16

INTRODUCTION TO MINE VENTILATING PRINCIPLES AND PRACTICES

FUNDAMENTALS OF AIRFLOW

The principles of airflow are:

1. Airflow in a mine is induced by pressure differ-
ences between intake and exhaust openings.

2. The pressure difference is caused by imposing
some form of pressure at one point or a series of points
in the ventilating system.

3. The pressure ‘created must be great enough to
overcome frictional resistance and shock losses.

4. Passageways, both intakes and returns, must be
provided to conduct the airflow.

5. Air always flows from a point of higher to lower
pressure.

6. Airflow follows a square-law relationship between
volumes and pressures, that is, twice the volume re-
quires four times the pressure.

7. Mine-ventilating pressures, with respect to atmos-
pherie pressures, may be either positive (blowing) or
negative (exhausting).

The common method of measuring ventilating pres-
sures producing circulation are either equivalent feet
of air column or equivalent inches of water. In most
calculations, pressures must be reduced to inches of
water, where 1 inch of water equals a pressure of
5.2 pst.

8. The pressure drop for each split leaving from a
common point and returning to a common point will
be the same regardless of the air quantity flowing in
each split.

PRESSURE LOSSES

Pressure losses are divided into two separate
groups:

1. Friction pressure losses caused by the resistance of
the walls on the airstream. Friction losses therefore
depend upon conditions and roughness of individual
wall surfaces and velocity of the air.

2. Shock pressure losses caused by abrupt changes in
the velocity of air movement. Shock losses therefore

are the result of changes in air direction or of airway
area, obstructions, and regulation.

All formulas for pressure losses include an
empirical factor, the value of which varies with
conditions and is determined by actual experi-
ment. This empirical factor, commonly called
K (the coefficient of friction), unless approxi-
mately correct for the conditions, will give
erroneous results.

Table 4 lists the Bureau of Mines schedule
of friction factors for mine airways.

The common pressure-loss formula for mine
ventilation is:

KLOV?
A

P: at standard density of 0.075 p.c.f.

where P is the pressure loss, pounds per square
foot,
& is the friction factor,
L is the length of airway in feet,
O is the perimeter of the airway feet,
V is the velocity, feet per minute,
Ais the cross-sectional area of the airway
in square feet.
To change the pressure loss P, pounds per
square foot, to H, pressure loss in inches water,
the formula is divided by 5.2 and becomes

2
i: KLOV". The formula is changed to in-
clude the quantity of airflow as follows: The
quantity of airflow q is equal to the velocity

Taste 4.—Bureau of Mines schedule of friction factors for mine airways '

‘Values of K?
Sinuous or curved
Straight
Irregularities of
‘Type of airway faces, areas, Slightly Moderately ‘High degree
and alinement

[Slightly] Moder-| \slightly| Moder-| [Slightly] Moder- Slightly) Moder-

Clean | ob- | ately ‘ob- | ately ‘ob- | ately ob- | ately

(basic | struct-| ob- | Clean | struct-| ob- | Clean | struct-} ob- | Clean | struct- | _ob-

values)| ed | struct- ed | struct- ed | struct- ed | struet-

ed ed ed ed
‘Smoothlined. 10 15 25 20 25 35 25 30 40 35 40 50
15 20 30 25 30 40 30 35 6 40 45 55
20 25 35 30 35 45 35 0 50 45 50 60
Sedimentary (or coal). ...| Minimum. 30 35 45 40 45 55 45 50 60 55 60 0
Average 55 60 20 65 0 80 70 75 85 80 85 95
Maximum 70 75 85 80 85 95 85 95} — 100 95] 100 10
‘Pimbered (5-foot centers).| Minimum. 80 85 95 90 95} 105 95} 100) 10] 105] 0 120
Average. 95] 100! 10} 105} 110| 120) 90] 5] 125] 120] 125, 135
Maximum | 105] 10) 120} 115] 120} 130) 20] 125] 135} 130) 135, 45
Igneous rock Minimum... 90 95} 195| 100/ 105] 115] 105/ 0} 120] us| 120 130
Average 145| 150) 160) 155] 160! 165) ~ 160] 165] 175] 170] 175, 195
Maximum 195} 200| 210) 205] 210} +220 | 210] 215] +225] 220] 225, 235

1 McElroy, G. E., Engineering Factors in the Ventilation of Metal Mines: Bureau of Mines Bull. 385, 1935,

p. 43.
# All values of Aare for air weighing 0.075 p.c.f. Values in table 4 are expressed in whole numbers but must be multiplied by 10-!# to obtain the

proper K value.


Shaft A

Intake

NATURAL-VENTILATION INFLUENCE OF AIR DENSITIES

El. 1,000’
Barometer, 28.95” Hg
Temperature
Wet bulb, 45° F.

Dry bulb, 50° F.

Shaft B

Return

Mine

El. 300’
Barometer, 29.80 Hg
Temperature
Wet bulb, 53° F.
Dry bulb, 59° F.

37

El. 1,200’
Barometer, 28.75” Hg
Temperature
Return, 63° F. sat.

El. 500’
Barometer, 29.60 Hg
Temperature
Wet bulb, 65° F.
Dry bulb, 65° F.

Ficure 25.—Example of Conditions Causing Natural Draft.


MECHANICAL VENTILATION

make up the total air quantity of 300,000 c.f.m.

Figure 22 shows parallel fans operating on
separate airways. This problem is more com-
plicated than the previous performance curves
for parallel operation.

This problem involves placing a new fan in
service and splitting the duty previously
handled alone by fan B. Fan A is installed with
a separate return segment connected to the
mine ventilating system at junction X.

The natural splitting and pressures at which
the fans will operate can be determined graph-
ically if the resistance factors for separate seg-
ments of the system are known or can be
calculated.

The procedure follows:

1. Plot both fan characteristic curves on one
graph.

33

2. Plot resistance curves or each segment
from formula H= RQ’.

3. Determine the amount of air desired, for
example, 200,000 c.f.m., and determine whether
this quantity of air will flow under prevailing
conditions.

A. Since segment c will handle all air
going to segments a and 6, from the plotted
resistance curve c determine the pressures
necessary to move 200,000 c.f.m. This is
4 inches water pressure. Should the
original air quantity desired have been
250,000 c.f.m., this quantity strikes the
resistance curve of segment ¢ at approx-
imately 6.7 inches water. Obviously, this
is beyond the limit of fan A at the char-
acteristic curve shown. Should the original
air quantity desired have been 150,000

200,000 c.f.m. 125,000 c.f.m.
— — Fan B
Segment c Segment b

Fan A

PRESSURE, INCHES OF WATER = (H)

Q, fanA

La, fanB
!
L |

A
Segment b Fan

1

100 150

250 300

VOLUME, THOUSAND C.F.M.

Ficure 22.—Parallel Fans Operating on Separate Airways.


42

Intakes:
lentry rake
= a= 18 (ase 1.5).
5 entries R= Gi=po=0.06.
Returns:
J entry BOK 1.85 (use 1.9).
4 entries=

Leakage factors must be considered. With
new development and two rows of stoppings
between intakes and returns, a leakage factor
of 80 percent of the air reaching the face is fair;
this, of course, could be substantially more or
ess.

The total air quantity required for five oper-
ating units will be:

INTRODUCTION TO MINE VENTILATING PRINCIPLES AND PRACTICES,

The simplest way to calculate an air-current
circuit is to use a standard computation sheet.

EXPLANATION OF COMPUTATION SHEET

Segment: Represents portion or increment of
ventilation split studied; established at points
of splitting or airway changes.

Distance: Distance of segment, in feet.

Number of entries: Number of entries in seg-
ment.

Area: Average area of individual entries listed
for reference.

R 1,000 feet: Resistance factor established for
segment per 1,000 feet.

R total: Total R factor; equals distance divided
by 1,000 times # factor per 1,000 feet.

Q,: Air quantity at beginning of segment, di-
vided by 100,000.

Q:: Air quantity at end of segment, divided by
100,000.

Gains or losses: Gains or losses from leakage
or other causes.

Average Q: Average air quantity flowing in
segment, in units of 100,000 ¢.f.m.

Chm. H segment: Pressure drop calculated for seg-
& units at 25,000 o£. per pit. 125, 000 ment H=total RX average Q’.
entilation, idle develo tts - , :
Leakage, 80 percent 545,000 118, 000 Total H: Sum of segments H to aggregate
Ventilation for shops. -__.. 10, 000 pressure drop. . .
ae Figure 26 shows the air quantities and is
Total. _..-----------.---------- ‘271,000 used to explain the computations. The South
' Round off to 270,000 c.f.m Main split was divided into segments A, B,
LEGEND
— Overcast
<+—~ Intake air idle N.
<+— Retum air =_—o
1 unit E
: 3 Lunit
3 i . & & 8
3 —E E 3 25 "o
: 3 =
3 38 38 a8 58
3 8 ES
8 3S aa
q 3
3 50,000 c.f.m. TM 70,000 c.f.m. \ a
—- cans = _
¥ 50,000c.fm.. J —_—— J }130,000c4m. 4 J] — 2 2
an) y C 95,000 c.f.m. 5 B 160,000 ¢.f.m. a) 100,000 c.f.m. ] =
5,000 c.f. 60,000 c.f.
45,000 c.f.m. ,000 c.f.m.
” Noe,
Leakage Leakage Leakage "ON
| 10,000 c.f.m. 20,000 c.f.m. 30,000 c.f.m. | 6,
1,500" 1,500 2,000" %
| | unit

Fieure 26.—Mine and Ventilation Plan.

Problem 1.


10

INTRODUCTION TO MINE VENTILATING PRINCIPLES AND PRACTICES

\
|

AY Az «—— As Z ++ + K
Ss b>
y + €
S| 3
ine 25" >| & + S
S|
LE as
A,=107 sq. ft. — Ap=92 sq. ft. A3=105 sq. ft. t
|
Average area= 101 sq. ft. SECTION SHOWING
QUARTER POINTS
Figure 7.—Example of Smoke-Cloud Measurement.
Total pressure
S|
Static
pressure
— Is
Y ‘ WZ

Ne

Ficcre 8.—Illustration of Pitot Tube.

the other, with the end bent at right angles to
the shaft. The inner tube is open on the end
and receives the total pressure of the airstream.
The outer tube is closed on the end and receives
the static pressure of the airstream through a
number of pinholes set back from the end. The
tip is tapered or made hemispherical to reduce
interference with the airstream, The opposite
end of the pitot tube is provided with two
fittings, one for each tube, to which the legs of
the water gage are attached. When the legs of
a water gage are attached to the pitot tube, the

difference in pressures between the inner and
outer tubes, as registered by the U-tube, is the
velocity pressure. The tapered end of the
pitot tube must be pointed against and parallel
to the airstream. Figure 1 shows how the pitot
tube and water gage are used to determine static
and total pressures in addition to velocity pres-
sures in ducts or fan housings. Since velocity
pressures ordinarily are small (a velocity of 2,000
f.p.m, at standard air density equals }-inch
water gage), the use of the pitot tube is limited
to special high-velocity determinations in ducts


52 INTRODUCTION TO MINE VENTILATING PRINCIPLES AND PRACTICES,

Points 7-8:

H=0.60 inch water from gradient,

q=7,500 c.f.m. average,

Points 8-9:
H=0.40 inch water from gradient,
a= 10,000 c.f.m. average,

0.40 _ 0.40

=@.107 0.017 0

Proposed new shaft, 10 feet in diameter, un-
lined in sedementary ‘rocks. K=85 (see table
5), 1=500 feet:

A=78.5 sq. ft.,
0=314 tt,

KLO_ _85X500X31.4
52M 5.2XT9XT9X79

CALCULATIONS

Air quantity required:
‘ace ventilation, 10,000X2__ _-
Leakage, 100 percent X 20,000
Ventilation old works

R:

=0.52.

Total

‘Segment

Riactor| Os

Qs | age

age | ge H

0.34 | 0.50 | 0.50 0.09
782) 240 | .38 05
255] 138] °35 35
z02| 135 | 25 +26
5.49 | 25] 120 229
107.00 | £19 | 05 “64
40.00 | 21%} °25 1.60
+62 | 8.50) 50 |. et

} AIT quantity 1 active face; other aetive ince assumed to be similar,
? Includes retin from 2 active
1 Jncludes al return and leakage,

The new horsepower will be:
hp.=3.41 X50,000--6,350= 26.8.

Calculate new mine resistance:

H=R@, .H=3.41, Q=0.50,
H_ 341
R= GW Apt OF

Using the new R factor (13.64) and the

H=RQ?, calculate several pressure values
using different quantities:
Volume Q R Pressure
H=R'
ie
13.64 Ev
4.13

Plot these data (fig. 31) to obtain the new
mine-characteristic curve.

The intersection of this curve with the present
fan characteristics shows the benefit derived by
decreasing the mine resistance. Instead of
45,000 c.f.m. at 5.5 inches, the fan would now
deliver approximately 60,000 c.f.m. at 4.90
inches. However, since the mine requires only
50,000 c.f.m. at 3.41 inches, the new fan speed
can be calculated by using the fan laws for
speed and pressure or quantity; that is, pres-
sure varies as the square of the fan speed, or
quantity varies directly as the fan speed.

Calculation with pressures:

H,_(S:\ 4.90_(1 00

Hy "341

S= (1,400)? OPT en
=1,400X0.834,

=1,170 r.p.m. (approx.),


INTRODUCTION TO MINE VENTILATING PRINCIPLES AND PRACTICES.

50

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+ TO €—2
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48 INTRODUCTION TO MINE VENTILATING PRINCIPLES AND PRACTICES

Influence of natural draft at 0.5 inch water

Volume, e.f.n. H,com- | H,n.v.p. | H, total
Dined
0.50 12.86
= 80 12.04
50 1Lo1
250 acy

Plotting these data forms the total pressure-
volume relationship of all fans, plus natural
draft (fig. 28).

The mine pressure H is 3 inches water at
100,000 c.f.m. at standard density. Therefore,
the pressure at a density of 0.070 p.cf. i

. is
0.070 :
3X9 975728 inches water.

Calculate the mine-resistance curves as
follows, from the formula H=RQ?, where
= 2.8:

Pressure,
inches

Air quantities, ¢.f.m ‘water

001 81
180,000 91
190,000 10.1
200,000. 112
210,000... = 123

Plot these data on figure 28. The operating
point where the mine-resistance curve intersects
the curve for combined fans and the natural
draft indicates the air quantity expected at the
fan pressure.

The total air quantity is 188,000 ¢.f.m. at a
combined pressure of 10 inches water:

Total air quantity, 188,000 c.f.m.

Total air quantity, 188,000 c.f.m.
Fan A.

Fan B- 37
Combined 95
Natural draft. - 5

Total water pressure.____---____.. 10.0

PROBLEM 4: IMPROVED FACE VEN-
TILATION BY MEANS OF NEW AIR
SHAFT

Figures 29 and 30 show the line diagram and
pressure gradient of a small slope mine as de-
termined by a pressure survey. Overall air-
leakage losses and resistance to airflow in the
returns are excessively high. Of the 45,000
c.f.m. intake air, only 10,000 c.f.m. is available
for face ventilation. Approximately 78 percent
of the pressure losses was found to be in return
aircourses and in the 6-foot-diameter unlined
upcast shaft.

The fan (fig. 31) was operating at 5.5 inches
water gage and a speed of 1,400 r.p.m., inducing
an airflow of 45,000 c.f.m. The air horsepower
was calculated to be 39. The input horsepower

was 63. Fan operating efficiency was 62 percent.

Assume that a new unlined airshaft, 10 feet
in diameter, will be sunk 500 feet through sedi-
mentary rocks to the coalbed at pressure-
gradient point 9. What fan fressure will be
necessary to provide 20,000 c.f.m. intake air to
the face areas to ventilate two active longwall
faces? What fan speed will deliver this quan-
tity under improved conditions?

