FLOW IN A CONSTANT-AREA DUCT WITH FRICTION
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FLOW IN A CONSTANT-AREA DUCT WITH FRICTION
(short ducts/pipes; insulated ducts/pipes)
(not isentropic, BUT constant area, adiabatic)
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Constant Area Duct Flow with Friction
Quasi-one-dimensional flow affected by: no area change, friction, no heat transfer, no shock
friction
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CONSTANT
AREA
FRICTION
CH
ADIABATIC
12.3
Governing Euations
• Cons. of mass
• Cons. of mom.
• Cons. of energy
• 2nd Law of Thermo.
(Ideal Gas/Const. cp,cv)Eqs. of State• p = RT• h2-h1 = cp(T2 – T1)• s = cpln(T2/T1)
- Rln(p2/p1)
{1-D, Steady, FBx=0 only pressure work}
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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
Propertyrelationsfor idealgas withcv and cp
constant
Cons. Of Mass
Cons. of Momentum
Cons. of Energy
2nd Law of Thermodynamics
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+ constant area, adiabatic = Fanno Flow
A1 = A2
RX only friction
No Q/dm term
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Constant area, adiabatic but friction
If know:
p1,1, T1,s1, h1,V1
and Rx
Can find:
p2,2, T2,s2,h2,V2
properties changed because
of Rx
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(T-s curve)
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T-s diagram for Fanno Line Flow
s2-s1 = cpln(T2/T1) – Rln(p2/p1)
p = RT; p2/p1 = 2T2/(1T1); R = cp-cv
s2-s1 = cpln(T2/T1) – Rln(p2/p1) = cpln(T2/T1) – [Rln(2/1) + (cp-cv)ln(T2/T1)] = – Rln(2/1) + cvln(T2/T1)
2V2 = 1V1; 2/1 = V1/V2
s2-s1 = cvln(T2/T1) – Rln(V1/V2)
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s2-s1 = = cvln(T2/T1) – Rln(V1/V2)
Energy equation (adiabatic): h + V2/2 = ho; V = (2[ho – h])1/2
Ideal Gas & constant cp; h = cpT V = (2cp[To – T])1/2
-ln[V1/V2] = -(1/2)ln[(To-T1)/(To-T2)] = (1/2)ln[(To-T2)/(To-T1)]s2-s1 = cvln(T2/T1) + ½Rln [(To-T2)/(To-T1)]
(p to to V to T)
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T1, s1, V1, …
Locus of possible states that can be obtained under the
assumptions of Fanno flow:
Constant areaAdiabatic
(ho = h1+V12/2 = cpTo)
x
To
s-s1 = cvln(T/T1) + ½Rln [(To-T2)/(To-T1)]
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CONSTANT
FRICTION
CH
TS curve propertiesADIABATIC
12.3
AREA
direction ?
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Note – can only move from
left to right because s2
> s1
non isentropic.(Friction, Rx, is what
is changing states from1 to 2 and it is not
an isentropic process.)
s-s1 = cvln(T/T1) + ½Rln [(To-T2)/(To-T1)]
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CONSTANT
FRICTION
CH
TS curve propertiesADIABATIC
12.3
AREA
where is sonic ?
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Properties at P
Where ds/dT = 0
s-s1 = cvln(T/T1) + ½Rln [(To-T2)/(To-T1)]
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d (s – s1) /dT = ds/dT = 0
ds/dT = cv/T+{(cp-cv)/2}[-1/(To-T)] = 0
1/T = {(k-1)/2}[1/(To-T)]
T(k-1) = 2(To – T)
s-s1 = cvln(T/T1) + ½Rln [(To-T2)/(To-T1)]
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T(k-1) = 2(To – T)
h + V2/2 = cpT + V2/2 = ho = cpTo
V = (2cp[To – T])1/2
2(To – T) = V2/cp
T(k-1) = V2/cp
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T(k-1) = V2/cp at PV2 = cp T (k-1) = cp T (cp/cv- cv/cv)
V2 = (cp/cv)T(cp-cv)V2 = kRT
For ideal gas and ds = 0: c2 = kRT
Therefore V = c at P, where ds/dT = 0
Soniccondition
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CONSTANT
FRICTION
CH
TS curve propertiesADIABATIC
12.3
AREA
how does V change ?
