1

Part 2: Tutorial Sheet 1 (SOLUTIONS)

(Q1)

SOLUTION (a) Accelerating upward From the lecture note:

� �> @ azx pzgaxap ��U�U� Since ax = 0 and az = 5 m/s2 The above equation becomes

� �> @ az pzgap ��U�

The total force on face AB = pressure at centre of pressure (Pc) x area of AB

> @� �101300)1)(g5(8.01000 ���uu� u2 u1.5 =339.44 kN

(b) Accelerating downward ax = 0 and az = -5 m/s2 The total force on face AB = > @� �101300)1)(g5(8.01000 ����uu� u2 u1.5

=315.45 kN

h=2m

A

B

Pa

Pc

1 m

2

(Q2)

(a) � �> @ azx pzgaxap ��U�U�

Since ax = 4 m/s2 and az = 0 m/s2

� �> @ ax pzgxap �U�U�

At B, x = 0, z =-0.3

� �> @ aB p)3.0(g0p ��U��

� � aB p)3.0(gp �U (1) At C, x = 0.5, z = -0.3

� �> @ ac p)3.0(g)5.0)(4(p ��U�U�

� � ac p)3.0(g)5.0)(4(p �U�U� (2) (1) – (2) gives

2CB m/kN2)5)(4)(1000()5.0)(4(pp U �

0.3

0.5

ax= 4 m/s2

A

B C

x

z

3

For point C to be atmospheric, g

a500300tan x� � T

Therefore ax = -5.886 m/s2

At B, x = 0, z =-0.3

� �> @ aB p)3.0(g0p ��U��

� � aB p)3.0(gp �U (3) At C, x = 0.5, z = -0.3

� �> @ ac p)3.0(g)5.0)(4(p ��U�U�

� � ac p)3.0(g)5.0)(886.5(p �U�U� (4) (3) - (4) gives

2CB m/kN943.2)5.0)(886.5(pp U �

0.3

0.5

A

B C

x

z

Isobar (atmospheric)

4

(Q3) Note: x-axis is parallel to the ground (a)

gaatan

z

x

�� T

ax = 4 cos 15 az = 4 sin 15

356.08.915sin4

15cos4tan � �

� T

Therefore, T =-19.62o

From geometry, it can be shown that the slope at which the water surface makes witht ehorizontal ground when it is about to spill is

T = tan -1(2.0/2.5) -15o = -23.65o

Since the slope 19.62o made by the surface is less than -23.65o, therefore the water will not spill when the acceleration is 4 m2/s. (b) Let the acceleration at which the water will just spill = a This occurs when T = -23.65o

8.915sina15cosatan�

� T

4 m/s2

15o

2.0 m

2.5 m

1.0 m

Water surface on horizontal ground

z

x T

5

or

)65.23(tan8.915sina

15cosatan � �

� T = -0.438

Therefore

acos15�0.438uausin15=0.438u9.8

a = 5.03 m/s2

6

(Q4) Let zo be the minimum reference level of the broken parabolic curve, and zA, zB and zC the liquid level above the base. From your lecture note:

o

22

o p2r)zz(gp �UZ��U�

At air-water interface, p=po

o

22

oo p2r)zz(gp �UZ��U�

o

22 z

g2rz �Z (1)

At A, r = 0.3m; at B, r = 0.1m and at C, r =0.5 m Substitute into equation 1, we get

o

22

o

221

A z81.9x260

50x2)3.0(z

g2rz �

¸¹·

¨©§ S

�Z

(2)

400 400 300

300

A B C

a

a

Z

zo

z

r

7

o

22

o

222

B z81.9x260

50x2)1.0(z

g2rz �

¸¹·

¨©§ S

�Z

(3)

o

22

o

223

C z81.9x260

50x2)5.0(z

g2rz �

¸¹·

¨©§ S

�Z

(4)

Before the rotation, the original zA + zB + zC = 0.9. Since there is no spillage, the volume of fluid remains the same