Because present returns are connected to old
workings, most of the present return aircourses
cannot be used as parallel intakes (see fig. 29).
The present upcast shaft, however, will become
downcast and will connect to the intake system
at point 3 through the old coal slope, which is
not connected to old workings.

To ventilate the old returns and bleed old
workings, approximately 5,000 c.f.m. intake air
will be regulated into the parallel returns on
each side of intake entries near point 3.
Normal air leakage is expected to be 100
percent of face requirements.

The air density will be at standard 0.075 p.c.f.

The first step is to determine resistance
factors for the proposed air circuit. From the
pressure gradient (fig. 30), both the air quan-
tities and pressure losses between points can
be determined.

CALCULATION OF R FACTORS

1. Intake through slope and rock slope to
point 3.

2. Intake from old shaft and coal slope to
point 3.

The resistance factors are calculated from

the common formula H= RQ’, RH, where Q

is the actual air quantity c.f.m. divided by
100,000.

From the pressure gradient, the pressure loss
H through the rock slope to point 3 is 0.12 inch
for 45,000 c.f.m.;
therefore

R. 0.12 0.12

(0-45) 0.2025

The resistance factor for the second intake
can be determined from the present return
system.

The resistance factor for segments 10-11 and
11-12 should be calculated separately and
added for total resistance.

Segment 10-11:
H=4.60—4.30=0.30 inch water,
q=45,000 c.f.m.,

0.30 __ 0.30
(0.45)? 0.2025

==0.59.

R: ==1,48-


20 INTRODUCTION TO MINE VENTILATING PRINCIPLES AND PRACTICES

1 52,500 c.f.m.
150,000 c.f.m.
A 150,000 c.f.m. B 2 48,000 c.f.m. D Fan
3 49,500 c.f.m.
——
0
~ A
g oi 1 All splits
o<
Js
Fare 1.5” B
Zo 2 ————_. =
an C2.5
gu
wo
ceo 3
az
7 D
4
PRESSURE GRADIENT
as
& Fan-pressure volume
= 4 characteristic
3 ai
a 3r
x Mine |
3 o| resistance A
al HH
& Bili8
gir alll
a SS
& t n 1iti t Ll n

4 8 12 16 20 24 26
VOLUME, THOUSAND C.F.M.

Figure 11—Example of Airflow Splitting.


38 INTRODUCTION TO MINE VENTILATING PRINCIPLES AND PRACTICES

Average density

1.327
="Fo9 X[29-70— 0.378X0.465)],

=0.00254 X 29.52,
=0.0750 p.c.f.

WEIGHTED AVERAGE OF DENSITIES
SHAFT A

200 feet of surface density
=200X0.0748= 14.96

700 feet of shaft density
= 700 X0.0755= 52.85

Average density =" To, 0753 p.c.f.,

00
column A.

SHAFT B

700 feet of shaft density
=700 X0.0733=51.31

200 feet of mine density
== 200 X0.0750= 15.00

66.31.

Average density =O) =0.0737 p.c.f.,

column B.

The difference between the average densities
of both columns:

Column A__--- p.c,f. 0. 0753
Column B-___-_ p.c.f. 0.0737

Difference__._ p.c.f. 0. 0016

The height of the air column was 900 feet.
900 X 0.0016= 1.44 pounds pressure;
1.44--5.2=0.28 inch water pressure.

The natural draft under conditions shown
would work with an exhaust fan. Under other
conditions of temperature this pressure could
work against the fan.

The amount of air that would flow froin this
natural draft would depend on resistances of
the mine and shafts. Assuming that the mine
resistance was 2, the flow would be calculated

by HRY, C=F

wie

@ O28 0. 140,

q=v0.140 X 100,000,
= 37,500 e.f.m.

Calculating the densities without considering
the effect of vapor pressures, the results are as
follows:

Shaft A: aie Sa temperature, 54.5° F.

Average Py ometer, 29.38 inches

5

Density= p72 x29. -38= 0.0758

p.cf.

Shaft B: Average temperature, 64° F. dry
ulb ;

Average barometer, 29.18 inche

hg;
Density =432" x 29.18=0.0739

p.cf.

Surface: Temperature, 50° F. dry bulb;
Average barometer, 28.85 inches hg;

Density=1222 728. 85=0.0751 p.c.f.

310
Mine: Average temperature, 62° F. dry bulb;

Average barometer, 29.70 inches hg;
1.327

Density=-355 X29. .70=0.0755 p.c.f.
The weighted average is:

Column A=200X0.0751=15.02 surface 68.08
700 X0.0758=53.06 shaft A _

68.08--900=0.0756 p.cf average density.
Column B= 5.10 mine
1.73 shaft BS °0-83-
66.83+-900=0.0742 p.cf. average density.
Column A- p.c.f. 0. 0756
Column B-._.- p.c.f. 0.0742
Difference_..___ p.c.f. 0. 0014

0.0014 <900= 1.26 pounds pressure;
1,.26--5.2=0.24 inch water pressure.

Considering all variables, this result is
probably as accurate as the computed 0.28
inch water pressure of the first calculation.
Both methods were given for illustration and
comparison. For mines deeper than 1,500 feet,
the influence of vapor pressures is important,
and must be considered.


INTRODUCTION TO MINE VENTILATING
PRINCIPLES AND PRACTICES’
By
D. S. KINGERY?*

Introduction

JHE IMPORTANCE of the ventilating system in coal mining is recognized

universally, and standards for ventilating requirements have been estab-

lished by State mine-inspection departments and the Federal Coal Mine
Safety Act.

Mine management, of course, is responsible for the effectiveness of the
ventilating system and for compliance with State and Federal laws. Few
companies employ ventilating engineers; the engineering department may plan
the mine development and the ventilating system, but it falls upon the mine
superintendents and the foremen to assure adequate ventilation for the working
sections.

The fundamentals of airflow are taught in mining schools, and every mine
supervisor must prove a working Knowledge of these fundamentals before he is
certified by the State. Most mine foremen, therefore, along with their practical
knowledge, have a working knowledge of certain basic ventilating formulas.
However, many ventilation problems require solution by more complex formu-
las, changing certain ventilation problems to mathematical exercises.

This bulletin is written to explain in layman language the basic laws and
fundamentals of mine airflow and their application to the solution of common
ventilating problems. It presupposes that the reader will possess at least an
elementary knowledge of mine ventilation and of arithmetic and algebra. Our
attempt is to reduce the complex and difficult ventilating formulas to simple,
practical fundamentals.

We have taken certain liberties with well-known theories of airflow princi-
ples, believing that it is unrealistic to apply higher mathematics and refinements
to ventilating problems, the data of which are based upon assumption or relative
field measurements of air quantities and pressures. For example, in our ve-
locity-pressure conversions to velocity, we will use 4,000 f.p.m., rather than the
accurate 4,008 feet, as equivalent to 1 inch water pressure. In calculations we
will use averages rather than integrate between limits.

This bulletin was prepared for use by mine personnel interested in and
responsible for mine-ventilation operations.

ACKNOWLEDGMENTS

The author has drawn from knowledge gained piecemeal from many
sources—books, periodicals, publications, Bureau of Mines records, and prac-
tical experience. Special assistance was given by members of the Ventilation
Group, Federal Bureau of Mines, and the several qualified ventilating engineers
who reviewed the manuscript and submitted constructive suggestions. ecial
acknowledgment also is given to G. E. McElroy, former Bureau of Mines
ventilation engineer, with whom the author worked closely for several years.

! Work on manuscript completed July 1959,
? Director, Health and Safety Research and Testing Center, Pittsburgh, Pa.


MECHANICAL VENTILATION

31

10 T T

Pressure, fan B

PRESSURE, INCHES OF WATER
a

Fan A

Pressure, fan A

Mine system

Combined

1

0 50 100

150 200 250

VOLUME, THOUSAND C.F.M.

Ficure 19.—Combined Characteristics of Two Different Fans Operating in Series.

system and the combined fan characteristics

letermine the pressure; and, since the pressure
must be the same for both fans, this pressure
determines the relative quantities that each
fan will produce.

If the mine-resistance curve does not intersect
the combined characteristics, the higher pressure
fan will handle more air alone than the two fans
together, and if an attempt is made to operate
them in parallel, the high-pressure fan will
reverse air through the low-pressure fan.

Fans that have steeply sloping characteristics
act together in parallel. Those fans that have
characteristics combining a comparatively flat
and steeply sloping part must be operating on
the steeply sloping part of the performance

range. Otherwise, sudden changes in_ the
system resistance could cause the higher
pressure fan to assume all the load.

A common method of installing fans in
parallel is to place them on separate airways.
Where the resistances of the separate airways
are comparatively large, little operating trouble
is experienced. However, when the separate
airway resistance is small, it is difficult. to
determine the resistance against which the fans
must operate. Consequently, the character-
istics of the separate, as well as of the combined,
airways must be considered. .

Figure 20 illustrates results with two identical
fans operating in parallel. Compared with a
single fan, pressures and quantities have


FUNDAMENTALS OF AIRFLOW 17

times the area (g=VA); V therefore equals

4 and substituting this in formula it now be-

KLO”.
5.2.A8

This formula is in general use; however, it is
difficult to use because the value of K (friction
factor) is usually expressed in decimals, such as
(100X10-"°) or 0.0000000100. Changing the
value of g by dividing it by 100,000, greatly
simplifies the formula and makes it much easier

LO. ¢
A®* (100,000)

q now becomes Q, the air quantity divided by
100,000, and the value of K, for example, can
be changed from 0.0000000100 to [0.0000000100
X (100,000)? 100], the easily used whole num-
ber given in table 4, when Q is expressed as the
decimal part of 100,000.

For general and easy application, the formula
can be reduced still further, and F the resistance

comes H==

touse. Now H=

(100,000)?;

factor of the airway substituted for the

now making the formula H=RQ.
where H is the pressure loss in inches water,
RF is the resistance factor of airway or
mine,
Qis the quantity of flow, c.f.m., expressed
in units of 100,000.
When working with air densities other than
standard, correction must be made for den-
sity variations, and the formula becomes

KL
3.24”

d

— Rex ——

H=ROXo 075

where

d=density of the air.

This formula, H=RQ’, is simple to use, and
once the individual is familiar with it, it can be
used practically for all calculations, as will be
shown by subsequent examples.

PARALLEL FLOW IN AIRWAYS

The advantage of multiple entries, both in-
take and return, to improve ventilation effi-
ciencies is understood by most mining engineers.
Actual comparison of pressure losses between
single and multiple entries can be shown from
KLO.
5.2A8
tional entries are of similar size, area, length,
and friction factor, by doubling the entries the
KL, 2(0)

52 *@ay

the formula R= Assuming that addi-

new resistance factor R? becomes =>

since the only changes are in doubling the

perimeter and area; therefore, the formula
becomes:

R= Kr 2(0)_KL,, 0 —K1Oy.

~ 8A 5.244? 5.248

The resistance of two entries is therefore one-
fourth that of a single similar entry.

The formula can be developed further to
apply in all instances by using n to represent the
number of entries.

KL, nO _KLO.,n

Re 3a Xa 52
KLO
R= =FaeX

KLO

Since the SG 5a

same for similar entries, the formula can be re-
1

duced to R= GR.
nr

resistance will be in ratio to the number of en-

@

EXAMPLE: What will be the resistance chang-
ing from a single entry to four entries?

or R theoretically will be the
This means that the new

tries squared, that is,

m=1,m=4. The ratio is 4 or 4,

1 1
Rye BXhi=ygXRi-

When multiple entries are used, the same
simple relationship applies.

EXAMPLE: If the resistance factor for three en-
tries is 2,, what will be the resistance factor for
five entries?

TM=3,%=5. The ratio is 5

3
(3)
BXAMPLE: If the situation were that five entries
are available but two must be stopped off, the

Br=aGX R= “ae ~(35x2)}

ratio would be 2

XR=t Xk

1 =
yas
25

R=

Gen)



24 INTRODUCTION TO MINE VENTILATING PRINCIPLES AND PRACTICES

right, 60-percent efficient inby the overcast;

total intake, 130,000 c.f.m.:
e.fim, through last

4 right split: breakthrough
30,000 c.f.m. X0.60__-------------- 18, 000
8 right split:
25,000 c.f.m. X0.60.
Development.__
Total effective air... -_---_-. 43, 000

43,000 + 130,000 X 100 = 33-percent. efficiency,
67-percent loss.

In this study the large leakage losses on the
mains were due to the high-pressure differential
between intake and return. Five center entries
were used for intake and seven entries as
returns.

MECHANICAL VENTILATION

The application of fans to induce airflow is
common to most coal mines. Modern mine
fans are the axial-flow type; however, there are
many centrifugal fans still in use. The general
fan laws are the same for either axial flow or cen-
trifugal fans. The only differences are indi-
vidual characteristics of power, pressure, and

air volumes.
FAN LAWS

1. Air quantity varies directly as fan speed;
quantity is independent of air density
(twice the volume requires twice the
speed).

2. Pressures induced vary directly as fan
speed squared, and directly as density
(twice the volume develops four times the
pressure).

3. The fan-power input varies directly as the
fan speed cubed and directly as the air
density (twice the volume requires eight
times the power).

4. The mechanical efficiency of the fan is inde-
pendent of the fan speed and density.

SELECTION OF FANS

The performance of a fan in a ventilating
system is determined by its characteristic
curve and the mine resistance if acting alone
on the system; and by its characteristic curve,
characteristics of other pressure sources, and
the resistance of individual zones of influence
if acting in combination with other pressure
sources. Fan characteristic curves are a matter
of design, which is controlled by the manu-
facturer. The resistance of the mine is a matter
of layout and maintenance of the ventilating
network and is controlled by the mine operator.
Figure 14 illustrates the relationship of the fan
characteristic and mine resistance.

The amount of airflow induced in a mine will
depend on the fan characteristic and mine resist-
ance. The pressure (H) required to pass a quan-

tity (q) through a mine or segment of a mine is
expressed by the common formula H=RQ
where R, the resistance factor, may be calcu-
lated from known pressure losses or from the
common formula previously _ discussed.
rE and Q is the air quantity expressed
in units of 100,000 ¢.f.m.

Mine fans are available that will suit most
conditions of mine resistance and desired
volume relationships. Modern fans are built
with variable pitch blades that permit a wide
range of application for the single fan. Figure
15 compares the characteristic curves for a
fan operating at different speeds and blade
settings. The various blade settings permit
the same fan to operate through a wide range
of mine-resistance relationships.

All fan-performance data are based upon a
standard density of 0.075 p.c.f. At altitudes
where the air density is substantially less than
standard, corrections must be applied on the
basis of the true density. Fan law No. 2 states
that pressures vary directly with density; there-
fore, pressures can be corrected from the ratio

aa where d is the actual air density.

The most practical and convenient method of
correcting fan-performance curves is to estab-
lish an average density for the fan elevation and
to place the corrected pressures directly on the
chart. For example, if the average density is
determined to be 0.060 p.c.f., the correction
ratio will be 0.060-+-0.075=0.80. Figure 16
shows the corrected fan-performance curve. As
can be seen, the same fan at different elevations
will operate at different pressures for the same
air quantity.

The efficiency of the fan is the relationship of
the power output, divided by the power input.
The power input is the actual horsepower used
to drive the fan rotor and equals the measured
motor horsepower, less any shaft or belt losses.



FUNDAMENTALS OF AIRFLOW 19

CALCULATIONS

. KLOo¢
The basic formula n=Klo¢ applies.

_ 5.243
U—“KLO

The pressure drop H, the conversion factor 5.2, and
the coefficient of friction K, will be the same for all
splits. Consequently, these’ can be canceled and the
formula becomes:

AB

es

Extracting the square roots and simplifying the
formula, it becomes:
z
=4y/Z5

This formula often is called the formula for
potential air splitting. For ease with calcula-
tions, L should be expressed in units of 1,000
feet; for example, a length of 3,200 feet becomes
3.2.

