
A.H.U. 09 • Stylization 37330 1225.6 1.78356
A.H.U. 10 • Recovery zone 9831 314.1 0.46971
A.H.U. 11 • Store 1-OP
• Store 2-OP
• Prep. -R- OP
• Hall near-OP
22043 815 1.0532
Pipe design 5

51. Pipe design 5
Pipe accessories
Pressure gauge (on chiller)
Temperature gauge
Gate valve
Pressure regulating valve
Butterfly valve
Balancing valve
Strainer

52. Pipe design 5
Air separator is removing the air from pipes system because air
create noise in the system and causes corrosion in pump
impellers.
Air Separator
Expansion tank
Expansion tanks are required in a closed loop chilled water
HVAC system to absorb the expanding fluid and limit the
pressure within a cooling system

53. Pipe design 5
Results of pipe calculating
Diameter of main pipe = 100 mm
Flow = 16.7 L/s
Return pressure loss= 4.18 bar
Supply pressure = 3.8 bar
Supply
Return

54. Pipe design 5
Pump Sizing
centrifugal pump designs which are the most common
types used in comfort air conditioning.
∆𝒑 = 𝜌𝑔ℎ
Total pressure drop = 7.98 bar
ℎ = 81.34 𝑚
Flow rate = 16.7 L/s

55. Pipe design 5
Pump selection

56. Terminal units 6
Fan-Coil Unit
fan-coil units used to control the volume and
temperature of air delivered to the space as required to
maintain occupant thermal comfort and/or ventilation.
the basic components of fan-coil units are a cooling
coil, filter, fan, and temperature control device

57. Terminal units 6
In our project we
used 14 FCU
According to loads
F.C.U. Spaces Q(W) V(L/s) mw(L/s)
F.C.U. 01 • Break 3
• Break 4
• Break 5
8688 416.8 0.4151
F.C.U. 02 • Break 1
• Break 2
• Store NR break
11651 557.6 0.55667
F.C.U. 03 • Monitoring 9967 309.6 0..47621
F.C.U. 04 • Dressing NR STR 4659 244.3 0.2226
F.C.U. 05 • Fixing NR STR
• Sub NR STR
6904 241 0.32986
F.C.U. 06 • Sub way 01 9621 470.1 0.45968
F.C.U. 07 • Sub way 02 4089 159.1 0.19537
F.C.U. 08 • Sub way 03 10204 400.7 0.48753
F.C.U. 09 • Replacing dressing
zone
6831 321 0.3264
F.C.U. 10 • Replacing NR CATH 7284 347.3 0.348
F.C.U. 11 • ICU services rooms 6242 313.15 0.29823
F.C.U. 12 • CATH services rooms 3984 342.17 0.19
F.C.U. 13 • Services room
• Electric room
• Hall with stair
12556 464 0.59991
F.C.U. 14 • Replacing zone 11052 396.5 0.528

58. Terminal units 6
Selection of FCU from catalogs
For FCU01 with load 8.688 Kw and flow 416 L/s
475 L/s

59. Terminal units 6
enclosed assembly consisting of a
fan or fans and other necessary
equipment to perform one or more
of the functions of circulating,
cleaning, cooling, humidifying,
dehumidifying.

60. Terminal units 6
The AHU is used to control the
following parameters of the space.
•Temperature
•Humidity
•Air Movement
•Air Cleanliness
Components of Air Handling Unit
Housing
Fan
Cooling Coil
Filters