81.9x260

50x2)3.0(2

2 ¸¹·

¨©§ S

+81.9x260

50x2)1.0(2

2 ¸¹·

¨©§ S

+81.9x260

50x2)5.0(2

2 ¸¹·

¨©§ S

+3zo=0.9

zo= 0.137

Substitute into equations (2), (3) and (4), we get

zA= 0.263, zB= 0.151, zC= 0.486

pC - pA=Ug (zC-zA) = (1000)(9.81)(0.486-0.263) = 2.188 kN/m2

8

(Q5)

From your lecture, the equation of pressure distribution in the liquid is given by

a

22 p

2rgzp �UZ�U�

Substituting the following boundary conditions into the above equation:

(a) at r =D, z =ho: p = pa , we get 222

oa cD21ghp �:U�U� (1)

(b) at r =D+d, z =ho+h: p = pa , we get � � � � 222

oa cdD21hhgp �:�U��U� (2)

Substract equation (2) from equation (1), and after some simple manipulation, we get

h ¸¸¹

·¨¨©

§¸¹·

¨©§�

:

222

Dd

Dd2

g2D

g

h

D d

ho

:

r

z

9

(Q6)

Consider the control volume abcd as shown above. Since the atmospheric pressure acts all over the control surface normal to it, its net effect is zero. The fluid entering and leaving at p1 and p2 exerts pressure

� � � � � � 2a021a01 ppppressuregaugeppp � � over areas A1 and A2, respectively. Use momentum equation

� �12 uumF � ¦ � (A) Consider x-direction: Force on the fluid by solid boundary in the x-direction Fx Body force in x-direction = 0 Surface force acting on CV in x-direction:

2211 ApAp �

Substitute the above forces into momentum equation (A), we get

Nozzle P01 A1 u1

a

P02 A2 u2

b

c

d

(1)

(2)

U

C.V.

x

10

x2211 FApAp �� = � �12 uum �� (1)

But 0Ap 22 (since the gauge pressure of atmopsheric pressure is zero)

Therefore equation (1) becomes

xF = � �1211 uumAp ��� � (2)

But 2211 uAuAm U U � which gives 12

12 u

AAu

Substitute the above into equation 2, therefore

xF = ¸̧¹

·¨̈©

§�U�� 11

2

11111 uu

AAuAAp

Force on the nozzle by the fluid, F = -Fx , Therefore

F = ¸̧¹

·¨̈©

§�U� 1

AAuAAp

2

121111 (3)

For inviscid flow, apply Bernoulli’s Equation between (1) and (2)

222021

2101 gzu

21pgzu

21p U�U� U�U� (4)

But p02=atmospheric pressure, and Ugz1=Ugz2

� � 22

210201 u

21u

21pp U U��

or

� � � �21

22atm01 uu

21pp �U �

� � ¸̧¹

·¨̈©

§�U �U 2

1212

2

21

121

22111 uu

AAA

21uuA

21Ap

or � � ¸̧¹

·¨̈©

§�U �U 1

AAuA

21uuA

21Ap 2

2

212

1121

22111 (5)

11

Substituting equation (5) into equation (3), we get

¸̧¹

·¨̈©

§�U�

°¿

°¾½

°̄

°®­

�¸̧¹

·¨̈©

§U 1

AAuA1

AAuA

21F

2

1211

2

2

1211

2

2

1211 1

AAuA

21F

¿¾½

¯®­

�U

12

(Q7) From continuity

332211 AVAVAV �

or ¸̧¹

·¨̈©

§ S�¸̧

¹

·¨̈©

§ S ¸̧

¹

·¨̈©

§ S4dV

4dV

4dV

23

3

22

2

21

1

Substitute all known values into the above equation, we get

s/m15.23V3 Now consider the control volume abcde as shown above. Since location 2 and 3 are opened to the atmosphere, therefore

p2 (gauge) = p3 (gauge) =0. Net surface force acting on it in

x-direction:

o3322 45cosApAp � (1)

0.5

0.2

0.18

V1=5

V2=10

45o

P1=25 kPa gauge

x

y

V3

C.V.

