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Chapter 12: INCOMPRESSIBLE FLOW

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1 Chapter 12: INCOMPRESSIBLE FLOW
Effect Of Area Change

2 One-Dimensional Compressible Flow
Rx 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 adiabatic
No 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


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