ME 262 BASIC FLUID MECHANICS Assistant Professor Neslihan ...

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ME 262 BASIC FLUID MECHANICS

Assistant Professor Neslihan Semerci

Lecture 6

(Bernoulli’s Equation)

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19. CONSERVATION OF ENERGY- BERNOULLI’S EQUATION

Law of Conservation of Energy: “energy can be neither created nor destroyed. It can be

transformed from one form to another.”

Potential energy

Kinetic energy

Pressure energy

In the analysis of a pipeline problem accounts for all the energy within the system.

Inner wall of the pipe

V

Centerline

z element of fluid

Reference level

An element of fluid inside a pipe in a flow system;

- Located at a certain elevation (z)

- Have a certain velocity (V)

- Have a pressure (P)

The element of fluid would possess the following forms of energy;

1. Potential energy: Due to its elevation, the potential energy of the element relative to some

reference level

PE = Wz

W= weight of the element.

2. Kinetic energy: Due to its velocity, the kinetic energy of the element is

KE = Wv2/2g

P

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3. Flow energy(pressure energy or flow work): Amount of work necessary to move

element of a fluid across a certain section aganist the pressure (P).

PE = W P/γ

Derivation of Flow Energy:

L

P

F = PA

Work = PAL = FL=P∀

∀= volume of the element.

Weight of element W = γ ∀volume

Volume of element ∀ = W

γ

Total amount of energy of these three forms possessed by the element of fluid;

E= PE + KE + FE

E = Wz + Wv2/2g + W P

γ

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Figure 19.1. Element of fluid moves from a section 1 to a section 2, (Source: Mott, R. L.,

Applied Fluid Mechanics, Prentice Hall, New Jersey)

Total energy at section 1: E1 = Wz1 + W V1

2

2g +W

P1

γ

Total energy at section 2: E2 = Wz2 + W V2

2

2g + W ×

P2

γ

If no energy is added to the fluid or lost between sections 1 and 2, then the principle of

conservation of energy requires that;

E1 = E2

Wz1 + W V1

2

2g +W

P1

γ = Wz2 + W

V22

2g + W ×

P2

γ

The weight of the element is common to all terms and can be divided out.

BERNOULLI’S EQUATION

z1 + v2

2g +

P1

γ = z2 +

v22

2g +

P2

γ

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19.1 INTERPRETATION OF BERNOULLI’S EQUATION

z1 + v2

2g +

P1

γ = z2 +

v22

2g +

P2

γ

Each term in Bernoulli’s equation is one form of the energy possessed by the fluid per unit

weight of fluid flowing in the system. The units in each term are “energy per unit weight”. In

the SI system the units are N.m/N and in the U.S. Customary System the units are Ib.ft/Ib.

The force unit appears in both the numerator and denominator and it can be cancelled. The

resulting unit is simply the meter (m) or foot (ft) and can be interpreted to be a height. In fluid

flow analysis, the terms are typically expressed as “head” referring to a height above a

reference level.

P/γ pressure head

z elevation head Summation of the terms called total head

V2/2g velocity head

Figure 19.2.Pressure head, elevation head, velocity head, and total head in a pipe flow.

(Source: Mott, R. L., Applied Fluid Mechanics, Prentice Hall, New Jersey)

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19.2. Restrictions on Bernoulli’s Equation

1. It’s valid only for incompressible liquids since the specific weight of the fluid is assumed

to be the same at two sections of interest.

2. There can be no mechanical devices between section 1 and 2 that would add energy to or

remove energy from the system, since the equation states that the total energy of fluid is

constant.

3. There can be no heat transfered into or out of the fluid.

4. There can be no energy loss due to friction.

Example 19.1. In Figure 19.2, water at 100C is flowing from section 1 to section 2. At section

1, which is 25 mm in diameter, the gage pressure is 345kPa and the velocity of flow is 3.0

m/s. Section 2, which is 50 mm in diameter, is 2.0 m above section 1. Assuming there are no

energy losses in the system, calculate the pressure P2.

Knowns;

P1 = 345 kPa P2 =?

V1 = 3 m/s V2 =?

∅1 = 25 mm ∅2 = 50 mm

Twater = 10 ℃

z2 − z1 = 2.0 m

Q = A1 × V1

(25 mm

1000 m)

2 π

4 × 3 m/s = 1.47 × 10−3 m3/s

From continuity Equation

Q = A1 × V1 = A2 × V2

(25 mm

1000 m)

2 π

4 × 3 m/s = (

50 mm

1000 m)

2 π

4 × V2

V2 = 0.75 m/s

P1 = 345 kPa = 345 kN/m2

z1 + v1

2

2g +

P1

γ = z2 +

v22

2g +

P2

γ

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z1 + 345 kN/m2

9.81 kN/m3 + (3.0)2

2×(9.81) = z2 +

(0.75)2

2×(9.81) +

P2

γ

P2

γ =

345 kN/m2

9.81 kN/m3 + (3.0)2

2×(9.81) −

(0.75)2

2×(9.81) + z1 − z2

P2

γ = 35.1 m + 0.458 −0.0286 – 2 m

P2

γ = 33.5 m Pressure Head

P2 = (9.81 kN/m3) × (33.5 m) = 328.9 kPa

19.3. Tanks, Reservoirs, and Nozzles Exposed to the Atmosphere

A siphon draws fluid from a tank and delivers it through a nozzle at the end of the pipe.

