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Unformatted text preview: CHG 2312
CHG
Fluid Flow
Bernoulli's Equation Outline
1.
2.
3. Energy balance: Steady, incompressible flow
Friction heating
Bernoulli equation
3.1. Zero flow (fluid static)
3.2. Head form of Bernoulli equation
Diffusers (gradual expansion)
5. Sudden expansion
6. Torricelli equation
7. Fluid flow measurement
a) Pitot tube
b) Pitotstatic tube
c) Venturi meter
d) Orifice meter
d) Orific
te
e) Rotameter
8. Cavitation
8.
9. Unsteady flow
10. Nonuniform flows 1. Energy balance:
Steady, incompressible flow
Bernoulli’s Equation: dW
⎛P
V2 ⎞ ⎛ P
V2 ⎞
dQ
⎜ u + + gz + ⎟ − ⎜ u + + gz + ⎟ = − n.f. −
⎜
ρ
2 ⎟in ⎜
ρ
2 ⎟out
dm dm
⎝
⎠⎝
⎠ ⎛P
V 2 ⎞ dWn.f. ⎛
dQ ⎞
⎟=
− ⎜ ∆u −
∆⎜ + gz +
⎟
⎟
⎜ρ
2⎠
dm ⎝
dm ⎠
⎝ University of Ottawa, CHG 2312, P. Mehrani 3 2. Friction heating
Examples in solids: smoking brakes, saw cutting wood, etc.
Temperature increases in fluid due to friction heating are
less, why?
o Am ount of frictional work per unit m ass is generally less
than in examples of solids above. (in solids energy is
concentrated in small volume, whereas in fluid it is
spread over a larger volume).
o Heat capacity of liquids is generally greater than that of
solids. University of Ottawa, CHG 2312, P. Mehrani 4 2. Friction heating
For constantdensity materials only: ∆u = friction h eating dQ
+
dm
dm ∆u − dQ
=F
dm F is the friction heating per unit mass and if it exist, it will be positive.
Usually referred to as friction loss! 5 University of Ottawa, CHG 2312, P. Mehrani 2. Friction heating
Incompressible flow vs. incompressible fluid:
o Incompressible flow: density changes unimportant. o All steady flows of liquid and most steady flows of gasses at
low velocities are considered incompressible. o Some unsteady flows of liquids and all steady flows of gases
at high velocities may not be considered incompressible.
P1Patm V2 (ft/s,calc) V2 (ft/s, actual) T2 (oF) 0.01 35 35 67.9 0.1 110 111 67.0 0.3 190 191 65.0 0.6 267 268 62.0 1.0 340 343 58.2 2.0 466 476 49.1 5.0 678 713 25.6 University of Ottawa, CHG 2312, P. Mehrani 6 3. Bernoulli’s equation:
⎛P
V 2 ⎞ dWn.f. ⎛
dQ ⎞
⎟=
− ⎜ ∆u −
∆⎜ + gz +
⎟
⎟
⎜ρ
2⎠
dm
dm ⎠
⎝
⎝ ⎛P
V 2 ⎞ dW n. f .
⎟=
∆ ⎜ + gz +
−F
⎜ρ
2⎟
dm
⎝
⎠
F is a positive number for all real flows.
Referred to as the mechanical energy balance. 7 University of Ottawa, CHG 2312, P. Mehrani 3.1. Zero flow (fluid statics)
z Bernoullli’s equation
l ⎛P
V 2 ⎞ dWn. f .
⎟=
∆⎜ + gz +
−F
⎜ρ
2⎟
dm
⎝
⎠ 2
1 Fluid at rest: ⎛P
⎞
∆⎜
⎜ ρ + gz ⎟ = 0
⎟
⎝
⎠ lim ∆z→ 0 1 ρ ∆P = − g∆z ∆P
dP
=
= − ρg
∆z
dz University of Ottawa, CHG 2312, P. Mehrani 8 3.2 Head form of Bernoulli’s equation
Divide both sides of Bernoulli's equation by gravitational
constant (g): ⎛P
V2
∆⎜
⎜ ρg + z + 2 g
⎝ ⎞ dW n. f . F
⎟=
−
⎟
gdm
g
⎠ Every term has the dimension of length [m]
From left: pressure head, gravity head, velocity head,
pump or turbine head and friction head loss. University of Ottawa, CHG 2312, P. Mehrani 9 4. Diffusers (gradual expansion)
One of the ways to slow down a fluid: gradual expansion.
of he
slow own luid radua xpansion. U niv ersity of Ottawa, CHG 2312, P. Mehrani 10 4. Diffusers
4. ffuser
⎛P
V 2 ⎞ dW n. f .
