Budapest University of Technology and Economics Faculty of Mechanical Engineering 2002

Dr. Mikls Blah

Selected Problems in Fluid Mechanics

1 Hydrostatics ............................................................................ 3 2 Kinematics .............................................................................. 8 3 Bernoulli Equation ................................................................ 10 4 Integral Momentum Equation............................................... 15 5 Hydraulics ............................................................................. 20 6 Compressible Flows.............................................................. 24

RESULTS 1 Hydrostatics .......................................................................... 27

atics ............................................................................ 29 ulli Equation ................................................................ 31 al Momentum Equation............................................... 34 ulics ............................................................................. 36 ressible Flows.............................................................. 39 2 Kinem3 Berno4 Integr5 Hydra6 Comp

1

1/

1/

1/

4

rtical section of a gas pipe. At the lower tap e of 500 Pa. How big is the overpressure at the

pipe.

kgK/J288Rair

=

s constant for m2000z0

6

d with water.

lar velocity is needed to

Pa108.0 5A = ? he A-B section and have pressure of

Pa in it?

y is negligible.

y is negligible.

]m/g

3

3

surface at standstill Hydrostatics 5

1/9 The vehicle is filled with oil.

[ ]20A3

oil

s/m?a

Pa0ppm/kg950

==

=

1/10 The tank wagon shown in the figure is taking a curve with a centripetal acceleration of 2s/m3a = . The tank is filled with water.

a.) How high will climb the water surface on the A-B side?

b.) How big force will affect the A-B side, when the vehicle is 1.6 m long?

1/11 Where are the both surfaces of the liquid situated if the pipe accelerates to the left

with an acceleration of 2ga = ?

Hydrostatics

1/13 The pipe is fillePa10p 50 =

How high angu

a.) reach p

b.) empty t

108.0 5

1/14 Effect of gravit

?ppmin/16000nm/kg800

0A

3

===

1/15 Effect of gravit

[Pa?ppm/kg800

k1000s/1100

0A

oil

water

===

=

1/12 min/11000n =

[ ]Pa?ppm/kg1000

0A

3water

==

1/16 What area does an ice-floe have, which can carry a person weighing 736 N? The thickness of the ice-floe is 10 cm and its density is 900 kg/m3 ?

.

1/17 The rope is weightless

N200G

mm300rm/kg1000

m/kg2300

Sphere

Sphere

3Water

3Cube

====

[ ]2m/s? a =

2 Kinematics

2/1 Pressure changes are negligible.

C80tC15t

s/m40q

2

1

3v

==

=

[ ][ ]s/m?v

s/m?v

2

1

==

2/2 Two dimensional flow:

( )[ ] [ ]s/1?vrotr10v

Az ==

2/3 Axisymmetric flow.

?vmean = Hydrostatics 7

1/18 A balloon is filled with hot air of 60C. Its diameter is 10 m. The environmental temperature is C0 . Pressure outside and inside the balloon is 105 Pa. The weight of the balloon material is can be neglected. Determine the buoyant force!

1/19 221 m/N20pp = 3

liquid m/kg800= [ ]= ? if an error of mm1 at the reading of the fluid column position causes %2 relative error of 21 pp .

1/20 After having been filled the pipe both taps were closed. During the rotation the surface in the left pipe section sinks to the point B as shown in the figure.

constT

Pa102p

Pa10p4

steamsaturated

50

==

=

[ ]s/1?=

vmax

sional flow.

nd convective acceleration in point 2/4 Unsteady, two dimen

2x

y

yt5v

0v

==

Calculate the local a'A' at s5.0t = .

9

3 Bernoulli Equation

3/1 Pa103p 5t =

[ ]s/m?vPa10p 50

==

3/2 s/m10v =

[ ]Pa?ppm/kg10

s/m4u

0A

33

===

3/3 Friction losses are negligible.

[ ]s/m?vm/kg2.1

2

3

==

Kinematics

2/5 Calculate the circulation along the dashed line.

