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57:020 of and Transport Processes Chapter 5 Professor Fred Stern Fall 2006 1 Chapter 5 , , and

Flow Rate and

1. cross-sectional oriented normal to vector (simple case where V ⊥ A)

U = constant: Q = volume = UA [m/s × m2 = m3/s] U ≠ constant: Q = ∫ UdA A Similarly the = m = ∫ρUdA A

2. general case

Q = ∫ V ⋅ndA CS = ∫ V cosθdA CS m = ∫ρ(V ⋅n)dA CS

57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2006 2 Q average velocity: V = A

Example: At low the flow through a long circular tube, i.e. pipe, has a parabolic velocity distribution (actually paraboloid of revolution). 2 ⎛ ⎛ r ⎞ ⎞ u = u ⎜1− ⎟ max ⎜ ⎜ ⎟ ⎟ ⎝ ⎝ R ⎠ ⎠ i.e., centerline velocity

a) find Q and V

Q = ∫ V ⋅ndA = ∫ udA A A

2π R ∫∫∫udA = u(r)rdθdr A 0 0 R = 2π∫ u(r)rdr 0 dA = 2πrdr 2π u = u(r) and not θ ∴ ∫ dθ = 2π 0

57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2006 3 2 R ⎛ ⎛ r ⎞ ⎞ 1 Q = 2π u ⎜1− ⎟rdr = u πR 2 ∫ max ⎜ ⎜ ⎟ ⎟ max 0 ⎝ ⎝ R ⎠ ⎠ 2 1 V = u 2 max

Continuity

RTT can be used to obtain an relationship expressing conservation of mass by defining the extensive property B = M such that β = 1.

B = M = mass β = dB/dM = 1

General Form of dM d = 0 = ∫∫ρdV + ρV ⋅dA dt dt CV CS or d ∫ρV ⋅dA = − ∫ρdV CS dt CV net rate of outflow rate of decrease of of mass across CS mass within CV

Simplifications: d 1. Steady flow: − ∫ρdV = 0 dt CV 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2006 4 2. V = constant over discrete dA (flow sections):

∫ ρV ⋅dA = ∑ρV ⋅ A CS CS

3. Incompressible (ρ = constant) d ∫∫VdA⋅=− dV conservation of volume CSdt CV

4. Steady One-Dimensional Flow in a Conduit: ∑ρV ⋅A = 0 CS

−ρ1V1A1 + ρ2V2A2 = 0

for ρ = constant Q1 = Q2

Some useful definitions:

Mass flux m = ∫ρV ⋅dA A

Volume flux Q = ∫ V ⋅dA A

Average Velocity V = Q / A

1 Average ρ = ∫ρdA A

Note: mQ ≠ρ unless ρ = constant 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2006 5 Example

*Steady flow

*V1,2,3 = 50 fps *@ u V varies linearly from zero at wall to Vmax at pipe center *find m 4, Q4, Vmax 3 0 *, ρw = 1.94 /ft d ∫ρV ⋅dA = 0 = − ∫ρdV dt CS CV m 4

i.e., -ρ1V1A1 - ρ2V2A2 + ρ3V3A3 + ρ ∫ V4dA4 = 0 A4 ρ = const. = 1.94 lb-s2 /ft4 = 1.94 slug/ft3

m 4 = ρ∫ V4dA4 = ρV(A1 + A2 – A3) V1=V2=V3=V=50f/s 1.94 π = ×50× (12 + 22 −1.52 ) 144 4 = 1.45 slugs/s

57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2006 6 3 Q4 = m 4 ρ = .75 ft /s

= ∫ V4dA4 A4 velocity profile r o 2π ⎛ ⎞ Q = r 4 ∫∫Vmax ⎜1− ⎟rdθdr 0 0 ⎝ ro ⎠ dA4 V4 ≠ V4(θ) r o ⎛ r ⎞ = 2π∫ Vmax ⎜1− ⎟rdr 0 r ⎝ o ⎠ 1 πr 2V Q 3 o max V4 = = r 2 2 o ⎡ r ⎤ A πro = 2πVmax ∫ ⎢r − ⎥dr 1 0 ⎣ ro ⎦ = V 3 max

r r ⎡r 2 0 r3 o ⎤ = 2πV ⎢ − ⎥ max 2 3r ⎣⎢ 0 o 0 ⎦⎥

2 ⎡1 1⎤ 1 2 = 2πVmaxro − = πro Vmax ⎣⎢2 3⎦⎥ 3 Q V = 4 = 2.86fps max 1 πr 2 3 o 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2006 7 Momentum Equation