G+G2+Gs= 150,000 ¢.f.m.,

=A,,/ Ar a vie
anAvy/ 745 =00 32x40 OV 28

=90Xy.703=75.4

~A,,| 42 — Vara or
oA] 75 =70 axe OV 24

210.3,
=70X¥.833=63.8
=4,,| 49 — 90 es
a Avy] T,0, OV 58 x40 OV
=90XV.625=71.1
75.4
N=979,3% 150,000 cfm.
=53,800 e.f.m.
63.8
%=3 79 3% 150,000 c.f.m. 150,000 ¢.f.m.
=45,500 c.f.m.
7A
= 349.3% 150,000 ef.m.
=50,700 ¢.f.m.

This problem illustrates free splitting and
shows how approximations of airflow can be
made for planning on the basis of projected
airway measurements without knowing pres-
sure values.

Problems involving regulation and ventila-
tion changes require pressure-volume relation-
ships for segments of the mine that will be
affected. Figure 11 shows a similar simple

roblem but with air quantities—the pressure
Iosses illustrated by a pressure gradient and the
fan characteristic. The problem now is to regu-
late entries 2 and 3 so that entry 1 will carry
70,000 c.f.m. The first step is to calculate the
resistance factors for each segment and split,

using the formula H=RQ*. no where the

given quantity Q is expressed in units of
100,000. H=pressure in inches water.

Segment A-B:
H=1.5 inch, g=150,000 c.f.m.,
Q=1.5,

15 1.5
R= py a957 0-87.

Segment B-C:
Split 1:

H=1.0 inch, g=52,500 c.f.m.,
Q=0.525

1 1

R= C535 -0.276

3.62.

Split 2:
H=1.0 inch, g=48,000 c.f.m.,
Q=0.480,

Split 3:
H=1.0 inch, g=49,500 c.f.m.,
Q=0.495,

1 1
R= 495)? 0.24518

Segment C-D:
H=1 inch, g=150,000 c.f.m.,
Q=1.500,

1
= 7950-44.



32

increased. Each fan is now operating at a
pressure of 3.85 inches at a combined air
quantity of 245,000 c.fm. Each fan theo-
retically will induce one-half of the total
quantity. This will not be so in actual

INTRODUCTION TO MINE VENTILATING PRINCIPLES AND PRACTICES

practice, since fan installations and mechanical
arrangements will vary slightly.

Figure 21 illustrates results with two different
fans operating in parallel at the same pressures;
each fan contributes different quantities to

5 T T T T T T T
Combined
atk 4
&
&
=
ra
5 3b 4
a
¥ <
a
$ Single fan rd
ST EP TT TTT 5
ul 2b ~ 7
3 H é
a | 5
ra al ‘9
= 21 é |
1 Mine resistance oy!
zl
a
I
a i in \ i
0 50 100 150 200 250 300 350 400
VOLUME, THOUSAND C.F.M.
Fravre 20,—Two Identical Fans Operating in Parallel.
7 T T T T T T T T
oan 4
we oF 4
¥
= Fan A.
5 Combined 4
9 46 —~------ __-~______
8
Ed
3 \
= i
wi ! 4
5 |
5
B 1
B
ra |
Fa
& |
1 4
1
I
f 4
Mine resistance ~a, |
1
1 i i 1
00

1
200

250 450

2

VOLUME, THOUSAND C.F.M.

Ficure 21.—Two Different Fans Operating in Parallel.


14

The formula for precise air-density determina-
tion is:

1,327
~ 460+7
The units are the same as the previous air-
density formula, except that (f), the influence
of the vapor pressure in inches of mercury at
the dewpoint, has been included. The use of
this formula requires that psychrometric tables
(table 3) or charts must be available.

EXAMPLE: What is the air density under the
following conditions: Wet bulb=60° F., dry
bulb=70° F., B=28.75 inches mercury, and

qe 1827
~ 460+70
(Ref., table 3.)

d (B—0.378 f).

(28.75—0.378 f).

INTRODUCTION TO MINE VENTILATING PRINCIPLES AND PRACTICES

The temperature of the dewpoint at 29 inches
mercury and dry-bulb temperature 70° F, and
10° F. wet-bulb depression=53° F. The vapor
pressure at 53° F.=0.402.

(28.75—(0.378X0.402)],

(28.75—0.15),

=0.0716 p.c.f.

Comparison between these calculations shows
that the influence of the vapor pressure can be
ignored for most mine-ventilating problems.

TABLE 2.—Approzimate air densities for dry air at different temperatures

Air densities, p.c.f.

o | ase ase | 90°
as F.] FP] | PR] OF

Barometer, inches
or | se | 10° | 15° | 20° | 25° | 0° | 35°
FY PR] rR] RB] Rp ROR | OR

31.0. o. 0875/0. 08660. 0857/0. 03480, 0839)0, 0831

30.5, 70] 0861) - 0852): 0843) . 0834] 8)

30.0. 0847|

25 0833) | 0824

29.0 0819) | 0810}

25. (0805) | 0796]

28.0. 0790) | 0782}

275 ‘0776 - 0768}

270. 0762] - 0754

26.5. ‘0748 ‘ 0740|

26.0. 0734] | 0726|

25.5. 0720) ‘ 0712]

25.0 0706) | 0698!

245 884

240 0677| | 0671

23.5 ‘0663| | 0857

23.0 156] | 0649} | 0643

22.5 0635) 06%

22.0. 0621) | 0615]

 a7salo.ozsso. 0748

10730):
0718) -

(0. o823[0. 0815]0. 0807/0. 0799]0. 0791/0. 07830. 0776)0. 07
(0808) : 0801! 0794) 0786] :
0788)



6 INTRODUCTION TO MINE VENTILATING PRINCIPLES AND PRACTICES

1,800, | T

MANUFACTURER'S.
CORRECTION FOR ANEMOMETER

Feet per minute Correction

200
300
400
500
600
700
800
1,000
1,200
1,400
1,600
1,800

1,500}-—

— 48= 1,552
— 60= 1,740

1,000;-—

TRUE VELOCITY, FEET PER MINUTE

500;—

30 t 4

200

A=30
B=0.95

I
800

1,300 1,800

REGISTERED VELOCITY, FEET PER MINUTE.

Figure 2.—Anemometer Calibration Curve.

and less at the sides. Approximate continuous
traverses, made by moving the anemometer
slowly through a plane at right angles to the
axis of the airway, are recommended. One
common method is to divide the airway into
two equal areas and traverse each area sep-
arately, keeping the body out of the airstream
being measured (figs. 3 and 4). The time
interval for conducting measurements varies
with the accuracy desired. If the half-area
traverse is employed, 1-minute measurements
of each area checked by '%-minute traverses
are common. All air measurements should be
checked by repeat measurements until a
satisfactory comparison of velocities is ob-
tained. With a little practice, check measure-

ments within 2 percent can be obtained under
most. conditions in coal mines.

Precise traversing methods with anemometers
are justifiable only on major air currents, in
connection with tests of fan performance, or to
determine pressure-quantity relations. They
should be applied only at sections of regular
area on straight airways and away from any
disturbing influence. Air-measuring stations
with located traverse points are necessary
when precise measurements are required. Tra-
verse points can be located by strings, wires,
or marks on guide frames at centers of equal
areas of cross sections. The number of areas
depends on the conditions of flow and should
not be less than 16 for a square section or less


MECHANICAL VENTILATION

29

TaBie 6.—Fan-test data sheet, fan blowing

Observed pitot readings in quadrant

Pitot position A B c D
VP TP vP TP vP TP VP TP
1 - 2.24 6.15 2.19 6.14 2.20 6.14 2.27 6.15
2 : 2.20 6. 06 2.22 6.05 2.18 6.08 2.21 6. 06
3 : 2.18 6.08 2.18 6. 06 2.11 6.03 2.16 6. 08
4 . 2.15 5.90 2.20 5.90 2.17 5.89 2.19 5.90
5. - 2.10 5.95 2.15 5.95 2.12 5.94 2.11 5.95
Organized data
Square root of VP Average | Average Remarks
VvP | TP
-| 1497] 1.480] 1.483] 1.507] 1.492] 6.15] Average VP=2.18 inches water
- 1. 483 1, 490 1, 476 1. 487 1. 484 6. 06,
- 1.476 1, 476 1, 453 1. 470 1. 469 6. 06
- 1. 466 1. 483 1. 473 1. 480 1. 476 5. 90
- 1. 449 1. 466 1, 456 1. 453 1. 456 5.95 | Average TP=6.02 inches water

SP=TP—VP=6.02 inches —2.18 inches =3.84 inches H20.

it can

Comparing fan pressures: From these tests
be seen that the total pressure of an ex-
haust fan compares with the static pressure of
a fan under a similar condition when blowing.
This comparison appears confusing, since
pressure differences, negative and positive,
measured directly against atmospheric pressures
do not appear consistent with the total-pressure
differences generated by the fan.
shows the pressure changes in ventilating sys-
tems for both blowing and exhausting fans.

flow of the

Figure 17

FANS IN SERIES

Fans operating in series are those operatin;
on the same flow circuit, each handling the total

circuit and each generating a part

of the total pressure required to balance the
ressure losses of the circuit.
fans can be used.
operating in series is that where resistance to
flow is high more efficient distribution and better
operating conditions may be obtained.

