85 5 x A I 492 ' v 1 000

60 m3/s

This agrees with the answer obtained in Example 14 of Note 27 . Work is done in moving air against a resistance and energy must be supplied in order to do

this work. The next question is then, how much energy must be supplied?

In Note 7 it was stated that power is the rate of doing work. A joule is the amount of work done when a force of I newton moves its point of application by I metre .

When a cubic metre of air passes through an opening of one square metre at such a velocity that the pressure loss is one newton per square metre (1 pascai) it is obvious that the amount of work done is equivalent to one newton metre or one joule because the force is a newton and the air is moved a distance of one metre. If the cubic metre of air, however, passed through a two square mtre hole with the same pressure drop, the air would only have to move a distance of 0,5 metre. The work done in each square metre of the hole is now only I x 0,5 = 0,5 J but the total work is again 2 x 0,5 = I Nm = 1 J .

The amount of work done is thus independent of the cross-sectional area through which the air passes. It depends only on the volume of air which is moved and on the amount of pressure or force which is required. Thus when 60 m of air is moved through a tunnel of any size over any ditance and the pressure drop is 1 N/m2 (1 Pa), the amount of work done is 60m1 x 1 N/m2 = 60Nm = 60J .

If the pressure drop is 2 N/m2 the work done is 60 m1 x 2 N/m2 = 1 20 Nm = 1 20 J.

If the pressure drop is 49 2 N/m2 then the work done is 60 m1 x 49 2 N/m2 = 29 5 20 Nm

= 29 5 20 J = 29 ,5 kJ

If 60 m1 of air is moved every second .and the pressure drop is 49 2 Pa then work has to be done at the rate of 29 5 20 J/s = 29 5 20 watts = 29 ,5 kW. Thus the power required to move 60 m1 Is of air at a density of 1 ,35 kg/m1 through a 4 m x 3 m rock tunnel I 500 m long (Example 19) is 29,5 kW and this is called air power. (See Note 7 for definition of watt.)

From the reasoning given above, a simplified formula can be derived for calculating air power:

Wa

where Wa Q p

pQ 1 000 air power (kW) volume flow (m1/s) pressure (Pa)

This is a very important formula which must be memorized.

72

NOTE 30

NATURAL VENTILATION

When air flows, work is done, and for work to be done energy (power) is required. This power can be supplied by a machine such as a fan or co!Tlpressor or by other sources such as heat or falling water or rock.

When air movement is caused by heat (normally supplied by the rock) or falling water (normally from a fissure in the rock) it is usually called natural \"entilation.

I I 11

Out 5tdc 2-1 I Aor I I 11 11 I ''-'""'-"""=i I

Adt

FIG. JO

_./Shaft

In a mine having a shaft and an adit into a hillside as in Figure 10, air will travel down the shaft during summer and up the shaft during winter because the temperature of the air in the mine shaft stays comparatively constant throughout the year due to the consistency of the rock temperature, while the temperature of the outside air changes with the seasons. During summer, when the outside air is hot, it is too light to balance the cool column of air in the shaft and consequently air moves down the shaft and out of the adit. Jn winter the cold outside air is heavier than that in the shaft and consequently air enters the adit and upcasts through the shaft.

,.,,,,.,,...""liliP""'

Once the flow of air has been started by one of these methods, the rock in the downcast shaft would be cooled more than in the upcast shaft and the air would continue flowing because it would also gradually become cooler in the downcast. Consider such a case as applied to a deep mine:

I Deplh Virgin Rock m Temperature 'C Top of Shafi 0 18 Middle of Shafi I 000 27 Bollom of Shaft 2000 36 Mean

TABLE 18

Downcast Shaft Barometric

Pressure Temperatures Density kPa 'C kg/m'

81 10/10 0,99 92 16/16 1,10

103 22/22 1,21 16/16 1,10

Upcast Shaft

Temperatures I Density 'C kg/m'

2/22 I 0,95 27/27 1,05 32/32 1, 15 27/27 J ,05 For ease of calculation the air has been assumed to be saturated throughout the circuit. In each shaft the mass of an air column with a cross-sectional area of I m2 will be given by

the mean density of the air multiplied by the depth of the shaft, thus 2 000 x I , JO = 2 200 kg in the downcast shaft and 2 000 x 1,05 = 2 JOO kg in the upcast shaft. To obtain the pressure exerted by these two columns of air the masses must be multiplied by g. the gravitational acceleration. Thus the downcast air exerts a pressure of 2 200 x 9,8 = 21 560 N/m2 or 21 560 Pa and the upcast air exerts a pressure of 2 100 x 9,8 = 20 580 Pa.

The difference between these two pressures is 980 Pa and the available natural ventilation pressure is thus 980 Pa. It is obvious that the same answer can be obtained more expeditiously by multiplying the difference between the two masses (2 200 - 2 100 = 100) by 9,8, thus 100 kg/m2 x 9,8 m/s2 = 980 kg m s2/m2

= 980 N/m2 = 980 Pa (See Note 26) The above method of calculation involves a considerable amount of work. In practical mine

ventilation work it often happens that an approximate value of the natural ventilation pressure is required quickly. In such cases the author has found the following method of calculation, which can be done mentally, to be very satisfactory.

It is only necessary to remember two facts. The first is that approximately 0,085 m of motive column is equivalent to 1 Pa. This simply means that an air column of standard density (1,2 kg/m3) and about 85 mm high exerts a pressure of 1 Pa. Secondly, the density of air at constant pressure is roughly proportional to the absolute dry. bulb temperature, the effect of humidity being relatively small.

In the case of the example given above, the mental calculation would be done as follows: The mean absolute dry bulb temperature in the downcast shaft is 273 + 16 = 289 K.

The difference between the mean dry bulb temperature in the two shafts is 27 - 16 = 11C The moti\e column will therefore be approximately

I I 289 x 2000 m I 26 x 2000m

76 m

11

The natural ventilation pressure is therefore approximately 76

OS5 - 900Pa

This answer is very near to the correct figure calculated above, but in cases where the one air column is more humid than the other. discrepancies will be much greater.

The amount of air flowing as a result of natural ventilation pressure will depend on the resistance of the circuit. In the above case. if the shafts and airways are large. it is quite possible that 150 m3/s will flow through the mine, which means that the equivalent of nearly 150 kW of power is supplied by natural sources.

w. 150 x 980 1000

0,5 m/s

Barometer 108,0 kPa

I

147 kW

Drive 1a0Ma z.s .. 2.sm

t 33/34"C -

0,9 m/s

30/30"c l 5,6 m'"/s

FIG. 12

52 lewel &,5 m/s

Raise 50m x 1,r. m 12.sm

I

The author once encountered the case ill.ustrated in Figure 12 in a deep mine.