RTT with B = MV and β = V d ∑[]FS + FB = ∫∫ρVdV + VρVR ⋅dA dt CV CS V = velocity referenced to an inertial frame (non-accelerating) VR = velocity referenced to FS = surface + forces (due to and viscous normal and shear stresses) FB = body (due to )

Applications of the Momentum Equation Initial Setup and Signs 1. Jet deflected by a plate or a vane 2. Flow through a nozzle 3. Forces on bends 4. Problems involving non-uniform velocity distribution 5. of a 6. Force on rectangular sluice gate 7.

Derivation of the Basic Equation General form for moving but dBsys d Recall RTT: = ∫βρdV + ∫βρVR ⋅dA non-accelerating dt dt CV CS reference frame

VR=velocity relative to CS=V – VS=absolute – velocity CS Subscript not shown in text but implied! i.e., referenced to CV Let, B = MV = linear momentum V must be referenced to β = V inertial reference frame 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2006 8 d(MV) d = ∑F = ∫ VρdV + ∫ VρVR ⋅dA dt dt CV CS Must be relative to a non-accelerating ’s 2nd law inertial reference frame where ΣF = vector sum of all forces acting on the control volume including both surface and body forces

= ΣFS + ΣFB ΣFS = sum of all external surface forces acting at the CS, i.e., pressure forces, forces transmitted through , shear forces, etc.

ΣFB = sum of all external body forces, i.e., gravity force

ΣFx = p1A1 – p2A2 + Rx ΣFy = -W + Ry

R = resultant force on fluid

in CV due to pw and τw free body diagram i.e., reaction force on fluid

Important Features (to be remembered)

1) Vector equation to get component in any direction must use

x equation Carefully define coordinate system with forces positive in d positive direction of ∑Fx = ∫ρudV + ∫ρuVR ⋅dA dt CV CS coordinate axes 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2006 9 y equation d ∑Fy = ∫ρvdV + ∫ρvVR ⋅dA dt CV CS

z equation d ∑Fz = ∫ρwdV + ∫ρwVR ⋅dA dt CV CS

2) Carefully define control volume and be sure to include all external body and surface faces acting on it. For example,

(Rx,Ry) = reaction force on fluid

(Rx,Ry) = reaction force on nozzle

3) Velocity V must be referenced to a non-accelerating inertial reference frame. Sometimes it is advantageous to use a moving (at constant velocity) reference frame. Note VR = V – Vs is always relative to CS.

i.e., in these cases V used for B also referenced to CV (i.e., V = VR) 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2006 10 4) Steady vs. Unsteady Flow

d Steady flow ⇒ ∫ρVdV = 0 dt CV

5) Uniform vs. Nonuniform Flow

∫ VρVR ⋅dA = change in flow of momentum across CS CS = ΣVρVR⋅A uniform flow across A 6) Fpres = −∫ pndA ∫∫∇fdV = f nds VS f = constant, ∇f = 0 = 0 for p = constant and for a closed surface

i.e., always use gage pressure

7) Pressure condition at a jet exit

at an exit into the jet pressure must be pa

Application of the Momentum Equation 1. Jet deflected by a plate or vane

Consider a jet of water turned through a horizontal angle

57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2006 11

CV and CS are for jet so that Fx and Fy are vane reactions forces on fluid

d x-equation: ∑Fx = Fx = ∫∫ρudV + ρuV ⋅dA dt CS

Fx = ∑ ρuV ⋅ A steady flow CS = ρV1x (−V1A1) + ρV2x (V2A 2 )

continuity equation: ρA1V1 = ρA2V2 = ρQ for A1 = A2 V1 = V2 Fx = ρQ(V2x – V1x) y-equation: ∑Fy = Fy = ∑ρvV ⋅A CS

