1. Chapter 12: INCOMPRESSIBLE FLOW
Effect Of Area Change

2. One-Dimensional Compressible FlowRx P2 P1
dQ/dt
Surface force
from friction
and pressure
heat/cool
(+ s1, h1)
(+ s2, h2)
Shock, heating, cooling, area change and friction affect 1-D flow.
Could be T(s), p(s), ..
Look at each one separately though in real world more than one may occur simultaneously.
What can affect fluid properties?
Changing area, heating, cooling, friction, normal shock

3. isentropic Only change in area No heat transfer No shock No friction
No friction ~ irreversible
IF SHOCK THEN IRREVERSIBLE AND FRCTION (?) AND NOT ADIABATIC

4. Rx = pressure force along walls, no friction
A + dA
(= 1V1A1 = 2V2A2 ) A
1-D, steady, isentropic flow with area change; no friction,
no heat exchange or change in potential energy or entropy

5. = ho s2 = s1 = const adiabaticNo
Friction
adiabatic
= ho
s2 = s1 = const C A L O R I Y I D E A L G S
CONSTANT
Isentropic so Q = 0, Rx is only pressure forces – no friction

6. Q = 0 S = constant S = constant
Stagnation enthalpy is constant so sum of enthalpy and kinetic energy is constant if entropy does not change.
(also effects of gravity are negligible, no shaft work, shear work or any other work but pressure work
S = constant

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

8. Total energy* of flowing fluid:
Stagnation enthalpy is constant so sum of enthalpy and kinetic energy is constant if entropy does not change.
(also effects of gravity are negligible, no shaft work, shear work or any other work but pressure work *
Total energy* of flowing fluid:
h + ½ V2 = constant = ho

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

10. For isentropic flow if the fluid accelerates
what happens to the temperature?
if V2>V1
then h2<h1

11. 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

12. Isentropic so: s = const = so and h + V2/2 = const = ho
Follows that stagnation properties are constant
for all states in an isentropic flow.

13. If ho and so constant then po To and o are constant
since anyone thermodynamic variable can be
expressed as a function of any two other.

14. 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

15. 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

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

17. ~ cs VxV  dA = Vx{-VxA} + FSx = cs VxV  dA
Differential form of momentum equation for steady, inviscid, quasi-
one-dimensional flow (Euler’s Equation)
from section
11-3.1 ~
FSx = cs VxV  dA
no body forces, frictionless, steady
FSx = (p + dp/2)dA + pA – (p+dp)(A + dA)
cs VxV  dA = Vx{-VxA} +
(Vx + d Vx)[( + d )(Vx + d Vx)(A + dA)
dp/ + d{Vx2/2} = 0
If density is a constant get Bernoulli’s Eq. Which assumes isentropic flow
Euler’s equation was derived from differential form of the general momentum equation in three dimensions, holds along a streamline.

18. cs VxV  dA = Vx{-VxA} +
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 VxV  dA = Vx{-VxA} +
(Vx + d Vx)[( + d )(Vx + d Vx)(A + dA)
pdA + (dp/2)dA + pA – pA – pdA – dpA – dpdA = -VxVxA + (Vx + d Vx)[VxVxA]
-Adp =dVxVxVxA
dp/ + d{Vx2/2} = 0
~ 0
~ 0
If density is a constant get Bernoulli’s Eq. Which assumes isentropic flow
Euler’s equation was derived from differential form of the general momentum equation in three dimensions, holds along a streamline.

19. dp/ + d{Vx2/2} = 0* NOTE: Changes in pressure and velocity
always have opposite sign
*True along streamline in steady,
inviscid flow (no body forces)

20. Rx = pressure force along walls, no friction
p + ½ dp/dx Rx
½ dA
{p + ½ dp/dx}dA = Fx
1-D, steady, isentropic flow with area change; no friction,
no heat exchange or change in potential energy or entropy

21. EQ b
EQ.12.1a
{d(AV) + dA(V) +dV(A)}/{AV} = 0
steady
isentropic, steady

22. isentropic, steady

23. EQ. 12.5
EQ.12.6
isentropic, steady

24. Although cannot use these equations for computation
since don’t know how M varies with A,
still can provide interesting insight
into how pressure and velocity change with area.

