control volume part 2

7/29/2019 Control Volume Part 2

1/36

MOMENT OF MOMENTUM EQUATION OR ANGULAR MOMENTUM EQUATION

Torque - moment of a force w.r.t an axis is important in engg problems

Moment of a linear

momentum with each

particle of fluid

Moment of resultant forces acting

on each particle of fluid=

particleFVVDt

D

Newtons second law of motion to a particle of fluid

V particle velocity measured in an inertial reference system

- density of the fluid particle

- infinitesimally small particle volume

- resultant external forces acting on the particle

V

particleF


2/36

particleFrVVDt

Dr

Dt

VVDrVVDtDrVVr

DtD

VDt

Dr0VV 0

VV

Dt

Dr

VVrDt

D

Dt

VVDr

particleFr

VVrDt

D

r is the position vector from the origin of the inertial coordinate system

Above equation is valid for every particle of a system. For a system (collection of

fluid particles), we need to use the sum of both side


3/36

syssys

FrVVrDt

D

where sysparticle FrFr

syssys

VVrDt

DVVr

Dt

D

sys

sys

Fr

VVrDt

D

Time rate of change of moment

of momentum of the systemSum of the external torques

acting on the system=


4/36

For a control volume that is instantaneously coincident with the system, the torques

acting on the system and on the control volume contents are identical

cvsys FrFr

cssys cv

dAn.VVrVdVrt

VdVrDt

D

Time rate of change of

moment of momentum of

the system

Time rate of change of

moment of momentum

of the contents of CV

Net rate of flow of the moment of

momentum through

the control surface

= +

For a fixed and non-deforming control volume

volumecontroltheofcontents

cscv

FrdAn.VVrVdVrt


5/36

APPLICATION OF THE MOMENT OF MOMENTUM EQUATION

1. Flows are one-dimensional (uniform distributions of average

velocity at any section)

2. Steady flows or steady-in-the-mean cyclical flows

0VdVr

t cv

volumecontrol

theofcontents

cs

FrdAn.VVr


6/36


7/36

r V = r V sin(-90o) = - r V

r

V

-90oV

r

-90o


8/36

r V = - r V

r

V

-90oV

r

-90o

In both the cases, the right hand rule application makes the thumb point into the

paper plane


9/36

cs

dAn.VVr

Flow into the control volume is negative

Flow out of the control volume is positive

dAn.V

inout

cs

22 mzeromVrdAn.VVr

out

cs

mVrdAn.VVr 22

Water enters the control volume axially, at this portion

0Vr


10/36

shaftvolumecontrol

theofcontents TFr

We have taken Tshaftas a positive quantity. This is equivalent to assuming that Tshaftis

in the same direction as rotation

shaft22 TmVr

Tshaft- being a negative quantity means that the shaft torque opposes the rotation

of the sprinkler arms

Tshaft, - shaft torque opposes rotation in all turbine devices

mVrTW 22shaftshaft

mVUW 22shaft

Negative shaft work is work out of the control volume i.e., work is done by the fluid

on the rotor and thus its shaft


11/36

Moment of momentum equation for a more general one dimensional flow through a

rotating machine

outoutoutinininshaft VrmVrmT

U and V - same direction; r V is POSITIVE

U and V - opposite direction; r V is NEGATIVE

Tshaftis positive ifTshaftis in the same direction as

outoutoutinininshaftshaftVrmVrmTW

outoutoutinininshaftshaft VUmVUmTW

shaftW is positive, power is into the control volume - PUMP

shaftW is negative , power is out of the control volume - TURBINE

mmm outin CONSERVATION OF MASS


12/36

r V = r V sin(-90o) = - r V

r

V

-90oV

r

-90o

U and V - opposite direction; r V is NEGATIVE

r V = r V sin(-90o) = + r V

r

V

90oV

r

90o

U and V - same direction; r V is POSITIVE

Direction of rotation

Direction of rotation


13/36

Water enters a rotating lawn sprinkler through its base at the steady rate of 60 lpm

as sketched in Fig. The exit area of each of the two nozzles is 30 mm2,and the flow

leaving each nozzle is in the tangential direction. The radius from the axis of rotation

to the centerline of each nozzle is 200 mm.

(a) Determine the frictional torque associated with the sprinkler rotating with a

constant speed of 500 rpm.

(c) Determine the speed of the sprinkler if no frictional torque is applied.


14/36

cscv

sysdAnbWVbd

ttD

DB

cscv

sys

dAn

WVdttD

DM

0

cs

dAnW

02

AWmdAnW incs

s/m.A

Q

A

mW inin 716

1030260

1060

6

3

2

s/m.W 7162


15/36

2222

UWVV

5107162

..V

s/m..rN

rU 5102060

5002

60

2

222

s/m.V 262

mVrTshaft 22

12620 ..Tshaftm

s/kg.Qm in 0160

10601000

3

m.N.Tshaft 241

When the sprinkler rotates at the

constant speed, the friction torque at

the sprinkler pivot just balances the

torque generated by the angular

momentum of the two jets

h d f h kl f f l l d


16/36

mVrTshaft 22

222 UWmrTshaft

The speed of the sprinkler if no frictional torque is applied.