Fy = ρV1y(– A1V1) + ρV2y(– A2V2) = ρQ(V2y – V1y)

for above geometry only where: V1x = V1 V2x = -V2cosθ V2y = -V2sinθ V1y = 0 note: Fx and Fy are force on fluid - Fx and -Fy are force on vane due to fluid

If the vane is moving with velocity Vv, then it is convenient to choose CV moving with the vane

57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2006 12 i.e., VR = V - Vv and V used for B also moving with vane x-equation: Fx = ∫ρuVR ⋅dA CS

Fx = ρV1x[-(V – Vv)1A1] + ρV2x[-(V – Vv)2A2]

Continuity: 0 = ∫ρVR ⋅dA

i.e., ρ(V-Vv)1A1 = ρ(V-Vv)2A2 = ρ(V-Vv)A Qrel Fx = ρ(V-Vv)A[V2x – V1x] Qrel on fluid V2x = (V – Vv)2x V1x = (V – Vv)1x

Power = -FxVv i.e., = 0 for Vv = 0

Fy = ρQrel(V2y – V1y)

2. Flow through a nozzle

Consider a nozzle at the end of a pipe (or hose). What force is required to hold the nozzle in place? CV = nozzle and fluid ∴ (Rx, Ry) = force required to hold nozzle in place 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2006 13 Assume either the pipe velocity or pressure is known. Then, the unknown (velocity or pressure) and the exit velocity V2 can be obtained from combined use of the continuity and Bernoulli equations.

1 2 1 2 Bernoulli: p + γz + ρV = p + γz + ρV z1=z2 1 1 2 1 2 2 2 2 1 2 1 2 p + ρV = ρV 1 2 1 2 2

Continuity: A1V1 = A2V2 = Q 2 A1 ⎛ D ⎞ V2 = V1 = ⎜ ⎟ V1 A2 ⎝ d ⎠ 4 1 ⎛ ⎛ D ⎞ ⎞ p + ρV2 ⎜1− ⎟ = 0 1 1 ⎜ ⎜ ⎟ ⎟ 2 ⎝ ⎝ d ⎠ ⎠ 1/ 2 ⎡ ⎤ ⎢ − 2p ⎥ Say p known: V = 1 1 1 ⎢ 4 ⎥ ⎢ρ⎜⎛1− ()D ⎟⎞⎥ ⎣ ⎝ d ⎠⎦ To obtain the reaction force Rx apply momentum equation in x- direction

d ∑Fx = ∫ uρdV + ∫ρuV ⋅dA dt CV CS =∑ρuV ⋅ A steady flow and uniform CS flow over CS

Rx + p1A1 – p2A2 = ρV1(-V1A1) + ρV2(V2A2) = ρQ(V2 - V1) 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2006 14

Rx = ρQ(V2 - V1) - p1A1

To obtain the reaction force Ry apply momentum equation in y- direction

∑∑Fy = ρvV ⋅A = 0 since no flow in y-direction CS

Ry – Wf − WN = 0 i.e., Ry = Wf + WN

Numerical Example: Oil with S = .85 flows in pipe under pressure of 100 psi. Pipe diameter is 3” and nozzle tip diameter is 1” Sγ ρ = =1.65 g V = 14.59 ft/s 1 D/d = 3 V2 = 131.3 ft/s 2 π⎛ 1 ⎞ Q = ⎜ ⎟ V2 Rx = 141.48 – 706.86 = −569 lbf 4 ⎝12⎠ 3 Rz = 10 lbf = .716 ft /s

This is force on nozzle

3. Forces on Bends

Consider the flow through a bend in a pipe. The flow is considered steady and uniform across the inlet and outlet sections. Of primary concern is the force required to hold the bend in place, i.e., the reaction forces Rx and Ry which can be determined by application of the momentum equation. 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2006 15

Rx, Ry = reaction force on bend i.e., force required to hold bend in place

Continuity: 0 = ∑ρV ⋅ A = −ρV1A1 + ρV2A2 i.e., Q = constant = V1A1 = V2A2 x-momentum: ∑∑Fx = ρuV ⋅ A p1A1 − p2A2 cosθ + R x = ρV1x (− V1A1 )()+ ρV2x V2A2 = ρQ(V2x − V1x ) y-momentum: ∑∑Fy = ρvV ⋅A

p2A2 sin θ + R y − w f − w b = ρV1y (− V1A1 )()+ ρV2y V2A2

= ρQ(V2y − V1y )

