NATIONAL HIGHER POLYTECHNIC INSTITUTE (NAHPI)
UNIVERSITY OF BAMENDA
CIVIL AND ARCHITECTURAL ENGINEERING
LEVEL 400
BUILDING SERVICES ASSIGNMENT
GROUP 5
10/12/2021
LECTURER: Mr NKENFACK ROGER
BUILDING SERVICES ASSIGNMENT GROUP 5 WORK
NAME OF MEMBER
DIANGHA MALCOM NGONGBI
MATRICULE NO
UBA19E0231
DIANGHA BETRAND TIMGUM
UBA19E0023
CHINDO MACARIUS KIBAM AJIMSEBOM
UBA19E0022
CHIAFIE NKEUMBOU CORENTIN
UBA21E3012
CHE CLINTON AKONGNWI
UBA21E3011
BUILDING SERVICES ASSIGNMENT
-
For Douala
Only Offices( no corridors)
External walls = 20cm. ( internal walls and external concrete plastering)
Partition in hollow block (partition wall)=10cm (Int and Ext concrete plastering)
Ceiling in poured concrete=20cm
Height= 3m(wall height)
200den doors 1m*2m= 2.5cm frame thickness
Window=1m×1.5m simple glass/ metallic frame/ Alluminium binds
Lighting = fluorescent tubes
Time = 1 pm
1 person/ m2 = population density
Ɵe= 32℃
Ɵi = 26℃
Relative humidity ( RHC)= 83%
Absolute humidity ( We) = 0.025kg/Kgda
Latitude 4N ( February)
Relative Humidity (RHi) = 51%
Absolute Humidity (We) = 0.0108Kg/Kgda
OFFICE 1 = OFFICE 2
OFFICE 3 = OFFICE 4
OFFICE 5
1) OFFICE 1 = OFFICE 2
Description
Volume (m3)
OFFICE 1
6m×4m×3m
OFFICE 2
72m3
Orientation
of the wall
Area (m2)
KLW/ m3
SE
NW
SW
NE
Ceiling
16.5
16
12
12
24
2.09
2.37
2.37
2.37
1.14
Intensity of
Thermal Radiation
(W)
206.91
113.76
85.32
85.32
246.24
Floor
Glass (SE)
Door (NW)
24
1.5
2
1.30
5.8
3.14
Q=KA∆Ɵ
Q1 SE = 2.09×16.5×(32-26)
= 2.09×16.5×6= 206.91
Q1NW= 2.37×16× (Ɵe-Ɵi-3℃)
= 2.37×16×3=113.76
Q1SW = 2.37×12×3=85.32
Q1NE =2.37×12×3= 85.32
Q1CEILING=24×1.14×(Qe-Qi+3)
Q1CEILING=24×1.14×9=246.24
Q1FLOOR=24×1.30×(20℃-Ɵi)
=24×1.30×-6=-187.2
Q1GLASS=1.5×5.8×(Ɵe-Ɵi)
=1.5×5.8×6=52.2
Q1DOOR=2×3.94×(Ɵe-Ɵi-3)
=2×3.94×3=23.64
Q1=∑Q
Q1 = 626.19
Q2 Contribution of solar radiation heat through walls
Q2RadW3=a. S. F. Rm
Where time = 1Pm =13Hr
Wall
Rm
SE
256
To find F where F = Solar Radiation factor
Thus F = 0.1
S = Area
→ S = 16.5m2
a= 0.4
thus Q2RadW= a. S. F. Rm
= 0.4×16.5×0.1×256=168.96
-187.2
52.2
23.64
Q3 Solar Radiation heat through glazing
Q3radg= a. S. F. RV
Wall
RV
SE
220
S= 1.5
g =0.45 , a= 0.86
thus Q3radg= 0.86×1.5×0.45×220=127.71W
Q4 Contribution of heat by renewal and infiltration
QSRW= 0.33qV(Qe-Qi) → ( for sensible heat)
QIRW= 0.84qV(We-Wi) → ( Latent heat)
Where qV= external air rate of renewal (m3/h)
