Chapter#05 Flow of Air Through Ducts and Mine openings Introduction: Quantity control:

The control of air movement, its direction and its magnitude is termed quantity control.

Ventilation:The principal air conditioning process

concerned with control of air circulation.

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Flow of Air Through Ducts and Mine openings Purpose:Supply enough air for human and product needs.For comfort airconditioning—20 cfm per manThis requirement multiplies for various other

reasons.Other functions served by ventilation are control

of gases, dusts, heat and moisture.Thus ventilation requirements may reach up to

200 cfm per man and occasionally 2000 cfm per man.

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Chapter#05 Flow of Air Through Ducts and Mine openings Energy Change in Fluid Flow.

General Energy Equation:Mine ventilation is example of steady flow

( in which the fluid is continually in motion in a liner direction through a conduit

Transition and losses are involved in such a process.

Energy changes are basic to the calculation of the mine quantity and of head.

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The total energy at any section in a moving fluid consist of the sum of the internal static, velocity, and potential energies at that section

Total energy1= (total energy)2 + (flow energy losses)1-2

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5

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Chapter#05 Flow of Air Through Ducts and Mine openings Total energy at section 1 and 2 will be P1/w +V1

2/2g+ Z1 = P2/w + V22/2g + HL

(Bernoulli equation)Where P/w is static energy, V2/2g is velocity

energy, Z is potential energy and HL it is is flow loss

The general energy equation and Bernoulli equation in term of heads

HT1 = HT2 + HL

HS1 + HV1 + HZ1 = HS2 + HV2 + HZ2 + HL , Where Hs is static head, Hv is velocity head, and Hz is elevation head or potential head, all head have the unit of in. water

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Modified Energy Equation :The elevation HZ can complicate the

calculation in mine ventilation because of sizable difference in elevation

For example an elevation difference of 70ft is equivalent to

(Z)(w/5.2) = 70*0.075/5.2 = 1 in. waterIt show that for each 70ft increases in

elevation the potential head term HZ increases 1 in static head will decreases 1 in.

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Chapter#05 Flow of Air Through Ducts and Mine openings

Example: straight duct place first in horizontal and then in vertical position, let the head loss HL b/w 1 and 2 equal 3 in, and 1 atm= 408 in

Other data: for horizontal position HS1=4 in, HV1= 1in, HZ1=0

HS2= 1in, HV2= 1in, HZ2= 0

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Chapter#05 Flow of Air Through Ducts and Mine openings

Putting the above data in general energy equation so we get

(4 + 408) + 1+0= (1+408) +1+0+3413=413 it is determined by using the

obsolete pressureUsing gage pressure the equation becomes4+1+0 = 1+1+0+35 = 5

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Chapter#05 Flow of Air Through Ducts and Mine openings

Given data for duct in vertical position HS1=4 in, HV1= 1in, HZ1=0 HS2= 1in, HV2= 1in, HZ2= 1 Putting the above data in general energy equation so we get

(using the absolute pressure) (4 + 408) + 1+0= (1+408) +1+0+3 413=413 Now using gage pressure 4+1+0 = 1+1+1+3 5 ≠ 6 Omit the HZ term from all calculation and employ a gage pressure

basis, then for horizontal and vertical position of duct both will give the same value

4+1 = 1+1+3 5 = 5

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Chapter#05 Flow of Air Through Ducts and Mine openings Head losses and Mine Heads: Head losses in fluid flow: Energy which is supplied to a steady flow process by

either natural or mechanical means and creates pressure difference, is consumed in overcoming flow losses, it represent by HL

Flow occurs due to pressure difference Head losses have two components1.Friction losses (Hf): it represent head losses in linear

flow through ducts of constant area 2.Shock losses (Hx): it resulting from change in direction

of flow or area of ducts3.HL = Hf + Hx

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Chapter#05 Flow of Air Through Ducts and Mine openings Over all or mine heads: The cumulative energy consumption termed as mine

heads Mine static head (mine Hs): it represent the

energy consumed in the ventilation system to overcome all flow losses

mine Hs = ∑HL = ∑(Hf +Hx)

Mine velocity head (mine Hv): Hv = V2/2g

Mine total heads (HT): it is sum of all losses in mine ventilation system

mine HT = mine Hs + Hv

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Chapter#05 Flow of Air Through Ducts and Mine openings Pressure Gradient:It represent the various heads components of

the general equation graphically

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Chapter#05

State of Air Flow in Mine Openings

Two type of states of flow with intermediate zone.1.Laminar flow2.Intermediate flow3.Turbulent flow identification of the state of low is necessary

while calculating in fluid flow.Because the fluid exhibits different

characteristics and head losses.

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State of Air Flow in Mine Openings

Criteria for identifying the boundaries State of Air Flow in Mine Openings

between the different states is Reynolds Number NRe.

Laminar flow exists up to NRe=2000Turbulant flow up to 4000 These boundaries are approximate and the region

between then is intermediate. Reynolds number is a function of the fluid

properties.