2220 UWmr

22UW

s/rad.

.

..r. 583

20

71620716

2

rpmNs/rad.N

79735860

2

rpmN 797


17/36

FIRST LAW OF THERMODYNAMICS THE ENERGY EQN

Time rate of increase of

the total stored energy

of the system

Net time rate of energy

addition by heat transfer

into the system

Net time rate of energy

addition by work

transfer into the system= +

syssysteminnetinnet

WQVdeDt

D

e total energy stored per unit mass for each particle in the system

gz2

Vue

2

u internal energy per unit mass for each particle in the system

V2/2 kinetic energy per unit mass

gz potential energy per unit mass

innetQ

Net rate of heat transfer into the system

innetW

Net rate of work transfer into the system

Work and heat

+ is into the system

- is coming out


18/36

cvinnetinnet

cscvsys

WQdAnVeVdet

dVetD

D

Time rate of increase

of the total stored

energy of the system

time rate of increase of total

stored energy of the

contents of the control

volume

Net time rate of

energy addition by

work transfer into the

system

=+

outininnetWWW

shaftshaft TW

When the control surface cuts through the shaft material, the shaft torque is exerted

by shaft material at the control surface


19/36

Normal stress = - p

Power transferred due to fluid normal stresses

cscsressstnormal

dAn.VpdAn.VW


20/36

csinnetshaft

innet

cscv

dAn.VpWQdAnVeVdet

gz2

V

ue

2

innetshaft

innet

cs

2

cv

WQdAnVgz2

VpuVde

t

Steady flow

inflow

2

outflow

2

cs

2

mgz2

Vpumgz

2

VpudAnVgz

2

Vpu

If there is only one stream entering and leaving the control volume

in

in

2

out

out

2

cs

2

mgz2

Vpumgz

2

VpudAnVgz

2

Vpu

A

Substituting this in eqn A


21/36

innet

innetinout

2in

2out

inout

inout WQzzg2

VVppuum

One Dimensional Energy Equation For Steady Flow Applicable for

Compressible And Incompressible Flow

puhEnthalpy

innet

innetinout

2in

2out

inout WQzzg2

VVhhm

Flow is steady throughout, one dimensional, only one fluid stream is involved, thenthe shaft work is zero

innetinout

2in

2out

inout

inout Qzzg2

VVppuum

Steam enters a turbine with a velocity of 30 m/s and enthalpy of 3348 kJ/kg (see


22/36

Steam enters a turbine with a velocity of 30 m/s and enthalpy, of 3348 kJ/kg (see

fig). The steam leaves the turbine as a mixture of vapor and liquid having a velocity

of 60 m/s and an enthalpy of 2550 kJ/kg. If the flow through the turbine is adiabatic

and changes in elevation are negligible, determine the work output involved per unit

mass of steam through-flow.

in

net

in

net12

21

22

12 WQzzg2

VVhhm

Adiabatic flow


23/36

2

VVhh

m

W

w21

22

12in

net

innetshaft

innetshaftoutnetshaft ww

2

VVhhw

2

2

2

121outnetshaft

10002

603025503348w

22

outnetshaft

kg/kJ35.1kg/kJ2550kg/kJ3348w outnetshaft

kg/kJ797w outnetshaft

Comparison of the Ener Eq ation and Berno lli Eq ation


24/36

Comparison of the Energy Equation and Bernoulli Equation

in

netinout

2in

2out

inout

inout Qzzg

2

VVppuum

One dimensional steady flow energy equation

Heat transfer per unit

mass

Bernoullis equation

0quu innetinout

innetinout

2in

2outinout

inout Qzzg2

VVppuum

INCOMPRESSIBLE FLOW

m

Q

qinnet

innet

innetinoutin

2inin

out

2outout quugz

2

Vpgz

2

Vp

Steady and Incompressible flow is frictionless

0quu innetinout Steady,Incompressible flow with friction

in

2

ininout

2

outout gz2

Vpgz

2

Vp


25/36

gz2

Vp2

USEFUL OR AVAILABLE ENERGY

Lossquu innetinout Loss of useful or available energy thatoccurs in an incompressible fluid flow

because of friction

lossgz2

Vpgz2

Vp in

2

ininout

2

outout


26/36

ONE DIMENSIONAL, INCOMPRESSIBLE, STEADY FLOW

WITH FRICTION AND SHAFT WORK

innet

innetinout

2

in

2

outinoutinout WQzzg2

VVppuum

in

netinout

innet

shaftin

2inin

out

2outout quuwgz

2

Vpgz

2

Vp

losswgz2

Vpgz

2

Vp

innetshaftin

2inin

out

2outout

Pump when the shaft work is into the control volume large amount of loss willresult in more shaft work being required for the same rise in available energy

