heat gain due to infiltration
TRANSCRIPT
Heat gain due to infiltration
The infiltration air is the air that enters a conditioned space through window cracks and
opening of doors. This is caused by the pressure differential between the room and the
ambient pressure. Infiltration is also a function of wind velocity and density variation of air
due to temperature difference between inside and outside air. The infiltration is estimated by
the crack length or air changes per hour method.
The air change method is standard out of the two and is explained here. According to this
method infiltrated air through windows and walls is:
Air flow rate =:
Where L = room length; W = room width , H = room height (all in meters) ; Ac = Air
changes per hour
The air changes per hour depend on type of room and its usage. Table 14.5 and Table 14.7
list air changes per hour and values for infiltrated air through doors.
Total room infiltration air for entire building =
This is because infiltration takes place from the windward side of the building. The door
infiltration values also need to be added to the infiltration from windows and walls to find
total building infiltration
Table 14.6 number of air changes per hour
S.N Type room / Building Number of ACH
(Air Changes/ Hr)
1 Rooms with no windows 0.5
2 Room with one wall exposed 1
3 Lounges 1.5
4 Reception Halls 2
5 Entrance Halls 3
6 Bathroom 2
Table 14.7 Door Infiltration
Usage of Door Infiltration Air in m3 for Freely
Revolving Door
Infiltrated Air for the door with brake
in m3
Infrequent 2.5m3 2.0m3
Average 2.0m3 1.75m3
Heavy 1.5m3 1.25m3
Heat Gain due to Ventilation
Ventilation or supply of outside air is provided to maintain air quality in the room. It is
provided to minimize odour, concentration of smoke, carbon dioxide and other undesirable
gases in air. The ventilation air adds sensible heat as well as latent heat. The quantity of
outside air used for ventilation should provide at least half of the air changes per hour in the
building. Also, ventilation air should be higher than infiltration air. Ventilation air quantity
depends on occupancy of smokers and duration of their stay in buildings. Table 14.8 lists the
outside air requirements for specific buildings and premises.
Table 14.8: Ventilation Loads (Outside Air)
S/N Application Smoking Recommended Outside
Air in m3/min/Person
1 Apartment Some 0.6
2 Bank Occasional 0.3
3 Bank Considerable 0.9
4 Departmental store Occasional 0.23
5 Drug Store Some 0.3
6 Factories None 0.3
7 Hospital None 0.9
8 Hotels Considerable 0.9
9 Offices None 0.45
10 Restaurant Considerable 0.45
11 Theatre None 0.23
12 Shops Occasional 0.3
Heat Gain from Occupants
Human body in a cooled space constitutes a load, which is both sensible and latent. The heat
gain from occupants for a building is estimated on the average number of people present in
the building. The heat load of the person depends on the activity of the person in the building.
Table 14.9 lists the heat gain from occupants. Note that total occupant heat:
Q= (no. of occupants) × (load per occupant)
The values listed in table 14.9 are based on 27degree Celsius dry temperature for the room.
Table 14.9 Heat Gain from Occupants
S.N Degree of Activity Application Area Sensible Load
kcal/hr
Latent load
1 Seated Theatre 45.4 37.8
2 Moderately Active Offices 49.4 57.7
3 Standing or walking Dept. Store 50.4 63
4 Dancing Dance hall, bar 61.7 152.5
5 Bowling Bowling Alley 117.2 248.2
6 Exercising Gymnasium 117.2 248.2
7 Light Work Factory 75.6 176.4
8 Moderate Work Hotel/appt 110.3 197.3
Table 14.10 Heat Gain from Appliances
S.N Name of Appliance Electric Gas Sensible Load kcal/hr Latent Load kcal/hr
1 Coffe brewer 226 55
2 Egg boiler 302 201
3 Fry Kettle 882 1260
4 Grill for meat 1184 630
5 Hair Dryer 579 100
6 Toaster 1285 327
7 Stove 1058 831
8 Hot plate 560 980
Heat Gain from Appliances
Appliances are daily use equipment used in conditioned spaces. They are electrical, gas fired,
oil fired or stea-heated. The heat gain from an appliance depends on its size, capacity, and
power consumption. Table14.10 lists sensible and latent load from some standard appliances.
Heat Gain from Lighting Equipment
Heat gained from electric lights depends on the rating of the lights in watts, use factor and
allowance factor.