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What else can we say about Fanno Line?
Sonic
Energy equation: h + V2/2 = constant = ho=cpToAs h goes down, then V goes up;
but h=cpT, so as T goes down V goes up; To = const
T goes down so V goes up
T goes up so V goes down
Subsonic ?
Supersonic ?
Tds > 0
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CONSTANT
FRICTION
CH
TS curve propertiesADIABATIC
12.3
AREA
how does M change ?
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What else can we say about Fanno Line?
Sonic
What does M do?
T goes down; V goes up
T goes up; V goes down
Subsonic
Supersonic
M = V/[kRT]1/2
h+V2/2 = ho
h = cpT
M increasing
M decreasing
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Note – friction causes an increase in velocity in subsonic flow!
Turns out that pressure dropping rapidly, making up for drag due to friction.
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CONSTANT
FRICTION
CH
TS curve propertiesADIABATIC
12.3
AREA
how does change
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What else can we say about Fanno Line?
Sonic
What does do?
V goes up, then goes down
V goes down, then goes up
Subsonic
Supersonic
V = constant
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CONSTANT
FRICTION
CH
TS curve propertiesADIABATIC
12.3
AREA
how does p change
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What else can we say about Fanno Line?
Sonic
What does p do?
T & goes down, p goes down
T & goes up, p goes up
Subsonic
Supersonic
p = R T
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What else can we say about Fanno Line?
Sonic
in summary
Subsonic
Supersonic
V = constant
p = R T
T goes down; V goes up
T goes up; V goes down
p and decreases
p and increases
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CONSTANT
FRICTION
CH
TS curve propertiesADIABATIC
12.3
AREA
how does o and po change
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What else can we say about Fanno Line?
po = oRTo
Since To is a constant(so To1 = To2 = To)then po and o must change the same way.
What do o and po do?
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What else can we say about Fanno Line?What do o and po do?
so2 – so1 = cpln(To2/To1) – Rln(po2/po1)
so2 – so1 = cpln(To2/To1) – Rln(o2/o1)
Since so2 > so1
then po2 and o2 must both decrease!
1
1
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CONSTANT
FRICTION
CH
TS curve propertiesADIABATIC
12.3
AREA
(summary)
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CONSTANT
FRICTION
CH
TS curve propertiesADIABATIC
12.3
AREA
(critical length)
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ab
a b
?c
c
M<1
M=0.2 M=0.5
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flow is choked
For subsonic flow can make adjustments upstream – mass flow decreases
M1 < 1
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For supersonic flow adjustments can not be made upstream
– so have shock to reduce mass flow
M1 > 1
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subsonic, supersonic, shock
M>1 M>1 M>1
M<1M<1M<1
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CONSTANT
FRICTION
CH
Fanno FlowADIABATIC
12.3
AREA
(examples)
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FIND Ve and Te
Example ~
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Basic equations for constant area, adiabatic flow:V = constantRx + p1A –p2A = (dm/dt)(V2 – V1)h1 + V1
2/2 = h2 + V22/2
= ho {= constant}s2 > s1
p = RTh = h1 – h2 = cp T; {To = constant}s = s2 – s1 = cpln (T2/T1) –R ln(p2/p1)Local isentropic stagnation propertiesTo/T = 1 + [(k-1)/2]M2
Given: adiabatic, constant area, choked (Me = 1), To = 25oC, Po = 101 kPa (abs)
Find: Ve, Te; include Ts diagram
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Given: adiabatic, constant area, choked (Me = 1), To = 25oC, Po = 101 kPa (abs)
Find: Ve, Te; include Ts diagram
Computing equations:
(1) To/Te = 1 + [(k-1)/2]Me2
(2) Ve = Mece = Me(kRTe)1/2
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To/Te = 1 + [(k-1)/2]Me2
Equation for local isentropic stagnation property of ideal gas,so assume ideal gas
Used the relation: To = constant from h + V2/2 = h0 = cpTo
Assumed that cp is constant; adiabatic flow, P.E. = 0; 1-D flow (uniform at inlet), steady, dWs/dt = dWshear/dt = 0
Ve = Mece = Me(kRTe)1/2
Ideal gas (experimentally shown that sound wave propagates isentropically)
ASSUMPTIONS / NOTES for EQUATIONS USED
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(1) To/Te = 1 + [(k-1)/2]Me2; (2) Ve = Mece = Me(kRTe)1/2
To constant so at exit know To and Me so use (1)to solve for Te
Given Me and having solved for Te can use (2) tocompute Ve
Te = 248K, Ve = 316 m/s
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T-s Diagram
(Me = 1)
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CONSTANT
FRICTION
CH
Fanno FlowADIABATIC
12.3
AREA
(examples)
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Example ~
? Pmin, Vmax ?Where do they occur?
