67243 cooling and heating & calculation

A.M.ATIQULLAH(Avey†gvnv¤§`
AvwZKzj¨v)
INSTRUCTOR(TECH)
RAC(BÝUªv±i/AviGwm)
DHAKA POLYTECHNIC INSTITUTE,TEJGAON Dhaka.
PresentedBy :A.M.Atiqullah,INSTRUCTOR(Tech)RACDHAKA POLYTECHNIC INSTITUTE,M: 01711-056403
welq†KvW t 67243,4_©-ce©
cÂg - Aa¨vq 5th

Kzwjs †jvW
K¨vjKz‡jk‡b AwZwi³
wnU †jvW m¤ú‡K©
Aa¨vq -05
Presented By : A.M.Atiqullah,INSTRUCTOR(Tech)RACDHAKA POLYTECHNIC INSTITUTE,M:01711-056403

G.O : 5. External And Internal Cooling
And Heating Load Calculation.
PresentedBy :A.M.Atiqullah,INSTRUCTOR(Tech)RACDHAKA POLYTECHNIC INSTITUTE,M: 01711-056403
Kzwjs†jvW K¨vjKz‡jk‡b AwZwi³wnU†jvW m¤ú‡K©
aviYv

S.O:5.1 Describe the Heat Gain by Conduction
Through Building Structure.
S.O:5.2 Explain the Solar Heat Gain Throughout
Side Wall and Roof.
S.O: 5.3 Describe Solar Heat Gain Through Glass Area.
S.O: 5.4 Calculate the Heat Gain Due to Infiltration
and Ventilation Load of Door and Window.
S.O:5.5 Solve the Problems Relating
to the External Heat Load.Presented By : A.M.Atiqullah,INSTRUCTOR(Tech)RACDHAKA POLYTECHNIC INSTITUTE,M:01711-056403

mvaviYZt Zvc(Heat) AeKvVv‡gvi Pvicv‡ki
evqygÛj n‡Z cwienY(Conduction)
cÖwµqvq Qv`,‡g‡S,‡`Iqvj,`iRv Ges Rvbvjv
cÖf„wZi gva¨‡g mÂvwjZ n‡q wbqwš¿Z Kÿ‡K
DËß evMig(Hot) K‡i|

S.O: 5.1 Describe the heat gain by
conduction through building structure.

Primary Heat Transport Modes are:
1.Conduction(cwienb) (heat flow on a molecular scale. Medium
at rest or moving);
2.Convection(cwiPjb) (heat conveyed as internal thermal energy of mass
that is displaced by mean or turbulent motion);
3. Radiation(wewKiY) (heat transfer by electromagnetic waves such
as infrared or visible light).
1. Conduction Heat Transfe : Conduction heat transfer problems
relevant to buildings include:
(a)Exterior wall conduction- • transient heat transfer responding to
climatic effects, such as temperature fluctuation, solar radiation, wind
and precipitation; thermal storage ... damping and lag effect; and cold-
bridge effect (two-dimensional and non-linear heat flow path).
(b) Interior mass conduction- • heat storage in partition walls,
floor/ceiling sandwich.
(c) Conversion from heat gain/loss to cooling and heating load.
(d) Ground heat loss from slab-on-grade floor and basement walls.
S.O: 5.1 Describe the Heat Gain by Conduction
Through Building Structure

S.O: 5.1 Describe the heat gain by conduction
through building structure.

3. Radiant Heat Exchange Radiation heat transfer is very
important in building application in the following a reas:
(a) Short wavelength radiation: • solar heat absorption on opaque
exterior surfaces,
• solar heat transmission through transparent surfaces,
• solar heat absorption and reflection by interior buildings Unlaces,
• absorption and reflection of solar heat by window glass.
(b) Long-wavelength radiation: • heat emission by the exterior surfaces
to the sky ,
• heat exchange among interior surfaces,
• heat exchange between interior surfaces and occupants,
• heat exchange between the lighting fixture and interior surfaces.
S.O: 5.1 Describe the heat gain by
conduction through building structure.

