from fox-suggested extra problems for midterm-2

8/7/2019 From Fox-Suggested Extra Problems for Midterm-2

1/18

Problem 4.66 [Difficulty: 2]

Given: Nozzle hitting stationary cart

Find: Value of M to hold stationary; plot M versu

Solution:

Basic equation: Momentum flux in x direction for the tank

Assumptions: 1) Steady flow 2) Incompressible flow 3) Atmospheric pressure throughout 4) Uniform flow 5) Exit velocity is V

Hence Rx M g= V V A( ) V cos ( ) V A( )+= V2

A cos ( ) 1( )= M V

2 A

g1 cos ( )( )=

When = 40o Ms2

9.81 m1000

kg

m3

10m

s

2

0.1 m2

1 cos 40 deg( )( )= M 238 kg=

0 45 90 135 180

1000

2000

3000

Angle (deg)

M(kg)

This graph can be plotted in Excel


2/18


Given: Large tank with nozzle and wire

Find: Tension in wire; plot for range of water depths

Solution:


Assumptions: 1) Steady flow 2) Incompressible flow 3) Atmospheric pressure throughout 4) Uniform flow

Hence Rx T= V V A( )= V2

A= 2 g y( ) d

2

4= T

1

2 g y d

2= T is linear with y!

When y = 0.9 m T

21000

kg

m3

9.81m

s2

0.9 m 0.015 m( )2

N s2

kg m= T 3.12N=

0 0.3 0.6 0.9

1

2

3

4

y (m)

T(N)

This graph can be plotted in Excel


3/18


Given: Water tank attached to mass

Find: Whether tank starts moving; Mass to just hold in place

Solution:


Assumptions: 1) Steady flow 2) Incompressible flow 3) Atmospheric pressure at exit 4) Uniform flow

Hence Rx V cos ( ) V A( )= V2

D

2

4 cos ( )=

We need to find V. We could use the Bernoulli equation, but here it is known that V 2 g h= where h = 2 m is theheight of fluid in the tank

V 2 9.81m

s2

2 m= V 6.26m

s=

Hence Rx 1000kg

m3

6.26m

s

2

40.05 m( )2 cos 60 deg( )= Rx 38.5 N=

This force is equal to the tension T in the wire T Rx= T 38.5 N=

For the block, the maximum friction force a mass of M = 10 kg can generate is Fmax M g = where is static friction

Fmax 10 kg 9.81m

s2

0.55N s

2

kg m= Fmax 54.0N=

Hence the tension T created by the water jet is less than the maximum friction Fmax; the tank is at rest

The mass that is just sufficient is given by M g Rx=

MRx

g = M 38.5 N

1

9.81

s2

m

1

0.55

kg m

N s2

= M 7.14kg=


4/18


Rx

y

xCS

Given: Water flow through elbow

Find: Force to hold elbow

Solution:

Basic equation: Momentum flux in x direction for the elbow

Assumptions: 1) Steady flow 2) Incompressible flow 3) Atmospheric pressure at exit 4) Uniform flow

Hence Rx p1g A1+ V1 V1 A1( ) V2 V2 A2( )= Rx p1g A1 V12

A1 V22

A2+=

From continuity V2 A2 V1 A1= so V2 V1

A1

A2

= V2 10ft

s

4

1= V2 40

ft

s=

Hence Rx 15lbf

in2

4 in2

1.94slug

ft3

10ft

s

2

4 in2

40ft

s

2

1 in2

+

1 ft

12 in

2

lbf s

2

slug ft= Rx 86.9 lbf=

The force is to the left: It is needed to hold the elbow on against the high pressure, plus it generates the large change in x momentum


5/18


Given: Water flow through nozzle

Find: Force to hold nozzle

Solution:


Assumptions: 1) Steady flow 2) Incompressible flow 3) Uniform flow

Hence Rx p1g A1+ p2g A2+ V1 V1 A1( ) V2 cos ( ) V2 A2( )+= Rx p1g A1 V22

A2 cos ( ) V12

A1+=

From continuity V2 A2 V1 A1= so

V2 V1

A1

A2

= V1

D1

D2

2

= V2 1.5m

s

30

15

2

= V2 6m

s=

Hence Rx 15 103

N

m2

0.3 m( )

2

4 1000

kg

m3

6m

s

2 0.15 m( )

2

4 cos 30 deg( ) 1.5

m

s

2 .3 m( )

2

4

N s2

kg+=

Rx 668 N= The joint is in tension: It is needed to hold the elbow on against the high pressure, plus it generates the largechange in x momentum