2r2v =

[ ]s/m? 2=

2/6

[ ] [ ]2Akonv1

s/m?a.const

s/m20v

===

3/4 Steady flow with

min/m1.0q 3V = . [ ]m?h =

12

ting en

of steady flow?

acceleration g the tap? Bernoulli Equation 11

3/5 Pa106.1p 51 =

[ ]s/m?q Pa102.1p 3V5

2

==

3/6 2sm12a =

[ ]s/m?q Pa105.0pPa10p

3V

5t

50

==

=

3/7 s/125= [ ]s/m?w =

(w: relative velocity)

Bernoulli Equation

3/9 250 m/N10p =

[ ]2AAA

s/m?a

s/m4v0p

===

3/10 Pa10p 50 = Pa109.0p 51 =

Friction losses are negligible. a.) How big is the star

acceleration a whopening the tap?

b.) [ ]m?H = in case

3/11 How big is the startingin point B when openin

3/8 s/m3w =

g acceleration at the

)rerpressu [ ]s/1?= (w: relative velocity)

3/12 How big is the startinend of the pipe?

0vove(m/N102p 24t

==

14

wo-dimensional flow.

city distribution has

raw a diagram!) Bernoulli Equation 13

3/13 s/m1v = 2s/m1a =

Friction is negligible. How big force is needed to push the piston?

3/14 h/km72u = s/m4v =

Friction is negligible.

a.) [ ]s/m?q 3V = b.) How big power is needed to move the pipe?

3/15 3alc m/kg800=

[ ]s/m?vm/kg2.1 3air

==

3/16 The inner diameter of an orifice flowmeter is mm200d = . Flow coefficient 7.0=

Bernoulli Equation

3/18 Irrotational, horizontal, t

s/m5vm8.0rm5.0r

0

2

1

===

a.) What kind of velodeveloped in the arc?

b.) [ ]Pa?pp BA = c.) ?

rrf

2v

pp

1

22

0

BA

=

(D

Compressibility factor 1= . The measured difference pressure is 2m/N600p = . 3m/kg3.1= .

[ ]s/m?q 3V =

diagram

3/17 Width of the flow is 1 m.

a.) Construct the velocity distribution along the vertical line over the outlet.

b.) Calculate the flow rate [ ]s/mq 3V !

4

4

4

4

16

e negligible. ting on the cone!

e negligible. of body G [N]! Integral Momentum Equation

/1 Calculate the horizontal force acting on the conical part of the pipe!

min/m5.3q 3V = Friction losses are negligible.

/2 s/m30v1 = s/m13u =

Friction losses are negligible.

a) [ ]s/m?v2 = b) Calculate the angle of deviation [ ] (angle

between 1v and 2v )!

c) Determine the force acting on the blade! d) How is the kinetic energy of 1kg water changing, when passing the blade?

/3 s/m10v = Friction and gravity are negligible. Calculate the force acting on the arc!

Integral Momentum Equation

4/5 s/m10v = 3

Hg m/kg13600= Friction and gravity arCalculate the force ac

4/6 24 m10A = s/m10v =

Friction and gravity arDetermine the weight

4/7 N1G = [ ]s/m?v0 =

Friction is negligible.

4/8 Two dimensional flow. s/m30v =

a) [ ]N?F =

egligible. 4/4 s/m10v = s/m2u =

Friction and gravity are negligible. Calculate the force acting on the moving conical body!

b) ?AA 21 =

4/9 Two dimensional flow. Friction and gravity are n

[ ]= ?

18

anges of the air because of pressure

in the pipe.

the lower and upper outlet s due to the rapid cross Integral Momentum Equation 17

4/10 Two dimensional flow. Friction losses are negligible.

[ ]N?G15

s/m10v

==

=

4/11 Friction losses are negligible. The cylinder is balanced by the water jet.

[ ]m?hN10G

==

4/12 s/m10v = s/m6u =

Friction is negligible. Calculate the power transmitted by the water jet to the wheel!

4/13 s/m20v = s/m6u =

Friction is negligible. Calculate the mean force acting on the wheel blades in

Integral Momentum Equation

4/15 s/m2v1 =

C273tC0t

m/kg29.1

2

0

30

==

=

Friction and density chchanges are negligible. [ ]s/m?q 3V =

4/16 s/m20v1 =

[ ]m?hm/kg1 3

==

4/17 There is no friction loss[ ]Pa?pp 01 =

4/18 The flow rate through is the same. The lossethe direction x and y!

section change at the upper pipe must be considered.