QSRW Sensible heat
P.D =1 Person/m2
qV= 25m3/h/person
No of person = γ× S
→ γ = 0.1
S = 24
No of person = 0.1×24= 3 persons
qVT= No of person×qV/h/person
= 3×25=75m3/h
Thus QSRW= 0.33(75)(32-26)
QSRW= o.33(75)(6) = 148.5W
Latent heat
QIRW= 0.84qV(We-Wi)
We=0.025
Wi=0.08
Thus QIRW QIRW=0.84×75×(0.025-0.0108)
QIRW=0.84×75×0.0142= 0.8946W
QIRW = 0.8946W
Q5 Heat du e to occupants
QSWOC= Ƞ× CSOC
( Sensible heat)
CSOC = 63
Thus CSOC = 3×63= 189
Latent heat
QLOC = Ƞ× LSOC
LSOC = 59
QLOC = 3×59= 177
Q6 Contribution of heat by lighting
QSED =1.25P f or Flourescent tubes
P = S×W
→ S= 24m2 ,
W = 16w/m2
Thus P= 24×16 = 384W
QSED = 1.25(384)
= 480W
Q7 Machines and equipment heat contribution
Equipment = Computer and Printer
Computer
sensible heat → →Q7Sens= 250
Latent heat
Q7Lat =0
Printers ( ink-jet printer)
Sensible heat → Q7Sens = 52W
QSens =Q1+ Q2 …. + QN
→ 626.19+ 168.96+ 127.71 +148.5+ 189+ 480+ (250+ 52)
= 2042.36W
QT Latent =0.8946+ 177+ 0 = 177.8946W
Thus QT = QTsens+ QT Latent
QT = 2042.36W+ 177.8946W= 2220.2546W
Refrigerative Power
2220.2546W → KW
Thus → 2.2202546kW= 2.22KW
OFFICE 3
Description
Volume (m3)
OFFICE 3
84m3
Orientation of
the wall
Area ( m2)
KLW/M3
SE
NW
SW
NE
ceiling
floor
Glass (NW)
Door
19
19.5
12
12
28
28
1.5
2
2.37
2.09
2.37
2.37
1.14
1.30
5.8
3.94
Intensity of
thermal
radiation (W)
135.09
244.53
85.32
85.32
287.28
-218.4
52.2
23.64
OFFICE 3 is basically the same as OFFICE 4 except for lack of a coffee pot and television in
OFFICE 4
Q = KA∆Ɵ
Q1SE = 19×2.37×(Ɵe-Ɵi-3℃)
= 19×2.37×3 = 135.09W
Q1NW =19.5×2.09 ×(Ɵe-Ɵi)
Q1NW =19.5×2.09×6 = 244.53W
Q1SW = 12×2.37×3 = 85.32W
Q1NE = 12×2.37×3 = 85.32W
Q1 Ceiling = 28×1.14×(Ɵe-Ɵi+3℃)
Q1 Ceiling = 28×1.14×9 = 287.28W
Q1 Floor = 28×1.30×(20℃-Ɵi)
= 28×1.30×-6 = -218.4W
Q1 Glass = 1.5×5.8×6 = 52.2W
Q1 Door = 2×3.94×(Ɵe-Ɵi-3℃)
Q1 Door = 2×3.94×3 = 23.64W
Q1 Tot = ∑Qi
→ 694.98W
Q2 ( Contribution of Solar Radiation heat through the walls
Q2RadW = a. S. F. R
Where time = 1 Pm
S = 19.5m2
F = 0.2 , a = 0.4
Wall
RM
NW
380
Q2RadW = 0.4×19.5×0.1×380= 296.4W
Q3 Contribution of Solar Radiation on Glazing
Q3Radg = a. S. F. RV
At time 13hr
Wall
RV
NW
326
a = 0.86 , g = 0.45 , S = 1.5
Q3Radg = 0.86×0.45×1.5×326 = 189.243W
Q4 Contribution of heat by Renewal and Infiltration
QSRW = 0.33qVT(Ɵe-Ɵi)
→ qVT = γ × S
Population density = 1 person = qV = 25
