Duct Design

(1)

1

DESIGN OF AIR DUCT DESIGN CONDITIONS:

Inside room conditions

Temperature --- 22 oCDB

--- 16 oCWB Outside room conditions

Temperature --- 34 oCDB

--- 27 oCWB

Temperature of supply air ---15 OC Mass of outside air,mo ---10 % of mS Mass of recirculated air, mR ---90%of mS

Air Properties

From ASHRAE PSYCHROMETRIC CHART, M.E.T.C, page 75 Outside room conditions, @32 OC DB&27oCWB

(2)

2 kg kJ h kg m kg kg W O O da v O 0.0199 ; 0.898 ; 85.6 3    

Inside room conditions, @ 22 OC DB&16oCWB

kg kJ h kg m kg kg W O O da v O 0.0089 ; 0.848 ; 45 3    

Mass of supply air, mS

(

)

(

r s

)

pa s S s r pa S s t t c Q m t t c m Q -= ; -= Where: S

m = mass of supply air

pa

c = specific heat of air, 1.0062kJ kg °K

r

t = inside room temperature, 22C

s

t = supply air temperature, 15°C

Therefore,





sec 66 . 32 15 -22 0062 . 1 037 . 230 kg K K kg kJ kW m O O S                

Mass of outside air, mO

R o S m m m m m = = ₄ ; + 1 Where: S

m = mass of supply air, 32.66kg s

R

m = mass of recirculated air, 90 % of mS Therefore,



1 0.9



32.66 kg sec



1 0.9



3.266 kg sec

m

(3)

3

Volume flow rate of outside air, Vo

O O O m V =  Where: O

m = mass of outside air, 3.266kg s

O

 = specific volume of outside air, 0.898m3 kg

Therefore, sec 93 . 2 898 . 0 sec 266 . 3 3 3 m kg m kg V_O _             

Mass of recirculated air, mR

O S R R o S m m m m m m   ;  Where: S

O

m = mass of outside air, 3.266kg sec Therefore, sec 39 . 29 sec 266 . 3 sec 66 . 32 kg kg kg m_R   

Volume flow rate of recirculated air, VR

i R R m υ V = Where: R

m = mass of recirculated air, 29.39 kg s

i

υ = specific volume of inside air, 0.848m3 kg

Therefore, sec 93 . 24 848 . 0 sec 39 . 29 3 3 m kg m kg V_R _             

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4

Temperature of air entering the AHU, t2

From the figure, considering point 2

By temperature balance, S R O o S R O o m t m t m t t m t m t m 4 2 2 4 ;     Where: O

m = mass of outside air, 3.266kg sec O

t = temperature of outside air, 34 OC

R

m = mass of recirculated air, 29.93kg s

4

t = inside air temperature, 22 oC

S

Therefore,

















C kg C kg C kg t O O O 56 . 23 sec 66 . 32 22 sec 93 . 29 34 sec 266 . 3 2   

Capacity of AHU, QAHU

(

4 - 1

)

=m h h

QCDA S

Where: S

4

h = enthalpy of air entering the AHU, kJ kg

1

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5

Enthalpy of air entering the AHU, h₂

From the figure, considering point 2

By heat balance, S R O o S R O o m h m h m h h m h m h m 4 2 2 4 ;     Where: O

m = mass of outside air, 3.266kg sec O

h = enthalpy of outside air, 85.6 kJ kg

R

m = mass of recirculated air, 29.93kg s

4

h = enthalpy of recirculated air, 45kJ kg

S

Therefore,





 





kg kJ kg kg kJ kg kg kJ kg h 49.8 sec 66 . 32 45 sec 93 . 29 6 . 85 sec 266 . 3 4   

Enthalpy of air leaving the AHU, h3





S T S T m Q h h h h m Q  ₄ - ₃ ; ₃  ₄ -Where: T

Q = total heat load, 476.228kWT

S

4

(6)

6 Therefore, kg kJ kg kW kg kJ h T 42 . 30 sec 66 . 32 228 . 476 -45 3   Thus,





kW or TOR kg kJ kg Q_AHU 49.8-30.42 632.9 179.98 sec 66 . 32        

Volume of supply air, VS

1 =m υ V_S _S Where:

