FLOW IN A CONSTANT-AREA DUCT WITH HEAT EXCHANGE
(Combustion Chambers, Heat Exchangers)
Frictionless Flow in a Constant Area Duct
with Heat Exchange
Q/dm
h1,
s1 ,
Rx= 0
h2 ,
s2 ,
Quasi-one-dimensional flow affected by:
area change, friction, heat transfer, shock
C
O
N
S
T
A
N
T
Governing Equations
• Cons. of mass
• Cons. of mom.
• Cons. of energy
• 2nd Law of Thermo.
Ideal Gas/Const. cp,cv
• p = RT
• h2-h1 = cp(T2 – T1)
• s = cpln(T2/T1)
- Rln(p2/p1)
{1-D, Steady, FBx=0
only pressure work}
A
R
E
A
H
N E
O A
T
F
R E
I X
C C
T H
I A
O N
G
CH E
12-4
Quasi-One-Dimensional, Steady, FBx = 0,
dWs/dt = 0, dWshear/dt = 0, dW/dtother = 0,
effects of gravity = 0, ideal gas*, cv, cp is constant
Cons. Of Mass
Cons. of Momentum
Cons. of Energy
2nd Law of Thermodynamics
Property
relations
for ideal
gas with
cv and cp
constant
Constant area, frictionless,
heat exchange = Rayleigh Flow
No Rx
Constant area, frictionless, heat exchange =
Rayleigh Flow
If know:
p 1,
 1,
T1,
s1,
h1,
V1
and
Q/dm
Can
find:
p2,
 2,
T2,
s2,
h2,
V2
breath
TS curve
C
O
N
S
T
A
N
T
N
O
A
R
E
A
F
R
I
C
T
I
O
N
H
E
A
T
E
X
C
H
A
N
G
E
Frictional, Constant Area,
Adiabatic Flow
Frictionless, Constant Area
with Heat Transfer
T
?
Rayleigh
Isentropic
Flow Line
s
Isentropic Flow
Fanno Line
dA0
No Frictional, Changing Area,
Adiabatic Flow
s2-s1 = cpln(T1/T2)-Rln(p2/p1)
Need p2/p1 in terms of T2 and T1
After manipulation eqs 12.30a – 12.30g
T
s
Rayleigh Line
For the same mass flow, each point on the
curve corresponds to a different value of
q added or taken away.
T
x
s
FLASHBACK - FANNO LINE, ADIABATIC & CONSTANT AREA
BUT FRICTION
To
h1 + V12/2 = h2 + V22/2
hO1 = hO2
x
T1, s1
cpTO1 = cpTO2
TO1 = TO2
s-s1 = cvln(T/T1) + ½Rln [(To-T2)/(To-T1)]
RALEIGH LINE, NOT ADIABATIC & CONSTANT AREA
BUT NO FRICTION
q+h1+V12/2 = h2+V22/2
q = hO2 – hO1
q = cp(TO2-TO1)
The effect of heat
addition is to directly
change the stagnation
(total) temperature
of the flow
TS curve properties
where is sonic ?
C
O
N
S
T
A
N
T
A
R
E
A
N
O
F
R
I
C
T
I
O
N
H
E
A
T
E
X
C
H
A
N
G
E
T
Rayleigh Line
B
dT/ds = 0
s
Properties:
at A – highest s
at B – highest T
A
ds/dT = 0
Want differential form of governing equations.
