Fluid flow throughFluid flow through ducts and air distribution

(Duct design)

131آب، 04

IntroductionIntroductionIn air conditioning systems that use air as the fluid in the thermal distribution system, it is essential to design the Air Handling Unit (AHU) properly. The 

primary function of an AHU is to transmit processed air from the air conditioning plant to the conditioned space and distribute it properly within 

the conditioned space A typical AHU consists of:the conditioned space. A typical AHU consists of: 1. A duct system that includes a supply air duct, return air duct, cooling and/or heating coils, humidifiers/dehumidifiers, air filters and dampers g p

2. An air distribution system comprising various types of outlets for supply air and inlets  for return air.

3.Supply and return air fans which provide the necessary energy to move the air throughout the system.

132آب، 04

Between any two points 1 and 2 in the flow field for irrotational flows, the Bernoulli’s equation is written as:

Each of the heads has units of length as explained before. The above equation can be written in terms of static velocity above equation can be written in terms of static, velocity, datum and total pressures as:

133آب، 04

Since all real fluids have finite viscosity, this is referred to h d l Th h d l ill th t t l tas head loss. The head loss will cause the total pressure to

decrease in the flow direction. If the head loss is denoted by H then Bernoulli’s equation can be modified to:HI then Bernoulli s equation can be modified to:

To overcome the fluid friction and the resulting head, a fan is i d i i diti i t Wh f i i t d drequired in air conditioning systems. When a fan is introduced

into the duct through which air is flowing, then the static and total pressures at the section where the fan is located risetotal pressures at the section where the fan is located rise. This rise is called as Fan Total Pressure (FTP). Then the required power input to the fan is given byq p p g y

134آب، 04

. Evaluation of FTP is important in the selection of a suitable fan for l I b l h h h l d a given application. It can be easily shown that when applied

between any two sections 1 and 2 of the duct, in which the fan is located the FTP is given bylocated, the FTP is given by

135آب، 04

Estimation of pressure loss in ducts:

The pressure drop due to friction is known as frictional d f i ti l ∆ Th dpressure drop or friction loss, ∆pf. The pressure drop

due to momentum change is known as momentum pressure drop or dynamic loss, ∆pd Thus the totalpressure drop or dynamic loss, ∆pd. Thus the total pressure drop ∆pt is given by:

136آب، 04

Evaluation of frictional pressure drop in ducts: The Darcy-Weisbach equation is one of the mostThe Darcy-Weisbach equation is one of the most

commonly used equations for estimating frictional pressure drops in internal flows. This equation is given by: p q g y

Where:f is the dimensionless friction factor.L is the length of the duct .D is the diameter in case of a circular duct and hydraulic

di i f i l ddiameter in case of a non-circular duct.

137آب، 04

Table1 :Average surface roughness of commonly used duct materialsmaterials.

138آب، 04

Taking GI as the reference material and properties of air at 20oC and 1 atm pressure the frictional pressure drop inat 20oC and 1 atm. pressure, the frictional pressure drop in a circular duct is given by:

Where: Qair is the volumetric flow rate of air in m 3/s.L is the length .

D i h i di f h d i D is the inner diameter of the duct in meters.

139آب، 04

Fig.1 :Chart for estimating frictional pressure drop in circular g g p pGI ducts.

1310آب، 04

For small changes in air density (ρ) and temperature oneFor small changes in air density (ρ) and temperature, one can use the following relation to obtain frictional pressure drop from the standard chart.

1311آب، 04

Rectangular ducts: Even though circular ducts require the least material for a

given flow rate and allowable pressure drop, rectangular ducts are generally preferred in practice as they fit easily into the building construction thus occupying less space and theythe building construction thus occupying less space, and they are also easy to fabricate. The ratio of the two sides ‘a’ and ‘b’of the rectangle (a/b) is called as aspect ratio of the duct.g ( ) p

A rectangular duct is said to be equivalent to a circular duct if the volumetric flow rate and frictional pressure dropduct, if the volumetric flow rate and frictional pressure drop per unit length ( ∆P. Qf/ L) are same for both.

Equating these two parameters for a rectangular duct g gand an equivalent circular duct, it can be shown that the equivalent diameter is given by:

1312آب، 04

Dynamic losses in ducts:Dynamic losses in ducts:In turbulent flows, the dynamic loss is proportional to square of velocity. Hence these are expressed as

h K i h d i l ffi iwhere K is the dynamic loss coefficient. 

Sometimes, an equivalent length Leq is defined to eq estimate the dynamic pressure loss through bends and fittings.

Where f is the friction factor L is the equivalent length

1313آب، 04

Leq is the equivalent length.

The dynamic pressure drop due to the elbow or 90o turn is:

1314آب، 04

Evaluation of dynamic pressure loss through various fittings: a) Turns, bends or elbows: The most common type of bends used in air conditioning ducts are 90o turns shown in Fig.

1315آب، 04

installing turning vanes in the bends reduces the dynamic pressure drop as it is equivalent to increasing W/H as shown in Fig drop as it is equivalent to increasing W/H, as shown in Fig.

