1. FLOW IN A CONSTANT-AREA DUCT WITH HEAT EXCHANGE
(Combustion Chambers, Heat Exchangers)

2. Frictionless Flow in a Constant Area Duct
with Heat Exchange
Q/dm
h1,
s1,
h2,
s2,
Rx= 0
Friction generates heat so can be adiabatic or isothermal / we’ll just look at adiabatic
Want to generate Ts diagram and provide insight as to how properties change along Fanno line -
Quasi-one-dimensional flow affected by:
area change, friction, heat transfer, shock

3. H N E C O A O T N F S R T I X A C C T N G R CH E 12-4
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}

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
Ws = shaft work
Wshear = work done by shear stresses at the control surface
dS >= dQ/T
Tds >= dQ/m = (dQ/dt) / (dm/dt)
In any chemical reaction process the composition of the reactants will be different than the products – hardly ideal gas yet in jet-engine combustion, fuel to air ratio is small so perfect gas assumption often reasonable.

5. Constant area, frictionless, heat exchange = Rayleigh Flow
No Rx

6. Constant area, frictionless, heat exchange = Rayleigh Flow
If know:
p1,
1,
T1,
s1,
h1, V1
and
Can find:
p2,
2, T2,
s2,
h2, V2
Friction is what is changing properties
Even if reversible (frictionless but heat would have to be added reversibly, can not calculate integral)
Also dQ/dt can be positive or negative so s can increase or decrease
Q/dm

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8. TS curve H E A T X C N G N O F R I C T C O N S T A A R E

9. Frictionless, Constant Area
Frictional, Constant Area,
Adiabatic Flow
Frictionless, Constant Area
with Heat Transfer ? T
Rayleigh Line
Isentropic Flow s
Isentropic Flow
Fanno Line
dA0
No Frictional, Changing Area,
Adiabatic Flow

10. 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

11. x s T 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

12. s-s1 = cvln(T/T1) + ½Rln [(To-T2)/(To-T1)]
FLASHBACK - FANNO LINE, ADIABATIC & CONSTANT AREA
BUT FRICTION To
h1 + V12/2 = h2 + V22/2
hO1 = hO2
cpTO1 = cpTO2
TO1 = TO2 x
T1, s1
s-s1 = cvln(T/T1) + ½Rln [(To-T2)/(To-T1)]

13. addition is to directly
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

14. TS curve properties where is sonic ? H E A N T O C O F X N R C S A I TG
where is sonic ? N O F R I C T C O N S T A A R E

15. s T Properties: at A – highest s at B – highest T Rayleigh Line B
dT/ds = 0 s
Properties:
at A – highest s
at B – highest T A
ds/dT = 0

16. pA – (p+ p)A = ( +  ) A(V +  V)2 - AV2
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

17. Momentum: dp/ = -VdV
Ideal gas: p = RT
dp = Rd(T) + RTd()
dp/p = dT/T+ d/
Continuity: V = constant
d/ + dV/V = 0

18. du = d(h-pv) = dh – pdv –vdp Tds = dh – vdp = dh – dp/
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)

19. 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)

20. 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

21. At A ds/dt = 0 A T s
MA = 1

22. At B ds/dt = 0 B T s
MB = ?

23. dT/ds = 1/{cp/T + (V/T) ( 1/ (T/V – V/R))}
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 (~ < 1)

24. At B ds/dt = 0 T s Subsonic on top. Supersonic on bottom.
MB= (1/k)1/2
MB < 1 B T s
Subsonic on top. Supersonic on bottom.
What happens as you add heat? subtract heat?

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26. TS curve properties how does s change with q ? H E A N T O C O F X N R

27. S = rev Q/T 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.

28. 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.

29. TS curve properties how does V change with q ? H E A N T O C O F X N R

30. 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?

31. 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

32. dT/dV = T/V – V/R (slide 19)
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

33. dV/V = {q/(cpT)}[1 – Ma]-1
RAYLEIGH – LINE FLOW
~0.715
dV/V = {q/(cpT)}[1 – Ma]-1 A B
s is +
s is -
Ts diagram for frictionless flow in a
constant-area duct with heat exchange

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35. TS curve properties how does T change with q ? H E A N T O C O F X N R

36. 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?

37. RAYLEIGH – LINE FLOW ? How does T change with q ?B
? How does T
change with q ?
Ts diagram for frictionless flow in a
constant-area duct with heat exchange

38. RAYLEIGH – LINE FLOW T increasing with qB
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

39. For supersonic flow, temperature increases with heating
(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.

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41. TS curve properties how does M change with q ? H E A N T O C O F X N R

42. 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?

43. For supersonic flow, M decreases with heating
(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

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45. TS curve properties how does p change with q ? H E A N T O C O F X N R

46. 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

48. For supersonic flow, pressure increases with heating
(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.

49. For supersonic flow, stagnation pressure decreases
(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.

50. p1A + (dm/dt)V1 = p2A + (dm/dt)V2
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.

51. 1V1 = 2V2 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.