V = ( +  )(V +  V)
pA – (p+ p)A = ( +  ) A(V +  V)2 - AV2
-p A= V A(V +  V) - AV2
-p A= V A  V
dp/ = -VdV
Momentum: dp/ = -VdV
Ideal gas: p = RT
dp = Rd(T) + RTd()
dp/p = dT/T+ d/
Continuity: V = constant
d/ + dV/V = 0
ds/dT = 0
dp/ = -VdV
Tds = du + pdv (1.10a)
du = d(h-pv) = dh – pdv –vdp
Tds = dh – vdp = dh – dp/
Ideal gas: dh = cpdT
Tds = cpdT – dp/
Tds = cpdT +VdV
ds/dT = cp/T + (V/T)(dV/dT)
ds/dT = cp/T + (V/T)(dV/dT)
Momentum: dp/ = -VdV
Ideal gas: dp/p =d/ + dT/T
Continuity: d/ + dV/V = 0
-VdV/p = d/ + dT/T
p = RT
-VdV/(RT) = d/ + dT/T
-VdV/(RT) = d/ + dT/T
-VdV/(RT) = -dV/V + dT /T
dV( 1/V – V/[RT]) = dT/T
dV/dT = (1/T)/(1/V –V/RT) = 1 / (T/V – V/R)
ds/dT = cp/T + (V/T)(dV/dT)
dV/dT = 1 / (T/V – V/R)
ds/dT = cp/T + (V/T) ( 1/ (T/V – V/R))
ds/dT = 0 = cp/T + (VA/T) ( 1/ (T/VA – VA/R))
-cp = VA / ([T/VA –VA/R])
-cpT/VA + cpVA/R = VA
VA (1 – cp/R) = -cpT/VA
VA2 = -cpT / (1 - cp/R) = cpT / (cp/R - 1)
VA2 = cpRT / (cp – R)
R = cp – cv; cp/cv = k
VA2 = kRT so MA = VA/(kRT)1/2 = 1
At A ds/dt = 0
A
T
s
MA = 1
At B ds/dt = 0
B
T
s
MB = ?
dT/ds = 1/ (ds/dT)
dT/ds = 1/{cp/T + (V/T) ( 1/ (T/V – V/R))}
0 = (TB /VB - VB /R)/[cp/VB – cpVB/(RTB) + VB/TB]
0 = TB /VB - VB /R
TB /VB = VB /R
VB = (RTB)1/2
MB = VB / (kRTB) = (RTB)1/2 / (kRTB)1/2
MB = (1/k)1/2 (~ 0.715 < 1)
At B ds/dt = 0
MB= (1/k)1/2
MB < 1
B
T
s
Subsonic on top. Supersonic on bottom.
What happens as you add heat? subtract heat?
breath
TS curve properties
how does s change with q ?
C
O
N
S
T
A
N
T
A
R
E
A
N
O
F
R
I
C
T
I
O
N
H
E
A
T
E
X
C
H
A
N
G
E
T
S = rev Q/T
s
Entropy, s, always increases with heating and decreases
with cooling. If continue to heat at A, can’t stay on Rayleigh
line since s must increase, so mass flow must change, get
new (lower mass flow) Rayleigh line.
EFFECTS OF HEATING / COOLING ON FLUID PROPERTIES
FOR RAYLEIGH FLOW
Heat addition increases disorder and hence always increases
entropy, whereas cooling decreases disorder and hence decreases
entropy. Independent of Ma.
cp ( To2 – To1 )
Heat addition increases stagnation temperature, cooling decreases
stagnation temperature. Independent of Ma.
TS curve properties
how does V change with q ?
C
O
N
S
T
A
N
T
A
R
E
A
N
O
F
R
I
C
T
I
O
N
H
E
A
T
E
X
C
H
A
N
G
E
To learn more about Rayleigh flow we
need to also consider the energy equation.
How does velocity change with Q?
How does temperature change with Q?
How does Mach number change with Q?
q = h + dh + (V + dV)2/2 – h – V2/2
q = dh + 2VdV/2 = dh + VdV
Ideal gas: dh = cpdT
q = cpdT + VdV
q/(cpT) = dT/T + VdV/(Tcp)
Ideal gas: cp = Rk/(k-1)
q/(cpT) = dT/T + (k-1)VdV/(RkT)
q/(cpT) = dV/V{[V/dV][dT/T] + (k-1)V2/(RkT)}
dV/V = {q/(cpT)} {[V/dV][dT/T] + (k-1)V2/(RkT)}-1
dV/V = {q/(cpT)} {[V/dV][dT/T] + (k-1)V2/(RkT)}-1
dT/dV = T/V – V/R (slide 19)
[V/dV][dT/T] + (k-1)V2/(RkT)
= (V/T)[T/V – V/R] + V2/(RT) – V2/(kRT)
= (V/T)[T/V – V/R + V/R] – Ma
= 1 - Ma
dV/V = {q/(cpT)}[1 – Ma]-1
RAYLEIGH – LINE FLOW
dV/V = {q/(cpT)}[1 – Ma]-1
~0.715
A
B
s is +
Ts diagram for frictionless flow in a
constant-area duct with heat exchange
s is -
breath
TS curve properties
how does T change with q ?