U f i i 90 b d ( lb ) Use of turning vanes in a 90 o bend (elbow) . 1316آب، 04

b)Branch take-offs:

1317آب، 04

Th d i d f th t ( )t The dynamic pressure drop from the upstream(u)to branch (b),p u-b is given by:

1318آب، 04

c) Branch entries: Branch entries are commonly used in return air ductsair ducts

1319آب، 04

d) Sudden enlargementd) Sudden enlargemente) Sudden contraction:

Sudden enlargement Sudden contraction

1320آب، 04

The pressure loss due to sudden enlargement, ∆Pd,enl is given by Borda Carnot equation asBorda-Carnot equation as:

Where v1 is the velocity before enlargement.A1 and A2 are the areas before and after enlargement ,respectively

1321آب، 04

Similar to sudden enlargement, the dynamic pressure loss due to sudden contraction ∆Pd con is given by:sudden contraction ∆Pd,con. is given by:

Where v2 is the velocity in the downstream.A1 and A2 are the areas at vena contract and after

t ti ti l contraction, respectively. The coefficient cc is known as contraction coefficient and is seen to be equal to area ratio A1/A2be equal to area ratio A1/A2

1322آب، 04

Filters, cooling and heating coils, dampers etc.:

The pressure drop across air handling unit equipment ,The pressure drop across air handling unit equipment , such as, air filters , dampers, cooling and heating coils depend on several factors. Hence, normally these values have to be obtained from the manufacturers data.

1323آب، 04

Static regain:

Whenever there is an enlargement in the cross-sectional area of the duct the velocity of air decreases andsectional area of the duct, the velocity of air decreases, and the velocity pressure is converted into static pressure. Theincrease in static pressure due to a decrease in velocity p ypressure is known as static regain. In an ideal case, when there are no pressure losses, the increase in static pressure (∆ ) i l l h d i l i(∆ps) is exactly equal to the decrease in velocity pressure(∆pv)and the total pressure (pt) remains constant as shown in figshown in fig.

1324آب، 04

Ideal enlargement.

1325آب، 04

Thus for the ideal case: 

1326آب، 04

for sudden enlargements or for other non-ideal enlargements the decrease in velocity pressure will beenlargements, the decrease in velocity pressure will be greater than the increase in static pressure, and the total pressure decreases in the direction flow due to pressure p plosses as shown in Fig. The pressure loss is due to separation of the boundary layer and the formation of eddies

Sudden enlargement. 1327آب، 04

for sudden or non ideal enlargementfor sudden or non-ideal enlargement:

The pressure loss due to enlargement ∆ploss is expressed in terms of a static Regain Factor R as:terms of a static Regain Factor, R as:

1328آب، 04

Where the static regain factor R is given by:

Thus for ideal enlargement the Static Regain Factor R is equal to 1 0 whereas it is less than 1 0 for non-ideal enlargement1.0, whereas it is less than 1.0 for non-ideal enlargement.

1329آب، 04

Example1: One m3/s of air is conveyed through a straight, h l d f f d l h f 40 If horizontal duct of uniform cross-section and a length of 40 m. If the velocity of air through the duct is 5 m/s, find the required fan power input when a) A circular duct is used and b) A rectangular power input when a) A circular duct is used, and b) A rectangular duct of aspect ratio 1:4 is used. Take the efficiency of the fan to be 0.7. If a GI sheet of 0.5 mm thick with a density of 8000 kg/m3 is y gused to construct the duct, how many kilograms of sheet metal is required for circular and rectangular cross section? Assume

d d di i d h i h i l d i f standard conditions and the static pressure at the inlet and exit of the duct to be same.

1330آب، 04

Example2 : Air at a flow rate of 1.2 kg/s flows through a fitting with sudden enlargement The area before and after thewith sudden enlargement. The area before and after the enlargement are 0.1 m2 and 1 m2 , respectively. Find the pressure drop due to sudden enlargement using Borda-Carnot p p g gEquation. What is the pressure drop if the same amount of air flows through a sudden contraction with area changing from 0 1 2 d 1 20.1 m2 and 1 m2.

1331آب، 04

Example3: Air at a flow rate of 1 m3/s flows through a fitting whose cross sectional area increases gradually fromfitting whose cross-sectional area increases gradually from 0.08 m2 to 0.12 m2. If the static regain factor (R) of the fitting is 0.8, what is the rise in static pressure (staticfitting is 0.8, what is the rise in static pressure (static regain) and total pressure loss as air flows through the fitting?

1332آب، 04

General rules for duct design: 1. Air should be conveyed as directly as possible to save space, power and material .2 Sudden changes in directions should be avoided When not2. Sudden changes in directions should be avoided. When not possible to avoid sudden changes, turning vanes should be used to reduce pressure loss. p3. Diverging sections should be gradual. Angle of divergence ≤ 20o. .4. Aspect ratio should be as close to 1.0 as possible. Normally, it should not exceed 4. 5 Air velocities should be within permissible limits to reduce5. Air velocities should be within permissible limits to reduce noise and vibration. 6. Duct material should be as smooth as possible to reduce pfrictional losses .

