Compressible Flow Properties

2 Review of Fundamentals

2.7 Compressible Flow Properties

REVIEW OF FUNDAMENTALS 93 These results for temperature and pressure at station 2 are higher by 126.6°R and 0.097 atm, respectively, than those obtained in Example 2.6 for air as a calori- cally perfect gas.

94 ELEMENTS OF PROPULSION

The temperature T is sometimes called the static t e m p e r a t u r e to distinguish it from the total temperature Tt. W h e n V = 0, the static and total temperatures are identical. F r o m these definitions, it follows that, for a calorically perfect gas,

A h t = c p A T t

Using these new definitions, we see that the steady flow energy equation in the absence o f gravity effects b e c o m e s

q - Wx = htout - htin (2.67)

or, for a calorically perfect gas,

q - Wx = c p ( T t o u t - Ttin) (2.68)

If q - Wx = 0, we see from Eqs. (2.67) and (2.68) that htout = htin and that, for a calorically perfect gas, Trout = Ttin.

Consider an airplane in flight at a velocity Va. To an observer riding with the airplane, the airflow about the wing o f the plane appears as in Fig. 2.12. W e mark out a control v o l u m e ~r as shown in the figure between a station far upstream from the wing and a station just adjacent to the w i n g ' s leading edge stagnation point, where the velocity o f the airstream is reduced to a negligibly small magnitude.

Stagnation point

Streamlines

a) A i r f l o w over a w i n g

//•///

^/

[ / IWin~ I

Coincident with ~ / / / / 7 / / / /streamlines .. ~ / ~ - ' ~ /

// / ~" k : V 7 / / Y,. j Stagnation point

~ " ~ /

- " ' - = ~ _ _ -~-~-'-- - - - ~1" - - (1)

(a) ~o" VI = 0

b) E n l a r g e d view o f flow in the n e i g h b o r h o o d o f the stagnation point Fig. 2.12 Control v o l u m e ~r with freestream inlet and stagnation exit conditions (reference system at rest relative to wing).

REVIEW OF FUNDAMENTALS 95 Applying the steady flow energy equation to the flow through or of Fig. 2.12, we have

q--Wx---- ( h + 1 - h + ~ g c a

o r

O = C p T l - C p T +

a

From this equation, we find that the temperature of the air at the stagnation point of the wing is

T1 = Ta_ ~ V2a __ Tta 2 gcCp

Thus we see that the temperature, which the leading edge of the wing "feels," is the total temperature Tta.

At high flight speeds, the freestream total temperature Tta is significantly different from the freestream ambient temperature Ta. This is illustrated in Fig. 2.13, where Tta - Ta is plotted against V~ by using the relation

( r t - : r ) a - - , (Vo)2o R 2 gcCp 12, 000

\H0/

with Va expressed in ft/s. Because the speed of sound at 25,000 ft is 1000 ft/s, a Mach number scale for 25,000 ft is easily obtained by dividing the scale for V~ in Fig. 2.13 by 1000. (Mach number M equals Va divided by the local speed of sound a.) Therefore, Mach number scales are also given on the graphs.

Referring to Fig. 2.13, we find that at a flight speed of 800 ft/s corresponding to a Mach number of 0.8 at 25,000-ft altitude, the stagnation points on an airplane experience a temperature that is about 50°R higher than ambient temperature. At 3300 ft/s (M = 3.3 at 25,000 ft), the total temperature is 900°R higher than ambient! It should be evident from these numbers that vehicles such as the X-15 air- plane and reentry bodies experience high temperatures at their high flight speeds.

These high temperatures are produced as the kinetic energy of the air imping- ing on the surfaces of a vehicle is reduced and the enthalpy (hence, temperature) of the air is increased a like amount. This follows directly from the steady flow energy equation, which gives, with q = wx = 0,

h + V2 V2

)in

o r

96 ELEMENTS OF PROPULSION

80

60 o....