61. Terminal units 6
Selection AHU criteria
load
flow rate
AHU 11 with 22.043 Kw and 815 L/s

62. Terminal units 6
A.H.U. Spaces Q(W) V(L/s) mw(L/s)
A.H.U. 01 • OP-Z1 15485 405.4 0.73985
A.H.U. 02 • OP-Z2 12230 423.5 0.58433
A.H.U. 03 • OP-Z3 13847 414.1 0.661586
A.H.U. 04 • OP-Z4 7689 344.9 0.36737
A.H.U. 05 • OP-Z5 18195 616.7 0.86933
A.H.U. 06 • I.C.U. 31777 876.75 1.5183
A.H.U. 07 • Cath heart zone
• PR-zone 2
33535 1118.03 1.6022
A.H.U. 08 • Cath zone
• Dirt unit
17100 673.63 0.817
A.H.U. 09 • Stylization 37330 1225.6 1.78356
A.H.U. 10 • Recovery zone 9831 314.1 0.46971
A.H.U. 11 • Store 1-OP
• Store 2-OP
• Prep. -R- OP
• Hall near-OP
22043 815 1.0532

63. Absorption chiller 8

64. Absorption chiller 8
Selection of
absorption chiller

65. Absorption chiller 8

66. Characteristics of different solar collectors (for solar cooling purposes)
Absorption chiller 8

67. Absorption chiller 8
Efficiency of evacuated
tube

68. Thermann TH-30 Collector Specification Source:[ Apricus ,technical manual, March2014]
Absorption chiller 8

69. Absorption chiller 8

70. 𝑨𝑪 =
𝑸𝑺
𝛈𝐬𝐨𝐥𝐚𝐫 𝐈𝐦𝐚𝐱
𝑨𝒄 = 𝒏𝑨𝒆 → 𝒏 =
𝑨𝒄
𝑨𝒆
Result
Item
369 kW
Qe
473.1 kW
Qg
333.2 Kw
Qs
504.9 𝑚2
Ac
792 𝑚2
Ag
180
Number of collector
5400
Number of tube
Absorption chiller 8

71. Type Size (mmXmm) No. Cost of unit $ Total Cost
Supply diffuser 600X600 10 55 550
Supply diffuser 450X450 82 55 4510
Return diffuser 450X450 45 55 2475
Exhaust grille 600X600 3 55 165
Exhaust grille 400X400 26 55 1430
Exhaust grille 300X300 4 55 220
TOTAL COST 170 9350
Cost of diffusers and grilles

72. F.C.U Actual Load (TR) standard Load
(TR)
Cost $
F.C.U. 01 2.4823 2.5 1875
F.C.U. 02 3.329 3.5 2625
F.C.U. 03 2.85 3 2250
F.C.U. 04 1.33 1.5 1125
F.C.U. 05 1.973 2 1500
F.C.U. 06 2.749 3 2250
F.C.U. 07 1.1683 1.5 1125
F.C.U. 08 2.915 3 2250
F.C.U. 09 1.952 2 1500
F.C.U. 10 2.08 2 1500
F.C.U. 11 1.7834 2 1500
F.C.U. 12 1.1383 1.5 1125
F.C.U. 13 3.5874 3.5 2625
F.C.U. 14 3.1577 3.5 2625
TOTAL COST 25875
Cost of F.C.U.

73. A.H.U. Actual Load
(TR)
Standard
Load (TR)
Cost $
A.H.U. 01 4.4243 4.5 6300
A.H.U. 02 3.494 3.5 4900
A.H.U. 03 3.9563 4 5600
A.H.U. 04 2.197 2.5 3500
A.H.U. 05 5.1986 5.5 7700
A.H.U. 06 9.08 9 12600
A.H.U. 07 10.153 10.5 14700
A.H.U. 08 4.886 5 7000
A.H.U. 09 10.666 11 15400
A.H.U. 10 2.81 3 4200
A.H.U. 11 6.298 6.5 9100
TOTAL COST 91000
Cost of A.H.U.

74. Duct every 1 m2
28 $
Total= 54000 $
Chiller every 1 TR 1700$ Total= 166600 $
Solar collector every piece 1200 $ Total=216000 $

75. Recommendations 8.2
▪ Using REVIT MEB for calculating load is effective most of things
considered in it .
▪ Using REVIT MEB for calculating duct and pipe losses is very
effective.
▪ Any one Using REVIT MEB should be precise in inserting data to
program from standards books for getting proper cooling loads
and losses .

76. THANKS FOR ATTENTION