Ch 4. Continuity, Energy, and Momentum Equation

Chapter 4 Continuity, Energy, and Momentum Equations

4.1 Conservation of Matter in Homogeneous Fluids

• Conservation of matter in homogeneous (single species) fluid → continuity equation

positive dA1 - perpendicular to the control surface negative

4.1.1 Finite control volume method-arbitrary control volume

- Although control volume remains fixed, mass of fluid originally enclosed (regions A+B)

occupies the volume within the dashed line (regions B+C).

Since mass m is conserved:

mmm  m (4.1)  ABBtttdt      Ctdt

m  m mmBBtdt   t A t  C  tdt  (4.2) dt dt

•LHS of Eq. (4.2) = time rate of change of mass in the original control volume in the limit

mmBB   m   tdt t B  dV (4.3) dt t t CV

4−1 Ch 4. Continuity, Energy, and Momentum Equation

where dV = volume element

•RHS of Eq. (4.2)

= net flux of matter through the control surface

= flux in – flux out

= qdA qdA nn12  where q = component of velocity vector normal to the surface of CV  q cos n    dV qdAnn12 qdA (4.4) CV CS CS t

※ Flux (= mass/time) is due to velocity of the flow.

• Vector form

    dV  qdA (4.5) t CV CS  where dA = vector differential area pointing in the outward direction over an enclosed control surface

     qdAq dA cos

 positive for an outflow from cv,   90    negative for inflow into cv, 90  180 

If fluid continues to occupy the entire control volume at subsequent times

→ time independent

4−2 Ch 4. Continuity, Energy, and Momentum Equation

 LHS:  dV  dV (4.5a) ttCV CV

Eq. (4.4) becomes

   dV q dA 0 CV CS t (4.6)

→ General form of continuity equation - integral form

[Re] Differential form

Use Gauss divergence theorem

F dV FdAi VA xi

Transform surface integral of Eq. (4.6) into volume integral  qdA  q dV CS CV

Then, Eq. (4.6) becomes

  qdV  0 CV  t (4.6a)

Eq. (4.6a) holds for any volume only if the integrand vanishes at every point.

  q  0 t (4.6b)

→ Differential form

◈ Simplified form of continuity equation

(1) For a steady flow of a compressible fluid

4−3 Ch 4. Continuity, Energy, and Momentum Equation

 dv  0 CV t

Therefore, Eq. (4.6) becomes   qdA0 (4.7) CS

(2) For incompressible fluid (for both steady and unsteady conditions)

 d  const. 0, 0 tdt

Therefore, Eq. (4.6) becomes   qdA0 (4.8) CS

[Cf] Non-homogeneous fluid mixture

→ conservation of mass equations for the individual species

→ advection - diffusion equation

cq = conservation of mass equation 0 tx

c + mass flux equation due to advection and diffusion qucD x

c c uc D 0 txx

cuc c  D txxx

4−4 Ch 4. Continuity, Energy, and Momentum Equation

4.1.2 Stream - tube control volume analysis for steady flow

There is no flow across the longitudinal boundary

∴ Eq. (4.7) becomes   qdA qdA  qdA 0 (4.9)  11 1 2 2 2

qdA  const.

If density = const.

q11 dA q 2 dA 2 dQ (4.10)

where dQ = volume rate of flow

(1) For flow in conduit with variable density

qdA V   → average velocity A

4−5 Ch 4. Continuity, Energy, and Momentum Equation

dQ    → average density Q

111VA  2 VA 22 (4.11)