Any number of
The advantage of fans

Fan total ~a|

Blower Exhauster G)
1
1
1
ae TP |
Lt static” ~~~ SP. Rated static
- Ta-~.] _ co
. TP Fan total 7
~~~ ll //
‘Atmospheric pressure’ Sa=-- IT S|
Measured

static

Ficure 17,—Pressure Changes in a Ventilating System for Blowing and Exhausting Fans.


25

MECHANICAL VENTILATION

O9T

OOUASISOY OUT PUL OSLOJORIEYD UBT UdEMJog drysuoNTeyY—pI auAI

OvT O@T

‘W'd'D GNVSNOHL ‘SWN10A uly
Oot 08 09

ov

0

(TS'T) eouejsises aul MeN —__ |

==
an

4

onsuayoeseyo uej

7 Wa
YY |__—- (gz) aoueysisay aul
ees ='dH
, ose'9-—— +h
Gg = OSE dy
98 | OxH |

Jamodasioy sly

aouRjsisal QUILU dI}slWa}IeIeYO Uey

gt@=z0+H=4¥
aunssaid Sayoul OTE Je “W'}'D COO'OZT Bulonpul uey

IS'T = Y Mau ‘02H Sayou! GG°Z= H “W'}9 COO‘OET = b mau _]
‘paonpal aouejsisad ‘papinod sKemule jeuoljppy ‘ajdwexy
‘uone|ad
aunssaid aWNJOA Je SaBueyo aduejsisas aul BulSueyD

204 =H

I
uaLVM JO S3HONI ‘AuNSSaud

+

(H)


REVIEW OF ELEMENTARY FUNDAMENTALS 3

the weight of 1 inch of water, it is equal to
62.5--12=5.2 psf. (fig. 1).

Pressure necessary to move air consists of
static, velocity, and, when combined, the total
pressure. (See fig. 1.)

Static Pressure (SP): Pressure, either nega-
tive or positive, exerted in all directions. May
be compared to atmospheric pressure.

Velocity Pressure (VP): Pressure exerted by
the kinetic energy of air movement.

Total Pressure (TP): Algebraic sum of the

Duct
0 Airflow 3
—_—
Pressure in inches,
water gage
= Water
Duct
Airflow
Positive pressure
f VP
TP "
14” ve
| tsP be
0.4”
Duct
Airflow
Negative pressure ))
sP uP Vp
" Y
1.4” 0.4” ve,

static pressure and velocity pressure, either
negative or positive.

Static Pressure Loss or Regain.—The total
pressure remains constant except for energy loss
caused by airflow. However, the velocity of
the airstream will vary with any changes in the
area of the airway; consequently, the velocity
pressure also will change. With velocity-pres-
sure changes, there must be a corresponding
change in the static pressure. For example,
if the area doubles in an airway, the air velocity

+

1 inch H20
oe bp 625+ 12=5.2
625 Ib. | Ib. /sa. ft
v 1
roy

(S P) Static pressure is pressure exerted
in all directions.

(VP) Velocity pressure is directional pressure
involving movement of kinetic energy.

(-P) Total pressure is the sum of both static
and velocity pressures.

Ficure 1.—Illustration of Static and Velocity Pressures,


SYMBOLS AND ABBREVIATIONS

_ cross-sectional area of airway, square feet.
barometric pressure, inches mercury.
cubic.
~ cubic feet per minute (of air).
~ diameter of duct or pipe.
density of dry mine air, pounds per cubic foot.
~ density of mine air and water vapor, pounds
per cubic foot.
_ vapor pressure at dewpoint, inches mercury.
linear feet.
feet per minute.
feet per second.
degrees Fahrenhei
~ influence of gravity, 32.2 f.p.s.
ventilating pressure in inches water.
~ horsepower.
water.
coefficient of friction (friction factor).
- length of airway in feet.
© pound, pounds.

minute, minutes.

natural ventilating pressure.

perimeter of airway.

ventilating pressure, pounds per square foot.

pounds per cubie foot.

pounds per square foot.

air quantity, cubic feet per minute.

air quantity, expressed in units of 100,000
cfm.

resistance factor of mine or airway.

statie pressure.

square feet.

temperature, dry bulb.

total pressure.

temperature, wet bulb.

velocity pressure.

relative average
uncorrected.

average velocity of airflow corrected.

velocity of airflow

Vr...

REVIEW OF ELEMENTARY FUNDAMENTALS

Air Quantity.—g is the amount of air flowing
through a mine or a segment of a mine, in
cubic feet per minute. Air quantity is the
product of the air velocity times the cross-
sectional area of the airway. g=VXA.

Velocity.—V is the rate of airflow in lineal
feet per minute and is measured by anemome-
ters or other instruments and the time it takes
for smoke to travel over measured distances.

Area.—A is the cross-sectional area of the
entry or duct through which the air flows,
expressed in square feet.

Perimeter.—0 is the linear distance in feet
of the airway perimeter rubbing surface at right
angles to the direction of the airstream.

Airflow.—When air flows slowly along the
walls of the containing airway and around ob-
structions without causing eddy currents or
turbulence, the flow is called streamline. The
rate of streamline flow will be approximately
10 f.p.m. or less. Such flow in mines is often
referred to as air seepage.

When the airspeed increases, the flow charac-
teristic changes from streamline to turbulent.
With turbulent airflow, the energy of air in
motion is great enough to bounce air particles
off the walls of the containing airway and ob-
structions increasing pressure losses due to
friction and eddy currents.

Normally, in mine airways and ducts airflow
is turbulent.

Pressures.—Some kind of pressure is neces-

2

sary to cause airflow between points in an air
circuit or mine.

The purpose of a fan is to supply mechanically
produced pressure at some point in the mine-
ventilating circuit. In addition to mechanical
pressure, atmospheric or barometric pressure
also influences airflow in a mine. The alge-
braic sum of the mechanical draft pressure and
the atmospheric natural ventilating pressure
comprises the mine ventilating pressure.

If the pressure in the mine or duct is at a
lower pressure than atmospheric or other base
pressure, the system is negative, or exhausting.
If the opposite is true, the system is positive,
or blowing.

The method of measuring pressures is usually
by water gage or other form of differential
pressure gage. For special surveys, precision
altimeters* or barometers are used with the
pressure data converted to inches water gage.

Water Gage.—The water gage or standard
U-tube is a common instrument that indicates
differential pressures in inches of water. The
pressure in inches of water can be converted
into pounds per square foot, P, by multiplying
the water gage reading by 5.2, the conversion
factor.

The conversion factor of 5.2 p.s.f. is derived
from the fact that 1 cubic foot of water weighs
62.5 pounds. If one-twelfth of this weight is

4 McElroy, G. E., and Kingery, D. 8., Makine Ventilatine-Pressure
Surveys With Altimeters: Bureau of Mines Inf. Circ. 7809, 1957, 20 pp.


AIR MEASUREMENTS

Tape 3.—Temperature of dewpoint, in degrees Fahrenheit (Pressure = 29.0 inches)

15

Depression of wet-bulb thermometer (tt)
Air tem- | Vapor
perature | pressure
t e {1.0}20/3.0}4.0] 5.0 | 60 | 7.0 | 8.0 | 9.0 | 10.0 | 11.0 | 12.0 | 13.0 | 14.0
20] 0.103} 17} 13) 8| 2] —5} —18
21) 108] 18} 14] 10] 4] —3} —14| —42
22) 113} 19] 15) 1} 6} —1] —10| —29
23) .118] 20/16) 12} 8} +1] —7| —22
24] 1124) 21] 18] 14] 9 3] -4]-17
25 | .130 | 22| 19 | 15 | a1 5] —2] -12] -36
26} . 136 | 23 | 20 | 16 | 12 7) +0] —9| —26
27} 143] 24 | 21 | 18 | 14 9) +3} —5 | -19
28) 150/25] 22} 19] 15] at 5| —2| —14| —45
29} .157| 26| 24) 20} 17] 12 7{ +0] —9] —29
30} 164) 27) 25) 22] 18} 14 9} +3] —5 | —20
31| .172| 29/26) 23] 20} 16} 11 5| —2] -14| —50
32} .180] 301 27] 24] 21} 17] 13 8] +1) —9| —29
33} .187| 31] 28} 25} 22} 19} 15| 10 3] —5 | —20
34] 195 | 32) 29] 27] 24] 20] 16} 12 6| —2| —14] —50
35 | 203 | 33} 30) 28] 25) 22) 18) 14 8] +1] —8| —28
36| 211 | 34] 31] 29/26} 23} 20] 15) 11 4| —4] -19
37| .219] 35/32] 30] 27] 24| 21] 17) 13 7) -1] -12
38 | .228] 36] 33] 31) 28| 26) 23} 19} 14 9| +3) -7
39| .237| 37/34] 32/29] 27} 24] 21) 16] 12 6| -3
40} 247} 38] 35] 33] 31} 28] 25) 22} 18] 14 8) +1 —35
41 | 256 | 39 | 37/34] 32} 29) 26} 23) 20) 16) 11 4 —21
42| 266 40] 38/35] 33} 30] 28; 25] 21) 17) 13 7 —13 | —59
43| .277| 41 | 39| 36) 34] 31) 29/ 26) 23] 19] 15} 10 -7{ —28
44] 287 | 42| 40) 38) 35] 32] 30} 27| 24| 21] 17) 12 —2| -17
45| .298/ 43] 41/39/36} 34] 31} 291 26] 22]| 19) 14 +2/ -9
46| .310| 44| 42] 40/37] 35) 32] 30] 27} 24] 20| 16 5] 4
47} 322) 45 | 43) 41] 39] 36) 34] 31) 28} 25] 22) 18 8| +0
48) 334 / 46 | 44/42] 40] 37] 35} 32} 30) 27] 23) 20 10] +4
49| (347 | 47| 45] 43} 41/ 39] 36] 34] 31] 28] 25| 21 13 7
50| .360| 48 | 46] 44) 42} 40] 37] 35] 32] 29] 27) 23 15 9
51] 373 | 49] 47/45/43] 41) 39] 36] 34] 31] 28| 25 17 12
52| .387]| 50| 48) 46) 44] 42/ 40/ 37] 35] 32] 29| 26 19 14
53 | .402] 51| 49) 47) 45 | 43) 41] 39) 36] 34] 31| 28 21 16
54| 417} 52] 50] 49/47} 44) 42] 40]; 38] 35] 32] 29 23 19
55| 432/53] 52} 50] 48| 46) 43] 41] 39] 36] 34] 31 24 21
56| 448 | 54) 53/51] 49) 47) 45] 43} 40] 38] 35] 32 26 23
57| 465 | 55 | 54] 52} 50} 48) 46] 44] 42] 39] 36| 34 28 24
58| 482/561 55/53/51) 49) 47] 45] 43] 40] 38] 35 29 26
59} .499 | 57 | 56 | 54) 52, 50] 48) 46 44} 42} 39] 37 31 28
60} 517] 58/57/55) 53} 51) 49| 47] 45| 43] 41| 38 32 29
61 -536 | 59| 58) 56] 54] 52] 51 49 46| 44/ 42/ 39 34 31
62} 555 | 60/ 59/57/55) 54) 52] 50; 48} 46] 43) 41 35 32
63 | .575|61| 60/58] 56] 55) 53| 51] 49] 47] 45) 42 37. 34
64| .595 | 62| 61/59) 58/ 56] 54| 52) 50; 48} 46) 44 38 36
65 | .616 | 63 | 62/60] 59} 57) 55] 53] 51] 49] 47| 45 40 37
66 | .638 | 64| 63/61} 60; 58) 56| 54] 53} 51] 48) 46 42 39
67) .661| 65 | 64/62/61) 59) 57] 56] 54] 52) 50} 48 43 40
68| .684 | 67| 65} 63/62| 6O) 58; 57| 55| 53] 51| 49 44 42
69| .707| 68} 66} 64/63} 61| 60} 58] 56) 54} 52) 50 46 43
70| .732 | 69 | 67 | 66 | 64 62 61 59] 57 55 | 53] 51 47 45
71) .757 | 70| 68| 67} 65) 63} 62] 60} 58| 57} 55] 53 49 46
72 783 | 71| 69/68} 66} 65} 63] 61] 60| 58) 56] 54 50 48
73 | .810 | 72} 70) 69} 67} 66) 64) 62) GI} 59) 57] 55 51 49
74] .838| 73] 71| 70} 68| 67] 65; 64] 62} 60| 58| 56 53 50
75| .866) 74/72) 71| 69] 68] 66] 65] 63} 61) GO| 58 54 52
76| .896|75|73| 72/70] 69] 67] 66; 64} 62) 61] 59 55 53
77 -926 | 76 | 74 | 731 71 70| 68] 67 65| 64) 62/ 60 56 54
78) .957| 77] 75|74| 72] 71) 69] 68] 66] 65) 63] 61 58 56
79 - 989 | 78 | 76 | 75 | 73 72 70| 69} 67] 66| 64 62 59 57
80} 1.022] 79! 77176175! 731 72] 70! 69] 67] 65! 64 60 58

1 Marvin, C. F., Psychrometric tables: U.S. Weather Bureau Bull. 235, 1915.


PART I
AIR MEASUREMENTS

The accuracy of the air quantities that are
determined will of course depend on many
factors. The most important is the care used
when selecting measuring stations, determin-
ing velocities, and cross-sectional areas. Most
air-quantity determinations for coal mines do
not need to be absolutely correct; in fact, care-
ful underground measurements probably will
be only 90 to 95 percent correct. The day-to-
day centerline measurements made by under-
ground foremen as relative or comparative
Measurements are satisfactory in most in-
stances. However, air traverses are necessary
when making air-quantity surveys, fan tests,
pressure surveys, and certain other investiga-
tions.

The air-quantity survey should be made
twice yearly to determine and locate air-leakage
losses and check fan operating efficiency. Air-
quantity surveys are necessary in conjunction
with pressure surveys to determine resistance
factors. These measurements should be made as
accurately as possible, otherwise the resultant
data may not reveal important deficiencies of
the ventilating system.

MODERATE-VELOCITY AIRFLOW
MEASUREMENTS

For measuring velocities from 120 to 2,000
f.p.m., ordinary commercial types of medium-
velocity vane anemometers are practical, con-
venient, and accurate. The vane anemometer
is a small windmill geared to a mechanical
counter through a small clutch, which is en-
gaged for recording revolutions. The common
commercial size is 4 inches in diameter. For
ordinary rough measurements, the anemometer
is held in the hand. For precise work, the ane-
mometer should be mounted on a shaft to re-
duce the effect of the hand, arm, and body on
the airstream. Anemometer readings are virtu-
ally independent of the air densities and air
temperatures encountered in underground work.
Another advantage is that readings are only
slightly influenced by nonparallel flow.

Jontrary to the opinions of some mining
engineers that the anemometer is not reliable
within the medium-velocity range, if used prop-

540577 0—60——2

erly the anemometer is the most practical in-
strument for determining such air velocities.

Vane anemometers register a velocity some-
where within plus or minus-10 percent of the
true velocity. Calibrations furnished by the
manufacturer with new instruments are usually
in the form of tables of plus-or-minus correc-
tions to be made to observed velocities.

The application of these correction factors
can be simplified by plotting the true velocity
(V7) in feet per minute against the registered
velocity (Vg) (anemometer reading in feet per
minute) and determining the equation Vr=A
plus B V, for the resulting line. In this general
equation, Vz is the true velocity, Ve is the
registered velocity, and A and B are constants
for the particular anemometer. Figure 2 shows
the plot of calibration data for a standard
anemometer. The plotted data fall approxi-
mately into a straight line; the constant A is
the increment that thisline crosses the true veloci-
ty ordinate (vertical line through zero, fig. 2)
above the zero point. The constant B is the
ratio of the true velocity to the registered
velocity or slope of the line. This formula.
Vr=30+0.95 TF, while not exact for all
points, is within the limit of error in use. For
example, at a registered velocity of 200 f.p.m.,
true velocity =30 +0.95 X 200=220 f.p.m., com-
pared with 210 f.p.m. for the anemometer cor-
rection table. For a registered velocity of 500
f.p.m., true _velocity=30+0.95 X 500=505
f.p.m., compared with 504 f.p.m. for the ane-
mometer correction table. This equation is
easy to use; each anemometer can be marked,
and corrected velocities can be determined at
the time of measurement. Manufacturer's cor-
rection cards often are lost, and when this
happens the value of the anemometer for more
precise measurements is destroyed unless cali-
brated by some method.

The common method of determining air
velocities in coal-mine entries is by one-point
measurements. That is, the anemometer is
held at the center of the section. For day-to-
day comparison, such one-point measurements,
provided they are taken at the same place, are
satisfactory. For more accurate determina-
tions, traversing methods should be employed,
since the air velocity is highest at the center

5


22

INTRODUCTION TO MINE VENTILATING PRINCIPLES AND PRACTICES

70,000 c.f.m.
A 135,000 c.f.m. B 35,000 c.f.m. R Cc 135,000 c.f.m. D
30,000 c.f.m.
R
0
~ A
BE 1
os 1.09” B
wa
So ?
raed) 177"
BS 3
az D
4 3.58”

PRESSURE GRADIENT

&5
wi
E Fan-pressure volume
fd 4L 3.79" characteristic
ar |
o
w 3 |
3 Mine 1
Z ,| resistance lg
Seat 18
ul ie
ce o
=) 109
Bib 7
B
w |
& L L tio t L
0 4 8 12 16 20 24 26

VOLUME, THOUSAND C.F.M.

Ficure 12.—Example of Airflow Splitting, With Regulation.

AIR LOSSES FROM LEAKAGE

Leakage losses are a serious detriment to the
efficiency of a mine ventilating system. When
the costs of fan installations, air shafts, over-
casts, stoppings, and other ventilating equip-
ment are compared with the small percentage
of air actually reaching the working face, the
importance of air leakage becomes evident.

ir leakage for operating mines in many in-
stances is as much as 70 percent between the
intake and the sum of the air volumes at the
last open breakthroughs. A leakage path is
simply a parallel return path to the fan.

The amount of leakage will be determined by
the pressure difference between intake and re-
turn and the condition of stoppings and over-
casts. The best stoppings are either brick or
