exercicios p2

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Problem 4.56

Problem 4.64

Problem 4.65

Problem 4.68is deflected by a hinged plate of length 2 m supported by a spring with spring constant k = 1 N/m and uncompressed length x0 = 1 m. Find and plot the deflection angle as a function of jet speed V. What jet speed has a deflection of 10?2

Given: Data on flow and system geometry Find: Deflection angle as a function of speed; jet speed for 10o deflection

Solution The given data are kg m3 2

999

A

0.005 m

L

2m

k

1

N m

x0

1m

Governing equation:

Momentum

Fspring

V sin

VA

But

Fspring

kx

k x0

L sin

Hence

k x0

L sin

V A sin

2

Solving for

asin kL

k x0 AV2

For the speed at which = 10o, solve

V

k x0

L sin

A sin

V

N ( 1 2 sin ( 10) ) m kg m m 2 kg 2 999 0.005 m sin ( 10) N s 3 m 1

V

0.867

m s

The deflection is plotted in the corresponding Excel workbook, where the above velocity is obtained using Goal Seek

Problem 4.68 (In Excel)A free jet of water with constant cross-section area 0.005 m2 is deflected by a hinged plate of length 2 m supported by a spring with spring constant k = 1 N/m and uncompressed length x 0 = 1 m. Find and plot the deflection angle as a function of jet speed V . What jet speed has a deflection of 10? Given: Geometry of system Find: Speed for angle to be 10o; plot angle versus speed Solution k x0 kL= xo = L = k = A = 999 1 2 1 0.005 kg/m3 m m N/m m2

The equation for is

asin

AV

2

To find when = 10o, use Goal Seek V (m/s) 0.867o ( )

10

V (m/s) 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4 1.5

(o ) 30.0 29.2 27.0 24.1 20.9 17.9 15.3 13.0 11.1 9.52 8.22 7.14 6.25 5.50 4.87 4.33

Deflection Angle vs Jet Speed35 30

(deg)

25 20 15 10 5 0 0 2 4 6 8 10 12 14 16 18 20

V (m/s)

Problem 4.69

Problem 4.70

Problem 4.71

Problem 4.72

Problem 4.73

Problem 4.74

Problem 4.75

Problem 4.76

Problem 4.77

Problem 4.78

Problem 4.79

Problem 4.80

Problem 4.81The horizontal velocity in the wake behind an object in an air stream of velocity U is given by u ( r) U 1 cos r 22

r

1

u ( r)

U

r

1

where r is the non-dimensional radial coordinate, measured perpendicular to the flow. Find an expression for the drag on the object. Given: Data on wake behind object

Find: An expression for the drag

Solution Governing equation:

Momentum

Applying this to the horizontal motion1

F

U

1 U0 1

2

u ( r)

2

r u ( r) dr

F

U

2

20

r u ( r) dr

2

1

F

U

2

1

20

r 1

cos

r 2

2 2

dr

1

F

U

2

1

20

r

2 r cos

r 2

2

r cos

r 2

4

dr

Integrating and using the limits

F

U

2

1

3 8

22

F

5 8

2

U

2

Problem 4.82

Problem 4.83

Problem 4.86

Problem *4.91

Problem 4.107

Problem 4.108

Problem 4.109A jet boat takes in water at a constant volumetric rate Q through side vents and ejects it at a high jet speed Vj at the rear. A variable-area exit orifice controls the jet speed. The drag on the boat is given by Fdrag = kV2, where V speed V. If a jet speed Vj = 25 m/s produces a boat speed of 10 m/s, what jet speed will be required to double the boat speed?

Given: Data on jet boat Find: Formula for boat speed; jet speed to double boat speed Solution

CV in boat coordinates

Governing equation:

Momentum

Applying the horizontal component of momentum

Fdrag

V

Q

Vj

Q

Hence

kV

2

Q Vj

QV

kV

2

QV

Q Vj

0

Solving for V

V

Q 2k

Q 2k

2

Q Vj k

Let

Q 2k

V

2

2

Vj

We can use given data at V = 10 m/s to find

V

10

m s

Vj

25

m s

10

m s

2

2 25

m s

2

50

10

2

100

20

2

10 m 3 s

Hence

V

10 3

100 9

20 V 3 j

For V = 20 m/s

20

10 3

100 9

20 V 3 j

100 9

20 V 3 j

70 3

Vj

80

m s

Problem 4.110

Problem 4.112

Problem 4.113

Problem 4.114

Problem *4.168

Problem *4.171Water flows in a uniform flow out of the 5 mm slots of the rotating spray system as shown. The flow rate is 15 kg/s. Find the torque required to hold the system stationary, and the steady-state speed of rotation after it is released. Given: Data on rotating spray system

