 V 1 2 / 2 + p 1 /  + gz 1 =  V 2 2 /2 + p 2 /  + gz 2 + h lT h lT = h l + h m HEADLOSSHEADLOSS

Convenient to break up energy losses, h lT, in fully developed pipe flow to major loses, h l, due to frictional effects along the pipe and minor losses, h lm, associated with entrances, fittings, changes in area,… Minor losses not necessarily < Major loss, h l, due to pipe friction.

Minor losses traditionally calculated as: h lm = KV 2 /2 (K for inlets, exits, enlargements and contractions) where K is the loss coefficient or h lm = (C pi – C p )V 2 /2 (C pi & C p for diffusers) where C p is the pressure recovery coefficient or h lm = f(L e /D)V 2 /2 (L e for valves, fittings, pipe bends) where L e is the equivalent length of pipe. Both K and L e must be experimentally determined and will depend on geometry and Re, u avg D/. At high flow rates weak dependence on Re.

h lT = h l + h m ; h lm = KV 2 /2 inlets, sudden enlargements & contractions; gradual contractions and exits  V 1avg 2 / 2 + p 1 /  + gz 1 =  V 2avg 2 /2 + p 2 /  + gz 2 + h lT

Minor losses due to inlets: h lm =  p/  = K(V 2 /2); V 2 = mean velocity in pipe If K=1,  p =  V 2 /2

 V 1avg 2 / 2 + p 1 /  + gz 1 =  V 2avg 2 /2 + p 2 /  + gz 2 + h lT h lT = h l + h m

Head is lost because of viscous dissipation when flow is slowed down (2-3) and in violent mixing in the separated zones For a sharp entrance ½ of the velocity head is lost at the entrance!

vena contracta unconfined mixing as flow decelerates separation K = 0.78 K = 0.04 r/D > 0.15 r D

h lT = h l + h lm ; h lm = KV 2 /2 inlets, sudden enlargements & contractions; gradual contractions and exits  V 1 2 / 2 + p 1 /  + gz 1 =  V 2 2 /2 + p 2 /  + gz 2 + h lT

h lm head losses are primarily due to separation Energy is dissipated deceleration after separation leading to violent mixing in the separated zones Minor losses due to sudden area change: h lm =  p/  = K(V 2 /2); V 2 = faster mean velocity pipe

NOTE SOME BOOKS (Munson at al.): h lm = K V 2 /(2g) our H lm !!!

AR < 1 V2V2 V1V1 h lm = ½ KV 2 fastest

Why is K contraction and K expansion = 0 at AR =1? AR < 1

h lT = h l + h lm ; h lm = KV 2 /2 inlets, sudden enlargements & contractions; gradual contractions and exits  V 1avg 2 / 2 + p 1 /  + gz 1 =  V 2avg 2 /2 + p 2 /  + gz 2 + h lT

Entire K.E. of exiting fluid is dissipated through viscous effects, V of exiting fluid eventually = 0 so K = 1, regardless of the exit geometry. h lm = KV 2 /2 Only diffuser can help by reducing V. Water, velocity = 14 cm/s, width of opening = 30 mm, Re = 4300 hydrogen bubbles

Which exit has smallest K expansion ? V 2 ~ 0

MYO K =1.0

h lT = h l + h lm ; h lm = KV 2 /2 inlets, sudden enlargements & contractions; gradual contractions and exits  V 1avg 2 / 2 + p 1 /  + gz 1 =  V 2avg 2 /2 + p 2 /  + gz 2 + h lT

GRADUAL CONTRACTION Where average velocity is fastest AR < 1

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 V 1avg 2 / 2 + p 1 /  + z 1 =  V 2avg 2 /2 + p 2 /  + gz 2 + h lT h lT = h l + h lm ; h lm =  p/  = (C pi – C p ) 1 / 2 V 1 2 gentle expansions ~ diffusers

DIFFUSERS good bad 3 cm/sec 20 cm/sec ugly

….. assume fully developed P1P1 P2P2 ?><=?><= Fully developed laminar flow, is: P 1 greater, less or equal to P 2 ? What if fully developed turbulent flow? What if developing flow?

P 1, V 1 P 2, V 2 P 1, V 1 P 3, V 3 Is P 2 greater, less than or equal to P 1 ? Is  P likely to be greater, less than or equal to  P?

DIFFUSERS Diffuser data usually presented as a pressure recovery coefficient, C p, C p = (p 2 – p 1 ) / ( 1 / 2  V 1 2 ) C p indicates the fraction of inlet K.E. that appears as pressure rise [ h lm =  p/  = (C pi – C p ) 1 / 2 V 1 2 ] The greatest that C p can be is C pi, the case of zero friction.

