LeakyDuct Worked Example, © MVA 1999-2006
Page 1 of 21
How to use LeakyDuct™
In the following worked example, air density is 1.2 kg/m3 unless otherwise noted.
•
Find the fan parabola constants for the auxiliary fan provided in the figure at the end of this
document (SDS GAL14 2*90 kW). What is the correlation coefficient? What does it mean?
Preview the curve.
Answer
Find the P and Q values for the maximum and minimum flows for this curve. These can be
either FSP or FTP values, but you must be consistent.
Select two additional points between these extremes at points of maximum deflection on the
curve. These points would be suitable.
FTP, Pa Q, m3/s
5400
26.5
5000
31.5
4000
37.5
750
48.3
Entering this data into the FTP input boxes of the FanCurveCalculator sheet within
LeakyDuct™ gives fan constants of a = -7.95844, b = 382.02719, c = 864.44534 for a
correlation coefficient, r2, of exactly 1.0000. Note that correlation coefficients above 0.9995 are
acceptable. Note that you must also enter the correct fan outlet diameter and the correct fan
curve air density.
You should end up with the following:
Figure 1
Fan Curve Fitting Calculator. Copyright MVA 2000-2004. All rights reserved.
Note: Pressures in Pa, volumes in m3/s. Enter the data from the curve supplied by the manufacturer, including the air density and the fan outlet diameter. You should normally use a standard density of 1.2
kg/m3. However, if the manufacturer's plot you are using is at a different density, then still enter the points directly from the curve, along with the air density of the plot. The calculated A, B and C values will
be at STANDARD DENSITY.
Select four points from fan curve. These MUST include the lowest and highest acceptable Q on the curve ('PtQmin' and 'PtQmax'), and then pick two points in between these at points where the curve is
changing slope (i.e. for best results, do NOT pick equi-distant points along the curve). Enter data in BLUE only. PtQmin should be at the surge (peak P) or stall point; PtQmax should be at the choke point
(or runaway condition, if no choke point). "Simulator" will not allow the fan to go outside these values of Q so choose them carefully.
The fan curve is given by P = A x Q2 + B x Q + C. The regression coefficient (r2 value) for the 'goodness of fit' of the curve is shown in PURPLE. A perfect fit is 1.0000. You should aim for better than
0.999 (preferably better than 0.9995). Curve fitting accuracy depends largely on how accurately you read the P and Q values from the supplied curve. You should ensure curves are CERTIFIED, are TO
SCALE, and are LARGE ENOUGH for P and Q values to be read to within 1% of the maximum P and Q values. If you still cannot get a correlation better than 0.999, then you may need to use a 3rd order
polynomial for curve fitting. However, this is unlikely and for most mine applications, a 2nd order (quadratic) polynomial is satisfactory. You should then PLOT the curve (see below) and make sure it is
correct by comparing it directly with the manufacturer's curve.
To find the A, B and C constants for two (or more) fans in series, or two (or more) fans in parallel, or series-parallel, put the correct number of fans in the marked cells (for one fan only, leave '1' in series
cell and '1' in parallel celll, for 2 fans in series put '2' in series cell and leave '1' in parallel cell', for 2 pairs of fans in parallel (4 fans), put '2' in series cell and put '2' in parallel cell etc. HINT: If you multiply
the no of fans in series x no of fans in parallel, you should get the no of fans in the installation. Hence a 1 fan installation will have 1 in each box (1 x 1 = 1). Two fans in parallel, will have 2 in the parallel box
and 1 in the series box (2 x 1 = 2), etc. You can then put the new A, B and C constants (for the fan combination) into the Main_Database as a new "FanName" by pressing the adjacent button. To find the
fan curve for dissimilar fans, use the "FanCurveCombiner".
For your own records, you should permanently store the P and Q values used to create your fan constants, along with the air density, in the Sheet called "P&Qpts". This can be done by pressing the
relevant button. It is the A, B and C constants in RED (the FTP values, NOT FSP values) that are put into the database, or used directly in the Simulator worksheet as "manual data" fan type. To view a plot
of the fan curve, press the correct "Plot Chart" button. Both FSP and FTP can be plotted. To print or preview a chart, click "Print or Preview Chart" button (after pressing "Plot chart"). If an error message
comes up when you input your P and Q values, then it means you have misread the data.
ParamPtQmin
eter
Use these rows if reading points
from a Fan Total Pressure Curve
Store P & Q points in
database
Plot
chart
Store fan constants
in database
Use these rows if reading points
from a Fan Static Pressure Curve
Store P & Q points in
database
Plot
chart
Store fan constants
in database
•
Pt 2
Pt 3
PtQmax
FSP
5 222
4 749
3 644
159
FVP
178
251
356
591
Q1.2
26.5
31.5
37.5
48.3
Q
2
Special data
1.400 m, fan outlet dia
A1.2
-8.21164
B1.2
C1.2
382.02719
2
1.20 kg/m3 density Correlation coefficient, r
-7.95844
382.02719
864.44534
1.0000
864.44534
702
992
1 406
2 333
No of fans in series
1
FTP
5 400
5 000
4 000
750
No of fans in parallel
1
FSP
3 600
2 900
1 900
940
FVP
187
276
385
490
Q1.2
27.2
33.0
39.0
44.0
740
1 089
1 521
1 936
No of fans in series
1
3 787
3 176
2 285
1 430
No of fans in parallel
1
Q
2
FTP
1.400 m, fan outlet dia
-3.17266
66.90161 4132.46838
2
1.20 kg/m3 density Correlation coefficient, r
-2.91946
0.9998
66.90161 4132.46838
View the Chart by selecting the “Plot chart” button relevant to the input values you have chosen
(either the top button for FTP input values, or the bottom button for FSP input values). When
the dialog box comes up, choose whether you want the chart to show FTP or FSP. You can plot
FTP values even when you used FSP inputs and vice versa.
17 August 2006
LeakyDuct Worked Example, © MVA 1999-2006
Page 2 of 21
Once the chart is displayed, you make enlarge it to full screen by pressing the “Print or Preview
Chart” button to obtain the following:
You can now check the curve against the manufacturer’s curve. To quit this window, pressure
the “Close” button and to get rid of the chart completely, press the “Clear Chart” button.
•
Once you are happy with your choice of points, save the P and Q points you used to create these
constants in the database by pressing the “Store P and Q points in database” button. You should
obtain the following dialog box:
17 August 2006
LeakyDuct Worked Example, © MVA 1999-2006
Page 3 of 21
Enter a suitable UNIQUE name for the fan and press OK. You will be given a message box and
then taken to the database to view the stored data. You can manually alter values in this database
at any time.
Return to the FanCurveCalculator sheet by clicking on the sheet tab.
•
Save the fan constants and other critical data in the database by clicking on the “Store fan
constants in database” button. You should obtain the following dialog box:
Enter a suitable UNIQUE name for the fan (usually the same name as for the P/Q points dialog
box above) and press OK. You will be given a message box and then taken to the database to
view the stored data. You can manually alter values in this database at any time.
•
Find suitable duct diameters (Protan or equiv duct) for a two fan two duct (one fan per duct)
forcing-only installation to achieve the necessary airflow at the face (250 kW) and in the truck
compartment (360 kW plus 250 kW) for both 1000 m of duct and 1600 m of duct. Use the SDS
fan you have entered above. Find the approximate re-entry time if the drive is 6 m high * 5 m
wide, the face advance is 3.7 m, the rock s.g. is 2.65 and the powder factor is 1.6 kg/m3? Is this
fan/duct a good combination for this job?
Answer
The 250 kW truck will require a minimum of 12.5 m3/s (at 0.05 m3/s per kW of engine power)
at the face and the truck will require 18.0 m3/s where it is working. If both are working together,
they would require 30.5 m3/s. For two ducts, this would mean 15.25 m3/s at the face per duct.
Using Protan duct for this long heading, with “typical” k factor and leakage factors of 0.003
Ns2/m4 and 48 respectively, then the following table can be constructed using Simulator.
17 August 2006
LeakyDuct Worked Example, © MVA 1999-2006
Page 4 of 21
Table 1
1000 m duct
(1100 m with shock losses)
Fan
airflow,
m3/s
1.02 m
duct (2
off)
1600 m duct
(1760 m with shock losses)
Face
airflow,
m3/s
Re-entry
time, mins
Not possible Not possible
Fan
airflow,
m3/s
Face
airflow,
m3/s
Re-entry
time, mins
Not possible Not possible
1.2 m duct 26.8 * 2
(2 off)
19.9 * 2
27 mins
Not possible Not possible
1.4 m duct 34.9 * 2
(2 off)
27.5 * 2
20 mins
32.7 * 2
21.4 * 2
35 mins
The solutions for the 1000 m duct (1.02 m φ and 1.2 m φ) are shown below:
Figure 2
LeakySimTM
Copyright MVA 2000-2006. All rights reserved. Enter data in blue only. NOTE: RED values indicate warnings or bad data or invalid/poor results
1. Assumes forcing or exhausting fan or fan combination at end of duct. Fan can be a series, parallel or "Y" combination (fan feeding twin ducts).
