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

CONSTANTAREACONSTANTAREA F R I C T I O N CH A D I A B A T I C 12.3 Governing Euations Cons. of mass Cons. of mom. Cons. of energy 2 nd Law of Thermo. (Ideal Gas/Const. c p,c v ) Eqs. of State p =  RT h 2 -h 1 = c p (T 2 – T 1 )  s = c p ln(T 2 /T 1 ) - Rln(p 2 /p 1 ) {1-D, Steady, F Bx =0 only pressure work}

Quasi-One-Dimensional, Steady, F Bx = 0, dW s /dt = 0, dW shear /dt = 0, dW/dt other = 0, effects of gravity = 0, ideal gas, c v, c p is constant Property relations for ideal gas with c v and c p constant Cons. Of Mass Cons. of Momentum Cons. of Energy 2 nd Law of Thermodynamics

+ constant area, adiabatic = Fanno Flow A 1 = A 2 R X only friction No  Q/dm term

Constant area, adiabatic but friction If know: p 1,  1, T 1, s 1, h 1, V 1 and R x Can find: p 2,  2, T 2, s 2, h 2, V 2 properties changed because of R x

(T-s curve)

T-s diagram for Fanno Line Flow s 2 -s 1 = c p ln(T 2 /T 1 ) – Rln(p 2 /p 1 ) p =  RT; p 2 /p 1 =  2 T 2 /(  1 T 1 ); R = c p -c v s 2 -s 1 = c p ln(T 2 /T 1 ) – Rln(p 2 /p 1 ) = c p ln(T 2 /T 1 ) – [Rln(  2 /  1 ) + (c p -c v )ln(T 2 /T 1 )] = – Rln(  2 /  1 ) + c v ln(T 2 /T 1 )  2 V 2 =  1 V 1 ;  2 /  1 = V 1 /V 2 s 2 -s 1 = c v ln(T 2 /T 1 ) – Rln(V 1 /V 2 )

s 2 -s 1 = = c v ln(T 2 /T 1 ) – Rln(V 1 /V 2 ) Energy equation (adiabatic): h + V 2 /2 = h o ; V = (2[h o – h]) 1/2 Ideal Gas & constant c p ; h = c p T V = (2c p [T o – T]) 1/2 -ln[V 1 /V 2 ] = -(1/2)ln[(T o -T 1 )/(T o -T 2 )] = (1/2)ln[(T o -T 2 )/(T o -T 1 )] s 2 -s 1 = c v ln(T 2 /T 1 ) + ½Rln [(T o -T 2 )/(T o -T 1 )] ( p to  to V to T)

T 1, s 1, V 1, … Locus of possible states that can be obtained under the assumptions of Fanno flow: Constant area Adiabatic (h o = h 1 +V 1 2 /2 = c p T o ) x ToTo s-s 1 = c v ln(T/T 1 ) + ½Rln [(T o -T 2 )/(T o -T 1 )]

CONSTANTCONSTANT F R I C T I O N CH TS curve properties A D I A B A T I C 12.3 AREAAREA direction ?

Note – can only move from left to right because s 2 > s 1 non isentropic. (Friction, R x, is what is changing states from 1 to 2 and it is not an isentropic process.) s-s 1 = c v ln(T/T 1 ) + ½Rln [(T o -T 2 )/(T o -T 1 )]

CONSTANTCONSTANT F R I C T I O N CH TS curve properties A D I A B A T I C 12.3 AREAAREA where is sonic ?

Properties at P Where ds/dT = 0 s-s 1 = c v ln(T/T 1 ) + ½Rln [(T o -T 2 )/(T o -T 1 )]

d (s – s 1 ) /dT = ds/dT = 0 ds/dT = c v /T+{(c p -c v )/2}[-1/(T o -T)] = 0 1/T = {(k-1)/2}[1/(T o -T)] T(k-1) = 2(T o – T) s-s 1 = c v ln(T/T 1 ) + ½Rln [(T o -T 2 )/(T o -T 1 )]

T(k-1) = 2(T o – T) h + V 2 /2 = c p T + V 2 /2 = h o = c p T o V = (2c p [T o – T]) 1/2 2(T o – T) = V 2 /c p T(k-1) = V 2 /c p

T(k-1) = V 2 /c p at P V 2 = c p T (k-1) = c p T (c p /c v - c v /c v ) V 2 = (c p /c v )T(c p -c v ) V 2 = kRT For ideal gas and ds = 0: c 2 = kRT Therefore V = c at P, where ds/dT = 0 Sonic condition

CONSTANTCONSTANT F R I C T I O N CH TS curve properties A D I A B A T I C 12.3 AREAAREA how does V change ?

