CHEN  3200   Fluid  Mechanics   Spring  2011      

Homework  4  –  Solutions    1.    (15  points)    Bernoulli’s  equation  can  be  adapted  for  use  in  evaluating  unsteady  flow  conditions,  such  as  those  encountered  during  start-­‐up  processes.    For  example,  consider  the  large  tank  below  that  is  initially  filled  with  water  to  a  depth  of  3  m.    A  pipe  is  attached  with  the  dimensions  shown,  and  initially  capped  at  point  2.    When  the  cap  is  removed,  water  will  begin  to  flow  out  of  the  tube,  and  the  velocity  of  the  water  in  the  tube  will  change  over  time.    We  can  modify  the  Bernoulli  equation  to  account  for  transient  effects  using    

!! +!2!!

! + !"!! = !! +!2!!

! + !"!! + !!"!" !"

!

!  

 where  the  last  term  describes  the  rate  of  change  of  the  fluid  momentum  as  we  travel  along  a  streamline.        

   

(a)    (5  points)  Show  that  the  integrand  on  the  right  hand  side  of  the  equation  has  units  of  pressure.  

 (b)  (10  points)  Making  the  assumptions  discussed  in  class,  we  saw  that  the  above  equation  can  

be  simplified  to  give    

!!!2!ℎ − !!!

=!"2!  

 Integrate  this  equation  to  find  an  expression  for  V2(t)  and  plot  the  result.    You  may  need  to  consult  a  table  of  integrals  (or  use  computational  tools)  to  evaluate  the  left  hand  side.  

         

CHEN  3200   Fluid  Mechanics   Spring  2011    

(a)    The  integrand  is  ! !"!"!",  where  !  has  units  of  kg/m3,  the  rate  of  change  of  the  velocity  !"

!"  has  

units  of  acceleration,  m/s2,  and  !"  is  the  differential  distance  traveled  along  the  path  s,  and  has  units  of  m.    Together,    

 kgm3

ms2m=

kg∙ms2

m2= Nm2=Pa    .  

 Thus,  the  integrand  has  units  of  pressure.    

 (b)  

 

80812 c06a.3d GGS 6/12/08 17:23

Example 6.9 Unsteady Bernoulli EquationA long pipe is connected to a large reservoir thatinitially is filled with water to a depth of 3 m. Thepipe is 150 mm in diameter and 6 m long. Determinethe flow velocity leaving the pipe as a function oftime after a cap is removed from its free end.

GIVEN: Pipe and large reservoir as shown.

FIND: V2(t).

SOLUTION:Apply the Bernoulli equation to the unsteady flow along a streamline from point 1 to point 2 .

Governing equation: p1 1 Vt

ds!! ! gz1 "

" 0(5) " 0(6)V 2

2#p2 2! ! gz2 !

V 2

2

2

# 1

Assumptions: (1) Incompressible flow.(2) Frictionless flow.(3) Flow along a streamline from 1 to 2 .(4) p1 " p2 " patm.(5) V2

1 ’ 0:(6) z2 " 0.(7) z1 " h " constant.(8) Neglect velocity in reservoir, except for small region near the inlet to the tube.

Then

gz1 " gh "V2

2

2!

Z 2

1

@V@t

ds

In view of assumption (8), the integral becomesZ 2

1

@V@t

ds !Z L

0

@V@t

ds

In the tube, V " V2 everywhere, so thatZ L

0

@V@t

ds "

Z L

0

dV2

dtds " L

dV2

dt

This is the rate of change over time of the momentum (per unit mass) within the pipe; in the long term it will approach zero.Substituting gives

gh "V2

2

2! L

dV2

dt

Separating variables, we obtain

dV2

2gh $ V22

"dt2L

Integrating between limits V " 0 at t " 0 and V " V2 at t " t,Z V2

0

dV2gh $ V2 "

1!!!!!!!!2gh

p tanh$1 V!!!!!!!!2gh

p" #$ %V2

0"

t2L

Since tanh$1(0) " 0, we obtain

1!!!!!!!!2gh

p tanh $ 1 V2!!!!!!!!2gh

p" #

"t

2Lor

V2!!!!!!!!2gh

p " tanht

2L

!!!!!!!!2gh

p& '!