S

1

υ = specific volume of supply air, m3 kg

From ASHRAE PSYCHROMETRIC CHART, M.E.T.C, page 75

kg kJ h DB C O 42 . 30 & 15 @ ₁  kg m3 1 0.824  Therefore,









sec 9 . 26 824 . 0 sec 66 . 32 3 3 m kg m kg V_S  

AIR DUCT SIZING DESIGN CONDITIONS:

Volume flow rate of supply air --- 26.9 m3 sec Duct dimensions --- W = 4H No. of units --- 2

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7

No. of branch take-off --- 5 No. of outlet per branch take off--- 4 Method of calculation--- Fitting Data

Elbow, inner & outer radii ratio --- 0.2 Angle, turns and branches--- O

90

Volume flow rate of supply air per branch take-off, QB

sec 38 . 5 5 sec 9 . 26 3 3 m m off Take Branch o Number V Q_B  S  

Volume flow rate of supply air per outlet, QO

sec 345 . 1 4 sec 38 . 5 - m3 m3 Outlets of Number off Take Branch per Air Supply of Rate Flow Volume Q_O   

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8

Duct Sizes

For the main duct, from the fan outlet to point O,

From figure 6-2, Pressure drop in straight, circular, sheet metal ducts, REFRIGERATION AND AIRCONDITIONING by Stoecker & Jones, page 107.

For a volume flow rate of supply air of, Q_FO 26.9m3 s and a

m Pa P1.5  , m Deq.f 1.436 .

From equation 6-8, REFRIGERATION AND AIRCONDITIONING by Stoecker & Jones, page 108

 





0.25 625 . 0 . 1.30 b a ab Deq f   Where:

a = H = height of the rectangular duct, m b = W = width of the rectangular duct, m

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9









 

H H H H H W H HW D_eq _f 2.0677 5 4 30 . 1 5 4 30 . 1 30 . 1 25 . 0 25 . 0 25 . 1 625 . 0 25 . 0 625 . 0 2 25 . 0 625 . 0 .      Thus, equation working D H 0.4836 eq.f 

For the height, HFO



m



m

HFO 0.4836 1.436  0.694

For the width, W_FO



m



m

WFO 4 0.694 2.78

For the size of duct, AFO

2 2 1930 93 . 1 78 . 2 694 . 0 mx m m mm W x H AFO  FO FO   

Run O to A, two branch take – off supplied

For a volume flow rate of supply air of, QOA m s 3 9 . 26  and a m Pa P1.5  , m Deq.f 1.436 .

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11 Therefore,



m



m H_OA 0.4836 1.436  0.694



m



m WOA4 0.694 2.78 2 2 1930 93 . 1 78 . 2 694 . 0 mx m m mm W x H AOA  OA OA   

Run A to B, two branch take – off supplied

For a volume flow rate of supply air of,



m s



 

m s Q_OB  5.38 3 4 21.52 3 m Pa P1.5  , m D_eq_._f 1.306 Therefore,



m



m HOB 0.4836 1.306  0.6316



m



m W_OB 4 0.324  2.52 2 59 . 1 52 . 2 6316 . 0 mx m m W x H A_OB  _OB _OB  

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11

Run B to C, 3 Branch take – off supplied,

For a volume flow rate of supply air of, s m s m Q_BA (5.38 3 )(3) 16.14 3 m Pa P1.5  , m Deq.f 1.186 Therefore,



m



m HBA 0.4836 1.186  0.5735



m



m WBC 4 0.5735  2.29 2 316 . 1 29 . 2 5735 . 0 mx m m W x H ABA  BA BA  

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12

Run C to D, 4 branch take – off supplied

From figure 6-2, Pressure drop in straight, circular, sheet metal ducts, REFRIGERATION AND AIRCONDITIONING by Stoecker & Jones, page 107.