(2) For a branching conduit    qdA0

11qdA 1 2 qdA 2 2   33 qdA 3 0 AA  A 12 3

111VA 2 VA 22 333 VA

4−6 Ch 4. Continuity, Energy, and Momentum Equation

[Appendix 4.1] Equation of Continuity

→ Infinitesimal (differential) control volume method

At the centroid of the control volume:  ,,,uvw

Rate of mass flux across the surface perpendicular to x is

 u dx flux in  u  dydz x 2

 u dx flux out  u  dydz x 2 u net flux = flux in flux out  dxdydz x

v net mass flux across the surface perpendicular to y  dydxdz y

w net mass flux across the surface perpendicular to z  dzdxdy z

dxdydz Time rate of change of mass inside the c.v. = t

4−7 Ch 4. Continuity, Energy, and Momentum Equation

Time rate of change of mass inside = The sum of three net rates

dxdydz u v  w    dxdydz txyz

By taking limit dV dxdydz

 uvw        qdivq tx  y  z   q 0 (A1) t

→ general point (differential) form of Continuity Equation

uvw   [Re]  qdivq xyz

By the way,   qq     q

Thus, (A1) becomes

  qq 0 (A2) t

1) For incompressible fluid

d 0 ( const.) dt   d q  0 tdt

4−8 Ch 4. Continuity, Energy, and Momentum Equation

Therefore Eq. (A2) becomes   qq 00    (A3)

In scalar form,

uvw  0 (A4) xyz

→ Continuity Eq. for 3D incompressible fluid

For 2D incompressible fluid,

uv 0 xy

2) For steady flow,

  0 t

Thus, (A1) becomes    qq     q 0

4−9 Ch 4. Continuity, Energy, and Momentum Equation

4.2 The General Energy Equation

4.2.1 The 1st law of thermodynamics

The 1st law of thermodynamics: combine continuity and conservation of energy → energy equation

– property of a system: location, velocity, pressure, temperature, mass, volume

– state of a system: condition as identified through properties of the system

The difference between the heat added to a system of masses and the work done by the system depends only on the initial and final states of the system (→ change in energy).

→ Conservation of energy

Q W

E

QWdE (4.14)

where Q = heat added to the system from surroundings

W = work done by the system on its surroundings

dE = increase in energy of the system

Consider time rate of change

QWdE  (4.15) dt dt dt

(1) Work

Wpressure = work of normal stresses acting on the system boundary

Wshear = work of tangential stresses done at the system boundary

on adjacent external fluid in motion

4−10 Ch 4. Continuity, Energy, and Momentum Equation

Wshaft = shaft work done on a rotating element in the system

(2) Energy

Consider e = energy per unit mass = E mass

eu = internal energy associated with fluid temperature = u

ep = potential energy per unit mass = gh

where h = local elevation of the fluid

q2 e = kinetic energy per unit mass = q 2

p u enthalpy 

q2 ee e e ugh (4.16) upq 2

• Internal energy

= activity of the molecules comprising the substance

= force existing between the molecules

~ depend on temperature and change in phase

4.2.2 General energy equation

QWdE  (4.15) dt dt dt

4−11 Ch 4. Continuity, Energy, and Momentum Equation

Consider work done

W WW W pressure shaft shear dt dt dt dt (4.15a)

 dA2

 q 2

W pressure = net rate at which work of pressure is done by the fluid on the surroundings dt

= work fluxout – work fluxin   = p qdA CS  

2 p = pressure acting on the surroundings = FA F/ L

 Positive for outflow into CV   qdA   Negative for inflow    qdAQ L3 / t

4−12 Ch 4. Continuity, Energy, and Momentum Equation

  FL3 p qdA 2 FL// t E t L t

Thus, (4.15a) becomes

W   Wshaft W p qdA shear CS  dt dt dt (4.15b)

Consider energy change

dE = total rate change of stored energy dt

= net rate of energy flux through C.V.

(@energytt t energy @) tt 

+ time rate of change inside C.V.