solid block set with mortar, anchored into the
rib, roof, and floor, and plastered. The com-
mon practice of stacking hollow blocks dry and
plastering the intake side to form a stopping is
conducive to large leakage losses.

Figure 13 illustrates leakage losses in a main
split of a large coal mine. The efficiency can
be calculated as follows—splits 4 right and 8


Bulletin 589
Bureau oF Mines

INTRODUCTION TO MINE
VENTILATING PRINCIPLES
AND PRACTICES

By D.S. Kingery

UNITED STATES GOVERNMENT PRINTING OFFICE
WASHINGTON : 1960


44 INTRODUCTION TO MINE VENTILATING PRINCIPLES AND PRACTICES

Problem 2.

Figure 27.—Mine and Ventilation Plan.

#2
¢
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goes a| 3
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8 INTRODUCTION TO MINE VENTILATING PRINCIPLES AND PRACTICES

than 24 for a section in which one dimension
is twice the other.

_ When measuring airways of divided sec-
tions, such as an entry that has been center-
posted, the air quantity for each division should

e determined separately because the posts
change the velocity contours of the airstream,
both upstream and downstream, making it
difficult to obtain consistent measurements,

LOW-VELOCITY AIRFLOW
MEASUREMENTS

Unless special low-speed anemometers are
available for determining direction and velocity
of air currents below 120 f.p.m., smoke-cloud
methods offer practical means. The accuracy
of the method is acceptable for most mine-air
measurements, Figure 5 shows the Bureau
of Mines type of smoke-cloud aspirator and
smoke tube.

To determine velocities by smoke-cloud

travel, measure a known distance along as
straight and uniform a section as can be found
in the airway. This distance will be determined
by how well the smoke cloud holds together
and how well it can be seen. Twenty-five
feet usually is suitable. Determine the average
area of the section along the measured dis-
tance. The average of three area measure-
ments is usually enough unless the cross-
sectional area is very irregular. Measurements
are taken by a man with the smoke generator
at the upstream point (fig. 6) and the timer
with a stopwatch at the downstream point.
Smoke clouds are released at quarter points
(fig. 7), and the time that it takes the smoke
cloud to travel the measured distance is
recorded. Each velocity determination sbould
be repeated several times. Abnormal highs
or lows should be discarded, and the remainder
averaged. With travel at about the quarter
points, readings will average about 10-percent,
high, Consequently, this correction must be

applied.

Figure 5.—Bureau of Mines-Type Smoke-Cloud Generator.


AIR MEASUREMENTS

Ficure 4,—One-Half Area Traverse, Right Side.


40 INTRODUCTION TO MINE VENTILATING PRINCIPLES AND PRACTICES

To simplify further, use Q in units of 100,000
cfm. The formula then becomes:

Area=—= id
3. Common ventilation formula HaLOL,

This formula is based upon the hydraulic law
for turbulent flow of a fluid through a straight
duct of uniform cross section, assuming that the
pressure loss varies as the square of the velocity
and is generally applicable either to liquids or
to gases.

ei is the hydraulic law for turbulent

flow,

h=pressure loss of flow, expressed in head
of feet of fluid flowing,

f=friction factor, based upon rubbing
surfaces,

L=length, in feet,

average velocity, feet per second,

acceleration due to gravity 32.2 f.p.s.,

r= hydraulic radius (equals the area of the
duct divided by the perimeter of air
or liquids filling duct).

4 =where A=area, and O=the airway
perimeter.

The formula then becomes:

SLV?
29

h= x2

A

The common ventilation formula
KLOV?
(PAS)

is derived as follows:

P (the pressure, pounds per square feet) equals
the head of the air column times the air den-
sity (d), expressed in pounds per cubic foot.

P.
P=hxXd, hag

therefore

eed with velocity in feet per second,

d
2
i=. =O withvelocityin feetperminute,

df V#LO
P=ax3,800 A

df

293,600 forms K, the ventilating friction factor.

Substituting K,

then
P= KxEOM,
q=VA,
v4
m(y)
then
Log.

H, pressure in inches water, equals P, pressure
in pounds per square foot, divided by 5.2:
Haz.

“3.2
P=HX52,
then
52H 108,
__ KLO¢.
A= 5.2.A8

4. Formula for total resistance of series and
parallel flow:

Series, flow, and the combined resistance for
any individual air split are found simply by
adding the individual resistance factors of the
increments making up the air split.

R total=R,+R2+Bst+... . Be
For parallel flow through two or more airways
with individual resistances of R,, Re, Rs,+Rn,
the equivalent resistance & representing all air-

ways involves the square relationship previously
discussed. The formula for total resistance yo

Rae AP) ABT ABT

TER Ra
and is derived as follows:

HRC, Oi Om
Total Q= Q+Qrt+Q+Q,
aint int nt ee

Total Q also equals V TER

parallel flow equals


28 INTRODUCTION TO MINE VENTILATING PRINCIPLES AND PRACTICES

TaBLe 5.—Fan-test data sheet, fan exhausting

V'P= Velocity pressure measured, inches water, always positive.

--- Model No, -..------- Date
Type .. 200-2 2ee eee Blade position -..---..-.

Observers ~..---.-----~-------------------0------

TP =Total pressure measured, inches water.

Observed pitot readings in quadrant
Pitot position A B c
VP TP vP TP vP TP vP TP
1 2.24 219) 3.95 2, 20 2.27 —3. 88
2 2.20 2:22} 3. 83 2.18 2.21 —3. 85
3. 2.18 218] ~3.88 211 2:16 —3, 92
4 2.15 2.20) 3.70 2.17 2.19 =3.71
5. 2.10 2.15) —3 80 2.12 211 —3. 84
Organized data
—
H Square root of VP Averare Final Average Final
VVP average 'P average
I 1. 497 1. 483 1. 507 1 —3, 92 -
2 1, 483 1. 476 1. 487 1 —3, 86 :
3 1 476 1. 453 1. 470 1 ~ .
4 1, 466 1 473 1. 480 1 - :
5 1, 449 1. 456 4 1 - :
7. 376 1.475 | —19. 24 —3. 85
i
CALCULATIONS
measuring point: 24.75 sq. ft ,000 1.475 =5,900 ft./min.
ity: 0.075 Ib./eu. ft. 900 X 24.75 = 146,000 e.f.m.
Air horsepower: 145,000%3.85-=88.5 hp. 1.475)?=2.18 H,0.
350
SUPPLEMENTARY FAN DATA
its X amperes X power factor X¥3_
Motor: 3-phase, 440 volt. 746
ating: 150 hp.
Ampcres: 150 measured. . ; 440 150 0.85% 1.732 _
Power factor: 0.85 medsured. Calculated input horsepower: hp. 746 130.
rp.m.: 1,175 rated. ir hp. 88.5
Fan Drive: V-belt, r.p.m. 1,060. Overall mechanical efficiency: —""P~_—=*-° 68 percent.
input hp. 130

These data do not include belt losses that
should be considered if the efficiency of the fan
only is to be determined.

Assuming a 5-percent loss due to belt slippage
and friction, the actual input horsepower to the

fan equals 130 times 0.95 or 123.50; actual fan

efficiency equals

8

8.5,
123.5

» or 72 percent.

For comparison, the following data are given
showing similar tests for a fan, blowing:


SOLUTION OF PROBLEM EXAMPLES 45

mine is operating with two main air splits,
each servicing four producing units. The mine
is laid out with a panel system, which operates
two units per pana (ref., fig. 27). Idle sections
will be ventilated from the belt entries, which
will be considered neutral but will carry 10,000
c.f.m. for main and 5,000 c.f.m. for panel belts.
Main entries will total 10—4 for intake, 5 for
return, and a belt entry; panel entries will
total 5—2 intake, 2 return, and 1 belt entry.
The average entry dimensions are 6 by 12 feet.
Outby point X, return entries are caved but
open; resistance to airflow is much greater than
for average entries.

The amount of air required and the air
distribution are as follows:

Total air quantity: 8 units at 25,000 ¢.f.m. per
unit. -
Leakage, 100 percen
Ventilation belts !...

Total____.--__-
Phe air distribution would be: Intake shaft, 420,000 .f.m.

Each split will have four units and a belt to
be ventilated. The North split will have all
production from panels, and the South split
will operate two units as development. Be-
cause of the reduced number of intakes and re-
turns in panels, and with all units operating
in panels the North split will have the highest
pressure losses and will be calculated:

Chm.
Intake North split, point A 210, 000
To belt 10, 000
To intakes 200, 000

The belt entry will be regulated to 10,000 c.f.m.
and will be considered neutral and not involved
in the intake calculations. This is not entirely
true, but to include the belt entry would compli-
cate the problem more than the resultant
accuracy would merit.

Intakes:

Split intake to panel 3-...---.--.--.---
Segment B-C (500 ft.):
Beginninj
Leakage.
End... - 70, 000
Ventilation to idle section by belt-... 10, 000
Segment C-D, intake to panel 4 (1,500 ft., 2

entries) :
Beginning- - - 70,000
Belt entry 5, 000

65, 000

Intakes—Continued
Segment C-D, intake to panel 4—Continued
Leakage - ------------------------ », 000

To producing units....------------

Returns:

Segment B-X (gain 70,000 from panel 3).
X is point where resistance to airflow increases:

Leakage.
End at return shaft__---.-...----- 210, 000

CALCULATION OF R FACTORS

Resistance factors for shafts: Both shafts are
18 feet in diameter, 300 feet deep, concrete
lined throughout, and moderately obstructed;

= 25:

KLO___25X300X57__9 og5,

R=SoAS 5D K254X254 XOB4

R factors for entries, 1,000 feet:
Intake entries single entry, K=80:
R= 80X1,000 X36
5.2X72X72 X72
Return entries, single entry, K=100:
100 1,00036

=1.50.

Ty 185 (use 1,90).
Return entries outby point X, K=200:
200% 1,000.36 _
RS OXTEXTEKTE

Intake 4 entries=+2=0.09.

15_

Intake 2 entries=—-=0.38.

Return 2 entries=—=

: 19
Return 5 entries= 570.08.

Return 5 entries outby point X: =f oo.5.


MECHANICAL VENTILATION 27

5 4.0
i
: he
BO | Nd 3
Se 4F 4320
wo 2
fo)
Be 3 Mine 424m
So resistance " 2
25 8
we 2 4166
Se faa
Bo
wo 1 4 8g
ae a
= of
0 200 300 400 500

VOLUME, THOUSAND C.F.M.

Ficure 16.—Corrected Fan-Performance Curve at Average Local Air Density.

The power output of the fan is measured in air
horsepower, which is the total fan pressure (P)
in pounds per square foot times the air quantity
in cubic feet. per minute, divided by 33,000.
Since 33,000 ft.-lb. equals 1 hp., air horsepower
xq
equals 33,000
measurements are in inches of water, the pressure
P is changed to H (pressure in inches of water);

: P
and since H=—5P=HX5.2. Substituting this in

HX5.2Xq sg.
33,000. Sim

plifying it further, air horsepower= 2% for

For convenience, since most

the formula, air horsepower=

practical applications.

Air horsepower, therefore, is based upon air
quantity and pressure measurements, which
have been discussed previously. Accuracy
limits also were discussed.

FAN TESTS

To determine accurately the efficiency of a
fan installation and to check the pressure-vol-
ume characteristic curve requires a test using
instruments to measure air volumes, air-density
pressures, speed, and electrical-input data.

Table 5 gives the field data and calculated
results of a 5-position, 20-point fan test of a
6-foot axial-flow fan. Velocity measurements
were made with a pitot tube using the measure-
ments shown in figure 10 and were always posi-
tive. Compiling the data and making calcula-
tions are relatively simple. This illustration
shows how the velocity pressures (VP) must be
handled by extracting square roots before they
are averaged. The fan static pressure, since
this is an exhausting fan, is negative and equals
the total pressure (7'P) less (VP) the velocity
pressure. (SP)=[7P—(+VP)]= —3.85—2.18
=6.03 inches water.


4 INTRODUCTION TO MINE VENTILATING PRINCIPLES AND PRACTICES

moving through the single airway changes.
The velocity automatically will be reduced
one-half for the same air quantity; the static
pressure theoretically increases corresponding
to the decrease in velocity pressure.

Air Temperatures and Psychometric Data.—-
Mine-air temperatures influence air densities;
consequently air temperatures and the moisture
content of the air (relative humidity) must be
determined before true air densities can be
calculated.

The measuring instrument is a sling psy-
chrometer—an assembly of two thermometers
mounted in a frame, equipped with a pivoted
handle to whirl the thermometers and obtain
the necessary movement of air, The wet bulb
of the thermometer used for wet-bulb readin;
is specially covered with muslin, which extends
into a water well across the bottom of the
psychrometer.

he psychrometer is used by wetting the
muslin cover of the wet bulb and then whirling
the psychrometer in the airstream. The whirl-

action causes evaporated cooling of the wet
bulb, which lowers the recorded temperature.
The dry bulb is not influenced and records the
sensible heat of the mine air. By using standard
psychrometric tables with wet- and dry-bulb
air-temperature measurements, the relative hu-
midity can be determined. Vapor pressure is
another pressure that affects air densities and
is influenced by the moisture content of the air.
Tt also can be determined from psychrometric
tables or charts.

Barometric Pressure.—Barometric pressures
usually are recorded in inches mercury. The
standard sea-level atmospheric pressure is 14.7
psi. at 29.9 inches mercury. At different
elevations above sea level, the barometer reads
less, since the air column above the instrument
is less.

Standard Air Density.—The standard density
of air for mine-ventilation work is considered
to be 0.075 p.c.f. This is based upon the weight
of 1 cubic foot of dry air at 70° F. at a sea-level
pressure of 29.9 inches mercury.


18 INTRODUCTION TO MINE VENTILATING PRINCIPLES AND PRACTICES.

SPLITTING AIR CURRENTS

Split ventilation is necessary for both
economic and safety requirements. Dividing
the mine ventilating system into multiple
splits provides separate ventilating districts in
the mine and permits better air control. Air
splits should be taken off as near the mine
intake as possible and brought together close
to the return opening to reduce air quantity
and pressure losses by full use of parallel flow.

Natural splits are those where the airflow
divides naturally; each split handles a volume
of air dependent on the pressure drop and
resistance factor.

Regulated splits are those where it is necessary
to control the volumes in certain low-resistance
splits to ‘cause enough air to flow into the
splits of high resistance. A regulator is artificial
resistance installed in a low-resistance split.
Regulators may be small openings in stoppings
controlled by slide doors or may be doors
latched partly open.

Split regulation, usually established by trial
and error, is used to control air quantities in
working sections, to equalize splits and to
control the operation. Such methods are
probably the only way that variable require-
ments can be satisfied. When planning ventila-
tion changes or checking ventilation efficiencies,
it is desirable to know the pressure loss-air
volume relationship for regulators. This rela-
tionship can be determined by the following
formula:

A=40X0+\H,
area regulator opening in square feet,
Q=air quantity, expressed in units of 100,000

EXAMPLES: (1) What size area will be needed
to pass 20,000 c.f.m. at a pressure drop of
2 inches water? (2) What will be the approxi-
mate pressure loss through a hole in a stopping
2 feet square, passing 8,000 c.f.m.?

Problem 1:

40. (20,000
V2’ *\ 100,000)”
40X0.2
A=

2
_ 8
1.414

A: =5.66 sq. ft.

Problem 2:
40 ./_ 8,000
2=75X(ro0:000)!

VH= 9X8 _99x0.08=1.56,

H=1.6?=2.56 inches water.

To illustrate the natural airflow potential for
splitting, based upon airflow fundamentals, the
following simple problem is worked out (fig. -0):

A mine has three main splits, all separating at point