There was a vertical difference of 45 m be1ween the levels and the virgin rock temperature was 40C. The barometric pressure was 108 kPa.

Considered purely as an air splitting problem, one would have expected much more air to have taken the short straight route along 52 level than the long roundabout way down to 53 level and back, but here just the opposite was happening. Calculation shows that in this case the N.V.P. is only 8,8 Pa but that is just sufficient to create the flow of 5,6 m3/s aloog the longer route.

t -

10 m3/s

- :_:::"!f:- 7, 5 ms/s

2r./32c

Z, 5 m3/s FIG. 13

75

l 21/z1c

10 ml/s

a

The opposite could also happen in the case of two cross-connected raises off a dry haulage if water were added to the air at the top as illustrated in Figure 13 .

Another interesting case experienced by the author was that of an abandoned vertical shaft which normally carried a dowhcast air volume of about 40 m3/s. This air served to keep an inclined shaft fresh and then joined the general mine upcast through old outcrop shafts and workings .

The downcast quantity was always less in summer than in winter, but during one dry summer period it suddenly stopped altogether, resulting in the inclined shaft being filled with smoke at blasting time. Two water hoses were used for spraying water down the shaft from surface. This immediately had the effect of starting the flow of air again and within half an hour the flow had increased to 40 m3 s. The water was then turned off and no further trouble was experienced .

In one old isolated inclined shaft there was a strong upcast in the ladderway compartment, probably due to heat from a compressed air main, while the other compartments were all downcasting .

A very common occurrence in hot mines is the flow of air in dead ends. Cold air flows in along the foot wall. gradually becomes warmer and flows out along the hanging. Such a flow of air has been seen to persist for over a thousand metres. Air temperatures near the hanging were about 3'C higher than near the footwall. Because flow of this type often occurs it is essential, when measuring small air quantities by means of the smoke method, not to rely on observations in the centre of the drive only, but to make check observations near the hanging and near the foot wall.

In a mine with wet downcast shafts and airways the fans get considerable assistance from natural pressure, while in a very dry mine it is even possible that the natural ventilation prc-ssure may act against the fans and that the flow may consequently reverse when the fans are stopped.

The effect of air cooling plants on natural ventilation pressures also depends on the placing of these plants. As a result of these last two factors, mines with wet downcast shafts or with surface cooling plants are much less affected by fan stoppages than mines with dry shafts and underground cooling plants .

A kilogram of water (or rock) falling down a shaft 2 000 m deep, loses 2 000 x I x 9,8 = 19 600 J of potential energy. If the water falls straight down, without hitting the sidewall or supports, all this energy, except for the small amount absorbed when it hits the shaft bottom, is transferred to the air through which it passes. If 2 litres of water fell down this shaft per

second, the energy imparted to the air would be of the order of 2 x 1 xi x 98 = 39,2 kW. Part of this energy would cause the air to downcast while the remainder would be converted into heat.

Rock which is tipped into an orepass system or discharged from a conveyor can similarly cause considerable amounts of air to flow because of the energy imparted to it, and in such cases it is usually advisable to reduce the distance of free fall of the rock to a minimum in order to prevent currents of dust-laden air.

76

NOTE 31

FANS

A fan is an appltance designed to cause a flow of air. There are two main types of fan -axial flow and centrifugal or radial flow

An axial flow fan consists of a shaft with a hub or boss to which is attached a number of blades. These blades can be fixed, which means that they are permanently attached to the hub at a certain angle, they can have adjustable pitch, which means that the angle of the blades can be changed by undoing the nuts and lock screws at their bases, or they can be of the variable pitch type, which means that the angle of all the blades can be altered by operating a suitable mechanis;n whilst the fan is in motion.

Shaft

Supports

Bell mouth Blade

Impeller

Diffuser or Evase

Outlet Fairing

Housing (Casing)

Construction of an axial-flow fan

FJGURE14A

77

FIGUREl48

Blade pitch adjustable

Adjustable pitch axial flow impeller

When an axial flow fan is revolved, the blades scoop up air on the one side of the impeller (the hub together with the blades) and push it to the other side, thereby causing a flow of air parallel to the shaft, or axis, from which fact it derives its name. At the same time the air is given a twist which causes it to leave the impeller with a spiral motion. To counteract this motion and to improve the efficiency of the fan, a set of stationary blades is usually installed on either the inlet or outlet side of the impeller. These fixed blades are the inlet or outlet guide nnes. Some fans have two or three impellers, one behind the other, each with its own set of guide vanes. They arc called two-stage or three-stage fans. Others have two impellers driven by separate motors and revolving in opposite directions, thereby obviating the need for guide vanes. These are called contrarotating fans.

Except in the case of small fans used purely for stirring up the air in a room, axial flow fans also have casings. In large, efficient fans, these normally consist of three parts - a cylindrical section round the impeller, an inlet section with a bell mouth to reduce the entrance losses, and a diverging outlet section or evase to reduce shock losses by allowing the air to slow down gradually. Finally, to improve the efficiency of the fan by ensuring smooth flow of the air through it, the fan has a short snub-nosed inlet fairing on the upstream side of the hub and a long, pointed outlet fairing on the downstream side.

The fan blades can te made from flat steel plate, in which case they are called laminar blades, or they can be cast or otherwise formed into a special aerodynamic shape in which case they are called aerofoil blades. In an aerofoil blade the leading edge is rounded and thick while the trailing edge is much thinner and sharp.

-

It is obvious that when the direction of rotation of an axial flo\\ fan is reversed, the direction of the air current will also be reversed. However, because the trailing edges of the blades now become the leading edges and the evase and outlet fairing are now on the inlet side, etc., the fan will naturally handle less air and be less efficient.

78

A centrifugal fan works on an altogether different principle to an axial flow fan. The word "centrifugal" means "fleeing from the centre". Anything which is revolved tends to leave the centre and will do so if allowed to. Thus a stone attached to a string and swung around by a man will move in circles around his hand, but the moment the string is released it will fly away.