constant mass flow
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Basic equations for constant area, adiabatic flow:V = constantRx + p1A –p2A = (dm/dt)(V2 – V1)h1 + V1
2/2 = h2 + V22/2
= ho {= constant}s2 > s1
p = RTh = h1 – h2 = cp T; {To = constant}s = s2 – s1 = cpln (T2/T1) –R ln(p2/p1)Local isentropic stagnation propertiesTo/T = 1 + [(k-1)/2]M2
P2
V2V1
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Computing equations: (1) p = RT(2) dm/dt = VA(3) To/Te = 1 + [(k-1)/2]Me
2
(2) Ve = Mece = Me(kRTe)1/2
P2
V2
V1
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p = RTIdeal gas ( point particles, non-interacting)
dm/dt = VAConservation of mass
ASSUMPTIONS / NOTES for EQUATIONS USED
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To/Te = 1 + [(k-1)/2]Me2
Equation for local isentropic stagnation property of ideal gas
Used the relation: To = constant from h + V2/2 = h0 = cpTo
Assumed that cp is constant; adiabatic flow, P.E. = 0; 1-D flow (uniform at inlet), steady, dWs/dt = dWshear/dt = 0
Ve = Mece = Me(kRTe)1/2
Ideal gas (experimentally shown that sound wave propagates isentropically)
ASSUMPTIONS / NOTES for EQUATIONS USED
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Computing equations: (1) p = RT; (2) dm/dt = VA(3) To/Te = 1 + [(k-1)/2]Me
2; (4) Ve = Mece = Me(kRTe)1/2
Know p1 and T1 so can solve for 1 from eq.(1) 1 = 0.5 lbm/ft3
Know dm/dt, 1 and A so from eq. (2) V1 = 229 ft/sec
Know T1 and V1 so from eq. (4) M1 = 0.201 < 1 subsonic
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SubsonicV increases
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Computing equations: (1) p = RT; (2) dm/dt = VA(3) To/Te = 1 + [(k-1)/2]Me
2; (4) Ve = Mece = Me(kRTe)1/2
Can get V2 from eq. (4) if know T2 since M2 = 1Can get T2 from eq. (3) if know To2 (= To1 = To)From eq. (3)
T2/ T1 = [(1+M12(k-1)/2)]/[(1+M2
2(k-1)/2)]T2 = 454R
From eq. (4) V2 = 1040 ft/secCan get p2 from eq.(1) if know 2
Can get 2 from eq.(2) since given dm/dt and A and have found V2; 2 = 0.110 lbm/ft3
Know 2 and T2 so can use eq. (1) to get p2, p2 = 18.5 psia.
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T-s Diagram
Mmax, Pmin
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CONSTANT
FRICTION
Fanno FlowADIABATIC
AREA
(knowledge of friction factor allows predictions of downstream properties based on knowledge
of upstream properties)
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fLmax/Dh = (1-M2)/kM2 + [(k+1)/(2k)] ln{(k+1)M2/[2(1+M2(k-1)/2]
T/T* = (T/To)(To/T*) = [(k+1)/2]/[1+(k-1)M2/2]
V/V* = M(kRT)1/2/(kRT)1/2 = /* = {[(k+1)/2]/[1+(k-1)M2/2]}1/2
p/p* = (RT)/(*RT*) = (1/M){[(k+1)/2]/[1+(k-1)M2/2]}1/2
po/po* = (po/p)(p/p*)(p*/po*) = (1/M) {[2/(k+1)][1+(k-1)M2/2]}(k+1)/(2(k-1))
Equations for ideal gas in duct with friction:
REMEMBER FLOW IS NOT ISENTROPIC