Duhok Polytechnic University
Technical College of Engineering
Refrigeration and Air-Conditioning Engineering Department
Refrigeration and Air-Conditioning Engineering
Cooling Load Estimation
Project
Students:Ajin Nawfal
Edris Salam
Lecturer: Mr.Badran
Date: 5-May-2018
Q/ Estimate the Cooling load capacity of a House of
149 m² showing in a figure below.
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located in Duhok /Kurdistan-Iraq
Plan Details
Livingroom: 33 m² Bathroom: 7 m²
Bedroom: 15 m² Bathroom: 4 m²
Bedroom: 17 m² Kitchen: 18 m²
Bedroom: 13 m² Hall: 14 m²
Porch: 9 m²
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1-ROOF CONSTRUCTION
𝐐 = 𝐔 ∗ 𝐀 ∗ (𝐂𝐋𝐓𝐃)𝐜𝐨𝐫𝐫𝐞𝐜𝐭𝐢𝐨𝐧
(CLTD)correction = [CLTD + LM] ∗ K + (25.5 − Tr) + (To − 29.4)
CLTD =19 -- From Table @152.4mm h.w concrete with25.4 insulation with suspended ceiling
U=0.71 -- From Table @152.4mm h.w concrete with25.4 insulation with suspended ceiling
LM=1.1 --From Table @32 latitude
K=1
Tr=22 -- minimum temperature
To=46 – maximum temperature
(CLTD)correction = [19 + 1.1] ∗ 1 + (25.5 − 22) + (46 − 29.4)
(CLTD)correction = 40.2
Q = 0.71 ∗ 149 ∗ 40.2
𝑸 = 𝟒𝟐𝟓𝟐. 𝟕𝟓𝟖 𝑾
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2-WALLS
A- not partition wall ( exposure wall)
Q = U ∗ A ∗ (CLTD)correction
(CLTD)correction = [CLTD + LM] ∗ K + (25.5 − Tr) + (To − 29.4)
LM from Table @32latitute
CLTD from table @ Group E WALLS
U =3.32 from Table @ H.W. concrete wall + finish .. 101.6 mm concrete
Direction wall Area CLTD LM
NORTH WALL 13.85*3 12 0.5
South WALL 13.85*3 19 6.6
EAST WALL 12.02*3 21 0
West WALL 12.02*3 27 0
I) NORTH WALL
(CLTD)correction = [12 + 0.5] ∗ 1 + (25.5 − 22) + (46 − 29.4)
(CLTD)correction = 32.6
Q = 3.32 ∗ (13.85 ∗ 3) ∗ 32.6
𝑸 = 𝟒𝟒𝟗𝟕. 𝟎𝟑𝟗𝟔 𝑾
II) SOUTH WALL
(CLTD)correction = [19 + 6.6] ∗ 1 + (25.5 − 22) + (46 − 29.4)
(CLTD)correction = 45.7
Q = 3.32 ∗ (13.85 ∗ 3) ∗ 45.7
𝑸 = 𝟔𝟑𝟎𝟒. 𝟏𝟑𝟐𝟐 𝑾
III) EAST WALL
(CLTD)correction = [21 + 0] ∗ 1 + (25.5 − 22) + (46 − 29.4)
(CLTD)correction = 41.1
Q = 3.32 ∗ (12.02 ∗ 3) ∗ 41.1
𝑸 = 𝟒𝟗𝟐𝟎. 𝟒𝟓𝟗𝟏𝟐 𝑾
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IV) West WALL
