Marymount University Fluid Kinematics & Control Volumes Worksheet I have homework 3 questions the third question need from you to copy from lectures I alre

Marymount University Fluid Kinematics & Control Volumes Worksheet I have homework 3 questions the third question need from you to copy from lectures I already upload lectures 5 and 4requirement blue pen and do it handwriting for all the question when you finish scan them and put them in one pdf file and send it to me thanks Chap. 4 Fluid Kinematics & Control
volumes
Bernoulli’s equation
1 2
?v + p + ?z = const.
2
Navier-Stokes Equation
?
( ?v) + (v ? ?)( ?v) + ( ?v)? ? v = ??p + ( µ + ? )?(? ? v) + µ? 2 v + ?g
?t
!
General equation of motion for a fluid
?a = ??p ? ?kˆ
Chap. 4. Kinematics
Chap. 4. Kinematics
Chap. 4. Kinematics
Remarks : generalizations of Bernoulli’s equation
(a) Compressible flow
?
dp
1 2
+ v + gz = const.
? 2
M
? = ? ( p) e.g. ideal gas ? = p
R0T
(b) Unsteady flow
– extra term because as contains ? t vs
=> generalized eqn.
(c) Viscous flow
– NO generalization !!
?vs
1 2
p + ?vs + ?z + ? ?
ds = const.
2
?t
Chap. 4.2 Streamline coordinate system
!
Streamlines : lines everywhere tangent to v
– normal lines ( ? to streamlines)
describe fluid motion along the s and n axis
Analogously,
! D !
a=
v = as sˆ + an nˆ
Dt
and also write
!
v = vs sˆ + vn nˆ
(moves in s direction)
Chap. 4. Kinematics
• Streamlines in 2-D flows
?u ?
! ? ?
v = ?v?
?0?
? ?
Slope=
and
u = u ( x, y )
v = v ( x, y )
dy v
=
dx u
dy v( x, y )
=
dx u ( x, y )
Integrate
=>
y (x)
Chap. 4. Kinematics
e.g.
! v0 ? xy ?
v = ?? 2 ??
A0 ? ? y ?
=>
dy ? y 2
y
=
=?
dx
xy
x
const.
1
=> ln y = ? ln x + c = ln + c
x
1 c
c
exp.
y = ?e
=> xy = e = ? = const.
x
i.e. Streamlines :
?
y=
x
?
= const.
dy
dx
? y = ?? x
Chap. 4 Kinematics
y
directionality: from
? =? 2
if
? =?1
!
v
v0 , A0 > 0
note : x, y axis are also streamlines
? =? 3
x
Streamlines do not uniquely determine velocity field
e.g.
! v0 ? x ?
v = ?? ??
L0 ? ? y ?
=>
!
v
dy
y
=?
dx
x
(Streamlines are paths of fluid elements, no direct information on speed on path)
Chap. 4.1.1 Lagrangian vs. Eulerian
description
!
x0 (1)
!
x0 (2)
!
x0 (3)
!
x1 (1)
!
x2 (1) ! (1)
v (t1 )
!
!
x3 (1) v ( 2) (t )
1
positions at time t1
positions at time t0
(1) “label” elements by initial position :
! (1) ! ! (1)
v = v ( x0 , t1 ) Lagangian description
Simple to take time derivatives, but have to keep track of labels.
(2) Explicity dependent on instantaneous position :
!
! !
v (1) = v ( x1(1) , t1 ) Eulerian description
Simple : only need local information
Difficulty : position
depends on time => material derivation
Chap. 4.1.4 Flow lines
Visualizations of flows :
(a) Streamlines : Tangent lines to
in experiment
! !
v ( x, t )
not usually directly visualized
az
(b) Pathlines : Lagrangian viewpoint, paths of (labeled) fluid elements.
– label fluid elements by dye (fluorescent, …)
– passive tracers in the flow (beads, bubbles, …)
analog : multiple-exposure photographs -> “connect the dots”
(c) Streaklines : label fluid elements as they pass a common location
Analog : single exposure photograph of multiple labeled elements
-> “connect the dots”
In STEADY flows, (a), (b), (c) are IDENTICAL
Example
Example
Chap. 4.3-4 Transport
– Material derivative
! D ! ?v ! ! !
a=
v=
+ (v ? ? )v
Dt
?t
convective
local
time derivative
Other quantities change like this as well,
e.g temperature : all intensive quantities ( have finite value as ?V ? > 0 )
intensive
? (density)
T (temperature)
e (energy per mass)
Temporal change of e.g
extensive
m = ? ?dV
Q = ? c?TdV
(mass)
(heat)
E = ? ?edV (energy)
! !
D?
?=
= ? t + (v ? ? ) ?
Dt
(
)
Chap. 4.2 Material Derivative
Can also happen in y-direction
Analogous for
a y az
a4 x = ?
?u
?u
? = ?y
?
?u
?u
?u
?u ?
? a x = ?t + u ?x + v ?y + w ?z ?
?
?
? a = ?v + u ?v + v ?v + w ?v ?