6/18


Rx

y

x

CS

Given: Water flow through orifice plate

Find: Force to hold plate

Solution:


Assumptions: 1) Steady flow 2) Incompressible flow 3) Uniform flow

Hence Rx p1g A1+ p2g A2 V1 V1 A1( ) V2 V2 A2( )+= Rx p1g A1 V22

A2 V12

A1+=

From continuity Q V1 A1= V2 A2=

so V1Q

A1

= 20ft

3

s

4

1

3ft

2

= 229ft

s= and V2 V1

A1

A2

= V1D

d

2

= 229ft

s

4

1.5

2

= 1628ft

s=

NOTE: problem has an error: Flow rate should be 2 ft3/s not 20 ft3/s! We will provide answers to both

Hence Rx 200lbf

in2

4 in( )

2

4 1.94

slug

ft3

1628ft

s

2 1.5 in( )

2

4 229

ft

s

2 4 in( )

2

4

1 ft

12 in

2

lbf s

2

slug ft+=

Rx 51707 lbf=

With more realistic velocities

Hence Rx 200

lbf

in2

4 in( )2

4 1.94slug

ft3 163

ft

s

2 1.5 in( )

2

4 22.9ft

s

2 4 in( )

2

4

1 ft

12 in

2

lbf s2

slug ft+=

Rx 1970 lbf=


7/18


8/18

Problem *4.91Problem *4.110 [Difficulty: 4]


9/18

Problem *4.111 [Difficulty: 4]

CS

Given: Air jet striking disk

Find: Manometer deflection; Force to hold disk

Solution:Basic equations: Hydrostatic pressure, Bernoulli, and momentum flux in x direction

p

V2

2+ g z+ constant=

Assumptions: 1) Steady flow 2) Incompressible 3) No friction 4) Flow along streamline 5) Uniform flow 6) Horizontal flow (gx = 0)

Applying Bernoulli between jet exit and stagnation point

p

air

V2

2+

p0

air

0+= p0 p1

2air V

2=

But from hydrostatics p0 p SG g h= so h

1

2air V

2

SG g=

air V2

2 SG g=

h 0.002377slug

ft3

225ft

s

2

1

2 1.75

ft3

1.94 slug

s2

32.2 ft= h 0.55 ft= h 6.6 in=

For x momentum Rx V air A V( )= air V2

D

2

4=

Rx 0.002377slug

ft3

225ft

s

2

0.5

12 ft

2

4

lbf s2

slug ft= Rx 0.164 lbf=

The force of the jet on the plate is then F Rx= F 0.164 lbf=


10/18

Problem *4.114 [Difficulty: 3]

CS

Given: Water jet striking disk

Find: Expression for speed of jet as function of height; Height for stationary disk

Solution:

Basic equations: Bernoulli; Momentum flux in z direction

p

V2

2+ g z+ constant=

Assumptions: 1) Steady flow 2) Incompressible flow 3) Atmospheric pressure in jet 4) Uniform flow

The Bernoulli equation becomesV0

2

2g 0+

V2

2g h+= V

2V0

22 g h= V V0

22 g h=

Hence M g w1 w1 A1( )= V2

A=

But from continuity V0

A0

V

A= so

V A

V0 A0=

Hence we get M g V V A= V0 A0 V02

2 g h=

Solving for h h1

2 gV0

2 M g

V0 A0

2

=

h1

2

s2

9.81 m 10

m

s

2

2 kg9.81 m

s2

m

3

1000 kg

s

10 m

4

25

1000 m

2

2

=

h 4.28 m=


11/18


12/18


CS (movesat speed U)

y

x

RxRy

Given: Water jet striking moving vane

Find: Force needed to hold vane to speed U = 10 m/s

Solution:

Basic equations: Momentum flux in x and y directions

Assumptions: 1) Steady flow 2) Incompressible flow 3) Atmospheric pressure in jet 4) Uniform flow 5) Jet relative velocity is

constant

Then Rx u1 V1 A1( ) u2 V2 A2( )+= V U( ) V U( ) A[ ] V U( ) cos ( ) V U( ) A[ ]+=

Rx V U( )2

A cos ( ) 1( )=

Using given data

Rx 1000kg

m3

30 10( )m

s

2

0.004 m2

cos 120 deg( ) 1( )N s

2

kg m= Rx 2400 N=

Then Ry v1 V1 A1( ) v2 V2 A2( )+= 0 V U( ) sin ( ) V U( ) A[ ]+=

Ry V U( )2

A sin ( )= Ry 1000kg

m3

30 10( )m

s

2

0.004 m2

sin 120 deg( )N s

2

kg m= Ry 1386N=

Hence the force required is 2400 N to the left and 1390 N upwards to maintain motion at 10 m/s