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State of Air Flow in Mine Openings1. NRe= (pDv)/µ = DV/v ----- 5-92. pv is= µ Where p is mass fluid density (lb-sec2/ft4), v is kinematic viscosity(ft2/sec), µ is absolute viscosity(lb-sec/ft2), D is diameter of conduit in ft, V is velocity in fps

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State of Air Flow in Mine Openings For air v= 1.6*10-4 ft2/sec Putting values in equation 5-9 NRe= 6250 DV --------5-10 The fluid velocity at NRe=4000 (lower

boundary of turbulent flow is called critical velocity Vc.

If the fluid velocity exceeds Vc, then flow is always turbulent.

Equation 5=11 show that turbulent flow will always prevail in mine openings

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State of Air Flow in Mine OpeningsSolving for Vc in ft/m and Setting

NRe=4000Vc=(60NRe/6250D)

Vc=(60*4000/6250D) = 38.5/DOr approximately Vc =40/D -------5-11In mining turbulent flow is desirable.This insures satisfactory dispersion and

removal of contaminants.

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State of Air Flow in Mine OpeningsFor a pipe having 1 ft diameter will have a

critical velocity of 40 fpm.Mine openings of 3 ft diameter will have a

velocity of 13 fpm and be a turbulent flow.

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Effect of state of flow on velocity distributionVelocity in a conduit varies as the Reynolds

number varies (Fig.5-9).Maximum velocity Vmax occurs at the center of

the conduit.in ventilation the concern is avg. velocity and

not the maximum velocity. The variation of V with Vmax is determined as

a function of Reynolds number (fig.5-10).This graph enables one to find the avg.

Velocity when only one measurement along the center line has been made.

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Effect of state of flow on velocity distributionHowever since,

Mine opening are non circularAnd many irregularities of the walls tend to

produce non- symmetrical flow patternThe Reynolds number generally exceed

10,000Approximate value is taken.i.e V= 0.8 Vmax.

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Calculation of head lossesoVelocity head: Velocity head represents energy which has

to be supplied to maintain flow. It is lost to the system at discharge. It must be taken at the discharge to

calculate the mine heads. Measurement of velocity head permit to

calculate velocity of air flow.

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Velocity head Derivation of equation: Starting with basic relation, Hv=V2/2g ----------5.2

Where V is in fps and Hv is in ft of fluid. Applying equation 2-1 i.e P= w1H1=w2H2

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Velocity head Hv =wV2/(5.2*64.4*602) =w(V/1098)2----5-13

Where V is in fpm and Hv in inches of water. For standard air w = 0.075 lb/cu ft

Hv =(V/4000)2 -------5-14 It states that velocity of 4000 fpm is

equivalent to 1 inch of water.

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Friction lossesAtkinson Equation

Friction losses constitutes 70 -90 % of the total losses.

These are of greater importance than shock lossesDeserves more care.A loss in static pressure.Occur in linear flow due to dragging action

between walls of the opening and fluid itself.Friction loss in a mine is a function of

1. Velocity of flow2. Interior surface characteristics of the conduit.3. Dimensions of the conduit.

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Friction lossesAtkinson Equation In mine ventilation systems the flow is always

turbulent.Fanning- Darcy equation for friction loss in a

circular conduit isHL = (f.L.V2)/(D.2g) ----------------5-15

Where HL is head loss in ft of fluid,L is length in ft.D is diameter in ft.V is velocity in fps.f is coefficient of friction

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Friction lossesAtkinson Equation Hydraulic radius RH is the ratio of the area to

the perimeter of the duct.RH = A/P

For circular ductA =(π/4) D2

P = πDPutting values

RH =D/4 ------------ 5-16

D=4RH

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Friction lossesAtkinson Equation Substituting equation 5-16 in equation 5-15HL = (f*L*V2)/(4*RH*2g) ---------5-17From this version the Atkinson equation for

mine ventilation can be derived as Hf= (f/5.2)*(L/4RH)*(0.075V2)/(2g*3600)

Hf= (KLV2)/(5.2RH)=(KPLV2)/(5.2A)----5.18

Where Hf =friction loss in inches of water.

V is the velocity in fpm) K is an empirical friction factor in lb.min2 /ft4

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This equation can also be written as;Hf = KSV2 / 5.2 A ---5.19

Where S (rubbing surface) = PLIf quantity of air is known then V = Q/AEquation can be written as Hf = KPLQ2 / 5.2A3 ---5.20

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The friction factor K in mine ventilation corresponds to the coefficient of friction f in general fluid flow.

MathematicallyK = (800) (10)-10 f ---5-21

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Determination of Airway Friction Factor The only accurate way to determine the

friction factor for a given airway is to calculate it by equation

Hf= (KPLV2)/(5.2A)

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Friction Factor K for Coal Mine Airways and Openings

Value of K x 1010

Straight Curved

Type of airway

clean Slightly obstructed

Moderately obstructed

Clean Slightly obstructed

Moderately obstructed

Smooth limed

25 28 34 31 30 43

Unlined(rock bolted)

43 49 61 62 68 74

Timbered 67 75 82 85 87 90

Observe the following precautions in selecting a value of friction factor from the table (5-1 from book) and

5-2 given in table above for use in calculations:To provide correct values of K, multiply the

numerical values obtained from the table by 10-10 , and attach units of lb.min2 /ft4

Employ a value of K determined or checked experimentally, if at all possible. This should be used on the results of actual tests conducted underground in the opening and ventilation duct involved.