~ 40

20

0 0

10,000

~" 1,000

I

100

1o

200 400 600 800 1000 o

V a (ft/s)

I t

100 1000 10,000 |00,000

Vc, fit/s)

I I I I I I I I I

0 0.2 0.4 0.6 0.8 1.0 0. l 1.0 l0 100

M @ 25,000 ft M @ 25,000 fl

Fig. 2.13 Total temperature minus ambient temperature vs flight speed and vs flight Mach number at 25,000 ft [gcCp is assumed constant at 6000 ft2/(s2-°R); therefore, these are approximate curves].

and

A T - ^{AV 2} 2gcCp

Thus a decrease in the kinetic energy of air produces a rise in the air temperature and a consequent heat interaction between the air and the surfaces of an air vehicle. This heat interaction effect is referred to as aerodynamic heating.

Example 2.9

The gas in a rocket combustion chamber is at 120 psia and 1600°R (Fig. 2.14). The gas expands through an adiabatic frictionless (isentropic) nozzle to 15 psia. What are the temperature and velocity of the gas leaving the nozzle? Treat the gas as the calorically perfect gas air with y = 1.4 and gcCp = 6000 ft2/(s 2. °R).

Solution: Locate the state of the combustion chamber gas entering control volume ~r on a T-s diagram like Fig. 2.11 in the manner depicted in Fig. 2.15.

Then, from the diagram, find Sl. Because the process is isentropic, s2 = Sl. The entropy at 2, along with the known value of P 2 , fixes the static state of 2. With 2 located in the T-s diagram, we can read T2 from the temperature scale as 885°R, and we can verify this graphical solution for T2 by using the isentropic relation (2.43) with "),=1.4. Thus T z = ( 1 6 0 0 ° R ) ( 1 5 / 1 2 0 ) ° z s 6 = 8 8 5 ° R (checks).

REVIEW OF FUNDAMENTALS 97

(1) a (2)

Combustion I

chamber I ,~ V 1=0 ~ V 2=?

I I

P1 = 120 psia P2 = 15 psia T] = 1600°R T2=?

Fig. 2.14 Rocket exhaust nozzle.

If, in addition to P2 and T2, the total temperature Tt2 o f the flowing gas at 2 is known, then the state o f the gas at 2 is c o m p l e t e l y fixed. F o r with P2 and T2 speci- fied, the values o f all t h e r m o d y n a m i c properties independent o f speed (the s t a t i c properties) are fixed, and the speed o f the gas is determined b y T2 and Tt2.

F r o m the steady flow energy equation (Fig. 2.15), we find that Tt2 = Tta = T t and, hence, V2 from the relation

= ~t2 - - T 2 2 & C p

W e see from this equation that the vertical distance Tt2 - T2 in the T - s d i a g r a m is indicative o f the speed of the gas at 2, which is V~ = 2(6000)(1600 - 885)ft2/s 2.

Thus V2 -- 2930 ft/s.

(1600°R) T 1

(885 °R) T 2

PI = (120 psia)

(1) ~1

(15 psia) i- - q = wx = 0 ~

I J 5 / ,'

- - . . .

, h2

i V] = 0 V~

/ (2) tW-__ _-" 2

V2 "1

I 0 = h : + "2 -ha

S 1 = S 2 2g c

he2 = htl = h 1

v~ = r ~ - r 2 2geCp

Fig. 2.15 Process plot for example rocket nozzle.

98 ELEMENTS OF PROPULSION

The series of states through which the gas progresses in the nozzle as it flows from the combustion chamber (nozzle inlet) to the nozzle exit is represented by path line a in the T-s diagram. The speed of the gas at any intermediate state y in the nozzle is represented by the vertical distance on the path line from 1 to the state in question. This follows from the relations

T r y = T 1 and V2Y - - Try - - T y = T 1 - - T y

2gcCp

2. 7.2 S t a g n a t i o n o r Total P r e s s u r e

In the adiabatic, no-shaft-work slowing of a flowing perfect gas to zero speed, the gas attains the same final stagnation temperature whether it is brought to rest through frictional effects (irreversible) or without them (reversible). This follows from the energy control volume equation applied to o" of Fig. 2.16 for a calo- rically perfect gas. Thus, from

q - - Wx : Cp(ry -- r l ) + - - V2y - V~

2 gc with q = Wx = 0 and Vy = O, Ty becomes

T y = T o = TI + - -

Vl

2gcCp

Because the energy control volume equation is valid for frictional or frictionless flow, Ty = To is constant and independent of the degree of friction between 1 and y as long as q = Wx = Vy = O.

