57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 1 Chapter 5 Finite Control Volume Analysis

5.1 Continuity Equation

RTT can be used to obtain an integral 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 Continuity Equation 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 2. V = constant over discrete dA (flow sections):

∫ ρV ⋅dA = ∑ρV ⋅A CS CS

57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 2 3. Incompressible fluid (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 Density ρ = ∫ρdA A

Note: mQ& ≠ ρ unless ρ = constant

57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 3 Example

*Steady flow

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

ρ ρ ρ ρ i.e., - 1V1A1 - 2V2A2 + 3V3A3 + ∫ V4dA4 = 0 A4 ρ = const. = 1.94 lb-s2 /ft4 = 1.94 slug/ft3

= ρ ρ m& 4 ∫ V4dA4 = V(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 2009 4 ρ = 3 Q4 = m& 4 .75 ft /s

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

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

= π 2 ⎡1 − 1⎤ = 1 π 2 2 Vmaxro ro 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 2009 5 5.2 Momentum Equation

Derivation of the Momentum Equation

Newton’s second law of motion for a system is

time rate of change = sum of external of the momentum forces acting on of the system the system

Since momentum is mass times velocity, the momentum of a small particle of mass is and the momentum of the entire system is . Thus,

Recall RTT:

With and ,

57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 6

Thus, the Newton’s second law becomes

where,

is fluid velocity referenced to an inertial frame (non- accelerating)

is the velocity of CS referenced to the inertial frame

is the relative velocity referenced to CV

∑ ∑ ∑ is vector sum of all forces acting on the CV

is body force such as gravity that acts on the entire mass/volume of CV

is surface force such as normal (pressure and viscous) and tangential (viscous) stresses acting on the CS

57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 7

Note that, when CS cuts through solids, may also include reaction force,

(e.g., the reaction force required to hold nozzle or bend when CS cuts through the bolts that are holding the nozzle/bend in place)

∑

∑

̂ ̂ = resultant force on fluid in CV due to and , i.e. reaction force on fluid

Free body diagram

57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 8 Important Features (to be remembered)

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

x equation Carefully define coordinate system with forces positive in = d ρ + ρ ⋅ positive direction of ∑Fx ∫ udV ∫ uVR dA dt CV CS coordinate axes y equation = d ρ + ρ ⋅ ∑Fy ∫ vdV ∫ vVR dA dt CV CS

z equation = d ρ + ρ ⋅ ∑Fz ∫ wdV ∫ wVR 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 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 9

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

4) Steady vs. Unsteady Flow

d Steady flow ⇒ ∫ρVdV = 0 dt CV

5) Uniform vs. Nonuniform Flow

ρ ⋅ ∫ V VR 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 atmosphere jet pressure must be pa

57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 10

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. Motion of a rocket 6. Force on rectangular sluice gate 7. Water hammer 8. Steady and unsteady developing and fully developed pipe flow 9. Empting and filling tanks 10. Forces on transitions 11. Hydraulic jump 12. Boundary layer and bluff body drag 13. Rocket or jet propulsion 14. Propeller

57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 11 1. Jet deflected by a plate or vane

Consider a jet of water turned through a horizontal angle

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

= = d ρ + ρ ⋅ x-equation: ∑Fx Fx ∫∫udV uV dA dt CS = ρ ⋅ Fx ∑ uV A steady flow CS ρ − + ρ = V1x ( V1A1) V2x (V2A2 )

ρ ρ ρ continuity equation: A1V1 = A2V2 = Q for A1 = A2 V1 = V2 Fx = ρQ(V2x – V1x)

= = ρ ⋅ y-equation: ∑ Fy Fy ∑ vV 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 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 12

If the vane is moving with velocity Vv, then it is convenient to choose CV moving with the vane i.e., VR = V - Vv and V used for B also moving with vane

= ρ ⋅ x-equation: Fx ∫ uVR dA CS

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

ρ ⋅ Continuity: 0 = ∫ VR 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 For coordinate system V1x = (V – Vv)1x moving with vane

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

Fy = ρQrel(V2y – V1y)

57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 13 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 Assume either the pipe velocity or pressure is known. Then,in place 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 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 14

= d ρ + ρ ⋅ ∑Fx ∫ u dV ∫ uV dA dt CV CS =∑ρuV ⋅ A steady flow and uniform CS flow over CS

Rx + p1A1 – p2A2 = ρV1(-V1A1) + ρV2(V2A2) = ρQ(V2 - V1) Rx = ρQ(V2 - V1) - p1A1

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

= ρ ⋅ = ∑∑Fy vV 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 V = 131.3 ft/s 2 π 2 ⎛ 1 ⎞ Q = ⎜ ⎟ V2 Rx = 141.48 – 706.86 = −569 lbf 4 ⎝12⎠ 3 Rz = 10 lbf = .716 ft /s

This is force on nozzle

57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 15 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.