4. Problems involving Nonuniform Velocity Distribution See text pp. 215− 216 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2006 16 5. Force on a rectangular sluice gate The force on the fluid due to the gate is calculated from the x- momentum equation:

∑∑Fx = ρuV ⋅ A

F1 + FGW − Fvisc − F2 = ρV1(− V1A1 )+ ρV2 (V2A2 )

usually can be neglected FGW = F2 − F1 + ρQ()V2 − V1 + Fvisc y y = γ 2 ⋅ y b − γ 1 ⋅ y b + ρQ()V − V 2 2 2 1 2 1 1 F = bγ()y2 − y2 + ρQ()V − V Q GW 2 2 1 2 1 V = 1 y b ρQ2 ⎛ 1 1 ⎞ 1 ⎜ − ⎟ b ⎜ y y ⎟ Q ⎝ 2 1 ⎠ V2 = y2b of Momentum Equation See text pp. 221 − 229

57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2006 17 Energy Equations

Derivation of the Energy Equation

The First Law of The difference between the added to a system and the done by a system depends only on the initial and final states of the system; that is, depends only on the change in energy E: principle of

∆E = Q – W

∆E = change in energy Q = heat added to the system W = work done by the system

E = Eu + Ek + Ep = total energy of the system

Internal energy due to molecular motion

The differential form of the first law of thermodynamics expresses the rate of change of E with respect to

dE = Q − W dt rate of work being done by system

rate of to system 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2006 18 Energy Equation for Fluid Flow The energy equation for fluid flow is derived from Reynolds transport theorem with

Bsystem = E = total energy of the system (extensive property)

β = E/mass = e = energy per unit mass (intensive property)

= uˆ + ek + ep dE d = ∫ ρedV + ∫ ρeV ⋅dA dt dt CV CS d   ˆˆ QW−=ρρ()() ue ++kp edV + ue ++ kp eVdA ⋅ dt ∫∫CV CS

This can be put in a more useable form by noting the following:

2 2 TotalKEof masswith velocityV ∆MV / 2 V 2 e = = = V = V k mass ∆M 2 Ep γ∆Vz e = = = gz (for Ep due to gravity only) p ∆M ρ∆V

dV⎛⎞⎛⎞22 V QW −= ρρ ++gzudVˆˆ + ++gzuVdA ⋅ ∫∫CV⎜⎟⎜⎟ Cs dt ⎝⎠⎝⎠22

rate of work rate of change flux of energy done by system of energy in CV out of CV (ie, across CS) rate of heat transfer to sysem 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2006 19    Rate of Work Components: W = Ws + Wf For convenience of analysis, work is divided into shaft work Ws and flow work Wf

Wf = net work done on the surroundings as a result of normal and tangential stresses acting at the control surfaces = Wf pressure + Wf shear

Ws = any other work transferred to the surroundings usually in the form of a shaft which either takes energy out of the system (turbine) or puts energy into the system (pump)

Flow work due to pressure forces Wf p (for system) Note: here V uniform over A

CS

System at time t + ∆t CV

System at time t Work = force × distance

at 2 W2 = p2A2 × V2∆t (on surroundings)  rate of work⇒ W2 = p2A2V2 = p2 V2 ⋅ A2 neg. sign since pressure at 1 W1 = −p1A1 × V1∆t  force on surrounding W1 = p1 V1 ⋅A1 fluid acts in a direction opposite to the motion of the system boundary 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2006 20 In general,

 Wfp = pV ⋅A

for more than one control surface and V not necessarily uniform over A:

 ⎛ p ⎞ Wfp = ∫CS pV ⋅dA = ∫CSρ⎜ ⎟V ⋅dA ⎝ ρ⎠    Wf = Wfp + Wfshear

Basic form of energy equation ⎛⎞p QW −− W −ρ VdA ⋅ sfshear∫CS ⎜⎟ ⎝⎠ρ dV⎛⎞⎛⎞22 V =+++++⋅ρρgz uˆˆ dV gz u V dA ∫∫CV⎜⎟⎜⎟ CS dt ⎝⎠⎝⎠22

dV⎛⎞2 QW −− W =ρ ++ gzudVˆ sfshear ∫CV ⎜⎟ dt ⎝⎠2 Usually this term can be ⎛⎞Vp2 eliminated by proper choice of + ρ +++gzuˆ VdA ⋅ ∫CS ⎜⎟ CV, i.e. CS normal to flow lines. ⎝⎠2 ρ Also, at fixed boundaries the velocity is zero (no slip h=enthalpy condition) and no shear flow work is done. Not included or discussed in text!