,, Delivery

Blades

Intake

A typical centrifugal fan (Left hand drive, top horizontal discharge)

FIGURE 14 C

79

.___

Discharge

Shaft

._,'+--- Inlet eye

Blade

Impeller (wheel)

Volute (Scroll)

Construction of a centrifugal fan FIGURE 140

The impeller of a centrifugal fan consists of two rings with blades fitted between them. The rings are attached to the shaft by means of spokes. When the fan is revolved, air is drawn parallel to the shaft into the open ends of the impeller (the eyes) and thrown out in a radial direction through the blades. If the air can enter on both sides of the impeller, it is called a double-inlet fan. If air entry is from one side only, it is called a single-inlet fan.

The blades can be either laminar or of aerofoil shape. They can be either radial, backward or forward inclined or curved as shown in Figure 14E .

It is obvious that when a centrifugal fan is rotated in the wrong direction the air will still enter at the eye and leave in a radial direction. Thus the air continues flowing in the same direction as before, but the air quantity and the efficiency of the fan are reduced .

Backward Curved Blades Radial Blades

FIGUREJ4 E

8 0

Shaft

Eye

Forward Inclined Blades

Normally a centrifugal fan does not have guide vanes or fairings, but it does have an evase and usually a short inlet cone or ring. The casing around the impeller is shaped like a spiral curve and is called the scroll or \'Olute of the fan. The point "here the scroll is nearest to the impeller is called the cut-off. In some special cases adjustable inlet control \anes are fitted. The angle of these vanes can be altered in order to change the output of the fan.

Centrifugal fans are not made with variat le pitch blades in the same sense as axial flo" fans . but in special cases \ariable tip-length blades are used to give flexibility of perforrr.ance. These consist of short fixed blades to which extensions of different lengths can be bolted as required.

While the air passes through an axial flow fan in a straight line unless a bend is specially added to it on either side, air is turned through ninety degrees on passing through a centrifugal fan. This is generally a nuisance because it increases the amount of space required for the installation, but sometimes it is an advantage when it happens to fit in with the remainder of the layout. The intake is invariably horizontal, but the discharge direction can easily be arranged to suit the customer. The most common cases are called: Top Horizontal Discharge, bottom horizontal discharge, top vertical discharge and bottom vertical discharge.

Centrifugal fans are also classed as either left or right-hand drive. This expression indicates on which side of the fan the motor is situated. When one stands behind the fan and faces the direction in which the air is discharged, it is a left-hand drive fan if the motor is on one's left and vice versa .

The fans on modern mines are nearly all driven by electric motors, but some old units are still driven by steam engines while some standby units are driven by diesel engines.

The fan can be direct driven, i.e. the motor shaft is in line with and directly connected to the fan shaft, or it can be indirectly drhen through gears. bell or a friction clutch .

81

NOTE 32

FAN CHARACTERISTIC CURVES

Every fan, just like any other machine or, for that matter, like every person, has certain characteristics or distinguishing qualities with respect to the work it is able to perform under different circumstances. If a Jong-distance runner and an experienced mountaineer competed in a series of races to see who could cover the most ground in one hour, one would probably find that the athlete would easily win on level ground, uut that the steeper the ground on which the race was run, the smaller the difference between the two. On very steep ground the mountaineer might win, while on a near \Crtical mountain cliff, the mountaineer would be able to proceed slowly while the athlete would not be able to move at all. The full story of the differences between the running characteristics of these two men would probably cover several pages, but it could also be told by means of the graphs in Figure I 5, which could then be described as characteristic curves of these two men.

900

\ \

c: \ - ' 03 500 ' I-

-0 -c: CJ)

:0 e? 300

(.!)

oo

0 5 10 15 Kilometres covered in an hour

FIG. 15

From these curves it is easy to determine which man should be sent to carry an urgent message if the gradient of the terrain is known.

82

Fan characteristic curves indicate how much air a fan can deliver at any particular pressure and how much power is required to drive the fan in each case. Figure 16 shows the characteristic curves of a fan.

As indicated, these curves are applicable to a particular fan when it is driven at a speed of 10 r Is and when it is handling air at a density of 1,2 kg/m3

Fnstltic ......... kP1

1,8 l---+---1----1---+-

I I +- + --t! 1,6 I---+,,.....-+--+-- ---+=-.....,--i'i------+--__r-. - -SWbcprmutt1 I I t

Eff-.:Y "

Input power kW

80

70

----'----

The higher the resistance of the mine. the less will be the quantity of air delivered and the greater the pressure put into it. until a point is reached (A) at which the fan delivers 92 m Is at a pressure of 1,6 kPa. If the mine resistance is increased still further, the fan will deliver still less air but at a lower pressure, as can be seen from the portion A-B of the curve. This is called the stall zone of the curve. What actually happens in this zone is that there is insufficient air to fully fill the space between the blade sections and that the air separates from the trailing edges of the blade. When a fan is "stalling" the sound tone changes markedly, and the manometer and motor ammeter can be seen to oscillate. Fans must never be operated in this stalled zone because vibrations which arc set up in the blades and in the shaft can cause mechanical failure .

The Fan Inpt Po:wer Curv, marked "input power" in the figure. shows the power required at t.he fan shaft with different air volumes being delivered. The motor driving the fan will have to dehver greatr power than this if there are any losses in the drive. In the case of a direct driven fan the.coupling and bearing losses are negligible. but there can te a loss of about 2 nt in a gear drive and a loss of up to 7 per cent in a belt drive.

!he shape of the input power curve is important. It will be noticed that in this particular case It starts at about 150 kW with no delivery, gradually increases to 230 kW at 130 m3/s and then .decreases to .110. kW at 175 m3/s. This is a non-overloading characteristic, which means tht if the. mtor is ?g enough to drive the fan at the normal design duty (about I 15 m/s in this case) 1t will be big enough for any duty of the fan. This is typical of all axial flow fans and of backward bladed centrifugal fans as well. Radial bladed centrifugal fans, however have overloading characteristics.as in Figure 17. In such a case a motor is also installed which

.is big enough for

_the normal design duty plus a reasonable margin of safety, but when the air volume is allowed to increase by the opening of a door or the breaking of a duct the motor will overload and may either trip or burn out.

'

.. Q.. :l

"' "' :;; D.. ;r:

0 c. '5 c.

.5

Air Volume m3/s

FIG. 17

84

The Efficiency Curve in Figure 16 is derived from the other two curves. The efficiency of any machine is defined as the ratio of the useful work outf111t to the energy input and it is usually expressed as a percentage .

The useful work done by a fan is to move air and its output is measured in terms of the volume of air it moves and the pressure it puts into this air. These two terms are combined into Air Power as described in Note 29. The power input into the shaft of the fan is the Fan Input Power. (Sometimes the electric input power is used, depending on whether the losses in the electric motor are considered separately or not).