2. Assumes air density is constant (compressibility ignored), leakage is influenced by static pressure across duct wall only, duct is of uniform diameter and quality (K and leakage factors are constant along duct [although
friction losses and leakage itself are NOT constant]), and velocity pressure at duct outlet (forcing fans) or duct inlet (exhausting fans) is small compared to fan total pressure.
3. When leakage exceeds 85% of fan airflow, result is unreliable.
4. Whilst very unlikely in most circumstances, the pressure loss to push the air back out of the heading may be significant on rare occasions and should be checked.
5. Note that a more powerful fan or lower leakage factor may not improve a solution, e.g. a lower leakage factor will increase the fan pressure perhaps to a point beyond the fan's pressure capability (i.e. goes into stall), and a
more powerful fan may have its surge point (min flow value) at a flow that is too high for the duct (i.e. it also goes into stall), etc. In both cases, the 'commonsense' solutions to improve the fan or duct performance may result in p
6. To find the airflow returning from the face at any point in the auxiliary compartment, use the "Predicted return air flow outside duct". This is helpful in determining how far diesels can go in a heading.
7. It is strongly recommended that you use real fan data and real duct friction, leakage and shock loss factors for any simulation.
Also you should consider a range of values (i.e. look at the sensitivity of the solution to different fan and duct characteristics) before making a final selection.
INPUTS
OUTPUTS (per duct)
DUCT INPUT DATA
Duct Length
Length with shock losses
Air density
Duct (NOT fan) diameter)
K factor of duct at 1.2 kg/m3
m
m
3
kg/m
m
Large diameter, 100 m lengths vent bag (Protan or equiv) - 0.00300
2
Ns /m
Duct leakage factor
OR manual entry
Calculate
4
Protan or equiv duct, 100 m lengths, typical - 48
2
2
mm leakage area per m area (including effects of duct joins, etc)
FAN INPUT DATA
GAL 14, 2 x 90 kW
Type of fan
Electrical power cost
$/(kW.hr)
Approx fan efficiency
% based on FSP, assumed constant
OR (instead of providing known fan type, provide fan curve constants)
Parabola A constant (from FTP curve, NOT FSP curve constants)
Parabola B constant (from FTP curve, NOT FSP curve constants)
Parabola C constant (from FTP curve, NOT FSP curve constants)
Maximum Q on curve, cms, (optional)
Minimum Q on curve, cms (optional)
Fan Total Pressure
Pa
5 182
1 000
Fan Static Pressure (FSP)
Pa
4 884
1 100
Fan Airflow per duct
18.2
m3/s
1.200
14.7
Fan Resistance at 1.2 kg/m3
Ns2/m8
12.6
1.02
Airflow out of/into duct
m3/s
0.00300
Leakage/fan Q
%
31%
3
Leakage per duct
5.6
0.0040
m /s
48
Leakage/Face flow
%
45%
100
Duct velocity pressure
Pa @ %FTP 142 @ 3%
Maximum FSP this fan
Pa
5 101
3
Fan free-delivery flow per duct
48.3
m /s
Fan Q as % of Qmax
%
38%
$ 0.08
FSP as % of FSPmax
%
96%
70%
Approx elec cost $/day/duct $/day/duct
117.97
3
3
-16.99
0.39
Approx cost/(m /s)
cents/(m /s)
Predicted return air flow outside duct
749.75
Distance from fan
m
992.76
50
3
Nett return this point/duct
38.5
17.7
m /s
Duct static pressure this pt
Pa
4,408
22.5
ERROR: Fan flow too low. Fan will be in stall zone (fan pressures not valid)
Note that an error message at the bottom of this screen will occur under several circumstances:
1. The duct resistance is too high for this fan and the fan effectively goes into stall.
2. The duct leakage is too high
3. The fan constants (a, b, c) you have calculated or entered manually do not describe a
valid fan curve
4. The solution is outside the range of valid flows and/or pressures for this fan
17 August 2006
LeakyDuct Worked Example, © MVA 1999-2006
Page 5 of 21
Figure 3
LeakySimTM
Copyright MVA 2000-2006. All rights reserved. Enter data in blue only. NOTE: RED values indicate warnings or bad data or invalid/poor results
1. Assumes forcing or exhausting fan or fan combination at end of duct. Fan can be a series, parallel or "Y" combination (fan feeding twin ducts).
2. Assumes air density is constant (compressibility ignored), leakage is influenced by static pressure across duct wall only, duct is of uniform diameter and quality (K and leakage factors are constant along duct [although
friction losses and leakage itself are NOT constant]), and velocity pressure at duct outlet (forcing fans) or duct inlet (exhausting fans) is small compared to fan total pressure.
3. When leakage exceeds 85% of fan airflow, result is unreliable.
4. Whilst very unlikely in most circumstances, the pressure loss to push the air back out of the heading may be significant on rare occasions and should be checked.
5. Note that a more powerful fan or lower leakage factor may not improve a solution, e.g. a lower leakage factor will increase the fan pressure perhaps to a point beyond the fan's pressure capability (i.e. goes into stall), and a
more powerful fan may have its surge point (min flow value) at a flow that is too high for the duct (i.e. it also goes into stall), etc. In both cases, the 'commonsense' solutions to improve the fan or duct performance may result in p
6. To find the airflow returning from the face at any point in the auxiliary compartment, use the "Predicted return air flow outside duct". This is helpful in determining how far diesels can go in a heading.
7. It is strongly recommended that you use real fan data and real duct friction, leakage and shock loss factors for any simulation.
Also you should consider a range of values (i.e. look at the sensitivity of the solution to different fan and duct characteristics) before making a final selection.
INPUTS
OUTPUTS (per duct)
DUCT INPUT DATA
Duct Length
Length with shock losses
Air density
Duct (NOT fan) diameter)
K factor of duct at 1.2 kg/m3
m
m
kg/m3
m
Calculate
Large diameter, 100 m lengths vent bag (Protan or equiv) - 0.00300
Ns2/m4
Duct leakage factor
OR manual entry
Protan or equiv duct, 100 m lengths, typical - 48
mm2 leakage area per m2 area (including effects of duct joins, etc)
FAN INPUT DATA
GAL 14, 2 x 90 kW
Type of fan
Electrical power cost
$/(kW.hr)
Approx fan efficiency
% based on FSP, assumed constant
OR (instead of providing known fan type, provide fan curve constants)
Parabola A constant (from FTP curve, NOT FSP curve constants)
Parabola B constant (from FTP curve, NOT FSP curve constants)
Parabola C constant (from FTP curve, NOT FSP curve constants)
Maximum Q on curve, cms, (optional)
Minimum Q on curve, cms (optional)
Fan Total Pressure
Pa
5 387
1 000
Fan Static Pressure (FSP)
Pa
5 050
1 100
Fan Airflow per duct
26.8
m3/s
1.200
7.0
Ns2/m8
Fan Resistance at 1.2 kg/m3
19.9
1.20
Airflow out of/into duct
m3/s
0.00300
Leakage/fan Q
%
26%
6.9
0.0040
Leakage per duct
m3/s
48
Leakage/Face flow
%
34%
Duct velocity pressure
Pa @ %FTP 186 @ 3%
100
Maximum FSP this fan
Pa
5 063
3
Fan free-delivery flow per duct
48.3
m /s
Fan Q as % of Qmax
%
55%
$ 0.08
FSP as % of FSPmax
70%
%
100%
Approx elec cost $/day/duct $/day/duct
193.22
3
3
0.40
-16.99
cents/(m /s)
Approx cost/(m /s)
Predicted return air flow outside duct
749.75
992.76
Distance from fan
m
50
26.1
38.5
Nett return this point/duct
m3/s
22.5
Duct static pressure this pt
Pa
4,558
Re-entry times are found using the Re-entry sheet as shown below for 1000 m and 21.5 m3/s to
the face:
Figure 4
Re-Entry Times after Blasting
Enter data in blue only.
1. This program assumes main explosive in ANFO. All ppm are by volume. See "comments" in each cell for Env Eng in S.A. mines values.
2. Program calculates time after vent re-established, NOT time after blast. If vent duct is leaky, then re-entry times will be somewhat less than these for forcing duct, and
somewhat more than these for exhausting system.
3. If forcing configuration, assumes vent duct is reasonably close to face, so that effective mixing occurs between blasting fumes and fresh air exiting the duct.
4. If exhausting configuration or overlap, assumes exhaust airflow exceeds forcing airflow.
5. Factors that increase production of noxious gases include: inadequate priming, insufficient water resistance, lack of confinement, inadequate burden, reaction of explosion
with rock (e.g. with sulphides to produce H2S), overcharging and incomplete detonation.
This program is for guidance and design purposes only and must be confirmed on site using proper gas detection equipment.
Note: The author and the supplier have no liability to the licensee or any other person or entity for any damage or loss, including special, incidential or consequential damages caused by this product
directly or indirectly. This software is provided "as is" without warranty of any kind, either expressed or implied. Warranties of merchantability or of fitness for any purpose are specifically excluded.