S = 28
Γ = 0.1
Thus No of persons = 0.1×28 = 3 people
qVT = N × qV
qVT = 3×25 = 75m3/h
Latent heat
We = 0.025
Wi = 0.0108
QIRW = 0.84qV(We- Wi)
= 0.84×75×(0.025-0.0108)
= 0.8946W
Q5 Heat due to Occupants
sensible heat (QSOC)
QSOC =Ƞ× CSOC
QSOC = 3×63 = 189W
Latent heat (QLOC))
QLOC = Ƞ× LOC
QLOC = 3× 59 = 177W
Q6 Contribution of heat by lighting
QSED = 1.25P
S = 28
W = 16W
Thus P = 28×16 = 448W
Thus QSED = 1.25×448 = 560W
Q7 ( Machines and equipment heat contribution)
1) Computer
Sensible heat = 250W
Latent heat = 0
2) Printer ( ink- jet printer)
Sensible heat = 52W
3) Television
Sensible heat = 175W
Latent heat = 0
4) Coffee pot
Sensible heat = 750W
Latent heat = 300W
Thus Q7Sens = 1227W
Q7Lat = 300W
QTot = Q1+Q2+Q3…+ Q7
Thus → QTot Sensible = 3305.12W
→ QTot Latent = 477.8946W
QTOT = QTot Sensible+ QTot Latent
QTOT = 3305.12W + 477.8946W = 3783.0146W
Refrigrerative Power
= 3.783KW
OFFICE 4
Description
Volume (m3)
Orientation of
the wall
Area ( m2)
KLW/M3
Intensity of
thermal
radiation (W)
OFFICE 4
84m3
SE
NW
SW
NE
ceiling
floor
Glass (NW)
Door
19
19.5
12
12
28
28
1.5
2
Q = KA∆Ɵ
Q1SE = 19×2.37×(Ɵe-Ɵi-3℃)
= 19×2.37×3 = 135.09
Q1NW =19.5×2.09 ×(Ɵe-Ɵi)
Q1NW =19.5×2.09×6 = 244.53W
Q1SW = 12×2.37×3 = 85.32W
Q1NE = 12×2.37×3 = 85.32W
Q1 Ceiling = 28×1.14×(Ɵe-Ɵi+3℃)
Q1 Ceiling = 28×1.14×9 = 287.28W
Q1 Floor = 28×1.30×(20℃-Ɵi)
= 28×1.30×-6 = -218.4W
Q1 Glass = 1.5×5.8×6 = 52.2W
Q1 Door = 2×3.94×(Ɵe-Ɵi-3℃)
Q1 Door = 2×3.94×3 = 23.64W
Q1 Tot = ∑Qi
→ 694.98W
Q2 ( Contribution of Solar Radiation heat through the walls
Q2RadW = a. S. F. R
Where time = 1 Pm
2.37
2.09
2.37
2.37
1.14
1.30
5.8
3.94
135.04
244.53
85.32
85.32
287.28
-218.4
52.2
23.64
Wall
RM
NW
380
S = 19.5m2
F = 0.2 , a = 0.4
Q2RadW = 0.4×19.5×0.1×380= 296.4W
Q3 Contribution of Solar Radiation on Glazing
Q3Radg = a. S. F. RV
At time 13hr
Wall
RV
NW
326
a = 0.86 , g = 0.45 , S = 1.5
Q3Radg = 0.86×0.45×1.5×326 = 189.243W
Q4 Contribution of heat by Renewal and Infiltration
QSRW = 0.33qVT(Ɵe-Ɵi)
→ qVT = γ × S
Population density = 1 person = qV = 25
S = 28
Γ = 0.1
Thus No of persons = 0.1×28 = 3 people
qVT = N × qV
qVT = 3×25 = 75m3/h
Latent heat
We = 0.025
Wi = 0.0108
QIRW = 0.84qV(We- Wi)
= 0.84×75×(0.025-0.0108)
= 0.8946W
Q5 Heat due to Occupants
sensible heat (QSOC)
QSOC =Ƞ× CSOC
QSOC = 3×63 = 189W
QLOC = Ƞ× LSOC
QLOC = 3× 59 = 177W