What else can we say about Fanno Line? Sonic Energy equation: h + V 2 /2 = constant = h o =c p T o As h goes down, then V goes up; but h=c p T, so as T goes down V goes up; T o = const T goes down so V goes up T goes up so V goes down Subsonic ? Supersonic ? Tds > 0

CONSTANTCONSTANT F R I C T I O N CH TS curve properties A D I A B A T I C 12.3 AREAAREA 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+V 2 /2 = h o h = c p T 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.

CONSTANTCONSTANT F R I C T I O N CH TS curve properties A D I A B A T I C 12.3 AREAAREA 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

CONSTANTCONSTANT F R I C T I O N CH TS curve properties A D I A B A T I C 12.3 AREAAREA 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

CONSTANTCONSTANT F R I C T I O N CH TS curve properties A D I A B A T I C 12.3 AREAAREA how does  o and p o change

What else can we say about Fanno Line? p o =  o RT o Since T o is a constant (so T o1 = T o2 = T o ) then p o and  o must change the same way. What do  o and p o do?

What else can we say about Fanno Line? What do  o and p o do? s o2 – s o1 = c p ln(T o2 /T o1 ) – Rln(p o2 /p o1 ) s o2 – s o1 = c p ln(T o2 /T o1 ) – Rln(  o2 /  o1 ) Since s o2 > s o1 then p o2 and  o2 must both decrease! 1 1

CONSTANTCONSTANT F R I C T I O N CH TS curve properties A D I A B A T I C 12.3 AREAAREA (summary)

CONSTANTCONSTANT F R I C T I O N CH TS curve properties A D I A B A T I C 12.3 AREAAREA (critical length)

a b ab ? c c M<1 M=0.2 M=0.5

flow is choked For subsonic flow can make adjustments upstream – mass flow decreases M 1 < 1

For supersonic flow adjustments can not be made upstream – so have shock to reduce mass flow M 1 > 1

subsonic, supersonic, shock M>1 M<1

CONSTANTCONSTANT F R I C T I O N CH Fanno Flow A D I A B A T I C 12.3 AREAAREA (examples)

FIND V e and T e Example ~

Basic equations for constant area, adiabatic flow:  V = constant R x + p 1 A –p 2 A = (dm/dt)(V 2 – V 1 ) h 1 + V 1 2 /2 = h 2 + V 2 2 /2 = h o {= constant} s 2 > s 1 p =  RT  h = h 1 – h 2 = c p  T; {T o = constant}  s = s 2 – s 1 = c p ln (T 2 /T 1 ) –R ln(p 2 /p 1 ) Local isentropic stagnation properties T o /T = 1 + [(k-1)/2]M 2 Given: adiabatic, constant area, choked (M e = 1), T o = 25 o C, P o = 101 kPa (abs) Find: V e, T e ; include Ts diagram

Given: adiabatic, constant area, choked (M e = 1), T o = 25 o C, P o = 101 kPa (abs) Find: V e, T e ; include Ts diagram Computing equations: (1) T o /T e = 1 + [(k-1)/2]M e 2 (2) V e = M e c e = M e (kRT e ) 1/2

T o /T e = 1 + [(k-1)/2]M e 2 Equation for local isentropic stagnation property of ideal gas, so assume ideal gas Used the relation: T o = constant from h + V 2 /2 = h 0 = c p T o Assumed that c p is constant; adiabatic flow,  P.E. = 0; 1-D flow (uniform at inlet), steady, dW s /dt = dW shear /dt = 0 V e = M e c e = M e (kRT e ) 1/2 Ideal gas (experimentally shown that sound wave propagates isentropically) ASSUMPTIONS / NOTES for EQUATIONS USED