m s



 

m s Q_OC  5.38 3 2 10.76 3 & P1.5Pa m, m Deq.f 0.966 Therefore,



m



m HOC 0.4836 0.966  0.467



m



m W_OC 4 0.467 1.868 2 873 . 0 868 . 1 467 . 0 mx m m W x H A_OC  _OC _OC  

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13

From figure 6-2, Pressure drop in straight, circular, sheet metal ducts, REFRIGERATION AND AIRCONDITIONING by Stoecker & Jones, page 107. s m Q_CD  5.38 3 & P1.5Pa m, m D_eq_._f 0.756 Therefore,



m



m H_CD 0.4836 0.756 0.3656



m



m W_CD 4 0.3656 1.46 2 5338 . 0 46 . 1 3656 . 0 mx m m W x H ACD  CD CD  

Branch outlet dimensions Note:

Since the number of outlets in each branch take – off are equal for the whole system, only one branch take – off will be considered in

computing for the duct dimensions and the values obtained will be replicated in all seven (7) branch outlets.

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14

Run A – 1, 4 outlets supplied

For a volume flow rate of supply air of, Q_A₁ 5.38m3 s

m Pa P1.5  , m D_eq_._f 0.756 Therefore,



m



m H_A₁ 0.4836 0.756  0.3656



m



m WA1 4 0.3656 1.4624 2 1 1 1 H xW 0.3656m x 1.4624 m 0.5347m AA  A A  

Run 1 – 2, 3 outlets supplied



m s



 

m s Q₁₂  1.345 3 3  4.035 3 m Pa P1.5  , m Deq.f 0.7 Therefore,



m



m H12 0.4836 0.7  0.33852



m



m W12 4 0.33852 1.354 2 12 12 12 H xW 0.33852m x 1.354m 0.4584m A   

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15



m s



 

m s Q₂₃  1.345 3 2 2.69 3 m Pa P1.5  , m D_eq_._f 0.58 Therefore,



m



m H₂₃ 0.4836 0.58 0.28



m



m W₂₃ 4 0.28 1.122 2 23 23 23 H xW 0.28 m x 1.122m 0.314 m A   

For a volume flow rate of supply air of,Q₃₄  1.345m3 s

m Pa P1.5  , m Deq.f 0.43 Therefore,



m



m H₃₄ 0.4836 0.43  0.208



m



m W34 4 0.208 0.832 2 34 34 34 H xW 0.208 m x 0.832 m 0.173m A   

(16)

16 Air Velocity, υ A Q υ υ A Q = ; = Where: υ = velocity of air, m s

Q= volume of supply air, m3 s

A = area of the air duct, m2

Branch take - off

From fan outlet to branch take – off O, FO

sec 937 . 13 93 . 1 sec 9 . 26 2 3 0 0 m m m A Q F F FA     Where: FO Q = 26.9 m3 s FO A = 1.93 2 m Run A-B, _AB sec 53 . 13 59 . 1 sec 52 . 21 2 3 m m m A Q AB AB AB     Where: AB Q = 21.52 m3 s

(17)

17 AB A = 1.59 m2 Run B -C, BC sec 26 . 12 316 . 1 sec 14 . 16 2 3 m m m A Q BC BC BC     Where: BC Q = 16.14 m3 s BC A = 1.316 m2 Run C - D, CD sec 325 . 12 873 . 0 sec 76 . 10 2 3 m m m A Q CD CD CD     Where: CD Q = 10.76 m3 s CD A = 0.873 2 m Run D - E, υ_DE sec 078 . 10 5338 . 0 sec 38 . 5 2 3 m m m A Q DE DE DE     Where: DE Q = 5.38 m3 s DE A = 0.5338 2 m

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18

Branch Outlets Note:

Since the number of outlets in each branch take – off are equal for the whole system, only one branch take – off will be considered in

computing for the duct velocities and the values obtained will be replicated in all seven (7) branch outlets.

