Gas Dynamics-rayleigh Flow

RAYLEIGH FLOW Constant area Duct flow with Heat Transfer and negligible friction is called Rayleigh flow (Simple diabatic flow) •Many compressible flow problems encountered in practice involve chemical reactions such as combustion, nuclear reactions, evaporation, and condensation as well as heat gain or heat loss through the duct wall •Such problems are difficult to analyze •Essential features of such complex flows can be captured by a simple analysis method where generation/absorption is modeled as heat transfer through the wall at the same rate •Still too complicated for introductory treatment since flow may involve friction, geometry changes, 3D effects •We will focus on 1D flow in a duct of constant cross-sectional area with negligible frictional effects GDJP

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RAYLEIGH FLOW

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RAYLEIGH FLOW- Assumptions To isolate the effects heat transfer we make the following assumptions:

1. The area of the flow passage or duct is constant. 2. The flow is steady and one-dimensional. 3. There is no work, body forces are negligible, and the effects of friction are negligible. 4. Heat transfer is the only driving potential.

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RAYLEIGH FLOW – Fundamental Equations

Continuity Equation

X-Momentum equation

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RAYLEIGH FLOW – Fundamental Equations Energy equation CV involves no shear, shaft, or other forms of work, and potential energy change is negligible.

For and ideal gas with constant cp, ∆h = cp∆T

Entropy change In absence of irreversibility's such as friction, entropy changes by heat transfer only

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RAYLEIGH LINE •Infinite number of downstream states 2 for a given upstream state 1 •Practical approach is to assume various values for T2, and calculate all other properties as well as q. •Plot results on T-s diagram Called a Rayleigh line

•This line is the locus of all physically attainable downstream states •S increases with heat gain to point a which is the point of maximum entropy (Ma =1) GDJP

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RAYLEIGH LINE

Various flow parameters on Rayleigh Line

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Variation of fluid Properties

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Rayleigh Flow Relations Pressure

Impulse Function

F2 = F1 Stagnation Pressure

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Temperature

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Cont.. Stagnation Temperature

Density & Velocity

Change of Entropy

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Heat Transfer

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Derivation Proof Momentum •

(p1 − p2 )A = m(c2 − c1 ) (p1 − p2 )A = ρ 2 Ac22 − ρ1 Ac12 ⇒ (p1 − p2 ) = ρ 2c22 − ρ1c12 p M 2γRT = pM 2γ We know that ρc = RT Therefore p1 − p2 = γM 22 p2 − γM12 p1 2

(1 + γM12 ) p1 = (1 + γM 22 ) p2 p2 (1 + γM12 ) = p1 (1 + γM 22 ) GDJP

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Derivation Proof Stagnation Pressure γ γ − 1   P 0 = p  1 + M 2  γ − 1 2   γ γ − 1   γ −1 2 M 2  1+ P 02 p 2  2  = γ − 1 P 01 p1  2  M + 1  1  2  γ γ − 1   γ −1 2 2 M + 1   1 + γM 1 2 P 02 2  = × γ − 1  P 01 2  1 + γ M 12 M + 1   1 2   GDJP

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Derivation Proof Temperature • c1 ρ2 From Continuity m = ρ1 Ac1 = ρ 2 Ac 2 ⇒ = eqn (1) c2 ρ1 From Mach number M 2 c2 a 2 c2 = = c1 a1 M1 c1 From eqn 1 & 2

γRT2

c 2  T1   = ×  c1  T2  γRT1

1/ 2

1/ 2 ρ 2 M 1  T1  c1   = = eqn (3)   ρ 1 M 2  T2  c2 We know that, p1= ρ 1 RT1 , p 2 = ρ 2 RT2 So

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T p2 ρ = 2× 2 ρ 1 T1 p1

eqn (4)

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eqn

(2)

Derivation Proof-Cont.. Substituin g from eqn (3) in (4) 1/ 2 M 1  T2  p2 T2    = ×  ⇒ = p1 M 2  T1  T1  p2 substitute in the above eqn p1 T2 T1

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 1 + γM 2   M2  1 = ×  2 M  1 + γM 1 2  

2

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p2 M 2 × p1 M 1

   

2

SLOPE OF RAYLEIGH LINE W .K .T p + G 2ν = constant For any two state points, p1 + G2ν 1 = p2 + G2ν 2 2

•  m p1 − p2 = −G 2 = −  ν 1 −ν 2  A    dp    = −G2 = tanθ = −( ρc)2 = − ρ 2a 2 M 2  dv  R (1)

Slope of S=cons. Line

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Cont..

(2)

1

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CONS. ENTHALPY LINE

Note:

pv = RT = constant ; for T = constant line Differenti ating yields , pdv + vdp = 0 dp p =− dv v

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Cont..

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Problem 1 In your opinion, which assumption(s) in the Rayleigh flow analysis may be potential source(s) of error in solving a real life problem? Answer: •Heat transfer causes the total temperature to change significantly in the flow, which leads to a large variation of static temperature. The perfect gas assumption (constant specific heats) may not be appropriate. At higher and higher temperatures, more and more energy modes are activated within the molecules. In general, this causes the specific heats to rise with temperature.