(Y)

Vl , V y = 0

T

~tl-

Ptl =- Pa

(Yr) (Yi) P I

/ (1)

~1 I~ s

S 1 = Sy r Syi = S 1

Fig. 2.16 D e f i n i t i o n o f t o t a l p r e s s u r e .

REVIEW OF FUNDAMENTALS 99 Although the gas attains the same final temperature To in reversible or irrevers- ible processes, its final pressure will vary with the degree of irreversibility associ- ated with the slowing down process. The entropy state and control volume equations for the flow through o-are

T,, P,,

Cp (~._z_~ _ R E ~ = Sy - - S 1 > 0 (2.69)

T1 P1 -

Since Ty = To = const from Eq. (3.5), the final value of Py depends on the entropy increase Sy - Sl, which in turn is a measure of the degree of irreversibility between 1 and y.

When the slowing down process between 1 and y is reversible, with sy - Sl = 0, the final pressure is defined as the total pressure Pt. The final state is called the total state tl of the static state 1. Using this definition of total pressure, we have, from Eq. (2.43),

_ [ T t \ 'y/(~'- 1)

p , =-- t ' t ~ ) (2.70)

These ideas are illustrated in the T-s diagram of Fig. 2.16. Let us imagine the flowing gas at station 1 to be brought to rest adiabatically with no shaft work by means of a duct diverging to an extremely large area at station y, where the flow velocity is zero. If the diverging duct is frictionless, then the slowing down process from 1 to y is isentropic with the path line C~r in the T-s diagram of the figure. If the diverging duct is frictional, then the slowing down process from 1 to y is irreversible and adiabatic (Syi > Sl) to satisfy the entropy control volume equation for adiabatic flow and is shown as the path line ai, in the T-s diagram.

The total pressure of a flowing gas is defined as the pressure obtained when the gas is brought to rest isentropically. Thus the pressure corresponding to state Yr of the T-s diagram is the total pressure of the gas in state 1. The state point Yr is called the total or stagnation state tl of the static point 1.

The concepts of total pressure and total temperature are very useful, for these two properties along with the third property (static pressure) of a flowing gas are readily measured, and they fix the state of the flowing gas.

We measure these three properties in flight with pitot-static and total tempera- ture probes on modern high-speed airplanes, and these properties are used to determine speed and Mach number and to provide other data for m a n y aircraft subsystems.

Consider a gas flowing in a duct in which P and T may change due to heat interaction and friction effects. The flow total state points tl and t2 and the static state points 1 and 2, each of which corresponds to flow stations 1 and 2, are located in the T-s diagram of Fig. 2.17. By definition, the entropy of the total state at any given point in a gas flow has the same value as the entropy of the static state properties at that point. Therefore, st~ = $1 and st2 = s2.

100 ELEMENTS OF PROPULSION

vl v2

(1) (2)

Pt2

Ptl 7~t2~11

, I P2

I - -

I /P1

J l L

I s

I,t $ 2 - - S 1 •

] • St2 -- Stl •

Fig. 2.17 Entropy change in terms of the stagnation properties Tt and P t .

From the entropy equation of state of a perfect gas, the entropy change between 1 and 2 is

$ 2 - - S 1 = Cp~¢¢~11

- R ( ~

The entropy change between total state points t~ and t 2 is

Tt2 Pt2

st2 - s , l = C p & w - - R~,~

Pt--7

ltl

^(2.71a)

Since Stl ~ S 1 and st2 ~ $2, w e have

St2 -- Stl ~ S 2 -- S 1

Therefore, the change of entropy between two states of a flowing gas can be determined by using total properties in place of static properties.

Equation (2.71a) indicates that in an adiabatic and no-shaft-work constant-Tt flow (such as exists in an airplane engine inlet and nozzle, or flow through a shock wave), we have

s2 - sl = - R ~ , ~ - - Pt2 (2.71b) Ptl

By virtue of this equation and the entropy control volume equation for adiabatic flow, s2 - sl > 0. Thus, in a constant-Tt flow,

et2 m < l

Ptl --

REVIEW OF FUNDAMENTALS 101 Hence the total pressure of air passing through an engine inlet and nozzle or a shock wave cannot increase and must, in fact, decrease because of the irreversible effects of friction.