⎟=
∆ ⎜ + gz +
−F
⎜ρ
2⎟
dm
⎝
⎠ P2 − P1 V 2 2 − V 12
+
= −F
ρ
2
P2 − P1 = ρ V 12
2 V 2 = V1 A1
A2 ⎛
A2 ⎞
⎜ 1 − 1 ⎟ − ρF
2
⎜
A2 ⎟
⎝
⎠
11 U niv ersity of Ottawa, CHG 2312, P. Mehrani 5. Sudden expansions A2
A1 P2 − P1 = ρ V12
2 − ρF
Separation regions: eddies
University of Ottawa, CHG 2312, P. Mehrani 12 6. Torricelli’s equation
Tankdraining problem z 1
h In addition to assumptions for Bernoulli:
o V2 >> V1 (V1 ≈ 0),
o P1 ≈ P2 = Patm
o W=F=0
o steadystate flow, i.e. constant tank level g ( z 2 − z1 ) + 2 V22
= 0 & z 2 − z1 = − h
2 V2 = (2gh)1 2
University of Ottawa, CHG 2312, P. Mehrani 13 7. Fluid flow measurement
a) Pitot Tube P 2 − P 1 V 12
−
= −F
ρ
2
University of Ottawa, CHG 2312, P. Mehrani 14 7. Fluid flow measurement
B.E. Equation: P2 − P1 ρ V 12
−
= −F
2 Inside the tube the fluid is not moving: P 2 = P a t m + ρ g (h 1 + h 2 ) From fluid statics: P1 = P a t m + ρ g h 2 Velocity at point 1: V1 = (2 g h 1 + 2F ) 1 2 V1 ≈ (2 gh1 ) University of Ottawa, CHG 2312, P. Mehrani 1 2 15 7. Fluid flow measurement
b) PitotStatic Tube ⎛ 2∆P⎞
V1 = ⎜
⎟
⎝ρ⎠ 1
2 University of Ottawa, CHG 2312, P. Mehrani 16 7. Fluid flow measurement
Pitot vs. PitotStatic Tube V1 ≈ (2gh1 )
⎛ 2∆P⎞
V1 = ⎜
⎟
⎝ρ⎠ 1 2 1
2 17 University of Ottawa, CHG 2312, P. Mehrani 7. Fluid flow measurement
c) Venturi Meter
Venturi eter 2 − V12 P2 − P1 V2
+
ρ
2 =0 ⎡ 2( P − P ) ⎤
1
2
⎢
⎥
ρ
⎢
⎥
V2 =
2
⎢⎛
A ⎞⎥
⎢ ⎜1 − 2 ⎟ ⎥
2
⎜
A1 ⎟ ⎥
⎢⎝
⎠⎦
⎣ University of Ottawa, CHG 2312, P. Mehrani 1 2 18 7. Fluid flow measurement University of Ottawa, CHG 2312, P. Mehrani 19 7. Fluid flow measurement
Differences exist between calculated and actual!
(actual < calc.)
Introduce coefficient of discharge Cv ⎡ 2(P − P ) ⎤
1
2
⎢
⎥
ρ
⎢
⎥
V2 = CV
2
⎢⎛
A ⎞⎥
⎢ ⎜1 − 2 ⎟ ⎥
2
⎜
A1 ⎟ ⎥
⎢⎝
⎠⎦
⎣ 1 2 University of Ottawa, CHG 2312, P. Mehrani 20 7. Fluid flow measurement
Cv depends only on Reynolds number Re =
1.
2.
3.
4.