A and joining again at point B, as shown by figure 10.
All entries have the same coefficient of frietion:

Split 1 is 3,200 ft. long, area— 90 sq. ft., perimeter = 40 ft.
Split 2 is 2,400 ft. long. area= 70 sq. ft., perimeter =35 ft.
Split 3 is 3,600 ft. long, area= 90 sq. ft., perimeter = 40 ft.

cf.m.,
H=pressure drop across regulator in inches EXAMPLE: What will be the natural splitting if the
water. intake to point A is 150,000 c.f.m.?
1
A 2 B Fan
150,000 c.f.m.
3
Entry 1. Length, 3,200 ft. Entry 2. Length, 2,400 ft. Entry 3. Length, 3,600 ft.
Area, 90 sq. ft. Area, 70 sq. ft. Area, 90 sq. ft.

Perimeter, 40 ft.

Perimeter, 35 ft.

Perimeter, 40 ft.

Ficure 10.—Airflow With Natural Splitting.


NATURAL-VENTILATION INFLUENCE OF AIR DENSITIES 35

ef.m., the following situation holds:
150,000 ¢.f.m. strikes the resistance-curve
segment c at 2.25 inches water pressure—
obviously not good for the operation of
fan B.

4. To determine the pressure necessary to
overcome the resistance of segments a and b
and the air-quantity distribution, 4 inches
water pressure must be subtracted from the
characteristic curves A and B. Where these
points intersect the segment resistance curves
a and 6, air quantities and pressures are
indicated.

5. If the sum of the air quantities indicated
by segment intersections @ and 6 do not total
the quantity tested in segment c, another trial
must be made until the quantities are correct.
In the example we started with a quantity of
200,000 c.f.m., moving up to intersect the re-
sistance curve c. This requires a pressure of
4 inches water, which must be provided by the
combined fans, but it applies only to segment c.
Therefore, we subtract 4 inches of water from
the characteristic curves of fans A and B and
see where this intersects segment curves @ and b.
In the illustration they intersected at g=75,000
c.fm., H=1.4 inches water for fan A; and at g
= 125,000 c.f.m., H=3.15 inches water for fan
B. The sum of these quantities equals 200,000
c.f.m., which satisfied segment ¢. The fan
pressures would then be:

A=4.00+1.40=5.40 inches water.
B=4.00+3.15=7.15 inches water.
This type of graphic solution quickly de-

termines whether two or more fans will work in
parallel and what performance can be expected.

FANS IN COMBINATION

Most multiple-fan ventilating systems in
coal mines use two or more fans in combination
rather than in series or parallel operation. By
combined operation, each fan, although it may
share an airway with another fan, has an indi-
vidual zone of influence with intakes and re-
turns. Figure 23 illustrates the combined
operation of a mine segment with two fans.
Both fans induced some air through the slope.
However, with fan A, the slope was the prin-
cipal intake. The principal intake for fan B
was, of course, the shaft. Both fans had indi-
vidual return systems and general zones of
influence. Figure 23 also shows the pressure

adient for this section of the mine. As can

e seen, intake air traveled to the main junc-
tion C from both the slope and intake shaft.
The pressure loss along the intake amounted
to 0.7 inch water. The intake air was split at
C—one split North and one split South. Part
of the air was returned through back entries to
points (b-b), at a pressure loss of 0.3 inch water.
From points (6-5), air traveled to both fan B
and fan A; the air returned to fan A began with
small air quantities, which were increased by
leakage and by air coming from bleeder entries
of both the North and South sections. An indi-
cation of the pressure drops for both main splits
are shown. To return, junction-X air will flow
from points b-b and the bleeders. From point
X to fan A, air may travel the returns or again
enter the North bleeder system. Two paths
are indicated to illustrate that, although air-
flow does not follow similar airways, each split
does not have the same volume of airflow, and
secondary splits through the gob and bleeders
will differ; pressure drops between points, such
as from point 6 to fan A, will be identical in
all splits.

NATURAL-VENTILATION INFLUENCE OF AIR DENSITIES

Natural ventilating pressures are induced by
differences of total weights of air columns for
the same vertical distance. Natural draft ma:
operate with or against the mechanical draft
(ge 24) or may be the only source of pressure.

atural draft pressures at near standard den-
sity can be estimated quickly as 0.030-inch
water gage for each 10° F. average temperature
difference per 100-foot increment of vertical
elevation. For accurate determinations the
average air densities of influencing air columns
must be calculated.

Temperature conditions are so variable in
mines that rigid mathematical solutions for

average temperatures and absolute pressures
are impracticable. For practical purposes it is
accurate enough to determine the average
densities for each column from averages of
temperatures and barometric pressures.

Figure 25 shows the conditions of natural
draft. This example is calculated as follows:

SHAFT A
Average temperatures:
Wet bulb=45° F.+53° F.+2=49.0° F.,
Dry bulb=50° F.+-59° F.+-2=54.5° F.


36 INTRODUCTION TO MINE VENTILATING PRINCIPLES AND PRACTICES

Natural draft with fan,
total quantity increased

Combined characteristic

H total
Fan characteristic

PRESSURE

{__Q Natural draft

Natural draft against fan,
total quantity decreased

Fan characteristic

PRESSURE

Combined
characteristic

T Qnatural draft 1

QUANTITY

Ficure 24.—Influence of Natural Draft to Mechanical-
Ventilation Performance.

Average barometric pressure= 28.95 inches hg,
+29.80 inches hg --2=29.38 inches hg.
Dry-bulb temperature minus wet-bulb tem-

perature equals wet-bulb depression.

54.5° F.—49.0° F.=5.5° F. wet-bulb depression.
From psychrometric tables (table 3), as pre-

viously explained for air densities: At air tem-

perature 54.5° F. and 5.5° depression, the

temperature at the dewpoint is 43.7° F. The

vapor pressure (Vp) at 43.7° F. is 0.284.

Average density

1.327
= Fg 77 B-0.378xVP),

1.327
= 5p 5 (29-38—-(0.378X0.284)],

=0.00258 X 29.27,
=0.0755 p.c.f.

SHAFT B

Average temperature:
Wet bulb, 65° F.+63° F.
Dry bulb, 65° F.+63° F.+2=64° F.
Average barometri¢ pressure=28.75 inches hg
-+29.60 inches hg +2=29.18 inches hg.
With saturated air, the temperature at the
dewpoints is the temperature of the air—in this

instance, 64° F.; the vapor pressure at 64° F. is
0.595.

Average density

1.327

= “Hyp X(29-18— 0.378 X0.595)],

=0.00253 X 28.96 inches hg,
=0.0733 p.c.f.

SURFACE DENSITY

The surface density must be determined
because shaft B is 200 feet higher than shaft A,
and this air column also acts on the system.
Temperature at surface: 45° F. wet bulb,

50° F. dry bulb.

Wet-bulb depression, 5°.

Temperature at dewpoint, 40° F.

Vapor pressure at dewpoint, 0.247.

Average barometer= 28.75 inches ++ 28.95 inches
~+2==28.85 inches hg.

Average density

1.327
= S10 — [28.85— (0.378 X0.247)],

= 0.0026 X 28.76,
=0.0748.

MINE DENSITY
Average temperature:
Wet bulb=53° F. +65+2=59°,
Dry bulb=59° F.+65+-2=62°.

Average barometer=29.80 inches+29.60
inches-+2=29.70 inches hg.

Wet-bulb depression, 3° F.

Temperature at dewpoint, 57° F.

Vapor pressure at dewpoint, 0.465.

Average barometer=29.70 inches hg.


SOLUTION OF PROBLEM EXAMPLES 51

6
p— Present operating point
Operating point
5 Present fan speed new
mine characteristic
Mine resistance Sy

“4
5 Fan characteristic
= Operating point 1,400 r.p.m.
if New fan speed new
° mine characteristic
g
5 3 _
z \e Fan characteristic
ul 1,170 r.p.m.
a
3 a +
8 y, New mine resistance
Pa
a2

1

0

20 30 40 50 60 70 80 90

VOLUME, THOUSAND C.F.M.
Fieure 31.—Graph of Fan Data. Problem 4.

Calculating other resistance factors—
Points 3-4:
H=0.05 inch water from gradient,
q=39,500 c.f.m. average,

0.05 _ 0.05
~ (0.395)? 0.156

Points 4-5:
H=0.20 inch water from gradient,

=0.321.

q= 28,000 c.f.m. average,

0.20 0.20
=(.28)* 0.0784

R =2.55.

Points 5-6:
H=0.10 inch water from gradient,
q=18,500 c.f.m. average,

0.10 _ 0.10
=(.185)°~ 0.03247

R:

Points 6~7:
H=0.10 inch water from gradient,
q= 13,500 c.f.m. average,

0.10 0.10

= (135) 0.18374


AIR LOSSES FROM LEAKAGE

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54 INTRODUCTION TO MINE VENTILATING PRINCIPLES AND PRACTICES

Calculation with quantities:

Si
a

60,000 1,400

50,000 S,”

_1,40050,000
60,000”

Sp:

S,=1,170 r.p.m. (approx.).

If a fan-characteristic curve for 1,200 r.p.m.
is not available, the necessary points of a curve
at 1,170 r.p.m., for example, can be calculated
by using the previously mentioned fan laws and
the 1,400-r.p.m. characteristic (fig. 31).

Pressure volume characteristic

Fan operating at 1,400 r.p.m.

‘Fan operating at 1,170 r.p.m.

‘Volume, c.f.m. | Pressure #2,

inches water

Volume, c.f.m. Pressure Hy |_-——————
q inches wat 1,170 1A170\3,

Fa) | (7am) x

pesos
peepee
SEserz

The intersection of this curve with the new
mine-resistance curve indicates the pressure and
quantity at which the fan will operate.

Figure 32 is the new line diagram showing the
proposed distribution of air.

VENTILATION REFERENCE MATERIAL

The following reference material is listed for those
desiring a more comprehensive reference to certain
phrases of mine ventilation:

1. Barensruc, A. W. T. Psychrometry and Psy-
chrometric Charts, ‘Transvaal and Orange Free
State Chamber of Mines, Johannesburg (South
Africa), 1955.

Bearp, J. T. Mine Gases and Ventilation of
Mines. "McGraw-Hill, New York, N:¥., 2d ed.,
1920, 433 pp.

3. BromiLow, Joun G. Ventilation of Deep Coal
Mines. ’Iron and Coal Trades Rev., vol. 170,
1955: Feb. 11, pp. 303-308; Feb. 18, pp. 367-376;

Feb. 25, pp. 427~

4. . Ventilation of Coat Mines. Inst. of
Heat. and Vent. Eng. Jour., vol. 19, No. 199,
March 1957.

5. Cauzen, A. C., and Smrru, C. M. Measurement of
Air Quantities and Energy Losses in Mine
Entries. Parts I, II, III. "Univ. Illinois Exp.
Sta. Bull. 158, 1926; Bull’ 170, 1927; Bull. 184,

1928.

6. CoLutery ENGINEERING. Velocity and Energy of
Mine Air. Vol. 19, No. 215, January 1942,
pp. $12.

7. Cooke, W. E. Natural Ventilation. Coll. Eng.,
vol. 11, No. 129, November 1934, pp. 375-378.

.. Ventilation Economics. Coll. Eng., vol.

35, No. 413, July 1958, pp. 289-294.

9, Fuvece-De Suir, R. A. H. Notes on Fan Pres-
sures. Jour. Chem. Met. Min. Soc. South
Africa, vol. 56, No. 9, March 1956, pp. 325-330.

10. Greenwatp, H. P., and McEtroy, G. E. Coal
Mine Ventilation Factors. Bureau of Mines
Bull. 285, 1929, 106 pp.

11. Huvsuey, F. B, The Determination of Pressure
Losses in’ Mine Shafts, Trans. Inst. of Min.
Eng,, vol. 108, 1949, pp. 613-627.

12, MacWiiutams, F, Fraser, Power Savings Made
by Careful ‘Study of Ventilating Plans. Coal
Age., vol. 44, September 1939, p. 35.

13. Mancna, Rarmonp._ Pitot Tube Field Tests of

Axial Flow Mine Fans. AIME Tech. Paper
1

14. Marvin, C. F. Psychrometric Tables for Obtain-
ing the Vapor Pressure, Relative Humidity, and
Temperature of the Dew-Point. U.S. Weather
Bureau Pub. 235, 1915.

15. McEtroy, G. E. Mine-Air Flow. Trans. AIME,
vol. 74, 1926, pp. 297-311

16. —..’ Engineering Factors in the Ventilation of
Metal Mines. Bureau of Mines Bul. $85, 1939,
1 .

17, ——— 'Historicol Résumé of Mine and Tunnel

Ventilation Studies, 1910-49. Bureau of Mines
Inf. Circ, 7556, 1950, 16 pp.

18. McEtroy, G. E., and Ricwarpson, A. S. Deter-

ing the Airflow Resistance of a Small Shaft
Mine by Natural Draft. Bureau of Mines Rept.
of Investigations 2890, 1928, 15 pp.

19. ~——. Experiments on Mine-Fan Performance.
Bureau of Mines Tech. Paper 447, 1929, 61 pp.

20. Peeve, Roserr. Mining Engineers’ Handbook.
John Wiley & Sons, New York, N.Y., 1941.

21. Penman, D., and Paymay, J.S. The Principles
and Practice of Mine Ventilation. C. Griffin
(London), 2d ed., 1947, 337 pp.

22. PropuctioN ENGINEERING. Guide to Fan Selec-
tion. Vol. 29, No. 17, Apr. 28, 1958.

23, Ranson, 8. R. ‘Elementary Notes on Flow of Air
and Atkinson's Formula. Jour. Min. Vent. Soc.
South Africa, vol. 11, No. 2, December 1958.

24, Roars, A. W. Features of Axial Flow Fans.
Air Cond. Heat. and Vent., vol. 55, January 1958.

25. Saxe, J. H. Value of Ventilation Surveys. Coal
Mining, vol. 24, January 1947.

26. Tonen, K. Fundamental Fan Laws. Power and
Works Eng., vol. 52, No. 609, March 1957.

27. TRICKLER, Fan Laws and System Curves.
Heat. Pip. and Air Cond., vol. 28, December
1956, pp. 105-106.

28. Uran Untversiry ENGINEERING Exrerment Sta-
tron. Fan Performance and Selection Using
Dimensionless Ratio, Bull. 66, May 1954.

29. Witutamson, J. N. How to Measure Pressure
Losses in Ventilation. Coal Age, vol. 37, August
1932, pp. 297-298.

30. ———. How to Make Air Surve
Results. Coal Age, vol. 37,
pp. 331-333.

and Calculate
ptember 1932,

U. 8, GOVERNMENT PRINTING OFFICE : 1960 0 ~ 540557


Iv

1. Velocity-pressure data converted to velocity_

2. Approximate air densities for dry air at dif-
ferent temperature:

3. Temperature of dewpoint in degrees Fahren-
heit (pressure=29.0 inches)__._---_--_.

CONTENTS

TABLES
Page
12 4, Bureau of Mines schedule of friction factors
for mine airways. -
145. Fan-test data sheet, fan exhausting.
5 & Ran-test data sheet, fan blowing.
1

Page

28
29


SOLUTION OF PROBLEM EXAMPLES

C, and D, as shown. Other segments are the
shaft and the slope. Computing intakes, we
begin at the slope—the first segment. This is
simple because there is no leakage. The R
factor has been established, so that pressure is
the intake quantity squared times the R factor.

Segment A~B begins the Main South split.
Three units to operate. The number of entries

and the unit R factor have been established.
The distance is 2,000 feet, so the total R is
twice the unit 2 per 1,000 feet. The air quan-
tity at the beginning is the total incoming air
(270,000 c.f.m.) less the following deductions
for the North split:

Chm.

50, 000
40, 000

43

Therefore,
1=270,000—110,000= 160,000 c.f.m.,and a=q
less assigned leakage along mains, unless actual
losses are known. Leakages are assigned on a
distance basis, so losses of 30,000 c.f.m. were
assigned ; therefore,

g2= 160,000—30,000= 130,000 c¢.f.m.

The average quantity is, of course, the aver-
age of 160,000 c.fm. and 130,000 c.f.m., or
145,000 cfm. The average air quantity is
then squared, and the pressure loss through
segment A~B is calculated from the resistance
factor.

The return air is calculated on the same basis,
except with returns the air volume is increasing
as the various losses and split returns build up.

Secondary splits to operating sections X and
Y were not calculated, since the drop through
section Z at 50,000 c.f.m. was obviously greater
than the drop through these sections.

Computation sheet

Actual quantity

H=R@ Q 100,000

Dis- | Num- R, Qe Gain | Aver- Seg-
Mine segment tance, | ber of| Area | 1,000 R,_ |begin-| Q:, or age @ =} ment | Total H

feet | en- feet | total | ning | end | losses H

tries
Slope......--.---.}| 1,500} 1] 180 )-..__...| 0.32 | 2.70] 2.70 }.--...} 2.70] 73) 234 |.-.-____
A-B. 2.59
B-C 7 2.69
c-D- 7 2.72
Drop across faces__ 12.97
RETURN

4) 72} 0.12 | 0.18] 0.50 | 0.60 |+.10 | 0.55} 0.3 | 0.05 )...-_-._
4) 72 12) 18} 195] 115 4.20] 105) 11) .20 0. 25
4| 72 12) 124] 1.30 | 160 |4+.30) 145] 21] .50 75
1| 256 |-.-2. 2. 02 | 2.70 | 2.70 |... 2.70) 7.3) 15 ae 30

‘Intake. ? Return, 2 Split.

Conclusions: These computations show that
approximately 4.0 inches water pressure will be
necessary. Over half of the system energy loss
is in the slope, which will have a velocity of
