1
8.
ORIFICES AND MOUTHPIECES (A) ORIFICES
8.1 Introduction to to Orifices An orifice is an opening provided in the side or bottom of a tank, through which a fluid flows and is used for for measuring the discharge. discharge. Orifices may be classified as follows: (a) Based on size (i) Small orifice : An orifice is termed small when its dimensions are small compared to the head causing flow. If the head of liquid from the centre of orifice is more than five times the depth of the orifice, the orifice is termed a small orifice. Velocity does not vary appreciably from top to bottom edge of orifice. orifice: An orifice is termed large when the head of liquid is less than (ii) Large orifice: five times the depth of the orifice. Velocity varies appreciably from top to bottom edge of orifice. (b) Based on shape: Circular, triangular, rectangular or square. (c) Based on shape of upstream edge of orifice: Sharp-edged, rounded, bell-mouthed. (d) Based on discharge conditions: Free discharging, drowned/submerged – further classified as fully submerged or partially submerged. 8.2 Discharge Through a Small Orifice The flow through a small orifice will be as shown in Fig 8.1. The point at which streamlines are straight and parallel to each other is called the vena-contracta .
H 1
C
2, Vena contracta
C Fig 8.1: Small Orifice
Consider the two points, 1 and 2 when the flow is steady and at a constant head H. Applying Bernoulli’s equation to points 1 and 2: p1 v12 p 2 v 22 + + z1 = + + z 2 where w 2g w 2g p1
∴
+
w Now,
p1
v12
=
2g
w
+
v 22 2g
(∵ z1
=
z2 )
p2
= 0 (atmospheric pressure), v 1 is very small compared to v 2, as the w w tank area is very large when compared to that of the jet of the liquid.
∴H +
0
Hence,
=
=
H and
p2
0+
v2
v 22 2g =
v th
=
2gH , (where vth = theoretical velocity)
2 Hydraulic coefficients: (i) Coefficient of velocity, C v – is the ratio of the actual velocity of the jet at vena contracta to the theoretical velocity of the jet. actual velocity of jet at vena contracta v Cv = = theoretica l velocity 2gH
Cv varies from 0.95 to 0.99 for different orifices depending on size, shape and head. For sharp edged orifice, C v = 0.98. (ii) Coefficient of contraction, C c – is the ratio of the area of jet at vena contracta to the area of the orifice. area of jet at vena contracta a c Cc = = area of orifice a Cc varies from 0.61 to 0.69 depending on shape, size and head of liquid, and may be generally taken as 0.64. (iii) Coefficient of discharge, Cd – is the ratio of the actual discharge from an orifice to the theoretical discharge. actual discharge Cd = theoretical discharge actual velocity × actual area =
=
theoretical velocity × theoretical area actual velocity actual area ×
theoretical velocity i.e. Cd
=
Cv
×
theoretical area
Cc , where Cd varies from 0.61 to 0.65, and generally may be
taken as 0.62. Experimental Determination of Coefficients (a) Determination of coefficient of discharge, C d By maintaining a constant head of water in the tank over the orifice, the jet of water discharged by the orifice is collected in a measuring tank. The rise in level of water in the measuring tank is measured in a known interval of time. (area of measuring tank × rise in water level) Actual discharge, Qa = interval of time Theoretical discharge, Qt = a 2gH ∴
Cd
=
Qa Qt
=
Qa a 2gH
The experiment is repeated a number of times and the average value of C d obtained. (b) Determination of coefficient of velocity, Cv Constant head of water is maintained over the orifice. Consider any point P on the centre line of the jet as shown in Fig 8.2. Let: v = velocity of the jet at the vena contracta, the horizontal and vertical ordinates of P being x and y with respect to the centre of the jet at venacontracta as origin. t = time taken by a particle of water to move from vena-contracta to P.