DIFFUSERS Diffuser data usually presented as a pressure recovery coefficient, C p, C p = (p 2 – p 1 ) / ( 1 / 2  V 1 2 ) C p indicates the fraction of inlet K.E. that appears as pressure rise [ h lm =  p/  = (C pi – C p ) 1 / 2 V 1 2 ] C p will get from empirical data charts. It is not difficult to show that the ideal (frictionless) pressure recovery coefficient is: C pi = 1 – 1/AR 2, where AR = area ratio

C pideal = 1 – 1/AR 2 p 1 + ½  V 1 2 = p 2 + ½  V 2 2 (BE - ideal) p 2 /  – p 1 /  = ½ V ½ V 2 2  A 1 V 1 =  A 2 V 2 (Continuity) V 2 = V 1 (A 1 /A 2 ) p 2 /  – p 1 /  = ½ V ½ [V 1 (A 1 /A 2 )] 2 p 2 /  – p 1 /  = ½ V ½ V 1 2 (1/AR) 2 (p 2 – p 1 )/( ½  V 1 2 ) = 1 – 1/AR 2 C pi = 1 – 1/AR 2 C p = (p 2 – p 1 ) / ( 1 / 2  V 1 2 ) AR = A 2 /A 1 > 1

Relating C p to C pi and h lm p 1 /  + ½ V 1 2 = p 2 /  + ½ V h lm (z 1 = z 2 = 0) h lm = V 1 2 /2 - V 2 2 /2 – (p 2 – p 1 )/  h lm = V 1 2 /2 {1 + V 2 2 /V 1 2 – (p 2 – p 1 )/ ( 1 / 2  V 1 2 )} A 1 V 1 = A 2 V 2 C p = (p 2 – p 1 )/ ( 1 / 2  V 1 2 ) (C p is positive & < C pi ) h lm = V 1 2 /2 {1 - A 1 2 /A 2 2 – C p } C pi = 1 – 1/AR 2 h lm = V 1 2 /2 {C pi – C p } Q.E.D. (see Ex. 8.10)

h lm = (C pi – C p )V 2 /2; C pi = 1 – 1/AR 2 CpCp

N/R 1 = 0.45/(.15/2) = 6 C p  0.62 Pressure drop fixed, want to max C p to get max V 2 ; minimum h lm * AR ~ 2.7

If flow too fast or angle too big may get flow separation. C p for Re > 7.5 x 10 4, “essentially” independent of Re

 V 1avg 2 / 2 + p 1 /  + z 1 =  V 2avg 2 /2 + p 2 /  + gz 2 + h lT h lT = h l + h lm ; h lm = f(L e /D)V 2 /2 valves and fittings

h lm = f(L e /D)V 2 /2

Head loss of a bend is greater than if pipe was straight (again due to separation).

Nozzle Problem

Neglecting friction, is flow faster at A or B or same? A

If flow at B did not equal flow at A then could connect and make perpetual motion machine. A A  V 1 2 / 2 + p 1 /  + z 1 =  V 2 2 /2 + p 2 /  + gz 2 + h lT V A 2 / 2 + p atm /  + d = V B 2 / 2 + p atm /  + d = 0

C d

V T 2 / 2 + p atm /  + d = V C 2 / 2 + p atm /  + d  V 1 2 / 2 + p 1 /  + z 1 =  V 2 2 /2 + p 2 /  + gz 2 + h lT = 0 C d 0

?  V 1 2 / 2 + p 1 /  + z 1 =  V 2 2 /2 + p 2 /  + gz 2 + h lT neglect friction

V T 2 / 2 + p atm /  + d = V C 2 / 2 + p atm /  + d  V 1 2 / 2 + p 1 /  + z 1 =  V 2 2 /2 + p 2 /  + gz 2 + h lT = 0 C d 0 Nozzle

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Pipe Flow Examples ~ Solving for pressure drop in horizontal pipe

 V 1avg 2 /2 + p 1 /  + gz 1 – (  V 2avg 2 /2 + p 2 /  + gz 2 ) = h lT =  h l +  h lm =  f [L/D][V 2 /2] +  f [L e /D][V 2 /2] +  K[V 2 /2] Laminar flow ~ f = 64/Re D Turbulent flow ~ 1/f 0.5 = -2.0 log{(e/D)/ /(Re D f 0.5 ) (f = 0.316/Re D 0.25 for Re < 10 5 ) p 2 - p 1 = ?; Know h lT, L, D, Q, e, , , z 2, z 1

p 2 - p 1 = ?; Know L, D, Q, e, , , z 2, z 1 Compute the pressure drop in 200 ft of horizontal 6-in-diameter asphalted cast-iron pipe carrying water with a mean velocity of 6 ft/s.*  V 1avg 2 /2 + p 1 /  + gz 1 -  V 2avg 2 /2 - p 2 /  - gz 2 =  f [L/D][V 2 /2] +  f [L e /D][V 2 /2] +  K[V 2 /2] p 1 /  - p 2 /  = f [L/D][V 2 /2] = h lm

p 1 /  - p 2 /  = f [L/D][V 2 /2] = h l p 2 - p 1 = ?; Know L, D, Q, e, , , z 2, z 1 f(Re, e/D); Re D = 270,000 & e/D = /f 0.5 = -2.0 log{(e/D)/ /(Re D f 0.5 ); f = f ~ 0.02 p 2 – p 1 =  h l = f(Re, e/D)  [L/D][V 2 /2] = 280 lbf/ft 2