V2"t#

1

2D = 150 mm

h = 3 m

L = 6 m

V2

z

Flow

6-6 UNSTEADY BERNOULLI EQUATION—INTEGRATION OF EULER’S EQUATION ALONG A STREAMLINE W-17

CHEN  3200   Fluid  Mechanics   Spring  2011    

 

 

80812 c06a.3d GGS 6/12/08 17:23

Example 6.9 Unsteady Bernoulli EquationA long pipe is connected to a large reservoir thatinitially is filled with water to a depth of 3 m. Thepipe is 150 mm in diameter and 6 m long. Determinethe flow velocity leaving the pipe as a function oftime after a cap is removed from its free end.

GIVEN: Pipe and large reservoir as shown.

FIND: V2(t).

SOLUTION:Apply the Bernoulli equation to the unsteady flow along a streamline from point 1 to point 2 .

Governing equation: p1 1 Vt

ds!! ! gz1 "

" 0(5) " 0(6)V 2

2#p2 2! ! gz2 !

V 2

2

2

# 1

Assumptions: (1) Incompressible flow.(2) Frictionless flow.(3) Flow along a streamline from 1 to 2 .(4) p1 " p2 " patm.(5) V2

1 ’ 0:(6) z2 " 0.(7) z1 " h " constant.(8) Neglect velocity in reservoir, except for small region near the inlet to the tube.

Then

gz1 " gh "V2

2

2!

Z 2

1

@V@t

ds

In view of assumption (8), the integral becomesZ 2

1

@V@t

ds !Z L

0

@V@t

ds

In the tube, V " V2 everywhere, so thatZ L

0

@V@t

ds "

Z L

0

dV2

dtds " L

dV2

dt

This is the rate of change over time of the momentum (per unit mass) within the pipe; in the long term it will approach zero.Substituting gives

gh "V2

2

2! L

dV2

dt

Separating variables, we obtain

dV2

2gh $ V22

"dt2L

Integrating between limits V " 0 at t " 0 and V " V2 at t " t,Z V2

0

dV2gh $ V2 "

1!!!!!!!!2gh

p tanh$1 V!!!!!!!!2gh

p" #$ %V2

0"

t2L

Since tanh$1(0) " 0, we obtain

1!!!!!!!!2gh

p tanh $ 1 V2!!!!!!!!2gh

p" #

"t

2Lor

V2!!!!!!!!2gh

p " tanht

2L

!!!!!!!!2gh

p& '!

V2"t#

1

2D = 150 mm

h = 3 m

L = 6 m

V2

z

Flow

6-6 UNSTEADY BERNOULLI EQUATION—INTEGRATION OF EULER’S EQUATION ALONG A STREAMLINE W-17

80812 c06a.3d GGS 6/12/08 17:23

For the given conditions,

!!!!!!!!2gh

p!

!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!2

"9:81 m

s2 "3 m

r! 7:67 m/s

and

t2L

!!!!!!!!2gh

p!

t2

"1

6 m"

7:67 ms

! 0:639t

The result is then V2 ! 7.67 tanh (0.639t) m/s, as shown:

Notes:! This problem illustrates use of the unsteady Bernoulli equation.! Initially the head available at state 1 is used to accelerate the fluid in

the pipe; eventually the head at state 2 equals the head at state 1 .! This problem is somewhat unrealistic except for the initial instants—the

asymptotic flow condition actually corresponds to a turbulent flow!The Excel workbook for this Example allows exploration of the effectof varying the parameters for this problem.

