Ideal Fluids in Motion
• Ideal Fluid:
Steady Flow, Incompressible Flow, Non viscous Flow, Irrotational Flow.
• The Equation of Continuity:
If a fluid is incompressible, its density ρ is constant throughout.
Thus the volume of fluid entering a tube at one end per unit of time
must be equal to the volume of fluid leaving the other end per unit
time. In the time Δt we have A1v1Δt = A2v2Δt.
ΔV = v2A2Δt
ΔV = v1A1Δt
A1v1 = A2v2 (continuity equation) RV = Av = volume flow rate = constant
Rm = ρRV = ρAv = mass flow rate = constant
• Bernoulli’s Equation:
Application of W = ΔKE + ΔU:
W = ( P1 A1 )Δx1 − ( P2 A2 )Δx2 = ( P1 − P2 )ΔV = ( P1 − P2 )
( P1 − P2 )
M
ρ
= 12 M (v22 − v12 ) + Mg ( y2 − y1 )
F1 = P1A1
P1 + 12 ρv12 + ρgy1 = P2 + 12 ρv22 + ρgy2
R. Field 10/29/2013
University of Florida
M
M
ρ
M
F2 = P2A2
Δx2 = v2Δt
Δx1 = v1Δt
PHY 2053
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Bernoulli’s Equation: Applications
• Bernoulli’s Equation: P + ½ρv2 + ρgy = constant
P1 + 12 ρv12 + ρgy1 = P2 + 12 ρv22 + ρgy2 (conservation of energy for a fluid)
v1 >> v2
• Constant Height (y1 = y2): P + ½ρv2 = constant
P1 + 12 ρv12 = P2 + 12 ρv22
P1 << P2
v1 P1
v2 P2 Air Foil
If the speed of a fluid element increases as the element travels
along a horizontal streamline, the pressure of the fluid must ΔP = P − P = 1 ρ (v 2 − v 2 )
2
1
1
2
2
decrease, and conversely.
• Example (velocity of efflux):
Area A1
y-axis
We can use Bernoulli’s equation to calculate the speed of
efflux, v2, from a horizontal orifice (and area A2) located a
depth h below the water level of a large talk (with area A1).
P1 + ρv + ρgy1 = P2 + ρv + ρgy2
1
2
2
1
P1 = P2 = Patm
v2 =
1
2
v1 = v2A2/A1
2
2
v1
(2)
v = v + 2 gh = ( A2 / A1 ) v + 2 gh
2
2
h
(1↔2)
2
1
2
2 gh
⎯⎯⎯→ 2 gh (Torricelli’s Law)
(1 − ( A2 / A1 ) 2 ) A2 << A1
R. Field 10/29/2013
University of Florida
(1)
PHY 2053
2
2
ρ
v2
y=0
Area A2
Page 2
Bernoulli’s Equation: Application
• Venturi Meter:
A Venturi meter is used to measure the flow of a fluid
in a pipe. The meter is constructed between two
sections of a pipe, the cross-sectional area A of the
entrance and exit of the meter matches the pipe’s
cross-sectional area. Between the entrance and exit,
the fluid (with density ρ) flows from the pipe with
speed V and then through a narrow “throat” of crosssectional area a with speed v. A manometer (with
fluid of density ρM) connects the wider portion of the
meter to the narrow portion. What is V in terms of ρ,
ρM, h, a, and A?
P1 + ρV = P2 + ρv
1
2
2
1
2
2
C
VA = va
(1↔2)
v = VA / a
⎛ A2 ⎞ 2
2 gh
V ⎜⎜ 2 − 1⎟⎟ = ( P1 − P2 ) =
(ρM − ρ )
ρ
⎝a
⎠ ρ
PC = P1 + ρg ( d + h) = P2 + ρgd + ρ M gh (C↔C)
2
P1 − P2 = ( ρ M − ρ ) gh
R. Field 10/29/2013
University of Florida
d
PHY 2053
⎛ρ
⎞
2 gh⎜⎜ M − 1⎟⎟
ρ
⎝
⎠
V=
2
A
⎛ ⎞
⎜ ⎟ −1
⎝a⎠
Page 3
Bernoulli’s Equation: Application
• Siphon:
The figure shows a siphon, which is a device for
removing liquid from a container. Tube ABC must
initially be filled, but once this is done, liquid will
flow until the liquid surface of the container is
level with the tube opening A. With what speed
does the liquid emerge from the tube at C? What
is the greatest possible height h1 that a siphon can
lift water?
Patm + ρV = PA + ρv − ρgd
1
2
2
2
1
2
VA = va
(S↔A)
V = va / A
⎛ a ⎞
2
1
Patm = PA + 12 ρv 2 ⎜⎜1 − 2 ⎟⎟ − ρgd ⎯a⎯
⎯
→
P
+
ρ
v
− ρgd
A
2
<< A
⎝ A ⎠
2
PA + 12 ρv 2 − ρgd = Patm + 12 ρv 2 − ρg (d + h2 )
1
2
ρv 2 = ρg (d + h2 )
A = Area of container
a = area of tube
S
y=0
V
(A↔C)
v
PB + 12 ρv 2 + ρgh1 = Patm + 12 ρv 2 − ρg (d + h2 )
v = 2 g (d + h2 )
( P − PB )
− (d + h2 )
h1 = atm
ρg
R. Field 10/29/2013
University of Florida
PHY 2053
(h1 ) max
(B↔A)
P
= atm − (d + h2 )
ρg
v
Page 4