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echnology and Economics ngineering 2002

Dr. Miklós Blahó

Selected Problems in Fluid Mechanics

statics ............................................................................ 3 atics .............................................................................. 8

ulli Equation ................................................................ 10 al Momentum Equation ............................................... 15 ulics ............................................................................. 20

Budapest University of T Faculty of Mechanical E

1 Hydro 2 Kinem 3 Berno 4 Integr 5 Hydra

6 Compressible Flows.............................................................. 24

RESULTS 1 Hydrostatics .......................................................................... 27 2 Kinematics ............................................................................ 29 3 Bernoulli Equation ................................................................ 31 4 Integral Momentum Equation ............................................... 34 5 Hydraulics ............................................................................. 36 6 Compressible Flows.............................................................. 39

[ ]Pa?pp 14 =−

ows a vertical section of a gas pipe. At the lower tap erpressure of 500 Pa. How big is the overpressure at the

w in the pipe.

kgK/J288R air

m/kg2

m/N 3

=

  

] ], erature is constant for m2000z0 <≤ .

2m/N

air m/kg25.

m/N0 3



  

A the temperature is constant for Azz0 <≤ .

1/8 The vehicle is filled with oil.

4 Pa10p 50 ≈ (for the calculation of ρ )

Outside (air): C0T1 o=

In chimney (smoke):   

≈

C250T

mmHg760p

2 o

[ ]Pa?pp 21 =−

[ ]Pa?pp s/m3a

m/kg950

3 oil

=− =

=ρ

1 Hydrostatics

For all relevant problems Kkg/J287R = , kg/N81.9g =

1/1 [ ]Pa?pp 0A =−

1/2 [ ]Pa?pp 21 =−

1/3 Section 1-2: 312 m/kg3.1=ρ

Section 3-4: 3m/kg1.1=ρ

Hydrostatics

1/5 The figure sh there is an ov upper tap?

There is no flo

gas

air

/kg7.0

m/kg2.1

=ρ

1/6

10p 0z



  

=ρ

= =

a.) [K?T0 = b.) [Pa?p A =

if the temp

1/7 5A 105.0p ⋅=

1p 0z



  

=ρ

= =

[ ]m?z = if

1/12 min/11000n =

[ ]Pa?pp m/kg1000

3 water

=− =ρ

water.

city is needed to

Pa105⋅ ?

section and have pressure of

it?

ligible.

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 ?

ightless.

m/g

m/g 3

surface at standstill

1/17 The rope is we

N200G

mm300r k1000

k2300

Sphere

Water

Cube

= =ρ

=ρ

[ ]2m/s? a =

Hydrostatics 5

1/9 The vehicle is filled with oil.

[ ]2 0A

3 oil

s/m?a

Pa0pp m/kg950

=− =ρ

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 2 g

a = ?

Hydrostatics

1/13 The pipe is filled with Pa10p 50 =

How high angular velo

a.) reach 8.0p A =

b.) empty the A-B

Pa108.0 5⋅ in

1/14 Effect of gravity is neg

?pp min/16000n m/kg800

=− = =ρ

1/15 Effect of gravity is neg

[ ]Pa?pp m/kg800

m/kg1000 s/1100

3 oil

3 water

=− =ρ

=ρ

=ω

1/1

1/2

Kinematics

negligible.

max

2/4 Unsteady, two dimensional flow.

2 x

yt5v

Calculate the local and convective acceleration in point 'A' at s5.0t = .

drostatics 7

8 A balloon is filled with hot air of 60°C. 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!

9 221 m/N20pp =− 3

liquid m/kg800=ρ [ ]°=α ? if an error of mm1± at the reading of the fluid

column position causes %2± relative error of 21 pp − .

0 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

Pa10p 4

steamsaturated

5 0

⋅=

[ ]s/1?=ω

2/1 Pressure changes are

C80t C15t

s/m40q

3 v

°= °=

[ ] [ ]s/m?v

s/m?v

= =

2/2 Two dimensional flo

( )[ ] [ ]s/1?vrot r10v

Az =

2/3 Axisymmetric flow.

? v v mean =

Bernoulli Equation

egligible.

3/4 Steady flow with

min/m1.0q 3V = .

[ ]m?h =

inematics 9

/5 Calculate the circulation along the dashed line.