[ ]m?h = 4/14 s/m2v1 =

C300ttC20t

m/kg2.1

2'1

1

31

===

=

Friction, gravity and density changes of the air because of pressure changes are negligible.

[ ]Pa?pp 21 =

4/19 Steady flow. [ ]m?h =

19

es with and

5 Hydraulics

5/1 The width of the gap is 100 mm (perpendicular to the paper plane).

[ ]N?Fms/kg1.0s/m5.0v

===

5/2 Friction loss in the confuser is negligible.

s/m10m/kg850s/m5.0v

25

31

===

[ ]Pa?pp 01 = 5/3 Friction loss of the transitional section is negligible.

s/m1014m/kg2.1s/m10v

26

31

===

[ ]Pa?pp 01 =

5/4 How do the Reynolds number and the pressure loss of a straight, smooth pipe depend on Integral Momentum Equation 4/20 Determine the quotient of the flow rat

without horizontal plate!

?q

qplatewith

V

platewithoutV =

diameter in case of laminar and turbulent flow, if the flow rate is constant?

5/5 How does a straight, smooth pipes pressure loss depend on the flow rate in case of laminar and turbulent flow?

of s/m102q 34V= has to be transported through a 10 m long straight pipe

s/m10, 243 = ). The available pressure difference is not more than termine the diameter [ ]mmD of the pipe! 5/6 Oil flow rate ( m/kg800=

Pa102 5 . De

22

s/m10 26

]s/

ly smooth pipe walls.

s/m10 26

[ ]Pa

ly smooth pipe walls.

mins/m10 26

[ ]Pa Hydraulics 21

5/7 h/m8000q 3V =

8.0025.0

m/kg2.1

D

3

===

[ ]Pa?pp 01 =

5/8 min/l1200qV =

[ ]m?hm/kg106.13 33Hg

==

5/9 The figure shows a part of a lubrication equipment, which has to transport an oil flow rate of s/m1005.0q 33V

= . For the calculation of the friction loss, it can be considered that the pipe is straight.

[ ]mm?ds/m10

m/kg80024

oil

3oil

===

Hydraulics

5/11 3.1water = [m?q 3V =

5/12 Hydraulical

s/l5q3.1

V

water

==

?pp 01 =

5/13 Hydraulical

/l180q3.1

V

water

==

?pp 01 = 5/10 The additional losses of the bends can be neglected. (It can be considered that the steel pipe is straight.)

5/14 Steady flow, hydraulically smooth pipe.

s/m1vs/m103.1

1

26water

==

[ ]Pa [ ]s/m?q s/m103.1 3V26

water

==

a) [ ]m?H =b) ?pp 01 =

Compressible Flows

bar1=

of state.

Pa10p 52 =

of state.

bar1=

of state. [ ]C

]C Hydraulics 23

5/15 What power is needed to drive the shaft of a glide bearing with min/12880 , when the shaft is mm 60 wide, 100mm long and the gap between bearing and shaft is 0.2 mm? ( ms/kg01.0oil = ) How is it possible to decrease this power?

5/16 a) Determine the confusers output diameter 2d , when the water jet is 12 m high!

b) Calculate the flow rate [ ]s/mq 3V through the pipe! Friction losses of the bends, the confuser and friction effects between the water jet and the air are negligible.

5/17 Water of h/m18q 3V = flow rate has to be transported by the equipment shown in the figure. a) How wide pipe do we need to fulfill this task? b) Determine the maximal dike height where the transport is possible? (theoretical answer)

6

6/1 p,bar5.1p 21 =

4.1

Kkg/J1000cK300T

p

1

===

Isentropic change[ ]s/m?v2 =

6/2 ,Pa103.1p 51 =

4.1Kkg/J287R

K273T1

===

Isentropic change[ ]s/kg?q m =

6/3 p,bar4.1p 21 =

4.1C20t1

==

Isentropic changea) ?t static2 =b) [?t total2 =

(temperature measured by the stagnation point

thermometer)

bar

f state.

