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1
Cotents
 Introduction equivalent orifice
 Determination of Area of Equivalent orfice
 Results and Conclusions
2
Calculation of Area
of Equivalent Orifice
PREPARED BY
AAMIR IQBAL
3
Introduction
to Equivalent
Orifice
 Mechanical definition:
An expression of fan performanc
e as the theoretical sharp-
edge orifice area which would of
fer the same resistance to flowas
the system resistance itself.
 The equivalent orifice is a
conception which gives a
convenient visual presentation
of the relative ease of a
ventilating through a mine
opening.
4
Cont….
 The equivalent orifice of a mine is defined
as;
 An opening A in a thin plate through
which same amount of air passes under
the effect of a pressure difference
between the sides of the plate, as would
pass through the mine opening under the
same head.
5
Determination of area of equivalent
Orifice
 Let us assume;
 Q (m3 /sec ) is the quantity of air flowing through the orifice.
 A (m2 ) is area of orifice [fig 8.8]
 Let p1 be the pressure on the left side of the plate
 And p2 be the pressure on the right side of
the plate (Kg/m2)
 So mine head in that case is given as .
h = p1 - p2
6
Determination of area of equivalent
Orifice
 For section 1-1,
 As it far away from the surface so v1 is neglected
 i.e. V1 =0
 And for section 2-2,
 At a point in the stream, where stream lone is parallel,
 We will use Bernoulli’s equation i.e.
 p1 + (v1
2/2g ) ϒ = p2 + (v2
2/2g) ϒ --------- (1)
7
Determination of area of equivalent
Orifice
 Now , as we know that.
 V1 = 0
 Equation (1) becomes,
 p1 - p2 = h = v2
2 ϒ /2g
 And consequently,
 v2 = 2𝑔ℎ/ϒ m/sec
 Now for quantity flowing through section 2-2, We know
 Q = v2A2
8
Determination of area of equivalent
Orifice
 Where, A2 is the cross section of stream in plane 2-2
 But A2 = φ A where,
 φ = the coefficient of discharge which on average has
a value of 0.65
 Thus,
 A = A2 / φ
 => Q/ φ. V2
 => Q/ φ. 2𝑔ℎ/ϒ
9
Determination of area of equivalent
Orifice
 => (1/0.65 19.62/1.2 )(Q/ ℎ)
 => 0.38Q/ ℎ
 The dimension of the equivalent orifice are;
 [A] = Q / 𝑔ℎ/ϒ (sqr.meter)
10
Result;
 The simple dimension of the equivalent orifice. Sqr.meters
enables the value of the orifice to visualize very simply and
thus to judge the degree of ease or difficulty of ventilating a
colliery or metal ore mine.
 The larger the value of A, the easier it is to ventilate
 For metal ore mines,
 If Equivalent Orifice Area A = 1 sq m, the mine mine will
narrow
 If A = 1 or 2 sq m, it is medium mine
11
Results cont….
 If A > 2 sq m, it means the mine wide or large.
 So all we need to do is fine out the mine head h and quantity
of air flowing the through the mine.
 As the equation shows .
 A is directly proportional to the quantity of air Q flowing
through the mine.
 And Area A is inversely proportional to the sqr.root of mine
head h.
12
Any queries? 13
Much appreciated 14

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ventilation

  • 1. 1
  • 2. Cotents  Introduction equivalent orifice  Determination of Area of Equivalent orfice  Results and Conclusions 2
  • 3. Calculation of Area of Equivalent Orifice PREPARED BY AAMIR IQBAL 3
  • 4. Introduction to Equivalent Orifice  Mechanical definition: An expression of fan performanc e as the theoretical sharp- edge orifice area which would of fer the same resistance to flowas the system resistance itself.  The equivalent orifice is a conception which gives a convenient visual presentation of the relative ease of a ventilating through a mine opening. 4
  • 5. Cont….  The equivalent orifice of a mine is defined as;  An opening A in a thin plate through which same amount of air passes under the effect of a pressure difference between the sides of the plate, as would pass through the mine opening under the same head. 5
  • 6. Determination of area of equivalent Orifice  Let us assume;  Q (m3 /sec ) is the quantity of air flowing through the orifice.  A (m2 ) is area of orifice [fig 8.8]  Let p1 be the pressure on the left side of the plate  And p2 be the pressure on the right side of the plate (Kg/m2)  So mine head in that case is given as . h = p1 - p2 6
  • 7. Determination of area of equivalent Orifice  For section 1-1,  As it far away from the surface so v1 is neglected  i.e. V1 =0  And for section 2-2,  At a point in the stream, where stream lone is parallel,  We will use Bernoulli’s equation i.e.  p1 + (v1 2/2g ) ϒ = p2 + (v2 2/2g) ϒ --------- (1) 7
  • 8. Determination of area of equivalent Orifice  Now , as we know that.  V1 = 0  Equation (1) becomes,  p1 - p2 = h = v2 2 ϒ /2g  And consequently,  v2 = 2𝑔ℎ/ϒ m/sec  Now for quantity flowing through section 2-2, We know  Q = v2A2 8
  • 9. Determination of area of equivalent Orifice  Where, A2 is the cross section of stream in plane 2-2  But A2 = φ A where,  φ = the coefficient of discharge which on average has a value of 0.65  Thus,  A = A2 / φ  => Q/ φ. V2  => Q/ φ. 2𝑔ℎ/ϒ 9
  • 10. Determination of area of equivalent Orifice  => (1/0.65 19.62/1.2 )(Q/ ℎ)  => 0.38Q/ ℎ  The dimension of the equivalent orifice are;  [A] = Q / 𝑔ℎ/ϒ (sqr.meter) 10
  • 11. Result;  The simple dimension of the equivalent orifice. Sqr.meters enables the value of the orifice to visualize very simply and thus to judge the degree of ease or difficulty of ventilating a colliery or metal ore mine.  The larger the value of A, the easier it is to ventilate  For metal ore mines,  If Equivalent Orifice Area A = 1 sq m, the mine mine will narrow  If A = 1 or 2 sq m, it is medium mine 11
  • 12. Results cont….  If A > 2 sq m, it means the mine wide or large.  So all we need to do is fine out the mine head h and quantity of air flowing the through the mine.  As the equation shows .  A is directly proportional to the quantity of air Q flowing through the mine.  And Area A is inversely proportional to the sqr.root of mine head h. 12