2141-365 - 2011 - hw2 solution - area vectors, …fmeabj.lecturer.eng.chula.ac.th/2141-365 hp/365...

2141-365 Fluid Mechanics for International Engineers HW #2: Area Vectors, Pressure and Pressure Force, RTT, and C-Mass Due: Fri, Sep 23, 2011 at the ISE office.

1

2141-365 Fluid Mechanics for International Engineers HW #2: Area Vectors, Pressure and Pressure Force, RTT, and C-Mass

Problem 1. Static Fluid [Fox et al., 2010, Problem 3.18, p. 82.] A partitioned tank as shown contains water and mercury. 1.1. What is the gage pressure in the air trapped in the left chamber? 1.2. What pressure would the air on the left need to be pumped to in order to bring the water and mercury free

surfaces to the same level?

Solution Assumptions

1. Static fluid 2. Fluids each has constant specific weight . w = 9,810 N/m3 , M = 132,925 N/m3.

Basic Equation: )( 1212 zzppdz

dp

Pamm

Nm

m

N

zzzzpp

zzzzpp

zzpp

zzpp

bcMabwca

bcMabwac

bcMbc

abwab

483,3)1.0(925,132)1(9810:

)3()()(:

)()()2()1(

)2()(

)1()(

33

The gage pressure in the air trapped in the left chamber is 3.48 kPa. ANS (1.1)

If the free surfaces are at the same level, we have ca zz .

Since the volume of water is conserved, we still have ba zz = 1 m.

Equation (3) can be evaluated to give

Pamm

Nm

m

N

zzzz

zzzzpp

baMabw

bcMabwca

115,123)1(925,132)1(9810

)()(

)()(

33

In order to bring the water and mercury free surfaces to the same level, the air on the left need to be pumped to 123 kPa gage. ANS (1.2)

a

b c z

1 m a

b

c

(1.1) (1.2)


2

Problem 2. Area/Surface as A Vector, Components of A (Uniform) Pressure Force, and The Net Pressure Force Due to Uniform Pressure

In order to facilitate the description and the quantification of a surface force (a force that is distributed over a

surface area), we invent the concept of area/surface as a vector. By convention, we consider the area vector as having the direction outward normal to the surface of the system

under consideration, i.e., pointing from the system to the surroundings.

In the following problems, we consider a system whose surface is subjected to an external uniform pressure

op , and we want to evaluate the net force due to this uniform pressure of the surroundings over the surface of

the system.

For each of the following system and the corresponding plate surface (a and b), we consider the right side of the surface as a system (e.g., we want to find the net external force on the system for amF

), thus the left side is

the surroundings. The surroundings exert a uniform pressure op on the surface of the system.

2.1. Find the net area vector A

of each surface. Are they equal? If so, why? 2.2. Find the net force due to uniform pressure op of the surroundings on each system.

The plates have a width w in the z direction.

Solution

2.1

Here, we shall integrate from 21 . xd

: The differential displacement vector xd

is given by

jdyidxxd ˆ)(ˆ)(

. (1)

Ad

: Thus, the differential area vector Ad

is given by

jdxidywjdyidxkwxwdkAd ˆ)(ˆ)(ˆ)(ˆ)(ˆ)(ˆ

. (2)

A

: The net area vector is then given by

x

y

z

h

(a)

l

op

(b)

op

Surface (b) is plotted here as dotted line in order to compare its size to (a).

Ad

Ad

1: ),( 11 yx

x

y

zAd

Ad

2: ),( 22 yx

xd

xd

system

surroundings Ad

system

surroundings Ad


3

jxxiyywjdxidywAdAx

x

y

yCA

ˆ)(ˆ)(ˆˆ1212

21:,

2

1

2

1

. (3)

Since both surfaces have the same end points, both have the same net area vector and it is equal to

jwliwhjxxiyywA ˆ)(ˆ)(ˆ)(ˆ)( 1212

. (4) ANS

[Both components point in the negative directions.]

