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Bernoulli's equation and its applications, Frictional losses, Losses in pipe flow

1.
Energy Conservation (Bernoulli’s Equation)

dp

Recall Euler’s equation: VdV gdz 0

Also recall that viscous forces were neglected, i.e. flow is invisicd

If one integrates Euler’s eqn. along a streamline, between two points , &

2 dp 2 2

We get : 1 1 VdV 1 gdz 0

Which gives us the Bernoulli’s Equation

2

p1 V1 p2 V22

gz1 gz 2 Constant

2 2

Flow work + kinetic energy + potential energy = constant

dp

Recall Euler’s equation: VdV gdz 0

Also recall that viscous forces were neglected, i.e. flow is invisicd

If one integrates Euler’s eqn. along a streamline, between two points , &

2 dp 2 2

We get : 1 1 VdV 1 gdz 0

Which gives us the Bernoulli’s Equation

2

p1 V1 p2 V22

gz1 gz 2 Constant

2 2

Flow work + kinetic energy + potential energy = constant

2.
Bernoulli’s Equation (Continued)

Flow Work (p/) :

It is the work required to move fluid across the control volume boundaries.

Consider a fluid element of cross-sectional area x

p

A with pressure p acting on the control surface

as shown. A

Due to the fluid pressure, the fluid element moves a distance x within

time t. Hence, the work done per unit time W/t (flow power) is:

W pAx p x p

A AV ,

t t t

p 1 W Flow work or Power

AV t

Flow work per unit mass

1/mass flow rate

p

pv Flow work is often also referred to as flow energy

Flow Work (p/) :

It is the work required to move fluid across the control volume boundaries.

Consider a fluid element of cross-sectional area x

p

A with pressure p acting on the control surface

as shown. A

Due to the fluid pressure, the fluid element moves a distance x within

time t. Hence, the work done per unit time W/t (flow power) is:

W pAx p x p

A AV ,

t t t

p 1 W Flow work or Power

AV t

Flow work per unit mass

1/mass flow rate

p

pv Flow work is often also referred to as flow energy

3.
Bernoulli’s Equation (Cont)

Very Important: Bernoulli’s equation is only valid for :

incompressible fluids, steady flow along a streamline, no energy loss due

to friction, no heat transfer.

2

p1 V1 p2 V22

z1 z2 , where g (energy per unit weight)

2g 2g

Application of Bernoulli’s equation - Example 1:

Determine the velocity and mass flow rate of efflux from the circular

hole (0.1 m dia.) at the bottom of the water tank (at this instant). The

tank is open to the atmosphere and H=4 m

p1 = p2, V1=0

1

V2 2 g ( z1 z2 ) 2 gH

2 * 9.8 * 4 8.85 (m / s )

m AV 1000 * (0.1) 2 (8.85)

4

2 69.5 (kg / s )

Very Important: Bernoulli’s equation is only valid for :

incompressible fluids, steady flow along a streamline, no energy loss due

to friction, no heat transfer.

2

p1 V1 p2 V22

z1 z2 , where g (energy per unit weight)

2g 2g

Application of Bernoulli’s equation - Example 1:

Determine the velocity and mass flow rate of efflux from the circular

hole (0.1 m dia.) at the bottom of the water tank (at this instant). The

tank is open to the atmosphere and H=4 m

p1 = p2, V1=0

1

V2 2 g ( z1 z2 ) 2 gH

2 * 9.8 * 4 8.85 (m / s )

m AV 1000 * (0.1) 2 (8.85)

4

2 69.5 (kg / s )

4.
Bernoulli’s Eqn/Energy Conservation (cont.)

Example 2: If the tank has a cross-sectional area of 1 m2, estimate the time

required to drain the tank to level 2.

1 First, choose the control volume as enclosed

by the dotted line. Specify h=h(t) as the water

h(t) level as a function of time.

2

4

4

3

water height (m)

h( t ) 2

1 0

2 h - 0.0443t

2.5e-007 4

0

0

0

20 40

t

60 80 100

100

t 90.3 sec

time (sec.)

Example 2: If the tank has a cross-sectional area of 1 m2, estimate the time

required to drain the tank to level 2.