`~iZ¡
‡ÿÎdj
Mig VvÛv
S.O: 5.1 Describe the Heat Gain by Conduction
Through Building Structure

Heat transfer through Conduction: Heat transfer through a
material takes place by conduction from warm to cold side. The
same process takes place in a building. Generally the thermal
conductivity of the building materials will be much lower. In solid
bodies including building components, thermal conduction takes
place when one part of the component is subjected to higher
temperature and the other part to a lower temperature condition.
Most cases of thermal conduction are usually analysed and
treated in their simplified form as one dimensional heat flow
cases, i.e. heat flow in directions other than the main direction is
neglected. Similarly, if the changes in atmospheric conditions
(inside and I or outside) are assumed to be very slow, neglecting
these changes, the process of heat transfer can be assumed to be
"Steady State Heat Transfer" in its simplified form.
S.O: 5.1 Describe the heat gain by conduction
through building structure.

S.O: 5.1 Describe the Heat Gain by Conduction Through Building Structure.

‡mŠiZvcwewKiYcÖwµqv2 fv‡M n‡q _v‡K-
(K) B‡¤úwiK¨vj c×wZ (Emperical Methods)|
(L) BKzBf¨v‡j›U †U¤úv‡iPviwWdv‡ibwkqvj c×wZ|
wWwµ‡g›Ud¨v±i A¨vÛUvBg j¨vM c×wZ t-
hw` †`Iqv‡ji Zvc aviYÿgZv AMÖvn¨ Kiv nq Zvn‡j 𝛉 mg‡q †`qv‡ji gva¨‡g K‡ÿ Zvc
mÂvj‡bi ZvrÿwYKcwigvb wb‡¤œv³ mgxKi‡Yi gva¨‡g cÖKvk nq-
Q𝛉 = A× 𝐔 × CLTD [ CLTD = Cooling Load Temperature Different ]
Solar Air Temp: evB‡ii evZv‡mi ZvcgvÎv Ges AvcwZZ †mŠiZvc
wewKi‡Yi †hvMdj‡K †mvjvi Gqvi ZvcgvÎv e‡&j| Q 𝜽 = fo (to - tso) + ∝ I.

In Summer the Solar Radiation Affects the
Outside Surface of Wall and Roof. The Absorbed
Radiation Increases the Temperature of the
Outside Surface to a Value that is Greater than
Outside Air Temperature. This Outside Surface
Temperature is Called Solar Air Temperature.
roof
side
wall

ROOF AND WALL HEAT GAIN (CLTD METHOD)
In CLTD/SCL/CLF method the heat gain through wall and roof is
Q = U*A*(CLTD) [ CLF = Cooling Load Factor ]
Q = Sensible Heat Gain through Wall or Roof
A = Surface Area of Wall or Roof
U = Overall U-Value for composite Wall or Roof
CLTD=Cooling load temperature difference from ASHRAE table for a
given
Latitude
Wall or roof type
Wall or roof exposure orientation
Hour of day
1
2
3

†mvjvi Gqvi ZvcgvÎv(Solar Air Temperature):
Avgiv mvaviYZt eywS †h, evB‡ii evZv‡mi ZvcgvÎv
(T0) Ges AvcwZZ †mŠiZvc wewKi‡Yi (I) GK‡Î
†hvMdj ev wµqvdj‡KB†mvjvi Gqvi ZvcgvÎv e‡j
∴ Q 𝛉 = fo
(to
– tso
) + ∝ 𝐈
GLv‡b, ∝= Z‡jiA¨vemicwUwfwU(Absorptivity).
Q 𝛉 = Pvwicvk¦© n‡Z evB‡ii Z‡j Zvc mÂvjb, tso
= evB‡ii Z‡ji ZvcgvÎv
Fo
= evB‡ii Z‡ji wdj¥ †Kv-Bwdwk‡q›U, 𝐈 = ZvcwewKi‡Yi me©‡gvU cÖLiZv

