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)
Constant Area Duct Flow with Friction
Quasi-one-dimensional flow affected by: no area change, friction, no heat transfer, no shock
friction
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}
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
+ constant area, adiabatic = Fanno Flow
A1 = A2
RX only friction
No Q/dm term
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
(T-s curve)
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)
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)
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)]
CONSTANT
FRICTION
CH
TS curve propertiesADIABATIC
12.3
AREA
direction ?
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)]
CONSTANT
FRICTION
CH
TS curve propertiesADIABATIC
12.3
AREA
where is sonic ?
Properties at P
Where ds/dT = 0
s-s1 = cvln(T/T1) + ½Rln [(To-T2)/(To-T1)]
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)]
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
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
CONSTANT
FRICTION
CH
TS curve propertiesADIABATIC
12.3
AREA
how does V change ?
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
CONSTANT
FRICTION
CH
TS curve propertiesADIABATIC
12.3
AREA
how does M change ?
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
Note – friction causes an increase in velocity in subsonic flow!
Turns out that pressure dropping rapidly, making up for drag due to friction.
CONSTANT
FRICTION
CH
TS curve propertiesADIABATIC
12.3
AREA
how does change
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
CONSTANT
FRICTION
CH
TS curve propertiesADIABATIC
12.3
AREA
how does p change
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
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
CONSTANT
FRICTION
CH
TS curve propertiesADIABATIC
12.3
AREA
how does o and po change
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?
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
CONSTANT
FRICTION
CH
TS curve propertiesADIABATIC
12.3
AREA
(summary)
CONSTANT
FRICTION
CH
TS curve propertiesADIABATIC
12.3
AREA
(critical length)
ab
a b
?c
c
M<1
M=0.2 M=0.5
flow is choked
For subsonic flow can make adjustments upstream – mass flow decreases
M1 < 1
For supersonic flow adjustments can not be made upstream
– so have shock to reduce mass flow
M1 > 1
subsonic, supersonic, shock
M>1 M>1 M>1
M<1M<1M<1
CONSTANT
FRICTION
CH
Fanno FlowADIABATIC
12.3
AREA
(examples)
FIND Ve and Te
Example ~
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
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
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
(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
T-s Diagram
(Me = 1)
CONSTANT
FRICTION
CH
Fanno FlowADIABATIC
12.3
AREA
(examples)
Example ~
? Pmin, Vmax ?Where do they occur?
constant mass flow
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
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
p = RTIdeal gas ( point particles, non-interacting)
dm/dt = VAConservation of mass
ASSUMPTIONS / NOTES for EQUATIONS USED
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
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
SubsonicV increases
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.
T-s Diagram
Mmax, Pmin
CONSTANT
FRICTION
Fanno FlowADIABATIC
AREA
(knowledge of friction factor allows predictions of downstream properties based on knowledge
of upstream properties)
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