1 First, choose the control volume as enclosed

by the dotted line. Specify h=h(t) as the water

h(t) level as a function of time.

2

4

4

3

water height (m)

h( t ) 2

1 0

2 h - 0.0443t

2.5e-007 4

0

0

0

20 40

t

60 80 100

100

t 90.3 sec

time (sec.)

5.
Energy exchange (conservation) in a thermal system

Energy added, hA

(ex. pump, compressor)

2

p1 V1

2 p2 V2

z1 z2

2g 2g

Energy extracted, hE Energy lost, hL

(ex. turbine, windmill) (ex. friction, valve, expansion)

hL

loss through

valves heat exchanger

hE

hA

pump turbine

hL, friction loss

through pipes hL

loss through

condenser elbows

Energy added, hA

(ex. pump, compressor)

2

p1 V1

2 p2 V2

z1 z2

2g 2g

Energy extracted, hE Energy lost, hL

(ex. turbine, windmill) (ex. friction, valve, expansion)

hL

loss through

valves heat exchanger

hE

hA

pump turbine

hL, friction loss

through pipes hL

loss through

condenser elbows

6.
Energy conservation(cont.)

If energy is added, removed or lost via pumps turbines, friction, etc.then we use

2

p1 V1 p2 V22

Extended Bernoulli’s Equation z1 hA hE hL z2

2g 2g

Example: Determine the efficiency of the pump if the power input of the motor

is measured to be 1.5 hp. It is known that the pump delivers 300 gal/min of water.

No turbine work and frictional losses, hence: hE=hL=0. Also z1=z2

6-in dia. pipe 4-in dia.pipe Given: Q=300 gal/min=0.667 ft3/s=AV

V1= Q/A1=3.33 ft/s V2=Q/A2=7.54 ft/s

1 2

pump kinetic energy head gain

zo V22 V12 (7.54) 2 (3.33)2

0.71 ft,

Z=15 in 2g 2 * 32.2

Looking at the pressure term:

p1 w zo m z p2 w zo w z

Mercury (m=844.9 lb/ft3) p2 p1 ( m w )z

water (w=62.4 lb/ft3)

1 hp=550 lb-ft/s (844.9 62.4) * 1.25 978.13 lb / ft 2

If energy is added, removed or lost via pumps turbines, friction, etc.then we use

2

p1 V1 p2 V22

Extended Bernoulli’s Equation z1 hA hE hL z2

2g 2g

Example: Determine the efficiency of the pump if the power input of the motor

is measured to be 1.5 hp. It is known that the pump delivers 300 gal/min of water.

No turbine work and frictional losses, hence: hE=hL=0. Also z1=z2

6-in dia. pipe 4-in dia.pipe Given: Q=300 gal/min=0.667 ft3/s=AV

V1= Q/A1=3.33 ft/s V2=Q/A2=7.54 ft/s

1 2

pump kinetic energy head gain

zo V22 V12 (7.54) 2 (3.33)2

0.71 ft,

Z=15 in 2g 2 * 32.2

Looking at the pressure term:

p1 w zo m z p2 w zo w z

Mercury (m=844.9 lb/ft3) p2 p1 ( m w )z

water (w=62.4 lb/ft3)

1 hp=550 lb-ft/s (844.9 62.4) * 1.25 978.13 lb / ft 2

7.
Energy conservation (cont.)

Example (cont.)

Pressure head gain:

p2 p1 978.13

15.67 ( ft )

w 62.4

p2 p1 V22 V12

pump work hA 16.38( ft )

w 2g

Flow power delivered by pump

P = w QhA (62.4)(0.667)(16.38)

681.7( ft lb / s)

1hp 550 ft lb / s

P 1.24hp

P 1.24

Efficiency = 0.827 82.7%

Pinput 1.5

Example (cont.)