2r 2

v =

[ ]s/m? 2=Γ

[ ] [ ]2Akonv

s/m?a .const

s/m20v

=ρ =

3/1 Pa103p 5t ⋅=

[ ]s/m?v Pa10p 50

= =

3/2 s/m10v =

[ ]Pa?pp m/kg10

s/m4u

=− =ρ

3/3 Friction losses are n

[ ]s/m?v m/kg2.1

= =ρ

ting en

of steady flow?

acceleration g the tap?

g acceleration at the

)rerpressu

3/8 s/m3w = [ ]s/1?=ω

(w: relative velocity)

3/12 How big is the startin end of the pipe?

0v ove(m/N102p 24t

= ⋅=

Bernoulli Equation 11

3/5 Pa106.1p 51 ⋅=

[ ]s/m?q Pa102.1p

3 V

5 2

⋅=

3/6 2sm12a =

[ ]s/m?q Pa105.0p

Pa10p

3 V

5 t

5 0

⋅=

3/7 s/125=ω [ ]s/m?w =

(w: relative velocity)

Bernoulli Equation

3/9 250 m/N10p =

[ ]2A A

s/m?a

s/m4v 0p

= =

3/10 Pa10p 50 =

Pa109.0p 51 ⋅=

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

acceleration ’a’ wh opening the tap?

b.) [ ]m?H = in case

3/11 How big is the starting in point B when openin

ipe?

eter is mm200d = . Flow coefficient 7.0=α

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

[ ]s/m?q 3V =

Bernoulli Equation 14

3/18 Irrotational, horizontal, two-dimensional flow.

s/m5v m8.0r m5.0r

= = =

a.) What kind of velocity distribution has developed in the arc?

b.) [ ]Pa?pp BA =−

c.) ? r r

2 v

2 2

BA  

  

 =

− (Draw a diagram!)

3/17 Width of the flow is 1 m.

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

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

Bernoulli Equation

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 p

3/15 3alc m/kg800=ρ

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

= =ρ

3/16 The inner diameter of an orifice flowm

egligible. on the cone!

egligible. body ’G’ [N]!

4/8 Two dimensional flow. s/m30v =

a) [ ]N?F = b) ?AA =

. e negligible.

/4 s/m10v = s/m2u =

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

4/9 Two dimensional flow Friction and gravity ar

[ ]°=α ?

4 Integral Momentum Equation

4/1 Calculate the horizontal force acting on the conical part of the pipe!

min/m5.3q 3V =

Friction losses are negligible.

4/2 s/m30v1 = s/m13u =

Friction losses are negligible.

a) [ ]s/m?v 2 = b) Calculate the angle of deviation [ ]°β (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?

4/3 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=ρ

Friction and gravity are n Calculate the force acting

4/6 24 m10A −= s/m10v =

Friction and gravity are n Determine the weight of

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

Friction is negligible.

the direction x and y!

ges of the air because of pressure

the pipe.

lower and upper outlet due to the rapid cross upper pipe must be

considered.

[ ]m?h =

4/14 s/m2v1 =

C300tt C20t

m/kg2.1

2'1

3 1

°== °=

=ρ

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

[ ]Pa?pp 21 =−

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

Integral Momentum Equation 17

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

[ ]N?G 15

s/m10v

= °=α

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

[ ]m?h N10G

= =

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 =

C273t C0t

m/kg29.1

3 0

°= °=

=ρ

Friction and density chan changes are negligible.

[ ]s/m?q 3V =

4/16 s/m20v1 =

[ ]m?h m/kg1 3

= =ρ

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

4/18 The flow rate through the is the same. The losses section change at the

Hydraulics

he gap is 100 mm (perpendicular to the paper plane).

the confuser is negligible.

f the transitional section is negligible.

s/2

eynolds number and the pressure loss of a straight, smooth pipe depend on diameter in case of laminar and turbulent flow, if the flow rate is constant?

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

5/6 Oil flow rate of s/m102q 34V −⋅= has to be transported through a 10 m long straight pipe

( s/m10,m/kg800 243 −=ν=ρ ). The available pressure difference is not more than Pa102 5⋅ . Determine the diameter [ ]mmD of the pipe!

Integral Momentum Equation 19 4/20 Determine the quotient of the flow rates with and

without horizontal plate!

