2 Review of Fundamentals
2.3 Steady Flow Energy Equation
68 ELEMENTS OF PROPULSION
V /' ~ . ~ . ~ 2 )
(1)
Fig. 2.2 One-dimensional flow through a convergent duct. The flow is uniform at sections 1 and 2, hence one-dimensional, even though the flow direction may vary elsewhere in the flow.
and V± is the velocity component normal to area A. This equation is known as the
conservation of mass equation.
For steady flows through any control volume, Eq. (2.10a) simplifies tothout ~ thin (2.11)
If the flow is steady and one-dimensional through a control volume with a single inlet and exit such as shown in Fig. 2.2, then by Eq. (2.10b)
th = pAV±
(2.12a)which is called the
one-dimensional mass flow equation,
andplA1V±~ = P2A2V_L2 (2.12b)
REVIEW OF FUNDAMENTALS 69
Vin
Zin
• •
i ~ 1 v°ut(out ) I
Zout (iln) ~ ~ Control surface o l
Fig. 2.3 Steady flow through control volume ~r.
where q and Wx are the heat and shaft-work interactions per unit mass flow through 0-. All of the terms in this equation have units of energy per unit mass.
Example 2.1
The first step in the application of the steady flow energy equation is a clear definition of a control surface 0-. This is so because each term in the equation refers to a quantity at the boundary of a control volume. Thus, to use the equation, one needs only to examine the control surface and identify the applicable terms of the equation.
In the application of Eq. (2.14) to specific flow situations, many of the terms are zero or may be neglected. The following example will illustrate this point.
The perfect gas equation (presented in detail later in this chapter) is used to model the relationship between change in enthalpy and temperature of the gas.
Consider a turbojet aircraft engine as shown in Fig. 2.4a. We divide the engine into the control volume regions:
o-1: inlet 0-4: turbine
0-2: compressor 0-5: nozzle 0-3: combustion chamber
Let us apply the steady flow energy equation to each of these control volumes.
In all cases, the potential energy change (gz/gc)out - (gz/gc)in is zero and will be ignored.
It is advisable in using the steady flow energy equation to make two sketches of the applicable control surface 0-, showing the heat and shaft work interactions (q, Wx) in one sketch and the fluxes of energy [h, V2/(2gc)] in the second sketch.
The term [h + V2/(2gc)]out is a flux per unit mass flow of internal energy
70 ELEMENTS OF PROPULSION
(o) (a)
rhf (~- Fuel tank "~
, t t U ~ ~ ~ ~ . - ~ - - ~ i I ., ~ e
i • . _ _ Turbine_ _ .
(1) O 2 a3 G4
Inlet Compressor . _ Combustor Nozzle
- -I- -I- (9)
(2) (3) (4) (5) (e)
Fig. 2.4a Control v o l u m e for a n a l y z i n g each c o m p o n e n t o f a turbojet engine.
e, kinetic e n e r g y vZ/(2gc), and flow w o r k Pv. W e will use the expression flux of energy to i n c l u d e the flow w o r k flux also.
a) Inlet and nozzle: or1 and 0-5. T h e r e are no shaft w o r k interactions at c o n t r o l surfaces cr I a n d ~r 5. H e a t interactions are n e g l i g i b l e and m a y be taken as zero. T h e r e f o r e , the steady flow e n e r g y equation, as d e p i c t e d in Fig. 2.4b, for the inlet or n o z z l e control surfaces gives the r e s u l t
0 = ( h + V2
)ou,
Nozzle: numerical example L e t the gases flowin~g through the n o z z l e control v o l u m e o-5 be p e r f e c t with Cpgc=6OOOft/(s-°R). D e t e r m i n e V9 for T5 = 1800°R, V5 = 4 0 0 f t / s , and T9 = 1200°R.
q = 0, Wx = 0 (In) (Out)
I I
-I "1
I I hin I , ] , hout
I a 1, % I I a~,% I
, , ( e ) ,, ,
I __ ] ~gc in I l ~gc out
Interactions ~ Net energy flux
0 "- ( h + V2 (h V2
~gc )oat - + ~gc )in Fig. 2.4b E n e r g y equation a p p l i e d to control v o l u m e s o" 1 a n d ors.