Algebraically, (for smooth surfaces) it can be seen that the net area vector depends on the positions of the end points only. Since both surfaces have the same endpoints, they have equal net area vectors.

Geometrically, recall that

The total area vector is the sum of its components: jAiAA yxˆˆ

.

Each component is the net projection of the surface along the corresponding axis. As the figure below illustrated, both surfaces have equal projection areas xA and yA . ANS

2.2.

Ad

: Consider an area element Ad

of the system.

ApdFd

: The infinitesimal pressure force on this area element is given by ApdFd

.

A

ApdF

: The net pressure force over the surface area A is then given by A

ApdFdF

.

Since opp is uniform over the surface, we have A

o

A

o AdAApAdpF

, .

Since both surfaces have equal net area vector jwliwhA ˆ)(ˆ)(

,

we find that they have equal net pressure force due to uniform op , jwliwhpApF ooˆ)(ˆ)(

. ANS

iAxˆ

iAxˆ

iAxˆ

jAyˆ

Effectively, only this portion contributes to yA .

jAyˆ

These two parts cancel since their y components are equal in

magnitude but opposite in direction.

jAyˆ

Ad

ApdFd

system


4

Note Due to the above results on the net area vector, and the net pressure force due to uniform pressure,

we find that – for example - all the plates below, which have the same end points, have equal net area vector

A

and equal net pressure force due to uniform op , jwliwhpApF ooˆ)(ˆ)(

.

op

1

2


5

Problem 3. Area/Surface as A Vector, Components of A (Uniform) Pressure Force, and The Net Pressure Force Due to Uniform Pressure

A two-dimensional, curved plate is subjected to uniform pressure lp on its left surface and another uniform

pressure lr pp on its right surface as shown below. The plate has the width w in the z direction.

Determine the vector expression in the given Cartesian coordinates for the net pressure force F

due to both

pressures on the plate. Solution Using the result from last problem,

jxxiyywA ˆ)(ˆ)( 1212

,

with uniform pressure we find

F

)ˆˆ)((

)ˆˆ(ˆˆ

1

11

jlihppw

jwlihwpjwlihwp

lr

lr

ANS

x

y

z

h lp

1l

2l3l

4l

rp


6

Problem 4. Resultant Pressure Force on A Curved Surface [Adapted from Fox et al., 2010, Problem 3.75, p. 87.] A dam is to be constructed across the Wabash River using the cross-section shown. Assume the dam width w = 50

m. For water height H = 2.5 m, calculate the resultant force vector due to fluid pressure at the water-dam interface (marked as red line). Note: Here, the resultant force in question is referred to the physical force that actually acts on that surface (and that

surface only).

Solution

mx

mx

mmm

mA

y

Bx

2.2

2.2

76.04.05.2

9.0

3

2

2

11

my

my

my

0

5.0

5.2

3

2

1

Assumptions

1. Static fluid 2. Constant specific weight . w = 9,810 N/m3.

3. Atmospheric pressure op = 101.325 kPa.

x

y

F

3

2

1

jYiXX ˆˆ

jYiXX ˆˆ

is the position vector that points

along the line of action of the resultant force F

.

x

y

Ad

ApdFd

Ax

Bxy

)(

H = 2.5 m

A = 0.4 m B = 0.9 m2

h

xd

3

2

1 op

jyixx ˆˆ

jyixx ˆˆ is the position vector that points to

the infinitesimal force on the infinitesimal area

element Ad


7

Resultant Force Vector Integrate from 1 3

jdyidxxd ˆ)(ˆ)(

jwdxiwdyxwdkAd ˆ)(ˆ)(

jwdxiwdyghpAdghpApdFd ooˆ)(ˆ)()()(

plate]flat verticalfrom component - theon tocontributi no[,0:

ln)()(:

))((::

ln)()(:

)ln()(:

,)()(:

,)()(:

)(:

))((::

:

)(2

1)()(:

))((::

)(2

1)()(:

2/)(:

,)()(:

)(:

))((::

:

ˆ))(())((ˆ))(())((

32

2

32323

2

1

21212

12

12

12

12

1

21

22

232323

2

21

221212

212

12

12

1

21

23

23

12

1231

3

2

2

1

2

1

2

1

2

1

2

1

3

2

2

1

2

1

2

1

2

1

3

2

2

1

3

2

2

1

yxx

Ax

AxBxxHgwxxwp

wdxghpF

Ax

AxBxxHgwxxwp

AxBHxgwxxwp

Ax

Bydx

Ax

BHgwxxwp

yHhdxyHgwxxwp

hdxgwxxwp

wdxghpF

FFF

yyyyHgwyywp

wdyghpF

yyyyHgwyywp

yHygwyywp

yHhdyyHgwyywp

hdygwyywp

wdyghpF

FFF

jwdxghpwdxghpiwdyghpwdyghp

ApdApdApdFdF

o

x

x

oy

o

xxo

x

x

o

x

x

o

x

x

o

x

x

oy

yyy

o

y

y

ox

o

yyo

y

y

o

y

y

o

y

y

ox

xxx

componenty

x

x

o

x

x

o

componentx

y

y

o

y

y

o

surfaceflatthefromonContributisurfacecurvedthefromonContributi

Hence, the resultant force vector is given by


8

jAx

AxBxxHgwxxwpiyyyyHgwyywp

jAx

AxBxxHgwxxwp

iyyyyHgwyywpyyyyHgwyywp

jFiFF

oo

o

oo

yx

ˆln)()(ˆ)(2

1)()(

ˆln)()(

ˆ)(2

1)()()(

2

1)()(

ˆˆ

1

21212

21

231313

1

21212

22

232323

21

221212

which is evaluated to be

MN

mmmmm

kNmmkPa

yyyyHgwyywpF ox

2.14

)5.20(2

1)5.2(5.25081.9)5.2(50325.101

)(2

1)()(

223

21

231313

MN

m

mmmmm

m

kNmmkPa

Ax

AxBxxHgwxxwpF oy

35.8

)4.076.0(

)4.02.2(ln9.0)76.02.2(5.25081.9)76.02.2(50325.101

ln)()(

23

1

21212

Thus, the resultant force vector is given by MNjiF ˆ35.8ˆ2.14

. ANS


9

Line of Action of F

: Resultant moment about the origin oM

)(3

1)(

2)(

2

1ln)()(

2)(

2

1

)(3

1)(

2)(

2

1)(

3

1)(

2)(

2

1

ln)()(2

)(2

1

)(3

1)(

2)(

2

1)(::

)(3

1)(

2)(

2

1:

32)(

2

1:

)()(2

1:

)(2

1)(::

,0:

ln)()(2

)(2

1)(::

ln)()(2

)(2

1:

)ln(2

)(2

1:

)()(2

1:

,)()(2

1:

)(2

1)(::