5. V1 D1 ρ µ Use V2 from equation with no Cv
Us
Compute Reynolds number
Read Cv from graph
Calculate new V2
Repeat steps 2, 3 and 4
until convergence achieved. maximum minimum
average 21 University of Ottawa, CHG 2312, P. Mehrani 7. Fluid flow measurement
maximum minimum Cv average Re =V1 D1 ρ/µ
University of Ottawa, CHG 2312, P. Mehrani 22 Fluid flow measurement
Orifice Meter ⎡ 2( P1 − P2 ) ρ ⎤
⎥
V2 = Cv ⎢
2
2
⎢ (1 − A2 A1 ) ⎥
⎣
⎦ 1 2 University of Ottawa, CHG 2312, P. Mehrani 20 7. Fluid flow measurement Cv Re =V2 D2 ρ/µ
University of Ottawa, CHG 2312, P. Mehrani 24 7. Fluid flow measurement
Orifice vs. Venturi University of Ottawa, CHG 2312, P. Mehrani 25 7. Fluid flow measurement
e) Rotameter
e)
o The float rises within the
tube to a location where
the float weight, drag
oat
ight, dr
force, and buoyancy
orce
forces balance each
orces
other. University of Ottawa, CHG 2312, P. Mehrani 26 7. Fluid flow measurement
Force balance:
Fgravity + F pressure from above = Fbuoyancy + F pressure from bottom 0= π
6 3
D o ρ ball g + P3 π
4 2
Do − π
6 3
Do ρ fluid g − P1 π
4 2
Do From B.E:
⎛V 2
⎛V 2 V 2 ⎞
P − P2 = ρ fluid ⎜ 2 − 1 ⎟ = ρ fluid ⎜ 2
1
⎜2
⎟
⎜2
2⎠
⎝
⎝
V2 = 2
⎞⎛ A2 ⎞
⎟
⎟⎜1 −
⎟⎜ A 2 ⎟
1⎠
⎠⎝ 4 Do g ρ ball − ρ fluid
•
ρ fluid
3 Q = A 2V 2
University of Ottawa, CHG 2312, P. Mehrani 27 7. Fluid flow measurement
e) Rotameter
e)
o The float rises within the
tube to a location where
the float weight, drag
oat
ight, dr
force, and buoyancy
orce
forces balance each
orces
other. University of Ottawa, CHG 2312, P. Mehrani 28 7. Fluid flow measurement
For gas flow rate measurements:
o From B.E: ⎛ ρball − ρ fluid
V∝ ⎜
⎜
ρ fluid
⎝
o ⎞
⎟
⎟
⎠ ⇒ Q1 ρ1 = Qo ρo Mass flow rate is constant through a system: Q1 ρ1 = Qo ρ o 29 University of Ottawa, CHG 2312, P. Mehrani 7. Fluid flow measurement
For finding the process flow rate from the rotameter
scale reading: Q 2 = Q0 T2
P2 P0 P1 M 0
T0T1 M 1 For setting the gas flow rate at the rotameter to
give a desired process gas flow rate:
Accessories Q0 = Q 2 P2
T2 T0T1 M 1
P0 P1 M 0 Process conditions:
Q2, T2, ρ2, M2
Rotameter conditions: Q1, T1, ρ1, M1 Reference (standard) conditions: Q0, T0, ρ0, M0
University of Ottawa, CHG 2312, P. Mehrani 30 8. Cavitation
(negative absolute pressures)
Bernoulli’s equation sometimes predicts negative absolute
pressures.
o In gas systems:
likely indicates very high gas velocities, thus Bernoulli’s
equation does not apply (see Chap. 8). o In liquid systems:
Occurs in very rare conditions.
When the absolute pressure on a liquid approaches the vapour
pressure of the liquid, liquid boils (twophase flow).
Therefore, the calculated flow is physically impossible. University of Ottawa, CHG 2312, P. Mehrani 31 8. Cavitation
(negative absolute pressures)
Examples: University of Ottawa, CHG 2312, P. Mehrani 32 8. Cavitation
(negative absolute pressures) University of Ottawa, CHG 2312, P. Mehrani 33 9. Unsteady flows
Bernoullli’s equation can be applied to some unsteady flows,
l
if the changes in flow rate are slow enough to be ignored.
Most problems in which unsteady flow cannot be neglected
involve suddenly stopping the flow with a valve or starting
the flow from rest.
These problems are more easily solved using the
momentum balance. University of Ottawa, CHG 2312, P. Mehrani 34 10. Nonuniform flows
Normally ignore difference in velocity perpendicular to flow.
Cannot ignore for shallow gravitydriven flows, e.g., weirs in
distillation columns, etc. Sieve plates Bubble caps University of Ottawa, CHG 2312, P. Mehrani 35 ...
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 Spring '11
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