Q lights= Wattage x Use Factor x Allowance factor
Wattage is the power of the light tube or bulb. Use factor is the ratio of actual wattage in use
to the installed wattage. For residences, commercial stores and shops, use factor is usually
one, but for industries it is 0.5. Allowance factor is used for fluorescent tubes and represents
power used for ballast. The value is usually taken as 1.25.
Heat Gain from Products
Heat emitted from products, which include fruits and vegetables, adds to the sensible as well
as latent cooling load. In case of cold storage, this load is very high. The load can be divided
in four parts. They are:
1. Cooling load above freezing = Q1
Q1=
Where m = mass of product
Cpm = mean specific heat of product
T1 = product temperature
T2= desired storage temperature
Tch = time of chilling
2. Cooling load below freezing = Q2
Q2=
Where T1/ = actual storage temperature of product
T2/= desired freezing temperature’
3. Freezing Load = Q3
Q3 =
Where m = mass of product
L = latent heat
Tf = time of freezing
4. Product respiration heat = Q4
Q4 = m × (Heat evolution rate) kg.hr
Where, total heat from products = Q = Q1 + Q2 +Q3+Q4
Products Q1+Q2+Q3+Q4
Table 14.11 lists the rate of heat production for some products.
Heat Gain from Power Equipment
Power Equipment such as fans add sensible heat in the air- conditioned space. The power
consumed by these devices is converted to heat. The power equipment is usually driven
by motors. The heat gain from the motors is given by ------
Qm= × Load Factor
The load factor is the fraction of total load at which a motor works.
Table 14.11 Rate of heat production from Agricultural Products
S/N Name of Commodity Storage Temperature( Degree Celsius) Heat Evolved Per Ton in
24hrs in kcal
1 Apples 15 1660
2 Bananas 20 2110
3 Carrots 5 875
4 Beets 0 670
5 Cherries 0 440
6 Lemons 15 520
7 Tomatoes 4.5 320
8 Mushrooms 0 1550
9 Potatoes 20 890
10 Oranges 4.5 330
11 Peppers 15 2135
12 Onions 10 500
13 Grapes 15 700
14 Pears 0 220
15 Strawberries 2.5 1665
16 Raspberries 2.5 1110
Heat Gain through Ducts
The heat gain due to supply duct is given by
QD= U ×Ad × (Ta-Ts)
Where U = overall heat transfer coefficient
AD= surface area of duct
Ta= Temperature of ambient air
Ts= Temperature of supply air
This heat gain depends on supply and ambient temperatures for the duct. Ducts, if located
in conditioned space, have zero heat gain. However, if ducts are located in unconditioned
spaces, there is heat gain and condensation in the duct. This is the reason why ducts are
insulated. The heat gain through duct as a rule of thumb is %% of room sensible heat.
Air Leakage from duct joints is of the order of 5-15%. Air leakages affect the cooling
capacity of the duct. For long runs, 10% leakage is assumed and for medium runs, 5%
leakage is assumed. Duct air leakages have to be considered in duct design. This
completes our review of important types of cooling loads in air conditioning.
2. ASHRAE. 1998. Fundamentals of Air System Design. Atlanta: American Society of
Heating, Refrigerating and Air-Conditioning Engineers, Inc.
3. ASHRAE.1998. Fundamentals of Heating Systems. Atlanta: American Society of Heating,
Refrigerating and Air-Conditioning Engineers, Inc.
4. ASHRAE. 1998. Fundamentals of Water System Design. Atlanta: American Society of
Heating, Refrigerating and Air-Conditioning Engineers, Inc.
Problem:
An air conditioning system is to be designed with the following data:
Outside design conditions = 40°C DBT, 28°C WBT
Inside design conditions = 25° DBT, 50% RH
Solar heat gain through walls, roof and floor = 5050 kcal/hr
Solar gain through glass = 4750 kcal/hr
Occupants = 25
Sensible heat gain per person = 50 kcal/hr
Latent heat gain per person = 50 kcal/hr
Internal lighting load = 15 lamps of 100W and 10 fluorescent tubes of 80W
Sensible heat gain from other sources = 10,000kcal/hr
Infiltrated air volume per minute = v1=¿15 m3/min
Assume 25% fresh air and 75% re-circulated air is mixed and passed through the conditioner
coil and draw the skeleton psychometric chart with the relevant points. The by-pass factor is
taken as 0.2. The following quantities are to be determined..
a. Amount of total air required in m3/hr
b. Dew point temperature of coil
c. Condition of supply air to room
d. Capacity of conditioning plant
Fig abc shows the psychometric chart. First point 1 is marked on the chart for 40°C DBT and
28°C WBT, followed by 25°C DBT and 50% RH as point 2. Point 1 and 2 is then joined.