)()()()(

)()(

ˆ

ˆˆˆˆ

,

31

33

21

23

21

23

1

212

21

22

21

22

32

33

22

23

22

23

31

32

21

22

21

22

1

212

21

22

21

22

32

33

22

23

22

234

31

32

21

22

21

22

3221

22

21

22

21

223

32

2

323

22

23

22

232

1

212

21

22

21

22

221

22

21

22

21

22

21

221

3

2

2

1

2

1

2

1

2

1

3

2

2

1

2

1

2

1

2

1

2

1

3

2

2

1

3

2

2

1

yyyyH

gwyywpAx

AxBAxxBxx

Hgwxxwp

yyyyH

gwyywpyyyyH

gwyywp

Ax

AxBAxxBxx

HgwxxwpM

yyyyH

gwyywpwydyghpM

yyyyH

gwyywp

yyHgwyywp

ydyyHgwyywp

hydygwyywpwydyghpM

xx

Ax

AxBAxxBxx

HgwxxwpwxdxghpM

Ax

AxBAxxBxx

Hgwxxwp

AxAxBxH

gwxxwp

xdxAx

BHgwxxwp

yHhxdxyHgwxxwp

hxdxgwxxwpwxdxghpM

wydyghpwydyghpwxdxghpwxdxghpM

wdyghpywdxghpx

ydFxdFdM

kydFxdF

jdFidFjyix

FdxMd

oo

oo

oo

o

y

y

oo

o

y

y

o

y

y

o

y

y

o

y

y

oo

o

x

x

oo

o

x

xo

x

x

o

x

x

o

x

x

o

x

x

oo

y

y

o

y

y

o

x

x

o

x

x

oo

oo

xyo

xy

yx

o

which is evaluated to be mMNM o 62.4 .

The line of action is then given by


10

mXXY

F

MX

F

FXY

MYFXF

kYFXFFXMFX

x

o

x

y

oxy

xyo

)33.059.0()(

)(

ˆ)(,

ANS


11

2145-213 Fluid Mechanics for Aerospace Engineering HW #3: Convection Flux, RTT, and C-Mass

Summary of Terminology (that differs from Fox et al.)

Current Terminology Fox et al.

Material Volume (MV) (Closed) System, Control Mass

Coincident material volume the system (or material volume) that instantaneously coincides with the control volume of interest.

Coincident control volume the control volume that instantaneously coincides with the system (or material volume) of interest

RTT

)(or )( is )(:

,)()(

)()(,)()()(

)(

/

)(

)()(

/

tCVtMVtV

AdVdVdt

d

dVtNAdVdt

tdN

dt

tdN

tCS

sf

tCV

tV

V

tCS

sfCVMV

CSCV

CV

CV

CS

CVSys

AdVdVt

dVNAdVt

N

dt

dN

)()(

,)(

Time rate of change of any property N of the system/MV: dt

dN MV , dt

dNSys

Time rate of change of any property N in the control volume CV:

CV

CV dVdt

d

dt

dN ,

CV

dVt

Net convection efflux (net rate of outflow) of N through the control surface: )(

/ )(tCS

sf AdV

Convection efflux (rate of outflow) of N through an open area A: A

sf AdV )( /

For example, the convection flux of kinetic energy [Energy/Time] through a cross sectional area A

=

22

/2

2

1,

2

1)(

2

1VmVNAdVV

A

sf


12

Problem 5. Convection Flux Through A Surface For a steady, incompressible, fully-developed, laminar flow in a circular pipe, the velocity profile at any axial

location is axisymmetric and paraboloid and is given by

2)/(1)(

RrU

ru ,

where )(ru is the axial velocity at radius r , U is the maximum velocity which occurs at the center of the pipe, r is

the radial coordinate, and R is the radius of the pipe. 5.1. Find the algebraic expression for the convection flux of kinetic energy [Energy/Time] through a cross section

of the pipe for a fluid with density . [Find A

AdVV )(2

1 2

.]

5.2. Is this flux equal to the kinetic energy flux of the corresponding uniform flow with the same mass flow rate? If not, what is the numerical value of the ratio ( ) of the kinetic energy flux of the pipe flow to that of the uniform flow?

Note: As we shall later see in this course, this ratio is referred to as the kinetic energy coefficient in the analysis for mechanical energy loss in a piping system.

5.3. If the fluid is water, what is the volume flowrate Q of water that would result in the kinetic energy flux

through a cross section of a pipe of 40 W, the power of a typical neon light? Assume that the average flow velocity V ( AQ / , where A is the cross sectional area of the pipe) is 5 m/s (and that the flow still remains

laminar).

Solution

5.1. Convection flux of kinetic energy

The convection flux ( NF ) of any extensive property N with intensive property mN /: is given by

A md

sfN AdVF

)( / . (A)

Thus, for the convection flux of kinetic energy

22

2

1/,

2

1VmNmVN , we have

A md

sf AdVVF

)(2

1/

2 . (1)

re

e

xe

x

A

U

dr

drAd

x

R r

re e

xe )(ru

U

Fluid density

Uniform flow with the same mass flow rate.