Pressure head gain:

p2 p1 978.13

15.67 ( ft )

w 62.4

p2 p1 V22 V12

pump work hA 16.38( ft )

w 2g

Flow power delivered by pump

P = w QhA (62.4)(0.667)(16.38)

681.7( ft lb / s)

1hp 550 ft lb / s

P 1.24hp

P 1.24

Efficiency = 0.827 82.7%

Pinput 1.5

8.
Frictional losses in piping system

2

p1 V1 p2 V22

Extended Bernoulli's equation, z1 hA hE hL z2

2g 2g

p1 p2 p

hL frictional head loss

P1

P2 R: radius, D: diameter

L: pipe length

Consider a laminar, fully developed circular pipe flow w: wall shear stress

[ p ( p dp)](R 2 ) w (2R)dx,

w Pressure force balances frictional force

p P+dp 2 w

dp dx, integrate from 1 to 2

R

F I F IF I 2

44w w L L L LV 2 V

HK HKG H JK

p p1 p2

Darcy’s Equation: hhLL f f

gg D D D D2 g 2 g

F IF I

2

f f VV 2 where f is defined as frictional factor characterizing

HK G

H JK

w

w

44 22 pressure loss due to pipe wall shear stress

2

p1 V1 p2 V22

Extended Bernoulli's equation, z1 hA hE hL z2

2g 2g

p1 p2 p

hL frictional head loss

P1

P2 R: radius, D: diameter

L: pipe length

Consider a laminar, fully developed circular pipe flow w: wall shear stress

[ p ( p dp)](R 2 ) w (2R)dx,

w Pressure force balances frictional force

p P+dp 2 w

dp dx, integrate from 1 to 2

R

F I F IF I 2

44w w L L L LV 2 V

HK HKG H JK

p p1 p2

Darcy’s Equation: hhLL f f

gg D D D D2 g 2 g

F IF I

2

f f VV 2 where f is defined as frictional factor characterizing

HK G

H JK

w

w

44 22 pressure loss due to pipe wall shear stress

9.
When the pipe flow is laminar, it can be shown (not here) that

64 VD

f , by recognizing that Re , as Reynolds number

VD

64

Therefore, f , frictional factor is a function of the Reynolds number

Re

Similarly, for a turbulent flow, f = function of Reynolds number also

f F(Re). Another parameter that influences the friction is the surface

roughness as relativeto the pipe diameter .

D

F I

H DDK

Such that ff FF Re,

Re, : Pipe frictional factor is a function of pipe Reynolds

number and the relative roughness of pipe.

This relation is sketched in the Moody diagram as shown in the following page.

The diagram shows f as a function of the Reynolds number (Re), with a series of

FI.

parametric curves related to the relative roughness

HDDK

64 VD

f , by recognizing that Re , as Reynolds number

VD

64

Therefore, f , frictional factor is a function of the Reynolds number

Re

Similarly, for a turbulent flow, f = function of Reynolds number also

f F(Re). Another parameter that influences the friction is the surface

roughness as relativeto the pipe diameter .

D

F I

H DDK

Such that ff FF Re,

Re, : Pipe frictional factor is a function of pipe Reynolds

number and the relative roughness of pipe.

This relation is sketched in the Moody diagram as shown in the following page.

The diagram shows f as a function of the Reynolds number (Re), with a series of

FI.

parametric curves related to the relative roughness

HDDK

10.

11.
Losses in Pipe Flows

Major Losses: due to friction, significant head loss is associated with the straight

portions of pipe flows. This loss can be calculated using the Moody chart or

F

G IJ

Colebrook equation. 1 2.0 log D 2.51 , valid for nonlaminar range

f H3.7 Re fK

Minor Losses: Additional components (valves, bends, tees, contractions, etc) in

pipe flows also contribute to the total head loss of the system. Their contributions

are generally termed minor losses.

The head losses and pressure drops can be characterized by using the loss coefficient,

KL, which is defined as hL p

K 1 , so that p K V

1

L 2

2

V / 2g 2 V

L 2 2

One of the example of minor losses is the entrance flow loss. A typical flow pattern

for flow entering a sharp-edged entrance is shown in the following page. A vena

contracta region is formed at the inlet because the fluid can not turn a sharp corner.

Flow separation and associated viscous effects will tend to decrease the flow energy;

the phenomenon is fairly complicated. To simplify the analysis, a head loss and the

associated loss coefficient are used in the extended Bernoulli’s equation to take into

consideration this effect as described in the next page.