Run A - 1, υ_A₁ sec 06 . 10 5347 . 0 sec 38 . 5 2 3 1 1 1 m m m A Q A A A     Where: 1 A Q = 5.38 m3 s 1 A A = 0.5347 2 m Run 1 - 2, υ₁₂ sec 802 . 8 4584 . 0 sec 035 . 4 2 3 12 12 12 m m m A Q     Where: 12 Q = 4.035 m3 s 12 A = 0.4584 2 m

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19 Run 2 - 3, υ23 sec 57 . 8 314 . 0 sec 69 . 2 2 3 23 23 23 m m m A Q     Where: 23 Q = 2.69 m3 s 23 A = 0.314 m2 Run 3 - 4, υ₃₄ sec 77 . 7 173 . 0 sec 345 . 1 2 3 34 34 34 m m m A Q     Where: 34 Q = 1.345 m3 s 34 A = 0.173 2 m

Determination of Pressure Drops Considering the Run 1 2 3 4

A A A A FOA

(20)

21



P m



m



Pa m



Pa L P_FO  _FO  4.6387 1.5 6.958  Run O, Turn

From figure 6 – 8, Pressure losses in rectangular elbows,

REFRIGERATION and AIRCONDITONING by: Stoecker & Jones, page 113

For a width to height ratio (W/H) equal to 4, inner and outer radius ratio of 0.20

( ) (

)

= 0.25 Δ 2 2 air OB TOB ρ v P Where: air

ρ = density of supply air, kg / m3 OB

v = velocity of air in run O, 13.937 m / sec

From table 6 – 2, Viscosity and Density of dry air at standard atmospheric pressure, REFRIGERATION and AIRCONDITONING by: Stoecker & Jones, page 106

Temperature O C Viscosity , Pa - s Density , kg / m3 10 20 17.708 18.178 1.2467 1.2041

Since the temperature of the supply air is at 15 0C, by interpolation, @ 15 C = 1.2254 kg / m 3 Therefore,



  

v



 

m





kg m



Pa P OB air OT 29.75 2 2254 . 1 sec 937 . 13 25 . 0 2 25 . 0 3 2 2     

Run OA, Straight Duct



P m



m



Pa m



Pa

L

P_OA  _OA  10.4360 1.5 15.654



(21)

21

   

2 2 0.25  air AT AT v P  Where: air

ρ = density of supply air, 1.2254 kg / m3 AT

v = velocity of air in run OA, 13.937 m / sec Therefore,



  

v



 

m





kg m



Pa P OB air AT 29.75 2 2254 . 1 sec 937 . 13 25 . 0 2 25 . 0 3 2 2     

Run AA1, Straight Duct



P m



m



Pa m



Pa L P AA AA1  1  2.1703 1.5 3.255 



P m



m



Pa m



Pa L P AA AA2  2  5.7749 1.5 8.6623 



P m



m



Pa m



Pa L

P_AA3  _AA3  9.613 1.5 14.4195





P m



m



Pa m



Pa L

P_AA4  _AA4  12.4634 1.5 18.6951

(22)

22 Therefore,





Pa P Pa P P P P P P P P P FOBA AA AA AA AA AT OA OT FO FOA 1439 . 127 6951 . 18 4195 . 14 6623 . 8 225 . 3 75 . 29 654 . 15 75 . 29 958 . 6 7 4 1 2 3 4                           

Considering the Run 1 2 3 4

B B B FOBB

Run FO, Straight Duct



P m



m



Pa m



Pa

L

P_FO  _FO  4.6387 1.5 6.958



Run O, Turn

   

2 2  0.25  air TO TOB v P  Where: air

(23)

23

TO

v = velocity of air in run OB, 13.937 m / sec

Therefore,



  

v



 

m





kg m



Pa P TO air TO 29.75 2 2254 . 1 sec 937 . 13 25 . 0 2 25 . 0 3 2 2     

Branch take – off A, Duct Straight Through

From Equation 6 – 16, REFRIGERATION and AIRONDITIONING by Stoecker & Jones, page 114

 

_{ }



 

_{ }

Pa P m P V V V P A DST A DST AB OA air OA A DST 038 . 0 937 . 13 53 . 13 1 4 . 0 2 2254 . 1 sec 53 . 13 1 4 . 0 2 2 2 2 2                     

Run OB, Turn



2

  