Solution The given data is 999 kg m D3

mflow

15

kg s

0.015 m

ro

0.25 m

ri

0.05 m

0.005 m

Governing equation: Rotating CV

For no rotation ( = 0) this equation reduces to a single scalar equation

Tshaft

r Vxyz

Vxyz dA

ro

ro

or

Tshaft

2ri

rV

V dr

2

V

2 ri

r dr

V

2

ro

2

ri

2

where V is the exit velocity with respect to the CVmflow

V

2

ro

ri

mflow

2

Hence

Tshaft

2

ro

ri

ro

2

ri

2

Tshaft

mflow 4

2

ro ro

ri ri

Tshaft

1 4

kg 15 s

2

m 999 kg

3

1 0.005 m

( 0.25 ( 0.25

0.05) 0.05)

Tshaft

16.9 N m

For the steady rotation speed the equation becomes

r

2

Vxyz

dV

r Vxyz

Vxyz dA

The volume integral term

r

2

Vxyz

dV must be evaluated for the CV.

The velocity in the CV varies with r. This variation can be found from mass conservation

For an infinitesmal CV of length dr and cross-section A at radial position r, if the flow in is Q, the flow out is Q + dQ, and the loss through the slot is Vdr. Hence mass conservation leads to (Q dQ) V dr Q 0

dQ

V

dr

Q ( r)

V

r

const

At the inlet (r = ri)

Q

Qi

mflow 2

Hence

Q

Qi

V

ri

r

mflow 2 2

mflow ro ri

ri

r

Q

mflow 2

1

ri ro

r ri

mflow 2

ro r ro ri

and along each rotor the water speed is v ( r)

Q A

mflow 2 A

ro r ro ri

Hence the term -

r

2

Vxyz

dV becomes

ro

ro

r

2

Vxyz

dV

4

Ari

r v ( r) dr

4ri

r

mflow 2

ro r dr ro ri

or

ro

r

2

Vxyz

dV

2 mflowri

ro r r dr ro ri

mflow

ro

3

ri 2 ri 3 ro 3 ro ri

2

Recall that

r Vxyz

Vxyz dA

V

2

ro

2

ri

2

Hence equation

r

2

Vxyz

dV

r Vxyz

Vxyz dA becomes

mflow

ro

3

ri 2 ri 3 ro 3 ro ri

2

V

2

ro

2

ri

2

Solving for

3 ro

ri3

V ri

2

ro 2 ri

2

ri

2

mflow ro

2

461 rpm

3 ro

Problem *4.172If the same flow rate in the rotating spray system of Problem 4.171 is not uniform but instead varies linearly from a maximum at the outer radius to zero at a point 50 mm from the axis, find the torque required to hold it stationary, and the steady-state speed of rotation. Given: Data on rotating spray system

Solution The given data is 999 kg m D3

mflow

15

kg s

0.015 m

ro

0.25 m

ri

0.05 m

0.005 m

Governing equation: Rotating CV

For no rotation ( = 0) this equation reduces to a single scalar equation

Tshaft

r Vxyz

Vxyz dA

ro

or

Tshaft

2ri

rV

V dr

where V is the exit velocity with respect to the CV. We need to find V(r). To do this we use ma conservation, and the fact that the distribution is linear

V ( r)

Vmax

ri ro ri mflow

r

and

2

1 V r 2 max o

ri

so

V ( r)

mflow

r ro

ri ri2

ro

Hence

Tshaft

2ri

r V dr

2

2

mflow

2

ro

r r rori

ri ri2

2

dr

Tshaft

mflow 6

2

ri ro

3 ro ri

Tshaft

1 6

15

kg s

2

m 999 kg

3

1 0.005 m

( 0.05 3 0.25) ( 0.25 0.05)

Tshaft

30 N m

For the steady rotation speed the equation b