Fan Efficiency ( %) Air Power x 100 Fan Input Power

It will be seen that when this fan is handling no air it produces a pressure of 2,0 kPa. The air power is then:

Air Power, W,

Efficiency

pxQ I 000

2000 x 0 I 000

0 Air Power

Fan Input Power 0

150 x JOO o

x 100

It requires 150 kW input power to keep the fan turning at 10 rls and to maintain a pressure of 2 kPa across the fan, but because there is no air movement no useful work is done and therefore the efficiency of the fan is nil. Energy cannot be destroyed and the 150 kW input is all converted into heat which raises the temperature of the fan and of the air .

At free delivery the fan moves a large volume of air, but the useful pressure is nil. The useful work is also nil and so is the efficiency .

At the design point, however, this fan delivers 115 m31s at l,37 kPa, requiring an input power of 225 kW.

Air Power, W, = 1,37 x 115

Efficiency

157,6kW

1576 x 100 = 700!o 225

The fan efficiency at various air volumes is calculated in this way and the efficiency curve is drawn by plotting the answers against fan quantity. It can now be seen that this is a fairly efficient fan when it is handling 115 m3ls, but if it is installed under conditions where it will handle only 60 m' Is or as much as 170 m'ls it will be a very inefficient installation .

85

NOTE 33

FAN LAWS

There are certain laws which apply to fans, by means of which it is possible to calculate characteristic curves for a fan at different speeds and ait densities once the curve at any one speed and density is known, and also to determine curves for similar fans of different sizes. These laws can be stated in many different ways, of which the following is perhaps the simplest.

For varying speed but with constant air density:

I. Quantity vaies directly as the speed. Q1 = Q1 X J 2 2. Pressure vanes as the square of the speed. P2 = P1 x (1 3. Power varies as the cube of the speed. power2 = power1 x 1 Y?,)3 4. Efficiency is constant. Eff2 = Eff1 \ 1

For 'arying density but with constant speed: 5. Quantity remains constant. Q1 = Q1 ( ) 6. Pressure varies directly as the density. P2 = P1 x :: (wi\ 7. Power varies directly as the density. power2 = power1 x \w) 8. Efficiency is constant. Eff2 = Eff1

For geometrically similar (or homologous) fans: 9. With the same linear velocity of the blade tips, the pressure supplied by two similar

fans will be the same when the quantity given by the larger is to the quantity given by the smaller as the square of the wheel diameter of the larger is to the square of the wheel diameter of the smaller.

JO. With the same linear velocity of the blade tips and producing the same pressure, the power varies as the square of the wheel diameters and the mechanical efficiency is the same.

Laws 9 and 10 will not be discussed in these Notes. They are of great importance to fan designers who manufacture small experimental fans and test them at any convenient speed and air density, and then use these laws to calculate the performance of larger fans at other speeds and densities. A geometrically similar fan can then be constructed of the correct size to give any required performance.

The curves in Figure 16 of Note 32 apply to a particular fan running at JO r/s and handling: air at a density of 1,2 kg/m3 Let us calculate new curves for this same fan when it is run at a speed of 13,3 r/s and handling air at a density of 1,04 kg/m1

At the original speed of 10 r/s and density of 1 ,2 kg/m1 the fan will, at its design point, deliver 1 1 5 m3/s at l,37 kPa using 225 kW and giving an efficiency of70,0 per cent.

At the new speed of 1 3,3 r/s and density of 1,04 kg/m1 the fan laws apply as follows:

New Quantity I 15 x 103 (Law 1) x I (Law 5) 153 m3/s

New Pressure 1,31 x (1i03f (Law 2) x \',';4 (Law 6) 2, 1 kPa

86

New Input Power

New Efficiency

Or as a check,

Efficiency

13 33 1 04 225 x (To-) (Law 3) x U (Law 7) , 459 kW 70,0 x l (Law 4) x 1 (Law 8) 70,00Jo

1 53 x 2,1 x 100 459 70,00fo

Similarly, various other points on the curves for the old conditions can be recalculated for the new conditions and these results can be plotted on a graph as shown in Figure J 8, in which the full lines represent the old conditions of 10 r/s and 1 ,2 kg/m1 density, while the dotted lines represent the new conditions of 13,3 r/s and 1,04 kg/m1 density.

0 Iii 100

75 0 g .. 50 0 ., a. c .. :; 0 25 a. g .5

so 100 150 200 250 Fan Volume m3/1

FIG. 18

87

When a fan is installed in a mine (or system) its performance will be.determined by the point at which the mine (or system) characteristic cuts the fan characteristic. The mine or system characteristic is a curve showing how the pressure drop across the mine or system varies as the quantity of air varies. A mine.characteristic curve is shown in the figure which cuts the 10 r/s fan characteristic at the design point, viz 115 m3/s at 1,37 kPa. Under these conditions the fan will handle 11 S m3/s if it is not affected by pressures from any other sources. If it attempts to handle more air, it finds it impossible to do so because it produces insufficient pressure to overcome the mine resistance. Likewise it could not handle less air because it would have pressure to spare.

If the 115 m3/s line is followed vertically up and down, it will be seen that the fan will consume 225 kW and that the efficiency will be 70,0 per cent.

It will be seen that this same mine characteristic curve cuts the 13,3 r/s fan characteristic at 144 m3/s.However, this fan characteristic is drawn for a density of I ,04 kg/m3 while the mine characteristic is drawn for a density of 1,2 kg/m3 If the 13,3 r/s fan curve is re-drawn for a density of 1,2 kg/m it will be seen that it now cuts the mine curve at 156 m3/s and this is the air volume that will be handled.

When no curve is available for a fan - either because the original curve has been lost and the type of fan is obsolete, or because some alterations have been made to the fan - it is sometimes necessary to construct a curve for the fan. This is done by making observations on the fan at various performance levels which are obtained by opening and closing doors if they are con: veniently available, or otherwise by building a bratticc in stages in the airway some distance from the fan. At each performance level the air quantity, the static pressure, and the power input must be measured .. It is seldom necessary in !!Ch a case to determine the complete curve. Determining the exact curve through the stall zone is not possible because it is a very unstable area and it is also dangerous because of the vibrations set up in the fan .