approximately 1,800 f.p.m. blowing dust into the
mine. The North split of two units must have
the same pressure drop, obtained by regulation.

PROBLEM 2: FAN PRESSURE NECES-
SARY TO INCREASE AIR QUANTITY
IN AN OLD MINE

A belt mine proposes to increase ventilation
at each operating unit to 25,000 c.fm. The


SOLUTION OF PROBLEM EXAMPLES 41

therefore

a _ fH, fA, fA, [x
Volk ViVi

All pressure losses H in a parallel circuit are

the same, so the symbol Vf can be removed;

therefore
1 1 1 1 1
Ss sete tetHS
VR VR, VR, VR; VR

SOLUTION OF PROBLEM EXAMPLES

The following problems are typical of those
often encountered in mines. Without the
application of ventilating principles, the solution
often is left to trial-and-error methods. We do
not maintain that solutions derived by calcula-
tion give the absolute answer; conditions in a
mine are not static, and neither are mining
methods. Calculated solutions, however, point
out with a relatively good degree of accuracy
what must be done to achieve certain desired
results and what can be expected of the present
ventilating system under various conditions.

PROBLEM 1: PRESSURE VOLUME
COMPUTATIONS FOR A NEW MINE

A new coal mine is projected, with entrance
by slope 1,500 feet long. Nine main entries
will be standard for development, five as intake
and four as return. One return shaft is planned
near the slope bottom, and an intake shaft and
portal are planned 5,000 feet north of the slope

ottom. The slope will have an area of 150
square feet and a perimeter of 55 feet. A belt
and supply track will be within the slope. The
return shaft will be circular with a diameter of
18 feet and will be 500 feet deep with an escape-
way. Two hundred feet of the shaft and 500
feet of the slope will be concrete lined.

Will it be possible to operate five units per
shift and drive the 5,000 feet to a proposed new
portal, using the slope for an intake? Two
main splits are planned with the maximum of
three units on any split. The amount of air
desired at each unit is 25,000 c.f.m.; a shop at
the slope bottom will require 10,000 ¢.f.m.

In the solution of this or any similar problem
the first step is to determine the resistance
factors.

CALCULATION OF R FACTORS

Slope area=150 sq. ft. Perimeter=55 feet.
Length= 1,500 feet.

The friction factor K must be established:

Five hundred feet of the slope is smooth lined,
moderately obstructed by belt; K=35 (ref.,
table 4).

The remaining 1,000 feet is in sedimentary
rock moderately obstructed by belt; K=85
(ref., table 4.)

SLOPE
Weighted average=500 X35= 17,500
102,500.
=1,000 X85=85,000

102,500-+-1,500=68;
therefore
K=68 average,

KLO 68X 1,500 X55

Rae 5A 52x 150 X150X 150

=0.32.

SHAFT

Diameter=18 feet. A=xX(9)?=254 sq. ft.
Perimeter=xD=n X 18=56.5 ft.
Determine K factor:
200 feet of concrete-lined shaft, obstructed
by escapeway; use maximum K=35 (ref.,
table 4);
300 feet of sedimentary rock, also ob-
structed by escapeway; K=85 (ref.,
table 4).

Weighted average=200X35= 7,000
32,500.
300 X 85=25,500
32,500+500=65;
therefore

K=65 average,

KLO___65X500X57
X 254 X 254 X 254

=0.02.

ENTRIES

With entries, R factors are calculated on a
basis of unit length per 1,000 feet; these entries
will average 6 by 12 feet.

Area=72 sq. ft. Perimeter=36 ft.

The friction factor will be 80 for intakes and
100 for returns, since intakes will be better
miaintained than returns. Moreover, this is
a new development. For old entries, the K
factors should be substantially greater. These
friction factors were based upon open field ex-
perience.


AIR MEASUREMENTS 11

or tubing. When measuring air velocities in
ducts or tubing, again the relative accuracy of
centerline versus traverse measurements be-
comes important. Consequently, the end use
of the data will determine the degree of accuracy
needed. For most mine work, other than spe-
cial tests, centerline measurements multiplied
by a correction factor of 90 percent is sufficiently
accurate. For special tests requiring a high
degree of accuracy, point traverses should be
made. Square ducts can be divided into a
number of imaginary squares, and determina-
tions made at the center of each square, measur-
ing the distance that the pitot tube penetrates
the duct. A number of holes in the duct
opposite the row of squares to be measured are

necessary. These holes can be plugged when
not used. With round ducts the problem of
obtaining average velocities is substantially
more difficult.

The problem is to divide the cross-sectional
area of the duct into equal areas, which can be
sampled. Figure 9* shows the position of
traverse points in a circular air-measurement
section for a 5-area, 20-point traverse. Four

‘McElroy, G. E., Engineering Factors in the Ventilation of Metal
‘Mines; Bureau of Mines Bull. 385, 1935, fig. 3-B, p. 11,

0.342 dia.

-—0.226 dia —-|
[0.146 __|

dia.

0.082 dia.—-] L

0.026 dia.} +

Quadrant A

a

holes are necessary in the horizontal and verti-
cal diameters of the circular section to make
the measurements.

Pitot-tube determinations of velocity pres-
sures in inches of water can be changed into
velocity of airflow, feet per minute, either
by calculation or by reference to conversion
tables. Conversion tables (see table 1) are simple
to use. For example, the velocity represented
by a velocity pressure of 0.78 inch water is
3,533 f.p.m. If tables are not available or do
not cover the specific measurement, calcula-
tions can be made direct from the velocity
pressures. The simplified formula for standard
air density of 0.075 p.c.f. is:

Velocity (f.p.m.)=4,000/ VP inches water.

EXAMPLE: What velocity is represented by a
velocity pressure of 1.2 inches water? If this
is a centerline reading, what is the corrected
velocity?

V=4,000 y1.2
The square root of 1.2= 1.096;
therefore

The velocity =4,000 X 1.096 =4,384 f.p.m.

2)3\ 4

Quadrant

wey,

Quadrant ©

Ficure 9.—Diagram Position of Traverse Points, Circular Air-Measurement Section.


53

SOLUTION OF PROBLEM EXAMPLES

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AIR MEASUREMENTS 13

The corrected velocity for a centerline measure-
ment is

4,384 X90 percent=3,946 f.p.m.

When’ determining the average of several
readings made with a pitot tube to determine
true velocities, the average velocity of a
section cannot be obtained by the direct
average of the velocity pressures obtained at
traverse points, because velocity pressures vary
as the square of the velocities. The simplest
method is to determine first the velocity
pressures of each traverse point, then the square
root, and average the square roots. Otherwise,
the velocities of the separate traverse points
must be determined and averaged.

EXAMPLE: What is the average velocity repre-
sented by the following velocity pressures at
traverse points across a circular section? (The
following velocities will not compare exactly
with velocities listed in table 1, since the square
roots of the velocity pressures were carried only
to two decimal places in the example, while
those shown by table 1 were calculated more
accurately. The results obtained are suffici-
ently accurate for most uses.)

Point | vP VVP | V,f.p.m.
0. 72 0. 85 3, 400
277 - 88 3, 520
83 291 3, 640
95 197 3, 880
84 192 3, 680
77 +88 3, 520
74 - 86 3, 440
Total. 6.27 25, 080
Average.--.._|--------_- . 896| 3, 583

Average velocity calculated, 4,000 0.896
=3,584 f.p.m., compared with 3,583 f.p.m. for
the average velocity.

INFLUENCE OF AIR DENSITY

The density of the air is considered unim-
portant when measuring with an anemometer
or with a smoke tube, as the density influence
with these devices is minor and within a normal
error of measurements. When measuring by
pitot tube, however, the air density should be
considered if a substantial variance exists from
standard air density of 0.075 p.c.f. 'Tempera-
tures, barometric pressures, and relative humid-
ity are variables that must be considered. The
correction factor to be used with air densities
other than standard is: The square root of 0.075
(standard density) divided by the true density

540577 O—60 —3

= 5. This correction factor, multiplied

by the velocity, equals the true velocity within
a practical degree of accuracy.

Exampe: What is the velocity represented
by a velocity pressure of 1.31 when the air
density is 0.062 p.c.f.?

(0.075
V=4,000V1.31X 5953"
=4,000V1.31 Xy1.21,
=4,000X1.145X1.10,
=4,000X 1.26,
=5,040 f.p.m.

From tables:
Velocity at velocity pressure 1.30=4,580,

0.075
0.062’
=4,580Xy1.21,
=4,580X1.10,
=5,038 f.p.m.

Air densities may be determined from charts
or can be calculated from available data. The
general formula is:

V7=4,580X

1,327

BOLT

B; d=density, p.c.f.,

T=temperature air, °F. dry bulb,
B=barometer, inches mercury,

460=conversion to absolute temp-
erature.

Absolute temperature is the theoretical tem-
perature at which the volume of a gas is reduced
to zero; consequently it ceases to exist as a gas.
On the Fahrenheit scale such a temperature is
—460°.

1.327=conversion factor.

EXAMPLE: What is the air density at 70°
F., with the barometer at 28.75 inches mercury?

1.327
d= F179 %78-75,

27
= 30 628-75,

=0.072 p.c.f.

The above formula is accurate for dry air
only but is generally used for density determina-
tion unless precise densities are required. Table
2 gives the approximate air densities for dry
air at different temperatures.


Page PART I—Continued
Introduction. __ - 1 a
: 1  Natural-ventilation influence of air densities—
2 Continued
: 2 Mine density.
Weighted average of densities.
Shaft A.
. Shaft B
Air measurements. 5
Moderate-veloc 5 PART II
Low-velocity airflow measurements 8
High-velocity airflow measurements_ 9 Application of airflow fundamentals.
Influence of air density... 13 Derivation of ventilating formulas
Fundamentals of airflow. - 16 Solution of problem examples
Pressure losses... - 16 Problem 1: Pressure volume computations
Parallel flow in airways_ _ a7 for a new mine... __ -
Splitting air currents. 18 Calculation of R factors
Air losses from leakage . 22 Explanation of computation sheet-
Mechanical ventilation - 24 Problem 2: Fan pressure necessary to i
Selection of fans_ - - 24 crease air quantity in an old mine.
Fan tests. . rrr Calculation of F factors.
Fans in series. 29 Problem 8: Practical application
Fans in parallel . 30 and air densities
Fans in combination _ 35 Computation
Natural-ventilation influence of air densities... 35 Problem 4: Improved face ventilation by
Shaft A 35 means of new air shaft... ..
Shaft B 36 Calculation of R factors.
Surface density 36 Ventilation reference material
ILLUSTRATIONS
Fig. Page Fig.
1. Illustration of static and velocity pressures_. 3 16. Corrected fan-performance curve at average
2. Anemometer calibration curve 6 local air density.
3. One-half area traverse, left sid - 7 O17. Pressure changes i a ventilating system for
4. One-half area traverse, right side. 7 lowing and exhausting fans.
5. Bureau of Mines-type smoke-cloud gener- 18. Com ined characteristics of two identical
ator _. : 8 fans operating in se:
19. Combined characteristics of two different
8, Smoke-cloud release 9 fans operating in series. -
7. Example of smoke-cloud measurement. 10 20. Two identical fans operating in parallel.
8. Illustration of pitot tube__._.-_.__ 10 21. Two different fans operating in parallel____
9, Diagram position traverse ‘points, circular- 22. Parallel fans operating on separate airways_
air-reasurement: section 11 23. Combined operation of two fans.
10. Airflow with natural splitting. - 18 24. Influence of natural draft to mechanical
11. Example of airflow splitting - 20 op ientilation performance aos jugat deat.
9 . Example of conditions causing natural
Be Tee of aiflow splitting, with regulation. 3226; Mine and ventilation plan: Problem 1
+ Mustration of air-leakage losses. - 3 97. Mine and ventilation plan: Problem 2_
14, Relationship between fan character 28. Plotted fan data: Problem 3.
mine resistance -- 25 29. Line diagram: Problem 4_
15. Comparison of characteristic curves for fan 30. Pressure gradient: Problem 4...
operating at different speeds and blade 31. Graph of fan data: Problem 4__.
settings n-------------. 26 32) Line diagram with new air shaft: Problem 4.

m


INTRODUCTION TO MINE VENTILATING PRINCIPLES AND PRACTICES

34

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FUNDAMENTALS OF AIRFLOW 21

Regulating splits 2 and 3 to cause 70,000
c.f.m. in split 1 will change the mine pressure-
volume relationship, since split 1 now will have
a pressure drop of more than the original 1.0
inch, owing to the increased flow against the
same resistance. The new pressure losses in
splits now will be:

Split 1:
q=70,000, and R=3.62,
calculated above;
Q=0.70,
RO*=3.62 X (0.70)?
=3.62 X0.49= 1.77 inches water.

This shows an increase of 0.77 inch water in
split 1. Each split with regulation must now
carry this same pressure drop.

With increased resistance of the mine system,
the air volume delivered by the fan, based upon
the pressure-volume characteristic, will be
substantially less. To determine the amount
of this reduction a trial-and-error approxima-
tion must be made.

Step 1.— Assume an air-volume reduction of 15
percent (150,000 85 percent=127,500 c.f.m.).
Then from the fan characteristic (fig. 11) a
volume of 127,500 ¢.f.m. will be induced at a
pressure of approximately 3.92 inches water.
This pressure consists of segments A-B, B-C,
and C-D; Segments A-B and C-D will vary
with the total quantity. Segment B-C at
70,000 c.f.m. is fixed at 1.77-inch water gage.

Step 2.—Calculate pressure drop, segments
A-B and C-D. Calculating A-B, R factor
67, new g=127,500 c.f.m., @=1.275. New
H=R(?=0.67 X (1.275)?= 1.09 inches water.

Calculating C-D, R factor=0.44, new ¢
=127,500 c.f.m., Q=1.275. New H=0.44
X (1.275)?=0.72 inch water.

Step 3.—Sum up new pressure losses, inches
water gage:

Inches

1.09
_ 177
we 72

-. 3.58

Total water gage...

(Bee fig 12)

Step 4.—Compare the pressure 3.58 inches
with the fan characteristic (fig. 11). This is a
delivered quantity of approximately 142,000
c.f.m.; so the true volume falls between 127,500
¢.f.m. and 142,000 ¢.f.m.

Recalculating: Steps 2 and 3 at 135,000 ¢.f.m.:
This air quantity at approximately 3.80 inches

340357 0604

water is accepted as within practical limits.
(See fig. 12.)

After establishing that with regulation the
new pressure-volume relationship would be
135,000 c.f.m. at 3.80 inches water, the size of
regulators can be calculated.

he total air to be divided by splits 2 and 3
is 135,000 c.f.m., less the 70,000 c.f.m. for split 1;
this equals 65,000 c.f.m., which can be split in
any quantity desired. In this instance, allow
35,000 c.f.m. for split 2 and 30,000 c.f.m. for
split 3.

Split 2, with 35,000 c.f.m. flowing, will have
a pressure loss equal to the resistance factor
X (0.35)? =4.35 X (0.385)? =4.35 X 0.122 =0.53
inch water.

Split 3, with 30,000 c.f.m. flowing, will have a
pressure loss equal to the resistance factor
X (0.30)? =4.08 X (0.30)?=0.37 inch water.

The required regulations for split 2 equals
the pressure loss through split 1, less the pres-
sure loss through split 2 with 35,000 c.f.m.,
that is, 1.77—0.53. The result: 1.24 inches
water pressure must be consumed through reg-

ulation. The formula a= applies, where
H=1.24 inches, Q= (0.35).

40X0.35_ 14
= == 12.6 sq. ft.
1.24 v1.24 4
Split 3, calculated the same way:

H=1.77—-0.37=1.40, g=30,000, Q=0.30,

30 12
=T7g7 102 sq ft.

