Propulsion 3 Rayleigh line and normal shock PDF

Preview

Summary

Rayleigh line and normal shock...

Description

Aerospace Eng., Propulsion 3

2.6 1-D, frictionless flow through a constant area duct with heat transfer Rayleigh Flow This flow is frictionless but there is heat transfer at the fluid/ duct wall interface. A schematic of the flow is shown below:

Q

u

control volume q, heat transfer per unit mass

There is heat transfer q per unit mass of the fluid. The flows are important for an understanding of jet pipe flows, and flows in heat exchangers. For the analysis the tools at our disposal are the Euler equation (since the flow is frictionless), the continuity equation, and the state equation for a perfect gas. The adiabatic energy equation cannot be used since heat is being added to the flow. We will therefore consider an initial analysis by examining the Euler and the continuity equations:

2 m˙ = ρ uA, dp + d(ρ u ) = 0 The Euler equation may be integrated to give

p + ρu

2

= const

65

Aerospace Eng., Propulsion 3 Eliminating u from the momentum equation using continuity, and using the state equation to eliminate the density term gives

2 m ˙  RT  p+  A p

= const

Hence for a given mass flow rate and duct area specific relationships between pressure and temperature variations exist. However, we know that for positive heat transfer entropy increases, and vice-versa. Therefore we can use the entropy change equation

Δs = Cp log e

T2 T1

− R log e

p2 p1

to force a natural direction to the process. Hence the states of a perfect gas satisfying a permitted entropy change direction, conservation of momentum and conservation of mass may be mapped out. When plotted on T~s axes the result is a system of Rayleigh lines, as shown below. subsonic branch

T

maximum temperature 1

maximum entropy

positive q negative q supersonic branch 2

s

Consider state 1 of the gas. The pressure, temperature and velocity of the gas are such that the flow is subsonic, and the state 1 is indicated on the diagram. The definition of entropy from classical thermodynamics is

ds =

dqrev T

Hence for positive heat transfer (i.e. to the gas) then ds is positive. The gas state therefore follows the subsonic branch of the Rayleigh line in the direction of increasing s, as indicated. Initially T increases, although T soon reaches a maximum value. Any further heat addition then causes T to fall , although s still rises. The branch of the Rayleigh line is then followed until maximum entropy is reached. Essentially the gas flow cannot sustain a heat transfer which will take it beyond the position of maximum entropy. If however heat

66

Aerospace Eng., Propulsion 3 transfer was negative (i.e. heat transfer from the gas) the ds is negative. The gas state therefore follows the subsonic branch of the Rayleigh line in the direction of decreasing s, as indicated. Note that T falls. Now consider state 2, where state 2 is supersonic. This is also indicated on the diagram. Consider that heat transfer is positive. Hence the supersonic branch of the Rayleigh line is followed in the sense indicated. Note that T rises. This process continues until maximum entropy is reached. The gas flow cannot sustain a heat transfer that would take it beyond this point. Now consider negative heat transfer from state 2. The supersonic branch of the Rayleigh line is followed in the direction of decreasing s. The significance of maximum entropy is as follows. From the definition of entropy we can write

dqrev = Tds = dh −

dp

ρ dp

From the energy equation, dq = dh + udu . Therefore at maximum entropy udu =

.

ρ du

From the continuity equation,

dρ = −

u

dp

, therefore at maximum entropy

2

dρ

ρ

For an isentropic process the expression

=u

dp

is equivalent to the speed of sound squared. dρ Hence at maximum entropy where ds=0 the gas speed is sonic. 2.6.1 Effect of heat addition Consider a subsonic entry speed to a duct. Heat is added to the flow. Hence the gas speeds up. The gas can only speed up until sonic conditions are reached at the duct exit. If the duct is longer than the sonic length, then the mass flow rate through the duct changes until sonic exit conditions are reached. If the duct is shorter than the sonic length then the exit conditions are subsonic, and the mass flow rate may be increased until sonic exit conditions are reached. Now consider a supersonic entry speed. The duct is of such a length that exactly sonic conditions are reached at the duct exit. If the duct is lengthened the mass flow rate of the duct alters such that sonic exit conditions are reached. If the duct is shortened the exit flow is supersonic and no choking occurs. 2.6.2 Analysis of effect of heat addition. We need to know the effect of heat transfer upon the gas properties as the gas flows along the duct. Like the Fanno flow and isentropic flow the gas property at any position along the length of the duct is best expressed in terms of its value at the sonic state *. We must therefore consider the fluid flow equations in their complete form. These are as follows: Energy equation:

67

Aerospace Eng., Propulsion 3 Heat added is equal to the change of total enthalpy of the gas flow. Hence the heat added q* per unit mass of gas flow to attain sonic flow from an inlet state 1 is given by

* * q = ho − ho1 where the o subscript indicates stagnation conditions. Therefore in general for a perfect gas * * q = C pT o − Cp T o

Hence the total temperature ratio for the gas flow is given by * q To * = 1− * Cp T o To

The total temperature is given by

 

To = T 1 +

γ −1 M 2

2



Euler equation: This is the inviscid momentum equation, and may be written as

p + ρu

2

= const

The above may be written as

p + γpM

2

= const

Hence the pressure ratio

p * is given by p

p 1+ γ * = p 1 + γM 2 Equation of state:

* T p ρ From the perfect gas law, * = * p ρ T

68

Aerospace Eng., Propulsion 3 From continuity,

ρ

* =

ρ T * = T

u u T , and . Therefore = M * * u u T*

 M(1 + γ ) 2  2  1 + γM 

Therefore the total temperature ratio is given by

T o = * T o

 M(1 + γ ) 2  2 + ( γ   2 γ  1 + γM  

− 1)M 2  +1

 

and the velocity ratio is given by

2 u M (1 + γ ) 2 * = 1 + γM u From the definition of total pressure, the total pressure ratio is given by

γ p

o = * p o

 (1 + γ )   2 + ( γ − 1)M 2    γ + 1 2   1 + γM  

γ −1

2.6.3 Example An estimate of Rayleigh line stagnation pressure loss is required for subsonic combustion taking place in a constant area duct. The air enters at a total pressure of 18atm, a total temperature of 755K and a Mach number of 0.2, and fuel is added to raise the total temperature to 1555K. Take γ =1 .4. The drop in total pressure along the combustion chamber is required. Therefore the exit Mach number is needed. To at inlet and exit are known, and we know that To* is constant. Hence the exit M can be found. At M=0.2, To/To* = 0.1736, ∴To* = 4349K, and po/ po* = 1.2346 At exit, therefore, To/To* = 0.3575. Therefore the exit M is 0.3058. At M=0.3058, po/po* = 1.1961. Hence the ratio of inlet total pressure to exit total pressure is 1.0322.

69

Aerospace Eng., Propulsion 3 The exit stagnation pressure is therefore 17.44 atm, and the loss in total pressure through the combustion chamber is therefore 0.56atm. A gas stream enters a constant area heater at M=0.3, po = 600kPa and To = 500K. There is a heat transfer of 500 kJ kg-1 into the gas. Determine the Mach number, total pressure and total temperature at the heater exit. (Take γ= 1.4, Cp = 1.004 kJ kg-1 K-1). The heat transfer q is known. The heat transfer q* to reach sonic conditions may be calculated directly from the tables. At M=0.3, po/po* = 1.1985, To/To* = 0.3469. Therefore To* = 1441K To reach sonic conditions, q* = 947.7kJ kg-1. q, however, is 500 kJ kg-1, therefore at the heater exit, q* = 447.7kJ kg-1. Therefore at the heater exit, To/To* = 0.6906, and therefore Mexit = 0.5. Hence at the heater exit po/po* = 1.1141, and therefore po at exit is 557.7kPa.

70

Aerospace Eng., Propulsion 3 TUTORIAL QUESTIONS: 1-D FLOW WITH HEAT TRANSFER AND REHEAT 1. A gas stream enters a 1-D duct. With reference to a Rayleigh line indicate what happens to the gas state under the following conditions: (i) (iii)

Subsonic inlet, heat transfer to the gas. (ii) Supersonic inlet, heat transfer to the gas. (iv)

Subsonic inlet, heat transfer from the gas. Supersonic inlet, heat transfer from the gas.

In each case indicate what the limiting state of the flow is, if applicable. 2. Show that the point of maximum entropy on a Rayleigh line is when the local gas speed is the speed of sound (hint: at maximum entropy ds=0, so flow is isentropic! Look at how expression for speed of sound was derived.) Also show that the position of maximum temperature is when the Mach number is given by 1 M = γ 3. The Rayleigh line applies to flow through a constant area duct. Consider the flow through a duct of varying area but with the gas pressure remaining constant. (i) Use the 1-D momentum equation to show that if the gas pressure is constant, then the gas velocity is also constant in a frictionless flow. (ii) From first principles show that the area variation dA required to hold the static pressure constant in a frictionless duct with heat transfer to the gas stream is given by the expression dA γ −1 2 dT o = 1+ 2 M A To

[

]

Note: start from 1-D continuity

ρ uA = const and use result that u is constant. (iii) Hence show that the Mach number variation is given by the expression 2 dM γ −1 2 dT o 2 = −1+ 2 M To M and therefore that the relationship between duct area and Mach number at two positions 1 and 2 is given by M2   1  A2 = A  2  1 M  2 (iv) Is the duct divergent or convergent for heat transfer to the gas?