Only fan static pressures are referred to in this book in order to keep the discussion simple. However, it was explained in Note 25 that air has both static and velocity pressure and that the algebraic sum of these two is the total air pressure. As a general rule fans impart both static and velocity pressure to the air. It is therefore important, especially when high air veloci ties are involved, say more than 10 m/s, and more particularly when there is a marked differ ence in the velocities at the fan outlet and the fan inlet, that the tubes installed for measuring the fan prure must be correctly placed and that the results must be correctly interpreted. In lieu of a detailed explanation, only the definitions of terms relating to fan performance and pressure measurements as they appear in the British Standards (BS 848) are given below:

Side tube or static pressure tube. A tube which allows the air to flow without disturbance past one or more small orifices having their axes at right angles to the direction of the air stream in which it is placed.

Facing tube or total pressure tube. An open-ended tube, the axis of which is coincident with the direction of the air stream in which it is placed, the open end facing upstream, i.e. against the direction of flow.

Fan total pressure. The algebraic difference between the mean total pressure at the fan outlet and the mean total pressure at the fan inlet .

Fan velocity_ pressure. The velocity pressure corresponding to the average velocity at the fan outlet based on the total outlet area without any deductions for motors, fairings, or other bodies.

Fan static pressure. The difference between the fan total pressure and the fan velocity pressure .

88

NOTE 34

FAN SELECTION

When it is decided to install an existing fan or to order a new fan for a specific purpose, one wants to know beforehand that the particular fan will be both effective and efficient. This means that the fan must deliver the required quantity of air at the least possible total cost. Considering the question of cost first, it is important to realise that the cost of the power which is required to drive the fan is usually greater than the actual cost of the fan itself - in some cases much greater. As a rough guide the cost of a large fan, including the cost of the motor and the cost of installation, can be estimated at approximately R250 per kW (air power). (This figure is influenced by a variety of factors and should therefore not be used indiscriminately.)

The cost of electric power on the South African goldficlds has risen sharply in recent years. For continuous running machinery it works out at Rl20 per kW per year (1978 cost, see Note 39) .

Assuming a fan efficiency of 75 per cent, power costs amount to about Rl60 per kW of air power per year. Thus within a period of two years the electric power which is used to

drive a fan may cost more than the total original cost of the fan itself, and mine fans are

usually installed to run for ten to twenty years. As a comparison, a RIO 000 motor car using I litre of petrol costing 40 cents per 9 kilometres, may have a life of 150 000 km during which the total fuel cost will be only R6 670 or two thirds of the original cost of the car. For this rea son mechanical efficiency is a much more important factor in the case of a fan than in the case

of a motor car. Take the example of a mine which requires a fan to handle ISO m3/s at 0,66 kPa for a

period of 15 years. A new fan to give this performance at not less than 75 per cent efficiency might cost about R25 000 installed and would use 132 kW costing about Rl6000 per annum. If the fan for which the characteristic curve is given in Figure 16 happens to be available on the mine, management would be sorely tempted to install it because it can provide exactly the required duty. However, this fan would be working at an efficiency of only 45 per cent and would be absorbing 220 kW at a cost of R26 000 per annum. Compared with this fan, the new fan would thus save RIO 000 per annum in power costs and would pay for itself within three .years.

When a single fan is required in a mine or a duct system, the selection of a suitable unit is comparatively easy if the mine characteristic is also known, as was shown in Note 33. However, the problem becomes rather more complicated when two or more fans are installed in the same mine in such a way that they have an effect on each other.

When two fans are so installed one behind the other that they handle the same air, they arc said to be in series. When two fans are installed side by side in such a way that they draw air from the same source and deliver it to the same destination they are said to be in parallel.

When two fans are in series the same volume of air passes through each and each adds a certain amount of pressure to it in the process. Thus in Figure 19, if fans A and B arc blowing air into a duct that is closed off at the end, neither will be handling any volume of air but A will raise the pressure by 1,25 kPa and B by another 0, 75 kPa and the final pressure will be 1,25 + 0, 75

89

= 2,0 kPa. When they arc allowed to handle 5 m3/s, A will increase the pressure by 0,9 kPa and B by another 0,59 kPa giving a total of 1,49 kPa. When handling 8,3 ml/s fan A will not be putting any pressure into the air, but fan B will still be adding 0,4 kPa. When B is handlin more .than 8'.3 1/s, A will not be of any assistance and will actually form a slight r1stancc. m the circwt. Curve C for the two fans in series is obtained by plotting these combined duties.

On the other hand, when two fans are installed in parallel they do not necessarily handle te same volume of air, but they must always produce the same pressure because they draw arr from a common point and deliver it to a common point. At a pressure of 0,75 kPa fan A would handle 5,9 m3 Is, but fan B would not handle any air. At a pressure of 0,2 kPa A would handle 7,? m1/s and B 10,7 m3/s, thus a total of 18,6 m3/s at 0,2 kPa. The curve D for the wo fans m parallel can be drawn by plotting and joining these points. If the pressure required is mor than 0,75 Pa fan B will not be able to provide this pressure and will not only be of no assistance but will actually be a hindrance because A will force some of its air back through B.

1,0 i---t----1r-i.-+--!-,1:.._-4---l----ll--- !I;

aBt-i:;:u,;-t--+---f---\,.+--!----4----1----ih.---l o.a t---t---A:::-li-\--:..__.p--..+,..L-1---

VOLUMEm3/1 FIG. 19

Wether. it wo11:d be better to use either one of these fans in a particular case, or to use the two fans m s:ns or m parallel would depend on the resistance of the duct system. Three system charactnstJcs are shown in Figure 19. X is a high resistance, Y is a medium resistance and z a low resistance system. The figure is already too congested to show the input power curves for the fans as well, so let it be assumed that fan A always consumes 7, 5 kW and that fan B always consumes 6, 0 k.W. The following figures can then be read off the graphs.

90

I System

x

y

z

Volume P"'5Sure

Input Power J Efficiency

Volume Pressure Input Power Efficiency

Volume I Pressure Input Power Efficiency

m'/s kPa kW

--.,, ml/s

kPa kW .,,

m'/s kPa kW %

TABLE19

Fan A FanB alone alone

5,4 4,6 0,84 0,61

7,S 6,0 61 47

7,2 7,3 0,47 0,48

7,5 6,0 45 58

8,0 9,8 0,18 0,28

7,S 6,0 13 46

I

Fans A & B Fans A & B in series in parallel

6,3 -1,18 -13,S 13,5

SS

8,0 I 8,7 0.58 I

0,68 13,S 13,S

34 44 - I 13,3 - I 0,52 13,5 13,S - SI

Using this tabulation it is now possible to decide in the case of each system which would be the most suitable installation.