Point A and F are exposed to atmosphere (they are not confined with a solid boundry).Gage

pressure of point A and F are zero.

(Source: Mott, R. L., Applied Fluid Mechanics, Prentice Hall, New Jersey)

Pressure Head at points open to the atmosphere :When the fluid at reference point is exposed

to the atmosphere, the pressure is zero and the pressure head term can be cancelled from

Bernoulli’s equation.

Velocity Head at points inside the tank or reservoir :The velocity head at the surface of the

tank or reservoir is considered to be zero and it can be cancelled from Bernoulli’s equation.

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Velocity Heads at points with the same c/s area:Points B,C, D and E are inside the pipes and

has a uniform flow area. The velocity will be same througout the pipe.

When the two points of reference for Bernoulli’s equation are both inside a pipe of the same

size, the velocity head terms at both sides of the equation are equal and can be cancelled.

Elevation Head at points on the same level:When two points of reference for Bernoulli’s

equation are both at the same elevation, the elevation head terms, z1 and z2 are equal and can

be cancelled.

Example: (Example 6.10 from Mott, R. L., Page 164)

A siphon draws fluid from a tank and delivers it through a nozzle at the end of the pipe.

Calculate the volume flow rate through the siphon and the pressure at points B, C, D, and E.

Diameter of siphon = 40 mm

Diameter of nozzle = 25 mm

Assume that there are no energy losses in the system.

Reference points A and F (most convenient points)

zA + vA

2

2g +

PA

γ = zF +

vF2

2g +

PF

γ

A is open to the atmosphere PA = 0

Velocity of point A is small, velocity head will be negligible.

Elevation difference between A and F = (1.8+1.2) = 3.0 m.

F is open to the atmosphere, PF = 0

zA − zF = 3.0 m

zF − zA = −3.0 m

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zA = zF + vF

2

2g

VF = √(zA − zF) × 2g

VF = √(3.0) × (2 × 9.81)

VF = 7.67 m/s

Q = A × V

= (25 mm

1000 m)

2 π

4 × 7.67 m/s Q = 3.77 × 10−3 m3/s

c/s area of nozzle

Determine the pressures at points B, C, D and E

Write Bernoulli’s Equation between A and B

zA + vA

2

2g +

PA

γ = zB +

vB2

2g +

PB

γ

zA = zB

Q = AB × VB

3.77 × 10−3 m3/s = (40 mm

1000 m)

2 π

4 × VB

VB = 3.0 m/s

0 = (3.0)2

2×(9.81) +

PB

γ

PB

9.81 kN/m3 = −0.459 m

PB = −4.50 kPa

Negative sign indicates the pressure is below

atmospheric pressure.

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Write Bernoulli’s Equation between A and C

zA + vA

2

2g +

PA

γ = zC +

vC2

2g +

PC

γ

vC = vB since pipe diameter is same.

3.0 m = 4.2 m + (3.0)2

2×(9.81) +

PC

9.81 kN/m3

PC = −16.27 kPa

Write Bernoulli’s Equation between A and D

zA + vA

2

2g +

PA

γ = zD +

vD2

2g +

PD

γ

zA = zD

0 = 32

2×(9.81) +

PD

9.81 kN/m3

PD = −4.50 kPa

Write Bernoulli’s Equation between A and E

zA + vA

2

2g +

PA

γ = zE +

vE2

2g +

PE

γ

3.0 m + 0 + 0 = 0 + (3.0)2

2×(9.81) +

PE

9.81 kN/m3

PE = 24.93 kPa

Summary of the Results of Example

1. Velocity from nozzle and therefore the volume flow rate delivered by siphon depends

on the elevation difference between the free surface of fluid and the outlet of nozzle.

2. The pressure at point B is below atmospheric pressure even though it is on the same

level as point A, which is exposed to the atmosphere. Bernoulli’s equation shows that

the pressure head at B is decreased by the amount of the velocity head. That is, some

of the energy is converted

3. The velocity of flow is the same at all points where the pipe size is the same, when

steady flow exists.

4. The pressure at point C is the lowest in the system because point C is at the highest

elevation.

5. The pressure at point D is the same as that at point B, because both are on the same

elevation and the velocity head at both points is the same.

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6. The pressure at point E is the highest in the systen because point E is at the lowest

elevation.

19.4 Venturi Meters and Other closed Systems with unknown velocities

Venturi meter is a device that can be used to measure the velocity of flow in a fluid in

the manometer. Analysis of venturi meters is based on the application of Bernoulli’s

Equation.

Example 6.11 from Mott, R. L., Page 168

The venturi meter shown in figure

carries water at 60℃ . The specific

gravity of the gage fluid in the

manometer is 1.25. Calculate the

velocity of flow at section A and the

volume flowrate of water.