13

Velocity field

22

1:,ˆ1ˆ),(R

rUVe

R

rUeVtxV xx

An infinitesimal area element Ad

on the cross sectional surface A at x = xedrrdAd ˆ)(

.

Since the surface is stationary 0

sV , and the local relative velocity of fluid wrt the surface is given by

VVVVVV ffsfsf

0/

Powermassenergykinetictimemass

UmUUADimension

RAAU

UR

drR

rdRrdUR

rdrR

rU

drrdR

rUAdVV

erdrdAdrdrdR

rU

AdeV

AdeVVmdV

AdeVAdVAdVmd

R

R

A md

sf

x

x

x

xsf

]/][/[:

]][[][][:

,,8

1

4

1

2

3

2

1

2,)/(,1

2

1

1)2(2

1

12

1)(

2

1

ˆ,12

1

]ˆ[2

1

])ˆ([2

1

2

1

)ˆ(

22

23

1

0

43232

22

1

0

332

0

3

2

23

0

2

0

3

2

23

/2

3

2

23

3

22

/

Thus, the convection flux of kinetic energy through a cross section is given by 3

8

1AU , 2RA . ANS

Since the mass flowrate and the average velocity ( AQV /: ) in this pipe flow are

2/UA

AdVmA

sf

,

2

UV

the kinetic energy flux in this pipe flow can be expressed in terms of the mass flowrate and the average velocity as 23

8

1VmAU . ANS

5.2. Comparison of the kinetic energy flux in pipe flow to that of the uniform flow with the same mass flowrate

From the mass flowrate in this pipe flow 2/:, UVAVm ,

the uniform flow with the same mass flowrate will have the uniform velocity = 2/UV .

This uniform flow has the kinetic energy flux 2

/2

2

1)(

2

1VmAdVV

A md

sf

,

which is not equal to that of the pipe flow in 1.1.

22/ flowrate mass same with theflow uniform offlux energy kinetic

flow pipe offlux energy kinetic:

2

2

Vm

Vm

That is, the kinetic energy flux in pipe flow is twice of that uniform flow with the same mass flowrate. ANS


14

5.3.

The kinetic energy flux in pipe flow = 22 VQVm ,

sLsm

s

m

m

kg

W

V

PQ

PVQ

/6.1/0016.0

5000,1

40 3

2

22

3

2

2

. ANS

This is the pipe flow in a pipe of diameter ~ 2 cm with the average velocity of 5 m/s.


15

Problem 6. Reynolds Transport Theorem (RTT) [Adapted from Munson et al., 2002, Problem 5.31, p. 281, and 2141-365-2008, HW#3.]

A nozzle is attached to a vertical pipe and steadily discharges 0.1 m3/s of water into the atmosphere. Assume that all properties in this flow field are steady. Consider the material volume MV (or the control mass) that instantaneously coincides with the volume CV in the nozzle. Determine the followings.

6.1. Time rate of change of any property N in the control volume CV,

CV

dVt

dt

dNCV .

6.2. Time rate of change of any property N (intensive property ) of the coincident system/MV,

dt

dN

dt

dNMVSys .

Where appropriate, refer to the inlet and the exit with the subscript 1 and 2, respectively.

6.3. Time rate of change of mass of the coincident system/MV,

dt

dM

dt

dMMVSys

.

6.4. Time rate of change of the linear x -momentum of the coincident system/MV,

dt

dP

dt

dP xMVxSys ,, .

6.5. Time rate of change of the linear y -momentum of the coincident system/MV,

dt

dP

dt

dP yMVySys ,,.

6.6. Time rate of change of the kinetic energy of the coincident system/MV, 2

2

1, mVN

dt

dN

dt

dNMVSys

.

