Page 90 - Computational Fluid Dynamics for Engineers

Problems 75

and taking the dot product of Eq. (2.4.15) with (P2.4.1) and noting that V has
2 2 2
the same direction as ds, show that, with V = v? + v + w , the left-hand side
of the resulting expressicn can be written as
2
- - dV fv "
(P2.4.2)
(V • V)V 'ds = V'—-ds = VdV = dl —

and Eq. (2.4.16) follows.

2-5. Derive Eq. (2.4.28).
Hint: Start with Eq. (2.4.26b). Assuming u : v to be small relative to V^, write

-K- = Voo + U, — = V (P2.5.1)
ox oy
Substitute Eq. (P2.5.1) into Eq. (2.4.24)

a +——[(V 00 + u) + v]=a OQ + T/ (P2.5.2)
2 °°
2
expand a^/a by the binomial theorem

l 1 2 u ii 2 + V 2
°° ~ = 1 ' V ^ M I 2-f- + 2 (P2.5.3)
~~V
and use the assumptions in Eqs. (2.4.27).

2-6. Starting from the full-potential equation, Eq. (2.4.26), derive the transonic
small disturbance (TSD) equation written in non-conservative form,

XX I <pyy = 0 (P2.6.1)
and, in conservative form,

(7 + ) M £ , 2
l
(1 - Ml)^ - 9 = 0 (P2.6.2)
2 Voo * + • 'vv
Hint: Use Eqs. (P2.5.1) and (P2.5.3), with the following small disturbance ap-
proximations:
u
< 1
Vo,

2-7. (a) With the definitions in Eq. (2.5.5), show that Eqs. (2.5.1) to (2.5.3)
can be written in the form given by Eqs. (2.5.6) to (2.5.8).
1 f f
(b) Show that with q' in Eq. (2.5.9) corresponding to u ', v and p , the continuity
and momentum equations given by Eqs. (2.5.6) to (2.5.8) can be written as