3
Inflow to maintain constant level H
C x
C
y P •
Fig 8.2
Horizontal displacement, x = vt Vertical displacement, y = ½gt² From equation (i) and (ii), y=
g x
2
2 v
2
⋅
gx
, or v =
2
2y
Theoretical velocity, v = ∴
...(i) ...(ii)
2gH
Coefficient of velocity, Cv =
v V
=
gx 2 2y
×
1
=
2gH
x2 4yH
The experiment may be repeated taking other points on the centre line of the jet and the values of Cv obtained averaged. (c) Determination of the coefficient of contraction, Cc (i) First method The area of the jet at vena contracta is measured using a micrometer, as shown in Fig 8.3. The instrument consists of a ring with four radial screw gauges, equally spaced. The screws are adjusted such that the sharp points just touch the jet. The instrument is then removed, and the spacings between the screws measured accurately. Shortcomings noted include: (i)-Jet is not absolutely circular (ii)It is not possible to adjust all four screws at the same time. (ii) Second method Coefficients of discharge and velocity are first determined. Coefficient of contraction is then found by dividing the coefficient of discharge by the coefficient of C velocity. ∴ C c = d Cv Screw gauge Water jet
Fig 8.3
4 8.3 Discharge Through a Large Orifice. In the case of a large orifice, the velocity is not constant over the entire cross-section of the jet. Consider a large rectangular orifice in one side of a tank, discharging freely into atmosphere under a constant head, H, as shown in Fig 8.4.
H1
h
H H2 dh
d
b Fig 8.4: Large Orifice
Let: H1 = height of liquid above top of the orifice. H2 = height of liquid above bottom edge of orifice. b = breadth of orifice. d = depth of orifice = (H 2 – H1) Cd = coefficient of discharge of the orifice. Consider an elementary strip of thickness dh, at a depth of h below the free surface of liquid in the tank as shown. Area of strip = b×dh. Theoretical velocity of water through the strip =
2gh
Discharge through the elementary strip, dQ = Cd×area of strip×velocity. =
Cd × b ⋅ dh × 2gh
=
Cd ⋅ b ⋅ 2g ⋅ h 2 dh
1
Total discharge through the orifice may be obtained by integrating between the limits, H1 and H2. ∴
∫
Q = dQ =
H2
∫ Cd
⋅
b ⋅ 2g h 1 2 dh
H1 H2
=
Cd ⋅ b ⋅
h3 2 2g 3 2 H
1
=
2
Cdb 2g H 32 2 3
−
H13 2
8.4 Discharge Through a Submerged Orifice A Submerged Orifice (or Totally Drowned Orifice) discharges at the outlet to another liquid of the same kind as shown in Fig 8.5. Let: H1 = height of liquid above top of orifice on upstream side. H2 = height of water above bottom of orifice. H = difference in water level. b = width of orifice Cd = coefficient of discharge.
5 Height of water above centre of orifice on upstream side, H 2 − H1 H1 + H 2 = H1 + = 2 2 Height of water above centre of orifice on downstream side, H1 + H 2 = −H 2
H H1 H2 1
2
Fig 8.5: Submerged Orifice
Applying Bernoulli’s equation at sections 1 and 2, we get 2 2 p1 v1 p 2 v 2 (∵ z1 = z 2 ) + = + w 2g w 2g p H + H 2 p 2 H1 + H 2 Now, 1 = 1 , = − H , and v1 is negligible. w 2 w 2 H1 + H 2 H1 + H 2 v 22 ∴ +0 = −H+ 2 2 2g 2
Or
H= v2
∴
v2 2g
=
2gH
Area of orifice = b × (H 2 − H 1 ) Discharge = Cd × area × velocity
Or
Q
=
Cd × b(H 2
=
Cd b(H 2
−
−
H 1 ) × 2gH
H 1 ) 2gH
8.5 Discharge Through a Partially Submerged Orifice This has the outlet partially submerged under the liquid as shown in Fig 8.6.
H1
H
H2 1
2
Fig 8.6: Partially Submerged Orifice
Total discharge is equal to that of the submerged portion and free portion.
6 Discharge through submerged portion is given by: Q1
=
Cd b(H 2 − H1 ) 2gH (submerged portion)
Discharge through the free portion is given by: 2 Q 2 = Cd b 2g (H 32 2 − H13 2 ) (free portion) 3 Total discharge, Q = Q1 + Q 2 2 3 2 3 2 Or Q = Cd b (H 2 − H 1 ) 2gH + Cd b 2g (H 2 − H 1 ) 3 8.6 Time of Emptying a Tank Through an Orifice (a) Cylindrical tank Consider a circular tank with an orifice at the bottom as shown in Fig 8.7. Let : A = area of tank top surface. a = area of orifice. H1 = initial height of liquid. H2 = final height of liquid T = time in seconds for level to fall from H 1 to H2.