2.7.3 T/Tt and P/Pt as Functions of Mach Number The speed of sound a in a perfect gas is given by

a = v/-ygcRT

Using this relation for the speed o f sound and Eq. (2.8), we can write the Mach number as

g 2 M 2 - - _ _

ygcRT

With the help of this expression for the Mach number, we can obtain many useful relations that give gas flow property ratios in terms of the flow Mach number alone. Two such relations for T/Tt ^and P/Pt are

= (1 - 1 -1

P ( l + Y - 1 2 \ - v / ( r - a )

(2.72)

(2.73)

Equations (2.71) and (2.72) appear graphically in Fig. 2.18 and are given in the isentropic flow functions of the G A S T A B program for user input 3' value. These equations show that for each freestream Mach number (hence, for each flight Mach number of an airplane), the ratios P/Pt and T/Tt ^have

unique values.

Both Fig. 2.18 and the corresponding equations provide Mach numbers and ambient temperatures for known values o f P, Pt, and Tt. For example, we are given the following in-flight measurements:

P = 35.4 kPa Tt ^{= 300 K} Pt ^{= 60.0 kPa}

From these data, P/Pt = 0.59. If we enter Fig. 2.18 with this value of P/Pt,

we find M = 0.9 and T/Tt = 0.86. Then we obtain the ambient temperature by using T = (T/Tt) Tt = 0.86(300) = 258 K.

Figure 2.18 shows that in a sonic (M = 1.0) stream o f gas with y = 1.4, P

-- = 0.528

Pt

102 ELEMENTS OF PROPULSION

T

Wt

P Pt

1.0-

0.8

0.6

0.4

0.2

O.C 0.0

T

I I I I [ I

0.5 1.0 1.5 2.0 2.5 3.0

Mach number

Fig. 2.18

P/Pt

^and

T/Tt

vs M a c h n u m b e r ( y = 1.4).

and

and for supersonic flow,

and

- - = 0.833 T r,

- - < 0.528 P Pt

- - < 0.833 T

r,

Consider the one-dimensional steady flow of a gas in a duct with Tt and Pt con- stant at all stations along the duct. This means a total temperature probe will measure the same value of Tt at each duct station, and an isentropic total pressure probe will measure the same value of Pt at each station. The path line of a of the flow is a vertical line in the T-s diagram. The state points on the path line can be categorized as follows.

Subsonic:

T > 0.883 Tt P > 0.528 Pt

REVIEW OF FUNDAMENTALS 103

V

Pt

P , = 0 5 2 s p ,

Subsonic Supersomc ~

Fig. 2.19 Subsonic and supersonic state points in an isentropic flow.

Sonic:

Supersonic:

T=O.883Tt P = 0.528Pt

T < 0.883 Tt P < 0.528 Pt

Figure 2.19 delineates the subsonic, sonic, and supersonic portions of path line c~.

The thermodynamic properties at the state point where M = 1 on a are denoted by P*, T*, W, etc. (read as P star, etc.) and the state is called the "star state." In addition, the cross-sectional flow area at the M = 1 point is indicated by A*. The magnitude of A* is determined by the relation

A* - rh (2.74)

WV*

2.7.4 Mass FIow Parameter

The stream properties at any general station of a calorically perfect gas are related by Eqs. (2.21), (2.36), (2.66), (2.70), (2.72), (2.73), and the one- dimensional mass flow equation

in = pAY

The mass flow per unit area at duct area A in the flow of a calorically perfect gas is given by the expression

a RT ~ ,v/RT V R PrY T ~

where the one-dimensional flow equation [Eq. (2.12a)], the perfect gas equation of state [Eq. (2.21)], the perfect gas speed of sound equation [Eq. (2.36)], and the

definition of Mach number have been used. Replacing the static/total property ratios with Eqs. (2.72) and (2.73) and rearranging, we obtain the grouping defined as the

mass flow parameter

^(MFP):

M F P -

APt

t h a t \/'@M(l_t 3,-1M2"~z(rv+-ll~

MFP(M) - - a P~- -- 2 /

(2.75)

(2.76)

0.7

Thus the MFP is a unique function of the Mach number in a calorically perfect gas. Values of the mass flow parameter are plotted in Fig. 2.20 for 3/= 1.4 and 3' = 1.3, and calculated in the G A S T A B program for user input values of 3, and mean molecular weight .///. From Fig. 2.20 and the G A S T A B program, one sees that the maximum value of the mass flow parameter occurs when the Mach number is unity. Thus, for a given total temperature and pressure

Tt

and

Pt,

the maximum mass flow rate per area corresponds to a flow Mach number of 1.