? q

q platewith

platewithout V =

5/1 The width of t

[ ]N?F ms/kg1.0 s/m5.0v

= =µ =

5/2 Friction loss in

s/m10 m/kg850 s/m5.0v

−=ν

=ρ

[P?pp 01 =−

5/3 Friction loss o

m1014 m/kg2.1

s/m10v

3 1

−⋅=ν

=ρ

[P?pp 01 =−

5/4 How do the R

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

s/m26−

]

smooth pipe walls.

s/m 26−

]Pa

smooth pipe walls.

in s/m 26−

]Pa

5/14 Steady flow, hydraulically smooth pipe.

s/m1v s/m103.1

26 water

= ⋅=ν −

[ ]m?H = [ ]Pa?

[ ]s/m?q

s/m103.1 3

26 water

⋅=ν −

b) pp 01 =−

Hydraulics 21

5/7 h/m8000q 3V =

8.0 025.0

m/kg2.1

=η =λ =ρ

[ ]Pa?pp 01 =−

5/8 min/l1200q V =

[ ]m?h m/kg106.13 33Hg

⋅=ρ

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?d s/m10

m/kg800 24

oil

3 oil

= =ν

=ρ −

Hydraulics

5/11 103.1water ⋅=ν

[ s/m?q 3V =

5/12 Hydraulically

s/l5q 103.1

water

= ⋅=ν

[?pp 01 =−

5/13 Hydraulically

m/l180q 103.1

water

= ⋅=ν

[?pp 01 =−

5/1

Compressible Flows

bar1

f state.

Pa1052 =

f state.

bar1

f state. ]C ]

ured by the stagnation point thermometer)

bar1

of state.

6/4 p,bar4p 21 ==

4.1 Kkg/J287R

K300T1

=κ = =

Isentropic change [ ]s/kg?q m =

draulics 23

5 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 =µ ) How is it possible to decrease this power?

6 a) Determine the confuser’s 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.

7 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)

6/1 p,bar5.1p 21 ==

4.1

Kkg/J1000c K300T

=κ

= =

Isentropic change o [ ]s/m?v 2 =

6/2 p,Pa103.1p 51 ⋅=

4.1 Kkg/J287R

K273T1

=κ = =

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

6/3 p,bar4.1p 21 ==

4.1 C20t1

=κ °=

Isentropic change o a) [?t static2 °= b) [ C?t total2 °= (temperature meas

, if

s/m180v = . 4.1=κ , Kkg/J287R = .

flows at a Mach number of 3.0Ma = .

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

Compressible Flows 26

6/11 Kkg/J287R = , Kkg/J1000c p = ,

4.1=κ . a) How wide should be the diameter 2d , if

the outflow needs to be isentropic? b) Calculate the thrust [ ]NF of the rocket

engine!

Kkg/J1000c p =

[ ]C?t A °=

6/10 An aircraft flies in air of C0t °= at a velocity s/m200u = . The relative velocity 2w in a definite point of the wing makes s/m250 . Kkg/J287R = , 4.1=κ . Calculate the Mach

number in this point.

Compressible Flows

6/5 bar1p,bar4p 21 ==

4.1 Kkg/J287R

C70t1

=κ =

°=

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

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

a) 99.0 p p

2 =

b) 6.0 p p

2 =

c) 4.0 p p

2 =

Isentropic change of state.

6/7 Air of temperature C40t °−= flows at a velocity Calculate the Mach number (Ma) !

6/8 Carbon-dioxide of the temperature C20t °= 3.1=κ , Kkg/J189R = .

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

b.) g p p

g dz dp

0 0

ρ−=ρ−=

z gdpA 0ρ−=

230 m/N

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:

ent potential:

2 12 2

ω =

gz r;

m. g z

Rz 2360011 =ω=

2 22

14300 2

m/N R

A = 

ω −

equation

const+ 

2 .

n points (surfaces in the left and the right section), the angular velocity can be calculated. a.) s/. 1421=ω

b.) s/. 1324=ω

const r

gz + 

  

 ω −ρ

written for the surface of the fluid:

22 0 ω

] 25202A m/N107.19rr ⋅=−

A p pp 00 ∫

A A z

p g

p p

ln 0

ρ −=

25107880 m/N.p A ⋅=

/7 m5650h =

1/14 Equation p −=

0 ρ−= r

p.const

[ 2

0A 2 pp

ω ρ=−

RESULTS

1 Hydrostatics

1/1 20 6200 m/Npp A =−

1/2 221 m/N12360pp =−

1/3 214 m/N392pp =−

1/4 221 m/N486pp =−

1/5 The overpressure at the upper tap is 600 Pa.

