FLOW IN A CONSTANT-AREA DUCT WITH HEAT EXCHANGE

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Presentation transcript:

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, 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

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}

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.

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

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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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

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

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

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

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)]

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

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

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

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

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

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)

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/{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 (~ 0.715 < 1)

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

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.

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 ? H E A N T O C O F X N R

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

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

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

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 ? 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

RAYLEIGH – LINE FLOW T increasing with q 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

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

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?

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

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

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.

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.

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.

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.

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

T Rayleigh – Line Flow s

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

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 (= 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 – 1)/2}M2

To/T = 1 + {(k – 1)/2}M2 Find: V1 (= 281 m/s) Calculating Equations: 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 – 1)/2}M2

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 = 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 – 1)/2}M2

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 = 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 – 1)/2}M12

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 = 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 – 1)/2}M12

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 = 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 – 1)/2}M12

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: 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 – 1)/2}M12

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: 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 – 1)/2}M12

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: 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 – 1)/2}M22

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: 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 – 1)/2}M22

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

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

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

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-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)

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*}

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}

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

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)

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]

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}

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)

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) 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

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)

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

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

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 = 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

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(2380-833)[R] =5.16x104[BTU/sec]

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