C
O
N
S
T
A
N
T
A
R
E
A
N
O
F
R
I
C
T
I
O
N
H
E
A
T
E
X
C
H
A
N
G
E
To learn more about Rayleigh flow we
need to also consider the energy equation.
How does velocity change with Q?
How does temperature change with Q?
How does Mach number change with Q?
RAYLEIGH – LINE FLOW
A
B
? How does T
change with q ?
Ts diagram for frictionless flow in a
constant-area duct with heat exchange
RAYLEIGH – LINE FLOW
T increasing with q
A
B
Note, between A and B heating
the fluid results in reducing the
temperature! Not surprising if
consider stagnation temperature
and fluid velocity changes in this
region.
Ts diagram for frictionless flow in a
constant-area duct with heat exchange
(from Rayleigh Line)
For supersonic flow, temperature increases with heating
and decreases with cooling.
For subsonic flow and M < 1/(k)1/2, temperature increases
with heating and decreases with cooling.
For subsonic flow and 1/(k)1/2 < M < 1, temperature
decreases with heating and increases with cooling.
breath
TS curve properties
how does M change with q ?
C
O
N
S
T
A
N
T
N
O
A
R
E
A
F
R
I
C
T
I
O
N
H
E
A
T
E
X
C
H
A
N
G
E
To learn more about Rayleigh flow we
need to also consider the energy equation.
M = V/c
How does velocity change with Q?
How does temperature change with Q?
How does Mach number change with Q?
(from Rayleigh Line)
For supersonic flow, M decreases with heating
and increases with cooling.
For subsonic flow, M increases with heating
and decreases with cooling.
<1
M = V/(kRT)1/2
breath
TS curve properties
how does p change with q ?
C
O
N
S
T
A
N
T
A
R
E
A
N
O
F
R
I
C
T
I
O
N
H
E
A
T
E
X
C
H
A
N
G
E
To investigate how p depends on heat transfer, first go back to:
(1st and 2nd laws, ideal gas, constant specific heats)
s2 – s1 = cpln(T2/T1) – Rln(p2/p1)
Keeping p constant
T = Toe(s-so)/cp
(from Rayleigh Line)
For supersonic flow, pressure increases with heating
and decreases with cooling.
For subsonic flow, pressure decreases with heating
and increases with cooling.
(from Rayleigh Line)
For supersonic flow, stagnation pressure decreases
with heating and increases with cooling.
For subsonic flow, stagnation pressure decreases with
heating and increases with cooling.
For supersonic flow,
pressure increases with
heating and decreases with
cooling.
For subsonic flow, pressure
decreases with heating and
increases with cooling.
+
p1A + (dm/dt)V1 = p2A + (dm/dt)V2
For supersonic flow, velocity decreases with heating
and increases with cooling.
For subsonic flow, velocity increases with heating
and decreases with cooling.
For supersonic flow, velocity decreases with heating
and increases with cooling.
For subsonic flow, velocity increases with heating
and decreases with cooling.
+
1V1 = 2V2
For supersonic flow, density increases with heating
and decreases with cooling.
For subsonic flow, density decreases with heating
and increases with cooling.
Rayleigh Flows
Influences can
propagate upstream
M<1
Influences can not
propagate upstream
M=1
M=1
M>1
SUBSONIC
SUPERSONIC
NORMAL SHOCK
T
Rayleigh – Line Flow
s
Steady, frictionless flow of air through a constant area pipe.