(CLTD)correction = [27 + 0] ∗ 1 + (25.5 − 22) + (46 − 29.4)
(CLTD)correction = 47.1
Q = 3.32 ∗ (12.02 ∗ 3) ∗ 47.1
𝑸 = 𝟓𝟔𝟑𝟖. 𝟕𝟕𝟒𝟑𝟐 𝑾
3-GLASSES
A- Conduction
𝐐 = 𝐔 ∗ 𝐀 ∗ 𝐂𝐋𝐓𝐃
CLTD=8 From Table
U=1
4 windows 1.5*1.5
Q = 1 ∗ (1.5 ∗ 1.5) ∗ 8
𝑸𝟏 = 𝟏𝟖 𝑾for each window *4 windows
𝑸𝟏 = 𝟏𝟖 ∗ 𝟒 = 𝟕𝟐 𝑾
3 windows 1.5*1.1
Q = 1 ∗ (1.5 ∗ 1.1) ∗ 8
𝑸𝟐 = 𝟏𝟑. 𝟐 𝑾for each window *3 windows
𝑸𝟐 = 𝟏𝟑. 𝟐 ∗ 𝟑 = 𝟑𝟗. 𝟔 𝑾
3 windows 1.5*0.5
Q = 1 ∗ (1.5 ∗ 0.5) ∗ 8
𝑸𝟑 = 𝟔 𝑾for each window *3 windows
𝑸𝟑 = 𝟔 ∗ 𝟑 = 𝟏𝟖 𝑾
QTotal = Q1 + Q2 + Q3
𝑸𝑻𝒐𝒕𝒂𝒍 = 𝟕𝟐 + 𝟑𝟗. 𝟔 + 𝟏𝟖 = 𝟏𝟐𝟖. 𝟔 𝑾
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B-Solar:
Q = A ∗ SC ∗ SHFG ∗ CLF
SC=0.92From Table @ 10mm Clear
SHFG From Table
CLF From Table
Direction OF GLASSES Area SHFG CLF
NORTH GLASSES 2(1.5*1.5)+(1.1*1.5) 148 0.91
EAST GLASSES 2(1.1*1.5) 710 0.8
South GLASSES 2(1.5*1.5)+2(0.5*1.5) 801 0.83
I) NORTH GLASSES
Q = (No. of windows ∗ A) ∗ SC ∗ SHFG ∗ CLF
Q = {(1.1 ∗ 1.5) + (2 ∗ 1.5 ∗ 1.5 )} ∗ 0.92 ∗ 148 ∗ 0.91
𝑸 = 𝟕𝟔𝟐. 𝟎𝟏𝟗𝟒𝟒 𝑾
II) EAST GLASSES
Q = A ∗ SC ∗ SHFG ∗ CLF *NO.OF WINDOWS
Q = (2 ∗ 1.1 ∗ 1.5) ∗ 0.92 ∗ 710 ∗ 0.8
𝑸 = 𝟏𝟕𝟐𝟒. 𝟒𝟒𝟖 𝑾
III) SOUTH GLASSES
Q = A ∗ SC ∗ SHFG ∗ CLF *NO.OF WINDOWS
Q = ((2 ∗ 1.5 ∗ 1.5) + (2 ∗ 0.5 ∗ 1.5)) ∗ 0.92 ∗ 801 ∗ 0.83
𝑸 = 𝟑𝟔𝟔𝟗. 𝟖𝟔𝟏𝟔 𝑾
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4-LIGHTS:
𝑄 = 𝑁𝑂. 𝑂𝐹𝐿𝐼𝐺𝐻𝑇𝑆 ∗ 𝑊𝐴𝑇𝑇𝐴𝐺𝐸 ∗ 𝐶𝐿𝐹
No.of light=15
CLF=0.93 From Table 17A
𝑄 = 15 ∗ 40 ∗ 0.93
𝑸 = 𝟓𝟓𝟖 𝑾
5-PEOPLE :
I) Sensible heat
𝑄𝑠 = 𝑁𝑂. 𝑂𝐹𝑃𝐸𝑂𝑃𝐿𝐸 ∗ 𝑆𝑒𝑛𝑠. 𝐻𝐺 ∗ 𝐶𝐿𝐹
No. of People=5
Sens.HG=75 From Table 18
CLF=0.97 From Table 19
𝑄𝑠 = 5 ∗ 75 ∗ (0.97)
𝑸𝒔 = 𝟑𝟔𝟑. 𝟕𝟓 𝑾
II) Latent heat
𝑄𝑙 = 𝑁𝑂. 𝑂𝐹𝑃𝐸𝑂𝑃𝐿𝐸 ∗ 𝐿𝑎𝑡𝑒𝑛𝑡. 𝐻𝐺
No. of people=5
Latent.HG=75 From Table 19
𝑄𝑙 = 5 ∗ 75
𝑸𝒍 = 𝟑𝟕𝟓 𝑾
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6-APPLIANCE:
I) Sensible heat
𝑄𝑠 = 𝑁𝑂. 𝑂𝐹𝐴𝑃𝑃𝐴𝑅𝐴𝑇𝑈𝑆 ∗ 𝑆𝑒𝑛𝑠. 𝐻𝐺 ∗ 𝐶𝐿𝐹
NO.of App=10
Sens.HG=550 From Table 21
CLF=0.73 From Table 22
𝑄𝑠 = 10 ∗ 550 ∗ (0.73)
𝑸𝒔 = 𝟒𝟎𝟏𝟓 𝑾
II) Latent heat
𝑄𝑙 = N𝑂. 𝑂𝐹𝐴𝑃𝑃𝐴𝑅𝐴𝑇𝑈𝑆 ∗ 𝐿𝑎𝑡𝑒𝑛𝑡. 𝐻𝐺
NO.of App=10
Latent.HG=100 From Table 21
𝑄𝑠 = 10 ∗ 100
𝑸𝒔 = 𝟏𝟎𝟎𝟎 𝑾