Point A is located by drawing vertical and horizontal lines from points 1 and 2.
From psychometric chart
vs1=0.914 m3/kg;
h1=21.48 kcal /kg
Enthalpy at point 2,
h2=12.1kcal /kg
ha=15.72 kcal /kg
Mass of air infiltrated = m1=v s
vs 1= 15.72m3/min
0.914 m3/kg = 16.41 kg/min
Sensible heat gain due to infiltrated air
= mass of infiltrated air at point 1 x enthalpy difference between point a and 2
= m1 ( ha−h2 )=16.41 kg /min (15.72−12.1 ) kcal / kg = 59.4 kcal/min
=3564 kcal/hr
Latent heat due to infiltrated air
= mass of infiltrated air at point 1 x enthalpy difference between point 1 and a
= m1 ( h1−ha)=16.41 kg/min (21.48−15.72 )kcal /kg = 59.4 kcal/min
= 5671 kcal/hr
Total sensible heat gain from occupants
= Sensible heat gain per occupant x no. of occupants.
= 50 kcal/occupant x 25 occupants
= 750 kcal/hr
Total latent heat gain from occupants
= Latent heat gain per occupant x no. of occupants
= 50 kcal/occupant x 25 occupants
= 750 kcal/hr
Sensible heat gain due to lighting load
= no of lamps of 100 x power/lamp + no of lamps of fluorescent tubes of 80W
= 15 x 100 + 10 x 80
= 32.5 kcal/min
= 1950 kcal/hr
Total room sensible heat = RSH
= solar heat gain from walls, roof, floor + solar heat gain from glass + infiltration load + heat
gain from occupants + lighting load + other sources
= 5050 + 4750 + 3564 + 750 + 1950 + 10,000 (kcal/hr)
= 26,040 kcal/hr
Total room latent heat gain = RLH
= latent heat gain from infiltrated air + latent heat from occupants
= 5670 + 750
= 6421 kcal/hr
Room sensible heat factor RSHF = RSH
RSH +LSH= 26,064
26,064+6421 = 0.802
From point 2, draw line 2-5 parallel to the alignment line (starting from aligning circle, 26°C
DBT and 50% RH to RSHF 0.802). This line 2-5 is the RSHF line. Since 25% fresh air and
75% re-circulated air is mixed and passed through the conditioner coil, point 3 is marked on
the chart such that length 2-3 = length 2-1 x 0.25
From the chart at point 3, DBT is T d 3=¿ 25°C and h3=14.4 kcal /kg
Point 4 has to be marked on the chart.
By-pass factor BPF ¿Td 4−T d 6
T 3−T 6
0.2 = T d 4−T d 6
28.8−T d 6
Trial and error is necessary to compute T d 6∧T d 4
T d 4=14.4°C and T d 6=¿ 10.8°C
Point 4 represents condition of air leaving the coil and entering the room
Point 6 represents the DPT (dew point temperature) of the coil. Locate point 4 on the chart
h4=7.8 kcal /kg
vs 4 =0.826 m3/kg
va=a mount of total air required∈m3 /kg
va=ma xv s 4
ma= total roomheatheat removed =
ESH+RLHh2−h4
=26064+642112.1−7.8 = 7554.6 kg/ hr
va=7554.6 x0.826=6240 m3
Dew point temperature of coil = T dp=T d6=10.8 ° C
Condition of humidity at point 4 = T d 4=¿14.4°C
Relative humidity RH Φd=88%
Duct Design Objectives
The aim of duct design is to work out the size of ducts in an air-conditioning system. All
measurements of the ducts need to be estimated appropriately. The ducts should carry the
essential volume of conditioned air from the fan outlet to the area of interest with minimal
bends, obstructions or area changes. Also, the arrangement of the ducts should be symmetric
as far as possible with aspect ratio between 4 and 8 for rectangular ducts. The velocity of air
in the duct should be high enough to trim down the size of the duct and small enough to
minimize n oise and pressure losses. There are three methods to design a duct. They are the
velocity reduction method, static regain method and equal pressure drop (equal friction)
method.