(1) T o /T e = 1 + [(k-1)/2]M e 2 ; (2) V e = M e c e = M e (kRT e ) 1/2 T o constant so at exit know T o and M e so use (1) to solve for T e Given M e and having solved for T e can use (2) to compute V e T e = 248K,V e = 316 m/s

T-s Diagram (M e = 1)

CONSTANTCONSTANT F R I C T I O N CH Fanno Flow A D I A B A T I C 12.3 AREAAREA (examples)

Example ~ ? P min, V max ? Where do they occur? constant mass flow

Basic equations for constant area, adiabatic flow:  V = constant R x + p 1 A –p 2 A = (dm/dt)(V 2 – V 1 ) h 1 + V 1 2 /2 = h 2 + V 2 2 /2 = h o {= constant} s 2 > s 1 p =  RT  h = h 1 – h 2 = c p  T; {T o = constant}  s = s 2 – s 1 = c p ln (T 2 /T 1 ) –R ln(p 2 /p 1 ) Local isentropic stagnation properties T o /T = 1 + [(k-1)/2]M 2 P2V2P2V2 V1V1

Computing equations: (1) p =  RT (2) dm/dt =  VA (3) T o /T e = 1 + [(k-1)/2]M e 2 (2) V e = M e c e = M e (kRT e ) 1/2 P2V2P2V2 V1V1

p =  RT Ideal gas ( point particles, non-interacting) dm/dt =  VA Conservation of mass ASSUMPTIONS / NOTES for EQUATIONS USED

T o /T e = 1 + [(k-1)/2]M e 2 Equation for local isentropic stagnation property of ideal gas Used the relation: T o = constant from h + V 2 /2 = h 0 = c p T o Assumed that c p is constant; adiabatic flow,  P.E. = 0; 1-D flow (uniform at inlet), steady, dW s /dt = dW shear /dt = 0 V e = M e c e = M e (kRT e ) 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) T o /T e = 1 + [(k-1)/2]M e 2 ; (4) V e = M e c e = M e (kRT e ) 1/2 Know p 1 and T 1 so can solve for  1 from eq.(1)  1 = 0.5 lbm/ft 3 Know dm/dt,  1 and A so from eq. (2) V 1 = 229 ft/sec Know T 1 and V 1 so from eq. (4) M 1 = < 1 subsonic

Subsonic V increases

Computing equations: (1) p =  RT; (2) dm/dt =  VA (3) T o /T e = 1 + [(k-1)/2]M e 2 ; (4) V e = M e c e = M e (kRT e ) 1/2 Can get V 2 from eq. (4) if know T 2 since M 2 = 1 Can get T 2 from eq. (3) if know T o2 (= T o1 = T o ) From eq. (3) T 2 / T 1 = [(1+M 1 2 (k-1)/2)]/[(1+M 2 2 (k-1)/2)] T 2 = 454R From eq. (4) V 2 = 1040 ft/sec Can get p 2 from eq.(1) if know  2 Can get  2 from eq.(2) since given dm/dt and A and have found V 2 ;  2 = lbm/ft 3 Know  2 and T 2 so can use eq. (1) to get p 2, p 2 = 18.5 psia.

T-s Diagram M max, P min

CONSTANTCONSTANT FRICTIONFRICTION Fanno Flow ADIABATICADIABATIC AREAAREA (knowledge of friction factor allows predictions of downstream properties based on knowledge of upstream properties)

fL max /D h = (1-M 2 )/kM 2 + [(k+1)/(2k)] ln{(k+1)M 2 /[2(1+M 2 (k-1)/2] T/T * = (T/T o )(T o /T*) = [(k+1)/2]/[1+(k-1)M 2 /2] V/V * = M(kRT )1/2 /(kRT )1/2 =  /  * = {[(k+1)/2]/[1+(k-1)M 2 /2]} 1/2 p/p * = (  RT)/(  *RT*) = (1/M){[(k+1)/2]/[1+(k-1)M 2 /2]} 1/2 p o /p o * = (p o /p)(p/p*)(p*/p o *) = (1/M) {[2/(k+1)][1+(k-1)M 2 /2]} (k+1)/(2(k-1)) Equations for ideal gas in duct with friction: REMEMBER FLOW IS NOT ISENTROPIC