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7-VENTILATION
L/S/PERSON
𝐋/𝐬
𝑽̇𝒇 = 𝑵𝑶. 𝑶𝑭𝑷𝑬𝑹𝑺𝑶𝑵 ∗
𝑷𝑬𝑹𝑺𝑶𝑵
ṁ𝒇 = 𝑽̇ ∗ 𝝆
No.of Person=5
L/s/Person =3.5 from Table 6.1
Ρ=1.2 Kg/m3
Cp=1.022 Kj/Kg*C
𝑳 𝒎𝟑
𝑽̇𝒇 = 𝟓 ∗ 𝟑. 𝟓 = 𝟏𝟕. 𝟓 = 𝟎. 𝟎𝟏𝟕𝟓
𝑺 𝒔
𝒌𝒈
ṁ𝒇 = 𝟎. 𝟎𝟏𝟕𝟓 ∗ 𝟏. 𝟐 = 𝟎. 𝟎𝟐𝟏
𝒔
1) Sensible heat
𝑄𝑠 = ṁ𝒇 ∗ 𝐶𝑃𝑚 ∗ ∆𝑇
𝑄𝑠 = 𝟎. 𝟎𝟐𝟏 ∗ 1.022 ∗ (46 − 22)
𝑸𝒔 = 𝟎. 𝟓𝟏𝟓𝟏 𝑲𝑾 = 𝟓𝟏𝟓. 𝟏𝑾
2) Latent heat
𝑄𝑙 = ṁ𝒇 ∗ ℎ𝑓𝑔 ∗ ∆𝑊
Hfg=2500 kj/kg
w2=0.011 from psychrometric chart
w1=0.0087 from psychrometric chart
𝑄𝑙 = 𝟎. 𝟎𝟐𝟏 ∗ 2500 ∗ (0.011 − 0.0087)
𝑸𝒍 = 𝟎. 𝟏𝟐𝟏 𝑲𝑾 = 𝟏𝟐𝟏 𝑾
3) Total
𝑄𝑡 = 𝑄𝑠 + 𝑄𝑙
𝑄𝑡 = 515.1 + 121
𝑸𝒕 = 𝟔𝟑𝟔. 𝟏 𝑾
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8-INFILTRATION
ACH (AIR CHANGE PER HOUR)
𝑽̇𝒇 = 𝑨𝑪𝑯 ∗ 𝑽𝑶𝑳𝑼𝑴𝑬
ṁ𝒇 = 𝑽̇ ∗ 𝝆
ACH=0.69 From Table 62
Volume=Area*L=149*3
Ρ=1.2 Kg/m3
Cp=1.022 Kj/Kg*C
𝟏𝟒𝟗 ∗ 𝟑 𝐦𝟑
𝑽̇𝒇 = 𝟎. 𝟔𝟗 ∗ ( ) = 𝟎. 𝟎𝟖𝟓𝟕 = 𝟖𝟓. 𝟕 𝐋/𝐬
𝟑𝟔𝟎𝟎 𝐬
ṁ𝐟 = 𝟎. 𝟎𝟖𝟓𝟕 ∗ 𝟏. 𝟐 = 𝟎. 𝟏𝟎𝟑 𝐤𝐠/𝐬
I) Sensible heat
Qs = ṁ𝐟 ∗ CPm ∗ ∆T
Qs = 𝟎. 𝟏𝟎𝟑 ∗ 1.022 ∗ (46 − 22)
𝑸𝒔 = 𝟐. 𝟓𝟐𝟐 𝑲𝑾 = 𝟐𝟓𝟐𝟐𝑾
II) Latent heat
Ql = ṁ𝐟 ∗ hfg ∗ ∆W
Ql = 𝟎. 𝟏𝟎𝟑 ∗ 2500 ∗ (0.011 − 0.0087)
𝑸𝒍 = 𝟎. 𝟓𝟗𝟐𝟐 𝑲𝑾 = 𝟓𝟗𝟐. 𝟐 𝑾
III) Total
Qt = Qs + Ql
Qt = 2522 + 592.2
𝑸𝒕 = 𝟑𝟏𝟏𝟒. 𝟐 𝑾
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Results Table
add 8%
Total Total Total heat gain
Q sensible Q latent for factor
ITEM heat heat gain Refrigeration
(W) (W) of safity
gain (W) (KW) Tons (RT)
(R.T)
Roof 4253 0 4253 4.253 1.209321 1.30607
Wall north 4497 0 4497 4.497 1.278701 1.381
Wall east 4921 0 4921 4.921 1.399264 1.5112
Wall south 6305 0 6305 6.305 1.792798 1.93622
Wall west 5639 0 5639 5.639 1.603424 1.7317
Glass conduction 129 0 129 0.129 0.03685 0.03808
Glass solar north 762 0 762 0.762 0.216671 0.234
Glass solar east 1725 0 1725 1.725 0.490496 0.52974
Glass solar south 3700 0 3700 3.7 1.052078 1.13624
Lights 558 0 558 0.558 0.158665 0.17136
People 364 375 739 0.739 0.210131 0.22694
Appliance 4015 1000 5015 5.015 1.425992 1.54007
Ventilation (L/s) 515 121 636 0.636 0.180844 0.19531
Infiltration 2522 593 3115 3.115 0.885736 0.95659
TOTAL 39900 2089 41989 41.989 11.9394 12.8945
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