REVIEW OF FUNDAMENTALS 71 Solution:
stations, respectively, we have
h9 q- ~ c : h5 q- - - and
From the steady flow energy equation with 5 and 9 as the in and out
2go
V9 = v/2gc(h5 - h9)-t- V 2 = v/2Cpgc(T5 - T9)-t- V 2
o r
= ~ / 2 ( 6 0 0 0 ) ( 1 8 0 0 - 1200) + 4002
SO
= 2700 ft/s
b) Compressor and turbine: o2 and o-4. The heat interactions at control surfaces o-2 and o-4 are negligibly small. Shaft work interactions are present because each control surface cuts a rotating shaft. The steady flow energy equation for the compressor or for the turbine is depicted in Fig. 2.4c and gives
V 2 V 2
- - W x : ( h - ~ g c ) o u t - - (h'q-'~gc)in
Compressor and turbine: numerical example For an equal mass flow through the compressor and turbine o f 185 lb/s, determine the compressor power and the turbine exit temperature T5 for the following conditions:
Cpgc : 6000 ft2/(s2-°R)
Compressor Turbine
T2 = 740°R, T3 = 1230°R T4 = 2170°R, T5 = ?
V2 = V3 V4 = V5
Wx < 0 for c 2 (compressor) (In) (Out)
Wx > 0 for 13 4 (turbine) I I
"1 1
I hin [ , I • hout
°2' °4 [ ~ Wx V2 [ 02' 04 I v2
' ( G ) i ' ' I ' ( G ) o
I n I I ut
q = 0
Interactions ~ N e t e n e r g y f l u x
-w x (h+~--~c)out- (h+ V2
~gc ) i n Fig. 2.4c E n e r g y e q u a t i o n a p p l i e d to c o n t r o l v o l u m e s 0.2 a n d 0" 4.
72 ELEMENTS OF PROPULSION
Solution: The compressor power Wc = (inWx)~2 is, with V2 = V3,
~VVc = -th(h3 - h2) -- -thcp(T3 - T2) 6000 (ft/s) 2
= - ( 1 8 5 lbm/s) 32.174 ft-lbf/(lbm-s z ) -- - 1 6 . 9 × 106 fl-lbf/s x l h p
550 ft-lbf/s
= - 3 0 , 7 0 0 h p
(1230 - 740)
The minus sign means the compressor shaft is delivering energy to the air in tr2.
The turbine drives the compressor so that the turbine power Wt = (rhWx)~4 is equal in magnitude to the compressor power. Thus Wt = -Wc, where, from the energy equation,
ff't = rh(h5 - h 4 ) and Wc --- - t h ( h 3 - h 2 )
Thus
; n c p ( T 5 - T4) = - & c p ( T 3 - T 2 ) and
= T4 - ( ~ - T2)
= 2170°R - (1230°R - 740°R) = 1680°R = 1220°F
c) Combustion chamber: ~r3. Let us assume that the fuel and air entering the combustion chamber mix physically in a mixing zone (Fig. 2.4d) to form what we will call reactants (denoted by subscript R). The reactants then enter a com- bustion zone where combustion occurs, forming products of combustion (sub- script P) that leave the combustion chamber. We apply the steady flow energy equation to combustion zone ira. Because the temperature in the combustion zone is higher than that of the immediate surroundings, there is a heat interaction between o'3 and the surroundings that, per unit mass flow of reactants, is negligi- bly small (q < 0 but q = 0). Also the velocities of the products leaving and of the reactants entering the combustion zone are approximately equal. There is no shaft work interaction for o'3. Hence the steady flow energy equation, as depicted in Fig. 2.4d, reduces to
hR3 = he4 (2.15 )
We must caution the reader about two points concerning this last equation. First, we cannot use the relation cpAT for computing the enthalpy difference between
REVIEW OF FUNDAMENTALS 73 Fuel
Air
I rh R
Mixing Ii J' zone I
(3) S ~ 3 (4)
Combustion zone
mp
"1
I I
I o 3 I
I I
I j
q=O, wx=O Interactions
(In) (Out)
I I
I
h R ~ I , hp
I
2g--'-c I l 2gc
vp=v R
= Net energy flux
_ -
+ 2-ggc )out-- )in Fig. 2.4d E n e r g y e q u a t i o n a p p l i e d to control v o l u m e s o'3.
two states of a system when the chemical aggregation of the two states differs.