T1 = 52oC; p1 = 60 kPa (abs); T2 = 45oC; M2 = 1.0
D = 100mm; dm/dt = 1.42 kg/sec
Find: 1, po1,To1,V1, M1, s2 – s1,
2, po2,To2,V2, Q/dm
Know: T1 = 52oC; p1 = 60 kPa (abs); T2 = 45oC;
M2 = 1.0; D = 100mm; dm/dt = 1.42 kg/sec
Find: 1
(= 0.643 )
Calculating Equations:
h1 + V12/2 + Q/dm = h2 + V22/2
dm/dt = VA
M = V/c;
c = (kRT)1/2
p1 = 1RT1
s2 – s1 = cpln (T2/T1) – Rln(p2/p1)
po/p = [ 1 + {(k – 1)/2}M2]k/(k-1)
To/T = 1 + {(k –
2
1)/2}M
Know: T1 = 52oC; p1 = 60 kPa (abs); T2 = 45oC;
1 = 0.643 kg/m3; M2 = 1.0; D = 100mm; dm/dt = 1.42 kg/sec
Find: V1
(= 281 m/s)
Calculating Equations:
h1 + V12/2 + Q/dm = h2 + V22/2
dm/dt = 1V1A
M = V/c;
c = (kRT)1/2
p1 = 1RT1
s2 – s1 = cpln (T2/T1) – Rln(p2/p1)
po/p = [ 1 + {(k – 1)/2}M2]k/(k-1)
To/T = 1 + {(k –
2
1)/2}M
Know: T1 = 52oC; p1 = 60 kPa (abs); T2 = 45oC; V1 = 281 m/s
1 = 0.643 kg/m3; M2 = 1.0; D = 100mm; dm/dt = 1.42 kg/sec
Find: M1
(= 0.778 < 1)
Calculating Equations:
h1 + V12/2 + Q/dm = h2 + V22/2
dm/dt = 1V1A
M1 = V1/c1; c1 = (kRT1)1/2
p1 = 1RT1
s2 – s1 = cpln (T2/T1) – Rln(p2/p1)
po/p = [ 1 + {(k – 1)/2}M2]k/(k-1)
To/T = 1 + {(k –
2
1)/2}M
T1 = 52oC; p1 = 60 kPa (abs); M1=0.78; T2 = 45oC; V1 = 281 m/s
To1 = 364K, 1 = 0.643 kg/m3; M2 = 1.0; dm/dt = 1.42 kg/sec
Find: To1
(= 364o K)
Calculating Equations:
h1 + V12/2 + Q/dm = h2 + V22/2
dm/dt = 1V1A
M1 = V1/c1; c1 = (kRT1)1/2
p1 = 1RT1
s2 – s1 = cpln (T2/T1) – Rln(p2/p1)
po/p = [ 1 + {(k – 1)/2}M2]k/(k-1)
To1/T1 = 1 + {(k –
2
1)/2}M1
T1 = 52oC; p1 = 60 kPa (abs); M1=0.78; T2 = 45oC; V1 = 281 m/s
To1 = 364K, 1 = 0.643 kg/m3; M2 = 1.0; dm/dt = 1.42 kg/sec
Find: po1
(= 89.5 kPa)
Calculating Equations:
h1 + V12/2 + Q/dm = h2 + V22/2
dm/dt = 1V1A
M1 = V1/c1; c1 = (kRT1)1/2
p1 = 1RT1
s2 – s1 = cpln (T2/T1) – Rln(p2/p1)
po1/p1 = [ 1 + {(k – 1)/2}M12]k/(k-1)
To1/T1 = 1 + {(k –
2
1)/2}M1
T1 = 52oC; p1 = 60 kPa (abs); M1=0.78; T2 = 45oC; V1 = 281 m/s, po1 = 89.5 Pa
To1 = 364K, 1 = 0.643 kg/m3; M2 = 1.0; dm/dt = 1.42 kg/sec
Find: V2