Knowns: Elevation difference between

A and B = 0.46 m.Diameters of A and B

is known. The manometer allows the

determination of the difference in

pressure between points A and B.

γwater = 9.65 kN/m3

γM = (1.25)(9.81 at 4 ℃) = 12.26 kN/m3

Manometer equation starting at

A: PA + γwatery + γwater × (1.18) − γg × (1.18) − γwater × y − γwater × (0.46)

PA − PB = γ × (0.46 − 1.18) + γg × (1.18)

PA−PB

𝛾 = −0.72 +

γg

γ × (1.18)

= −0.72 + (12.26

9.65) × (1.18)

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PA−PB

𝛾 = 0.78 m

Q = AA × VA

AA = (300

1000)

2 ×

𝜋

4 AB = (

2001000

)2

× 𝜋

4

AA = 0.0706 m2 AB = 0.031 m2

VA = Q

AA VB =

Q

AB

AA × VA = AB × VB

0.0706 m2 × VA = 0.031 m2 × VB

VB = 2.25 VA

VB2 = 5.06 VA

2

VB2 − VA

2 = 5.06 VA2 − VA

2

= 4.06 VA2

zA + vA

2

2g +

PA

γ = zB +

vB2

2g +

PB

γ

PA

γ−

PB

γ + zA − zB =

vB2

2g−

vA2

2g

0.78 – 0.46 = 4.06 VA

2

2𝑔

VA = 1.24 m/s VB = 2.25 × (1.24)

= 2.79 m/s

Q = 0.0706 × (1.24 m/s)

= 0.08765 m3/s

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20. Application of Bernoulli’ s Equation

20.1 Velocity of flow from tank with a constant water depth

Figure 20.1 Flow from a tank (Source: Mott, R. L., Applied Fluid Mechanics, Prentice

Hall, New Jersey)

Bernoullis’ s equation between a reference point on the fluid surface and a point in the jet

issuing from the nozzle:

z1 + V1

2

2g +

P1

γ = z2 +

V22

2g +

P2

γ

P1 = 0 and P2= 0 (open to the atmosphere), V1 = 0 (velocity at the surface of the tank)

V2 = √2g(z1 − z2)

h= z1-z2

V2 = √2gh

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Example 20.1 (Example 6.12, Mott, R. L., Page 171) Compute the velocity of flow from a

nozzle for a fluid depth of 3.0 m.

V2 = √2 × (9.81) × (3 m) = 7.67 m/sec

Example 20.2 (Example 6.13, Mott, R. L., Page 172) Compute the velocity of flow from

the nozzle and the volume flowrate for a range of depth from 3.0 m to 0.5 m in steps of 0.50

m. The diameter of the jet at the nozzle is 50 mm.

Q = V × A

A=Area of the jet = (50

1000)

2×

π

4=1.963x10

-3 m

2

h V2(m/s) 𝑄(m3/s) = V × A

3.0 7.67 1.51 × 10−2

2.5 7.0 1.38 × 10−2

2.0 6.26 1.23 × 10−2

1.5 5.42 1.07 × 10−2

1.0 4.43 0.87 × 10−2

0.5 3.13 0.61 × 10−2

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20.2 Velocity and flow from a tank due to falling head (time required to empty a tank)

(Source: Mott, R. L., Applied Fluid Mechanics, Prentice Hall, New Jersey)

VJ = √2gh when water depth is equal to h.

The volume flowrate from the nozzle Q = AJ × VJ

In a small amount of time, dt, the volume of water flowing through the nozzle;

Volume flowing = Q(dt) = AJ × VJ(dt)

As the fluid leaving the tank, depth of fluid in the tank will decrease. For time interval dt, the

fluid level drops a small distance, dh

Volume removed = -Atdh

AJ × VJ(dt) = −At(dh)

Solving for time dt dt = −(At/AJ)

VJ × dh =

−(At/AJ)

√2gh × dh

VJ = √2gh

Volume of fluid

flowing

volume leaving

the tank

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dt = −(At/AJ)

√2gh × dh

Rewriting to separate the terms involving h gives

dt = −(At/AJ)

√2gh−1/2 × dh

∫ dtt2

t1 =

−(At/AJ)

√2g ∫ h−1/2 dh

h2

h1

t2 −t1 = −(At/AJ)

√2g ×

h21/2−h1

1/2

1/2

t2 −t1 = 2(At/AJ)

√2g ( h1

1/2 − h21/2)

Time requirement initial water depth final water depth

Example 20.3: (Example 6.15, Mott, R. L., Page 172) Find the time required to drain the

tank from a level of 3.0 to 0.5 m. The tank has a diameter of 1.50 m and the nozzle has a

diameter of 50 mm.

At = π(1.5)2/4 = 1.767 m2

AJ = π(50 mm/1000)2/4 = 0.001963 m2

At

AJ=

1.767

0.001963

t2 − t1 = 2(At/AJ)

√2g × (h1

1/2 − h21/2)

= 2(900)

√19.62 × (31/2 − 0.51/2)

t2 − t1 = 417 sec