AsK t GKwU †Ìq‡ji evB‡ii ZvcgvÎv 35 wW.‡m. Ges evB‡ii Z‡ji Zvc mÂvjb mnM
23w/m2 k , ‡Ìqv‡ji gva¨‡g me©‡gvU cÖZ¨ÿ I cwie¨ß †mŠiZvc wewKi‡Yi cwigvb
260 w/m2 Ges A¨veRicwUwfwU 0.9 n‡j †mvjviGqvi ZvcgvÎv wbY©q Ki|
mgvavb t †Ìqv Av‡Q,
I = 260 w/m2 , ∝ = 𝟎. 𝟗 , 𝐟 𝟎 = 𝟐𝟑
𝐖
𝐦 𝟐
° 𝐊 , 𝐚𝐧𝐝 𝐭𝟎 =
𝟑𝟓 𝟎 𝐂
∴ ‡mvjvi Gqvi ZvcgvÎv, Te = ?
∴ Avgiv Rvwb,
‡mvjvi Gqvi ZvcgvÎv, Q 𝛉 = fo (to - tso) + ∝ I. or Te = to +
∝𝐈
𝐟 𝐨
Dc‡iv³m~‡Îgvbewm‡qcvB,Te = (35 +
𝟎.𝟗×𝟐𝟔𝟎
𝟐𝟑
) =45.17O C Answer.
∴ Te = 45.17℃

cÖZ¨ÿ m~h©¨iwk¥ Kv‡Pi ga¨ w`‡q AwZµg ev hvevi
mgq KvP KZ©„K †mvjvi GbvwR©i cwigvb wbf©i K‡i
wbw`ó ai‡bi Kv‡Pi ‰ewkó¨ I¸bv¸‡bi Dci|
mvaviY Kv‡Pi †kvlY mn‡Mi gvb(∝) Lye †ewk bq| wKšÍ
we‡klfv‡e wbwg©Z KvP ch©vß cwigvb Zvc †kvlY K‡i,
hvi d‡j Kv‡Pi ZvcgvÎv e„w× cvq Ges Zvc cwienb
cÖwµqvq Zvc Kv‡Pi Dfq Zj w`‡q cÖevwnZ nq, d‡j
iæ‡giZvcgvÎv (Temperature)e„w× cvq|

‡Pw½sm Gqvi †cvU©
wmsMvcyi

Shading Coefficient(SC)
Here, 𝜆(Lamda) is the Wave Lengh of Radiation and θ is the
Angle of Incidence. “T” is the Transmissivity of the Glass,
“A” is its Absorptivity, and “N” is the Fraction of Absorbed
Energy that is re-emitted into the space. The Overall Shading
Coefficient is thus given by the Ratio:
S.C = F(𝜆, 𝜃)1 /F(𝜆, 𝜃)0
Mmmmmmmmmmmmmkkkkk,,m
Here “T”(𝜆) is the spectral Transmittance at a given Wave
Length in Nanometers and E(𝜆) is the Incident Solar
Spectral Irradiance.

Bbwdj‡Uªkb (infiltration ) ev evZvm AbycÖ‡ek ej‡Z Avgiv mvaviYZ
eywS †h, `iRv †Lvjvi Rb¨ Ges `iRv I Rvbvjvi Pvwicv‡k¦© dvUj Ges wQ‡`ªi
gva¨‡g wbqwš¿Z ¯’v‡b evB‡ii evZv‡mi (Out side Air) cÖ‡e‡ki d‡j
evZv‡mi Aby‡gq Zvc (Sensible) Ges myßZvc (Latent) Df‡qB wdj‡Uªkb
†jv‡Wi AšÍfz©³ ev e„w× cvq| wb‡¤œv³ Dcv`vb¸‡jvi Kvi‡Y wj‡KR ev evZv‡mi
AbycÖ‡ek N‡U, †hgb-
1| evZv‡mi Pvc
2| evZvmcwieZ©b
3| ÷¨vK B‡d±(Stack Effect.) |
µ¨vK c×wZ (Crack System)
Gqvi †PÄ c×wZ (Air Change System)