6.7. Is the kinetic energy of the coincident system/MV increasing or decreasing as it flows through the CV. Is this consistent with your ‘physical intuition’ of flow through a nozzle, and how so?

x

y


16

Solution

Control volume The control volume CV is stationary and non-deforming. It includes only the water volume in the

nozzle. Material Volume We consider a material volume MV that instantaneously coincides with the control volume CV. Assumptions

1. All properties are steady. 2. The property is uniform at each cross section.

3. The velocity is axial and uniform at each cross section. 4. Incompressible flow. 5. Water density = 1,000 kg/m3.

Basic Equation:

RTT: )()(

/ )()(,)()()(

tV

V

tCS

sfCVMV dVtNAdVdt

tdN

dt

tdN

(A)

C-Mass: )()(

/ )()(,)()()(

0tV

V

tCS

sfCVMV dVtMAdVdt

tdM

dt

tdM

(B)

Calculations: for Q = 0.1 m3/s : for Q = 0.01 m3/s :

mA

D

mA

D

smm

sm

A

QV

smm

sm

A

QV

skgs

m

m

kgQm

113.04

159.04

/1001.0

/1.0

/502.0

/1.0

/1001.0000,1

22

11

2

3

22

2

3

11

3

31

mA

D

mA

D

smm

sm

A

QV

smm

sm

A

QV

skgs

m

m

kgQm

113.04

159.04

/101.0

/01.0

/5.002.0

/01.0

/1001.0000,1

22

11

2

3

22

2

3

11

3

31

jVV ˆ11

)ˆsinˆ(cos22 jiVV

x

y


17

6.1. Unsteady Term: Time rate of change of any property N in the control volume CV,

dt

dNCV .

0

]0)(

steady, are and[,)0(

deforming]-non and stationary is[,)(

)(

tdV

CVdVt

dVdt

d

dt

dN

CV

CV

CV

CV

Since and are steady and the CV is stationary and non-deforming, the total amount of N in the

CV ( CVN ) does not change with time. Thus, 0/ dtdNCV . ANS

6.2. Time rate of change of any property N (intensive property ) of the coincident system/MV,

dt

dN

dt

dNMVSys .

Where appropriate, refer to the inlet and the exit with the subscript 1 and 2, respectively.

Net Convection Efflux Term CS

sf AdV )( /

12

12

12

)(:,)(:,

section] crosseach over uniform is[,)()(

)()()(

/1/21122

/1/2

///

A

sf

A

sf

A

sf

A

sf

A

sf

A

sf

CS

sf

AdVmAdVmmm

AdVAdV

AdVAdVAdV

Since the flow is incompressible ( is steady and uniform) and the velocity is axial and uniform over

each cross section, we also have

11

22

/1111/1

/2222/2

:,)(:

:,)(:

A

sf

A

sf

A

sf

A

sf

AdVQQVAAdVm

AdVQQVAAdVm

Thus, we also have

11122211221122/ )( VAVAQQmmAdVCS

sf

RTT

With the above, the RTT gives

)1(

)()()(

11122211221122

/

VAVAQQmm

AdVdt

tdN

dt

tdN

CS

sfCVMV

ANS The C-Mass also gives mmm :12 and QQQ :12 . Thus,

)()()()(

121212 iiMV VAQmdt

tdN . (2) ANS

6.3. Time rate of change of mass of the coincident system/MV,

dt

dM

dt

dMMVSys

.

According to the physical law of conservation of mass, we have 0dt

dM MV . ANS


18

6.4. Time rate of change of the linear x -momentum of the coincident system/MV,

dt

dP

dt

dP xMVxSys ,, .

With Eq. (2), we have

Ns

m

s

kg

VmVmVVmdt

tdP

o

xxxMV

86630cos10100

cos)0cos()()(

2212,

. ANS

The linear x -momentum of the coincident material volume increases as it flows through the nozzle.

6.5. Time rate of change of the linear y -momentum of the coincident system/MV,

dt

dP

dt

dP yMVySys ,,.


Ns

m

s

kg

VVmVVmdt

tdP

o

yyyMV

0530sin10100

)sin()()(

1212,

. ANS

The linear y -momentum of the coincident material volume does not change as it flows through the

nozzle.

