Integral Relation for a Control Volume

Department of Mechanical Engineering MEHB223 Mechanics of Fluids 1: Lecture 8: Integral Relations for CV (Part 4)

Lecture 8 Integral Relation for a Control Volume ((Part 4))


Chapter Summary Review of 1 f st Law of Thermodynamics f y Derivation of the Energy Equation in CV Steady Flow Energy Equation Incompressible Flow in a Single Pipeline Bernoulli Equation Revisited Kinetic Energy Correction Factor Kinetic Energy Correction Factor Application of Energy, Momentum and Continuity Equation in Combination • Concept of Hydraulic and Energy Grade Line

• • • • • • •

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7.1. Review of 1st Law of Thermo • The 1st law of thermodynamics is basically the statement of the conservation of energy • The expression of 1st law for a SYSTEM is : ΔE = Q − W

Where : ΔE = The change of energy of the system Q = Heat transfer TO the system W = Work done BY the system


7.1. Review of 1st Law of Thermo

The corresponding rate form :dE dt

•

•

•

= E = Q− W

E can be splitted into 3 main components which are the Kinetic Energy, Ek, the Potential Energy, Ep and the Internal Energy, Eu which is associated with the motion of the molecules. •

•

•

•

E = Ek + E p + Eu •

•

E p = m gz = ρVA( gz )

•

Ek = •

1 • 2 ⎛1 ⎞ m V = ρVA⎜ V 2 ⎟ 2 ⎝2 ⎠ •

•

E u = U = m u = ρVA(u )

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7.2. Derivation of EE in Control Volume • Consider the general CV Equation : • • dX C dX S = + ∑ Xi − ∑ Xo dt dt

Let X be the energy, E : • • dEC dES = + ∑ Ei − ∑ Eo dt dt

B 1st Law By L off Thermo Th :

So the General EE :

dES • • = Q− W dt

• • dEC • • = Q− W + ∑ E i − ∑ E o dt


7.2. Derivation of EE in Control Volume… • Work, W can be divided into shaft work, Ws and flow work, Wf : •

•

•

W = Ws + W f

Flow work is the work done by the system ON the surrounding due to the pressure forces as the system moves through the surrounding •

•

•

W f = ∑ W f ,out − ∑ W f ,in • ⎛ • p⎞ ⎛ • p⎞ W f = ∑ ⎜⎜ m ⎟⎟ − ∑ ⎜⎜ m ⎟⎟ ⎝ ρ ⎠ out ⎝ ρ ⎠ in

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7.2. Derivation of EE in Control Volume… • Substitution into the general EE gives : • dEC • • ⎛• = Q− W s + ∑ ⎜ E i + W dt ⎝

f ,i

• ⎞ ⎛• ⎟ − ∑⎜ Eo +W ⎠ ⎝

f ,o

⎞ ⎟ ⎠

• ⎛1 • ⎛1 dEC • • p⎞ p⎞ = Q − W s + ∑ mi ⎜⎜ V 2 + gz + u + ⎟⎟ − ∑ mo ⎜⎜ V 2 + gz + u + ⎟⎟ dt 2 ρ 2 ρ ⎝ ⎠i ⎝ ⎠o

⎛1 ⎛1 dEC • • p⎞ p⎞ = Q − W s + ∑ (ρVA)i ⎜⎜ V 2 + gz + u + ⎟⎟ − ∑ (ρVA)o ⎜⎜ V 2 + gz + u + ⎟⎟ dt 2 ρ 2 ρ ⎝ ⎠i ⎝ ⎠o

For general nonnon-uniform flow :

⎛1 ⎛1 dEC • • p⎞ p⎞ = Q − W s + ∑ ∫ (ρVA)i ⎜⎜ V 2 + gz + u + ⎟⎟ dA − ∑ ∫ (ρVA)o ⎜⎜ V 2 + gz + u + ⎟⎟ dA dt ρ ⎠i ρ ⎠o ⎝2 ⎝2 A A


7.3. Steady Flow Energy Equation • For steady flow :

dEC =0 d dt

Steady Flow Energy Equation (SFEE) : • • • ⎛1 • ⎛1 p⎞ p⎞ Q − W s = ∑ mo ⎜⎜ V 2 + gz + u + ⎟⎟ − ∑ mi ⎜⎜ V 2 + gz + u + ⎟⎟ ρ ⎠o ρ ⎠i ⎝2 ⎝2

