Duct Pressure Loss Calculation
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Transcript of Duct Pressure Loss Calculation
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7/29/2019 Duct Pressure Loss Calculation
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To find pressure loss
1.duct size is known.2.flow rate is known.
S.1 Compute the loss pressure for each branch of simple
Duct system shown in figure using loss coefficient
method.
Plenum
1
400cfm
50ft, 10in
200cfm
50 ft, 9in
120cfm
40 ft, 6 in
a2
3
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Solution:
For section 1-a
(i)
for plenum exit loss
C = 0.85 ( from T-4 )
ft/min385.733)12/10)(4/(
400
A
QV
2
u
uu ===
22
ua-1
4000733.3850.85
4000VCH
=
=
For straight line 1-a ( 50 ft )
100
0.09
L
P
in10
400cfm=
( ) waterofin0.04550100
0.09P a1 ==
Plenum
1
400cfm
50 ft, 10 in
a
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waterofin0.07350.02850.045a-1Runforlosspressuretotal =+=
section a- 2
( )( )ft/min633.79
129
4
280AQV
2
d
dd ===
waterofin0.002Pft/min633.79V
ft/min733.385Vfit
d
u =
=
=
For straight run
waterofin100
08.0
L
P
in9
cfm280=
waterofin0.0450100
0.08Pst ==
waterofin0.0420.0020.042-aRunforlosspressuretotal =+=
280 cfm
50 ft, 9in
a 2
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Section a-3
( ) ( )ft/min611.154
126
4
120
A
QV
22
b
bb ===
0.48C
offtake45loss)(branch
0.833733.385
611.154
V
V
bu
b
=
==
o
( ) waterofin0161.04000
733.3850.48
4000
VCH
22
ubb =
=
=i
a
3
VU Vd
Vb
120cfm
40ft, 6 in
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( )
losselbow45 oii
)2-Tfrom1DR
assume(0.37C==
waterofin0.00864000
611.1540.37
4000
V0.37H
22
be =
=
=
( ) losselbow90 oiii
waterofin0.0086He =
( ) )ft(40lossRunStraightiv
waterofin100
0.12
L
P
in6
cfm120=
( ) waterofin0.04840100
0.12P st ==
waterofin0.0813
0.0480.00860.00860.01613-aRunforlosspressuretotal
=
+++=
total pressure loss for section 1-2 is
pa28.75950waterofin0.11550.0420.0735P 2-1 ==+=
Total pressure loss for section 1-3 ispa38.54waterofin0.15480.08130.0735P 31 ==+= =
+==========================+
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Equivalent Length Method
S.2 Compute the loss pressure for each branch of simpleduct system shown in figure usingEquivalent Length
approach.
Section 1-a
ft85(35)50L a1 =+=
ftwater/100ofin0.09L
P
in10:dia
cfm400=
=
waterofin0.076585
100
0.09L
L
PP a-1 ==
=
Plenum
1
400cfm
50ft, 10in
200cfm
50 ft, 9in
120cfm
40 ft, 6 in
a2
3
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Section a-2
ft55)5(50L 2-a =+=
waterofin1000.08
LP
in9:diacfm200 =
=
waterofin0.04455100
0.08P 2-a ==
Section a-3
ft90)10()5()35(40L 3-a=+++=
waterofin0.12L
P
in6:dia
cfm120=
=
waterofin0.10890100
0.12P 3a == =
The pressure loss for section 1 to 2 is
Pa30.00
waterofin0.12050.0440.0765PPP 2-aa-12-1
=
=+=+=
The pressure loss for section 1 to 3 is
pa45.9
waterofin0.18450.1080.0765PPP 3aa-13-1
=
=+=+= =
+======================+
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TO FIND DUCT SIZE
Equal Friction MethodS-3 Select duct sizes for the simple duct system of shown
in figure using Equal Friction Method. The system is
for a residence. The total pressure for the duct system
is 0.12 in of water and the loss in total pressure for
each diffuser at the specified flow rate is 0.02 in ofwater.
Plenum
o
a b
150cfm
150cfm
200cfm
1
5
2
3
4
20ft
15ft
17ft
30ft
10ft
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ft16730)105355(35101720Le123 =++++++++=
ft15230)555535(301720Le124 =++++++++=
ft16030)105035(1520Le15 =+++++=
Le123 is design line.
waterofin0.10.02-0.12P ==
water/ftofin100.5980.000598167
0.1
L
P
L
P 3-
design123
===
=
cfm500150200150cfm1 =++=
cfm350200150cfm2 =+=
line.ofallforequalisft100water /ofin0.0598L
P
design
=
ft/min660Vin,9.11Dcfm500cfm 111 ===
ft /min600Vin,2.10Dcfm350cfm 222 ===ft/min490Vin,5.7Dcfm150cfm 333 ===
ft/min525Vin,4.8D200cfmcfm 444 ===
ft/min490Vin,5.7Dcfm150cfm 555 ===
+===========================+
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Balance Capacity Method
S-4 Select duct sizes for the simple duct system of shownin figure using Balance Capacity Method. The system
is for a residence. The total pressure for the duct
system is 0.12 in of water and the loss in total pressure
for each diffuser at the specified flow rate is 0.02 in of
water.
Plenum
o
a b
150cfm
150cfm
200cfm
1
5
2
3
4
20ft
15ft
17ft
30ft
10ft
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ft16730)105355(35101720Le123 =++++++++=
ft15230)555535(301720Le124 =++++++++=
ft16030)105035(1520Le15 =+++++=
Le123 is design line.
waterofin0.10.02-0.12P ==
water/ftofin100.5980.000598167
0.1
L
P
L
P 3-
design123
===
=
cfm500150200150cfm1 =++=
cfm350200150cfm2 =+=
2,3.1,lineforlinedesignisft100water /ofin0.0598L
P
design
=
ft/min660Vin,9.11Dcfm500cfm 111 ===
ft /min600Vin,2.10Dcfm350cfm 222 ===
ft/min490Vin,5.7Dcfm150cfm 333 ===According to the theory, the pressure of all line are equal.