57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2006 21 Simplified Forms of the Energy Equation

Energy Equation for Steady One-Dimensional

Consider flow through the pipe system as shown

Energy Equation (steady flow) ⎛⎞Vp2 QW −= ρ +++gzuVdAˆ ⋅ s ∫CS ⎜⎟ ⎝⎠2 ρ 3   ⎛⎞pV111ρ QW−+s + gzu11111 +ˆ ρ VA + dA 1 ∫∫AA⎜⎟ 11⎝⎠ρ 2 ⎛⎞pVρ 3 =+++222gz uˆ ρ V A dA ∫∫AA⎜⎟22222 2 22⎝⎠ρ 2

*Although the velocity varies across the flow sections the streamlines are assumed to be straight and parallel; consequently, there is no normal to the streamlines and the pressure is hydrostatically distributed, i.e., p/ρ +gz = constant.

57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2006 22 *Furthermore, the internal energy u can be considered as constant across the flow sections, i.e. T = constant. These quantities can then be taken outside the integral sign to yield ⎛⎞pV3 QW −+ 11 +gzu +ˆ ρρ VdA + dA s ⎜⎟11∫∫AA 11 1 ⎝⎠ρ 112 ⎛⎞pV3 =++22gz uˆ ρρ V dA + dA ⎜⎟22∫∫AA 22 2 ⎝⎠ρ 222

Recall that Q = ∫ V ⋅dA = VA So that ρ∫ V ⋅dA = ρVA = m

3 2 V3 ρV A V Define: ρ∫ dA = α = α m A 2 2 2 K.E. flux K.E. flux for V=V =constant across pipe 3 1 ⎛ V ⎞ i.e., α = ∫⎜ ⎟ dA = kinetic energy correction factor A A⎝ V ⎠ ⎛⎞⎛⎞22 ppVV12 QW −+ ⎜⎟⎜⎟12 +gzu ++ˆˆαα m = +gzu ++ m ⎜⎟⎜⎟11 122 2 2 2 ⎝⎠⎝⎠ρρ 22 1   pp12VV12 ()QW−++++ gzu11ˆˆαα 1 =+++ gzu 2 2 2 m ρρ22

Nnote that: α = 1 if V is constant across the flow section α > 1 if V is nonuniform

α = 2 turbulent flow α = 1.05 ∼ 1 may be used 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2006 23 Shaft Work Shaft work is usually the result of a turbine or a pump in the flow system. When a fluid passes through a turbine, the fluid is doing shaft work on the surroundings; on the other hand, a pump does work on the fluid      Ws = Wt − Wp where Wt and Wp are ⎛ work ⎞ magnitudes of ⎜ ⎟ ⎝ time ⎠ Using this result in the energy equation and deviding by g results in

W pV22W p Vuuˆˆ− Q p +++11zzαα =t +++ 2 221 + − mg γγ1122mgg 2 2 m g

mechanical part thermal part

Note: each term has of length Define the following:

W W W h = p = p = p p m g ρQg γQ

W h = t t m g

uuˆˆ− Q hheadloss=−=21 L gmg 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2006 24 Head Loss In a general fluid system a certain amount of mechanical energy is converted to thermal energy due to viscous . This effect results in an increase in the fluid internal energy. Also, some heat will be generated through energy and be lost (i.e. -Q ). Therefore the term from 2nd law

uuˆˆ− Q represents a loss in h =−>21 0 mechanical energy due L ggm to viscous stresses

Note that adding Q to system will not make hL = 0 since this also increases ∆u. It can be shown from 2nd law of thermodynamics that hL > 0.

Drop ⎯ over V and understand that V in energy equation refers to average velocity.