dh H1 h H2 Orifice Fig 8.7: Cylindrical Tank
Let at any time, height of liquid above orifice be h, and liquid surface fall by dh in time dT. Volume of liquid leaving tank in time dT = A×(-dh) …(i) (negative sign indicates level is falling) Theoretical velocity through orifice, v =
2gh
Discharge through orifice in time dT, dQ = Cd×area of orifice×theoretical velocity =
C d ⋅ a ⋅ 2gh ⋅ dT
…(ii)
As volume of liquid leaving tank is equal to volume of liquid flowing through orifice in time dT, A(− dh ) = C d ⋅ a ⋅ 2gh ⋅ dT Or
dT =
−
h 1 / 2 dh A where k = − = k C ⋅ a ⋅ 2g 2gh C d ⋅ a ⋅ 2g d
Adh
Cd ⋅ a ⋅
−
By integrating the above equation between limits H 1 to H2, total time T is obtained:
7 H2
H2
T
∫
∫
T = dT = k h 0
Or
T
T
∴
=
=
1 / 2
−
H1
k × 2 H
12 2 −
2A [H 11 2
−
h1 2 dh = k 1 2 H
=
k [2h 1 2 ]H1
H2
1
H 12 1
H 12 2 ]
Cd ⋅ a ⋅ 2g
To empty tank completely, H 2 = 0 2A H 11 2 ∴ T= Cd ⋅ a ⋅ 2g (b) Hemispherical tank Consider a hemispherical tank as shown in Fig 8.8 Let R = radius of tank a = area of orifice. H1 = initial height of water in tank. H2 = height of water in tank after time T h = head of liquid over orifice at any instance of time, x being the radius of the liquid surface.
O R
R dh
C
D
H1
h H2 Orifice Fig 8.8: Hemispherical Tank
Area of liquid surface, A = πx² Theoretical velocity of liquid through orifice,v =
2gh
Let liquid level fall down by dh in time dT. Volume of liquid leaving tank in time dT, = A(-dh) = -πx²dh Also, volume of liquid flowing through orifice, = Cd×area of orifice×velocity
… (i)
= Cd ⋅ a ⋅ 2gh Volume of liquid flowing through orifice in time dT, = Cd ⋅ a ⋅ 2gh dT Equating equations (i) and (ii) ∴
πx
2
( −dh ) = Cd ⋅ a ⋅ 2gh dT (negative sign introduced since with increase of T, h decreases)
… (ii)
8
∴
−
2
πx dh
=
Cd ⋅ a ⋅ 2gh dT
…(iii)
From figure and considering ∆OCD, OC = R, and DO = R-h ∴
CD = x
or x 2
R2
=
OC2
= −
OD2
−
( R − h)
2
=
R2
=
R2
−
(R
( R − h)
−
2
+
h2
−
2
2Rh ) = 2Rh − h2
Substituting value of x² in equation (iii): −
or
π(2Rh − h )dh 2
dT = −
Cd ⋅ a ⋅ 2gh ⋅ dT
=
π
Cd ⋅ a ⋅ 2g π
= −
π
= −
Cd ⋅ a ⋅ 2g
2
Cd ⋅ a ⋅ 2g
(2Rh
12
−
(2Rh − h )h
12
−
dh
h 3 2 )dh
Total time T required to bring level from H 1 to H2 is obtained by integrating the above equation between limits H 1 to H2: H2
T
=
∫
H1
π
−
Cd ⋅ a ⋅ 2g
where k = −
h3 2 T = k 2 32
Or
=
−
π
Cd ⋅ a ⋅
−
h
32
)dh = k ∫ (2Rh
12
−
h 3 2 )dh ,
H1
π
Cd ⋅ a ⋅ 2g H2
5 2 H
h
52
1
4 k R ( H 32 2 3
=
(2Rh
H2
12
−
H 13 2 ) −
2
(H 5
4 R ( H13 2 2g 3
−
5 2 − 2
H15 2 )
H32 2 ) −
2
(H 5
52 − 1
H52 2 )
For complete emptying of tank, H 2 = 0; π 2 5 2 4 32 ∴ − T= RH H1 1 5 Cd ⋅ a ⋅ 2g 3 (c) Time of Emptying a Circular Horizontal Tank Consider a circular horizontal tank as shown in Fig 8.9, of length L, and radius R, containing liquid up to a height H 1.