Second, we must measure the enthalpy of each term in the equation relative to the same datum state. To place emphasis on the first point, we have introduced the additional subscripts R and P to indicate that the chemical aggregations of states 3 and 4 are different.
To emphasize the second point, we select as our common enthalpy datum a state d having the chemical aggregation of the products at a datum temperature Td. Then, introducing the datum state enthalpy (he)d into the last equation, we have
hR3 -- hpd ~ hP4 -- hpa (2.16)
Equation (2.16) can be used to determine the temperature of the products of com- bustion leaving an adiabatic combustor for given inlet conditions. If the combus- tor is not adiabatic, Eq. (2.16), adjusted to include the heat interaction term q on the left-hand side, is applicable. Let us treat the reactants and products as perfect gases and illustrate the use of Eq. (2.16) in determining the temperature of the gases at the exit of a turbojet combustion chamber via an example problem.
Combustion chamber: numerical example For the turbojet engine combus- tion chamber, 45 Ibm of air enters with each 1 Ibm of JP-8 (kerosene) fuel. Let us assume these reactants enter an adiabatic combustor at 1200°R. The heating value her of JP-8 is 18,400 Btu/lbm of fuel at 298 K. [This is also called the
74 ELEMENTS OF PROPULSION
lower heating value (LHV) of the fuel.] Thus the heat released (AH)298 K by the fuel per 1 Ibm of the products is 400 Btu/lbm (18,400/46) at 298 K. The follow- ing data are known:
Cpe = 0.267 Btu/(lbm°R) and Cpl¢ = 0.240 Btu/(lbm°R) Determine the temperature of the products leaving the combustor.
Solution: A plot of the enthalpy equations of state for the reactants and the products is given in Fig. 2.5. In the plot, the vertical distance hR - he between the curves of hR and he at a given temperature represents the enthalpy of combustion AH of the reactants at that temperature (this is sometimes called the heat o f com- bustion). In our analysis, we know the enthalpy of combustion at T d = " 298 K (536.4°R).
States 3 and 4, depicted in Fig. 2.5, represent the states of the reactants enter- ing and the products leaving the combustion chamber, respectively. The datum state d is arbitrarily selected to be products at temperature Td. State d' is the reac- tants' state at the datum temperature Td.
In terms of Fig. 2.5, the left-hand side of Eq. (2.16) is the vertical distance between states 3 and d, or
hR3 - - hed = hR3 -- hRa, + hRd, -- hpa
h,
Fig. 2.5 Enthalpy perfect gases.
R p
I I I j.
T d T 3 T 4 T
vs temperature for reactants and products treated as
REVIEW OF FUNDAMENTALS 75
and since
A h l ¢ = c p R A T and hRd, - hpd = (AH)ra then
hR3 -- hpa = cpR(T3 -- Td) + ( A H ) ~ (i)
Similarly, the right-hand side of Eq. (2.16) is
hp 4 - hea = cpp(T4 - Td) (ii)
Substituting Eqs. (i) and (ii) in Eq. (2.16), we get
CpR(T3 -- Td) + (AH)Td = Cpp(T4 - Ta) (2.17)
We can solve this equation for T4, which is the temperature of the product gases leaving the combustion chamber. Solving Eq. (2.17) for T4, we get
T4 cpR(T3 - Ta) + (AH)Ta
= ~-Ta
Cpp
_ 0.240(1200 - 536.4) + 400
+ 536.4 0.267
= 2631°R (2171°F)
This is the so-called adiabatic flame temperature of the reactants for a 45 : 1 mixture ratio of air to fuel weight. For the analysis in portions of this book, we choose to sidestep the complex thermochemistry of the combustion process and model it as a simple heating process. The theory and application of thermo- chemistry to combustion in jet engines are covered in many textbooks, such as the classic text by Penner (see Ref. 14).