6.6. Time rate of change of the kinetic energy of the coincident system/MV, 2

2

1, mVN

dt

dN

dt

dNMVSys

.


kWs

m

s

kg

VVmdt

tdKEMV

75.35102

1100

)2

1

2

1(

)(

2

222

21

22

. ANS

6.7. The kinetic energy of the coincident material volume increases as it flows through the nozzle. This is consistent

with our physical intuition that flow through a nozzle accelerates, hence resulting in increase in velocity and kinetic energy. ANS


19

Problem 7. Conservation of Mass [Munson et al., 2002, Problem 5.15, p. 279, and 2141-365 – 2008, HW #3.]

Air at standard conditions enters the compressor at a rate of 10 ft3/s. It leaves the tank through a 1.2-in.-diameter pipe with a density of 0.0035 slug/ft3 and a uniform speed of 700 ft/s.

7.1. Determine the rate (slug/s) at which the mass of air in the tank is increasing or decreasing. 7.2. Determine the average time rate of change of air density within the tank.

. Solution

Control Volume Control volume is stationary and non-deforming. It includes volume of air in the tank as well as

the portions in the inlet and exit pipes, upto the points where we have data (e.g., density, velocity).

Assumptions 1. Uniform density over each cross section. (Different cross section may have different density.) 2. Velocity is axial and uniform at each cross section. 3. Uniform density in CV.

Basic Equations

C-Mass: )()(

/ )(,)()()(

0tCV

V

tCS

sfCVMV dVtMAdVdt

tdM

dt

tdM

(A)

7.1. Determine the rate (slug/s) at which the mass of air in the tank is increasing or decreasing.

Net Convection Flux Term, )(

/ )(tCS

sf AdV

12

12

)(:,)(:,

)()()(

/1/212

///

A

sf

A

sf

A

sf

A

sf

CS

sf

AdVmAdVmmm

AdVAdVAdV

With uniform density and velocity over each cross section, we also have

12

/1/2

111222112212/

:,::

,)(

A

sf

A

sf

CS

sf

AdVQAdVQ

VAVAQQmmAdV

C-Mass The C-Mass then gives

2 1


20

21

)(

/ )()(

mm

AdVdt

tdM

tCS

sfCV

: sslugs

ft

ft

slugQAdVm

A

sf /0238.01000238.0)(:3

311/1

1

: sslugfts

ft

ft

slugAVAdVm

A

sf /01924.012

2.1

47000035.0)(: 2

2

3222/2

2

Thus,

sslugsslugmmdt

tdM CV /00456.0/)01924.00238.0()(

21 .

Neglect any contribution of the part of the CV that is not in the tank (CV = Tank + Pipes), we find that the mass of air in the tank increases at the rate of 0.00456 slug/s. ANS

7.2. Determine the average time rate of change of air density within the tank.

Unsteady Term,

)(

)()(

tCV

CV dVdt

d

dt

tdM

dt

dVV

dt

d

CVdVdt

d

dVdt

d

dt

tdM

CVCVCVCV

CV

CV

CV

]in uniform is steady,not though ,[,

)()(

.

Again, neglect any contribution of the part of the CV that is not in the tank (CV = Tank + Pipes), we have the time rate of change of air density in the tank

s

ftslug

ft

sslug

dt

d

sslugdt

dV

dt

tdM

CV

CVCV

CV

34

3

/1028.2

20

/00456.0

/00456.0)(

. ANS


21

Problem 8. Conservation of Mass and Reynolds Transport Theorem (RTT) [Adapted from Munson et al., 1998, Problem 4.54, p. 206, and 2141-365 – 2008 Mid Term.]

A rectangular elbow with a cross section of 1 0.2 m2 is mounted horizontally (gravity is in the z direction). Air

enters the elbow with a uniform speed but leaves non-uniformly. In fact, there is a region of separation or reverse flow

as shown in the figure. Assume that the flow of air in the elbow is steady.