O dh A h
R
L Fig 8.9: Circular Horizontal Tank
B
C H1 H2
9 Let a = area of orifice at bottom of the tank. Time required to bring level from H 1 to H2 is obtained as follows: Let at any time, height of liquid over orifice be h, and in time dT, let height fall by dh. Let at this time width of liquid surface = L×AC But AC = 2 × 2AB = 2 AO 2
∴
OB 2
−
(R
=
2 R2
=
L × AC = L × 2
Surface area, A
−
2
h2
+
2 R2
=
−
−
(R − h )2
2Rh ) = 2
(2Rh − h ) 2
(2Rh − h ) 2
Volume of liquid leaving tank in time dT, A(-dh) = 2L 2Rh − h 2 × ( −dh) Also volume of liquid flowing through orifice in time dT, = Cd×area of orifice×velocity in time dT =
… (i)
2gh
But velocity of liquid at time considered =
Volume of liquid flowing through orifice in time dT,
∴
=
… (ii)
Cd ⋅ a ⋅ 2gh × dT
Equating equations (i) and (ii): 2L 2Rh − h
2
×
(- dh ) = Cd ⋅ a ⋅
2gh × dT
(negative sign introduced as with increase of T, height h decreases) ∴
dT
= −
2L 2Rh − h
2
Cd ⋅ a ⋅ 2gh = −
2L
(2R − h )
Cd ⋅ a ⋅ 2g H2
Total time, T
=
∫
H1
−
12
Cd ⋅ a ⋅ 2g 2L
H2
12 ( − h ) dh = k (2R − h ) dh 2R ∫ ∫ 12
H1
(2R − h )3 2 = k 32 T=
dh
H2
Cd ⋅ a ⋅ 2g
Or
(taking h as common)
dh
2L(2R − h )
− =
dh
4L 3Cd ⋅ a ⋅ 2g
H1 H2
× −1 H
[(2R
1
−
For complete emptying of the tank, 4L (2R )3 2 T= 3Cd ⋅ a ⋅ 2g
[
8.7 8.10.
H2 )
−
3 2
−
(2R − H 1 )3 2 ]
(2R − H 1 )3 2 ]
Time of Flow From One Vessel to Another Consider two tanks with communicating orifice in the common wall as shown Fig
10
h Area, A1 Area, A2 a,v
Fig 8.10
Let: A1 and A2 be the areas of the two tanks a = area of the orifice. Let at any instant the water level in the tank of area A 1 be h units above the water level of the other tank, i.e. h = head causing the flow. Let in a small interval of time dT the fall in water level of the first tank be dh 1, and the corresponding rise in water level of the other tank be dh 2. Quantity discharged in dT seconds, = A1dh1 = A2 dh2 A ∴ dh 2 = 1 dh 1 A2 ∴ Change in the head causing the flow,
∴
dh
=
dh 1
=
dh 1
+
dh 2
=
dh 1
A1
+
A2
dh 1 = 1 +
A 1 dh 1 A 2
dh
A1 1 + A 2
Since the quantity discharged from the first tank equals the quantity that has passed through the Orifice, A 1dh 1 ∴
=
dT =
=
=
Cda 2gh dT A 1dh 1 Cda 2g A1
h
12
−
dh
Cda 2g A 1 1 + A 2 A
A Cda 2g 1 + 1 A 2
h
h
12
−
12
−
dh
Suppose initially the difference of water levels in the tanks is H 1. Let after time T the difference of water level in the tanks be H 2. Integrating the above equation between the lower and upper limits of h,
11
T
=
=
H1
A1
∫h A
Cda 2g 1 + 1 H A 2 2A1
(
H1
−
H2
12
−
dh
2
)
A Cda 2g 1 + 1 A 2 Duration of flow: For this condition, put H 2 = 0. ∴
T=
2A1 H1
A Cda 2g 1 + 1 A
8.8
2
Discharge From a Tank With Inflow Consider a tank receiving an inflow as shown in Fig 8.11. Q dh H1 h H2 q Fig 8.11
Let: A = area of tank a = area of orifice Q = uniform inflow of water into the tank q = rate of outflow through the orifice h = head of water over orifice at an instant of time dh = change in level in a small interval of time, dT Amount of inflow = QdT Amount of outflow = qdT =