6/4 1p,bar4p 21 ==

4.1Kkg/J287R

K300T1

===

Isentropic change o[ ]s/kg?q m =

26

Kkg/J1000cp = ,

d be the diameter 2d , if

s to be isentropic? ust [ ]NF of the rocket Compressible Flows 25

6/5 bar1p,bar4p 21 ==

4.1Kkg/J287R

C70t1

==

=

Isentropic change of state. [ ]mm?d min =

6/6 What kind of formula can be used to calculate 2v , if

a) 99.0pp

1

2 =

b) 6.0pp

1

2 =

c) 4.0pp

1

2 =

Isentropic change of state.

6/7 Air of temperature C40t = flows at a velocity s/m180v = . 4.1= , Kkg/J287R = . Calculate the Mach number (Ma) !

6/8 Carbon-dioxide of the temperature C20t = flows at a Mach number of 3.0Ma = . 3.1= , Kkg/J189R = .

Calculate the velocity of the flow! [ ]s/m

Compressible Flows

6/11 Kkg/J287R = , 4.1= .

a) How wide shoul

the outflow needb) Calculate the thr

engine!

6/9 A rocket flies in air of C23t = at a velocity of s/m400u = . Kkg/J1000c =

s/m200= . The relative velocity 2w in a Kkg/J287 , 4.1= . Calculate the Mach p

[ ]C?t A =

6/10 An aircraft flies in air of C0t = at a velocity udefinite point of the wing makes s/m250 . R =number in this point.

1

1

1

1

1

1

1

28

2310 m/N

m

N

e left side is situated at the left lower corner, the other surface in the right t a height of 100 mm.

same in standstill and rotation:

1

lent potential:

212 2

=gz

r;

n:

m.gz

Rz 2360011 ==

222

143002

m/NRz A =

equation

const+2

.

wn points (surfaces in the left and the right section), the angular velocity can RESULTS

Hydrostatics

/1 20 6200 m/NppA =

/2 221 m/N12360pp =

/3 214 m/N392pp =

/4 221 m/N486pp =

/5 The overpressure at the upper tap is 600 Pa.

/6

a.) KR

pT 290

0

00 ==

Results

1/8 0 237.ppA =

1/9 2452 s/m.a =

1/10 a) 422.0h =b) F 1400=

1/11 The surface at thvertical section a

1/12 Volumes are the2

02

21 zrzR =

Points of equiva22

1 02= rzg

After substitutio

zgzzR 221

21

02

=

0 gppA =

1/13 After writing the

rgzp =

2

2

for the both knob.) gppg

dzdp

00

== p

zgdpA

0=

be calculated. a.) s/. 1421= b.) s/. 1324=

constrgz +

2

22

written for the surface of the fluid:

2

22

] 25202A m/N107.19r = Ap pp 00

AA z

pg

ppln

0

0

0

= 25107880 m/N.pA =

1/7 m5650h =

1/14 Equation p =

00 = rp.const

[20A r2pp =

30

( )1.0100

yvAintpoat

yx

x2yx41xy

x

22

43222

=

++

( ) s/15.471.010050y

vx =+=

lar coordinates:

s/..rrrr

c

A

154710

1515102

110 ===+=+

7

has to be divided into rings of elementary width 'dr'. Integrate the rate through the rings as follows:

778.097

vv

v97

921v

rr

91v2

rrd

rr1v

rr2

rrd

rrv

rr2dr)r(v

max

meanmaxmax

1

0

9

0max

0

1

0

7

0max

0

1

0 000

===

=

=

=

=

2nn

vv

max

mean

n

+=

2s Results 29

1/15 Apply the equation constrgzp +

=

2

22

at first for the oil-filled part and then for the

water filled part of the pipe. It can be written then:

( ) ( )[ ] 2422water22oil20A m/N1025.91.015.005.01.02pp =+= 1/16 257 m.A =

1/17 m.a 30=

1/18 N1200F =

1/19 mm55.281.9800

20h ==

mm5002.0

mm1l ==

=== 92051050552 ...sin

1/20 s/. 1881=

2 Kinematics

2/1 s/m9.6v;s/m10v 21 ==

Results

x10

xv

4 2

y+

=

( )[ ]xv

vrot yAz =

Solution with po

( )[ ]drdccrot Az =

2/3

=

0max r

r1vv

The cross sectionelementary flow

rr

21v2

r2r1v

2

0max

r

020

mean

0

=

In general:

rr1vv0

max

=

2/4 [ ]5.0t

1ylocal /m5a ===2/2 Solution with Cartesian coordinates: ( ) yx

yyyrxvcosvv;

ryvsincv ==== 0a convective =

2/5 s/m61.2sdv 2== v

4 22

y

4 22x

yxx10

rx10

rxr10v

yx10

r10

rr10v

+===

+===

( )1.050

xv

)0,1.0(y,x:Aintpoatyx

x2yx41xyx

10x

v y22

43224 22

y ==

+++

=

2/6 rvr 212

1 =

22

11 r1rvv =

32

2rhg

2v 222

22 +

on the water surface on an arbitrary radius r1 , point 2 at the upper end of

sd

tv

m3

s/

Ba5.75 =+

2

lli-Equation has to be written between the surface point (1) and the pipes Results 31

[ ] 2542

Aconvective

5

41

21

convective

32

11

s/m1328.0

05.0075.0

05.0202a

xr

rrv2

xvva

xr

r2rv

xr

rv

xv

===

=

=

=

3 Bernoulli Equation

3/1 hgp2vp 0

2t ++=

s/m8.19v =

3/2 ( ) Pa108.1uv2

pp 420A ==

3/3 s/m4.7v150

100v2

hg4

2water =

=

3/4

m141.0g2

Aq

hV

==

3

Results

( )2

r2r 221

21 =

Point 1 is situated the pipe.

s/m8.10v2 =

3/8 s/124=

3/9 ++=A

0

2A0 hg2

vp

2A

AA

A

0

s/m1.24a

alasdtv

===

3/10 a.) [ ] m55.6a 0t ==b.) m52.1H =

3/11 BB

A 20510asd

tv

=

[ ] 20tB s/m31.1a ==

3/12 [ ] 0t2 s/m94.7a ==3/13 N451F =

3/14 a) The Bernou3/5 s/m793.0q V =

3/6 ( )2vphagp

20t +=++

outlet point (2), in a co-ordinate system moving with the pipe. It means that s/m24v1 = . From the Bernoulli-equation:

s/m116.0qs/m4.23v 3V == necessary to lift the water and to increase its kinetic energy. The change y must be calculated with the absolute velocity v.

kW85.82

v 212

2 =

.

s/ s/m00589.0q 3v =

3/7 Observing in an absolute co-ordinate system, the flow is irrotational ( 0vrot = ). In a co-ordinate system rotating with the pipe, = 2wrot , so the term sdwrotw is equal to

sdw2 , the Coriolis force term. ( w relative velocity) The Bernoulli equation can be written after simplifying the terms above:

2

b) the power isof the kinetic energ

vhgqP V

+=

3/15 m36p2vair

==

34

tegral Momentum Equation

rnoulli equation for points situated upstream and downstream the blade

n 45 from the horizontal plane (Northeast)

tum equation written for a control surface including only the plate and jet:

vv00 t the lower surface of the control surface. rnoulli equation: Results 33

3/16 s/m67.0p24

dq2

V ==

3/17 Because the stream lines leaving the outlet are straight and parallel, there is only a hydrostatic pressure variation along the vertical axis. It follows that the outlet velocity is constant.

s/m15.3q 3V = .

3/18 a) in the arc rKv = , because 0vrot = .

b) 1

2

12

r

r12mean r

rln

rrKdr

rK

rr1v

2

1== Because of continuity: 0mean vv =

( )2.3

rr

ln

rrvK

1

2

12mean ==

s/m4.6rKv,s/m4

rKv

1B

2A ====

From the Bernoulli-equation:

( ) Pa1025.1vv2

pp 42A2

BBA == c.)

