student handout 24 2014
TRANSCRIPT
CHEE 3363Spring 2014Handout 24
Reading: Fox, Chapter 11
2
)R
3
p = ρRT
cv ≡
(
∂u
∂T
)
v
h ≡ u + p/ρ
)c cp
c cp
4
)
5
∆S ≡
∫
rev
δQ
T
∫
δQ
T≤ 0
T ds = du + p dv
)
6
= d(h − pv) + p dv = dh − p dv + v dp + p dv = dh − v dp
ds =
du
T+
p
Tdv = cv
dT
T+ R
dv
v
ds =
dh
T−
v
Tdp = cp
dT
T− R
dp
p
)c cp
7
s2 − s1 = cv ln
T2
T1
+ R ln
v2
v1
s2 − s1 = cp ln
T2
T1
− R ln
p2
p1
s2 − s1 = cv ln
p2
p1
+ cp ln
v2
v1
= cv ln
p2v2
p1v1
+ (cp − cv) ln
v2
v1
= cv ln
p2
p1
+ cv ln
v2
v1
+ (cp − cv) ln
v2
v1
s2 − s1 = cv ln
T2
T1
+ R ln
v2
v1
)
8
Tvk−1
=
T
ρk−1= constant
Tp1−k
k = constant
pvk
=
p
ρk
= constant
M =
V
c =
stationary frame inertial frame moving with sound wave
c
ρ�Vx = 0
pressure p
ρ + ρ�x
pressure +
9
∂
∂t
∫
CV
ρ dV +
∫
CS
ρv · dA = 0 (−ρcA) + [(ρ + dρ)(c − dVx)A] = 0
x
ρ =
)
FSx+ FBx
=
∂
∂t
∫
CV
Vxρ dV +
∫
CS
Vxρv · dA
FSx= pA − (p + dp)A = −A dp
∫
CS
Vxρv · dA = c(−ρcA) + (c − dVx) [(ρ + dρ)(c − dVx)A]
E
dp =
(
∂p
∂ρ
)
s
dρ +
(
∂p
∂s
)
ρ
ds =
(
∂p
∂ρ
)
s
dρ
Ev =
dp
dρ/ρ= ρ
dp
dρ
pvk
=
p
ρk
= constant
T ρ s V) V
p0 T0 ρ0 0 0 s0
s = s s02 = s2)
s = s = s02 = s2
8
p
ρ+
V 2
2
+ gz = constant
p0 = p +
1
2
ρV 2
x
(−ρVxA) + [(ρ + dρ)(Vx + dVx)(A + dA)] = 0
∂
∂t
∫
CV
ρ dV +
∫
CS
ρv · dA = 0
FSx+ FBx
=
∂
∂t
∫
CV
Vxρ dV +
∫
CS
Vxρv · dA
FSx= dRx + pA − (p + dp)(A + dA)
(
p +
dp
2
)
dA
FSx=
(
p +
dp
2
)
dA + pA − (p + dp)(A + dA) = −A dp
−A dp = Vx(ρVxA) + (Vx + dVx) [(ρ + dρ)(Vx + dVx)(A + dA)]
−A dp = (−Vx + Vx + dVx)(ρVxA)
p = Cρk ρ = p1/kC−1/kpvk
=
p
ρk
= constant
−
∫
0
V
d
(
V 2
2
)
= C1/k
∫
p0
p
p−1/kdp
C1/k
= p1/k/ρ
p0
p=
[
1 +
k − 1
k
ρV 2
2p
]
k/(k−1)
p = ρRT
c =
√
kRT
pvk
=
p
ρk
= constant
p0
p=
(
ρ0
ρ
)
k
ρ0
ρ=
(
p0
p
)
1/k
T0
T=
p0
p
ρ
ρ0
=
p0
p
(
p0
p
)
−1/k
=
(
p0
p
)
(k−1)/k
M
M p T ρ*)
p0
p∗=
[
k + 1
2
]
k/(k−1)
T0
T ∗
=
k + 1
2
ρ0
ρ∗=
[
k + 1
2
]
1/(k−1)
V ∗
= c∗ =
√
kRT ∗
=
√
2k
k + 1
RT0