Cda 2gh ⋅ dT
=
kh 1/2 dT , where k = Cd a 2g
Increase of liquid in tank, 1/2 1/2 = Adh = QdT-kh dT = (Q-kh )dT Adh ∴ dT = (Q − kh 1/2 ) Let T be the interval of time in which the head of water increases from H 1 to H2; ∴
H2
∴
T=
∫Q
H1
Put:
Q − kh 1/2
=
z
Adh −
kh 1/2
12
h
∴
dh
∴
or
1/2
∫Q
dh −
kh 1/2
=
=
=
∫
=
=
∴
or h k − 2(Q − z )dz
=
(Q − z )2 2
k
k 2 − 2(Q − z )dz k 2 z
2 Q − 1dz k 2 z −2 [Qlog e z − z] 2 k −2 [Qlog e (Q − kh 1/2 ) − (Q − kh 1/2 )] 2 k −
=
Q−z
∫
-2A
H2
Qlog e ( Q - kh1/2 ) - ( Q - kh1/2 ) 2 H k -2A = Qloge Q - k H 2 - Q - k H 2 k 2
T=
1
{
or
-2A T = 2 Qlog e k
) (
(
)} - {Qlog e (Q - k
H - Q-k H 1 1
) (
)}
Q - k H2 + k ( H 2 - H1 ) Q-k H 1
Examples 8.1. A circular vertical tank of diameter 4m contains water up to a height of 5m, and has an orifice of diameter 0.5m at the bottom. Determine the time required: (a) for water to fall from 5m to 2m (b) to completely empty the tank. Take Cd = 0.6 Solution
Diameter of tank, D = 4m,
∴
Diameter of orifice, d = 0.5m
π
A= ∴
4
a
=
×
4
π
2
×
4
=
12.566m
0.5
2
=
2
0.1963m
2
Initial height of water, H 1 = 5m Case 1 : When final height of water, H 2 = 2m; 2A (H11 2 − H12 2 ) T= Cd ⋅ a ⋅ 2g 2 × 12.566
) = 39.58s 0.6 × 0.1963 × 2 × 9.81 Case 2 : When final height of water, H 2 = 0; 2 × 12.566 (51/2 − 0) = 107.7s T= 0.6 × 0.1963 × 2 × 9.81 Examples 8.2. A hemispherical tank of diameter 4m contains water up to a height of 1.5m. An orifice of diameter of 0.05m is provided at the bottom. Find the time required: (i) for the water level to fall from 1.5m to 1.0m (ii) to completely empty the tank. Take Cd = 0.6. =
(5
1/2
−
2
1/2
13 Solution Diameter of hemispherical tank, D = 4m, Radius, R = 2m π Diameter of orifice, d = 0.05m ∴ a = × 0.05 2 = 0.001963m 2 4 Initial height of water, H 1 = 1.5m; Cd = 0.6 Case 1: (H2 = 1m) π 2 5/2 4 3/2 3/2 5/2 T= R (H1 − H 2 ) − (H1 − H 2 ) 5 Cd ⋅ a ⋅ 2g 3
2 4 3/2 3/2 5/2 5/2 ) ( ) × 2(1.5 −1 − 1.5 −1 3 5 2 × 9.81
π
=
0.6 × 0.001963 × = 921.4s = 15min 21.4s Case 2: (H2 =0) π 2 5/2 4 3/2 − T= RH H1 1 5 Cd ⋅ a ⋅ 2g 3 =
4 3/2 × 2 × 1.5 2 × 9.81 3
π
0.6 × 0.001963 × = 2286.33s = 38min 6.33s
−
2 5
×
1.5 5/2
(B) - MOUTHPIECES 8.9 Introduction to mouthpieces A mouthpiece is a short length of pipe which is 2 to 3 times the diameter in length, fitted into the side of a tank or vessel containing liquid and it is used for measuring the discharge. Mouthpieces are classified as follows: Based on position of tank where fitted – Internal or external mouthpiece. Based on shapes – Cylindrical, Convergent or convergent-divergent. Based on nature of discharge (for internal mouthpieces only): (Bordas or reentrant mouthpieces) - Running full or Running free (jet not touching the sides of the mouthpiece). 8.10 Flow through external cylindrical mouthpiece Consider an external cylindrical mouthpiece as shown in Fig 8.12.
