Page 369 - Bird R.B. Transport phenomena

§11.4 Use of the Equations of Change to Solve Steady-State Problems 351

to a subsonic value, while both the pressure and the density rise. These changes take place in
an extremely thin region, which may therefore be considered locally one-dimensional and
laminar, and they are accompanied by a very substantial dissipation of mechanical energy.
Viscous dissipation and heat conduction effects are thus concentrated in an extremely small
region of the nozzle, and it is the purpose of the example to explore the fluid behavior there.
For simplicity the shock wave will be considered normal to the fluid streamlines; in practice,
much more complicated shapes are often observed. The velocity, pressure, and temperature
just upstream of the shock can be calculated and will be considered as known for the pur-
poses of this example.
Use the three equations of change to determine the conditions under which a shock wave
is possible and to find the velocity, temperature, and pressure distributions in such a shock
wave. Assume steady, one-dimensional flow of an ideal gas, neglect the dilatational viscosity
к, and ignore changes of fx, k, and C with temperature and pressure.
p

SOLUTION The equations of change in the neighborhood of the stationary shock wave may be simpli-
fied to

Continuity: T pVx = ° (11.4-58)

Motion: Р ^ Т + Ш^Г
Ax Ax 3 Ax \ Ax
AT _ A ( , A T \ , _dp , 4 Jdv \2
Ener y: P C,, - = £ [k %) + v x± + | ^ , x (П.4-60)
V
g
The energy equation is in the form of Eq. / of Table 11.4-1, written for an ideal gas in a steady-
state situation.
The equation of continuity may be integrated to give
pv = p v (11.4-61)
x y x
in which рт and v are quantities evaluated a short distance upstream from the shock.
x
In the energy equation we eliminate pv by use of Eq. 11.4-61 and dp/dx by using the
x
equation of motion to get (after some rearrangement)

We next move the second term on the right side over to the left side and divide the entire
equation by p^v v Then each term is integrated with respect to x to give

CJ + \v\ = —^— 4~ (CT + (fPr)^) + C, (11.4-63)

in which C, is a constant of integration and Pr = C fx/k. For most gases Pr is between 0.65 and
p
0.85, with an average value close to 0.75. Therefore, to simplify the problem we set Pr equal to
J. Then Eq. 11.4-63 becomes a first-order, linear ordinary differential equation, for which the
solution is
C T + \v\ = С + С„ exp[(p,C v,/k)x] (11.4-64)
P p
Since C T + ^v\ cannot increase without limit in the positive x direction, the second integra-
p
tion constant C must be zero. The first integration constant is evaluated from the upstream
u
conditions, so that
C T + \v\ = C T, + \v\ (11.4-65)
P p
Of course, if we had not chosen Pr to be J, a numerical integration of Eq. 11.4-63 would have
been required.