2

A

2

BBA vvpp

Results

4 In

4/1 N12100Fx =

4/2 After writing the Bewe get the result:

12 vv =

4/3 N510F = , directio

4/4 N109F =

4/5 N57F =

4/6 N14G =

4/7 The integral momenthe upper end of the

AvAG 2 ==with v, the speed aAccording to the Be

hg2vv 20 = s/m55.4v0 = ( )22

3

0020

n1n

nln1n...

...vvv

2+=

=

= 4/8 Write the integral momentum equation for both

directions x and y: a) N 636F =

8.52 =

ucting the momentum rate

red that 2

22 vAv + ) with 1

2

rrn =

b) A/A1

Solution with constrvectors: (It has to be conside

12

0 AvA =

36

( )232CBCB vv2pwherep

hg =+ (Borda-Carnot-loss)

Hydraulics

const64Results 35

4/9 a1

aarcsin =

4/10 N52G =

4/11 m1h =

4/12 W302)uv(vAuP ==

4/13 N280FF yx ==

4/14 )vv(vpp 1'111'11 =

Pa123pp

)vv(2

pp

21

2'1

22

22'1

==

4/15 2112121 v2hg)(pp =

s/m51q

)vv(vpp3

V

121121

==

Results

0221

21 p

2vp

2v ++=+

12 vv = and 0v3 = .

4/18 m8.0h =

4/19 m1h =

4/20 2q

qplatewith

V

platewithoutV =

5

5/1 N5.7dydvAF ==

5/2 Pa72400pp 01 =

5/3 Pa1500pp 01 =

5/4 d

const

4d

dqRe 2

v ==

2 Lq4/16 )vv(vA)pp(A 1222212 = mm5.6h =

424v

lam dd

constd16

d2p ==

5

2

dconst

dconst316.0Lq 4/17 The Bernoulli-equation between point 1 and 2 (point 2 is situated at the outflow end of the

pipe):

++=+hgp

2vp

2v 0

221

21 because the area of cross section of the pipe is constant,

=const, 12 vv = An other solution can be the Bernoulli equation between point 1 and 3 (point 3 is situated on the water surface):

4

24v

turb d16

d2p =

38

02.0 , s/m827.0602.0

m05.0m200

s/m81.92m3v2

pipe =+

=

024.0102.3 4 = ration step, s/m755.0vpipe = , and the iteration can be finished. m , the necessary velocity at the confusers outlet must be:

s/m3.5

mm11mm50s =

s/3

locity without friction loss can be calculated: s/m7.7m3g2videal == ,

24

3

m105.6s/

s/=

eter is in this case 29 mm. Because of friction losses, we need a pipe of We start the iteration with 02.0= and mm50d = :

mm52dm102.21As/m36.214

s/m1 242 ===++

018.01045.9 4 == s number we consider that the pipe is hydraulically smooth) ion step with 018.0= and mm52d = we get the new diameter of Results 37

5/5 Vv

2

2v

lam qconst

Adq

64dL

Aq

2p =

=

75.1V

4 v2

2v

turb qconst

Adq

316.0dL

Aq

2p =

=

5/6 Considered that the flow will be laminar and using the formula Re/64= , we get mm4.13d = .

The Reynolds number is 189 which is less than 2300, so the flow is laminar.

5/7 Pa143pp 01 =

5/8 mm17h =

5/9

+=dL1

2vhg

2

Considering laminar flow, the result will be mm3.19d = . 230033Re

40

77.0=

N108.9v 322 = Results 39

6 Compressible Flows

6/1 s/m260v2 =

6/2 s/kg274.0200m/kg37.1m10vAq 323222m ===

6/3 a) C42t static2 = b) C20t total2 +=

6/4 833.01

2TT

1

*

=+=

s/kg018.0Avq

m/kg9.2TT

vs/m316aTTa

s/m346TRa

***m

31

11

1

**

*1

1

**

11

===

=

===

==

6/5 s/kg25.0vAq 222m ==

mm3.17dd

m1034.2v

qA

*min

24**

m*

====

6/6 a) ( )212 pp2v =

Results

6/9 C56t A =

6/10 Ma,K262T 22 =

6/11 a) mm138d =b) AF 22 =

b)

= 1

1

2

1

12 p

p1p1

2v

c)

+

=1

21p1

2v1

12

6/7 59.0Ma =

6/8 s/m80v =