Turbine when the shaft work is out of the control volume larger loss will result in

less shaft work out for the same drop in available energy


27/36

PUMP

gm

Loss

gm

W

gzg2

V

g

pz

g2

V

g

p innetshaft

in

2inin

out

2outout

gm

W

innetshaft

in

2inin

out

2outout gz

g2

V

g

pz

g2

V

g

p

gm

Loss

gm

W

gmLoss

gm

W

gm

W

gzg2

V

g

pzg2

V

g

p

innetshaft

innet

shaft

innetshaft

in

2

ininout

2

outout


28/36

TURBINE

gm

Loss

gm

W

gzg2

V

g

pz

g2

V

g

p innetshaft

in

2inin

out

2outout

gm

W

OUTnetshaft

out

2

outoutin

2

inin zg2

V

g

pgzg2

V

g

p

gmLoss

gm

LOSS

gm

W

gm

W

zg

V

g

pgz

g

V

g

p

gm

W

OUTnetshaf t

OUTnetshaf t

OUTnetshaf t

outoutout

ininin

22

22

A pump delivers water at a steady rate of 1135 lpm as shown in Fig Just upstream of


29/36

A pump delivers water at a steady rate of 1135 lpm as shown in Fig. Just upstream of

the pump [section (1)] where the pipe diameter is 90 mm, the pressure is 1.24 bar.

Just downstream of the pump [section (2)] where the pipe diameter is 25 mm, the

pressure is 4 bar . The change in water elevation across the pump is zero. The rise in

internal energy of water, associated with a temperature rise across the pump is 280

J/kg. If the pumping process is considered to be adiabatic, determine the power (hp)required by the pump.

D1 =

90

mm

P1 = 1.24 barP2 = 4.0 bar

280 J/kg

D2 = 25 mm

Q = 1135

lpm


30/36

D1 =

90

mm

P1 = 1.24 barP2 = 4.0 bar

280 J/kg

D2 = 25 mm

Q = 1135

lpm

gm

W

gm

Loss

gm

W

gm

W

gzg2

V

g

pzg2

V

g

p

innetshaft

innetshaft

innetshaft

in2inin

out2outout

3310113


31/36

s

m.lpmQ

3

3

3

10921860

1011351135

s

m

.

.

D

Q

A

Q

V 97421090

4

109218

4

23

3

2

11

1

s

m.

.

D

Q

A

QV 5438

1025

4

109218

4

23

3

2

22

2

gmgm

W

z.

.

.

.z

.

.

.

innetsha ft

inout

280

8192

9742

8191000

10241

8192

5438

8191000

1042525

gmgm

W

....innet

shaf t

2804510641271757840

W 280


32/36

gm

W

.innet

shaf t

280

4103

2804103 gm.Winnet

sha ft

28081991184103 ...Winnet

sha ft

2802519181.Winnet

shaf t

W.Winnet

shaf t2519461

W.HP 77451

HP.Winnet

shaf t126

An axial-flow ventilating fan driven by a motor that delivers 0.4 kW of power to the


33/36

An axial flow ventilating fan driven by a motor that delivers 0.4 kW of power to the

fan blades produces a 0.6-m-diameter axial stream of air having a speed of 12 m/s.

The flow upstream of the fan involves negligible speed. Determine how much of the

work to the air actually produces a useful effect, that is, a rise in available energy

and estimate the fluid mechanical efficiency of this fan.

211

222 VpVp


34/36

111

222

innetshaft gz

2

Vpgz

2

Vplossw

1

211

2

222

innetshaft gz

2

Vpgz

2

Vplossw

Atmospheric pressure

zero

kg/m.N722

122

Vlossw

22

2

innetshaft

124

6.0225.1

10004.0

72

AV

W

lossw

m

W

lossw

w

lossw

2innetshaft

innetshaft

innetshaft

innetshaft

innetshaft

innetshaft

752.0

8.95

72

VV 22


35/36

losswgz2

Vpgz

2

Vp

innetshaftin

2inin

out

2outout

losswzg2

V

pzg2

V

pinnet

shaftin

2in

inout

2out

out

g is the specific weight of the fluid. Each term involves the energy per unit volume

Lsin

2inin

out

2outout hhz

g2

Vpz

g2

Vp

gm

W

g

wh

innetshaft

innetshafts

g

LosshL

Turbine head; hT = - (hs+ hL)T;

Actual head drop across the turbine = work head out of the turbine + head loss in

the turbine

Pump; hP = (hs- hL)P;

Actual head rise across the pump = shaft work head into the pump head loss within

the pump

APPLICATION OF ENERGY EQUATION TO NON-UNIFORM FLOWS


36/36

Q

losswgz

2

Vpgz

2

Vp

innet

shaftin

2inin

out

2outout

If the velocity profile is not uniform

2

V

2

VmdAn.V

2

V2

inin2

outout

CS

2

dAn.V2

V

2

Vm

A

22

2Vm

dAn.V2

V

2A

2

losswgz

2

Vpgz

2

Vp

innet

shaftin

2ininin

out

2outoutout

- KINETIC ENERGY COEFFICIENT

control volume part 2

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