H
C
Vena contracta 1
A
C
1
Fig 8.12
Let: H = Height of liquid above centre of mouthpiece vc = velocity of liquid at vena contracta ac = area of flow at vena contracta a1 = area of mouthpiece at outlet
14 Cc = Coefficient of contraction Applying Continuity equation at C-C and 1-1: a v v1 a c × v c = a 1 v1 or v c = 1 1 = ac a c /a 1 But a c /a 1
=
vc
=
∴
Cc ≈ 0.62 v1
0.62 Due to sudden enlargement at 1-1, there will be head loss h L, given as, hL But v c
=
=
(vc − v1 )2 2g
v1 /0.62
v1 − v1 0.62
2 2
v12 1 v12 ∴hL = = − 1 = 0.375 2g 2g 0.62 2g Apply Bernoulli’s equation to point A and 1-1; pA
v 2A
p1
v12
+ + zA = + + z1 + h L w 2g w 2g But zA = z, vA is negligible, and p 1 /w = atmospheric pressure = 0 gauge pressure, pA /w = H. v12 v12 ∴H + 0 = 0 + + 0.375 2g 2g
H = 1.375
2g
2gH
= 0.855 2gH 1.375 Theoretical velocity of liquid at outlet is, ∴
v1 =
v12
vT ∴
=
2gH
Coefficient of velocity for mouthpiece, Cv =
actual velocity theoretica l velocity
=
0.855 2gH 2gH
=
0.855
Cc for mouthpiece = 1 as the area of the jet at inlet is equal to area at outlet. Thus Cd = Cc×Cv =1×0.855 = 0.855 Since Cd for mouthpiece is > Cd for orifice, the discharge will be more. Discharge, Q = Cda1v1
=
0.855a 1 2gH
Pressure head at vena contracta: Apply Bernoulli’s equation at A and C-C: 2 p A v 2A pc vc + + zA = + + zc w 2g w 2g
15 pA
But
=
w
Ha + H, v A
Ha + H + 0 = Hc +
∴
But v c
0, z A
v c2 2g
=
Zc ,
pc
= Hc w where Ha = atmospheri c pressure =
or Hc = Ha + H −
2g
v1
=
0.62 2
v 1 v Hc = Ha + H − 1 /2g = Ha + H − 1 × 2g (0.62)2 0.62
∴
Also H
=
1.375
v12 2g
or
v12 2g
=
2
H/1.375 = 0.7272H
Hc = Ha + H − 0.7272H ×
∴
or H C ∴
v c2
HC
=
Ha + H − 1.89H
=
Ha − 0.89H
1
(0.62 )2
8.11 Flow Through a Convergent-Divergent Mouthpiece Consider a convergent-divergent mouthpiece as shown in Fig 8.13.
C
1
H a1, v1 C
1
Fig 8.13: Convergent-Divergent Mouthpiece
Apply Bernoulli’s equation to the free water surface in the tank and section C-C. p v2 pc vc2 + +z = + + zc w 2g w 2g Assume datum at centre of mouthpiece: pc p = Ha, v = 0, z = H, = Hc, z c = 0 w w v c2 ∴ Ha + 0 + H = Hc + +0 …(i) 2g Or ∴
vc
v c2 2g =
=
Ha + H − Hc
2g(Ha + H − Hc )
…(ii)
16 Now, Apply Bernoulli’s equation at section C-C and 1-1; p c v c2 p1 v12 + + zc = + + z1 w 2g w 2g But z c
p1
z 1 and
=
=
w
Ha
2
2
vc
= Ha + 2g 2g Also from equation (i), 2 vc Hc + = Ha + H 2g ∴
Hc +
v1
2
∴
v1
Ha +
=
Ha + H
=
2gH
2g
or
v1
…(iii)
By continuity equation, a c vc a1
or
ac
=
=
a 1 v1 or
a1
=
ac
vc
2g(Ha + H − Hc)
=
v1
2gH
=
Hc Ha +1− H H
Ha − Hc 1 + H
…(iv)
Discharge, Q = Cd×area×velocity, with Cd = 1 for this mouthpiece. ∴
Q
=
a c 2gH
…(v)
Examples 8.3: A convergent-divergent mouthpiece having a throat diameter of 40mm discharges water under a constant head of 2m. Determine: (i) the outlet diameter for maximum discharge. (ii) the maximum discharge . Take Ha = 10.3m of water and Hsep = 2.5m of water (absolute). Solution (i) Given: d c = 40mm, H=2m, Ha = 10.3m, Hsep = Hc = 2.5m absolute π 2 4 2 a c = × 0.04 = 1.2566 × 10 m 4 Ratio of areas at outlet and throat is given by: π 2 2 d1 d 1 d1 4 ∴ = = 2.2135 or = 1.4877 π 2 d c dc dc 4 ∴ d 1 = 1.4877 d c = 1.4877 × 40 = 50.5mm −
(ii) Q max
=
a c 2gH
=
1.2566 × 10
4
−
×
2 × 9.81 × 2
=
3
0.00787m /s
8.12 Re-Entrant on Borda’s Mouthpieces (or Internal Mouthpieces) Borda’s mouthpiece running free Consider Borda’s mouthpiece running free as shown in Fig 8.14.