PP 43 =
( ) ft90301053510Le3 =++++=
waterofin0.0503890L
PP
design3
=
=
( ) ft753055530Le4 =++++=
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ftwater/100ofin0.07176waterofin100.717675
0.0538
L
P 3-
4
===
ft/min560Vin,1.8D
200cfm
ftwater/100ofin07176.0L
P
44
4
4 ==
=
=
Section a-5
o4o5 PP = .. equ:(1)
b4aboaa5oa PPPPP ++=+
b4aba5 PPP += equ:(2)
ft75LL 4b4 ==
ft22517LL 2ab =+==
waterofin01315.022100.598LL
PPP 3-222ab ==
==
waterofin0538.075107176.0LL
PPP 34
4
4b4 ==
==
waterofin0.06690.013150.0538PP a55 =+==
( ) ft10530105015La5 =+++=
water/ftofin10637.0ft105
waterofin0669.0
L
P
L
P 3
a5
a5
5
==
=
ft/min489Vin,7.5D150cfmft,water/100ofin0637.0L
P55 ==
==
+========================+
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S-5 Design the duct system of shown in figure by using
the equal friction method. The velocity in the duct attached
to the plenum must not exceed 900 ft/min and the overallloss in total pressure should not exceed about 0.32 in of
water. Total pressure losses for the diffusers are all equal to
0.04 in of water. The lengths shown are the total equivalent
length of each section. Use English unit.
For main duct velocity = 900 ft/min
cfm1 = 100+250+250+100+100 =800
Plenum
o a
b
cd
1 2 3 4
5
6
78
9
55 ft 25 ft 40ft 45ft
95ft
100
ft
100
ft
80
ft
100
ft
250 cfm 250 cfm
100cfm
100cfm 100 cfm
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from chart,
ftwater/100ofin0.096L
P
ft/min890Vin,9.12Dcfm
ityduct velocmain
design
11
1
=
==
lines.allforequalisL
P
design
From chart,
ft/min870Vin,12.2D700100100250250cfm 222 ===+++=
ft/min780Vin,4.10D450250100100cfm 333 ===++=
ft/min700Vin,5.9D350250100cfm 444 ===+=
ft/min530Vin,9.5D100cfm 555 ===
ft/min660Vin,2.8D250cfm 666 ===
ft/min530Vin,9.5D100cfm 777 ===
ft/min530Vin,9.5D100cfm 888 ===
ft/min660Vin,2.8D250cfm 999 ===
+===========================+
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S-6 Design the duct system of shown in figure by using
balance capacity method. The velocity in the duct attached
to the plenum must not exceed 900 ft/min and the overallloss in total pressure should not exceed about 0.32 in of
water. Total pressure losses for the diffusers are all equal to
0.04 in of water. The lengths shown are the total equivalent
length of each section. Use English unit.
For main duct velocity = 900 ft/min
cfm1 = 100+250+250+100+100 =800
Plenum
o a
b
cd
1 2 3 4
5
6
78
9
55 ft 25 ft 40ft 45ft
95ft
100
ft
100
ft
80
ft
100
ft
250 cfm 250 cfm
100cfm
100cfm 100 cfm
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from chart,
ftwater/100ofin0.096L
P
ft/min890Vin,9.12Dcfm
ityduct velocmain
design
11
1
=
==
ft2609545402555Le12345 =++++=
ft2458045402555Le12346 =++++=
ft220100402555Le1237 =+++=
ft1801002555Le129 =++=
ft15510055Le18 =+=
waterofin0.2496260100
096.0P 5-o ==
6-o5-o PP =
6-dd-o5-dd-o PPPP +=+
waterofin0912.095100
096.0L
L
PP 5
design
5-d ==
=
waterofin0912.0PP 5-d6-d ==
waterofin1014.1800912.0
LP
LP 3
6
6-d
6
===
ft/min690Vin,2.8D
250cfm
114.0ft100L
P
66
6
6 ==
=
=
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7-o6-o PP =
7-cc-o6-dd-4-cc-o PPPPP +=++
water.ofin0432.045100096.0L
LPP 4
design
d-4-c ==
=
waterofin0912.0PP 66-d ==
waterofin1344.00912.00432.0PPP 6-dd-4-c7-c =+=+=
3
7-c
7-c
7
10344.1
100
1344.0
L
P
L
P ==
=
ft/min600Vin,6.5D
100cfm
0.1344ft100L
P
77
7
7 ==
=
=
9-o7-o PP =
7-cc-bb-o9-bb-o PPPPP ++=+
7-cc-b9-b PPP +=
waterofin0384.040100
096.0L
L
PP 3
design
c-b ==
=
waterofin1728.01344.00384.0PPPP 7-cc-b9-b9 =+=+==
water /ftofin10152.1150
1728.0
L
P 3
9
==
ft/min700Vin,0.8D
250cfm
1152.0ft100L
P
99
9
9 ==
=
=
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9o8-o PP =
9bb-aa-o8-aa-o PPPPP =++=+
9-bb-a8-a PPP +=
waterofin024.025100
096.0L
L
PP 2
design
ba ==
=
waterofin1968.01728.0024.0P 8-a =+=
twater /fofin10968.1
100
1968.0
L
P 3
8-a
==
ft/min700Vin,2.5D
100cfm
ftwater/100ofin1968.0ft100L
P
88
8
8 ==
=
=
+=========================+