Infiltration is described as outside air that leaks into a building structure.
These leaks could be through the building construction or through entry
doors. Infiltration heat gains are found by the following equations. These
equations are discussed more in the Psychrometrics Section
The first equation is the total heat gains using enthalpy. In this equation, the
volumetric flow rate of the infiltration or ventilation air must be known. This
value is converted and multiplied by the difference in enthalpy between the
outdoor air conditions and the indoor air conditions.
The following two equation split the total heat gain into the sensible and
latent heat loads.
Sensible Heat Gains are calculated by multiplying the CFM of the infiltrated
air by the difference in the temperatures of the indoor and outdoor air

How to Calculate Heat Load :
Calculating a heat load is necessary before the installation of a radiant heating system can begin,
since different types of radiant heating systems have different BTU output values.
A typical heat load calculation consists of surface heat loss calculation and heat loss due to air
infiltration. Both should be done separately for every room in the house, so having a floor plan
with dimensions of all walls, floors, ceiling, as well as doors and windows is a good place to
start.
Below is a sample 5-step manual to surface heat loss calculation:
Step 1 – Calculate Delta T (Design Temperature):
Delta T is a difference between indoor design temperature (T1) and outdoor design temperature
(T2), where indoor design temperature is typically 68-72℉ depending on your preference, and
outdoor design temperature is a typical low during the heating season. The former can be
obtained by calling your local utility company.
Assuming that T1 is 72℉ and T2 is –5℉, Delta T = 72℉ - (-5℉) = (72℉ + 5℉) = 77℉
Step 2 – Calculate surface area:
If the calculation is done for an outside wall, with windows and doors, the calculations for the
window and door heat loss should be done separately.
Wall Area = Height x Width - Door Surface - Window Surface
If, Wall Area = 8 ft x 22 ft – 24 sq ft - 14sq ft = 176 sq ft – 38 sq ft = 138 sq ft
Step 3 – Calculate U-value:
Use "Typical R-values and U-values" guide to obtain the wall R-value.
U-value = 1 / R-value
U-value = 1 / 14.3 = 0.07
Step 4 – Calculate wall surface heat loss:
Surface heat loss can be calculated using the formula below:
Surface Heat Loss = U-value x Wall Area x Delta T
If, Surface Heat Loss = 0.07 x 138 sq ft x 77 °F = 744 BTUH is
(U-value is based on assuming a 2x4 wood frame wall with 3.5" fiberglass insulation)

Step 5 – Calculate total wall heat loss:
Follow the steps 1 through 4 to calculate heat loss separately for windows, doors, and ceiling.
If, Door Heat Loss = 0.49 x 24sq ft x 77F = 906 BTUH
(U-value is based on assuming a solid wood door)
If, Window Heat Loss = 0.65 x 14sq ft x 77F = 701 BTUH
(U-value is based on assuming a double-panel window)
If, Ceiling Heat Loss = 0.05 x 352sq ft x 77F = 1355 BTUH
(U-value is based on assuming a 6" fiberglass insulation. Ceiling surface is 22ft x 16ft)
Now, add all the number together:
Total Wall heat loss = (Wall loss + Window Loss + Door loss + Ceiling loss)
Total Wall heat loss = 744 BTUH + 906 BTUH + 701 BTUH + 1352 BTUH = 3703 BTUH
Air infiltration rates should always be taken into consideration.
The following formula can be used to calculate heat loss for a room
due to air infiltration:
Air Infiltration Heat Loss = Room Volume x Delta T x Air Changes per Hour x 0.018
Where Room Volume = Length x Width x Height
Air Changes per Hour accounts for air leakage into the room.
For example: Air Infiltration Heat Loss = (22ft x 16ft x 8ft) x 77F x 1.2 x 0.018 = 4683 BTUH
Note: For actual calculations, contact your contractor or system designer.