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

= ρ ⋅ = −ρ + ρ Continuity: 0 ∑ V A V1A1 V2A2 = i.e., Q = constant = V1A1 V2A2

= ρ ⋅ x-momentum: ∑Fx ∑ uV A − θ + = ρ (− )+ ρ ( ) p1A1 p2A2 cos R x V1x V1A1 V2x V2A2 ρ ( − ) = Q V2x V1x

= ρ ⋅ y-momentum: ∑Fy ∑ vV A 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 16 θ + − − = ρ (− )+ ρ ( ) p2A2 sin R y w f w b V1y V1A1 V2y V2A2 ρ ( − ) = Q V2y V1y

4. Force on a rectangular sluice gate The force on the fluid due to the gate is calculated from the x- momentum equation:

= ρ ⋅ ∑Fx ∑ uV A

+ − − = ρ (− )+ ρ ( ) F1 FGGW Fvisc F2 V1 V1A1 V2 V2A2

= − + ρ ( − )+ usually can be neglected FGW F2 F1 Q V2 V1 Fvisc y y = γ 2 ⋅ y b − γ 1 ⋅ y b + ρQ(V − V ) 2 2 2 1 2 1 = 1 γ( 2 − 2 )+ ρ ( − ) FGW b y2 y1 Q V2 V1 = Q 2 V1 y1b = Q V2 y2b 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 17 ρ 2 Q ⎛ 1 − 1 ⎞ ⎜ ⎟ b ⎝ y2 y1 ⎠

5. Application of relative inertial coordinates for a moving but non-deforming control volume (CV)

The CV moves at a constant velocity V CS with respect to the absolute inertial coordinates. If V R represents the velocity in the relative inertial coordinates that move together with the CV, then: =− VVVRCS

Reynolds transport theorem for an arbitrary moving deforming CV: dB d SYS =∀+⋅βρdVndA βρ ∫∫R dt dt CV CS For a non-deforming CV moving at constant velocity, RTT for incompressible flow: dB ∂β syst =∀+⋅ρρβdVndA dt∫∫∂ t R CV CS

1) Conservation of mass = β = BMsyst , and 1: dM =⋅ρ VndA dt ∫ R CS For steady flow: VndA⋅=0 ∫ R CS 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 18

2) Conservation of momentum BMVV=+()β ==+dB dM V V syst R CS and syst R CS

+∂+ dMVV[]()RRCS() VV CS ==F ρρdVVVndA ∀++() ⋅ dt∑ ∫∫∂ t RRCS CV CS For steady flow with the use of continuity: FVVVndA=+⋅ρ () ∑ ∫ RRCS CS 0 =⋅+⋅ρρVV ndA V V ndA ∫∫RRCS R CS CS FVVndA=⋅ρ ∑ ∫ RR CS

Example (use relative inertial coordinates):

Ex) A jet strikes a vane which moves to the right at constant velocity on a

frictionless cart. Compute (a) the force required to restrain the cart and (b) the power delivered to the cart. Also find the cart velocity for which (c) the

force is a maximum and (d) the power is a maximum.

57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 19

Solution:

Assume relative inertial coordinates with non-deforming CV i.e. CV moves at constant translational non-accelerating

̂ ̂ ̂

=− then VVVR CS . Also assume steady flow with and neglect gravity effect.

Continuity: VndA⋅=0 ∫ R CS Bernoulli without gravity: 0 +=1 ρ 2 0 + 1 ρ 2 p1 VpR12 VR2 2 2 = VVR12R ρρ= Since VARR11 VA 2 2 == A12AAj

Momentum:

∑

0

0

57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 20

cos 1cos

1cos

1cos, 0 0 2 1cos 2 1cos

4 3 1cos 0 3 4 0

4 16 12 4 2 6 6

3

2 1cos 3 3 4 1cos 27

57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 21 5.3 Energy Equation

Derivation of the Energy Equation

The First Law of Thermodynamics The difference between the heat added to a system and the work 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 conservation of energy

ΔE = Q – W

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

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

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 time

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

rate of heat transfer to system 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 22 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 −=ρρˆˆ ++ + ++ ⋅ QW& & ()() uekp edV ue kp eVdA dt ∫∫CV CS

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

TotalKEof masswith velocityV ΔMV2 / 2 V2 e = = = V 2 = V k mass ΔM 2 Ep γΔVz e = = = gz (for Ep due to gravity only) p ΔM ρΔV

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 2009 23 = + Rate of Work Components: W& W& s W& f 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 + Δt CV

System at time t Work = force × distance

at 2 W2 = p2A2 × V2Δt (on surroundings) = = ⋅ rate of work⇒ W& 2 p2A 2V2 p2 V 2 A 2

− × Δ neg. sign since pressure at 1 W1 = p1A1 V1 t = ⋅ force on surrounding W& 1 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 2009 24 In general,

= ⋅ W& fp pV A

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

= ⋅ = ρ⎛ p ⎞ ⋅ W& fp ∫CS pV dA ∫CS ⎜ ⎟V dA ⎝ ρ⎠ = + W& f W& fp W& fshear

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 stress flow work is done. Not included or discussed in text!