INPUTS
OUTPUTS
Fume throw-back (approximate figure only)
m
Minimum possible re-entry time assuming plug flow mins
Re-entry time based on most critical gas mins
NO2
CO2
NH3
INDIVIDUAL GAS VALUES
NO
DEVELOPMENT HEADING
Distance flow-thru to face
Height
Width
m
m
m
3
m /s
Face airflow
Type of ventilation Forcing only, fan at devt entry, duct outlet near face
Distance of duct outlet from face
m
1000
6.0
5.0
21.5
15.0
BLASTING
Advance per round incl butts
Rock density
Explosive powder factor
Maximum limit this mine for NO2
Maximum limit this mine for NO
Maximum limit this mine for NH3
m
tonnes per cubic meter
3
kg/m rock blasted
ppm
ppm
ppm
3.70
2.65
1.60
3
25
25
Maximum limit this mine for CO2
Maximum limit this mine for CO
NO2 produced per t ANFO
NO produced per t ANFO
NH3 produced per t ANFO
CO2 produced per t ANFO
CO produced per t ANFO
ppm
ppm
kg
kg
kg
kg
kg
5000
50
3.9
2.5
6.0
180.0
5.0
Re-entry time, min
Conc after blast in fume throw-back, ppm
Conc when fume reaches flow-past
Conc on reentry, ppm
Current ACGIH TLV, ppm
55
152
13
3
3
21
149
12
3
25
38
631
52
12
25
21
7310
604
145
5000
83
23
55
CO
21
319
26
6
50
IMPORTANT: Immediately after a blast, most nitrous fumes appear as NO. This is then slowly oxidised to NO2, with about
50% converted after 2 hours. If re-entry time is short, then use the NO re-entry time in the above table; if it is long, use the
NO2 re-entry time Using the NO2 values will always give the most pessimistic times (compared to NO). Also note that about
half the total gases remain in the muckpile until bogging starts.
The "minimum re-entry time assuming plug flow" is if the entire blasting fumes moved out as a plug.
Warning: Uncalculated data!
2.12
2.08
8.81
102.09
4.46
Calculate
outputs
Clearly the 1.02 m duct is not sufficient for either the 1000 m or 1600 m lengths. The 1.2 m duct
is satisfactory for 1000 m but will not be sufficient to see the job out to 1600 m. However, 1.4 m
duct (the next off-the-shelf size available) could be argued as being too big as it requires 66 m3/s
at the fan and puts 42 m3/s to the job, when only 30 m3/s is required. However, twin ducts are
best for long headings, as if one duct needs repair, or a fan is not operational, at least some air
can be sent to the face. In addition, 25 mins is a good upper limit for a re-entry time, especially if
this is a priority face or a single heading. It might be better to look a smaller fan, e.g. the single
GAL12.5 2*55 kW; however, it is probably better to take the shorter re-entry time and wear the
17 August 2006
LeakyDuct Worked Example, © MVA 1999-2006
Page 6 of 21
cost of more fan power. The actual decision will depend on what management is trying to
achieve on this heading
•
Construct the fan curve for two of these 180 kW fans in series. Now find the minimum duct
diameter that will deliver the face volume flow using only one duct. Find the approximate reentry time.
Answer
Using FanCurveCombiner and putting two GAL14 2*90 kW fans in series and calculating the
combined FTP curve gives constants of a = -15.9169, b = 764.0544 and c = 1728.8907 (see
below). Note that the “Calculate values and chart” button must be pressed after entering new
data.
Figure 5
Fan curve combiner Copyright MVA 2000. All rights reserved. Enter data in blue only.
1. Select "Fan combination type required".
2. Select Curve type required: FTP means output constants are for fan total pressure, FSP means ouput constants are for fan static pressure
3. Select fan(s). Note "manual entry" of fan constants etc is not possible on this sheet. Fan must already exist in database.
4. To create multiple combinations of fans, add the first combination, then store the resulting fan constants in the database, then add this as a new fan to another set, etc.
5. To print or preview the chart, select the button.
Series
1
Fan combination type
MaxQ1
48.3
1341
160
1341
S
1
Fan TOTAL Pressure
Series
Calculate
Curve type required?
1st
qtrcurve: FTP
42.9
3087
2157
3087
FTP
Fan
Fan
Series
configuration.
Density
=
1.20 kg/m3.
3
Air density for inputs, kg/m
Fan and
STATIC 2nd qtr
37.4
4374
3666
4374
Parallel
1.20
Values
3
Air density for chart plot, kg/m
Y (feeding twin ducts) Chart
3rd 12000
qtr
32.0
5205
4688
5205
1.20
55
MinQ1
L 14, 2 x 90
10000
MaxQ2
1st qtr
Print or
Store
2nd qtr
Preview
Combo fan
8000
Chart
in d'base
3rd qtr
P-Q curve for c MinQ2
1 L 14, 2 x6000
90 kW
MaxQ3
Warning: Uncalculated data!
1st qtr4000
Pa
Fan 1 at input air density
Fan name GAL 14, 2 x 90 kW
a1 constant
-7.9584
b1 constant
382.0272
c1 constant
864.4453
Max Q1
48.3
Min Q1
26.5
P at Max Q1
750
P at Min Q1
5399
Fan diameter, m
1.40
Fan 2 at input air density
GAL 14, 2 x 90 kW
Fan name
a2 constant
-7.9584
b2 constant
382.0272
c2 constant
864.4453
Max Q2
48.3
Min Q2
26.5
P at Max Q2
1643
P at Min Q2
10771
Fan diameter, m
1.40
55
2nd qtr
3rd qtr
2000
MinQ3
26.5
0.0
48.3
42.9
37.4
32.0
26.5
5577
5222
5577
1341
3087
4374
5205
5577
160
2157
3666
4688
5222
1341
3087
4374
5205
5577
48.3
42.9
2091
5708
910
4778
2091
5708
FVP
591
465
37.4
32.0
26.5
8395
10151
10977
7686
9634
10621
8395
10151
10977
354
258
178
Fan combination at input density
a combined
-15.9169
b combined
764.0544 Fan curve: F
0
0
c combined
1728.8907
0
0
10
Max Q combined
48.3
Min Q combined
26.5
0
P at Max Q combined
1500
GAL 14, 2 x 90 0
kW
P at Min Q combined
10799
20
30
40
50
GAL 14, 2 x 90 kW
P-Q curve for combination
This combined curve can now be plotted using the “Print or Preview Chart” button using a
similar procedure to above.
Once you are satisfied with the curve, press the “Store combo fan in d’base” button to store this
combination as a new fan type. You will need to enter a new fan name.
This new fan is now available from the Simulator sheet. Click on the Simulator sheet tab to go
to the sheet. The solution for the new fan at 1600 m using 1.4 m φ duct is shown below:
For a 1.4 m duct 1600 m long (1760 m with shock losses), the air delivered to the face is 27
m3/s (for a fan flow of 40 m3/s), see below. This does not meet the requirement of 30.5 m3/s, but
would still meet the requirement of 0.04 m3/s per kW (if this was in operation). Therefore this
“solution” would probably be acceptable. However, it suffers from the problems of single vent
17 August 2006
60
m3/s
LeakyDuct Worked Example, © MVA 1999-2006
Page 7 of 21
lines with long headings, i.e. no redundancy. If a 1.5 m duct is used, the face flow is still only 29
m3/s, but at 1.6 m φ, this increases to 31.4 m3/s.
LeakySimTM
Copyright MVA 2000. All rights reserved. Enter data in blue only. NOTE: RED values indicate warnings or bad data or invalid/poor results
1. Assumes forcing or exhausting fan or fan combination at end of duct. Fan can be a series, parallel or "Y" combination (fan feeding twin ducts).
2. Assumes air density is constant (compressibility ignored), leakage is influenced by static pressure across duct wall only, duct is of uniform diameter and quality (K and leakage factors are constant along duct [although
friction losses and leakage itself are NOT constant]), and velocity pressure at duct outlet (forcing fans) or duct inlet (exhausting fans) is small compared to fan total pressure.
3. When leakage exceeds 85% of fan airflow, result is unreliable.
4. Whilst very unlikely in most circumstances, the pressure loss to push the air back out of the heading may be significant on rare occasions and should be checked.
5. Note that a more powerful fan or lower leakage factor may not improve a solution, e.g. a lower leakage factor will increase the fan pressure perhaps to a point beyond the fan's pressure capability (i.e. goes into stall), and a
more powerful fan may have its surge point (min flow value) at a flow that is too high for the duct (i.e. it also goes into stall), etc. In both cases, the 'commonsense' solutions to improve the fan or duct performance may result in p
6. To find the airflow returning from the face at any point in the auxiliary compartment, use the "Predicted return air flow outside duct". This is helpful in determining how far diesels can go in a heading.
7. It is strongly recommended that you use real fan data and real duct friction, leakage and shock loss factors for any simulation.
Also you should consider a range of values (i.e. look at the sensitivity of the solution to different fan and duct characteristics) before making a final selection.