(= 357 m/s)
Calculating Equations:
h1 + V12/2 + Q/dm = h2 + V22/2
dm/dt = 1V1A
M2 = V2/c2; c2 = (kRT2)1/2
p1 = 1RT1
s2 – s1 = cpln (T2/T1) – Rln(p2/p1)
po1/p1 = [ 1 + {(k – 1)/2}M12]k/(k-1)
To1/T1 = 1 + {(k –
2
1)/2}M1
T1=52oC; p1=60 kPa (abs); M1=0.78; T2= 45oC; V1=281 m/s, po1= 89.5Pa
To1=364K, 1=0.643 kg/m3; M2=1.0; dm/dt =1.42 kg/sec; V2=357m/s
Find: 2
(= 0.51 kg/m3)
Calculating Equations:
h1 + V12/2 + Q/dm = h2 + V22/2
dm/dt = 2V2A
M2 = V2/c2; c2 = (kRT2)1/2
p1 = 1RT1
s2 – s1 = cpln (T2/T1) – Rln(p2/p1)
po1/p1 = [ 1 + {(k – 1)/2}M12]k/(k-1)
To1/T1 = 1 + {(k –
2
1)/2}M1
T1=52oC; p1=60 kPa (abs); M1=0.78; T2= 45oC; V1=281 m/s, po1= 89.5Pa,
2=0.51 kg/m3, To1=364K, 1=0.643 kg/m3; M2=1.0; dm/dt =1.42 kg/sec;
V2=357m/s
Find: p2
(= 46.2 kPa)
Calculating Equations:
h1 + V12/2 + Q/dm = h2 + V22/2
dm/dt = 2V2A
M2 = V2/c2; c2 = (kRT2)1/2
p2 = 2RT2
s2 – s1 = cpln (T2/T1) – Rln(p2/p1)
po1/p1 = [ 1 + {(k – 1)/2}M12]k/(k-1)
To1/T1 = 1 + {(k –
2
1)/2}M1
T1=52oC; p1=60 kPa (abs); M1=0.78; T2= 45oC; V1=281 m/s, po1= 89.5Pa,
2=0.51 kg/m3, To1=364K, 1=0.643 kg/m3; M2=1.0; dm/dt =1.42 kg/sec;
V2=357m/s, p2 = 46.kPa
Find: To2
(= 382oK)
Calculating Equations:
h1 + V12/2 + Q/dm = h2 + V22/2
dm/dt = 2V2A
M2 = V2/c2; c2 = (kRT2)1/2
p2 = 2RT2
s2 – s1 = cpln (T2/T1) – Rln(p2/p1)
po1/p1 = [ 1 + {(k – 1)/2}M12]k/(k-1)
To2/T2 = 1 + {(k –
2
1)/2}M2
T1=52oC; p1=60 kPa (abs); M1=0.78; T2= 45oC; V1=281 m/s, po1= 89.5Pa,
2=0.51 kg/m3, To1=364K, 1=0.643 kg/m3; M2=1.0; dm/dt =1.42 kg/sec;
V2=357m/s, p2 = 46.kPa, To2=382oK
Find: po2
(= 87.5 kPa)
Calculating Equations:
h1 + V12/2 + Q/dm = h2 + V22/2
dm/dt = 2V2A
M2 = V2/c2; c2 = (kRT2)1/2
p2 = 2RT2
s2 – s1 = cpln (T2/T1) – Rln(p2/p1)
po2/p2 = [ 1 + {(k – 1)/2}M22]k/(k-1)
To2/T2 = 1 + {(k –
2
1)/2}M2
T1=52oC; p1=60 kPa (abs); M1=0.78; T2= 45oC; V1=281 m/s, po1= 89.5Pa,
2=0.51 kg/m3, To1=364K, 1=0.643 kg/m3; M2=1.0; dm/dt =1.42 kg/sec;
V2=357m/s, p2 = 46.kPa, To2=382oK, po2 =87.5 kPa
Find: Q/dm
(=?)