? y ?t
?x
?y
?z ?
?
?w
?w
?w
?w ?
?a z =
+u
+v
+w ?
?t
?x
?y
?z ??
??
Chap. 4.2 Material Derivative
Shorter in vector notation
! ?v ! ! !
! ! ! ! D !
a=
+ (v ? ? )v = ? t v + (v ? ? )v =
v
?t
Dt
! !
D ?
Material derivative :
= + (v ? ? )
Dt ?t
? u ? ? ? / ?x ?
?
! ! ? ? ?
?
?
?
v ? ? = ? v ? ? ? ? / ?y ? = u + v + w
?x
?y
?z
? w ? ? ? / ?z ?
? ? ?
?
Newton’s 2nd
!
!
!
!
! !
!
Dv
?a = ? (? t v + (v ? ?)v ) = ?
= ??p ? ?kˆ
Dt
Euler’s equation
Chap. 4.2 Material Derivative
• Temporal change of extensive quantities
Dm
e.g. What s
?
Dt
? D?
?
Dm D ?
? ? ?dV ? =
? ? ?dxdydz ?
=
? Dt ?
?
Dt Dt ?? V
?
?V
?
If volume V is fixed (time-independent)
Dm
D
= ? (?dxdydz )
=>
Dt V Dt
=?
V
D?
D
dxdydz + ? ?
(dxdydz )
Dt
Dt
V
D ?
?
?
?
= +u +v + w
Dt ?t
?x
?y
?z
Chap. 4.2 Material Derivative
D?
D
=?
dxdydz + ? ?
(dxdydz )
Dt
Dt
V
V
?D ?
?D ?
?D ?
? dx ?dydz + dx? dy ?dz + dxdy? dz ?
? Dt ?
? Dt ?
? Dt ?
= udydz + vdxdz + wdxdy
? ?u ?v ?w ?
= ?? + + ??dxdydz
? ?x ?y ?z ?
D?
? ?u ?v ?w ?
=?
dxdydz + ? ? ?? + + ??dxdydz
Dt
?y ?z ?
V
V ? ?x
! !
= divergence ? ? v
Chap. 4.2 Material Derivative
! ! ?
Dm
? D?
= ??
+ ? (? ? v ) ?dV
Dt V ? Dt
?
! ! ?
? ?? ! !
= ? ? + (v ? ?) ? + ? (? ? v ) ?dV
?t
?
V?
!
!
? ? ( ?v )
!
!
Dm ?
= ? ?dV + ? ? ? ( ?v )dV
Dt ?t V
V
Chap. 4.2 Material Derivative
• Gauss’ theorem
!
! !
? ? ? ? dV = ? ? ? nˆdA
V
n? b? = B per volume
Example
Example
4.21 Determine the acceleration field for a three-dimensional
flow with velocity components u = ? x, v = 4 x 2 y 2 , and _ w = x ? y
Chap. 4.2 Control volumes
• Control volume
Volume of interest for measurements typically not the entire fluid
volume could be moving / time-dependent
One possible control volume (CV) or fixed CV
If(a portion of) the surface of a control volume is
moving, need to modify the transport theorem
Chap. 4.2 Control volumes
Often written as
DBsys
Dt
=
Time change of B in
system
(physical fluid molecules)
?
!
BCV + ? ?bw ? nˆdA
?t
CS
Explicit
change in CV
=> Like a fixed CV, but with fluid velocity relative to the moving CV :
! ! !
w = v ? vCV
relative velocity
!
w
and this is used in transport theorem
DB ?
!
= B + ? ?bw ? nˆdA
Dt ?t
CS
control surface
Chap. 4.2 Control volumes
heat system gains by moving through A2
heat system gains by moving through A1
v1, 2 uniform across A1, 2
Chap. 4.2 Control volumes
Example with B=m
b=1
v1 = v p
n?
? = const.
vp
A1
CV
A2
v2
n?2
n?
Dmsys
!
?
= mCV + ? ?1w ? nˆdA (uniform velocities)
Dt
?t
CS
?
= mCV + A1 ? ? (v1 ? v p )(?1) + A2 ? ? (v2 ? 0)(+1)
?t
m! out
0
Dmsys
Here (B=m)
= 0 (mass conservation)
Dt
?
mCV = ? A2 ?v2
=>
?t
Chap. 4.2 Control volumes
dBsys
dBCV
Discuss the differences between
and
dt
dt
if B represents mass
?
?
d ? ? ?dV ?
?
?
dBsys dmsys
sys
?
=
= ?
dt
dt
dt
?
?
?
d ? ? ?dV ??
dBCV dmCV
?
=
= ? CV
dt
dt
dt
Chap. 4.2 Control volumes
If mass is to be conserved (one of the basic laws governing
fluid motion), mass of the fluid in the system is constant
?
?
d ? ? ?dV ?
?
?
dBsys dmsys
sys
? =0
=
= ?
dt
dt
dt
Some of the fluid has left the control
volume through the nozzle. The amount of
mass within the tank decreases with time
?
?
d ?? ? ?dV ??
dBCV dmCV
?
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