INPUTS
OUTPUTS (per duct)
DUCT INPUT DATA
Duct Length
Length with shock losses
Air density
Duct (NOT fan) diameter)
K factor of duct at 1.2 kg/m3
m
m
3
kg/m
m
Large diameter, 100 m lengths vent bag (Protan or equiv) - 0.00300
2
Ns /m
Duct leakage factor
OR manual entry
4
Protan or equiv duct, 100 m lengths, typical - 48
2
2
mm leakage area per m area (including effects of duct joins, etc)
FAN INPUT DATA
Two GAL 14, 2 x 90 kW in series
Type of fan
Electrical power cost
$/(kW.hr)
Approx fan efficiency
% based on FSP, assumed constant
OR (instead of providing known fan type, provide fan curve constants)
Parabola A constant (from FTP curve, NOT FSP curve constants)
Parabola B constant (from FTP curve, NOT FSP curve constants)
Parabola C constant (from FTP curve, NOT FSP curve constants)
Maximum Q on curve, cms, (optional)
Minimum Q on curve, cms (optional)
Fan Total Pressure
Pa
6 946
1 600
Fan Static Pressure (FSP)
Pa
6 546
3
39.8
1 760
Fan Airflow per duct
m /s
3
2
8
1.200
4.1
Ns /m
Fan Resistance at 1.2 kg/m
3
26.7
1.40
Airflow out of/into duct
m /s
0.00300
Leakage/fan Q
%
33%
3
0.0020
Leakage per duct
13.1
m /s
48
Leakage/Face flow
%
49%
7
Duct velocity pressure
Pa, %FTP 180, 3%
Maximum FSP this fan
Pa
10 898
3
50.2
Fan free-delivery flow per duct
m /s
Fan Q as % of Qmax
%
79%
$ 0.08
FSP as % of FSPmax
%
60%
70%
Approx elec cost $/day/duct $/day/duct
335.14
3
3
0.52
-16.99
Approx cost/(m /s)
cents/(m /s)
Predicted return air flow outside duct
749.75
992.76
Distance from fan
m
500
3
32.9
38.5
Nett return this point/duct
m /s
22.5
Duct static pressure this pt
Pa
3,094
Using the 1.4 m duct (27 m3/s to the face) and the Re-entry sheet, gives a re-entry time of 43
minutes at 1000 m and 57 minutes at 1600 m (shown below). These re-entry times may be
excessive, depending on the application.
Enter data in blue only.
Re-Entry Times after Blasting
1. This program assumes main explosive in ANFO. All ppm are by volume. See "comments" in each cell for Env Eng in S.A. mines values.
2. Program calculates time after vent re-established, NOT time after blast. If vent duct is leaky, then re-entry times will be somewhat less than these for forcing duct, and
somewhat more than these for exhausting system.
3. If forcing configuration, assumes vent duct is reasonably close to face, so that effective mixing occurs between blasting fumes and fresh air exiting the duct.
4. If exhausting configuration or overlap, assumes exhaust airflow exceeds forcing airflow.
5. Factors that increase production of noxious gases include: inadequate priming, insufficient water resistance, lack of confinement, inadequate burden, reaction of explosion
with rock (e.g. with sulphides to produce H2S), overcharging and incomplete detonation.
This program is for guidance and design purposes only and must be confirmed on site using proper gas detection equipment.
Note: The author and the supplier have no liability to the licensee or any other person or entity for any damage or loss, including special, incidential or consequential damages caused by this product
directly or indirectly. This software is provided "as is" without warranty of any kind, either expressed or implied. Warranties of merchantability or of fitness for any purpose are specifically excluded.
INPUTS
OUTPUTS
Fume throw-back (approximate figure only)
m
Minimum possible re-entry time assuming plug flow mins
Re-entry time based on most critical gas mins
NH3
NO2
CO2
NO
INDIVIDUAL GAS VALUES
DEVELOPMENT HEADING
Distance flow-thru to face
Height
Width
m
m
m
3
m /s
Face airflow
Type of ventilation Forcing only, fan at devt entry, duct outlet near face
Distance of duct outlet from face
m
1600
6.0
5.0
27.0
15.0
BLASTING
•
Advance per round incl butts
Rock density
Explosive powder factor
Maximum limit this mine for NO2
Maximum limit this mine for NO
Maximum limit this mine for NH3
m
tonnes per cubic meter
3
kg/m rock blasted
ppm
ppm
ppm
3.70
2.65
1.60
3
25
25
Maximum limit this mine for CO2
Maximum limit this mine for CO
NO2 produced per t ANFO
NO produced per t ANFO
NH3 produced per t ANFO
CO2 produced per t ANFO
CO produced per t ANFO
ppm
ppm
kg
kg
kg
kg
kg
5000
50
3.9
2.5
6.0
180.0
5.0
Re-entry time, min
Conc after blast in fume throw-back, ppm
Conc when fume reaches flow-past
Conc on reentry, ppm
Current ACGIH TLV, ppm
57
152
8
3
3
28
149
8
3
25
36
631
33
12
25
28
7310
378
145
5000
57
CO
28
319
16
6
50
IMPORTANT: Immediately after a blast, most nitrous fumes appear as NO. This is then slowly oxidised to NO2, with about
50% converted after 2 hours. If re-entry time is short, then use the NO re-entry time in the above table; if it is long, use the
NO2 re-entry time Using the NO2 values will always give the most pessimistic times (compared to NO). Also note that about
half the total gases remain in the muckpile until bogging starts.
The "minimum re-entry time assuming plug flow" is if the entire blasting fumes moved out as a plug.
Warning: Uncalculated data!
2.12
2.08
8.81
102.09
4.46
Calculate
outputs
Assuming a single GAL14 2*90 kW fan, find the fan curve if this fan installation is used to feed
twin ducts to the face. What is the minimum duct size to achieve the required face flow?
Answer
Using FanCurveCombiner and putting a single GAL14 2*90 kW fans in a “Y” configuration
and calculating the combined FTP curve gives constants of a = -33.550, b = 825.205 and c =
341.529 (see below).
17 August 2006
83
30
LeakyDuct Worked Example, © MVA 1999-2006
Page 8 of 21
Pa
Fan curve combiner Copyright MVA 2000. All rights reserved. Enter data in blue only.
1. Select "Fan combination type required".
2. Select Curve type required: FTP means output constants are for fan total pressure, FSP means ouput constants are for fan static pressure
3. Select fan(s). Note "manual entry" of fan constants etc is not possible on this sheet. Fan must already exist in database.
4. To create multiple combinations of fans, add the first combination, then store the resulting fan constants in the database, then add this as a new fan to another set, etc.
5. To print or preview the chart, select the button.
Y (feeding twinYducts)
3
Fan combination type
MaxQ1
48.3
1341
160
1341
1
Fan TOTAL Pressure
Series
Calculate
Curve type required?
FTP
1st qtrcurve: FTP
42.9
2157 fan curves
3087 shown
Fan
Fan
- Single fan 3087
with two ducts,
3
Air density for inputs, kg/m
Fan and
STATIC 2nd qtr
1.20
Parallel
37.4
4374 = 1 20
3666
Values
for
one duct Density
kg/m3 4374
3
Air density for chart plot, kg/m
1.20
Y (feeding twin ducts) Chart
3rd 6000
qtr
32.0
5205
4688
5205
55
MinQ1
26.5
5577
5222
5577
Fan 1 at input air density
Fan name GAL 14, 2 x 90 kW
L 14, 2 x 90
0.0
5000
a1 constant
MaxQ2
40.0
-118029
-118839
-118029
-7.9584
b1 constant
382.0272
1st qtr
37.5
-102283
-102995
-102283
Print or
Store
c1 constant
2nd 4000
qtr
35.0
-87603
-88224
-87603
864.4453
Preview
Combo fan
Chart
in d'base
Max Q1
48.3
3rd qtr
32.5
-73991
-74526
-73991
Min Q1
P-Q curve for e MinQ2
30.0
-61446
-61902
-61446
26.5
3000
P at Max Q1
1
FVP
750
P at Min Q1
5399
MaxQ3
24.2
898
603
898
148
Warning: Uncalculated data!
Fan diameter, m
1st qtr
21.4
2738
2505
2738
116
1.40
2000
0
2nd qtr
18.7
4109
3932
4109
89
Fan 2 at input air density
Fan name
3rd qtr
16.0
5011
4882
5011
65
1000
a2 constant
MinQ3
13.3
5444
5355
5444
44
Fan combination at input density
-85.6168
b2 constant
317.1250
a combined
-31.8338
c2 constant
b combined
764.0544 Fan curve: F
0
5867.7000
0
Max Q2
c combined
864.4453
0
40.0
0
10
20
30
40
Min Q2
30.0
Max Q combined
24.2
P at Max Q2
Min Q combined
13.3
0
1643
m3/s
P at Min Q2
10771
P at Max Q combined
750
0 2 x 90 kW
GAL 14,
P-Q curve for each duct
Fan diameter, m
P at Min Q combined
5399
1.40
For two 1.4 m ducts each 1600 m long (1760 m with shock losses), the air delivered to the face
is 14.9 * 2 = 30 m3/s (see below). Therefore this “solution” would probably be acceptable. An
alternative solution would be to use two such fans in series, then in Y configuration. With 1.4 m
duct, this would deliver 2 * 15.8 (32) m3/s to the face. As the addition of the extra 180 kW of
fan power only delivers an additional 2 m3/s to the face, this arrangement is probably not
warranted
LeakySimTM
Copyright MVA 2000. All rights reserved. Enter data in blue only. NOTE: RED values indicate warnings or bad data or invalid/poor results
1. Assumes forcing or exhausting fan or fan combination at end of duct. Fan can be a series, parallel or "Y" combination (fan feeding twin ducts).