Calculating Equations:
h1 + V12/2 + Q/dm = h2 + V22/2
dm/dt = 2V2A
M2 = V2/c2; c2 = (kRT2)1/2
p2 = 2RT2
s2 – s1 = cpln (T2/T1) – Rln(p2/p1)
po2/p2 = [ 1 + {(k – 1)/2}M22]k/(k-1)
To2/T2 = 1 + {(k –
2
1)/2}M2
T1=52oC; p1=60 kPa (abs); M1=0.78; T2= 45oC; V1=281 m/s, po1= 89.5Pa,
2=0.51 kg/m3, To1=364K, 1=0.643 kg/m3; M2=1.0; dm/dt =1.42 kg/sec;
V2=357m/s, p2 = 46.kPa, To2=382oK, po2 =87.5 kPa
Find: Q/dm
(= 18 kJ/kg)
Calculating Equations:
h1 + V12/2 + Q/dm = h2 + V22/2
dm/dt = 2V2A
M2 = V2/c2; c2 = (kRT2)1/2
p2 = 2RT2; h1 = cpT1; h2 = cpT2
s2 – s1 = cpln (T2/T1) – Rln(p2/p1)
po2/p2 = [ 1 + {(k – 1)/2}M22]k/(k-1)
To2/T2 = 1 + {(k –
2
1)/2}M2
T1=52oC; p1=60 kPa (abs); M1=0.78; T2= 45oC; V1=281 m/s, po1= 89.5Pa,
2=0.51 kg/m3, To1=364K, 1=0.643 kg/m3; M2=1.0; dm/dt =1.42 kg/sec;
V2=357m/s, p2 = 46.kPa, To2=382oK, po2 =87.5 kPa
Find: Q/dm
(= 18 kJ/kg)
h1 + V12/2 + Q/dm = h2 + V22/2
Q/dm = h02 – ho1
dm/dt = 2V2A
M2 = V2/c2; c2 = (kRT2)1/2
p2 = 2RT2; h01 = cpT01; h02 = cpT02
s2 – s1 = cpln (T2/T1) – Rln(p2/p1)
po2/p2 = [ 1 + {(k – 1)/2}M22]k/(k-1)
To2/T2 = 1 + {(k – 1)/2}M22
T1=52oC; p1=60 kPa (abs); M1=0.78; T2= 45oC; V1=281 m/s, po1= 89.5Pa,
2=0.51 kg/m3, To1=364K, 1=0.643 kg/m3; M2=1.0; dm/dt =1.42 kg/sec;
V2=357m/s, p2 = 46.kPa, To2=382oK, po2 =87.5 kPa; Q/dm = 18 kJ/kg
Find: s2-s1
(= 0.0532 kJ/kg-K)
h1 + V12/2 + Q/dm = h2 + V22/2
Q/dm = h02 – ho1
dm/dt = 2V2A
M2 = V2/c2; c2 = (kRT2)1/2
p2 = 2RT2; h01 = cpT01; h02 = cpT02
s2 – s1 = cpln (T2/T1) – Rln(p2/p1)
po2/p2 = [ 1 + {(k – 1)/2}M22]k/(k-1)
To2/T2 = 1 + {(k – 1)/2}M22
T
s
To reduce labor in solving problems,
it is convenient to derive flow functions
for property ratios in terms of local
Mach numbers and the reference state,
where the local Mach number is 1.
For example p/p* = f(M), /* = f(M)
Momentum equation
Use critical state, M=1, as reference
Want to develop expressions like p/p*, t/T*, ect..
pA –p*A = (dm/dt) (V* - V)
pA + (dm/dt)V = p*A + (dm/dt)V*
p +AVV/A = p* + *A*V*V*/A
p + V2 = p* + *V*2
p {1 + V2/p} = p*{1 + *V*2/p*}
Momentum equation
Use critical state, M=1, as reference
Want to develop expressions like p/p*, t/T*, ect..
p {1 + V2/p} = p*{1 + *V*2/p*}
p = RT; /p = 1/RT
V2/p = V2/(RT) = kV2/(kRT) = kM2
p* = *RT*; */p* = 1/RT*
*V*2/p* = V2/(RT*) = kV*2/(kRT*) = k
p {1 + kM2} = p*{1 + k}
p/p* = {1 + kM2}/{1 + k}
p/p* = (1 + k) / (1 + kM2)
Continuity equation