Latent Heat Gains are calculated by multiplying the CFM of infiltrated air by
the difference in the humidity ratio of the indoor air and the outdoor air.
It is important to note that these loads
are not seen directly by the cooling coil.
These are indirect loads that occur in
each air conditioned space. Ventilation
air is seen directly at the coil and thus
this air must be cooled down to the
supply air distribution temperature
which is much lower than the room
condition air

Cooling Load Calculation

mgm¨vt 22RybZvwi‡L MÖx®§KvjxbmgqweKvj4.30NwUKvq (at 16.30 hr) m~‡h©iA¨vwRgv_wbY©q Ki|
(cÖ‡qvRbxqZ_¨vw`tjÛ‡bi L= 51° N, d = ± 23.5°Gesh = 60°)
mgvavb t
‡`IqvAv‡Q, h = 60°, L= 51°,d = ±23.5° , m~‡h©i A¨vwRgv_(Z) = ?
Avgiv Rvwb,
Tanz =
𝐬𝐢𝐧𝐡
𝐬𝐢𝐧𝐋 𝐜𝐨𝐬𝐡 −𝐜𝐨𝐬𝐋 𝐭𝐚𝐧𝐝
or =
𝐬𝐢𝐧𝟔𝟎°
𝐬𝐢𝐧𝟓𝟏° 𝐜𝐨𝐬𝟔𝟎° −𝐜𝐨𝐬𝟓𝟏° 𝐭𝐚𝐧𝟐𝟑.𝟓°
[gvb ewm‡q ]
=
𝟎.𝟖𝟔
(𝟎.𝟕𝟕 × 𝟎.𝟓 − 𝟎.𝟔𝟐 × 𝟎.𝟒𝟑)
or, =
𝟎.𝟖𝟔
𝟎.𝟑𝟖𝟓 −𝟎.𝟐𝟔𝟔
tanz =
𝟎.𝟖𝟔
𝟎.𝟏𝟏
or, tanz = 𝟕. 𝟐𝟔
∴ z = tan-1 7.26 = 82.16°
∴ Z = 82.16° `wÿY-cwðgw`‡K (West of South). Answer.
∴ m~‡h©i A¨vwRgv_ (Z) = 82.16°(W-S).

QUESTION ?

5g- Aa¨v‡qi cÖkœvejx t
1. KvVv‡gvi wfZi w`‡q cwien‡bi gva¨‡g mÂvwjZ ZvccÖevneY©bvKi|
2. ‡mŠiZvcwewKiY cÖwµqvq evB‡ii †Ìqvj Ges Qv‡ìgva¨‡g ZvcAR©b cÖYvjx†jL?
3. KvP KZ©„K ‡mŠiZvcAR©b eY©bv Ki ?
4. evZvm AbycÖ‡ek RwbZ Zvc AR©b eY©bvKi ?
5. AsK t RyjvB gv‡m weKvj wZb NwUKvq (at 15.00 hr. sun time) c~e©gyLx †Ìqv‡ji gva¨‡g
2.4m × 6.0m × 2m gv‡ci GKwU K‡ÿi Zvc AR©b wbiƒcY Ki|K‡ÿi ZvcgvÎv 22℃.
cÖ‡qvRbxq Z_¨vw` t
‡Ìqv‡ji cyiæZ¡, ∆x = 15cm; ‡Ìqv‡ji NbZ¡, 𝝆 = 1200 kg/m3; c~e©w`‡Ki
†Ìqv‡ji ( 2.4m × 2.6m) Rvbvjvi †ÿÎdj, A = 3m2; ‡Ìqv‡ji Zvc mÂvj‡bi
mvwe©K ¸Yv¼, U= 0.6w/m2°k; Kÿ ZvcgvÎv, t1= 22℃; UvBg j¨vM, ∅ = 5
N›Uv; wWwµ‡g›U d¨v±i, 𝝉 = 0.65|
evwoi KvR(Home Work)

01 FEBRUARY'2020Presented By : A.M.Atiqullah INSTRUCTOR(Tech)RACD.P.I
33

67243 cooling and heating & calculation

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