0

8

6

4

2

01 2 3 4 5

t (s)

V2

(m/s

)

V2 = 7.67 tanh (0.639 t)

W-18 CHAPTER 6 / INCOMPRESSIBLE INVISCID FLOW

CHEN  3200   Fluid  Mechanics   Spring  2011    2.    (15  points)  Fluid  approaches  a  submerged  cylinder  with  velocity  !! = 10  m/s.    The  cylinder  has  a  radius  of  ! = 55  cm.    Using  boundary  layer  theory,  it  is  possible  to  describe  how  the  fluid  velocity  changes  near  the  cylinder  surface  as  a  function  of  r  and  θ,  with  θ  measured  as  shown  below.    At  the  surface  of  the  cylinder,  the  fluid  speed  is  determined  to  be  ! = 2!! sin ! .        

   Calculate  as  and  an  at  point  A  on  the  surface  of  the  cylinder,  where  θ =  60  °.    

 

!!" !"

!"!"

CHEN  3200   Fluid  Mechanics   Spring  2011    

     

CHEN  3200   Fluid  Mechanics   Spring  2011    3.    (20  points)  An  incompressible,  one-­‐dimensional  fluid  flows  from  left  to  right  through  the  circular  nozzle  shown  below.    The  velocity  entering  the  nozzle  is  given  by  ! = !! + !! sin!",  where  !! = 20  m/s,  !! = 2  m/s,  and  ! = 0.3  rad/s.    The  nozzle  is  1  m  in  length,  0.4  m  in  diameter  at  the  entrance,  and  0.2  m  in  diameter  at  the  exit.  

 (a)  (10  points)  Determine  an  equation  for  the  acceleration  at  the  exit  of  the  nozzle  as  a  

function  of  time.  (b)  (5  points)  Plot  the  acceleration  versus  time  for  one  complete  cycle.  (c)  (5  points)  Now,  plot  the  acceleration  at  the  channel  exit  if  the  nozzle  has  a  constant  

diameter  of  0.4  m  (i.e.,  it  is  now  a  cylindrical  tube).    Explain  the  difference  between  the  two  plots.  

 (a)  

        = 12800+ 2560 sin!" + 128 sin2!" + 2.4 cos!"  

Problem 5.63 [3] Part 1/2

CHEN  3200   Fluid  Mechanics   Spring  2011        (b,  c)  

 

Problem 5.63 [3] Part 2/2

CHEN  3200   Fluid  Mechanics   Spring  2011    4.    (20  points)  A  fluid  velocity  field  is  given  by  ! = !" − ! !− !" !,  where  ! = 0.2  s!!  and  ! = 0.6  m/s.    If  x  and  y  have  units  of  meters:  

 (a)    (5  points)  Find  a  general  expression  for  the  acceleration  vector  (a)  as  a  function  of  x  and  y.      (b)    (10  points)  Find  the  acceleration  (a),  magnitude  of  the  acceleration  (|a|),  and  the  angle  the  

acceleration  vector  makes  with  the  x-­‐axis  (θ)  for  each  of  the  points  (0,1.33),  (1,2)  and  (2,4).  (c)    (5  points)  Find  an  expression  for  the  streamlines,  in  the  form  ! !,! = !,  where  C  is  a  

constant.    Plot  this  function  for  values  of  ! = ±0.2,±0.4  and  ± 0.8.    Draw  the  acceleration  vectors  from  (b)  on  your  plot.  

 (a)  

          = 0.04! − 0.12 !+ 0.04! !    (b)  The  acceleration  at  each  point  is  just  a  evaluated  at  the  x  and  y  coordinates:    

   The  magnitude  of  a,  given  by   ! = !!! + !!! .       At  (x,y)     =   (0,  1.33),    |a|  =  0.131  m/s2    

      =   (1,  2),    |a|  =  0.113  m/s2  

 =   (2,  4),    |a|  =  0.164  m/s2  

 The  angle  that  a  makes  with  the  x-­‐axis  is  given  by  ! = tan−1

!!!!.    Therefore,    

      At  (x,y)     =   (0,  1.33),    ! = tan−1 .!"##

!.!"= −23  °  

 

      =   (1,  2),    ! = tan−1 .!"!!!.!"!!