(Calculation should be checked for errors.)

Segment A-B: 35,000 cfm, R=0.67,
=0.67 X (1.35)?=1.22 inches
water.
Segment B-C 1.77, previously calculated.
Segment C-D 35,000 cfm. R=0.44,
H=0.44& (1.35"?=0.80 inch
water.

The total pressure of the system then equals
1.22 plus 1.77 plus 0.80 equals 3.79 inches
water.

This checks reasonably close with the fan
characteristi¢ for 135,000 c.f.m.; therefore, cal-
culations are assumed to be correct. Should
more air than 135,000 c.f.m. be required, with
70,000 for split 1, the fan must be altered either
by changing the blade position or increasing
the speed. Figure 12 illustrates the completed
airflow pattern and pressure gradient.


UNITED STATES DEPARTMENT OF THE INTERIOR
Fred A. Seaton, Secretary

BUREAU OF MINES
Marling J. Ankeny, Director

This publication has been cataloged as follows:

Kingery, Donald Samuel.
Introduction to mine ventilating principles and practices.
Washington, U.S. Govt. Print. Off., 1960.

iv, 54 p. illus, tables. 26cm. (U.S. Bureau of Mines. Bulletin
589)

Bibliography: p. 54.
1, Mine ventilation. I. Title. (Series)

TN23.U4 no. 589 622.06173
U.S. Dept. of the Int. Library

For sale by the Superintendent of Documents, U.S. Government Printing Office
Washington 25, D.C. - Price 40 cents


46

INTRODUCTION TO MINE VENTILATING PRINCIPLES AND PRACTICES

Computation sheet

Distance] Num- R, 1,000] R, | Q:,be-} @Q2, | Gains | Aver- Seg-
Mine segment feet | ber of | Area feet total | gin- | end or age Q | ment | Total
entries| ning losses H
INTAKE
-|--------|---- 24! ~-+2--]-----x-- 10. 005 | 4.2 17.6 j-.---- 0.09
-| 2,000 72| 0.09} .18 | 20 3.06 | 0.55 - 64
500 4) 72 -09} .05 | .80 56 |. 03 - 67
Cc-D - . | 1,500 72 - 38 | . 57 65 - 34] 219 . 86
Drop through
sections,
0.25 inch
Oa a es Pees Dns DOs ReneS Li
RETURN
D-C__... ._. __| 1,500 2 72 0. 48 | 0. 72 |20, 55 | 0. 70 |+0. 15 | 0.63 | 0. 40 | 0. 29 1. 40
C-B_ 500 5 72 -08 | .04] .80] .90] +.10} .85) .72] .03 1.43
i 500 5| 72 208 | 204] 1.60) 1.75 | 4.15) 1.68 | 282] .11 1. 54
X- - 1, 500 5 72 -15] 22] 1.75 | 2.10] +.35 | 1.93] 3.72] . 82 2, 36
Return sh es ee ee ee . 005) 4.2 | 4.2 _----| 42 (17.6 . 09 2.45
X=point where higher R factor is necessary for caved returns. Total H=2.45. This indicates a pressure of

approximately 2.45 inches water necessary to operate 8 units and provide air for belt ventilation.

+ 5,000 c-f.m. for belt entry not calculated in intake-entry pressure

losses.
2 5,000 ¢.f:m. from helt entry to return entries flows in return entries; calculated in return pressure losses.

PROBLEM 3: PRACTICAL APPLICA-
TION OF FAN LAWS AND AIR
DENSITIES

This problem illustrates the application of
fan laws, influence of air densities, and natural
draft to fan performance.’ The problem is en-
tirely hypothetical.

The following data are available on the per-
formance of two different fans:

Fan A Fan B
_ 700 800
of 0.070 0.075
Pressure H, inches water at following
air volumes, c.f.m.:

4.90 6.55
462 6.05
416 5.45
3.64 4.70
3.04 3.85
229 © 280

These two fans are to be installed in series
at an air density of 0.070 p.c.f.; both fans will
operate at speeds of 800 r.p.m.

_ Mine resistance has been established at 3
inches water pressure for 100,000 c.f.m. at a
density of 0.075 p.c.f.

The natural ventilating pressure is 0.5 inch
water assisting fans.

What volume will the fans deliver? At what
pressure will each fan operate?

5 Problem similar to one in article by Beadle, D. G., and Quilliam,
J. H.: Jour. Mine Vent. Soc. of South Africa, vol. 10, October 1957, p. 239.

COMPUTATION

The first step is to correct the pressure-
volume data for each fan to 800 r.p.m., and a
density of 0.070 p.c.f.

Fan A:
Fan law No. 1, correction for speed:

. : 800
Multiply given volumes by Ani
Fan law No. 2, correction for pressure:
. : '800\?
Multiply given pressure by () :

Fan B:
No correction necessary for speed.
Correction for pressure:
: . 0.070
Multiply given pressure by 0.075

Fan A

[Air density, 0.070 p.c.f.]

R.p.m,, 800

Pressure,

inches
water

Volume, c.f.m.



30

The flow conditions for series operations are
determined by mine resistance and the com-
bined pressure characteristic of the fans. The
combined pressure characteristic is obtained by
plotting the sum of the pressures for the same
quantity.

Figure 18 shows the combined characteristics
of two identical fans operating in series. With
a single fan, the quantity was 145,000 c.f.m. at
4 inches pressure; with the fans combined, the
quantity increased to 183,000 c.f.m. at a
pressure of 6.6 inches, divided equally at 3.3
inches for each fan. Figure 19 shows the
combined characteristics of two different fans
operating in series. The total air volume
delivered is 162,500 c.f.m. at 6.6 inches pressure.
The pressure will be supplied as follows: 1.9
inches by fan A and 4.7 inches by fan B. Unless
the two fans are properly selected for the work
to be done in series operation, they will not
operate together at maximum efficiency, and
the results will be disappointing. The mine-

INTRODUCTION TO MINE VENTILATING PRINCIPLES AND PRACTICES

resistance curve must intercept the combined
characteristic curve of both fans.

FANS IN PARALLEL

Fans may be used in parallel to increase the
air volume against changed mine resistance.
As a mine is developed and more air is needed,
a common practice is to install an available
fan in parallel with the operating fan.

Fans may be placed together to operate on
the same airway or they may be on separate
airways. When operating on the same airway,
they act together on the total flow. The
primary condition is that the pressure generated
by each fan must equal the pressure loss
sustained by the total volume they produce.
The combined performance characteristic of
fans acting in parallel at the same point in the
system is obtained by plotting the sum of the
quantities handled at the same pressures
against the pressures. The intersection of the

Combined

|
&
Bt :
z 1
co} 1
8 |
jr |
z
wh i
2 4 Single ___ 4
g 1
g
2 1 1
& LN
+ ©. Y r 4
r 1
H '
Mine resistance I |
ab 1 H 4
\ \
! \
! \
| 4
IF i
| !
| |
1 1
i L {y {4 1
) 30 100 150 200 250

300

VOLUME, THOUSAND C.F.M.

Fieure 18.—Combined Characteristics of Two Identical Fans Operating in Series.


PART II
APPLICATION OF AIRFLOW FUNDAMENTALS

The second part of this bulletin is devoted to
the practical application of airflow fundamen-
tals. Several problems are discussed; a general
knowledge of ventilation requirements and
methods used in mines and an understanding
of simple algebra are presupposed. The deri-
vation of certain basic ventilating formulas is
given.

DERIVATION OF VENTILATING
FORMULAS

1. Formula for conversion of velocities and
velocity pressures:

V=4,000 ¥VP
where

VP equals the velocity pressure in inches water.
This formula is derived from the basic formula
Vex 2gh where V= velocity Eps. A
=/2gh ravity of 32.2 f.p.s.
e fa feieht in foot.
Changing the velocity from feet per second
to feet per minute, the formula becomes:

V=60X2gh.

Extracting the square root of 2g, or 232.2,
and removing from under radical

V=8.02X60 vi.
=481.2 Vi,

The height of air column h can be converted
to equivalent inches of water at standard air
density as follows:

1 foot of air column at 0.075 p.c.f. = 0.075
-> by 5.2, or 0.0144 inch water pressure;

1 inch of water pressure then equals 1
+ by 0.0144, or 69.5 feet of air column.

The formula then becomes:

V=481.2 ¥69.5X VP

where VP is expressed in inches water pressure:
V=481.2X8.34 JVP

V—4,000 yVP for practical application.

2. Formula for computing the areas or pres-
sure losses through regulators:
A=2X2e.
vH
A=area regulator, square feet,

uantity c.f.m. in units of 100,000,
H=pressure loss through regulator inches water.

This formula is also derived from V?=2gh:
V?=2gh,
V= 29h velocity f.p.s.,

V=60X 2gh velocity fp.m. (From funda-
mental laws the air quantity q equals the
velocity V times the area in square feet.)

q=VA.
V=4,000 VH (from the previous formula) ;

therefore
q equals 4,000 VHXA.

With box-type regulators and sharp-edged
orifices a correction must be made for the Vena
Contracta, which in effect reduces the area of
the orifice. The Vena Contracta is a condition
that occurs whenever there is a substantial
increase in air velocity caused by abrupt area
restrictions, such as caused by regulators.
The accelerated air momentum does not
completely fill the area of the restriction. The
percentage of area filled will be influenced by
several factors, the most important being the
airspeed through the regulator and the ratio ot
the regulator area to entry area. The correc-
tion for Vena Contracta usually is 65 percent
of the regulator area.

q=4,000 A ¥HX0.65,
q=2,600 A VH,
la
A=—2 _ >=e
2,600 Vii 2,600"

yA .00049,
=0.00039% T="



26 INTRODUCTION TO MINE VENTILATING PRINCIPLES AND PRACTICES

T T T
Axial flow fan performance
No. 1 blade setting 4
Air density, 0.075 Ib. per cu. ft.

4- 600 r.p.m.

PRESSURE, INCHES OF WATER

0 l ! | L
100 150 200 250 300 350
VOLUME, THOUSAND C.F.M.

800 r.p.m.

Axial flow fan performance
No. 5 blade setting

600 r.p.m.

PRESSURE, INCHES OF WATER

0 ! 1 | 1
200 250 300 350 400 450
VOLUME, THOUSAND C.F.M.

Ficure 15.—Comparison of Characteristic Curves for Fan Operating at Different Speeds and Blade Settings.


AIR MEASUREMENTS 9

Example of observed gmokerelowd measurement
ta

(Distance, 25 feet, in seconds)

‘Measurements
Quadrant Average,
seconds
1}2};afla|s

Upper right | nu} 2] nu} 10 10.6
Lower righ “} 13{ um] a2] out] 12 120
Upper left. | ml} my} iw] 2} 0 10.8
Lower left ‘| | 15] 13] 3] 3 13.6
‘Total 47.0

CALCULATIONS

Final average=47-+4=11.8 seconds.

Velocity =77"5 x 25= 127.1 f.p.m. Velocity equals
the distance of smoke travel divided by the time in
seconds, multiplied by 60 to convert the final velocity
to feet per minute.

Correction for flow, 10 percent; [127.1X90 percent
=114 f.p.m.=Vr, the true velocity after corrections.

Area: Average of three measurements (fig. 7)=101
sq. ft.

Area quantity= 4X Vr=101X H4=11,500 ef.m.

HIGH-VELOCITY AIRFLOW
MEASUREMENTS

The use of the ordinary vane anemometer in
velocities over 2,000 f.p.m. may crack the bear-
ings. Special high-speed anemometers, usually
with half as many vanes as the standard type,
are made for this work but, except for special
tests, seldom are required for mine work. The
high velocities most often encountered are those
flowing in ducts or tubing where measurements
by anemometer is difficult. For such measure-
ments the most practical instrument is the pitot
tube (see fig. 8), which can be inserted through
a small hole in the duct or tubing, thereby

ermitting velocity pressures to be determined
by a U-tube water gage or some other differ-
ential pressure gage.

The pitot tube is a primary standard instru-
ment for determining velocities. Commercial
types generally are accurate to within 1
percent, and specially made types can be
accurate to within 0.01 percent. The usual
form (fig. 8) is two concentric tubes, one within

Figure 6.—Smoke-Cloud Release.


SOLUTION OF PROBLEM EXAMPLES

Fan B
Pressure, inches water
R.p.m., 800; at density
volume, e.f.m. = _—
0.075 0.070
70°
(Ce) XH,
6.55 6.11
6.05 5.65
545 | 500
4.70 4.39
3.85 3.50
2.80 2.61

The curves plotted can be used to obtain
the data necessary to determine the combined

Combined curve data, fans A and B

Density, 0.070 p.c.t.

R.p.m., 800;
volume, e.t.m, tana’ | H,tanB | #H,com-
bined
611
5.65
5.09
439
359

1 Pressure-volume data, fan A, taken from curve of fan A (fig. 28)
plotted from calculated data,

pressure-volume relationship of the fans Plotting these data forms the combined
(ref., fig. 28.) pressure-volume curve of both fans.

13 T T T T T T T T T T

With natural draft
od Combined New mine resistance 7
d=0.070
lle 7
top — —-— — — SN Operating point 4
9.5"

PRESSURE, INCHES OF WATER.
©
T

6 Po NN PLOTTED DATA T
Speed 800 r.p.m.
. at 5.78” Density 0.070 Ib. “cu. ft.
AS | at 3.72" 4d
. at 9.50"+0.50"
Total 188,000 c.f.m. at 10” Fan A
ab 4
|
3 1 i L n 1 im L L L L
130 140 150 160 170 180 190 200 210 220 230 240

VOLUME, THOUSAND C.F.M.

Ficure 28,—Plotted Fan Data. Problem 3.


SOLUTION OF PROBLEM EXAMPLES

49

LEGEND '
lat 10,000 c.f.m.
Overcast Longwall face 7
N. << Intake air PX7AWY75 000 em, 8
<=—~< Retum air K Sa ~~ nem
52
R Regulator IN Oe Ce x
A \ GD S|
K N q
TINY TRY IBV ITINWW7 TRY TIRWY 77 AWNY. > mes
Z <8 3
, WeN"S t6
> Mined out \ a4 on
K S
g <8 ] é
NZNO NACI. NOL. WUT NU WE: SULESSILISSLIANEISS S
30,000 |< —_—
c.f.m,
145,000 Slope 45,000c.f.m.!L —-34,000c.fm. 4, 22,000)
a2 c.f.m. 2 te 4 Pci.mo .
—~=—~< 45,000 c.f.m. Old coal slope 15,000 c.f.m. .——< 8,000 c.f.m. VE
i0 I I J gS
Shaft WTI 7TR WY TIRWY TRY TRY TAY NW IW TRY TIN)
fo}
K 7 si
> q
K Mined out Z
Z S
K
gq
ZA NLASSIZASNSIAKASSSULA\ WU LISSNLULISSS LLNS SLLISSNULISSSLIBS

Ficure 29.—Line Diagram. Problem 4.

Segment 11-12:
H=5.50—4.60=0.90 inch water,

q=45,000 c.f.m,
0.30 0.30 |
R= a5 0.0008

Total R, points 10-12=1.48+4.44—5.92.

Parallel flow is determined from the con-
ductance formula

1

We TR Te,

1h 1

We

therefore
R, equals resistance through rock slope to

point 3; R, equals resistance points 10-12;
points 10 and 3 will become the same.

|-
- B-

a5

,

5!

pr}

5.9:

I~ Rl-
8

1
Vr 0: eat Ta a3

VR
R= pin. 585s
R=(0.585)?=0.342.
The combined F factor 0.342 is for the coal

and rock slopes from the surface to point 3 of
the ventilating pressure gradient.

1.304+-0.41=1L71,


12

Vp=Velocity pressure, inches water.

INTRODUCTION TO MINE VENTILATING PRINCIPLES AND PRACTICES

Tasie 1.—Velocity-pressure data converted to velocity

V=Velocity, feet per minute; also 4,000 ¥VP.

Vp Vv Vp Vv Vp Vv Vp Vp Vv
0. 10 1, 265 0. 69 3, 323 4, 525 6, 274
1, 327 -70 3, 346 1.29 4, 533 6, 286
1, 386, 3, 370 1.30 4, 561 1.89 6, 299
1, 442 3, 394 4, 580 1. 90 2,49 6, 312
1, 497 3, 418 4, 596 2.50 6, 325
1, 549 3, 441 4, 613 . 6, 337
1, 600 3, 464 4, 630 6, 350
1, 649 3, 487 4, 648 6, 362
1, 697 3, 510 4, 665 6, 375
1, 743 3, 533 4, 682 6, 387
. 20 1, 789 , 555 4, 699 6, 4
1, 833 - 80 3, 578 4,716 6, 412
1, 876 , 600 1.40 4, 733 6, 425,
1, 918 3, 622 4, 750 2. 00 6, 437
1, 960 3, 644 4, 767 2. 60 6, 450
2,000 3, 666 4, 783 6, 462
2,040 3, 688 4, 800 6, 475
2,078 3, 709 4,816 6, 487
2, 116 3, 731 4, 833 6, 499
29 2, 154 3, 752 4, 850 6, 512
a) 2, 191 - 89 3, 774 4, 866 6, 524
2, 227 +90 3, 795 1.49 4, 883 6, 536
2, 263 3, 816 1.50 4, 899 2.09 6, 548
2) 298 3, 837 4,915 2.10 2. 69 6, 560
2, 332 , 857 4,932 2.70 6,573
2, 366° 3, 878 4, 948 6, 585
2, 400 3, 899 4, 964 6,597
2, 433 3,919 4, 980 6, 609
2, 466 3, 940 4, 996 6, 621
2, 498 3, 960 5, 012 6, 633
- 40 2, 530 3, 980 5, 028 6, 645
2,561 1.00 4, 000 5, 044 6, 657
2, 592 4, 020 1. 60 5, 060 6, 669
2, 623 4, 040 5, 075 2,20 6, 681
2, 653 4, 060 5, 081 2. 90 6, 693
2) 683 4,079 5, 107 6, 705
2,713 4,099 5, 122 6, 717
2, 742 4, 118 5, 138 6, 729
2771 4, 138 5, 154 6, 741
. 49 2; 800 4, 157 5, 169 6, 753
- 50 2; 828 1.09 4,176 5, 185 , 765,
2, 857 1.10 4, 195 1,69 5, 200 6, 776
2, 884 4, 212 1.70 5, 215 2,29 6, 788
2,912 4, 233 5, 231 2. 30 2.89 6, 800
2, 939 4, 252 5, 246 2. 90 6, 812
2, 966 4,271 5, 261 6, 823
2, 993 4, 290 5, 276 6, 835
3, 020 4, 308 5, 292 6, 847
3, 046 4, 327 5, 307 6, 859
3,072 4, 345 5, 322 6, 870
- 60 3, 098 4, 363 5, 337 6, 882
3, 124 1.20 4, 382 5, 352 6, 893
3, 150 4, 400 1.80 5, 367 , 905,
3, 175 4,418 5, 381 2.40 6, 917
3, 200 4, 436 5, 396 3.00 6, 928
3, 225 4, 454 5,411
3, 250 4,472 5, 426
3, 274 4, 490 5, 441
3, 299 4, 508 5, 455