*/ = V/V* = (Mc)/(M*c*) = Mc/c* = M[T/T*]1/2
Ideal gas
T/T* = (p/p*)(*/)
T/T* = {(1 + k)/(1 + kM2)}{M[T/T*]1/2}
{T/T*}2 = {(1 + k)/(1 + kM2)}2 {M}2{T/T*}
T/T* = {M(1 + k)/(1 + kM2)}2
T/T* = {M(1 + k)/(1 + kM2)}2
*/ = M[T/T*]1/2 = M2(1 + k)/(1 + kM2)
Local Isentropic Stagnation Properties
for Ideal Gas
po/p = [ 1 + {(k – 1)/2}M2]k/(k-1)
To/T = 1 + {(k – 1)/2}M2
2
k/(k-1)
o/ = [ 1 + {(k – 1)/2}M ]
Local Isentropic Stagnation Properties for Ideal Gas
po/p = [ 1 + {(k –
2
To/T = 1 + {(k – 1)/2}M
2
k/(k-1)
1)/2}M ]
p*o/p* = [ 1 + {(k – 1)/2}]k/(k-1)
k/(k-1)
p*/p*o = 1 / [(k+1)/2]
T*o/T* = 1 + {(k – 1)/2}
T*/T*o = 1 / [(k+1)/2]
p/p* = (1 + k) / (1 + kM2)
2
T
/T
=
1
+
{(k
–
1)/2}M
o
T/T* = {M(1 + k)/(1 + kM2)}2 T*/T*
o = 1 / [(k+1)/2]
2
2
*/ = M (1 + k)/(1 + kM )
Dimensionless stagnation temperature:
To/To* = (To/T) (T/T*) (T*/To *)
= (1 + {(k – 1)/2}M2)({M(1 + k)/(1 + kM2)}2)
x (1 / [(k+1)/2])
= {2(k+1)M2(1+ M2 (k-1)/2)} / {(1+kM2)2}
p/p* = (1 + k) / (1 + kM2)
T/T* = {M(1 + k)/(1 + kM2)}2
*/ = M2(1 + k)/(1 + kM2)
po/p = [ 1 + {(k – 1)/2}M2]k/(k-1)
p*/p*o = 1 / [(k+1)/2]k/(k-1)
Dimensionless stagnation pressure:
po/po* = (po/p) (p/p*) (p*/po *)
= ([1 + {(k – 1)/2}M2]k/(k-1)) ((1 + k)/(1 + kM2))
x (1 / [(k+1)/2]k/(k-1) )
= {(1+k)/(1+kM2)}{(2/(k+1))(1+ M2 (k-1)/2)}k/(k-1)
example
Equations for Rayleigh Flow
(steady, one-dimensional, frictionless, ideal gas, constant specific heats)
p/p* = (1 + k) / (1 + kM2)
T/T* = {M(1 + k)/(1 + kM2)}2
*/ = M2(1 + k)/(1 + kM2)
po/po* = {(1+k)/(1+kM2)}{(2/(k+1))(1+ M2 (k-1)/2)}k/(k-1)
To/To* = {2(k+1)M2(1+ M2 (k-1)/2)} / {(1+kM2)2}
po/p = [ 1 + {(k – 1)/2}M2]k/(k-1)
To/T = 1 + {(k – 1)/2}M2
M = V/c;
c = (kRT)1/2; p = RT; h = cpT
s2 – s1 = cpln (T2/T1) – Rln(p2/p1)
h1 + V12/2 + Q/dm = h2 + V22/2
Q/dm = h02 – ho1
Equations for Rayleigh Flow
(steady, one-dimensional, frictionless, ideal gas, constant specific heats)
M2 = 0.90
To1 = 333K
p2 = ?
p1 = 1.1 Mpa (abs)
M1 = 0.5
h1 + V12/2 + Q/dm = h2 + V22/2
Q/dm = h02 – ho1
dm/dt = 2V2A
M2 = V2/c2;
c2 = (kRT2)1/2
p2 = 2RT2; h01 = cpT01; h02 = cpT02
s2 – s1 = cpln (T2/T1) – Rln(p2/p1)
po2/p2 = [ 1 + {(k – 1)/2}M22]k/(k-1)
To2/T2 = 1 + {(k – 1)/2}M22
Equations for Rayleigh Flow
(steady, one-dimensional, frictionless, ideal gas, constant specific heats)
To1 = 333K
p1 = 1.1 Mpa (abs)
M1 = 0.5
M2 = 0.90
p2 = (p2/p*) (p*/p1)(p1)
p2 = (2.4/2.134)(1.35/2.4)(1.1 Mpa)
p2 = 696 kPa
p2/p* = (1 + k) / (1 + kM22)