1/6

a.) K R

p T 290

0 0 =ρ =

Results

1/8 0 1237.pp A ⋅=−

1/9 2452 s/m.a =

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

1/11 The surface at the vertical section a

1/12 Volumes are the

1 2

0 2

2 1

zrzR π=π

Points of equival 22

1 02 =

ω −⋅

r zg

After substitution

z gz

zR 2

2 1

0 2

ω =

0 zgpp A  

 ρ−=−

1/13 After writing the

r gzp  

 ω −ρ−=

for the both know

2/2 Solution with Cartesian coordinates:

( ) yx r x

vcosvv; r y

vsincv =α⋅=−=α−=

( ) 1.0100

y v

Aintpoat yx

x2yx 4 1

xyx x

4 3

2222

−= ∂ ∂

⇒ +

+−+ −

( ) s/15.471.010050 y

v x v

xy =+= ∂ ∂

− ∂

∂ =

ith polar coordinates:

s/. .rrrr

c dr dc

1547 10

151510 2

1 10 ===+=

  

  

  

  

 −

0r r

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

778.0 9 7

v v

v 9 7

9 2

1v r r

9 1

v2 r r

r r

d r r

1v r r

2 r r

d r r

v r r

2dr)r(vr2

max

mean maxmax

0 max

0 000

==⇒=  

   −=

  

  

  

 −





= 

  



  



  

  

  

 −=



  

  

  

 =π ∫∫ ∫

2n n

v v

r r

max

mean

0 + =⇒

  

  

  

 −

2/4 [ ] 2 5.0t

1ylocal s/m5a ===

0a convective =

s/m1 2

4 22 y

4 22 x

yx x

10 r

x 10

r x

r10v

yx y

10 r

y 10

r y

r10v

+ ===

+ −=−=−=

( ) 1.050

x v

)0,1.0(y,x:Aintpoat yx

x2yx 4 1

xyx 10

x v y

4 3

224 22

y = ∂

∂ ⇒=

+−+ =

∂

∂ −

2/5 6.2sdv −==Γ ∫

2/6 vrvr 21 2

1 π=π

2 2

11 r 1

rvv =

Results 29

1/15 Apply the equation const r

gzp + 

  

 ω −ρ−=

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

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

( ) ( )[ ] 2422water22oil 2

0A m/N1025.91.015.005.01.02 pp ⋅=−ρ+−ρ

ω =−

1/16 257 m.A =

1/17 m.a 30=

1/18 N1200F =

1/19 mm55.2 81.9800

20 h =

⋅ =

mm50 02.0

mm1 l =

± ±

°=α⇒==α 920510 50 552

.. .

sin

1/20 s/. 1881=ω

2 Kinematics

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

Results

10 x

v 4

y = ∂

∂

( )[ ]vrot Az Solution w

( )[ ]crot Az

2/3

  

 = max 1vv

The cross s elementary

2 1

r 1

max

2 0

mean

  

  



π =

In general:

1vv max   

 =

3/6 ( ) 2 vp

hag p 20t +

ρ =⋅++

2 r

hg 2

v 222 2 2 ω−⋅+

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

∂ ∂

sd t v

m3⋅

Ba5.75 = 

li-Equation has to be written between the surface point (1) and the pipe’s 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 3V2 =⇒=

s necessary to lift the water and to increase its kinetic energy. The change y must be calculated with the absolute velocity ’v’.

kW85.8 2

v 21 2

2 = 

− .

