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