chapter 12: incompressible flow
DESCRIPTION
Chapter 12: INCOMPRESSIBLE FLOW. Effect Of Area Change. One-Dimensional Compressible Flow. R x. P 2. P 1. dQ/dt. Surface force from friction and pressure. heat/cool. (+ s 1 , h 1 ). (+ s 2 , h 2 ). What can affect fluid properties? - PowerPoint PPT PresentationTRANSCRIPT
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Chapter 12: INCOMPRESSIBLE FLOW
Effect Of Area Change
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What can affect fluid properties? Changing area, heating, cooling, friction, normal shock
One-Dimensional Compressible Flow
dQ/dt
heat/cool
Rx
Surface forcefrom frictionand pressure
P1P2
(+ s1, h1) (+ s2, h2)
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Only change in area
No heat transferNo shock No friction
isentropic
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A
A + dA
1-D, steady, isentropic flow with area change; no friction, no heat exchange or change in potential energy or entropy
Rx
Rx = pressure force along walls, no friction
(= 1V1A1 = 2V2A2 )
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0
0
s2 = s1 = const
Isentropic so Q = 0, Rx is only pressure forces – no friction
IDEALGAS
0
CALORICALLY
CONSTANT
adiabatic = ho
NoFriction
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Q = 0
S = constant
S = constant
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h = cp T, so processes plotted on Ts diagram will look similar on hs diagram
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Total energy* of flowing fluid:h + ½ V2 = constant = ho
*
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For isentropic flow if the fluid accelerates what happens to the temperature?
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For isentropic flow if the fluid accelerates what happens to the temperature?
if V2>V1
then h2<h1
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For isentropic flow if the fluid accelerates what happens to the temperature?
if V2>V1
then h2<h1
if h2<h1
then T2<T1
Temperature Decreases
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Isentropic so: s = const = so and h + V2/2 = const = ho
Follows that stagnation properties are constant for all states in an isentropic flow.
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If ho and so constant then po To and o are constantsince anyone thermodynamic variable can be
expressed as a function of any two other.
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Seven nonlinear coupled equations, if isentropic and know for example: A1, p1,T1, h1, u1, 1 and A2
then could solve for p2, T2, h2, u2, 2, s2 and Rx
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Hard to solve (nonlinear & coupled) so will develop property relations in terms of local stagnation conditions and critical
conditions. But first lets look at general relationships e.g. how does V & p change with area
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Question? As dA varies, what happens to dV and dp?
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FSx = cs VxV dA no body forces, frictionless, steady FSx = (p + dp/2)dA + pA – (p+dp)(A + dA)
cs VxV dA = Vx{-VxA} + (Vx + d Vx)[( + d )(Vx + d Vx)(A + dA)
dp/ + d{Vx2/2} = 0
from section 11-3.1 ~
Differential form of momentum equation for steady, inviscid, quasi-one-dimensional flow (Euler’s Equation)
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FSx = (p + dp/2)dA + pA – (p+dp)(A + dA)
cs VxV dA = Vx{-VxA} + (Vx + d Vx)[( + d )(Vx + d Vx)(A + dA)
pdA + (dp/2)dA + pA – pA – pdA – dpA – dpdA = -VxVxA + (Vx + d Vx)[VxVxA]
-Adp =dVxVxVxA
dp/ + d{Vx2/2} = 0
Differential form of momentum equation for steady, inviscid, quasi-one-dimensional flow (Euler’s Equation)
~ 0 ~ 0
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dp/ + d{Vx2/2} = 0*
*True along streamline in steady, inviscid flow (no body forces)
NOTE: Changes in pressure and velocity
always have opposite sign
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1-D, steady, isentropic flow with area change; no friction, no heat exchange or change in potential energy or entropy
Rx
Rx = pressure force along walls, no frictionp + ½ dp/dx
½ dA
{p + ½ dp/dx}dA = Fx
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EQ. 11.19b EQ.12.1a
isentropic, steady
{d(AV) + dA(V) +dV(A)}/{AV} = 0
steady
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isentropic, steady
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EQ. 12.5
EQ.12.6isentropic, steady
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Although cannot use these equations for computation since don’t know how M varies with A,
still can provide interesting insightinto how pressure and velocity change with area.