17 Let: H = Height of liquid above centre line of mouthpiece. a = area of mouthpiece ac = area of contracted jet vc = velocity of jet through mouthpiece
H
1 vc
1 Fig 8.14: Borda’s Mouthpiece Running Free
Pressure force on entrance to mouthpiece, = w ⋅a ⋅h = w ⋅a ⋅H …(i) where h = distance of centre of gravity of area ‘a’ from free surface=H. The net force is equal to the rate of change of momentum: Mass of fluid flowing/s = ρ·ac·vc Liquid in tank is initially at rest, (v = 0), but final velocity is v c Rate of change of momentum, = mass of fluid flowing/s ×change of velocity (final-initial velocity) = ρ·ac·vc×(vc-0) 2 = ρ ⋅ ac ⋅ v c …(ii) Equate equations (i) and (ii): w ⋅ a ⋅ H = ρ ⋅ ac ⋅ v 2c =
w 2 ⋅ ac ⋅ v c g
(since ρ
=
w ) g
…(iii)
Applying Bernoulli’s equation to free surface of liquid and section 1-1, p v2 p1 v12 + +z = + + z1 w 2g w 2g Taking datum at centre of mouthpiece, p w ∴
∴
=
Ha
=
0 (gauge pressure), v
0+0+H = 0 + v c
=
v 2c 2g
+
0
Or H =
=
0, z
=
H, p1
=
0, v1
=
v 2c 2g
2gH
Substitute v c in equation (iii), w ⋅a ⋅H = ∴
ac a
=
w ⋅ ac ⋅ 2gH , Or a g
=
2 ac for the mouthpiece
0.5
Since there is no loss of head, Cv = 1.0 ∴ Cd = Cc × Cv = 0.5 × 1.0 = 0.5 Discharge, Q = Cd ⋅ a ⋅ 2gH
=
0.5a 2gH
vc , and z1
=
0
18
Borda’s mouthpiece funning full: Consider Borda’s mouthpiece as shown in Fig 8.15. Let H = height of liquid above centre of mouthpiece v1 = velocity at outlet, 1-1 a = area of mouthpiece ac = area of flow at C-C vc = velocity of liquid at vena-contracta
C H
1
C
1
Fig 8.15: Borda’s Mouthpiece Running Full
Since jet enlarges suddenly after C-C, there is loss of head due to sudden enlargement. ∴
hL
=
(v c − v 1 ) 2
(derived in chapter 9)
2g From continuity equation, ac ⋅ v c
=
a 1v1 or v c
vc
=
2v 1
∴
=
a1
⋅
ac
v1
=
v1 a c /a1
=
v1 Cc
=
v1
…(i)
( ∵ Cc
0.5
=
0.5)
Substitute the value in equation (i); ∴
hL
=
(2v1 − v1 )2
v12
=
2g 2g Applying Bernoulli’s equation to free surface of liquid in tank and section 1-1; p1 v12 p v2 + +z = + + z1 + h L w 2g w 2g Taking datum at centre line of mouthpiece,
0+0+H = 0+ ∴
v1
=
v12 2g
+
0+
v12 2g
or H =
v12 2g
+
v12
=
2g
v12 g
gH
where v1 is the actual velocity since losses have been considered. ∴
vT
=
2gH
Coefficient of velocity, Cv =
v1 vT
=
gH 2gH
=
1 2
=
0.707
Since area of jet at outlet = area of mouthpiece, Cc = 1 ∴ Cd = Cc × Cv = 1.0 × 0.707 = 0.707
Discharge, Q = Cd ⋅ a ⋅ 2gH or
Q = 0.707a 2gH