2. Assumes air density is constant (compressibility ignored), leakage is influenced by static pressure across duct wall only, duct is of uniform diameter and quality (K and leakage factors are constant along duct [although
friction losses and leakage itself are NOT constant]), and velocity pressure at duct outlet (forcing fans) or duct inlet (exhausting fans) is small compared to fan total pressure.
3. When leakage exceeds 85% of fan airflow, result is unreliable.
4. Whilst very unlikely in most circumstances, the pressure loss to push the air back out of the heading may be significant on rare occasions and should be checked.
5. Note that a more powerful fan or lower leakage factor may not improve a solution, e.g. a lower leakage factor will increase the fan pressure perhaps to a point beyond the fan's pressure capability (i.e. goes into stall), and a
more powerful fan may have its surge point (min flow value) at a flow that is too high for the duct (i.e. it also goes into stall), etc. In both cases, the 'commonsense' solutions to improve the fan or duct performance may result in p
6. To find the airflow returning from the face at any point in the auxiliary compartment, use the "Predicted return air flow outside duct". This is helpful in determining how far diesels can go in a heading.
7. It is strongly recommended that you use real fan data and real duct friction, leakage and shock loss factors for any simulation.
Also you should consider a range of values (i.e. look at the sensitivity of the solution to different fan and duct characteristics) before making a final selection.
INPUTS
OUTPUTS (per duct)
DUCT INPUT DATA
Duct Length
Length with shock losses
Air density
Duct (NOT fan) diameter)
K factor of duct at 1.2 kg/m3
m
m
3
kg/m
m
Large diameter, 100 m lengths vent bag (Protan or equiv) - 0.00300
2
Ns /m
Duct leakage factor
OR manual entry
4
Protan or equiv duct, 100 m lengths, typical - 48
2
2
mm leakage area per m area (including effects of duct joins, etc)
FAN INPUT DATA
GAL 14, 2 x 90 kW feeding twin ducts, per duct
Type of fan
Electrical power cost
$/(kW.hr)
Approx fan efficiency
% based on FSP, assumed constant
OR (instead of providing known fan type, provide fan curve constants)
Parabola A constant (from FTP curve, NOT FSP curve constants)
Parabola B constant (from FTP curve, NOT FSP curve constants)
Parabola C constant (from FTP curve, NOT FSP curve constants)
•
Fan Total Pressure
Pa
2 159
1 600
Fan Static Pressure (FSP)
Pa
2 035
3
22.2
1 760
Fan Airflow per duct
m /s
3
2
8
1.200
4.1
Ns /m
Fan Resistance at 1.2 kg/m
3
14.9
1.40
Airflow out of/into duct
m /s
0.00300
Leakage/fan Q
%
33%
3
0.0020
Leakage per duct
7.3
m /s
48
Leakage/Face flow
%
49%
Duct velocity pressure
Pa, %FTP
56, 3%
7
Maximum FSP this fan
Pa
5 449
3
Fan free-delivery flow per duct
25.1
m /s
Fan Q as % of Qmax
%
88%
$ 0.08
FSP as % of FSPmax
%
37%
70%
Approx elec cost $/day/duct $/day/duct
58.08
3
3
0.16
-16.99
Approx cost/(m /s)
cents/(m /s)
Predicted return air flow outside duct
749.75
992.76
Distance from fan
m
500
Find the solution using two separate 2 * 90 kW fans and using an overlap system (see below for
example of overlap system). Find the approximate re-entry time. What is the % recirculation in
the system?
17 August 2006
LeakyDuct Worked Example, © MVA 1999-2006
Page 9 of 21
One possible overlap system of the many variations that can be designed (Plan view)
← to RAR
Walls of drive
Flexible duct
Rigid
Face →
← to RAR
Flexible duct
Not to scale
Twin Howden 2 * 90 kW
Face →
Answer
Assuming a 60 m overlap at the mid point between the entry to the heading and the face, the fan
at the mid point delivering to the face has to deliver 30.5 m3/s over a distance of 1600 / 2 m + 30
m = 830 m (913 m as equivalent length). A single GAL14 2*90 kW fan feeding a 1.4 m duct
would deliver 31.8 m3/s to the face and take in 37.1 m3/s at the fan. Assuming 30% overlap
volume, the return fan [the fan at the mid point delivering air back to the development entry)
(also 830 m) would need to pick up 130% * 37.1 = 48 m3/s at the fan. This clearly couldn’t be
done using the same fan/duct combination as the face fan, but iterations produce the following
combinations:
•
Single GAL14 2*90 kW fan feeding twin 1.4 m ducts through Y-piece: 46 m3/s (fan)
and 39 m3/s (return)
•
Twin GAL14 2*90 kW fans in series feeding twin 1.4 m ducts through Y-piece: 48 m3/s
(fan) and 41 m3/s (return)
Twin GAL14 2*90 kW fans each feeding a 1.4 m duct: 74 m3/s (fans) and 64 m3/s
(returns)
It is likely that if there is room to fit twin ducts in the return leg, then it would be better to put
two 2 * 90 kW fans each with its own duct (74 m3/s at the fan) rather than bolting them in series
and then feeding twin ducts (48 m3/s).
Re-entry time would be the time for the face fan to rid the face compartment of fumes (as the
return fan then picks up the fumes and puts them into a return airway). This would take 31
minutes (down from 57 minutes) and is another reason why the overlap system is sometimes
preferred to the forcing only system.
•
•
See if the same vent duct you have selected for the “return” section of the drive can also be used
for the initial drive development up until the overlap system is installed? How far could
development proceed using this fan/duct combination?
Answer
Consider a single GAL14 2*90 kW feeding twin 1.4 m ducts through a Y-piece. For the initial
development (before the overlap system is installed), 30.5 m3/s is required at the face (15.25
m3/s per duct). This can be achieved up to a distance of 1500 m (see below), which is almost the
full requirement of 1600 m. This could potentially result in a design change!
17 August 2006
LeakyDuct Worked Example, © MVA 1999-2006
LeakySimTM
Page 10 of 21
Copyright MVA 2000. All rights reserved. Enter data in blue only. NOTE: RED values indicate warnings or bad data or invalid/poor results
1. Assumes forcing or exhausting fan or fan combination at end of duct. Fan can be a series, parallel or "Y" combination (fan feeding twin ducts).
2. Assumes air density is constant (compressibility ignored), leakage is influenced by static pressure across duct wall only, duct is of uniform diameter and quality (K and leakage factors are constant along duct [although
friction losses and leakage itself are NOT constant]), and velocity pressure at duct outlet (forcing fans) or duct inlet (exhausting fans) is small compared to fan total pressure.
3. When leakage exceeds 85% of fan airflow, result is unreliable.
4. Whilst very unlikely in most circumstances, the pressure loss to push the air back out of the heading may be significant on rare occasions and should be checked.
5. Note that a more powerful fan or lower leakage factor may not improve a solution, e.g. a lower leakage factor will increase the fan pressure perhaps to a point beyond the fan's pressure capability (i.e. goes into stall), and a
more powerful fan may have its surge point (min flow value) at a flow that is too high for the duct (i.e. it also goes into stall), etc. In both cases, the 'commonsense' solutions to improve the fan or duct performance may result in p
6. To find the airflow returning from the face at any point in the auxiliary compartment, use the "Predicted return air flow outside duct". This is helpful in determining how far diesels can go in a heading.
7. It is strongly recommended that you use real fan data and real duct friction, leakage and shock loss factors for any simulation.
Also you should consider a range of values (i.e. look at the sensitivity of the solution to different fan and duct characteristics) before making a final selection.
INPUTS
OUTPUTS (per duct)
DUCT INPUT DATA
Duct Length
Length with shock losses
Air density
Duct (NOT fan) diameter)
K factor of duct at 1.2 kg/m3
m
m
3
kg/m
m
Large diameter, 100 m lengths vent bag (Protan or equiv) - 0.00300
2
Ns /m
Duct leakage factor
OR manual entry
4
Protan or equiv duct, 100 m lengths, typical - 48
2
2
mm leakage area per m area (including effects of duct joins, etc)
FAN INPUT DATA
GAL 14, 2 x 90 kW feeding twin ducts, per duct
Type of fan
Electrical power cost
$/(kW.hr)
Approx fan efficiency
% based on FSP, assumed constant
OR (instead of providing known fan type, provide fan curve constants)
Parabola A constant (from FTP curve, NOT FSP curve constants)
Parabola B constant (from FTP curve, NOT FSP curve constants)
Parabola C constant (from FTP curve, NOT FSP curve constants)
Maximum Q on curve, cms, (optional)
Minimum Q on curve, cms (optional)
•
In the overlap region, four ducts cannot fit across the back. If the overlap is to be 60 m long,
find the diameter of steel duct that is equivalent to twin 1.4 m Protan ducts for each fan
installation? How much higher must the back be stripped in this region?