25. isentropic, steady, ~1-D M<1 If M < 1 then [ 1 – M2] is +,
Subsonic Nozzle
Subsonic Diffuser
M<1
If M < 1 then 1-M2 is + and dA and dp are the same sign but dV and dA are opposite signs
If M > 1 then 1-M is - and dA and dp are opposite signs but dV and dA are the same
If gas is ideal then h(T) so if V increases, the temperature must decrease no matter is M is > or < 1.
If also incompressible then c = , M = 0; and d = EQ.12.2a dV/V = -dA/A
If M < 1 then [ 1 – M2] is +,
then dA and dP are same sign;
and dA and dV are opposite sign
qualitatively like incompressible flow

26. isentropic, steady, ~1-D M>1 If M > 1 then [ 1 – M2] is -,
Supersonic Nozzle
Supersonic Diffuser
M>1
If M < 1 then 1-M2 is + and dA and dp are the same sign but dV and dA are opposite signs
If M > 1 then 1-M is - and dA and dp are opposite signs but dV and dA are the same
If gas is ideal then h(T) so if V increases, the temperature must decrease no matter is M is > or < 1.
If also incompressible then c = , M = 0; and d = EQ.12.2a dV/V = -dA/A
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

27. Both dV and dA can be same sign because d can be opposite sign
Note on counterintuitive supersonic results:
If M < 1 then 1-M2 is + and dA and dp are the same sign but dV and dA are opposite signs
If M > 1 then 1-M is - and dA and dp are opposite signs but dV and dA are the same
If gas is ideal then h(T) so if V increases, the temperature must decrease no matter is M is > or < 1.
If also incompressible then c = , M = 0; and d = EQ.12.2a dV/V = -dA/A
Both dV and dA can be same sign
because d can be opposite sign

28. e.g. in a supersonic nozzle both dV and dA can be
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.
If M < 1 then 1-M2 is + and dA and dp are the same sign but dV and dA are opposite signs
If M > 1 then 1-M is - and dA and dp are opposite signs but dV and dA are the same
If gas is ideal then h(T) so if V increases, the temperature must decrease no matter is M is > or < 1.
If also incompressible then c = , M = 0; and d = EQ.12.2a dV/V = -dA/A

29. Flow is
continuously accelerating, dV is always positive
So when M<1, dA must be negative so –dA is positive
So when M>1, dA must be positive so –dA is negative

30. Flow is
continuously decelerating, dV is always negative
So when M>1, dA must be negative so -dA is positive
So when M<1, dA must be positive so –dA is negative

31. Same shape, but in one case
accelerating flow, and in the
other decelerating flow

32. If M = 1 then I have a problem,
isentropic, steady ~ 1-D
If M = 1 then I have a problem,
Eqs 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!!!
If M < 1 then 1-M2 is + and dA and dp are the same sign but dV and dA are opposite signs
If M > 1 then 1-M is - and dA and dp are opposite signs but dV and dA are the same
If gas is ideal then h(T) so if V increases, the temperature must decrease no matter is M is > or < 1.
If also incompressible then c = , M = 0; and d = EQ.12.2a dV/V = -dA/A

33. Note: if incompressible, c =  & M = 0 and Eq 12.6 becomes:
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).
If M < 1 then 1-M2 is + and dA and dp are the same sign but dV and dA are opposite signs
If M > 1 then 1-M is - and dA and dp are opposite signs but dV and dA are the same
If gas is ideal then h(T) so if V increases, the temperature must decrease no matter is M is > or < 1.
If also incompressible then c = , M = 0; and d = EQ.12.2a dV/V = -dA/A

34. Same shape, but in one case
accelerating flow, and in the
other decelerating flow
dA = constant

35. Strategy: if know M1 & p1, can calculate p01;
Computations are tedious, but because s = 0
use reference stagnation and critical conditions
po/p =
[1 + (k-1)M2/2]k/(k-1)
To/T =
1 + (k-1)M2/2
o/ =
[1 + (k-1)M2/2]1/(k-1)
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

36. Need two reference states because the reference stagnation
State does not provide area information
(mathematically the stagnation area is infinite.)

37. Why is T* (critical temperature) less than D E
Why is T* (critical temperature) less than
To (stagnation temperature)?