13/18



14/18


CS (moves to

left at speed Vc)

y

x

Rx

Vj + Vc

Vj + Vc

t

R

Given: Water jet striking moving cone

Find: Thickness of jet sheet; Force needed to move cone

Solution:

Basic equations: Mass conservation; Momentum flux in x direction

Assumptions: 1) Steady flow 2) Incompressible flow 3) Atmospheric pressure in jet 4) Uniform flow 5) Jet relative velocity is

constant

Then V1 A1 V2 A2+ 0= Vj Vc+( ) Dj

2

4 Vj Vc+( ) 2 R t+ 0= (Refer to sketch)

Hence t

Dj2

8 R=t

1

84 in( )2

1

9 in=t 0.222 in=

Using relative velocities, x momentum is

Rx u1 V1 A1( ) u2 V2 A2( )+= Vj Vc+( ) Vj Vc+( ) Aj Vj Vc+( ) cos ( ) Vj Vc+( ) Aj+=

Rx Vj Vc+( )2

Aj cos ( ) 1( )=

Using given data

Rx 1.94slug

ft3

100 45+( )ft

s

2

4

12 ft

2

4 cos 60 deg( ) 1( )

lbf s2

slug ft= Rx 1780 lbf=

Hence the force is 1780 lbf to the left; the upwards equals the weight


15/18



16/18



17/18


Given: Data on system

Find: Jet speed to stop cart after 1 s; plot speed & position; maximum x; time to return to origin

Solution: Apply x momentum

Assumptions: 1) All changes wrt CV 2) Incompressible flow 3) Atmospheric pressure in jet 4) Uniform flow 5) Constant jet area

The given data is 999kg

m3

= M 100 kg= A 0.01 m2

= U0 5m

s=

Then arf M u1 V U+( ) A[ ] u2 m2+ u3 m3+=

where arfdU

dt= u1 V U+( )= and u2 u3= 0=

HencedU

dt M V U+( )

2 A= or

dU

dt

V U+( )2

A

M= which leads to

d V U+( )

V U+( )2

A

Mdt

=

Integrating and using the IC U= U0 at t= 0 U VV U0+

1 A V U0+( )

Mt+

+=

To find the jet speed Vto stop the cart after 1 s, solve the above equation for V, with U= 0 and t= 1 s. (The equation becomes a

quadratic in V). Instead we use Excel's Goal Seekin the associated workbook

From Excel V 5m

s=

For the position x we need to integratedx

dtU= V

V U0+

1 A V U0+( )

M

t+

+=

The result is x V tM

Aln 1

A V U0+( )M

t+

+=

This equation (or the one for Uwith U= 0) can be easily used to find the maximum value ofx by differentiating, as well as the time for x

to be zero again. Instead we use Excel's Goal Seekand Solverin the associated workbook

From Excel xmax 1.93 m= t x 0=( ) 2.51 s=

The complete set of equations is U VV U0+

1 A V U0+( )

Mt+

+= x V tM

Aln 1

A V U0+( )M

t+

+=


18/18

The plots are presented in the Excel workbook:

(s) x (m) U (m/s) To find V for U = 0 in 1 s, use Goal Seek

0.0 0.00 5.00

0.2 0.82 3.33 (s) U (m/s) V (m/s)

0.4 1.36 2.14 1.0 0.00 5.00

0.6 1.70 1.25

0.8 1.88 0.56 To find the maximum x , use Solver

1.0 1.93 0.00

1.2 1.88 -0.45 s x m

1.4 1.75 -0.83 1.0 1.93

1.6 1.56 -1.15

1.8 1.30 -1.43 To find the time at which x = 0 use Goal Seek

2.0 0.99 -1.67

2.2 0.63 -1.88 (s) x (m)

2.4 0.24 -2.06 2.51 0.00

2.6 -0.19 -2.22

2.8 -0.65 -2.37

3.0 -1.14 -2.50

Cart Speed U vs Time

-3

-2

-1

0

1

2

3

4

5

6

0.0 0.5 1.0 1.5 2.0 2.5 3.0

t (s)

U(m/

s)

Cart Position x vs Time

-1.5

-1.0

-0.5

0.0

0.5

1.0

1.5

2.0

2.5

0.0 0.5 1.0 1.5 2.0 2.5 3.0

t (s)

x(m)

from fox-suggested extra problems for midterm-2

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