p1/p* = (1+k) / (1 + kM12)
p*/p1 = (1 + kM12) / (1 + k)
To1 = 333K
p1 = 1.1 Mpa (abs)
M1 = 0.5
M2 = 0.90
p2 = 696 kPa
example
Find: T2, p2, 2, M2, po2, To2 and Q/dt
First find: T1, p1, 1, M1, po1, To1,
And T*, p*, *, To*, po*
Find: T2, p2, 2, M2, po2, To2 and Q/dt
p/p* = (1 + k) / (1 + kM2)
T/T* = {M(1 + k)/(1 + kM2)}2
*/ = M2(1 + k)/(1 + kM2)
po/po* = {(1+k)/(1+kM2)}{(2/(k+1))(1+ M2 (k-1)/2)}k/(k-1)
To/To* = {2(k+1)M2(1+ M2 (k-1)/2)} / {(1+kM2)2}
po/p = [ 1 + {(k – 1)/2}M2]k/(k-1)
To/T = 1 + {(k – 1)/2}M2
M = V/c;
c = (kRT)1/2; p = RT; h = cpT
s2 – s1 = cpln (T2/T1) – Rln(p2/p1)
h1 + V12/2 + Q/dm = h2 + V22/2
Q/dm = h02 – ho1
Given: T1, p1, M1, V2, Area
Find: T2, p2, 2, M2, po2, To2 and Q/dt
First find: T1, p1, 1, M1, V1, po1, To1, T*, p*, V*, To*, po*
1 = p1/(RT1) = 0.661 lbm/ft3
V1 = M1c1 = M1(kRT1)1/2 = 420.6 ft/sec
To1/T1 = 1 + {(k – 1)/2}M12 ; To1 = 832.7o R
po1 /p1 = [ 1 + {(k – 1)/2}M12]k/(k-1) ; po1 = 212.9 psia
Given: T1, p1, M1, V2, Area
Find: T2, p2, 2, M2, po2, To2 and Q/dt
First find: T1, p1, 1, M1, V1, po1, To1, T*, p*, V*, To*, po*
p1 /p* = (1 + k) / (1 + kM12); p* = 93.8 psia
T1/T* = {M1(1 + k)/(1 + kM12)}2; T* = 2001oR
*/1 = M2(1 + k)/(1 + kM2) {get  from 2 = 1(V1/V2)}
V1 /V* = M12(1 + k)/(1 + kM12) ; V* = 2193 ft/sec
po1 /po* = {(1+k)/(1+kM12)}{(2/(k+1))(1+ M12(k-1)/2)}k/(k-1)
po* = 177.6 psia
To1/T0* = {2(k+1)M12(1+ M12(k-1)/2)} / {(1+kM12)2}
To* = 2400o R
Given: T1, p1, M1, V2, Area Find: T2, p2, 2, M2, po2, To2 and Q/dt
Found: T1, p1, 1, M1, V1, po1, To1, T*, p*, V*, To*, po*
Find: M2,T2, p2, 2, po2, To2 and Q/dt
V2 /V* = M22(1 + k)/(1 + kM22) ; M2= 0.9
T2/T* = {M2(1 + k)/(1 + kM22)}2; T2 = 2049o R
p2 /p* = (1 + k) / (1 + kM22); p2 = 105.5 psia
po2 /po* = {(1+k)/(1+kM22)}{(2/(k+1))(1+ M22(k-1)/2)}k/(k-1)
po2 = 178.3 psia
To2/T0* = {2(k+1)M22(1+ M22(k-1)/2)} / {(1+kM22)2}
To2 = 2381o R
2 = 1(V1/V2) = 0.139 lbm/ft3
Given: T1, p1, M1, V2, Area Find: T2, p2, 2, M2, po2, To2 and Q/dt
Found: T1, p1, 1, M1, V1, po1, To1, T*, p*, V*, To*, po*
Find: M2,T2, p2, 2, po2, To2 and Q/dt
Q/dt = (dm/dt)(Q/dm)
= (1V1A )(Q/dm)
h1 + V12/2 + Q/dm = h2 + V22/2
Q/dm = h02 – ho1 = cp(To2 – T01)
Q/dt = 0.66[lbm/ft2]x420.6[ft/sec]x0.5[ft2]
x[0.24[BTU/lbm-R]x(2380-833)[R]
=5.16x104[BTU/sec]
p1 = 200 psia, T1 = 818o R, M1 = 0.3, T01 = 832.7o R, Po1 = 212.9 psia
p2 = 106 psia, T2 = 2050o R, M2 = 0.9, T02 = 2380o R, Po2 = 1777.6 psia