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 t hrough 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
3
Infiltration Air in m for Freely Infiltrated Air for the door with brake 3
Revolving Door
in m
Infrequent
2.5m3
2.0m3
Average
2.0m3
1.75m3
Heavy
1.5m
3
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 3
in m /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 Latent load
kcal/hr 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 st andard 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 = Q 3
Q3 = Where m = mass of product L = latent heat Tf = time of freezing 4. Product respiration heat = Q 4 Q4 = m × (Heat evolution rate) kg.hr Where, total heat from products = Q = Q 1 + 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 t ypes 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 = 15 /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 /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
/kg;
Enthalpy at point 2,
Mass of air infiltrated =
= 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 = ( ) ( ) = 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 = ( ) ( ) = 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 =
= 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 25°C and Point 4 has to be marked on the chart. By-pass factor BPF
0.2 =
Trial and error is necessary to compute
°C and 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
=0.826 /kg =
=
= 7554.6 kg/ hr
Dew point temperature of coil = Condition of humidity at point 4 = 14.4°C Relative humidity RH %
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 (P ) = R (
)
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 r egain. 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 t urn 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 = 100 / min and = 400 m/min From friction chart for duct,
column Since a = 58 cm and 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 = 40 / min and = 0.4mm of column = (5/100 x 8) From friction chart for duct,
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 = 25 / min and = 0.4mm of column = (5/100 x 8) Pressure drop in BD for 20m = pressure drop in BC for 5m = 0.4 mm O column Pressure drop in BD for 100 m = 100/20 x 0.4/1 = 2 mm O column Q = 25 / min and = 2mm From friction chart for duct,
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 O column Pressure drop in BE for 100 m = 100/20 x 0.4/1 = 2 mm O column Q = 35 / min and = 2mm From friction chart for duct,
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 O column ]