8.1. Find the inlet speed. 8.2. Find the time rate of change of the y -linear momentum of the material volume (or control mass) that

instantaneously coincides with the volume of air in the elbow. [Find

dt

dP

dt

dP yMVySys ,,.]

Take air density to be 1.23 kg/m3.

. Solution

Control Volume The control volume is stationary and non-deforming. It includes air in the elbow only.

Material Volume We consider the material volume that instantaneously coincides with the control volume.

Assumptions

1. Incompressible flow ( is steady and uniform).

2. Velocity field is steady.

3. Uniform velocity at section 1.

Basic Equations

RTT: )()(

/ )()(,)()()(

tV

V

tCS

sfCVMV dVtNAdVdt

tdN

dt

tdN

(A)

C-Mass: )()(

/ )()(,)()()(

0tV

V

tCS

sfCVMV dVtMAdVdt

tdM

dt

tdM

(B)

8.1. Find the inlet speed.

1V

2

1

x

y

smVC /15smVB /5

L = 1m

L = 1 m is the width of the channel w = 0.2 m is the span width of the

x

y 1V


22

C-Mass: )()(

/ )()(,)()()(

0tV

V

tCS

sfCVMV dVtMAdVdt

tdM

dt

tdM

(B)

Unsteady Term: 0)(

CV

CV dVdt

d

dt

tdM . [ CV is stationary and non-deforming, is steady.]


sf AdV

/

wLAAVVV

A

AVVV

wL

AVxVVVL

xw

AVdxVVVL

xw

VAVjwdxjV

VVVL

xxV

L

VV

x

VxV

AdVAdVAdV

CB

CB

L

BBC

L

BBC

L

y

BBCyBCBy

A

sf

A

sf

CS

sf

11

11

11

0

2

1211

0

111

0

2

22

///

,2

2

)(2

]: flow ibleincompress[)()(

1]at uniform are and[),(ˆ)(ˆ

)()(00

)(:

2at Velocity:12

C-Mass

C-Mass then gives

smsmVV

V

VVV

A

CB

CB

/5/2

)15(5

2

20

1

1

ANS

8.2. Find the time rate of change of the y -linear momentum of the material volume (or control mass) that

instantaneously coincides with the volume of air in the elbow. [Find

dt

dP

dt

dP yMVySys ,,.]

RTT: )()(

/ )()(,)()()(

tV

V

tCS

sfCVMV dVtNAdVdt

tdN

dt

tdN

(A)

CS

sfyyCVyMV

AdVVdt

tdP

dt

tdP)(

)()(/

,,

Unsteady Term dt

tdP yCV )(,.

0)()(,

CV

yyCV

dVVdt

d

dt

tdP [ CV is stationary and non-deforming, velocity and

density fields are steady.]


sfy AdVV )( /


23

N

s

mm

m

kg

VVVVAVVVVVV

A

VVVVVV

wLxVxL

VVVx

L

VVw

dxVLxVVVLxVVwdxVLxVVw

dxVw

jwdxjVV

AdVVAdVVAdVV

BBCCBCBBCC

BBBCBC

L

BBBCBC

L

BBBCBC

L

BBC

L

y

L

yy

A

sfy

A

sfy

CS

sfy

4.14

5)5)(15()15(2.023.13

1

3

1

332

3

)(3

)()(

3

)(

/)(2/)(/)(

)ˆ(ˆ

)()()(

2222

3

2222

22

0

2232

2

0

2222

0

2

0

22

0

22

///

12

RTT

The RTT then gives

NNdt

tdP yMV4.14)4.14(0

)(, . ANS

At the inlet, the air stream does not have the y momentum. However, as it flows through the elbow,

it gains a net negative y momentum. Thus, the time rate of change of the y momentum for the

coincident MV is negative.

2141-365 - 2011 - hw2 solution - area vectors, …fmeabj.lecturer.eng.chula.ac.th/2141-365 hp/365...

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