Answer
Using DuctEquiv, for a single 60 m length of steel duct (k = 0.0025 Ns2/m4) to be equivalent to
twin 1.4 m Protan ducts (k = 0.003 Ns2/m4), the steel duct would have to be 1.78 m diameter
(see below). This would require the back to be stripped an additional 0.4 m in this region, which
would be expensive and probably operationally difficult. If the overlap was reduced in length to
40 m, then the equivalent steel duct diameter would be 1.64 m, which would still require almost
250 mm to be stripped from the back. There appears to be little alternative to this.
Actual duct
installation
•
Fan Total Pressure
Pa
2 110
1 500
Fan Static Pressure (FSP)
Pa
1 984
22.2
1 650
Fan Airflow per duct
m3/s
3
2
8
1.200
4.0
Fan Resistance at 1.2 kg/m
Ns /m
3
15.4
1.40
Airflow out of/into duct
m /s
0.00300
Leakage/fan Q
%
31%
3
0.0020
Leakage per duct
6.8
m /s
48
Leakage/Face flow
%
44%
7
Duct velocity pressure
Pa, %FTP
60, 3%
Maximum FSP this fan
Pa
5 449
3
25.1
Fan free-delivery flow per duct
m /s
Fan Q as % of Qmax
%
89%
$ 0.08
FSP as % of FSPmax
%
36%
70%
Approx elec cost $/day/duct $/day/duct
58.73
3
3
0.16
-16.99
cents/(m /s)
Approx cost/(m /s)
Predicted return air flow outside duct
749.75
992.76
Distance from fan
m
500
3
18.5
38.5
Nett return this point/duct
m /s
22.5
Duct static pressure this pt
Pa
882
Equivalent duct
installation
Ducts in parallel
number
2.0
1.0
Ducts in series
number
1.0
1.0
K factor
N.s /m
0.00300
0.00250
Diameter
m
1.40
1.78
Length (per duct)
m
60.0
60.0
Circumference
m
4.398
5.596
Area (x-sectional)
m
1.539
2.492
Resistance
N.s /m
0.05426
0.05426
Flow (Q)
m /s
50.0
50.0
Pressure drop, Pd
Pa
135.6
135.6
2
4
2
2
3
8
Select value
to be found
Calculate the fan/duct performance if a GAL 14 2*90 kW fan has 1000 m of 1.2 φ Protan duct
in typical condition on its discharge and a steel induction tube 100 m long and 1.2 φ on its inlet.
Answer
17 August 2006
LeakyDuct Worked Example, © MVA 1999-2006
Page 11 of 21
Using DuctEquiv, the length of 1.2 m φ Protan duct equivalent to 100 m of 1.2 m φ steel tube is
78 m (see below). Therefore the true equivalent length of Protan is 1000 m + 78 m (+ shock
losses if any) and the problem can then be solved simply using Simulator.
Equivalent duct calculator
Copyright MVA 2000. All rights reserved.
1. Input data in blue only.
2. Check the button next to the value that you want to allow to change. You need to do this FIRST.
3. To find an equivalent SINGLE duct installation to a SINGLE duct installation, put '1' as the number of ducts in parallel and '1' as the number
of ducts in series. To find an equivalent DOUBLE (parallel) duct installation to a single duct, put '2' as the number of ducts in parallel and '1' as
the number of ducts in series', etc. It would be rare to use anything other than '1' as the number of ducts in series . For further instructions on
parallel/series combinations, refer to instructions on sheet "FanCurveCalculator".
4. Fill in blue data for actual (existing or known) duct installation. 'Q' (flow) is optional; if provided, the pressure drop along the duct as
calculated is correct. If not provided, then the duct resistances will be the same, but the pressure drops (whilst identical) will be unknown.
5. Fill in blue data for equivalent (proposed or unknown) duct installation. The value next to the check button is the one that will be varied so as
to ensure the ducts have equivalent resistance (and hence pass the same airflow and have the same pressure drop, the actual airflow is irreleva
Note: Program assumes no leakage in any duct; hence is not correct where leakage is significant. To compare leaky ducts, use "DuctCalculato
Actual duct
installation
•
Equivalent duct
installation
Ducts in parallel
number
1.0
1.0
Ducts in series
number
1.0
1.0
K factor
N.s /m
0.00275
0.00350
Diameter
m
1.20
1.20
Length (per duct)
m
100.0
78.6
Circumference
m
3.770
3.770
1.131
1.131
0.71665
0.71665
50.0
50.0
1791.6
1791.6
2
4
2
Area (x-sectional)
m
Resistance
N.s /m
Flow (Q)
m /s
Pressure drop, Pd
Pa
2
8
3
Select value
to be found
ERROR: All cells in blue for actual duct
installation (except Q) must have data
If the measured asperity height on a ventilation tube is 2 mm for a 600 mm diameter tube, what
is the e/D ratio and k factor at 1.3 kg/m3 for the 600 mm tube and for a 1.2 m and 1.5 m tube
constructed in the same way using the same material? What is the k factor at standard density
for the same duct? Comment on the change in k factor with diameter.
Answer
The KfactorCalc sheet is used for this exercise. The area and perimeter of the duct are
calculated as 0.2827 m2 and 1.885 m respectively. The e/D ratio is therefore 0.0033. The k
factor at a density of 1.3 kg/m3 is 0.00438 Ns2/m4 and at standard density of 1.2 kg/m3 is
0.00404 Ns2/m4.
The k factors for the other duct sizes are calculated as follows:
Duct dia e/D
k1.2
k1.3
600 mm
0.0033 0.00404 0.00438
1.2 m
0.0017 0.00335 0.00363
1.5 m
0.0013 0.00317 0.00343
It is clear that the k factor has changed from 0.00438 to 0.00343 (i.e. 22%) when the duct
diameter has changed. The point is that care needs to be taken in using old k factors in larger
diameter ducts (or vice-versa), as the asperity height is probably the thing that changes least;
therefore the k factor will not always be constant as the duct diameter changes
•
A 1.2 m diameter fan force feeds a 500 m long 1.2 m diameter duct with 100 m of additional
shock losses. The static pressure in the duct at the fan discharge is 1800 Pa and the airflow at the
fan and the duct discharge are 22 m3/s and 13 m3/s respectively. Air density is 1.2 kg/m3. Find
the duct k factor and leakage factor?
17 August 2006
LeakyDuct Worked Example, © MVA 1999-2006
Page 12 of 21
Answer
This problem is solved using kFactorCalc. Entering the data, and selecting the first option (to
calculate k and leakage) and using a blowing fan calculates that the k1.2 factor is 0.0042 Ns2/m4
and the leakage factor is 267. See below.
Duct resistance and leakage calculator Copyright MVA 2000. All rights reserved.
Key assumptions:
Air density along duct is constant; leakage is not influenced by velocity pressure; ducting is of uniform diameter and quality, FVP is small compared to FTP. All
pressures and flows are treated (and entered) as positive numbers, whether blowing or exhausting.
This program calculates true K factors, etc using the data you provide at the density you provide. It does NOT correct these to std density.
Instructions:
1. Select calculation option (check mark) and fan type option (check mark).
2. Change values in BLUE only. Values in BLACK are outputs.
3. Click "Calculate" button. Sometimes, program may indicate no solution is possible. Press "Calculate" again or try different seed values.
3
kg/m
Air density (assumed constant along duct)
Duct diameter
m
Fan diameter (optional)
m
Duct static pressure at fan (Point 1)
Pa
Duct static pressure downstream (Point 2)
Pa
m
True duct length between measuring points on duct
m
Equivalent length between measuring points on duct
3
m /s
Volume flow (higher value)
3
m /s
Volume flow (lower value)
Fan Static Pressure
Pa
Fan Velocity Pressure
Pa
Fan Total Pressure
Pa
t
m
Pressure Increase Ratio, PIR
dimensionless
Volume Increase Ratio, VIR
dimensionless
Leakage in duct
% of air through fan
2
8
Ns /m per 100 m
Duct resistance per 100 m
2
8
Ns /m per 100 m
Leakage resistance per 100 m - per formula A
2
8
Ns /m per 100 m
Leakage resistance per 100 m - per formula B
2
8
Ns /m per 100 m
Leakage resistance per 100 m - per formula C
Asperity height (roughness) of walls ('e' as in 'e/D' ratio)
mm
2
4
3
Ns /m , or kg/m
Friction 'K' factor for duct @ 1.20 kg/m3
2
2
mm /m
Leakage factor (use this value in worksheet 'Simulator')
2
4
3
Ns /m , or kg/m
Friction 'K' factor for duct @ 1.20 kg/m3
2
2
mm /m
Specific leakage area, As, at Ko of 0.6
2
2
mm /m
Leakage factor (use this value in "Simulator")
•
1.20
1.200
1.200
1 800
0
500
600
22.0
13.0
1 800
227
2 027
1.490
1.646
1.692
41%
1.09
247
158
165
0.0045
0.0030
48
0.0042
211.1
Point 1
Point 2
Point 1
Point 2
Note: These values are provided for
comparison purposes only, and
backward compatability.
