1
2
U
2
1 þ c P T 1 ¼
1
2
U
2
2 þ c P T 2 K 3
ð3:12fÞ
with the equation of state for an ideal gas.
p 1 ¼ ρ 1 RT 1 and p 2 ¼ ρ 2 RT 2 ,
then the system of Eqs. (3.12d, 3.12e and 3.12f) can be solved in terms of the
constants K 1 , K 2 and K 3 . Solving for U 1 or U 2 by eliminating ρ 1 , p 1 and T 1 (or ρ 2 , p 2
and T 2 ) we have
ρ 1 ¼
K 1
U 1
and Eq. (3.12e) gives
ρ 1 U
2
1 þ ρ 1 RT 1 ¼ K 2 :
Dividing across by ρ 1 and rearranging the latter equation gives
RT 1 ¼
K 2
ρ 1
À U
2
1 :
and using ρ 1 ¼ K 1 /U 1 in this latter equation yields
T 1 ¼
K 2
K 1 R
U 1 À
U
2
1
R
and substituting this result in Eq. (3.12f) we obtain the following quadratic equation
for U 1 ;
c P
R
À
1
2
U
2
1 À
c P
R
K 2
K 1
U 1 þ K 3 ¼ 0;
ð3:13Þ
clearly, the same equation is obtained with U 2 in place of U 1 . The solution of the
latter equation is given by
U 1 , U 2 ¼
c P K 2 =K 1
ð
Þ
c P þ c V
1 Æ
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi
1 À
4K 3 K
2
1 R
2
c 2
P K
2
2
c P
R
À
1
2
s
"
#
¼
c P K 2 =K 1
ð
Þ
c P þ c V
1 Æ Δ
½
Š
ð3:14Þ
where
3.5 A Very Weak Shock
95