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, ω= 2wrot , so the term sdwrotw∫ × is equal to sdw2∫ ω× , the Coriolis force term. ( w – relative velocity) The Bernoulli equation can be

written after simplifying the terms above:

b) the power i of the kinetic energ

v hgqP V 



 +⋅⋅ρ=

3/15 m36 p2

v air

= ρ ∆⋅

Results 31

[ ] 2 5

Aconvective

4 1

2 1

convective

3 2

s/m132 8.0

05.0 075.0

05.0202 a

x r

r rv2

x v

x r

r 2

rv x r

r v

x v

−= ⋅

−=

∆ ∆

−= ∂ ∂

∆ ∆

 

  −=

∂ ∂

= ∂ ∂

3 Bernoulli Equation

3/1 hg p

2 vp 0

2 t ⋅+

ρ +=

s/m8.19v =

3/2 ( ) Pa108.1uv 2

pp 420A ⋅=− ρ

=−

3/3 s/m4.7v1 50

100 v

2 hg

4 2

water =⇒   



  

 −

 

  ρ

=⋅⋅ρ

3/4

m141.0 g2

A q

h V

3/5 s/m793.0q 3=

Results

( ) 2

r 2 r 221

2 1 =

ω −

ω−

Point 1 is situated o the pipe.

s/m8.10v 2 =

3/8 s/124=ω

3/9 ∫+⋅+=ρ A

2 A0 hg 2

2 A

s/m1.24a

alasd t v

=⋅= ∂ ∂ ∫

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

3/11 B B

A 20 5

10asd t v

 

= ∂ ∂ ∫

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

3/12 [ ] 20t2 s/m94.7a ==

3/13 N451F =

3/14 a) The Bernoul

( ) 22

2 0

1n1n ...

... vvv

2 +−

= 

 

− 

 

= ρ

tegral Momentum Equation

rnoulli equation for points situated upstream and downstream the blade

45° from the horizontal plane (’Northeast’)

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

vv 00 ⋅⋅

the lower surface of the control surface. rnoulli equation:

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

2 2 vAv ⋅⋅ρ+⋅ )

nnln

with 1

r r

n =

b) A/A1

Solution with constr vectors: (It has to be conside

1 2

0 AvA ⋅ρ=⋅⋅ρ

Results 33

3/16 s/m67.0 p2

4 d

q 2

V =ρ ∆π

ε⋅α=

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 r K

v = , because 0vrot = .

b) 1

r12 mean r

r ln

rr K

dr r K

rr 1

v 2

1 −

= −

= ∫ Because of continuity: 0mean vv =

( ) 2.3

r r

rrv K

12mean = −

=⇒

s/m4.6 r K

v,s/m4 r K

v 1

B 2

A ====⇒

From the Bernoulli-equation:

( ) Pa1025.1vv 2

pp 42A 2

BBA ⋅=− ρ

=−

c.)

22 vvpp −

Results

4 In

4/1 N12100Fx =

4/2 After writing the Be we get the result:

12 vv =

4/3 N510F = , direction

4/4 N109F =

4/5 N57F =

4/6 N14G =

4/7 The integral momen the upper end of the

AvAG 2 ⋅ρ=⋅⋅ρ=

with v, the speed at According to the Be

hg2vv 20 ⋅⋅−=

s/m55.4v 0 =

4/16 )vv(vA)pp(A 1222212 −⋅ρ=− mm5.6h =

( )232CBCB vv2 pwhere

p hg −

ρ =∆

ρ ∆

+⋅ − − (Borda-Carnot-loss)

Hydraulics

424

2 v

lam d const

d const

64 d L

16 d q

2 p =

π ρ

=∆

5d const

d const 316.0

≈

4/17 The Bernoulli-equation between point 1 and 2 (point 2 is situated at the outflow end of the pipe):

ρ ⋅⋅ρ+

+= ρ

+ hgp

2 vp

2 v 0

2 21

2 1 because the area of cross section of the pipe is constant,

ρ =const, 12 vv =

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

2 v

turb d L

16 d q

2 p

π ρ

=∆

Results 35

4/9 a1

a arcsin

− =α

4/10 N52G =

4/11 m1h =

4/12 W302)uv(vAuP =−⋅⋅⋅ρ⋅=

4/13 N280FF yx ==

4/14 )vv(vpp 1'111'11 −⋅ρ=−

Pa123pp

)vv( 2

2 '1

2 2

2 2'1

=−

− ρ

=−

4/15 21 1

2121 v2 hg)(pp

ρ −⋅⋅ρ−ρ=−

s/m51q

)vv(vpp 3

121121

−⋅ρ=−

Results

0 2 21

2 1 p

2 vp

2 v

+ ρ

+= ρ

12 vv = and 0v 3 = .