Using the above definitions in the energy equation results in (steady 1-D ) p V2 p V2 1 + α 1 + z + h = 2 + α 2 + z + h + h γ 1 2g 1 p γ 2 2g 2 t L

form of energy equation used for this course!

57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2006 25 Comparison of Energy Equation and Bernoulli Equation

Apply energy equation to a stream tube without any shaft work

Infinitesimal stream tube ⇒ α =α =1 1 2

p V2 p V2 Energy eq : 1 + 1 + z = 2 + 2 + z + h γ 2g 1 γ 2g 2 L

•If hL = 0 (i.e., µ = 0) we get Bernoulli equation and conservation of mechanical energy along a streamline

•Therefore, energy equation for steady 1-D pipe flow can be interpreted as a modified Bernoulli equation to include viscous effects (hL) and shaft work (hp or ht)

Summary of the Energy Equation

The energy equation is derived from RTT with

B = E = total energy of the system

β = e = E/M = energy per unit mass

57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2006 26 1 = uˆ + V2 +gz 2

internal KE PE

dE d st = ∫ρedV + ∫ρeV ⋅dA = Q − W from 1 Law of dt dt CV CS Thermodynamics heat work add done Neglected in text presentation     W = Ws + Wp + Wv shaft work done on or pressure Viscous stress by system work done work on CS (pump or on CS turbine )

 Wp = ∫ pV ⋅dA = ∫ρ()p ρ V ⋅dA CV CS

   Ws = Wt − Wp    d Q − Wt + Wp = ∫ρedV + ∫ρ()e + p e V ⋅dA dt CV CS 1 eu=+ˆ V2 +gz 2

For steady 1-D pipe flow (one inlet and one outlet): 1) Streamlines are straight and parallel ⇒ p/ρ +gz = constant across CS 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2006 27 2) T = constant ⇒ u = constant across CS

3 1 ⎛ V ⎞ 3) define α = ∫ ⎜ ⎟ dA = KE correction factor A CS⎝ V ⎠

3 2 ρ 3 ρV V ⇒ ∫ V dA = α A = α m 2 2 2 Thermal mechanical energy energy p V2 p V2 1 + α 1 + z + h = 2 + α 2 + z + h + h γ 1 2g 1 p γ 2 2g 2 t L h = W m g Note: each term p p has units of length  h t = Wt m g V is average velocity  (vector dropped) and uuˆˆ21− Q hL =−= corrected by α g m g head loss > 0 represents loss in mechanical energy due to 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2006 28 Concept of Hydraulic and Energy Grade Lines

p V2 p V2 1 + α 1 + z + h = 2 + α 2 + z + h + h γ 1 2g 1 p γ 2 2g 2 t L p Define HGL = + z point-by-point γ application is p V2 graphically EGL = + z + α γ 2g displayed

HGL corresponds to pressure tap measurement + z EGL corresponds to stagnation tube measurement + z EGL = HGL if V = 0 EGL1 = EGL 2 + hL L V2 for hp = ht = 0 hL = f D 2g i.e., linear variation in L for D, V, and f constant

f = factor f = f(Re)

p pressure tap: 2 = h γ h = height of fluid in p V2 stagnation tube: 2 + α 2 = h tap/tube γ 2g

EGL1 + hp = EGL2 + ht + hL EGL2 = EGL1 + hp − ht − hL

abrupt L V 2 f change due D 2g to hp or ht 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2006 29 Helpful hints for drawing HGL and EGL

1. EGL = HGL + αV2/2g = HGL for V = 0

L V2 2.&3. h = f in pipe means EGL and HGL will slope L D 2g

downward, except for abrupt changes due to ht or hp

2 2 p1 V1 p2 V2 + z1 + = + z 2 + + h L γ 2g γ 2g

HGL2 = EGL1 - hL 2 V h L = for abrupt expansion 2g 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2006 30 4. p = 0 ⇒ HGL = z

L V2 5. for h = f = constant × L L D 2g i.e., linearly increased for f V2 EGL/HGL slope downward increasing L with slope D 2g 6. for change in D ⇒ change in V

i.e. V1A1 = V2A2 2 2 change in distance between πD1 πD2 V1 = V2 ⇒ HGL & EGL and slope 4 4 change due to change in h 2 2 L V1D1 = V1D2