0.6847

/

= .

0.6

0.5 M F P x ~ g ~

0.4

0 3 Pt A

0.2

0.1

104 ELEMENTS OF PROPULSION

0.0 I I I I I I

0.0 0.5 1.0 1.5 2.0 2.5 3.0

Mach number

F i g . 2 . 2 0 M a s s flow p a r a m e t e r vs M a c h n u m b e r ('y = 1.4 a n d y = 1.3).

REVIEW OF FUNDAMENTALS 105

2.7.5 Isentropic Area Ratio A/A*

The ratio of the one-dimensional flow area A at any flow station to the one- dimensional area for that same flow rate at sonic velocity A* is a unique function of the Mach number M at the flow station and the ratio of specific heats y. Con- sider the isentropic flow of a calorically perfect gas in an isentropic duct from p, M, P, T, and A to the sonic state where the properties are p*, M* = 1, P*, 7", and A*. Because both states have the same mass flow, we write

Rewriting gives A p* V*

However,

& = pAV = p*A*V*

P* T 1 a* 1 P* / P 1 P* / P T* PM a - - M T * / T ~ - - M 4'-T-~/T

T T/Tt [ 2 ( 1 - ~ - )I-1

r -r-Vf,-7 i +

A* pV

⁽ⁱ⁾

(ii) and

P P/Pt [ 2 ( T ~ )] -7/(7-1)

P ~ - - ~ t - - ~ 1 + M 2 (iii)

Substitution of Eqs. (ii) and (iii) into (i) gives

A 1 [ 2 ( "~--12"~-] (y+I)/[2(y-1)]

A*--M-- y - ~

I + ~ - - M ) J (2.77) and

A/A*, PIP,

and

T/T,

are plotted vs Mach number in Fig. 2.21 for 3' = 1.4 and calculated in G A S T A B for user input y value.

Example 2.10

Air at a total temperature of 300 K and total pressure of 1 atm flows through a 2-m ~ duct at a Mach number of 0.5.

1) What must the area of the duct be for the flow to isentropically accelerate to Mach 1?

2) What is the mass flow rate o f air through the duct?

Solution:

From G A S T A B for M = 0.5 and y = 1.4,

A/A*

= 1.3398 and MFP = 0.0301563.

1) The area of the duct at M = 1 is then

A 2

A* . . . . 1.4928 m 2

A/A*

1.3398

3.0

2.5

2.0

1.5

1.0

0.5

I

0.5 1.0 1.5 2.0 2.5 3.0

0.0 0.0

106 ELEMENTS OF PROPULSION

Mach number

Fig. 2.21

A/A*, P/Pt,

and

T/Tt

vs M a e h n u m b e r ( y = 1.4).

2) The mass flow rate can be calculated by using the MFP as follows:

PtA 101,325 x

rh --- ~/It~MFP = 30~/.f6_0~2N 0.0301563 k g - ~ / ( N , s) = 352.83 kg/s

2.7.6

Impulse Function I

The impulse function I is defined by

I =- PA + &V/gc (2.78)

Using one-dimensional mass flow equation (rh = pAV) and equations for a calorically perfect gas, Eq. (2.78) can be rewritten as

I = PA(1 + y M 2) (2.79)

From Eq. (2.20) for steady state, the streamwise axial force exerted on the fluid flowing through a control volume is/exit -- /entry, while the reaction force exerted by the fluid on the control volume is Ie,try -/exit.

REVIEW OF FUNDAMENTALS 107 The impulse function makes possible almost unimaginable simplification of the evaluation of forces on aircraft engines, rocket engines, and their components.

For example, although one could determine the net axial force exerted on the fluid flowing through any device by integrating the axial components of pressure and viscous forces over every infinitesimal element of internal wetted surface area, it is certain that no one ever has. Instead, the integrated result of the forces is obtained with ease and certainty by merely evaluating the change in impulse function across the device.

Compressible flow functions tabulate the impulse function as a ratio to that at the star state (M = 1) or I / F . We note that

I P A I + T M 2

(2.80)

I* P ' A * 1 + y

2.8 One-Dimensional Gas DynamicsmFinite Control Volume