•In cases where combustion occurs, chemical composition of the constituent gases changes significantly. Reactant species will be consumed and product species will be produced. Their relative ratio changes as combustion proceeds. Values like gas constant R, and specific heat ratio, will no longer be constant but depend on the extent of combustion.

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Problem 2 Unchoked case: Air at 250 K and 1.0 bar is moving at 100 m/sec towards the entrance of a combustion chamber. Determine the exit conditions if 300 kJ/kg is added to the flow during the combustion process. 1 P= 1.0 bar T= 250 K V= 100 m/s

2

Heat addition

Solution: For T 1 = 250 K, V 1 = 100 m/s. This gives an inlet Mach number From

M

the isentropic

From which GDJP

of

1

= 0 . 3156

flow table we

T 01 = 255 K Anna University

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obtain

T1 = 0 . 9805 T 01

Problem 2- Cont.. From the Rayleigh flow table, we obtain T01 T1 P1 0 . 3763 0 . 4427 = = = 2.106 * * * T01 T1 P1 From conservati on of energy, we have

q = C p (T02 − T01 ) ⇒

T02 q = +1 T01 C p T01

T02 T 02 T01 T02 T02 T02* At station (2), = ⇒ = × * = 2.172 × 0.3763 = 0.8173 * * T01 T01 T01 T01 To1 T02 From the rayleigh flow table, this correspond s to an exit Mach no. of T P M 2 = 0.5985 (subsonic) , and 2* = 0.9152 , 2* = 1.598 T2 P2 T2 T2 T2* T2 0.9152 The exit state is ⇒ = ⇒ = ⇒ T2 = 517 K * T1 T1 T1 250 0.4427 P2 P2 P2* P2 1.598 = ⇒ = ⇒ P2 = 0.7588 bar * P1 1 2.106 P1 P1 GDJP

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Problem 2- Cont.. Just choked case: How much more heat that can be added without changing the conditions at the entrance to the cumbustor? Solution: * * For the just choked case, M 2 = 1 . This implies T02 = T01 = T02

From the energy equation, we have q = C p (T02 − T01 ) * * = T02 but, for max. heat addition T0 max = T0* = T01 where M = 1 *   T * 01  − 1  Therfore, q = C p ( T0 − T01 ) = C p T01   T01   1  = 1004 × 255  − 1  = 424 kJ / kg  0 .3763 

Hence, we can add an extra 124 kJ/kg into the flow before we choke it thermally. At this choked condition, the stagnation temperatur e at station (2) is T02 = GDJP

T02*

=

T01*

T01*  1  = T01 =   ( 255 ) = 678 K T01  0 .3763  Anna University

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Problem 2- Cont.. Choked case: Let us add sufficient fuel to the system so that the exit stagnation temperature is raised to 1500 K now. Assume that the receiver pressure is very low. What do you expect to happen in the system? Describe the flow both qualitatively and quantitatively. Solution: • In this case T02 = 1500 K > 678 K (choking condition)

• The original flow cannot accommodate this large amount of heat. Something has to happen in order to take in so much heat addition. In other words, it cannot stay on the same Rayleigh line. •Recall that the upstream state can always communicate with the downstream states in a subsonic flow by means of pressure waves. GDJP

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Problem 2- Cont.. • “Sensing” the super-critical heat addition downstream, the flow decelerates

from the free stream to the inlet. Spillage occurs ahead of the inlet. It is shown schematically as follows: Heat addition ∞ M=0.3156 P=1.0 bar T=250 K V=100 m/s

1

2

Spillage

•With a smaller flow rate in the combustion chamber, the flow moves to a different Rayleigh line with a smaller mass flow rate/A value. •Since the receiver (back) pressure is very low, we can assume that the flow is choked at the station (2), i.e. M2 = 1 GDJP

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Problem 2- Cont.. With M 2 = 1 we conclude that To 2 = From

* T 01

=

= 1500

* T 02

the

above

by refering

K ;

This

leads

to

T 01

255 = = 0 . 17 * 1500 T 01

temperatur e ratio ,

Rayleigh

table,

we obtain

M 1 = 0 . 1977 < 0 . 3156 = M ∞ The flow decelerate s from M ∞ = 0 .3156 to M 1 = 0 . 1977 at the inlet With M 1 = 0 . 1977 we obtain from the isentropic flow table T1 = 0 . 9922 T 01

P1 = 0 . 9731 P 01

The inlet state is T1 T1 = T∞ T∞ P P1 = 1 P∞ P∞ GDJP

T 01 0 . 9922 = 0 . 9805 T0∞

We know that T ∞ = 250 K

P01 0 . 9731 = 0 . 932 P0 ∞

We know that P ∞ = 1 .0 bar Anna University

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∴ T1 = 253 K ∴ P1 = 1 . 04 bar

Problem 2- Cont.. higher ‘s’ due to more heat addition

T o determine exit condition P1 T1 @ M1 = 0.1977; = 0.2024& = 2.275 * * P1 T1 Recall that the exit state

inlet

is the referencestate due to choking, we can concludethe exit conditionto be,

exit exit

upstream

* 1 T T2 = T2* = T1* = 1 × T1 = × 253 T1 0.2024 = 1250 K

• m smaller A

* P 1 * * 1 P2 = P2 = P1 = × P1 = × 1.04 = 0.457 bar P1 2.275 GDJP

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