• • ⎛1 ⎛1 p⎞ p⎞ Q − W s = ∑ (ρVA)o ⎜⎜ V 2 + gz + u + ⎟⎟ − ∑ (ρVA)i ⎜⎜ V 2 + gz + u + ⎟⎟ ρ ⎠o ρ ⎠i ⎝2 ⎝2 • • ⎛1 ⎞ ⎛1 ⎞ Q − W s = ∑ (ρVA)o ⎜ V 2 + gz + h ⎟ − ∑ (ρVA)i ⎜ V 2 + gz + h ⎟ ⎝2 ⎠o ⎝2 ⎠i • • ⎛1 ⎞ ⎛1 ⎞ Q − W s = ∑ ∫ (ρVA)o ⎜ V 2 + gz + h ⎟ dA − ∑ ∫ (ρVA)i ⎜ V 2 + gz + h ⎟ dA ⎝2 ⎠o ⎝2 ⎠i A A

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7.4. Incompressible Flow in a Single Pipeline Turbine

2 •

P2, V2, u2

Wt P1, V1, u1

•

Q •

1

Wp

Pump

• Applying SFEE : • • • ⎛1 p ⎞ • ⎛1 2 p ⎞ 2 Q − W s = m2 ⎜⎜ V2 + gz2 + u2 + 2 ⎟⎟ − m! ⎜⎜ V1 + gz1 + u1 + 1 ⎟⎟ ρ ⎠ ρ⎠ ⎝2 ⎝2 •

•

•

m1 = m2 = m

•

•

•

Ws = Wt − W p


7.4. Incompressible Flow in a Single Pipeline • Let :

•

Q

q=

•

= Heat transfer per unit mass of moving fluid

m •

wt =

Wt •

= Turbine work done per unit mass of moving fluid

m •

wp =

Wp •

= Pump work input per unit mass of moving fluid

m

S b Substituting into SSFEE : ⎛1 2 p ⎞ ⎛1 2 p ⎞ q − wt + w p = ⎜⎜ V2 + gz 2 + u 2 + 2 ⎟⎟ − ⎜⎜ V1 + gz1 + u1 + 1 ⎟⎟ ρ ⎠ ⎝2 ρ⎠ ⎝2

Rearranging : 1 V12 + gz1 + p1 + wp = 1 V2 2 + gz2 + p2 + wt + [(u2 − u1 ) − q] 2

ρ

2

ρ

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7.4. Incompressible Flow in a Single Pipeline

The term in the square bracket represent the thermal energy As fluid flows some mechanical energy will be energy. changed into thermal energy through viscous action. This cause increase in internal energy (temperature rise) and eventually dissipated to the surrounding in term of heat. Thus this represent loss.

Let :

Eqn becomes :

The above equation physically express the energy balance. Specific energy input = specific energy output

eL = Energy loss per unit mass of moving fluid

1 2 1 2 p p V1 + gz1 + 1 + w p = V2 + gz 2 + 2 + wt + eL 2 ρ 2 ρ


7.4. Incompressible Flow in a Single Pipeline

It is common to write the equation in term of head (energy per unit weight) in meter meter. 2

2

p V p V1 + z1 + 1 + h p = 2 + z 2 + 2 + ht + hL 2g ρg 2g ρg

Where : h p = Head supplied by the pump

ht = Head given up in turbine

hL = Head loss in pipe

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7.5. Bernoulli Equation Revisited

Recall that in BE we assumed that Frictionless F i i l fluid, fl id hL = 0 No external work, wp = wt = 0 If the above assumptions are used in the pipeline eqn : 2

2

p1 V1 p V + + z1 = 2 + 2 + z 2 ρg 2 g ρg 2 g p1 +

1 1 ρV12 + ρgz1 = p2 + ρV2 2 + ρgz 2 2 2

Therefore BE is a statement of conservation of energy for frictionless fluid with no external work


Example 1 A pump draws water through a pipe p and discharges g it 20--cm p 20 through a 15 15--cm pipe in which the velocity is 5 m/s. The 15 15--cm pipe discharges horizontally into air at point C. To what height h above the water surface at A can the water be raised if 35 kW is delivered to the pump? Assume that the pump operates at 70% efficiency and that the head loss in the pipe between A and C is equal to 2Vc2/2g.