Q6 Contribution of heat by lighting
QSED = 1.25P
S = 28
W = 16W
Thus P = 28×16 = 448W
Thus QSED = 1.25×448 = 560W
Q7 ( Machines and equipment heat contribution)
1) Computer
S. h = 250W
L. h = 0
2) Printer ( ink- jet printer)
S. h = 52W
QTot = 2078.12W + 250W + 52W = 2380.12W
Refrigerative Power
→ 2.38012KW
OFFICE 5
Description
OFFICE 5
Q1 = KA∆Ɵ
Volume ( m3)
120m3
Orientation
of the wall
Area ( m2)
KLW/m3
SE
NW
SW
NE
Ceiling
Floor
Glass (SE)
Door ( NW)
27
26
12
12
40
40
3
4
2.09
2.37
2.09
2.37
1.14
1.30
5.8
3.94
Intensity of
thermal
radiation (W)
338.58
184.86
150.4
85.32
410
-312
104.4
47.28
Q1SE = 27×2.09×(Ɵe-Ɵi)
Q1SE = 27×2.09×6 = 338.58W
Q1NW = 26×2.37×3 = 184.86W
Q1SW = 12×2.09×6 = 150.48W
Q1NE = 12×2.37×3 = 85.32W
Q 1Ceiling = 40×1.14(Ɵe-Ɵi+3℃)
Q 1Ceiling= 40×1.14×9 = 410.4W
Q 1Floor = 40×1.14(20℃-Ɵi)
Q 1Floor = 40×1.14×-6 = -312W
Q 1Glass = 3×5.8×6 = 104.4W
Q 1Door = 4×3.94×3 = 47.28W
QT = ∑Qi
Thus QT = 1009.32W
Q2 ( Contribution of Solar Radiation of walls)
Q 2RadW= a. SW. F. Rm
Wall
Rm
SE
256
SW
256
a = 0.4 , S = 39 , F = 0.1 ,
Rm = ( 256 + 256)
Q2RadW = 0.4×39×0.1×512 = 798.72W
Q3 Contribution of Solar Radiation of Glazing
Q 2Radg = a. S. g. RV
Wall
RV
SE
220
a = 0.86 , g = 0.45 , S = 3
Q3radg = 0.86×3×0.45×220 = 255.42W
Q4 ( Contribution to heat by renewal and infiltration)
→Sensible heat
QSRW = 0.33qV(Ɵe-Ɵi)
→ 1 person/m2 = 25m3/h
No of persons = γ × S
→ γ = 0.1 and S = 40
No of persons = 0.1×40 = 4
qVT = N× 25
qVT = 4× 25 = 100m3/h
QSRW = 0.33×100×(32-26)
QSRW = 0.33×100×6 = 198W
Latent heat
QSRW = 0.84qV(We-Wi)
We =0.025
Wi = 0.0108
→ 0.84×100×(0.025-0.0108)
= 1.1928W
Q5 Heat due to Occupants
Sensible heat
QSOC = Ƞ × CSOC
CSOC = 63
→ QSOC = 6×34 = 252W
Latent heat (QLOC)
QLOC = Ƞ × LSOC
→ LSOC = 59
Thus QLOC = 4×59 = 236W
Q6 Contribution of heat by lighting
QSED = 1.25P
From table 17, we get W = 16W
P=S×W
W = 16W, S = 40
P = 40×16 = 640W
Thus QSED = 1.25P
= 1.25(640) = 800W
Q7 Contribution of heat due to machines and equipment
Computer
Sensible heat = 250W
Latent heat = 0
Printer (ink- jet printer)
Sensible heat = 52W
Photocopy Machine
Sensible heat = 750W
Latent heat = 300W
Coffee pot
Sensible heat = 750W
Latent heat = 300W
Q7Tot Sens = 1802W
Q7Tot Lat = 600W
QT =Q1 + Q2 … + Q7
Thus QTot = 5115.46W (sensible heat)
QTot =837.1928W
QT =QT1 + QT2
QT = 5952.6528W
Refrigerative Power
5.953KW