57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2006 31 7. If HGL < z then p/γ < 0 i.e., cavitation possible

condition for cavitation:

N p = pva = 2000 m2

N gage pressure pva,g = pA − patm ≈ −patm = −100,000 m2

p va,g ≈ −10m γ

9810 N/m3 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2006 32

57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2006 33

57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2006 34 Application of the Energy, Momentum, and Continuity Equations in Combination

In general, when solving problems, one should use all available equations in order to derive as much information as possible about the flow. For example, consistent with the approximation of the energy equation we can also apply the momentum and continuity equations

Energy: p V2 p V2 1 + α 1 + z + h = 2 + α 2 + z + h + h γ 1 2g 1 p γ 2 2g 2 t L

Momentum: 2 2 one inlet and ∑Fs = ρV2 A2 − ρV1 A1 = ρQ(V2 − V1 ) one outlet Continuity: ρ = constant A1V1 = A2V2 = Q = constant 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2006 35 Abrupt Expansion Consider the flow from a small pipe to a larger pipe. Would like to know hL = hL(V1,V2). Analytic solution to exact problem is extremely difficult due to the occurrence of flow separations and . However, if the assumption is made that the pressure in the separation region remains approximately constant and at the value at the point of separation, i.e, p1, an approximate solution for hL is possible:

Apply Energy Eq from 1-2 (α1 = α2 = 1) p V2 p V2 1 + z + 1 = 2 + z + 2 + h γ 1 2g γ 2 2g L

Momentum eq. For CV shown ( neglected)

∑Fs = p1A2 − p2A2 − Wsin α = ∑ρuV ⋅A

=ρV1(−V1A1) + ρV2 (V2A2 ) ∆z =ρV2A − ρV2A γA2L 2 2 1 1 L W sin α next divide momentum equation by γA2 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2006 36 2 2 2 p1 p2 V2 V1 A1 V1 A1 ⎛ A1 ⎞ ÷ γA 2 − − ()z1 − z2 = − = ⎜ −1⎟ γ γ g g A2 g A2 ⎝ A2 ⎠

from energy equation ⇓ 2 2 2 2 V2 V1 V2 V1 A1 − + h L = − 2g 2g g g A2

2 2 V2 V1 ⎛ 2A1 ⎞ h L = + ⎜1− ⎟ 2g 2g ⎝ A2 ⎠

1 ⎡ 2 2 2 A1 ⎤ continutity eq. h L = V2 + V1 − 2V1 ⎢ ⎥ V1A1 = V2A2 2g ⎣ A2 ⎦ A V 1 = 2 A V −2V1V2 2 1

1 h = []V − V 2 L 2g 2 1

If VV21 , 1 h= V2 L12g 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2006 37 Forces on Transitions

Example 7-6 Q = .707 m3/s V2 head loss = .1 2 2g (empirical equation)

Fluid = water p1 = 250 kPa D1 = 30 cm D2 = 20 cm Fx = ?

First apply momentum theorem

∑∑Fx = ρuV ⋅ A

Fx + p1A1 − p2A2 = ρV1(−V1A1) + ρV2(V2A2)

Fx = ρQ(V2 − V1) − p1A1 + p2A2

force required to hold transition in place

57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2006 38

The only unknown in this equation is p2, which can be obtained from the energy equation.

2 2 p1 V1 p2 V2 + = + + h note: z1 = z2 and α = 1 γ 2g γ 2g L

2 2 ⎡V2 V1 ⎤ p2 = p1 − γ⎢ − + h L ⎥ drop in pressure ⎣ 2g 2g ⎦

⎡ ⎛ V2 V2 ⎞⎤ ⎜ 2 1 ⎟ ⇒Fx = ρQ()V2 − V1 + A2 ⎢p1 − γ⎜ − + h L ⎟⎥ − p1A1 ⎣ ⎝ 2g 2g ⎠⎦

p2 (note: if p2 = 0 same as nozzle)

In this equation, continuity A1V1 = A2V2 A V = 1 V V1 = Q/A1 = 10 m/s 2 A 1 V = Q/A = 22.5 m/s 2 2 2 i.e. V > V V2 2 1 h = .1 2 = 2.58m L 2g

Fx = −8.15 kN is negative x direction to hold transition in place