   = e qdA  edV CS CV t (4.15c)   e E/; mass qdA mass / time

  eEt qdA  /

Substituting (4.15b) and (4.15c) into Eq. (4.15) yields

Q Wshaft W   shear  p qdA  dt dt dt CS 

   e qdA edV CSt CV

QWW shaft shear dt dt dt

4−13 Ch 4. Continuity, Energy, and Momentum Equation

p     e  qdA edV (4.17) CS CV  t

Assume potential energy eghp  (due to gravitational field of the earth)

q2 Then eu gh  2

Then, Eq. (4.17) becomes

Q W W shaft shear dt dt dt pq2    ugh  qdA edV (4.17) CS CV  2 t

◈ Application: generalized apparatus

At boundaries normal to flow lines → no shear

→ Wshear  0

4−14 Ch 4. Continuity, Energy, and Momentum Equation

2 Q Wshaft pq   ugh  qdA edV (4.19) CS CV dt dt 2  t

For steady motion,

2 Q Wshaft pq  ugh   qdA (4.20) CS  dt dt  2

◈ Effect of friction

~ This effect is accounted for implicitly.

~ This results in a degradation of mechanical energy into heat which may be transferred away

(Q , heat transfer), or may cause a temperature change

→ modification of internal energy.

→ Thus, Eq. (4.20) can be applied to both viscous fluids and non-viscous fluids (ideal frictionless processes).

4.2.3 1 D Steady flow equations

For flow through conduits, properties are uniform normal to the flow direction.

→ one - dimensional flow

1 2

V1 V2

Integrated form of Eq. (4.20) = ②-①

22 Q Wshaft pV pV ugh QQ  ugh    dt dt 22②①

4−15 Ch 4. Continuity, Energy, and Momentum Equation

V 2 where average kinetic energy per unit mass 2   Section1: qdA  Q mass flow rate into CV 1     Section2: qdA  Q mass flow from CV 2 

Divide by Q (mass/time)

22 heat transfer Wshaft pV pV ughugh     mass mass 22②①

Divide by g

22 heat transfer Wshaft  up V up V   hh    weight weight  gggg22②① 

(4.21)

◈ Energy Equation for 1-D steady flow: Eq. (4.21)

~ use average values for p,,, huand V at each flow section

~ use Ke (energy correction coeff.) to account for non-uniform velocity distribution over flow cross section

  1 mV 2 KVQ22 qdQ ---- kinetic energy/time  e 22 2 t

  qdQ2 K 2 1 e  VQ2 2

4−16 Ch 4. Continuity, Energy, and Momentum Equation

W 22 heat transfershaft pVpV u21 u hKee   hK  weight weight22gg②① g (4.23)

Ke = 2, for laminar flow (parabolic velocity distribution)

1.06, for turbulent flow (smooth pipe)

For a fluid of uniform density 

p V22 p VW heat transfer u u 1122hK   hK shaft   21 12ee1222g g weight weight g

(4.24)

→ unit: m (energy per unit weight)

For viscous fluid;

heat transfer u u 21H L12 weight g

→ loss of mechanical energy

~ irreversible in liquid

Then, Eq. (4.24) becomes

pVpV22 1122hK   hK  H H (4.24a) 12eeML121222gg

where H M = shaft work transmitted from the system to the outside

HH H H (4.24b) 12 M L12

where H12, H = weight flow rate average values of total head

4−17 Ch 4. Continuity, Energy, and Momentum Equation

◈ Bernoulli Equation

Assume

① ideal fluid → friction losses are negligible

② no shaft work → H M  0

③ no heat transfer and internal energy is constant → H  0 L12

pVpV22 1122hK   hK (4.25) 12ee1222gg

H12 H Potential head Pressure head

If KK1, then Eq. (4.25) reduces to Velocity head ee12

pVpV22 Hh1122 h (4.26) 1222gg

~ total head along a conduct is constant

◈ Grade lines

4−18 Ch 4. Continuity, Energy, and Momentum Equation

1) Energy (total head) line (E.L) ~ H above datum

2) Hydraulic (piezometric head) grade line (H.G.L.)

p ~  h above datum 

For flow through a pipe with a constant diameter

22 VV12 VV12  22gg

1) If the fluid is real (viscous fluid) and if no energy is being added, then the energy line may never be horizontal or slope upward in the direction of flow.