The performance of each fan when the two are running in combination can be read off the graph by reversing the original process by which the fan curves were added. Take the example of system Y. When the two fans are running in parallel and handling 8,7 m3/s at 0,68 kPa, each of them must be producing 0,68 kPa, and moving horizontally along this pressure line we find that fan A will handle 6,3 m3/s and fan B only 2,4 m3/s. However, when running in series they both handle 8,0 m3/s but by moving vertically along this volume line we find that fan A is producing only 0, 16 kPa, while fan Bis producing 0,2 kPa.

The same method flculation is appli9Ah.!nwo fans are installed tjther in series or in parallel.

.

Natural Ventilation Pressure (N.V.P.) can be regarded as an additional fan, usually in series with existing fans. N.V.P. depends on the density difference between the downcast and upcast air and on the depth of the mine. It is only indirectly affected by the volume of air ftowing in so far as a change of volume may cause a change in the prevailing temperatures. It is therefore depicted graphically as a straight horizontal line .

91

1,6

1 ,4

1,2

1,0

0,8 w a: ::i Ill Ill w a: ... 0,6

0.4

0,2

FIG. 20 0

0

- --- - - - IO..q .... ,., .,. ,., - i.'.p

' .... :..._ ..... ...

.... ,,

- -"'- IV

N.V.P .

VOLUME m3/s 25 50

' '

75

\ \ \ \

100

When the N.V.P. tends to move the air in the same direction as the fan it is added to the fan pressure and when it tends to move the air in the opposite direction to the fan it is subtracted from the fan pressure, as is shown in Figure 20 . In this case the fan would handle 75 m3/s at 0,8 kPa when there is no N.V.P., 83 m3/s at 0,68 kPa when being assisted by 0,3 kPa N.V.P., and 65 m3/s at 0,9 kPa when 0,3 kPa N.V.P. is working against it. This explains why mine fans often handle more air at a lower pressure in winter than in summer, because during winter N.V.P. is at its maximum value .

When using the method described above, care must be taken when more than one fan is involved. It is best first to combine the curves of all the fans - in parallel or in series as the case may be - and then as a final step to add or subtract the N.V.P. as the case may demand, otherwise there is a danger that N.V.P. may be added twice. To avoid this danger, an alternative method can be used, viz. not to alter the fan curves, but to add the N.V.P. to the mine characteristic curve when N.V.P. is opposing the fan or subtract the N.V.P. from the mine characteristic curve when N. V.P. is assisting the fan. The final result is exactly the same.

92

NOTE 35

FANS PARTIALLY IN PARALLEL OR SERIES

In actual mining practice fan combinations are unfonunatcly not always as simple as described in the previous chapter. Fans arc not always either strictly in series or strictly in parallel. Fans are partially in series when most of the air from one fan passes through the other, but some air is added or removed between the two positions. Similarly fans are partially in parallel when they draw their air from the same place .and discharge it to the same place but instead of being near to the fans these common points are some distance away. A fairly simple example of each of these two cases will be shown using the fan residual pressure method.

Consider the case where two small fans P and Q in ducts F and G both deliver their air through duct H. Figure 2 1 shows how to calculate the results of this combination, given the various characteristic curves.

oo

l....o!==::::t2::::=,-t-.....J-+ei---c!9

VOLUME

FIG. 21

93

First consider fan P in duct F. When 2 m3/s is flowing, fan P produces a pressure of 2 300 Pa and the pressure required by duct F is only 200 pascals. The residual pressure of the fan at junction J, i.e. the pressure which has not been used up in duct F, is thus 2 300 - 200 = 2 100 Pa. Similarly when 4 m3/s is flowing, the residual fan pressure is I 750 - 810 = 940 Pa and when S m3 /s is flowing the residual pressure i I 260 - I 260 = 0 Pa (Curves P and F intersect at this point and thus have equal values, which means that all the pressure produced by fan P is absorbed in duct F and there is no residual pressure left at point J). When these residual pressures are plotted and joined by a curve, this curve K represents the difference between curves P and F, i.e. K is a residual fan pressure curve.

In exactly the same way curve G can be subtracted from curve Q to give curve L which represents the residual pressure of fan Q after it has forced its air through duct G. Next, curves K and L are added as for fans in parallel to give curve M. The operating point is the point where curve M cuts the curve for the common duct H, viz. 6,8 m3/s at 600 Pa. By drawing a horizontal line from this point to where it cuts curves K and L and vertical Jines up from these points until they cut the fan curves, we find that fan P handles 4,4 m3/s at I 575 Pa of which 975 Pa is lost in duct F while 600 Pa is Jost in the common duct H, and fan Q handles 2,4 m3/s at 1 350 Pa of which 750 Pa is Jost in duct G and also 600 Pa in duct H.

The calculation is perhaps a little more confusing when fans are partially in series as shown in Figure 22 . This would occur in practice if fan X was drawing air from both ducts A and B, but it then became necessary to increase the volume of air drawn from A. If a smaller fan Y was available, the obvious solution would be to install this fan in duct A, but first it would be advisable to calculate the outcome of this plan.

Normally when two fans are in series, their curves are added together on the assumption that the same quantity of air is passing through each and that the pressures produced by the two individual fans at each quantity arc simply added together.

In Figure 22 this is obviously not possible because an unknown quantity of air can either enter or leave the system through duct B, depending on the pressure existing at the junction J.

The pressure at 1 must always be the same for duct A as for duct B, and if air is entering B, ducts A and B arc in parallel, so what we need to do is to plot the two duct curves in parallel. But because duct A has fan Y assisting the air along it, the curve of this fan must first be subtracted from the curve of the duct to determine the residual system curve.

At 2,65 m3/s where curves A and Y intersect, the fan pressure exactly overcomes the duct resistance and the residual pressure is nil. At 3 m3 /s the duct requires a pressure of l 150 Pa while the fan pressure is only 750 Pa, thus the fan only overcomes part of the resistance of duct A and its residual pressure is I ISO - 750 = 400 Pa. At 4,05 m3/s the fan pressure is nil and the residual pressure is equivalent to the total duct pressure of 2 050 Pa. When less than 2,65 m3/s is flowing along duct A, the fan pressure is more than the duct pressure and the residual duct pressure is thus a negative value.

94

1000

500

cf LIJ a: Dl 0 2 " 5 I/) LIJ I a: VOLUME m3/s, a.