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isentropic, steady, ~1-D
M<1
If M < 1 then [ 1 – M2] is +, then dA and dP are same sign;
and dA and dV are opposite signqualitatively like incompressible flow
Subsonic Nozzle Subsonic Diffuser
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isentropic, steady, ~1-D
M>1
If M > 1 then [ 1 – M2] is -, then dA and dP are opposite sign;and dA and dV are the same sign
qualitatively not like incompressible flow
Supersonic Nozzle Supersonic Diffuser
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Note on counterintuitive supersonic results:
Both dV and dA can be same sign because d can be opposite sign
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Note on counterintuitive supersonic results:
e.g. in a supersonic nozzle both dV and dA can be
same sign because d is the opposite sign and large.
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Flow iscontinuously accelerating, dV is always positive
So when M<1, dA must be negative so –dA is positiveSo when M>1, dA must be positive so –dA is negative
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Flow iscontinuously decelerating, dV is always negative
So when M>1, dA must be negative so -dA is positiveSo when M<1, dA must be positive so –dA is negative
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Same shape, but in one caseaccelerating flow, and in the other decelerating flow
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isentropic, steady ~ 1-D
If M = 1 then I have a problem, Eqs. 12.5 and 12.6 blow up!
Only if dA 0 as M 1 can avoid singularity.
Hence for isentropic flows sonic conditions can only occur where the area is constant!!!
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isentropic, steady~ 1-D
Note: if incompressible, c = & M = 0
and Eq 12.6 becomes: AdV + VdA = 0
dV and dA have opposite signs or d(AV) = 0 or AV = constant
(continuity equation for incompressible flow).
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Same shape, but in one caseaccelerating flow, and in the other decelerating flow
dA = constant
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Strategy: if know M1 & p1, can calculate p01;But p01 = p02, so if know M2 can calculate p2.
Need to know how M changes with A
Computations are tedious, but because s = 0use reference stagnation and critical conditions
po/p = [1 + (k-1)M2/2]k/(k-1)
To/T = 1 + (k-1)M2/2o/ = [1 + (k-1)M2/2]1/(k-1)
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Need two reference states because the reference stagnation State does not provide area information
(mathematically the stagnation area is infinite.)
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Why is T* (critical temperature) less thanTo (stagnation temperature)?
ASIDE
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Why is T* (critical temperature) less thanTo (stagnation temperature)?
Isentropic so ~ h0 = h* + V*2/2 ho = cpTo and h* = cpT* (ideal and constant cp)
cpTo = cpT* + c2/2 To = T* + c2/(2cp)
ASIDE
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(11.20) To /T = 1 + M2(k-1)/2 so To = T* (1 +(k-1)/2)k = 1.4 for ideal gas so To > T*
ASIDE
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Back to problem, want to come up with easier way of manipulating these equations.
Strategy: use isentropic reference conditions
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isentropic, steady, ideal gas
EQ. 11.19b
+
p / k = constant
EQ. 11.12c
Eqs. 11.20a,b,c = Eqs. 12.7a,b,c(po, To refer to
stagnation properties)
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If M = 1 the critical state; p*, T*, *….
EQ. 11.17 c = [kRT]1/2 c* = [kRT*]1/2
isentropic, steady, ideal gas
Local conditions related to stagnation
Critical conditions related to stagnation
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Missing relation for area since stagnation state does not provide area information. So to get area informationuse critical conditions as reference.
Can use stagnation conditions to go from1, p1, T1, c1 to
2, p2, T2, c2; but not A.
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EQ.11.17 c = [kRT]1/2
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EQ. 12.7b
EQ. 11.21b
EQ. 12.7c
EQ. 11.21c
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AxAy = Ax+y
1/(k-1) + ½ = 2/2(k-1) +(k-1)/2(k-1)= (k+1)/(2(k-1)
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isentropic, ideal gas, steady,only body forces
EQs. 12.7a,b,c,d
Provide property relations in terms of local Mach numbers,critical conditions, andstagnation conditions.
NOT COUPLED LIKEEqs. 12.2, so easier to use.
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Isentropic Flow of Ideal Gas
0
1
2
3
4
5
0 0.5 1 1.5 2 2.5 3 3.5
Mach Number (M)
Area
Rati
o A/A
*k=cp/cv=1.4
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Isentropic Flow of Ideal Gas
0
1
2
3
4
5
0 0.5 1 1.5 2 2.5 3 3.5
Mach Number (M)
Area
Rati
o A/A
*
accelerating
In practice this is not shape of windtunnel. To reduce chance of separation,divergence angle must be less severe.