38. Why is T* (critical temperature) less than
To (stagnation temperature)? A S I D E
Isentropic so ~ h0 = h* + V*2/2
ho = cpTo and h* = cpT* (ideal and constant cp)
cpTo = cpT* + c2/2
To = T* + c2/(2cp)

39. (11.20) To /T = 1 + M2(k-1)/2 so To = T* (1 +(k-1)/2) A S I D E
(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*

40. Back to problem, want to come up with
easier way of manipulating these equations.
Strategy: use isentropic reference conditions

41. stagnation properties)
EQ b +
EQ c
p / k = constant
Eq b: differential formof momentum equation for steady, inviscid, quasi-one-dimensional flow
Eq c: 1st and 2nd laws, reversible process, ideal gas, constant specific heats
Eqs a,b,c = Eqs. 12.7a,b,c
(po, To refer to
stagnation properties)
isentropic, steady, ideal gas

42. If M = 1 the critical state; p*, T*, *….
Critical conditions
related to stagnation
Local conditions
related to stagnation
EQ c = [kRT]1/2
c* = [kRT*]1/2
isentropic, steady, ideal gas

43. Can use stagnation conditions to go from
1, p1, T1, c1 to
2, p2, T2, c2; but not A.
Missing relation for area
since stagnation state does
not provide area information.
So to get area information
use critical conditions as
reference.

44. EQ c = [kRT]1/2
I used stagnation condition oAoVo = 0 so not much help.

45. EQ. 12.7c
EQ b
EQ. 12.7b
EQ c

46. 1/(k-1) + ½ = 2/2(k-1) +(k-1)/2(k-1)
AxAy = Ax+y
1/(k-1) + ½ = 2/2(k-1) +(k-1)/2(k-1)
= (k+1)/(2(k-1)

47. isentropic, ideal gas, steady,
EQs. 12.7a,b,c,d
Provide property relations in
terms of local Mach numbers,
critical conditions, and
stagnation conditions.
NOT COUPLED LIKE
Eqs. 12.2, so easier to use.
isentropic, ideal gas, steady,
only body forces

48. k=cp/cv=1.4
Looks like converging-diverging section for accelerating subsonic flow to supersonic – but not. For example diverging section too large an angle, may result in separation. Then no longer isentropic.

49. In practice this is not shape of wind tunnel. To reduce chance
of separation,
divergence
angle must be
less severe.
accelerating

50. 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.
accelerating

51. Two values of M# for one value of A/A* One value of A/A* for each
value of M
Accelerating
Decelerating

52. 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

53. p2 = po2/(1 + {[k-1]/2}M22)k/(k-1) po1 = 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}M22)
To1 = To2
p2 = po2/(1 + {[k-1]/2}M22)k/(k-1)
po1 = po2

54. T2 = To2/(1 + {[k-1]/2}M22) To1 = To2 p2 = po2/(1 + {[k-1]/2}M22)
Know:
Ma1=0.5, p1=250 kPa,
T1=300oC, Ma2=2.6
T2 = To2/(1 + {[k-1]/2}M22)
To1 = To2
p2 = po2/(1 + {[k-1]/2}M22)
po1 = po2
Then find:
Po2, p2, T2
po1 = po2 = p1(1 + ((k-1)/2)Ma11)k/(k-1) = 250[ (0.5)2]3.5
~ 297 kPa
p2 = p02/(1 + ((k-1)/2)Ma21)k/(k-1) = 297/[ (2.6)2]3.5
~14.9 kPa
To1 = To2 = T1(1 + {[k-1]/2}M12) = 573[1 +0.2((0.5)2]
~ 602oK
T2 = T02/(1 + {[k-1]/2}M22) = 602/[1 +0.2((2.6)2]
~ 256oK

55. Class 13 - THE END

56. FLASHBACK (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)
CONSTANT

57. (p2/2k) = (p1/1k) = constant
If isentropic, s2 – s1 = 0
CONSTANT
(p2/2k) = (p1/1k) = constant

58. So s = ln{(T2/T1)Cp/(p2/p1)R} Cp/Cv = k; R = Cp –Cv; p = RT
FLASHBACK
EQ. 12.1g
So s = ln{(T2/T1)Cp/(p2/p1)R}
Cp/Cv = k; R = Cp –Cv; p = RT
s = 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/1k) = const

59. s = (Cv)ln{(T2/T1)k / (p2/p1)k-1} If isentropic s = 0
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

60. (T2)k/(p2)k-1 = (T1)k/(p1)k-1 = constant (T2)k(p2)1-k = (T1)k(p1)1-k
FLASHBACK
(T2)k/(p2)k-1 = (T1)k/(p1)k-1
= constant
(T2)k(p2)1-k = (T1)k(p1)1-k
{(T2)k(p2)1-k = (T1)k(p1)1-k
= constant}1/k
(T2)(p2)(1-k)/k = (T1)(p1)(1-k)/k
= constant’

61. (T2)(p2)(1-k)/k = (T1)(p1)(1-k)/k = const
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) = (p11/k/1) = constant
(p2/2k) = (p1/1k) = const. QED