I -500 I I < I I /v /I -1000 A.

I , " 'i"

L(,"-.,, ,,, ...... Q;v ---1 500

FIG. 22

All these points are now joined by a curve C which represents the residual resistance of the system, i.e. duct A - Fan Y.

Now this residue of the duct resistance as represented by curve C is in parallel with duct B. The curves of parallel ducts can be added in the same way as the curves of parallel fans. The pressure Jost in the two ducts must be equal because both intake ends and both discharge ends arc at the same pressure. Thus to add curves B and C we add the quantities corresponding to equal pressures. At nil pressure no air will pass through duct B, but 2,65 m3/s will pass through duct A (curve C). At 500 Pa, J , l m3/s will pass through duct B and 3,1 m3/s through duct A (curve C), a total of 4,2 m3/s. At l 000 Pa, 1,6 m3/s will pass through B and 3,45 m3/s through A (curve C), a total of 5,05 m3/s. Joining these points gives us curve D which represents the two parallel ducts A and B including fan Y. The operating point of fan X is the point where

9S

its curve cuts curve D, viz. 4,9 m3/s at 925 Pa. Going back horizontally on this pressure line we find that 3,4 m3/s will come from duct A (curve C) and 1 ,5 m3/s from duct B. Following the 3,4 m3/s line vertically downwards we find that fan Y will deliver this air volume at 550 Pa .

On large mines with several fans various combinations occur of fans partially in parallel with each other and partially in series with others. Common sense is the main ingredient required in the solution of these problems, of which Figure 23 is an interesting example .

A mine is represented with a downcast shaft with characteristic curve P, two underground circuits with curves Q and R and with fans C and D in them partially in parallel, an upcast shaft with curve S and two parallel fans at its top which are only partially in series with the underground fans because 23 m3 /s leaks through the shaft headgear. In addition there is a natural ventilation pressure of 500 Pa assisting the fans. In order to reduce congestion, only the combined curves for fans A and B in parallel and for the shafts P and S in series are shown on the graph.

2DOO !--.>..---+----... ... _ _

I \ ' \ I I I \ \ ' ' I ' 1 : \ ' \ I ' \ ' ' ,: \ \ : \ I

I , .. I \ \0 , " \ \ ,c.. 1 . .. " 1P ' ' " \c:. \ \ -I

- ...O t-----+-7----11-----+----tt----+---'i--l----+----+---1-----I

25 SJ 76 100 125 151)

V"'morril/o FIG. 23

96

In order to avoid further complications, the density differences between the surface and

underground fans are ignored in this example:-

Briefly the method of solution is as follows: . 1. Subtract the resistance of airway Q from the pressure of its fa

n C (C - Q = n.

2. Subtract airway R from its fan D (D - R = ). 3. Add the residual curves T and U as for fans in parallel (T +

U = V).

4. Add 500 Pa N. v .P. in series to fans A and B combined in parallel (A + B + 500 Pa = E).

5. Subtract 23 m/s at each pressure level from E (E - 23 = F).

6. Add the pressure of the surface fans at the residual volumes in series with the residual

pressures of the underground fans (F + V = W) .

7. The mine operating point is represented by t?e. interstion_ of this combined residual cur:e for all the fans and N.V.P. (W) with the remammg senes resistance (

P + S), name_ly 136 m /s

at 1 100 Pa. This means that 136 ml/s will flow through the mine and that the pressure loss

in the downcast and upcast shafts combined will be 1 100 Pa .

8. Moving up from this point to curve F, then right to crve E an00d tphen down to curve A + B,

we find that fans A and B together will handle 1 59 m Is at 1 1 a .

9. Moving down on the 1 36 ml/s volume line to curve V and then left to curves U and T and

from these points up to D and Q respectively, we find that:

Fan C handles 38 m3/s at 810 Pa . Fan D handles 98 m3/s at 320 Pa. The pressure lost in system Q is 1 290 Pa . The pressure lost in system R is 830 Pa .

10. It is also possible to deduce that: . The air volume flowing through the mine would drop to 124 m

/s 1f fans C and D were

removed. h fi h d The volume would increase to 139 m3/s if the leakage at the upc

ast s a t ea gear were

eliminated .

97

NOTE 36

SURFACE AND UNDERGROUND FANS

A problem which is very confusing to many people concerns the relative merits of installing fans on surface or underground in a mine. The underground density in a deep mine can be very different from the surface air density. It has been shown in Note 27 that the pressure drop in an airway, for the same volume flow, is directly proportional to the air density. It was also shon in Note 33 that the volume delivered by a fan is not affected by air density, but that fan pressure and power consumption vary directly as the density. The combination of these factors is best studied by means of a simplified example which is given below because of the interesting answers obtained, although the actual calculation is rather beyond the scope of beginners.

Assume that a certain mine has a vertical shaft 2 500 m deep and that the air density is 1 ,0 kg/m on surface, 1 , 3 kg/m underground and l , 1 5 kg/m at a point approximately halfway down the shaft. Assume also that a 2 500 m long 800 mm diam. leakless ventilation column is available, which requires a pressure of 1 500 Pa when 3 ml/s of air at surface density is flowing through it. Lastly, assume that two identical fans of known characteristics are available.

The volume of air flowing in seven different cases will be calculated. These seven cases arc represented diagrammatically in Figure 24 and the relevant characteristic curves are given in Figure 25.

The fan curve S at an air density (w) of 1 , 0 kg/m3 is given. From this the fan curve T for w = 1 , 3 kg/m3 can be calculated by multiplying the pressure at the same volume by : ::

TABLE 20 Fan characteristic curves

Pressure at Pressure at Volume w = 1,0 kg/m' w = 1.J kg/m'

m'Js Pa Pa

2,0 1 930 2 510 2,S I 780 2 3 10 3,0 I 500 1 950 3 , S I 060 I 380 4 , 0 320 420

The curve U for the 2 500 m ventilation column at surface density is calculated from the information supplied by means of the square law, e.g. :

Pressure for 3 ml/s = 1 500 Pa 4 2 Pressure for 4 ml /s = 1 500 x ( T)

= 2 670 Pa

From these figures curve V, for a density of l , 15 kg/m3 is plotted by multiplying the I 15 pressure at each volume by T,o Curve W for w = 1,3 kg/m is calculated in the same way.

