Modern Compressible Flow Chapter 5 Notes

==Introduction==
We will be dealing with nozzles and diffusers. We will assume that flow properties only vary axially along the length of a duct. Changes in the axial flow properties result from a cross-sectional area change in the duct. In chapter 3, we dealt with a constant cross-sectional area duct.
We relax the constant cross-sectional area constraint by allowing the stream-tube area, A, to vary with distance, x.
:A = A(x)
:p = p(x)
:\rho = \rho(x)
:T = T(x)
:V = u = u(x)
Is the flow in the stream-tube truly one-dimensional? No, but for most engineering problems, the quasi-one-dimensional results are usually sufficient (ex. flow through wind tunnels, rocket engines, converging-diverging nozzles).
==Governing Equations==
The quasi-one-dimension assumption is actually three-dimensional, and the exact solution requires a three-dimensional numerical simulation. The quasi-one-dimensional assumption is an approximation to the actual flow physics; however, the governing equations will exactly enforce the mass conservation, Newton's second law (momentum), and the first law of thermodynamics (energy) for the quasi-one-dimensional flow. This is accomplished by utilizing the integral forms of the conservation equations (discussed in Chapter 2).
Let us consider a control volume for a quasi-one-dimensional flow (see figure 5.6, p. 196, Anderson).
===Mass Conservation (Continuity)===
The continuity equation for steady, quasi-one-dimensional flow is 
(see integral equation from previous notes) 
Why do we not include the integral taken over the sides of the control volume between locations 1 and 2? The velocity is oriented along the surface; no flow goes over the surface.
===Momentum Conservation===
The integral form of the momentum equation is 
(see integral equation from previous notes) 
The integral represents the pressure on the sides of the control surface between locations 1 and 2.
===Energy Conservation===
The integral form of the energy equation is 
(see integral equation from previous notes) 
If we assume steady, adiabatic flow with no body forces, the quasi-one-dimensional integral energy equation reduces to
:-\left(-\rho_1 u_1 A_1 + \rho_2 u_2 A_2 \right) = \rho_1 \left(e_1 + \frac{u_1^2}{2}\right)\left(-u_1 A_1\right) + \rho_2 \left(e_2 + \frac{u_2^2}{2}\right)\left(u_2 A_2\right)
Rearranging,
:p_1 u_1 A_1 + \rho_1 u_1 A_1\left(e_1 + \frac{u_1^2}{2}\right) = p_2 u_2 A_2 + \rho_2 u_2 A_2\left(e_2 + \frac{u_2^2}{2}\right)
Dividing by the continuity equation,
:\frac{p_1}{\rho_1} + e_1 + \frac{u_1^2}{2} = \frac{p_2}{\rho_2} + e_2 + \frac{u_2^2}{2}
Recalling that h = e + p/ρ, the energy equation for steady, adiabatic, quasi-one-dimensional flow is
It follows that ht = const along the flow. This is a general result in any steady adiabatic flow, the total enthalpy is constant along a streamline.
===Differential Forms of the Governing Equations===
Now, we will express the conservation laws in differential form. The continuity equation is
:\rho u A = \text{const}
which can be written as
:d(\rho u A) = 0
For the momentum equation and the energy equation, the infinitesimal control volume is given in Figure 5.7 (p. 198, Anderson).
Applying the momentum equation to this control volume gives, 
(eq)
Dropping all second-order terms with products of differentials, we get,
:Ad\rho + Au^2d\rho + \rho u^2 dA + 2\rho u A du = 0
Expanding the continuity equation and multiplying by u, and subtracting the result from the previous equation gives the Euler equation,
:dp = -\rho u du
The differential form of the energy equation is obtained from,
:h+\frac{u^2}{2} = \text{const}
thus,
:dh + udu = 0
It is import to note that the previous equations are exact; however, they have been applied to the simplified model of quasi-one-dimensional flow.
==Area-Velocity Relation==
From the differential form of the continuity equation we can write that,
:\frac{d\rho}{\rho} + \frac{du}{u} + \frac{dA}{A} = 0
From Euler's equation,
:\frac{dp}{\rho} = \frac{dp}{d\rho} \frac{d\rho}{\rho} = -udu
We are considering an adiabatic, inviscid flow (isentropic flow). Therefore, we can write,
:\frac{dp}{d\rho} = \left(\frac{dp}{d\rho}\right)_s = a^2
Substituting we get,
:a^2 \frac{d\rho}{\rho} = -udu
(some stuff)
Substituting we get the area-velocity relation, 
This tells us that
#For M→0, Au = constant. This is the expected result for an incompressible flow.
#For subsonic flow, an increase in velocity (positive du) corresponds to a decrease in area (negative dA), and vice-versa.
#For supersonic flow, an increase in velocity corresponds to an increase in area, and vice-versa.
#For sonic flow, dA/A = 0, which corresponds to a maximum or minimum in the area. For isentropic compression from supersonic to subsonic speeds, it must flow through a convergent-divergent duct, with a throat where the Mach number is equal to 1. Thus we can say that dA/A = 0 always corresponds to a minimum in area.
Notes:
#A convergent-divergent nozzle is sometimes called a de Laval (or Laval) nozzle.
#The derivation of the area velocity relation utilized only the basic conservation relations, so no assumptions were made about the type of gases; therefore, the area-velocity relation is a general relation that holds for real gases and chemically reacting gases, as long as the flow is isentropic.
==Nozzles==
Variable-area duct flow analysis generally requires numerical solutions; however, we can obtain closed-form results for calorically perfect gases.
===Isentropic Subsonic-Supersonic Flow of a Perfect Gas through Nozzles===
Consider a converging-diverging duct with sonic flow at the throat, as shown in Figure 5.12 (p. 203, Anderson).
From the continuity equation:
:\rho^* u^* A^* = \rho u A
From this,
:\frac{A}{A^*} = \frac{\rho^* u^*}{\rho u} = \frac{\rho^*}{\rho_t}\frac{\rho_t}{\rho}\frac{a^*}{u}
Recall that ρt is the stagnation density, and it is constant throughout the flow.
From equation 3.31 in Anderson, we know that: 
(equation)
At sonic condition (M=1), 
(equation)
Recall the definition of M*, equation 3.37 in Anderson: 
(equation)
Squaring equation 5.17 and using equations 3.31, 5.18, and 5.19, it can be shown that the area-Mach number relationis, 
It tells us that:
#M = f(A/A^*,\gamma)
#stuff
#stuff
#stuff
#There are two values of M that correspond to a subsonic and supersonic value
(some stuff)
As the exit pressure is reduced, the normal shock will move closer to the nozzle exit. The shock will stand at the nozzle exit when the pressure immediately behind the shock is at the design Mach number.
(some stuff)
When the pressure downstream of the nozzle, pB, is decreased so that pe6 < pB < pe5, the flow inside the nozzle (downstream of the throat) is fully supersonic. The increase of the exit pressure from pe6 to pB takes place across an oblique shock that is attached to the nozzle exit.
==Diffusers==
==Wave Reflection from a Free Boundary==
==Summary==