52. Rayleigh Flows Influences can propagate upstream Influences can not
SUBSONIC SUPERSONIC NORMAL SHOCK

53. T
Rayleigh – Line Flow s

54. Find: 1, po1,To1,V1, M1, s2 – s1, 2, po2,To2,V2, Q/dm
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

55. To/T = 1 + {(k – 1)/2}M2 Find: 1 (= 0.643 ) Calculating Equations:
Know: T1 = 52oC; p1 = 60 kPa (abs); T2 = 45oC;
M2 = 1.0; D = 100mm; dm/dt = 1.42 kg/sec
Find: 1 (= )
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 – 1)/2}M2

56. To/T = 1 + {(k – 1)/2}M2 Find: V1 (= 281 m/s) Calculating Equations:
Know: T1 = 52oC; p1 = 60 kPa (abs); T2 = 45oC;
1 = 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 – 1)/2}M2

57. To/T = 1 + {(k – 1)/2}M2 Find: M1 (= 0.778 < 1)
Know: T1 = 52oC; p1 = 60 kPa (abs); T2 = 45oC; V1 = 281 m/s
1 = kg/m3; M2 = 1.0; D = 100mm; dm/dt = 1.42 kg/sec
Find: M1 (= < 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 – 1)/2}M2

58. To1/T1 = 1 + {(k – 1)/2}M12 Find: To1 (= 364o K)
T1 = 52oC; p1 = 60 kPa (abs); M1=0.78; T2 = 45oC; V1 = 281 m/s
To1 = 364K, 1 = 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 – 1)/2}M12

59. To1/T1 = 1 + {(k – 1)/2}M12 Find: po1 (= 89.5 kPa)
T1 = 52oC; p1 = 60 kPa (abs); M1=0.78; T2 = 45oC; V1 = 281 m/s
To1 = 364K, 1 = 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 – 1)/2}M12

60. To1/T1 = 1 + {(k – 1)/2}M12 Find: V2 (= 357 m/s)
T1 = 52oC; p1 = 60 kPa (abs); M1=0.78; T2 = 45oC; V1 = 281 m/s, po1 = 89.5 Pa
To1 = 364K, 1 = 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 – 1)/2}M12

61. To1/T1 = 1 + {(k – 1)/2}M12 Find: 2 (= 0.51 kg/m3)
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:  (= 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 – 1)/2}M12

62. To1/T1 = 1 + {(k – 1)/2}M12 Find: p2 (= 46.2 kPa)
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: p (= 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 – 1)/2}M12

63. To2/T2 = 1 + {(k – 1)/2}M22 Find: To2 (= 382oK) Calculating Equations:
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: To (= 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 – 1)/2}M22

64. To2/T2 = 1 + {(k – 1)/2}M22 Find: po2 (= 87.5 kPa)
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: po (= 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 – 1)/2}M22

65. To2/T2 = 1 + {(k – 1)/2}M22 Find: Q/dm (=?) Calculating Equations:
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 – 1)/2}M22

66. To2/T2 = 1 + {(k – 1)/2}M22 Find: Q/dm (= 18 kJ/kg)
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 – 1)/2}M22

67. To2/T2 = 1 + {(k – 1)/2}M22 Find: Q/dm (= 18 kJ/kg)
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

68. To2/T2 = 1 + {(k – 1)/2}M22 Find: s2-s1 (= 0.0532 kJ/kg-K)
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-s (= 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

69. T s

70. 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)

71. pA –p*A = (dm/dt) (V* - V) pA + (dm/dt)V = p*A + (dm/dt)V*
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*}

72. p {1 + V2/p} = p*{1 + *V*2/p*} p = RT; /p = 1/RT
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}

73. p/p* = (1 + k) / (1 + kM2) T/T* = (p/p*)(*/)
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

74. Local Isentropic Stagnation Properties
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
o/ = [ 1 + {(k – 1)/2}M2]k/(k-1)

75. Local Isentropic Stagnation Properties for Ideal Gas
po/p = [ 1 + {(k – 1)/2}M2]k/(k-1)
To/T = 1 + {(k – 1)/2}M2
p*o/p* = [ 1 + {(k – 1)/2}]k/(k-1)
p*/p*o = 1 / [(k+1)/2]k/(k-1)
T*o/T* = 1 + {(k – 1)/2}
T*/T*o = 1 / [(k+1)/2]

76. Dimensionless stagnation temperature: To/To* = (To/T) (T/T*) (T*/To *)
p/p* = (1 + k) / (1 + kM2)
To/T = 1 + {(k – 1)/2}M2
T*/T*o = 1 / [(k+1)/2]
T/T* = {M(1 + k)/(1 + kM2)}2
*/ = M2(1 + k)/(1 + kM2)
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}

77. Dimensionless stagnation pressure: po/po* = (po/p) (p/p*) (p*/po *)
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)

78. example

79. 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

80. 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 = ?
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

81. 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)

82. M2 = 0.90
p2 = 696 kPa
To1 = 333K
p1 = 1.1 Mpa (abs)
M1 = 0.5

83. example

84. Find: T2, p2, 2, M2, po2, To2 and Q/dt
First find: T1, p1, 1, M1, po1, To1,
And T*, p*, *, To*, po*

85. 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

86. 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) = lbm/ft3
V1 = M1c1 = M1(kRT1)1/2 = ft/sec
To1/T1 = 1 + {(k – 1)/2}M12 ; To1 = 832.7o R
po1 /p1 = [ 1 + {(k – 1)/2}M12]k/(k-1) ; po1 = psia

87. 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* = psia
To1/T0* = {2(k+1)M12(1+ M12(k-1)/2)} / {(1+kM12)2}
To* = 2400o R

88. Find: M2,T2, p2, 2, po2, To2 and Q/dt
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 = psia
po2 /po* = {(1+k)/(1+kM22)}{(2/(k+1))(1+ M22(k-1)/2)}k/(k-1)
po2 = psia
To2/T0* = {2(k+1)M22(1+ M22(k-1)/2)} / {(1+kM22)2}
To2 = 2381o R
2 = 1(V1/V2) = lbm/ft3

89. Find: M2,T2, p2, 2, po2, To2 and Q/dt
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( )[R]
=5.16x104[BTU/sec]

90. p1 = 200 psia, T1 = 818o R, M1 = 0.3, T01 = 832.7o R, Po1 = 212.9 psia