2) Vertical drop in energy line represents the head loss or energy dissipation.

4−19 Ch 4. Continuity, Energy, and Momentum Equation

4.3 Linear Momentum Equation for Finite Control Volumes

4.3.1 Momentum Principle

Newton's 2nd law of motion      dqdmq  dM Fmam   (4.27) dt dt dt   M  linear momentum vector = mq  F  external force   boundary (surface) forces: normal to boundary - pressure, Fp  tangential to boundary - shear, Fs  body forces - force due to gravitational or magnetic fields, Fb

 dM FFFpsb (4.28) dt  Ff  dv , where f = body force per unit mass bbCV b

4.3.2 The general linear momentum equation



dA1  dA2 positive

negative

4−20 Ch 4. Continuity, Energy, and Momentum Equation

Consider change of momentum  dM = total rate of change of momentum dt

= net momentum flux across the CV boundaries

+ time rate of increase of momentum within CV

   qqdVqdA (4.29) CSt CV    where q qdA  = flux of momentum = velocity mass per time  dA = vector unit area pointing outward over the control surface

Substitute (4.29) into (4.28)

      FpsbFF qqdA  qdV (4.30) CSt CV

For steady flow and negligible body forces     FFps q qdA (4.30a) CS  

• Eq. (4.30)

1) It is applicable to both ideal fluid systems and viscous fluid systems involving friction and energy dissipation.

2) It is applicable to both compressible fluid and incompressible fluid.

3) Combined effects of friction, energy loss, and heat transfer appear implicitly in the magnitude of the external forces, with corresponding effects on the local flow velocities.

• Eq. (4.30a)

1) Knowledge of the internal conditions is not necessary.

2) We can consider only external conditions.

4−21 Ch 4. Continuity, Energy, and Momentum Equation

4.3.3 Inertial control volume for a generalized apparatus

• Three components of the forces

     x dir.: Fpsb F F uqdA u dV xx xCSt CV

     ydirF.: psb F F vqdA vdV yyyCSt CV

     z dir.: Fpsb F F wqdA w dV (4.32) zzzCS CV t

• For flow through generalized apparatus

  x dir.: Fpsb F F u dQ u dQ u  dV xxx21t CV

 For 1 - D steady flow, qdV 0 t CV

~ Velocity and density are constant normal to the flow direction.   x dir.: F F F F V Q V Q psbxxx xx 21 x

4−22 Ch 4. Continuity, Energy, and Momentum Equation

VQ VQQVV  Q VV xx2122 11  xx21  xxout in   QQQ  (4.12) ydir.: F VQ VQ 11 2 2  y  yy21 

zdir.: F VQ VQ (4.33)  zz 21 z where V = average velocity in flow direction

If velocity varies over the cross section, then introduce momentum flux coefficient

 V  q

    qKV  VA  qdA m     qdQKVQ    m  qdQ K    m VQ

where

V = magnitude of average velocity over cross section =QA  V = average velocity vector

Km = momentum flux coefficient  1

= 1.33 for laminar flow (pipe flow)

1.03-1.04 for turbulent flow (smooth pipe)

4−23 Ch 4. Continuity, Energy, and Momentum Equation

[Cf] Energy correction coeff.

  qdQ2 K  2 e   VQ 2

F KV Q KV Q  xmxmx21 

F K VQ KVQ  y  my21 my 

F KV Q KV Q (4.34)  zmzmz21 

[Example 4-4] Continuity, energy, and linear momentum with unsteady flow

2 , 2 ,  A1  20 ft A2  0.1 ft h0  16

At time t 0 valve on the discharge nozzle is opened. Determine depth h , discharge rate

Q , and force F necessary to keep the tank stationary at t  50 sec.