Pipe Flow Examples ~ Solving for pressure drop in non-horizontal pipe

p 2 - p 1 = ?; Know L, D, Q, e, , , z 2, z 1 Oil with  = 900 kg/m 3 and = m 2 /s flows at 0.2 m 3 /s through 500m of 200 mm-diameter cast iron pipe. Determine pressure drop if pipe slopes down at 10 o in flow direction.  V 1avg 2 /2 + p 1 /  + gz 1 -  V 2avg 2 /2 - p 2 /  - gz 2 =  f [L/D][V 2 /2] +  f [L e /D][V 2 /2] +  K[V 2 /2] p 1 /  + gz 1 - p 2 /  - gz 2 = f [L/D][V 2 /2] = h lm

p 1 /  + gz 1 - p 2 /  - gz 1 = f [L/D][V 2 /2] p 2 - p 1 = ?; Know L, D, Q, e, , , z 2, z 1 f(Re, e/D); Re D = 128,000 & e/D = /f 0.5 = -2.0 log{(e/D)/ /(Re D f 0.5 ) f = f ~ p 2 – p 1 =  h l -  g500(sin 10 o ) = 265,000 kg/m-s 2

If know everything but pressure drop or L then can use Moody Chart without iterations

Pipe Flow Examples ~ Solving for V in horizontal pipe

Q = ?; Know L, D, Q, e, , , z 2, z 1, p 1, p 2 Compute the average velocity in 200 ft of horizontal 6-in-diameter asphalted cast-iron pipe carrying water with a pressure drop of 280 lbf/ft 2.  V 1avg 2 /2 + p 1 /  + gz 1 -  V 2avg 2 /2 - p 2 /  - gz 2 =  f [L/D][V 2 /2] +  f [L e /D][V 2 /2] +  K[V 2 /2] p 1 /  - p 2 /  = f [L/D][V 2 /2]

V = (0.7245/f) e/D = Guess fully rough regime f ~ 0.19

f 1 = 0.19; V 1 = (0.7245/f 1 ) 1/2 = 6.18 ft/s Re D1 = 280,700 f 2 = 0.198; V 2 = (0.7245/f 1 ) 1/2 = 6.05 ft/s

Pipe Flow Examples ~ Solving for D in horizontal pipe

Q = ?; Know L, D, Q, e, , , z 2, z 1, p 1, p 2 Compute the diameter of a 200 m of horizontal pipe, e = mm, carrying 1.18 ft 3 /s, = ft 2 /s and the head loss is 4.5 ft.  V 1avg 2 /2g + p 1 /  g + z 1 -  V 2avg 2 /2g - p 2 /  g - z 2 =  f [L/D][V 2 /2g] +  f [L e /D][V 2 /2g] +  K[V 2 /2g] p 1 /  - p 2 /  = f [L/D][V 2 /2] f = function of Re D and e/D H lT = h l /g = [length] = 4.5 ft

Re D = VD/ = 4Q/(  D ) or Re D = 136,600/D f [L/D][V 2 /2] = h l f = h l [D/L]2/[(4Q/  D 2 ) 2 ] f ={  2 /8}{h l D 5 /(LQ 2 )} = 0.642D or D = f 1/5 e/D = /D

(1) Re D = 136,600/D (2) D = f 1/5 (3) e/D = /D Guess f ~ 0.03; then from (2) get D ~ 1.093(0.03) 1/5 ~ ft From (1) get Re D ~ 136,600/0.542 ~ 252,000 From (3) get e/D = /0.542 ~ 7.38 x f new ~ from plot; D new ~ 0.498; Re Dnew ~274,000; e/D ~8.03 x f newest ~ from plot D newest ~ 0.499

Solve for V in vertical pipe with minor losses, h lm

Assume D >> d; turbulent flow; Atm press. at top & bottom f = 0.01 Find V e as a function of g & d  V 1 2 /2 + p 1 /  + gz 1 – (  V 2 2 /2 + p 2 /  + gz 2 ) = h lT =  h l +  h lm =  f [L/D][V 2 /2] +  f [L e /D][V 2 /2] +  K[V 2 /2]

 V 1 2 /2 + p 1 /  + gz 1 – (  V 2 2 /2 + p 2 /  + gz 2 ) = h lT =  h l +  h lm =  f [L/D][V 2 /2] +  f [L e /D][V 2 /2] +  K[V 2 /2]  V 0 2 /2 + p atm /  + gz 0 -  V 2 2 /2 - p atm /  - gz 2 = f [L/D](V 2 2 /2)+(K 1 +K 2 +f[L/d])(V 2 2 /2)  V 0 2 /2 = 0;  V 2 2 /2 = 1V 2 2 /2 gz 0 –gz 2 –V 2 2 /2={f [L/D]+K 1 +K 2 )}(V 2 2 /2) V 2 2 = 2g(z 0 -z 2 )/[1 + K 1 + K 2 + f(L/d)] V 2 2 =2g140d/( x100) 1/2 V 2 = (80gd) 1/2

laminar transitionalturbulent

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