I. Velocity Reduction Method
Using this approach, the velocities of the air in the duct are assumed such that they
progressively decrease as the flow continues in the duct. The pressure drops are calculated for
air velocities of the branches and the main duct. The duct sizes are determined for assumed
velocities and known quantities of air supplied through ducts. The pressure at the outlet of the
duct can be varied by dampers at the exit. The fan is designed to overcome pressure losses
along any single run, including losses in branches, elbows, enlargements and contractions.
II. Equal Pressure Drop ( Equal Friction) Method
Using this method, the size of the duct is decided such that it gives equal pressure drop
(friction loss) per meter length in all the ducts. For a symmetrical duct design, this method
offers equal pressure drop in all runs of the duct and no dampers are needed to balance the
pressure. But, for a non-symmetrical duct design, the shortest run has the minimum loss and
highest pressure at the exit. Dampers would have to be employed to reduce the higher
pressure or the fan would have to supply air at a higher velocities. Clearly, increasing the
velocity may lead to noise and it is the shortcoming of such a method. One way to overcome
the problem is to use noise absorbing outlets in the ducts. The velocities of air in this method
are automatically reduced in the branch ducts as the flow is decreased. This method does not
self-balance if the branches are of different lengths. For such cases, it will be necessary to use
dampers to balance the pressure.
III. Static Regain Method
For the last method, the size of the duct is decided to give equal pressure values at all outlets
for prefect balancing of the duct layout. The friction loss in each branch is made equal to the
gain in pressure due to reduction in velocities. The gain in pressure (static regain) due to
changes in velocity is given by:
Static regain = R (Pv1−¿ Pv 2¿) = R ( v12−v2
2
2 g)
It may not possible to design long run of the duct branches and branches near the fan for
complete regain. In such cases, the main duct is designed first for complete regain. The outlet
pressure is then kept same at all outlets from the main duct for the branches. This method
allows self balancing for ducts but reducing noise increase duct sizes, which in turn affect the
economy of the system.
From all the methods described, an example of the equal friction method will be adopted.
The sizes of the various ducts shown on fig def is to be calculated. Find the maximum
pressure loss. The velocity in duct AB should not exceed 400m/min and ducts are rectangular
in section. One side of all rectangular ducts is 60cm.
Section AB
The quantity of air flowing through duct
AB = Q = 100m3/ min and vAB= 400 m/min
From friction chart for duct,
DAB=58 cm∧hf =8 mmof H 2Ocolumn
Since a = 58 cm and DAB=58 cm
From equivalent chart for a/D= 1.03, we get a/b ratio as 0.83.
Therefore, b = 50cm
Equivalent section of AB = 60cm x50cm
Section BC
Q = 40m3/ min and h f= 0.4mm of H 2O column = (5/100 x 8)
From friction chart for duct,
DBC=40 cm∧vbc=300 m /min
As a = 60 cm, from a/D ratio of 1.50
We get a/b ratio= 2.5 from equivalent length chart
b = 24cm
Equivalent section of BC = 60cm x 24 cm
Section BD
Q = 25m3/ min and h f= 0.4mm of H 2O column = (5/100 x 8)
Pressure drop in BD for 20m = pressure drop in BC for 5m = 0.4 mm H 2O column
Pressure drop in BD for 100 m = 100/20 x 0.4/1 = 2 mm H 2O column
Q = 25m3/ min and h f= 2mm
From friction chart for duct,
DBD=41 cm∧vBD=145 m /min
From equivalent chart, a = 60 cm,
For a/D ratio =1.46 we get a/b ratio =2.58
b = 23.2 cm
Equivalent section of BD = 60cm x 23.2 cm
Section BE
Pressure drop in BE for 20m = pressure drop in BC for 5m = 0.42 mm H 2O column
Pressure drop in BE for 100 m = 100/20 x 0.4/1 = 2 mm H 2O column
Q = 35m3/ min and h f= 2mm
From friction chart for duct,
DBE=44 cm∧v BE=160 m /min
From equivalent chart, a = 60 cm,
For a/D ratio =1.36 we get a/b ratio =2.09
b = 28.6 cm
Equivalent section of BD = 60cm x 28.6 cm
Maximum pressure loss possible = loss in AB + loss in BE = 15/100 x 8+0.4 = 1.6 mm H2O
column
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