4−24 Ch 4. Continuity, Energy, and Momentum Equation i) Continuity equation, Eq. (4.4)

 dV qnn dA12  q dA t CV

dV A11 dh,0 qn dA (because no inflow across the section (1))

 h  A122dh V A t 0

dh A VA (A) 122dt

ii) Energy equation, Eq. (4.19)

~ no shaft work

~ heat transfer and temperature changes due to friction are negligible

Q W W  shaft  shear dt dt dt

pq2    ugh  qdA edV CS CV  2 t II I q2 e = energy per unit mass = ugh 2

pq2   I = ugh   qdA CS   2

pq22 pq ugh VA22 ugh   VA 11 2221

pq2  ugh  VA22 V1  0  2 2

4−25 Ch 4. Continuity, Energy, and Momentum Equation

q2 II = edV  ugh dV CV CV  tt2 A1 dh

∵ nearly constant in the tank except near the nozzle

 h  A  ugh dh 1 t 0

2 pq  h  0 ugh VA22 A 1  ugh  dh 0  2 2 t

Assume  const. ,pp22atm 0 , h 0 (datum)

Vdhdh2 0 uV A2 V A  uA  A gh (B) 222 22 1dt 1 dt

Substitute (A) into (B)

V 2 0  uV A 2 VA uVA  ghVA 22 2 22 22 22

V 2  2 VA ghVA 2 22 22

Vgh2  2 (C)

Substitute (C) into (A)

dh AghA2  21dt

dh A  2 2gdt h A1

4−26 Ch 4. Continuity, Energy, and Momentum Equation

Integrate

1 htdh A  h 1 h  2 2   2gdt h dh  2 hh0    0 0 2h h h A1 0

2 1 2 A2 2g h  ht0  A1 2

2 0.1 232.2 h  16  t 20 2

2  4 0.0201t

2 At th50sec ,40.020150  8.98 ft

Vgh2 2 232.2 8.98  24.05 fps

Q2 VA2 24.05 0.1  2.405 cfs

I II iii) Momentum equation, Eq. (4.30)

      FFps  Fb qqdVqdA CSt CV

II = Time rate of change of momentum inside CV is negligible

if tank area  A1  is large compared to the nozzle area  A2  .

   I = qqdA  qnn qdA21 q nn qdA  VVA222 CS 

 FVVAVQpx 222 2 2

Fpx 1.94 24.05 2.405 112 lb

4−27 Ch 4. Continuity, Energy, and Momentum Equation

4.4 The Moment of Momentum Equation for Finite Control Volumes

4.4.1 The Moment of momentum principle for inertial reference systems

Apply Newton's 2nd law to rotating fluid masses   → The vector sum rF  of all the external moments acting on a fluid mass equals the

  time rate of change of the moment of momentum (angular momentum) vector rM  of the fluid mass.

 d  TrF  rM  (4.35) dt  where r = position vector of a mass in an arbitrary curvilinear motion  M = linear momentum

[Re] Derivation of (4.35)   dM Eq. (4.27): F  dt  Take the vector cross product of r

4−28 Ch 4. Continuity, Energy, and Momentum Equation

  dM rF r dt

By the way,   ddrdM rM M r dt dt dt

I   dr  dr  IMqmassq 0   q dt dt   qq qqsin0  0   dM d    r  rM dt dt

d rF rM  dt   where rM = angular momentum (moment of momentum)

[Re] Torque  TrF translational motion → Force – linear acceleration rotational motion → Torque – angular acceleration

[Re] Vector Product    Vab    Magnitude = V a b sin = area of parallelogram   direction = perpendicular to plane of a and b → right-handed triple

4−29 Ch 4. Continuity, Energy, and Momentum Equation

 ba ab  ka b k a  b   abc   ab  ab 

• External moments arise from external forces

Forces Moment     Boundary (surface) force FFp , s TTp , s   Body force Fb Tb

 d   rFpsb rF rF rM  dt    T T p T s b d   TTTpsb  rM  dt (4.36)  where TTTp ,,sb = external torque