[

]

4. A gas stream enters a constant area duct at a temperature of 600K, a pressure of 3 bar and a Mach number of 0.2. 800 kJ kg-1 of heat is added. Use Rayleigh line tables to find the values of the sonic temperature and pressure, and how much heat * * * −1 transfer is required to reach the sonic condition. ( T = 2904K; p = 1.32 bar ; q = 2891kJkg ) Hence find the exit Mach number, pressure and temperature, and the loss in total pressure along the duct. p * * M2 = 0.33; p2 = 2.748bar ; T = 1370 K ; o2 = 0.96 2 po1 5. For question 4, determine the heat transfer to the duct if the exit Mach number is 0.5. Hence find the exit pressure, temperature and the total pressure loss. p −1 p2 = 2.347bar ;T 2 = 2295K; o2 = 0.9;q = 1811kJkg po1

71

Aerospace Eng., Propulsion 3

6. A ramjet powered missile flies at M=3.0 in an atmosphere at 0.15 bar and 217K. Assuming that the ramjet diffuser is isentropic calculate the total pressure loss in the constant area combustion chamber for an inlet Mach number of 0.2 and a heat transfer of 1490 kJ kg-1. (Take γ = 1.4, Cp = 1.004 kJ kg-1 K-1). 7. A scramjet (supersonic combustion ramjet) powered vehicle is flying at an altitude where the ambient temperature is 200K. Inside the engine combustion starts at M=2.5 and 1000K. The static temperature upper limit is 1800K. Assuming an isentropic diffuser and a constant area combustion chamber calculate the following: (i) The flight Mach number, (ans: Mach 7.16) (ii) The Mach number at the combustion chamber exit, -1 (ans: Mach 1.64) (iii) The heat added, ans: 522 kJ kg (iv) The percentage loss in total pressure through the combustion chamber. (46% loss) 8. Describe the construction of a reheat chamber for thrust augmentation. How do the gas properties change along the reheat chamber? When reheat is applied, why does the nozzle need to be of variable area. 9. (a) below.

A turbojet is fitted with a constant area reheat chamber and a variable area nozzle, as shown nozzle

q

1

reheat chamber

2

3

The nozzle geometry is adjusted such that the nozzle is always choked at exit. Assuming that the flow through the nozzle is isentropic, show that the ratio of the nozzle exit area to the reheat chamber inlet area is given by A3 p T MFP1 where MFP is the mass flow parameter. = o1 ⋅ o2 ⋅ * p A1 T MFP o2 o1 A Hence for an inlet Mach number to the reheat chamber of M1=0.4 determine the area ratio 3 for the A1 * T o2 : 1, 1.307, 1.436, 1.777, and T o2 = T o following values of T o1 (Take γ= 1.4, R=0.2869 kJ kg-1 K-1 ) (b) and

(i)

The rate of fuel mass flow in the reheat duct is 1.04 kgs-1. Given that To1 = 800K find To2

A3 . (Jet A-1 fuel yields the equivalent of 42850 kJ kg-1 of energy as heat and the air mass flow rate A1

through the engine is 30 kgs-1.) (ii) For p1=3bar find the additional thrust due to reheat for take off in sea level standard air.

72

Aerospace Eng., Propulsion 3

Compressible flows with heat transfer: Worked answer to question 9 This question uses the mass flow parameter, MFP given by

MFP =

˙ To m po A

learn this! The first part is simply algebraic manipulation. Assume that the nozzle is isentropic (because it doesn't say otherwise). Hence the total pressure is constant through the nozzle. Therefore the ratio of nozzle inlet area to nozzle exit area is given by

A3

=

A2

MFP 2 * MFP

since the exit flow is sonic. In the reheat duct the flow is not isentropic, and the total temperature is not constant. However, the area is constant. Therefore MFP is given by