Partyka & Ivarsson, 2001, eqn 4
267.0
A second installation using the same duct (different fan) is subsequently measured as follows.
The duct is 800 m long with 150 m of shock losses and has 22 m3/s entering the duct and 15
m3/s leaving the duct. What is the leakage factor on this duct installation? What is the maximum
bursting pressure required by the duct for this application?
Answer
The same DuctCalculator sheet is used. However, the option selected is to calculate the leakage
factor knowing air volumes in and out and the k factor (third button). Enter the k factor as
0.0042 Ns2/m4. After calculation, the program states the leakage factor is 91 and the static
pressure in the duct at its start is 3.3 kPa.
17 August 2006
LeakyDuct Worked Example, © MVA 1999-2006
Page 13 of 21
Select Type of Calculation
Duct resistance and leakage calculator Copyright MVA 2000. All rights reserved.
Key assumptions:
Air density along duct is constant; leakage is not influenced by velocity pressure; ducting is of uniform diameter and quality, FVP is small compared to FTP. All
pressures and flows are treated (and entered) as positive numbers, whether blowing or exhausting.
This program calculates true K factors, etc using the data you provide at the density you provide. It does NOT correct these to std density.
Instructions:
1. Select calculation option (check mark) and fan type option (check mark).
2. Change values in BLUE only. Values in BLACK are outputs.
3. Click "Calculate" button. Sometimes, program may indicate no solution is possible. Press "Calculate" again or try different seed values.
3
kg/m
Air density (assumed constant along duct)
Duct diameter
m
Fan diameter (optional)
m
Duct static pressure at fan (Point 1)
Pa
Duct static pressure downstream (Point 2)
Pa
m
True duct length between measuring points on duct
m
Equivalent length between measuring points on duct
m 3/s
Volume flow (higher value)
m 3/s
Volume flow (lower value)
Fan Static Pressure
Pa
Fan Velocity Pressure
Pa
Fan Total Pressure
Pa
t
m
Pressure Increase Ratio, PIR
dimensionless
Volume Increase Ratio, VIR
dimensionless
Leakage in duct
% of air through fan
Ns2/m8 per 100 m
Duct resistance per 100 m
Ns2/m8 per 100 m
Leakage resistance per 100 m - per formula A
Ns2/m8 per 100 m
Leakage resistance per 100 m - per formula B
2
8
Ns /m per 100 m
Leakage resistance per 100 m - per formula C
Asperity height (roughness) of walls ('e' as in 'e/D' ratio)
mm
Ns2/m4, or kg/m3
Friction 'K' factor for duct @ 1.20 kg/m3
2
2
mm /m
Leakage factor (use this value in worksheet 'Simulator')
Ns2/m4, or kg/m3
Friction 'K' factor for duct @ 1.20 kg/m3
mm 2/m2
Specific leakage area, As, at Ko of 0.6
2
2
mm /m
Leakage factor (use this value in "Simulator")
•
1.20
1.200
1.200
3 294
0
800
950
22.0
15.0
3 294
227
3 522
1.166
1.422
1.467
32%
1.09
1 912
1 311
1 415
0.0046
0.0042
48
0.0042
72.0
Find duct K and leakage factors, by measuring static
pressure in duct at fan outlet (blowing) or fan inlet
(exhausting), and volume flow entering and leaving
duct
Find duct K and leakage factors, by measuring static
pressure and flow rate at two positions in duct
(leakage between positions should be less than 15%
and pressure ratio should be > 2.5)
Find duct leakage factor, by measuring air entering
and leaving the duct and knowing the K factor
Point 1
Point 2
Point 1
Point 2
Find the fan total and static pressure required to
provide a given amount of air at the end of a duct of
known K and leakage factors
Select Fan Type
Blowing (forcing) Fan
Exhausting Fan
WARNING: Duct exit velocity pressure (106 Pa)
too high (>5% fan pressure). Calculations may be
inaccurate.
Calculate
Note: These values are provided for
comparison purposes only, and
backward compatability.
Warning: Uncalculated data!
Partyka & Ivarsson, 2001, eqn 4
91.1
You have been asked to quickly examine the practicality of developing a 2.4 km decline to a
new orebody using a forcing-only system. You need 30 m3/s of air at the end of the duct and can
fit two ducts in the heading. What fan pressure and duct diameters would be required to do this,
assuming a low-leakage duct (k factor of 0.003 N.s2/m4 and leakage of 48 mm2/m2)? Assume an
average air density of 1.25 kg/m3 and allow for 5% shock losses. Assume the duct construction
is rated to 7.5 kPa. Try at least two different duct diameters? Could the job be done with a single
duct? What diameter would it need to be? Find the approximate re-entry time if the drive has the
same parameters as (d) above.
Answer
Again, we can use DuctCalculator sheet to quickly look at the required fan pressures. These are
as follows (per duct).
Duct dia FSP, kPa Q (each) at face, m3/s Q (each) at fan, m3/s
1.2 m
9487
15
32.4
1.3 m
5834
15
30.1
1.4 m
3756
15
28.2
1.6 m
6988
30
51.7
1.5 m
10041
30
53.6
Clearly, the 1.3 and 1.4 m ducts (twin) would be acceptable. If only one duct was possible, then
it would need to be 1.6 m as the pressures for 1.5 m are unacceptable.
Assuming 30 m3/s at the face, the re-entry time would be 102 minutes. Again, this reinforces the
need for airflow in these long headings that exceeds the diesel requirement and/or the use of an
overlap system.
•
You are faced with an urgent auxiliary ventilation job needing two GAL 12 2*37 kW fans in
series (150 kW total) feeding a 1200 mm ordinary vent bag. However, you have one a GAL 12
55 kW and one GAL 12.5 55 kW fan (110 kW total). Create a fan curve for this fan
combination (in series) and duct combination and calculate how far you could use this set-up to
17 August 2006
LeakyDuct Worked Example, © MVA 1999-2006
Page 14 of 21
ventilate a 250 kW loader requiring 0.05 m3/s per kW. Is this a particular suitable fan solution
for this problem?
Answer
The loader requires 12.5 m3/s at the face. We use FanCurveCombiner to first produce the
combined fan curve for two GAL 12 2*37 kW in series (see below). The duct can be up to 1200
m long (with an additional 10% shock losses) before the combination fan goes into stall. At this
length, it is still producing 15.8 m3/s at the face.
The fan curve for the GAL 12 55 kW and GAL 12.5 55 kW in series is now produced (see
below). With ordinary 1220 mm vent bag, the distance that could be gone (assuming no shock
losses and a density of 1.2 kg/m3) is 320 m. At this point, the fan combination is taking 30.4
m3/s of air and delivering 29.3 m3/s. This flow is well in excess of requirements; however,
increasing the duct length puts the weaker fan (the GAL 12) into stall. This is not a suitable fan
selection as the range of overlap flows at which both fans (when in series) are operating
correctly. This can be seen by the wide range of blue flows in the diagram below but the narrow
range of purple flows (the combined fan curve).
LeakySimTM
Copyright MVA 2000. All rights reserved. Enter data in blue only. NOTE: RED values indicate warnings or bad data or invalid/poor results
1. Assumes forcing or exhausting fan or fan combination at end of duct. Fan can be a series, parallel or "Y" combination (fan feeding twin ducts).
2. Assumes air density is constant (compressibility ignored), leakage is influenced by static pressure across duct wall only, duct is of uniform diameter and quality (K and leakage factors are constant along duct [although
friction losses and leakage itself are NOT constant]), and velocity pressure at duct outlet (forcing fans) or duct inlet (exhausting fans) is small compared to fan total pressure.
3. When leakage exceeds 85% of fan airflow, result is unreliable.
4. Whilst very unlikely in most circumstances, the pressure loss to push the air back out of the heading may be significant on rare occasions and should be checked.
5. Note that a more powerful fan or lower leakage factor may not improve a solution, e.g. a lower leakage factor will increase the fan pressure perhaps to a point beyond the fan's pressure capability (i.e. goes into stall), and a
more powerful fan may have its surge point (min flow value) at a flow that is too high for the duct (i.e. it also goes into stall), etc. In both cases, the 'commonsense' solutions to improve the fan or duct performance may result in p
6. To find the airflow returning from the face at any point in the auxiliary compartment, use the "Predicted return air flow outside duct". This is helpful in determining how far diesels can go in a heading.
7. It is strongly recommended that you use real fan data and real duct friction, leakage and shock loss factors for any simulation.
Also you should consider a range of values (i.e. look at the sensitivity of the solution to different fan and duct characteristics) before making a final selection.