= −45  °    

=   (2,  4),    ! = tan−1 .!"#!.!"!!

= −76  °        

Problem 5.54 [3]

Problem 5.54 [3]

CHEN  3200   Fluid  Mechanics   Spring  2011        (c)  

Problem 5.54 [3]

CHEN  3200   Fluid  Mechanics   Spring  2011            5.    (15  points)  From  Newtonian  physics,  a  particle  launched  with  velocity  V0  at  an  angle  θ  with  respect  to  the  ground  will  travel  in  a  parabolic  path  given  by  ! = − !/ 2!!! cos2 ! !! + tan ! !.    The  pathline  for  a  stream  of  water  leaving  a  nozzle  is  shown  on  the  last  page.  

   (a)    (10  points)  Use  the  end  of  the  nozzle  as  the  origin,  and  find  the  coordinates  of  several  

points  along  the  pathline  for  the  stream  of  water.    Plot  the  coordinates,  and  fit  a  curve  to  the  data  to  show  that  the  shape  of  the  pathline  is  parabolic,  with  the  general  form  ! = !!!! + !!!  

   (b)    (5  points)  Use  the  above  equation,  and  your  values  of  c1  and  c2,  to  calculate  the  angle  

(with  respect  to  the  x-­‐axis)  that  the  water  leaves  the  nozzle,  and  the  initial  speed  V0.      (a)    Points  along  the  pathline  for  the  stream  of  water  are  measured  as  shown  on  the  figure  below.    Converting  the  units  to  meters,  plotting  the  data  and  fitting  a  parabolic  model  to  the  points  gives    

   The  high  value  of  R2  indicates  good  agreement  between  the  data  points  and  the  parabolic  model.  

!"#$%&'()*+,-.&/&&&'&(&)&*&+&,&-&.'/'&'''(')'*'+

01234 51234/ /

/6//('' /6//&'&/6//+-) /6//(''/6/&/-+ /6//))'/6/&)-- /6//*+(/6/&., /6//*'(/6/')&' /6//)-(/6/',,) /6//(+'/6/(/.+ /6//')&/6/((,, /6//&'&/6/(+.. 7/6///-/6/)/+& 7/6//)/'/6/))+( 7/6//,'(/6/),)) 7/6/&//*/6/*/'* 7/6/&'-+/6/*'+, 7/6/&+/-/6/**-- 7/6/'/&/6/*-, 7/6/')&'/6/+&*& 7/6/'-*)/6/+)(' 7/6/(((,/6/++,) 7/6/(,,./6/+.** 7/6/)(-'/6/,&.+ 7/6/)-+)/6/,(-- 7/6/*&/'/6/,)(, 7/6/*(),/6/,*.' 7/6/**&

51817&-6'''0'191/6+).&01:;181/6...),1

7/6/+1

7/6/*1

7/6/)1

7/6/(1

7/6/'1

7/6/&1

/1

/6/&1

/1 /6/&1 /6/'1 /6/(1 /6/)1 /6/*1 /6/+1 /6/,1 /6/-1

!!"#

$!

"!"#$!

CHEN  3200   Fluid  Mechanics   Spring  2011    

   

 (b)  From  the  model  fit  in  Excel,  we  obtain  the  coefficients  !! = −18.2  m-­‐1  and  !! = 0.65.    From  the  equation  given  in  the  problem  statement,       !! = 0.65 = tan!    thus,  ! = 33  °,  which  appears  reasonable  in  comparison  with  the  figure  above.    We  can  now  find  the  velocity  from    

  !! = −18.2 !m= − !

!!!! cos2 !  

 or    

  !! = !.!"  m/s2

! !".!  m!! cos2 !!°= 0.62  m/s