4/18 m8.0h =

4/19 m1h =

4/20 2 q

q platewith

platewithout V =

5/1 N5.7 dy dv

AF =⋅µ⋅=

5/2 Pa72400pp 01 =−

5/3 Pa1500pp 01 =−

5/4 d

const

4 d

dq Re

2 v = ν

π ⋅

5/14 a) m2H =

b) Pa40000pp 01 =−

with 02.0=λ , s/m827.0 602.0

m05.0 m200

s/m81.92m3 v

pipe = +

⋅⋅ =

024.0102.3 103.

05.827. 4 6 =λ⇒⋅=⋅

⋅ −

e next iteration step, s/m755.0v pipe = , and the iteration can be finished.

h m12h = , the necessary velocity at the confuser’s outlet must be:

s/m3.15hg2 =⋅⋅

mm11mm50 s/m3.15 s/m755.0

=⋅

s/m1047 33−⋅

rst the velocity without friction loss can be calculated: s/m7.7m3g2videal =⋅⋅= ,

m105.6 s/m7.7

s/m 3600 18

−⋅==

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

mm52dm102.21As/m36.2 1402.0

m05. m14

s/m81.92m 24 2

=⇒⋅=⇒= ++

⋅⋅ −

018.01045.9 103.1

052.036. 4 6 =λ⇒⋅=⋅

⋅ −

Reynolds number we consider that the pipe is hydraulically smooth) ext iteration step with 018.0=λ and mm52d = we get the new diameter of

mm2.51 . The iteration can be finished.

b) If the dike is higher, the pressure in the pipe can reach the pressure of saturated steam. In this case, the water column is going to break. The lowest pressure appears after the valve, at

oint of the dike. From the equation

 

 

ζ+λ +

+ ρ

− d

LL 1v

2 h 212max

culated.

5/15 W77P = The power can be decreased by sinking the oil viscosity and by increasing the gap.

5/16 The resultant height loss is m3m12m15h res =−= .

  

  

ζ+λ=⋅ 2 d L

2 v

hg 2

res

the upper right p

⋅⋅ρ−= gpp 0min

maxh can be cal

Results 37

5/5 V v

2 v

lam qconst

A dq

64 d L

A q

2 p ⋅=

ν⋅

ρ =∆

75.1 V

4 v 2

2 v

turb qconst

A dq

316.0 d L

A q

2 p ⋅=

ν⋅

ρ =∆

5/6 Considered that the flow will be laminar and using the formula Re/64=λ , 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   

  

λ+=⋅ d L

1 2 v

hg 2

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

230033Re <= , so the flow is really laminar.

5/10 s/m23.0q 3v =

5/11 s/m0817.0q 3v =

5/12 Pa10900pp 01 =−

5/13 Pa28500pp 01 =−

Results

Starting

1 0

Re =

After th

To reac

v 2 =

d 2 =

.1q v =

5/17 a) At fi

and A

So the p larger d

3 v =

2 Re =

(At this In the n

mpressible Flows

s/kg274.0200m/kg 3 =⋅

212 ρ

b)   

  

 

  

 −

ρ−κ κ

= κ −κ 1

1 2 p

p 1

2 v

Results 40

6/9 C56t A °=

6/10 77.0Ma,K262T 22 ==

6/11 a) mm138d =

b) N108.9vAF 32222 ⋅=⋅⋅ρ=



c)  

 

+κ −

ρ−κ κ

= 1

2 1

p 1

2 v

1 2

6/7 59.0Ma =

6/8 s/m80v =

Results

6 Co

6/1 s/m260v 2 =

6/2 37.1m10vAq 23222m ⋅=ρ= −

6/3 a) C42t static2 °−=

b) C20t total2 °+=

6/4 833.0 1

2 T T

= +κ

/kg018.0Avq

m/kg9.2 T T

vs/m316a T T

s/m346TRa

*** m

3 1

1 1

* *

* 1

* *

=⋅ρ⋅=

=ρ 

  

 =ρ

===

=⋅⋅κ=

−κ

6/5 s/kg25.0vAq 222m =⋅ρ⋅=

mm3.17dd

m1034.2 v

q A

* min

24 **

⋅= ρ⋅

= −

6/6 a) ( )pp2v −=