Major Losses: due to friction, significant head loss is associated with the straight

portions of pipe flows. This loss can be calculated using the Moody chart or

F

G IJ

Colebrook equation. 1 2.0 log D 2.51 , valid for nonlaminar range

f H3.7 Re fK

Minor Losses: Additional components (valves, bends, tees, contractions, etc) in

pipe flows also contribute to the total head loss of the system. Their contributions

are generally termed minor losses.

The head losses and pressure drops can be characterized by using the loss coefficient,

KL, which is defined as hL p

K 1 , so that p K V

1

L 2

2

V / 2g 2 V

L 2 2

One of the example of minor losses is the entrance flow loss. A typical flow pattern

for flow entering a sharp-edged entrance is shown in the following page. A vena

contracta region is formed at the inlet because the fluid can not turn a sharp corner.

Flow separation and associated viscous effects will tend to decrease the flow energy;

the phenomenon is fairly complicated. To simplify the analysis, a head loss and the

associated loss coefficient are used in the extended Bernoulli’s equation to take into

consideration this effect as described in the next page.

12.
V1

Minor Loss through flow entrance

V2 V3

V 2

p gz

2

(1/2)V22 (1/2)V32

KL(1/2)V32

pp

2

p1 V1 p3 V32 V32

Extended Bernoulli's Equation : z1 hL z3 , hL K L

2g 2g 2g

2

p1 p3 p , V1 0, V3 1 ( 2 g ( z1 z3 ) gh

1 KL 1 KL

Minor Loss through flow entrance

V2 V3

V 2

p gz

2

(1/2)V22 (1/2)V32

KL(1/2)V32

pp

2

p1 V1 p3 V32 V32

Extended Bernoulli's Equation : z1 hL z3 , hL K L

2g 2g 2g

2

p1 p3 p , V1 0, V3 1 ( 2 g ( z1 z3 ) gh

1 KL 1 KL

13.
Energy Conservation (cont.)

Let us now also account for energy transfer via Heat Transfer, e.g.

in a heat exchanger

The most general form of conservation of energy for a system can be

written as: dE = dQ-dW where (Ch. 3, YAC)

dE Change in Total Energy, E

and E = U(internal energy)+Em(mechanical energy) (Ch. 1 YAC)

E = U + KE (kinetic energy) + PE(potential energy)

mechanical

dW Work done by the system where energy

W = Wext(external work) + Wflow(flow work)

dQ = Heat transfer into the system (via conduction, convection &

Convention: dQ > 0 net heat transfer into the system (Symbols Q,q..)

dW > 0, positive work done by the system

Q: What is Internal Energy ?

Let us now also account for energy transfer via Heat Transfer, e.g.

in a heat exchanger

The most general form of conservation of energy for a system can be

written as: dE = dQ-dW where (Ch. 3, YAC)

dE Change in Total Energy, E

and E = U(internal energy)+Em(mechanical energy) (Ch. 1 YAC)

E = U + KE (kinetic energy) + PE(potential energy)

mechanical

dW Work done by the system where energy

W = Wext(external work) + Wflow(flow work)

dQ = Heat transfer into the system (via conduction, convection &

Convention: dQ > 0 net heat transfer into the system (Symbols Q,q..)

dW > 0, positive work done by the system

Q: What is Internal Energy ?

14.
Energy Conservation (cont.)

U = mu, u(internal energy per unit mass),

KE = (1/2)mV2 and PE = mgz

Flow work Wflow= m (p/)

It is common practice to combine the total energy with flow work.

F

G I

HJK

V2 pp

Energy flow rate: m(u +

gz ) plus Flow work rate m

2

p V2 p V2

Flow energy in m in (u gz )in , Energy out = m out (u gz )out

2 2

The difference between energy in and out is due to heat transfer (into or out)

and work done (by or on) the system.

U = mu, u(internal energy per unit mass),

KE = (1/2)mV2 and PE = mgz

Flow work Wflow= m (p/)

It is common practice to combine the total energy with flow work.

F

G I

HJK

V2 pp

Energy flow rate: m(u +

gz ) plus Flow work rate m

2

p V2 p V2

Flow energy in m in (u gz )in , Energy out = m out (u gz )out

2 2

The difference between energy in and out is due to heat transfer (into or out)

and work done (by or on) the system.