57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 25 Simplified Forms of the Energy Equation

Energy Equation for Steady One-Dimensional Pipe Flow

Consider flow through the pipe system as shown

Energy Equation (steady flow) ⎛V 2 p ⎞ Q& −W& = ρ + gzzu++uˆ VdA⋅ s ∫CS ⎜ ⎟ ⎝ 2 ρ ⎠ ⎛ p ⎞ ρ V 3 Q& −W& + 1 + gz + uˆ ρ V A + 11 dA s ∫A ⎜ ρ 1 11⎟ 1 1 ∫A 1 1 ⎝ ⎠ 1 2 ⎛ p ⎞ ρ V 3 = 2 + gz + uˆ ρ VA+ 22dA ∫A ⎜ ρ 2 2 ⎟ 22 2 ∫A 2 2 ⎝ ⎠ 2 2

*Although the velocity varies across the flow sections the streamlines are assumed to be straight and parallel; consequently, there is no acceleration 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 2009 26 *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 ⎛ p ⎞ V 3 Q& −W& + 1 + gz+ uˆ ρ VdA + ρ 1 dA s ⎜ ρ 11⎟ ∫A 1 1 ∫A 1 ⎝ ⎠ 1 1 2 ⎛ p ⎞ V 3 = 2 ++gz uVuˆ ρ V dA + ρ 2 dA ⎜ ρ 2 2 ⎟ ∫A 22 ∫A 2 ⎝ ⎠ 2 2 2

Recall that Q = ∫ V ⋅dA = VA So that ρ∫ V ⋅dA = ρVA = m& mass flow rate

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 ⎠ 2 2 ⎛ p 1 ⎞ ⎛ p 2 ⎞ − ++1 + +α V = 2 + + +α V Q& WWg& ⎜ gzu11ˆ 1 ⎟m& ⎜ gz22 uˆ 2 ⎟m& ⎜ ρ 2 ⎟ ⎜ ρ 2 ⎟ ⎝ ⎠ ⎝ ⎠ 2 2 1 p V 1 p V 2 ( − ) + 1 ++ˆ +α = 2 + + ˆ +α Q& W& gz1 u11 gzu2 2 2 m& ρ 2 ρ 2

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

laminar flow 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 2009 27 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 = − W& s W& t W& p where W& t and W& p are ⎛ work ⎞ magnitudes of power ⎜ ⎟ ⎝ time ⎠ Using this result in the energy equation and deviding by g results in

W& pV22W p Vuuˆˆ− Q p +++11αα =&t +++ 2 221 + −& zz11 2 2 m&&g γγ22mggm &g

mechanical part thermal part

Note: each term has dimensions of length Define the following:

W& W& W& = p = p = p h p m& g ρQg γQ

= W& t h t m& g

uuˆˆ− Q =−=21 & hheadlossL gmg& 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 28 Head Loss In a general fluid system a certain amount of mechanical energy is converted to thermal energy due to viscous action. This effect results in an increase in the fluid internal energy. Also, some heat will be generated through energy dissipation and be lost (i.e. -Q& ). Therefore the term from 2nd law

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

Note that adding Q& to system will not make hL = 0 since this also increases Δu. 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 incompressible flow) 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 2009 29 Comparison ofo 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 2009 30 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& W& s W& p W& v shaft work done on or pressure Viscous stress by system work done work on CS (pump or on CS turbine )

= ⋅ = ρ()ρ ⋅ W& p ∫ pV dA ∫ p V dA CV CS

= − W& s W& t W& p − + = d ρ + ρ()+ ⋅ Q& W& t W& p ∫ edV ∫ 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 2009 31 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

= Note: each term h p W& p m& g has units of length = h t W& t m& g V is average velocity uuˆˆ− Q (vector dropped) and =−=21 & hL head loss corrected by α g m& g > 0 represents loss in mechanical energy due to viscosity 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 32 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 = friction 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 V2 change due f D 2g to hp or ht 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 33 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 p V p V 1 + + 1 = 2 + + 2 + z1 z2 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 2009 34 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 π 2 π 2 change in distance between D1 = D2 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 2009 35 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 2009 36

57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 37 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 38 Application of the Energy, Momentum, and Continuity Equations in Combination

In general, when solving fluid mechanics 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: = ρ 2 − ρ 2 = ρ ()− one inlet and ∑Fs V2 A2 V1 A1 Q 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 2009 39 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 turbulence. 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 (shear stress neglected)

= − − α = ρ ⋅ ∑Fs p1A2 p2A2 Wsin ∑ uV A ρ − + ρ = V1( V1A1) V2 (V2A2 ) Δ γ z =ρV2A − ρV2A A2L 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 2009 40 2 2 2 p1 p2 V2 V1 A1 V1 A1 ⎛ A1 ⎞ ÷ γA 2 − − ()z − z = − = ⎜ −1⎟ γ γ 1 2 ⎜ ⎟ g g A2 g A2 ⎝ A2 ⎠

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

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 2009 41 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

= ρ ⋅ ∑Fx ∑ uV 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 2009 42

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

⎡ ⎛ V2 V2 ⎞⎤ = ρ ()− + − γ⎜ 2 − 1 + ⎟ − ⇒Fx Q V2 V1 A2 ⎢p1 ⎜ 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