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Isentropic Flow of Ideal Gas
0
1
2
3
4
5
0 0.5 1 1.5 2 2.5 3 3.5
Mach Number (M)
Are
a R
ati
o A
/A*
accelerating
• For accelerating flows, favorable pressure gradient, the idealization of isentropic flow is generally a realistic model of the actual flow behavior.• For decelerating flows (unfavorable pressure gradient) real fluid tend to exhibit nonisentropic behavior such as boundary layer separation, and formation of shock waves.
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Isentropic Flow of Ideal Gas
0
1
2
3
4
5
0 0.5 1 1.5 2 2.5 3 3.5
Mach Number (M)
Area
Rati
o A/A
*
Accelerating Decelerating
Two values ofM# for one value of A/A*
One value of A/A* for each value of M
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Example ~Air flows isentropically in a duct. At section 1: Ma1 = 0.5, p1 = 250kPa,
T1 = 300oC At section2: Ma2= 2.6
Then find: T2, p2 and po2
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Example ~Air flows isentropically in a duct. At section 1: Ma1 = 0.5, p1 = 250kPa, T1 = 300oC At section2: Ma2= 2.6 Then find: T2, p2 and po2
Equations: T2 = To2/(1 + {[k-1]/2}M2
2)To1 = To2
p2 = po2/(1 + {[k-1]/2}M22)k/(k-1)
po1 = po2
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T2 = To2/(1 + {[k-1]/2}M22)
To1 = To2
p2 = po2/(1 + {[k-1]/2}M22)
po1 = po2
po1 = po2 = p1(1 + ((k-1)/2)Ma11)k/(k-1) = 250[1 + 0.2(0.5)2]3.5
~ 297 kPap2 = p02/(1 + ((k-1)/2)Ma2
1)k/(k-1) = 297/[1 + 0.2(2.6)2]3.5 ~14.9 kPaTo1 = To2
= T1(1 + {[k-1]/2}M12) = 573[1 +0.2((0.5)2]
~ 602oKT2
= T02/(1 + {[k-1]/2}M22) = 602/[1 +0.2((2.6)2]
~ 256oK
Then find:Po2, p2, T2
Know:Ma1=0.5, p1=250 kPa, T1=300oC, Ma2=2.6
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Class 13 - THE END
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CONSTANT
(from 1st and 2nd Laws got Tds = du + pdv)(define h = u + pv; Tds = dh – pdv –vdp + pdv)
(ideal gas so h = cpT; calorically perfect so dh = cpdT)(ideal gas so v/T = R/p)
ds = cpdT/T – Rdp/p(calorically perfect)
FLASHBACK
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CONSTANT
(p2/2k) = (p1/1
k) = constant
If isentropic, s2 – s1 = 0
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EQ. 12.1g
So s = ln{(T2/T1)Cp/(p2/p1)R}Cp/Cv = k; R = Cp –Cv; p = RTs = ln{(T2/T1)Cp / (p2/p1)Cp-Cv}s = (Cv)ln{(T2/T1)Cp / (p2/p1)Cp-Cv}1/Cv
s = (Cv)ln{(T2/T1)k / (p2/p1)k-1}
To show for s = 0: (p2/2k) = (p1/1
k) = const
FLASHBACK
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s = (Cv)ln{(T2/T1)k / (p2/p1)k-1}
If isentropic s = 0
So 0 = ln{(T2/T1)k / (p2/p1)k-1} (T2/T1)k/(p2/p1)k-1 = 1 (T2/T1)k = (p2/p1)k-1
(T2)k/(p2)k-1 = (T1)k/(p1)k-1 = constant
FLASHBACK
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(T2)k/(p2)k-1 = (T1)k/(p1)k-1 = constant
(T2)k(p2)1-k = (T1)k(p1)1-k = constant
{(T2)k(p2)1-k = (T1)k(p1)1-k = constant}1/k
(T2)(p2)(1-k)/k = (T1)(p1)(1-k)/k = constant’
FLASHBACK
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(T2)(p2)(1-k)/k = (T1)(p1)(1-k)/k = const
p = RT; T = p/(R)
(p2/2R)(p2)(1-k)/k = (p1/1R)(p1)(1-k)/k = const
(p21/k/2) = (p1
1/k/1) = constant
(p2/2k) = (p1/1
k) = const. QED
FLASHBACK