15.
Energy Conservation (cont.)

Hence, a system exchanges energy with the environment due to:

1) Flow in/out 2) Heat Transfer, Q and 3) Work, W

This energy exchange is governed by the First Law of Thermodynamics

2 Heat in, Q =dQ/dt

p V p V2

m in (u gz )in m in (u gz )out

2 2

system

Work out dW/dt

From mass conservation: m in m out m

From the First law of Thermodynamics (Energy Conservation):

dQ p V2 p V2 dW

m (u gz )in m (u gz ) out , or

dt 2 2 dt

dQ V2 V2 dW

m( h

gz )in m(h

gz )out

dt 2 2 dt

p

where h u is defined as "enthaply"

“Enthalpy”

Hence, a system exchanges energy with the environment due to:

1) Flow in/out 2) Heat Transfer, Q and 3) Work, W

This energy exchange is governed by the First Law of Thermodynamics

2 Heat in, Q =dQ/dt

p V p V2

m in (u gz )in m in (u gz )out

2 2

system

Work out dW/dt

From mass conservation: m in m out m

From the First law of Thermodynamics (Energy Conservation):

dQ p V2 p V2 dW

m (u gz )in m (u gz ) out , or

dt 2 2 dt

dQ V2 V2 dW

m( h

gz )in m(h

gz )out

dt 2 2 dt

p

where h u is defined as "enthaply"

“Enthalpy”

16.
Conservation of Energy – Application

Example: Superheated water vapor enters a steam turbine at a mass flow rate

1 kg/s and exhausting as saturated steam as shown. Heat loss from the turbine is

10 kW under the following operating condition. Determine the turbine power output.

P=1.4 Mpa From superheated vapor tables:

T=350 C hin=3149.5 kJ/kg

V=80 m/s

z=10 m dQ V2 V2 dW

10 kw m (h gz )in m (h gz )out

dt 2 2 dt

dW

( 10) (1)[(3149.5 2748.7)

dt

80 2 50 2 (9.8)(10 5)

]

2(1000) 1000

P=0.5 Mpa 10 400.8 1.95 0.049

100% saturated steam

392.8( kW )

V=50 m/s

z=5 m

From saturated steam tables: hout=2748.7 kJ/kg

Example: Superheated water vapor enters a steam turbine at a mass flow rate

1 kg/s and exhausting as saturated steam as shown. Heat loss from the turbine is

10 kW under the following operating condition. Determine the turbine power output.

P=1.4 Mpa From superheated vapor tables:

T=350 C hin=3149.5 kJ/kg

V=80 m/s

z=10 m dQ V2 V2 dW

10 kw m (h gz )in m (h gz )out

dt 2 2 dt

dW

( 10) (1)[(3149.5 2748.7)

dt

80 2 50 2 (9.8)(10 5)

]

2(1000) 1000

P=0.5 Mpa 10 400.8 1.95 0.049

100% saturated steam

392.8( kW )

V=50 m/s

z=5 m

From saturated steam tables: hout=2748.7 kJ/kg

17.
Q, q … ?!%

Q – total heat transfer (J)

Q – rate of total heat transfer (J/s, W)

q – heat transfer per unit mass (J/kg)

q – Heat Flux, heat transfer per unit area (J/m2) Back

Internal Energy ?

•Internal energy, U (total) or u (per unit mass) is the sum of all

microscopic forms of energy.

•It can be viewed as the sum of the kinetic and potential energies of the

• Due to the vibrational, translational and rotational energies of the moelcules.

•Proportional to the temperature of the gas. Back

Q – total heat transfer (J)

Q – rate of total heat transfer (J/s, W)

q – heat transfer per unit mass (J/kg)

q – Heat Flux, heat transfer per unit area (J/m2) Back

Internal Energy ?

•Internal energy, U (total) or u (per unit mass) is the sum of all

microscopic forms of energy.

•It can be viewed as the sum of the kinetic and potential energies of the

• Due to the vibrational, translational and rotational energies of the moelcules.

•Proportional to the temperature of the gas. Back