98

Air Volume m'ls

2, 0 2, 5 I 3 , 0 3 , 5 4 , 0

ease l

TABLE 21 Duct Pressun losses

Prasure I Pressure loss al loss ! w s l,O k1/ml I w = 1.1s q/ml Pa Pa 670 770

I 040 I 1 200

I 500 1 720 2 040

I 2 3SO

2 670 3 070

Prasure loss!

" a 1,3 kc/ml P 870

1 350 1 950 2 6SO 3 470

In this case the ventilation column is installed horizontally on surface at w = 1 ,0 kg!ml. One fan draws air through it. The fan operating point is point I at which curve S for the fan at w = 1 ,0 kg/ml cuts curve U for the duct at w = 1 ,0 kg/m3, thus 3 ml/s at 1 500 Pa.

Case II Here the column is installed horizontally underground at w = 1 ,3 kg/ml and one fan

again draws air through it. The fan operating point is at point II where curve T cuts curve W, viz., 3 ml/s ai l 950 Pa.

case m

A

' 8

c

Case I

Surface w = 1,0 kg/m3

w = 1 ,15 kg/m3

Case II D

Underground w = 1 ,3 kg/m3 FIG. 24 (a)

B E K

Case m; G Case nr J

F H

In this case the column is installed vertically in the shaft and one fan draws air up through it This case is more difficult to calculate because, while the fan is handling air at w = 1 ,0 kg/, the duct is carrying air at a mean density of 1 , 1 5 kg/m,3. The intersection of two curves at different densities has no meaning and, therefore, we proceed as follows:

99

When the fan on surface is handling 3 m3/s at 1 500 Pa the amount of air flowing in the duct halfway down the shaft where w = 1,1; kg/m3 will be 3 x /,'15 = 2,61 m3/s (the same mass is flowi.ng, but because of the higher density the volume is smaller) and the equivalent

pessure at this point is l 500 x \. = I 725 Pa (note that the air power must bl! the same at any point in the circuit to which the values of pressure and volume are referred if there is no source of energy other than the fan, and therefore 3 x l 500 = 2,61 x l 725) .

In this way various points are calculated from which curve X is drawn to represent the effect of the surface fan E at the point G halfway down the shaft:

TABLE 22

CURVE S CURVE X

Fan at Effect of 1 , 0 kg/m' Fan ,,, = t,O kg/m3 at w = 1,15 kg/ml

Volume Pressure Volume Prnsure m'/s Pa m'/s Pa

2,0 1 930 I , 74 2 220 2 , 5 1 780 2 , 1 7 2 050 3 ,0 I 500 2 , 61 1 730 3 ,5 I 060 3,04 1 220 4 , 0 320 3 ,48 370

Curves X and V intersect at the pointlIL viz. 2,82 m3/s at l 520 Pa and this represents the air flowing at point G. The fan, however, will be handling 2,82 x \ 5 = 3,24 m3/s at 1 520 x 1

1 .. 5 = I 320 Pa which is represented by point m'on Curve S. o'nly 2,45 m3/s enters

the bottom end of the pipe at F . Case !Y

In this case the column is again installed vertically in the shaft, but this time the fan is at the bottom forcing air up. Once again it is necessary to convert the fan curve T to curve Y which represents the effect of the fan at w = 1 ,3 kg/m3 at the point where w = I , 1 5 kg/m3

Curves Y and V intersect at point IY which shows that 3,13 m3/s is flowing at J halfway up the shaft, and when converted to underground conditions it is found that fan H is handling 2,77 m3/s at 2 125 Pa while 3,60 m3/s leaves the pipe on surface at K .

It will l:e noticed that the fan at H is handling less air than the fan at E but this is purely because it is handling denser air and in actual fact the mine is better ventilated in CaselY than in Case m because at the same depth it has more air flowing. However, it must not be deduced from this result that it is always better to install a fan underground than to install it on surface. Although the underground fan is more effective in producing a large flow of air, it is not necessarily more efficient. Bzcause it is handling denser air, it will also consume more power and its relative efficiency will depend on the shape of the input power curve. It may happen in certain cases that a fan which works quite satisfactorily on surface, will overload and burn out its motor when installed at the underground end of the same column, and it may even be less effective due to running in a stalled condition .

ICO

The remaining three cases all have two fans i n series, one at either end of te pipe column.

Curve SS represents two fans in series on surface. It is obtained simply by dubhg the rsure

values of curve s at each volume. Similarly, curve TT represents two fans m senes at \\ -1 , 3

kg/m'. Curve X Y i s obtained by adding curves X and Y in sc.ries an:! i t represents the effect

halfway down the shaft of one fan on surface plus one fan underground.

Case Y . . . In this case the column is horizontal on surface and the cornbmcd operating pomt of the

two fans is where curve SS cuts curve U, viz. 3,52 ml/s at 2 060 Pa. Each fan handles 3,52 rn1 Is

at I 030 Pa.

Case E . h With the two fans in the horizontal underground column, the operatmg pomt JS w ere

curve TI cuts W, viz. 3,52 m1/s at 2 660 Pa, el!_h fan producing I 330 Pa.

Case YII This system operates at point YU where curve XY cuts curve V, viz. 3,40 m3/s at 2 220 Pa

at w = 1,15 kg/ml. Fan p handles 3,40 x '.5 = 3,90 ml/s at 470 Pa, while fan Q handles 3 40 x = 3,00 m1/s at I 940 Pa. ' 1 , 3

Case Jr

surface w = 1,0 kg/m3

w = 1, 15 kg/m3

CaselZI

Underground w = 1 ,3 kg/m3

M 8 (

0 8 <

FIG . 24 (b)

101

p

Case:mI R

Q

4 VO L U IA E m S/s

FIG. 25

102

Summarising all these answers we find :

Case I Posltio or No. Column I

l[ m I On Surface Underground In Shaft ll: I in Shaft

I Slcnd In Shaft

TABLE 23

Situation of fan

Volume m'/s

I on surface 3, 00 I underground 3, 00 I on surface 3, 24 at top

i Fan Pressure Pa

I 500 1 950 I 320

! : :: tom 1 I underground 2, 77 at bottom I 1 2 on surface 3, 52

2 underground 3 , 52 I on surface 3, O(I I underground I 3, 00

2 1 25 I 030r.ach I 340 cach

470 1 940

These answers are not quite exact because certain complicating factors have been ignored, such as the volume changes caused by the pressure of the fans and the natural ventilation pressure which will probably exist due to density differences between the air inside and outside the pipe in the case of the vertical column. However, this is an easy method by which the effects of installing a fan in different positions in the same circuit can be compared.

103