Chapter 12: INCOMPRESSIBLE FLOW

Effect Of Area Change

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)

Only change in area

No heat transferNo shock No friction

isentropic

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 )

0

0

s2 = s1 = const

Isentropic so Q = 0, Rx is only pressure forces – no friction

IDEALGAS

0

CALORICALLY

CONSTANT

adiabatic = ho

NoFriction

Q = 0

S = constant

S = constant

h = cp T, so processes plotted on Ts diagram will look similar on hs diagram

Total energy* of flowing fluid:h + ½ V2 = constant = ho

*

For isentropic flow if the fluid accelerates what happens to the temperature?

For isentropic flow if the fluid accelerates what happens to the temperature?

if V2>V1

then h2<h1

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

Isentropic so: s = const = so and h + V2/2 = const = ho

Follows that stagnation properties are constant for all states in an isentropic flow.

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.

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

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

Question? As dA varies, what happens to dV and dp?

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)

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

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

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

EQ. 11.19b EQ.12.1a

isentropic, steady

{d(AV) + dA(V) +dV(A)}/{AV} = 0

steady

isentropic, steady

EQ. 12.5

EQ.12.6isentropic, steady

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.

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

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

Note on counterintuitive supersonic results:

Both dV and dA can be same sign because d can be opposite sign

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.

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

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

Same shape, but in one caseaccelerating flow, and in the other decelerating flow

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!!!

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).

Same shape, but in one caseaccelerating flow, and in the other decelerating flow

dA = constant

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)

Need two reference states because the reference stagnation State does not provide area information

(mathematically the stagnation area is infinite.)

Why is T* (critical temperature) less thanTo (stagnation temperature)?

ASIDE

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

(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

Back to problem, want to come up with easier way of manipulating these equations.

Strategy: use isentropic reference conditions

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)

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

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.

EQ.11.17 c = [kRT]1/2

EQ. 12.7b

EQ. 11.21b

EQ. 12.7c

EQ. 11.21c

AxAy = Ax+y

1/(k-1) + ½ = 2/2(k-1) +(k-1)/2(k-1)= (k+1)/(2(k-1)

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.

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

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.

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.

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

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

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

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

Class 13 - THE END

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

CONSTANT

(p2/2k) = (p1/1

k) = constant

If isentropic, s2 – s1 = 0

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

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

(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

(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