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Tables 5-1 and 5-2 list values of K based on standard air density. Since K is proportional to w, correct K for actual w by formula

corrected K = (table K) (w/0.0750) ---5-22before using Eqs 5-18 to 5-20

Select K carefully for the conditions (rock type, straightness, cleanliness, irregularities etc) prevalent in the airway.

If the airway is timbered and sets are spaced on other

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than 5ft centers, modify K according to Fig 5-11. If roof bolting is used in place of timbering, assume an unlined airway.

Example 5-3. Select the friction factor for a highly sinuous, slightly obstructed drift in igneous rock.

Solution: From table 5-1, read K=120 to 225 x 10-

10 lb.min2 /ft4 for minimum to maximum irregularities. If no other information is available, and assuming standard air density, use average value K = 175 x 10-10 lb.min2 /ft4

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Example 5-4. Select the friction factor for a straight, clean, unlined entry in coal, if w = 0.065 lb/ft3.

Solution: From table 6-2, read K = 43 x 10-10 lb.min2 /ft4 Corrected K = (43 x 10-10 )(0.065 / 0.075)

= 37 x 10-10 lb.min2 /ft4

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Determination of Friction Factor for Vent PipeFriction Factor to use with different types of

ventilation pipe or tubing vary with the material and its condition. The following are satisfactory for routine calculations ( based on w = 0.075 lb/ft3 )

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Friction Factor, K x 1010 lb.min2 /ft4 (kg / m3 )

Pipe or Tubing Good, New Average, Used

Steel, wood, fiberglass (rigid)

15 (0.0028) 20 (0.0037)

Jute, canvas, plastic (flexible)

20 (0.0037) 25 (0.0046)

Spiral-type canvas 22.5 (0.0042) 27.5 (0.0051)

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Calculation of Friction Loss by FormulaFriction Loss in a mine duct or airway can be

calculated by eq: Hf= (KLV2)/(5.2RH)=(KPLV2)/(5.2A)

And

Hf = KPLQ2 / 5.2A3

A separate calculation is necessary for each airway of different characteristics ( K) or cross-sectional dimensions (A, P) and for each different airflow (V or Q).

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For different airflows in a given airway, solve KPL/5.2A3 separately and multiply by Q2.

For different airway lengths or friction factors, multiply Hf by the ratio of lengths or friction factors, respectively.

Example 5- Determine the friction loss in a mine airway, by formula under the following conditions: unlined airway in coal, curved, moderately obstructed. Size 4 x 12 ft , L = 3000 ft, Q = 48,00 cfm, w = 0075 lb/ft3 .

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Solution: Using equation 5-20Select K = 80 x 10-10

Hf = (80)(32)(3000)(4800)2 / (5.2)(48)3(10)10

= 0.0308 in Example 5-6. Determine the friction loss in mine

vent tubing, by formula under following conditions; jute vent tubing, average condition; D= 18 in, L = 3000 ft, Q = 4800 cfm, w = 0.075 lb/ cu.ft.

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Solution: Using Equation

Hf = KPLQ2 / 5.2A3

= (25)(4.71)(3000)(4800)2 / (5.2)(1.77)3 (10)10 = 28.4 in.

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Compressibility EffectThe effect of the compressibility of air has

been ignored in the discussion to date.The effect is small with the usual pressure

heads existing in mine ventilation systems. The mine static head is usually less than 10

in water which means the error in assuming a constant air quantity is approx 2%.

In long ventilation pipe, however the change in pressure may be measured in psi, and compressibility can no longer be ignored.

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In determining the friction loss it may be necessary to apply the formula for compressed air flow, which for ventilation calculation is expressed as follows:

p12 – p2

2 = Kc Q2 L/ D5 ----- 5-23

where p absolute pressure in psiQ quantity of free air in cfmD diameter in inchesKc compression friction factor

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Shock LossShock loss occur in addition to friction losses These losses are caused by changing the

direction of air flow or the area of the duct.Obstruction cause shock loss by in effect

reducing the duct area.While generally constituting only 10% to 30%

of the total head loss in mine ventilation systems, shock losses should always be considered in exact calculations in major airways or in short length of duct with many bends or area changes.

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For a given source of shock loss; the pressure drop varies as the square of the velocity or directly as the velocity head.

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Calculation of Sock LossCalculation of shock loss can be carried out

in several ways.1.Calculation of Sock Loss directly:

The amount of shock loss Hx in in water can be calculated from the velocity head Hr

Hx = XHr Where x is shock-loss factor. This factor roughly comparable to the friction

factor K in friction loss is a constant only for a given constant set of conditions.

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2. Calculation of Shock Loss by Increase in Friction Factor:

Either a calculated or estimated increment in the friction factor K may be applied to allow for shock losses in each airway in a mine ventilation system.

Table 5-1 includes several allowances for shock loss which gives equivalent lengths for various sources of shock loss in feet.

It gives appropriate

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