INPUTS
OUTPUTS (per duct)
DUCT INPUT DATA
Duct Length
Length with shock losses
Air density
Duct (NOT fan) diameter)
K factor of duct at 1.2 kg/m3
m
m
3
kg/m
m
Large diameter, 100 m lengths vent bag (Protan or equiv) - 0.00300
2
Ns /m
Duct leakage factor
OR manual entry
4
Protan or equiv duct, 100 m lengths, typical - 48
2
2
mm leakage area per m area (including effects of duct joins, etc)
FAN INPUT DATA
GAL 12, 2 x 37 kW, 2 fans in series (150 kW)
Type of fan
Electrical power cost
$/(kW.hr)
Approx fan efficiency
% based on FSP, assumed constant
OR (instead of providing known fan type, provide fan curve constants)
Parabola A constant (from FTP curve, NOT FSP curve constants)
Parabola B constant (from FTP curve, NOT FSP curve constants)
Parabola C constant (from FTP curve, NOT FSP curve constants)
Maximum Q on curve, cms, (optional)
Minimum Q on curve, cms (optional)
17 August 2006
Fan Total Pressure
Pa
3 708
1 200
Fan Static Pressure (FSP)
Pa
3 480
22.0
1 320
Fan Airflow per duct
m3/s
1.200
7.2
Ns2/m8
Fan Resistance at 1.2 kg/m3
15.8
1.20
Airflow out of/into duct
m3/s
0.00300
Leakage/fan Q
%
28%
3
6.3
0.0020
Leakage per duct
m /s
48
Leakage/Face flow
%
40%
7
Duct velocity pressure
Pa, %FTP 117, 3%
Maximum FSP this fan
Pa
7 773
3
Fan free-delivery flow per duct
25.8
m /s
Fan Q as % of Qmax
%
86%
$ 0.08
FSP as % of FSPmax
%
45%
70%
Approx elec cost $/day/duct $/day/duct
105.44
3
3
0.28
-16.99
cents/(m /s)
Approx cost/(m /s)
Predicted return air flow outside duct
749.75
992.76
Distance from fan
m
500
3
17.9
38.5
Nett return this point/duct
m /s
22.5
Duct static pressure this pt
Pa
1,184
LeakyDuct Worked Example, © MVA 1999-2006
Page 15 of 21
Pa
Fan curve combiner Copyright MVA 2000. All rights reserved. Enter data in blue only.
1. Select "Fan combination type required".
2. Select Curve type required: FTP means output constants are for fan total pressure, FSP means ouput constants are for fan static pressure
3. Select fan(s). Note "manual entry" of fan constants etc is not possible on this sheet. Fan must already exist in database.
4. To create multiple combinations of fans, add the first combination, then store the resulting fan constants in the database, then add this as a new fan to another set, etc.
5. To print or preview the chart, select the button.
Series
1
Fan combination type
MaxQ1
40.0
845
-657
845
S
1
Fan TOTAL Pressure
Series
Calculate
Curve type required?
FTP
1st qtr
31.8
1393
447
Fan
Fan curve: FTP
- Series configuration.
Density
= 1393
1.20 kg/m3.
3
Air density for inputs, kg/m
Fan and
STATIC 2nd qtr
1.20
Parallel
23.5
1834
1316
1834
Values
3
Air density for chart plot, kg/m
1.20
Y (feeding twin ducts) Chart
3rd 3500
qtr
15.3
2168
1949
2168
18
MinQ1
7.0
2394
2348
2394
Fan 1 at input air density
Fan name Korfmann AL 12 550 5 d, 55 kW, single
12 550 5 d,
0.0
3000
a1 constant
MaxQ2
43.8
1635
110
1635
-1.2591
b1 constant
-9.8110
1st qtr
40.3
1838
543
1838
Print or
Store
2500
c1 constant
2nd qtr
36.9
1994
910
1994
2501.0000
Preview
Combo fan
Chart
in d'base
Max Q1
40.0
3rd qtr
33.4
2105
1214
2105
2000
Min Q1
P-Q curve for c MinQ2
30.0
2170
1453
2170
7.0
FVP
P at Max Q1
1 AL-12.5 55 kW (single)
94
P at Min Q1
2371
MaxQ3
40.0
1948
673
1948
637
1500
Warning: Uncalculated data!
Fan diameter, m
1st qtr
37.5
2331
1211
2331
560
1.20
19
2nd qtr
35.0
2675
1699
2675
488
Fan 2 at input air density
1000
Korfmann AL-12.5 55 kW (single)
Fan name
3rd qtr
32.5
2979
2138
2979
421
a2 constant
MinQ3
30.0
3243
2526
3243
359
Fan combination at input density
-2.3315
500
b2 constant
103.7096
a combined
-3.5906
c2 constant
b combined
93.8986
Fan curve: F
0
798.1506
0
Max Q2
c combined
3299.1506
0
43.8
Min Q2
30.0
Max Q combined
40.0
0
10
20
30
40
50
P at Max Q2
Min Q combined
30.0
1643
Korfmann AL 12 55005 d, 55 kW, single
Korfmann AL-12.5 55 kW (single)
m3/s
P at Min Q2
10771
P at Max Q combined
1310
0
P-Q curve for combination
Fan diameter, m
P at Min Q combined
2885
1.25
17 August 2006
LeakyDuct Worked Example, © MVA 1999-2006
Page 16 of 21
The following screen (Figure 6) should be shown when first selecting the “BranchingDucts” tab.
17 August 2006
LeakyDuct Worked Example, © MVA 1999-2006
Page 17 of 21
Figure 6
Press the “Remove errors” button. You should obtain the following screen (Figure 7).
Figure 7
Change the end condition of the first branch to “Tee” rather than blank by selecting “Tee” from the
drop down menu on the L19 cell (Figure 8). After a delay of 20 to 30 seconds, you should see the
following screen (Figure 9). Note that the message box should change to “Valid solution”. Note that
the “branch discharge” condition can be selected from: “Blank” (blanked or tied off), “Discharge”
(open to atmosphere), “N/a”, “Inline” (allows two ducts of different diameters or other parameters
to be joined), “Tee” (a branch), “Offtake” (a branch) and “Y” (a branch). At present, selecting
“Tee”, “Offtake” and “Y” are identical, although this may be changed in future versions to reflect
the different shock losses that may be applicable for these types of connections.
17 August 2006
LeakyDuct Worked Example, © MVA 1999-2006
Figure 8
Figure 9
Figure 10
17 August 2006
Page 18 of 21
LeakyDuct Worked Example, © MVA 1999-2006
Page 19 of 21
Now change any of the “branch data” values (data in blue) and the “branch end” conditions (the
drop down menu at the end of each branch) to create the system of branches and discharges that you
want. You can also select a different type of fan from the fan drop-down menu at the top of the
screen. If you want a fan that is not in the list, enter a new fan curve using the “Fan curve
calculator” or “Fan curve combiner” worksheets.
If an error develops at any point, press the “Remove errors” tab. This is usually due to either the fan
not being suitable for the system you have installed (or vice-versa). Pressing “Remove errors”
basically blanks off the first branch but leaves all the other data alone. With the errors removed, you
can then select a different type of fan and/or change the duct parameters. Then just “open up” the
branch end for the first duct again and wait for the calculation to end. If you do not get a valid
solution, then press “Remove errors” and try another combination. If at any time the program stops
calculating, press “Force calculation”. If the calculation time is too long, look at the “Calculation”
settings in Tools>Options (Figure 11) and set the “maximum change” to 0.001. Making this number
smaller (e.g. 0.000001) will force the program to iterate until the errors are much smaller, but this
will take much longer.
Figure 11
Example: Find the flows for the following system:
A fan has 8 branches feeding each of 8 drawpoints (1 to 8) in a sub-level cave. The fan is a GAL
12.5 (2 x 55 kW). Ducts A and B are 80 m long each with A being 960 mm Φ and B being 1085
mm Φ. Ducts C and D are joined at point E with a change in size (only). Duct C is 1.22 m Φ and
duct D is 1.085 m Φ. All ducts have a k factor of 0.004 Ns2/m4 and a leakage factor of 451. The
drawpoints are 15 m apart and the fan is 10 m from the first drawpoint.
What is the flow through the system if ducts 3 and 8 are open and all other ducts are tied off.
17 August 2006
LeakyDuct Worked Example, © MVA 1999-2006
Page 20 of 21
Figure 12
8
7
6
5
4
3
2
1
B
B
B
B
A
A
A
A
D
E
C
The network can be redrawn as follows. Note that the distance between the fan and the first Tee
(#3) is now 40 m, from that Tee to the change in size (from 1.4 to 1.22 Φ) is 15 m and it is then a
further 60 m of 1.22 m Φ duct to meet drawpoint #8.
Figure 13
8
3
B
A
D
E
C
The solution is shown in Figure 14. The fan draws 34 m3/s at 1570 Pa FSP with 19 m3/s exiting the
first open duct (drawpoint #3) and 11 m3/s exiting the last drawpoint (#8).
17 August 2006
LeakyDuct Worked Example, © MVA 1999-2006
Figure 14
17 August 2006
Page 21 of 21