1333آب، 04

Classification of duct systems:

1-Low pressure systems: Velocity ≤ 10 m/s, static pressure ≤ 5 cm H2O (g)

2-Medium pressure systems: Velocity ≤ 10 m/s,static pressure ≤ 15 cm H2 O (g)

3 High pressure systems: Velocity > 10 m/s3-High pressure systems: Velocity > 10 m/s, static pressure 15<p s≤ 25 cm H2 O (g)

1334آب، 04

High velocities in the ducts results in: 1. Smaller ducts and hence, lower initial cost and lower space1. Smaller ducts and hence, lower initial cost and lower space requirement. 2. Higher pressure drop and hence larger fan power consumption .3. Increased noise and hence a need for noise attenuation. Recommended air velocities depend mainly on the applicationRecommended air velocities depend mainly on the application and the noise criteria. Typical recommended velocities are: Residences: 3 m/s to 5 m/s Theatres: 4 to 6.5 m/s Restaurants: 7.5 m/s to 10 m/s If nothing is specified, then a velocity of 5 to 8 m/s is used for main ducts and a velocity of 4 to 6 m/s is used for the branches. The allowable air velocities can be as high as 30 m/s in shipsThe allowable air velocities can be as high as 30 m/s in ships and aircrafts to reduce the space requirement.

1335آب، 04

duct design methods :duct design methods : the following methods are most commonly used for i l l t h th h i Fisimpler lay-outs such as the one shown in Fig..

1. Velocity method .2 Equal Friction Method2. Equal Friction Method .3. Static Regain method .

1336آب، 04

T i l i di i i d lTypical air conditioning duct lay-out.

1337آب، 04

Velocity method:

The various steps involved in this method are: i S l t it bl l iti i th i d b h d ti. Select suitable velocities in the main and branch ducts. ii. Find the diameters of main and branch ducts from airflow rates and velocities for circular ductsrates and velocities for circular ducts. iii. find the frictional pressure drop for main and branch ducts using friction chart or equationiv. find the dynamic pressure losses for all the bends and

fittings. S l t f th t id ffi i t FTP f th i dv. Select a fan that can provide sufficient FTP for the index run.

vi. Balancing dampers have to be installed in each run.

1338آب، 04

. Equal friction method: In this method the frictional pressure drop per unit length in theIn this method the frictional pressure drop per unit length in the main and branch ducts (∆pf/L) are kept same, i.e.,

the stepwise procedure for designing the duct system is as follows: i. Select a suitable frictional pressure drop per unit length (∆ /L) th t th bi d i iti l d i t(∆pf/L) so that the combined initial and running costs are minimized. ii Then the equivalent diameter of the main duct (A) isii. Then the equivalent diameter of the main duct (A) is obtained from the selected value of (∆pf/L) and the airflow rate.

1339آب، 04

As shown in Fig., airflow rate in the main duct (QA)is equalAs shown in Fig., airflow rate in the main duct (QA)is equal to the sum total of airflow rates to all the conditioned zones, i.e.,

1340آب، 04

the equivalent diameters of the other duct runs, B to I are obtained from the equation:

1341آب، 04

the total frictional pressure drop of that run is obtained by multiplying the frictional pressure drop per unit length and the length, i.e.,

vi. Next the dynamic pressure losses in each duct run are obtained based on the type of bends or fittings used in that run. vii. Next the total pressure drop in each duct run is obtained by summing up the frictional and dynamic losses of that runby summing up the frictional and dynamic losses of that run, i.e.,

1342آب، 04

viii. Next the fan is selected to suit the index run with the highest pressure loss. Dampers are installed in all the duct runs to balance the total pressure loss.

1343آب، 04

Static Regain Method: In this method the static pressure is maintained same. In this method the static pressure is maintained same

before each terminal or branch. The procedure followed is as given below: gi. Velocity in the main duct leaving the fan is selected first. ii. Velocities in each successive runs are reduced such that th i i t ti d t d ti i l itthe gain in static pressure due to reduction in velocity pressure equals the frictional pressure drop in the next duct section Thus the static pressure before each terminalduct section. Thus the static pressure before each terminal or branch is maintained constant.

1344آب، 04

Then using the static regain factor, one can write:

Where∆p and ∆p are the frictional and dynamic losses between∆p f,2and ∆pd,2 are the frictional and dynamic losses between 1 and 2, p 1 and p 2 are the velocity pressures at 1 and 2pv,1 and pv,2 are the velocity pressures at 1 and 2 respectively.

1345آب، 04

Principle of static regain method

1346آب، 04

iii. If section 1 is the outlet of the fan, then its dimensions are known from the flow rate and velocity (initially selected), iv. The procedure is followed in the direction of airflow, and th di i f th d t d t bt i dthe dimensions of the downstream ducts are obtained. v. As before, the total pressure drop is obtained from the pressure drop in the longest run and a fan is accordinglypressure drop in the longest run and a fan is accordingly selected.

1347آب، 04

Variation of total pressure drop with flow rate for a given duct system.

1348آب، 04

Fan and duct performance curves and balance points.

1349آب، 04