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7.6. Kinetic Energy Correction Factor

Recall the general single pipeline equation : p p 1 2 1 2 V1 + gz1 + 1 + w p = V2 + gz 2 + 2 + wt + eL 2 ρ 2 ρ

In the above equation, it was assumed that the velocity profiles are uniform. In reality, because of no slip condition at the walls, the velocity is not uniform. To take this into account,, Kinetic Energy gy Correction Factor, α, is introduced such that equation becomes: α1

2

2

V1 p V2 p + gz1 + 1 + w p = α 2 + gz 2 + 2 + wt + eL 2 ρ 2 ρ

α1

2

2

V2 p V1 p + z1 + 1 + h p = α 2 + z 2 + 2 + ht + hL 2g ρg 2g ρg


7.6. Kinetic Energy Correction Factor …

α is defined such that : •

(

)V2 = ∫ ρV2dA 2

E k = α ρV A

3

A

So :

3

α=

1 ⎛V ⎞ ⎜ ⎟ dA A ∫A ⎝ V ⎠

For laminar flow with parabolic velocity profile, α is equal to 2. For turbulent flow, α will be approximately 1.05. In most common practice, when the flow is turbulent, α is normally assumed to be unity.

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Example 2

A microchannel is being designed to transfer fluid in a MEMS application. The channel is 200 micrometers in diameter and is 5 cm g Ethyl y alcohol is driven through g the system y at the rate off 0.1 long. microlitres/s with a syringe pump which is essentially a moving piston. The pressure at the exit of the channel is atmospheric. The flow is laminar, so α=2. The head loss in the channel is given by hL = 32μLV/gD2. Find the pressure in the syringe pump. The velocity head associated with the motion of the piston in the syringe pump is negligible.


Example 3

I this In hi system, d = 25 cm, D = 40 cm and d the h hhead d lloss from f the h venturii 2 meter to the end of the pipe is given by hL = 0.9V /2g, where V is the velocity in the pipe. Neglecting all other head losses, determine what head, H will first initiate cavitation if the atmospheric pressure is 100 kPa. What will be the discharge at incipient cavitation ?

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7.7. Application of EE, FME and MCE Combined

Head loss in an Abrupt Expansion

MCE :

FME :

V1 A1 = V2 A2 p1 A2 − p2 A2 − ρgA2 L sin α = ρV2 A2 − ρV1 A1 2

2


7.7. Application of EE, FME and MCE Combined 2

EE :

Combining :

2

p1 V1 p V2 + α1 + z1 = 2 + α 2 + z 2 + hL ρg 2g ρg 2

hL =

(V1 − V2 )2 2g

=

2 V1 ⎛ A ⎞ ⎜⎜1 − 1 ⎟⎟ 2 g ⎝ A2 ⎠

2

Special case of discharging from a pipe into a reservoir : hL =

V2 2g

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7.7. Application of EE, FME and MCE Combined

Forces on a Pipe Bend

Water flows in a bend at a rate of 5 m3/s and the pressure at the inlet is 650 kPa. If the head loss in the bend is 10m, what will the pressure be at the outlet of the bend? Estimate the force of the anchor block on the bend in the x direction to hold the bend in place.


7.8. EGL and HGL

EGL and HGL are representations of the changes in energy gy and transformation f off energy gy from f one form f to another along a pipeline system. In general (in term of head/energy per unit weight) : ⎛ p ⎞ V2 h total = ⎜⎜ + z ⎟⎟ + ⎝ ρg ⎠ 2g Total Head

Piezometric Head

Dynamic Head

EGL (Energy Grade Line) gives the locus of Total Head along the pipe as measured by total pressure probe HGL (Hydraulic Grade Line) gives the locus of Piezometric Head along the pipe as measured by a piezometer

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7.8. EGL and HGL …

EGL – HGL = Dynamic Head Consider a simple pipeline system as below :

Note that as the pipe diameter is constant, the velocity is constant and hence the spacing between EGL & HGL is constant throughout the length of the pipe.



Pump - In a pump energy is added .

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Turbine - In a turbine energy is given up .



Change in HGL and EGL due to flow through a nozzle – the spacing increase since the velocity increase. There will be an extra losses as well at the transition.

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Change in HGL and EGL due to change in the pipe diameter – the spacing increase since the velocity increase. The gradient increase due to the larger frictional forces causes larger rate of loss of energy



Subatmospheric pressure when pipe is above HGL – Gage Pressure is negative. The gradient and the spacing between EGL and HGL are constant since velocity is constant throughout the pipe

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Example 4

Sketch the EGL and HGL for the above pipe system.


Example 5

(1) Indicate which is HGL and which is EGL (2) Are all pipes are of the same size? If not which one is smaller? (3) Is there any region in pipe where pressure is below atmospheric (4) Where is the maximum pressure (5)Where is the minimum pressure (6) What do u think located at the end of the pipe (7) What do u think located at point B? (8) Is the pressure of the air in the tank above or below atmospheric?

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End of Lecture 8

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Integral Relation for a Control Volume

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