4.4.2 The general moment of momentum equation  dM    qqdVqdA dtCS t CV

d        rMrqqdA   rq dV dtCS t CV

     TTT rqqdA   rq dV(4.37) CS CV psb t

4−30 Ch 4. Continuity, Energy, and Momentum Equation

  xdir.:rq yz rqyz yz sin  rqcos 2 yz

angle betweenqryz and yz

     TTT rqcosqdA   rq cos dV px sx bx CSyz CV yz t

(4.39a)

    y dir.: Tpy T sy T by rqcosqdA  rq cos dV CSzxt CV zx

    zdirT.: pz T sz T bz rqcosqdA  rq cos dV CSxyt CV xy

4.4.3 Steady-flow equations for Turbomachinery

■ Turbomachine

~ pumps, turbines, fans, blows, compressors

~ dynamic reaction occurs between a rotating vaned element (runner) and fluid passing through the element

If rotating speed is constant → Fig. 4.14

TrVdQrVdQcos cos  – Euler's Eq. (4.40) r AA22 2 11 1 21

Where = radii of fluid elements at the entrance and exist of the runner rr12,

VV12, = velocities relative to the fixed reference frame (absolute)

= angles of absolute velocities with tangential direction 12, 

A12, A = entrance and exit area of CV which encloses the runner

4−31 Ch 4. Continuity, Energy, and Momentum Equation

Tr = torque exerted on fluid by the runner

• Tr positive for pump, compressor

negative for turbine

If rV22cos 2and rV 11 cos 1 are constant,  is constant

TQr   rV22cos 2 rV 11 cos 1 (4.41)

If rV22cos 2= rV 11 cos 1

QrVcos Qrv const.

~ zero torque, constant angular momentum

If   0 → free vortex flow

4−32 Ch 4. Continuity, Energy, and Momentum Equation

Homework Assignment # 2

Due: 1 week from today

4-11. Derive the equation for the volume rate of flow per unit width for the sluice gate

shown in Fig. 4-20 in terms of the geometric variable b , y1 , and cc . Assume the pressure

in hydrostatic at y1 and cbc and the velocity is constant over the depth at each of these sections.

4-12. Derive the expression for the total force per unit width exerted by the above sluice gate on the fluid in terms of vertical distances shown in Fig. 4-20.

4-14. Consider the flow of an incompressible fluid through the Venturi meter shown in Fig. 4-

22. Assuming uniform flow at sections (1) and (2) neglecting all losses, find the pressure difference between these sections as a function of the flow rate Q , the diameters of the sections, and the density of the fluid,  . Note that for a given configuration, Q is a function of only the pressure drop and fluid density. The meter is named for Venturi, who investigated its principle in about 1791. However, in 1886 Clemens Herschel first used the meter to measure discharge, and he is usually credited with its invention.

4−33 Ch 4. Continuity, Energy, and Momentum Equation

4-15. Water flows into a tank from a supply line and out of the tank through a horizontal pipe as shown in Fig. 4-23. The rates of inflow and outflow are the same, and the water surface in the tank remains a distance h above the discharge pipe centerline. All velocities in the tank are negligible compared to those in the pipe. The head loss between the tank and

the pipe exit is H L . (a) Find the discharge Q in terms of h , A , and H L . (b) What is the

horizontal force, Fx , required to keep the tank from moving? (c) If the supply line has an area A' , what is the vertical force exerted on the water in the tank by the vertical jet?

4-28. Derive the one-dimensional continuity equation for the unsteady, nonuniform flow of

an incompressible liquid in a horizontal open channel as shown in Fig. 4-29. The channel

has a rectangular cross section of a constant width, b. Both the depth, y0 and the mean

4−34 Ch 4. Continuity, Energy, and Momentum Equation velocity, V are functions of x and t.

4−35