2 0.25  air TOB TOB v P  Where: air

ρ = density of supply air, 1.2254 kg / m3 TOB

v = velocity of air in run OB, 13.53 m / sec Therefore,



 

v





 

m





kg m



Pa P TOB air TOB 28.04 2 2254 . 1 sec 53 . 13 25 . 0 2 25 . 0 3 2 2     

(24)

24

Run OB, Straight Duct



P m



m



Pa m



Pa

L

POB  OB  18.136 1.5  27.204



Run BB1, Straight Duct



P m



m



Pa m



Pa L

P_BB1  _BB1  2.3628 1.5 3.5442





P m



m



Pa m



Pa L P BB BB2  2  5.9674 1.5 8.9511 



P m



m



Pa m



Pa L

P_BB3  _BB3  9.8055 1.5 14.708





P m



m



Pa m



Pa L

P_BB4  _BB4  12.6559 1.5 18.9838



(25)

25





Pa P Pa P P P P P P P P P P FOBA BB BB BB BB OB TOB DSTA TO FO B B B FOBB 1771 . 138 9838 . 18 708 . 14 9511 . 8 5442 . 3 204 . 27 04 . 28 038 . 0 75 . 29 958 . 6 7 4 3 2 1 4 3 2 1                              

C C C FOCC



P m



m



Pa m



Pa

L

P_FO  _FO  4.6387 1.5 6.958



Run O, Turn

   

ρ = density of supply air, 1.2254 kg / m3 TO

(26)

26 Therefore,



  

v



 

m





kg m



Pa P TO air TO 29.75 2 2254 . 1 sec 937 . 13 25 . 0 2 25 . 0 3 2 2     

 

_{ }



 

_{ }

Branch take – off B, Duct Straight Through

 

_{ }



 

_{ }

Pa P m P V V V P B DST B DST BC OB air BC B DST 03245 . 0 53 . 13 26 . 12 1 4 . 0 2 2254 . 1 sec 26 . 12 1 4 . 0 2 2 2 2 2                     

Run OC, Turn

(27)

27



2

  

2  0.25  air TOC TOC v P  Where: air

ρ = density of supply air, 1.2254 kg / m3 TOC



 

v





 

m





kg m



Pa P TOC air TOC 23.0234 2 2254 . 1 sec 26 . 12 25 . 0 2 25 . 0 3 2 2     

Run OC, Straight Duct



P m



m



Pa m



Pa

L

P_OC  _OC  25.836 1.5 38.754



Run CC1, Straight Duct



P m



m



Pa m



Pa L P CC CC1  1  2.4153 1.5  3.623 



P m



m



Pa m



Pa L

P_CC2  _CC2  6.0199 1.5 9.02985





P m



m



Pa m



Pa L

P_CC3  _CC3  9.858 1.5 14.787





P m



m



Pa m



Pa L

P_CC4  _CC4  12.7084 1.5 19.0626

(28)

28 Therefore, Pa P Pa P P P P P P P P P P P C C C FOCC CC CC CC CC OC TOC DSTB DSTA TO FO C C C FOCC 0583 . 145 0626 . 19 787 . 14 02985 . 9 623 . 3 754 . 38 0234 . 23 03245 . 0 038 . 0 75 . 29 958 . 6 4 3 2 1 4 3 2 1 4 3 2 1                                       

D D D FODD



P m



m



Pa m



Pa

L

P_FO  _FO  4.6387 1.5 6.958



Run O, Turn

   

2 2  0.25  air TO TOB v P  Where:

(29)

29

air



  

v



 

m





kg m



Pa P TO air TO 29.75 2 2254 . 1 sec 937 . 13 25 . 0 2 25 . 0 3 2 2     

 

_{ }



 

_{ }

 

_{ }



 

_{ }

Pa P m P V V V P B DST B DST BC OB air BC B DST 03245 . 0 53 . 13 26 . 12 1 4 . 0 2 2254 . 1 sec 26 . 12 1 4 . 0 2 2 2 2 2                     

Branch take – off C, Duct Straight Through

(30)

31

 

_{ }



 

_{ }

Pa P m P V V V P C DST C DST BC OC air OC C DST 001046 . 0 26 . 12 325 . 12 1 4 . 0 2 2254 . 1 sec 325 . 12 1 4 . 0 2 2 2 2 2                     