MFP2 = MFP1

T o2 po1 T o1 po2

Hence we get the result

A3 A1

=

p o1 po2

⋅

T o2 T o1

⋅

MFP1 * MFP

To evaluate the area ratio we need the total temperature ratio and the total pressure ratio between the reheat duct exit and entry. We have been given values of total temperature ratio, hence total pressure ratio must be worked out for each temperature ratio. It should be apparent that the greater the heat addition the greater the fall in total pressure. The total pressure ratio is found by evaluating M2 for each total temperature setting as if the reheat duct was *

isolated; ignore the nozzle!!!!! We note that po extracted from Rayleigh flow tables relates *

to a Rayleigh line only. Therefore do not confuse the po from the Rayleigh part of the calculation with the sonic flow at the nozzle exit; they are not related. Hence for the various values of total temperature ratio we get the following values of M2,

p o1 po2

T o2

=1, this is a trivial case, since no heat is added.

T o1

73

and

A3 A1

:

Aerospace Eng., Propulsion 3 Detailed working for

T o2

=1.307:

T o1

MFP1=0.02542, MFP*=0.04043 For M1=0.4,

T o2

* To2

Hence

Hence

T o1

* T o1

=0.5290

    T T o1 o2 ⋅ =  since * T T   o1  o1

T o2 * To2

p o1

* * T o2 = T o1

=0.6914, and therefore M2=0.5.

p p =1.1566, and o2 * =1.1141, and therefore o1 =1.0381, po1 po2 po2 *

A3 ∴

=0.746

A1

The rest of the answers are as follows:

T o2

=1.4360, M2=0.55,

T o1

=1.05722,

po2

T o1 T o2

p o1

=1.777, M2=0.75,

p o1 po2

=1.1228,

A3

=0.7966;

A1 A3

=0.9411;

A1

* T o2 = T o2 is a trivial case.

Ignoring the attempt at humour in the last part of the question, the method is to find To2 for a given area ratio by interpolating between the above answers, which gives To2~1285K. Hence q=487 kJ kg-1, and the rate of fuel wasted is 1.04 kgs-1.

74

Aerospace Eng., Propulsion 3

2.7 Normal shock waves Consider a compressible gas flow that is both frictionless and adiabatic. Hence both the Euler equation and the adiabatic steady flow energy equation apply. If either equation is considered separately, then the possible gas states lie anywhere on a Fanno line, or anywhere on a Rayleigh line. However, since both sets of equations apply, then the possible gas states are where Fanno and Rayleigh lines for the same mass flow rate coincide. On the T~s plot the Fanno and Rayleigh lines for the same mass flow rate appear as below: Rayleigh line

T Fanno line

subsonic branches

2

sonic states

supersonic branches

1 s

Notice that the two lines intersect at two points. Therefore a frictionless. adiabatic, compressible flow has two possible states, one of which is subsonic, and the other of which is supersonic, and the gas may change state. However notice that the entropy at the supersonic intersection is lower than that of the subsonic intersection. Hence if the gas changes state at all, it can only do so from state 1 to state 2 (i.e. from supersonic to subsonic), and not in the other direction. This is because for an adiabatic flow the entropy cannot possibly decrease. We appear to have an anomaly, in that a frictionless, adiabatic flow has experienced an increase of entropy. In this case we have to relinquish the idealization of reversibility, even though the flow is regarded as frictionless. What may be regarded as happening is that the flow may undergo a transition from state 1 to state 2 in the limit as the friction becomes infinitesimally small, or that the flow is frictionless except over an infinitesimally small region. Another anomaly caused by the use of the Rayleigh/ Fanno line analogy is how the gas undergoes the transition from state 1 to state 2. Neither the Fanno line nor the Rayleigh line can be followed. In reality the transition from state 1 to state 2 occurs over only a few mean free paths, which for the present purposes may be modelled as a discontinuity.

75

Aerospace Eng., Propulsion 3 2.7.1 Analysis of normal shock wave behaviour. Consider a normal shock wave in a 1-D duct.

stationary, normal shock wave

gas state 1

gas state 2

As the gas flows through the shock wave there is a discontinuous change in the gas state from state 1 to state 2. The flow is adiabatic and frictionless, hence the following expressions apply: Continuity:

m˙ = ρ uA = const

Euler equation:

p + ρu

Energy equation (1st Law): T o = T +

Perfect gas assumption:

p

u

2

= const

2 = const

2Cp u = R = const , M =

γRT

ρT

The important expressions are the three conservation laws, i.e. mass, momentum and energy. With the perfect gas assumption the Euler equation reduces to