Run OD, Turn



2

  

2  0.25  air TOD TOD v P  Where: air

ρ = density of supply air, 1.2254 kg / m3 TOD

v = velocity of air in run OD, 12.325 m / sec Therefore,



 

v





 

m





kg m



Pa P TOD air TOD 23.268 2 2254 . 1 sec 325 . 12 25 . 0 2 25 . 0 3 2 2     

Run OD, Straight Duct



P m



m



Pa m



Pa

L

P_OD  _OD  33.536 1.5 50.304



Run DD1, Straight Duct



P m



m



Pa m



Pa L

P_DD1  _DD1  1.9931 1.5  2.9896

(31)

31



P m



m



Pa m



Pa L P DD DD2  2  5.904 1.5 8.856 



P m



m



Pa m



Pa L

P_DD3  _DD3  8.6386 1.5 12.9579





P m



m



Pa m



Pa L P_DD4  _DD4  11.1831 1.5 16.7746  Therefore, Pa P Pa P P P P P P P P P P P P D D D FODD DD DD DD DD OD TOD DSTD DSTB DSTA TO FO D D D FODD 932 . 151 7746 . 16 9579 . 12 856 . 8 9896 . 2 304 . 50 268 . 23 001046 . 0 03245 . 0 038 . 0 75 . 29 958 . 6 4 3 2 1 4 3 2 1 4 3 2 1                                         

E E E FOEE

(32)

32



P m



m



Pa m



Pa

L

PFO  FO  4.6387 1.5 6.958



Run O, Turn

   

v = velocity of air in run OB, 13.937 m / sec

Therefore,



  

v



 

m





kg m



Pa P TO air TO 29.75 2 2254 . 1 sec 937 . 13 25 . 0 2 25 . 0 3 2 2     

(33)

33

 

_{ }



 

_{ }

 

_{ }



 

_{ }

Pa P m P V V V P B DST B DST BC OB air OB B DST 03245 . 0 53 . 13 26 . 12 1 4 . 0 2 2254 . 1 sec 26 . 12 1 4 . 0 2 2 2 2 2                     

Branch take – off C, Duct Straight Through

 

_{ }



 

_{ }

Pa P m P V V V P C DST C DST CD OC air OC C DST 001046 . 0 26 . 12 325 . 12 1 4 . 0 2 2254 . 1 sec 325 . 12 1 4 . 0 2 2 2 2 2                     

(34)

34

 

_{ }



 

_{ }

Pa P m P V V V P D DST D DST DE OD air OD D DST 827 . 0 325 . 12 078 . 10 1 4 . 0 2 2254 . 1 sec 078 . 10 1 4 . 0 2 2 2 2 2                     

Run OE, Turn



2

  

2  0.25  air TOE TOE v P  Where: air

ρ = density of supply air, 1.2254 kg / m3 TOE

v = velocity of air in run OD, 10.078 m / sec Therefore,



 

v





 

m





kg m



Pa P TOE air TOE 15.557 2 2254 . 1 sec 078 . 10 25 . 0 2 25 . 0 3 2 2     

Run OE, Straight Duct



P m



m



Pa m



Pa

L

P_OE  _OE  41.236 1.5 61.854

(35)

35

Run EE1, Straight Duct



P m



m



Pa m



Pa L P EE EE1  1  2.1971 1.5 3.295 



P m



m



Pa m



Pa L

P_EE2  _EE2  5.5437 1.5 8.316





P m



m



Pa m



Pa L

P_EE3  _EE3  8.4626 1.5 12.6939





P m



m



Pa m



Pa L P_EE4  _EE4  11.20005 1.5 16.8  Therefore,

Pa

P

Pa

P

E E E FOEE EE EE EE EE OE TOE DSTD DSTC DSTB DSTA TO FO E E E FOEE

122 .

156

8 .

16 6939

.

12

316 .

8

295 .

3

854 .

61

557 .

15

827 .

0 001046

.

0 03245

.

0

038 .

0

75 .

29

958 .

6

4 3 2 1 4 3 2 1 4 3 2 1

































