P2 A2 P3, A3 a

b c

d

e

Fy

Fx

g

13

y-direction: o

3311 45sinApAp �� (2)

The force exerted on the fluid in the control volume in x-direction: Fx (3)

y-direction: Fy (4)

The body force of the fluid in the control volume in

x-direction: 0 (5) y-direction: - UgVinside the bend (6)

The net momentum rate leaving the control volume in x-direction:

� � 0Vm45cosVm 22o

33 �� �� (7)

y-direction:

� �11o

33 Vm()45sinVm �� ��� (8)

Applying momentum equation

x-direction:

xo

3322 F45cosApAp �� = � �22o

33 Vm45cosVm �� � (9)

y-direction:

o3311 45sinApAp �� +Fy - Ug.1= 11

o33 Vm45sinVm �� �� (10)

Min Mout

Mout Min

14

Substituting all known quantities into equation (9) and (10) keeping in mind that p2 and p3 are zero, we get

N5.9360Fx

N27.7718Fy

Therefore, the reaction force on the elbow

N5.9360FR xx � �

N27.7718FR yy � �

15

(Q8) To simply the problem, it is better to travel with the aeroplane as it makes the problem steady. Note that location 1 and location 4 are so far away from the propellor such the pressure is atmospheric.

55.5560x60

1000x200V1 m/s s/m33.8360x60

1000x300V4

Apply momentum equation to control volume ABCD. The force acting on the fluid is

� �12 uumF � ¦ � (A) Consider x-direction: Force on the fluid (air) by solid boundary (propellor) in the x-direction Fx Body force in x-direction = 0 Surface force acting on CV in x-direction = 0 (since atmospheric pressure is acting all along the control surface)

The net momentum rate leaving the control volume in the

� �14 VVQ �U

Substitute into equation (A), we get the force exerted on air by the propeller

� �14x VVQF �U

2 3 1 4

Fx

V1 V4 x

A B

C D

16

From your lecture note:

� �s/m44.69

233.8355.55

2VV

V 142

�

�

and s/m86.34044.69x)5.2(4

Q 32 S

Therefore Fx = 1.283 x 340.86 (83.33-55.55)= 12.15 kN

(a) Thrust on propeller =12.15 kN

(b) H.P. ouput = 12.15 x 55.55 =674.93 kW

(c) Propeller efficiency=V1/V2= 55.55/69.44 = 0.80 = 80%

(d) Velocity through the plane of propeller (V2) =69.44m/s

(e) Pressure difference across the propeller disc

� � � �1423 VVAVApp �U �

� � � �55.5533.8344.69x283.1pp 23 � � = 2.48 kN/m2

17

(Q9) Equation of the sloping part of the wake profile is:

y)3/c(Uu

3/cU5.0 oo

��

� � oo Uy)3/c(

3/cU5.0

u ��

Consider the control volume ABCD

Mass flow rate per unit depth OUT of control volume (BC)

dycy31U5.0U2

3c

0oo³ »

¼

º«¬

ª¸¹·

¨©§ ��U

oU6c5

U (1)

Mass flow rate per unit depth IN of control volume (AD)

c= 60

15

Uo=100 mm/s

Uo

Umax -Uo =0.5Uo c/3

c/3

oscillating

oscillating

x

y

y u

u-Uo

Control volume

A B

C D

L

18

oLUU (2)

From continuity (1) = (2)

oo U6c5LU U U

L=6c5

Momentum OUT of the control volume > @dyu23

c

0

2³U

dycy1U

232

23

c

0o³ »

¼

º«¬

ª¸¹·

¨©§ �U

dycy

cy21U

492

3c

02

22o³

»»¼

º

««¬

ª¸̧¹

·¨̈©

§��U

2oU

18c19¸¹·

¨©§U (3)

Momentum in of the control volume > @dyu26

c5.2

0

2³U

dyU26

c5.2

0

2o³U

2oU

6c5U (4)

19

Consider x-direction: Net force in the x-direction is

� �x12x uumF � ¦ �

Force on the fluid by oscillating airfoil in the x-direction Fx Body force in x-direction = 0

Surface force acting on CV in x-direction = 0 (since atmospheric pressure is acting all along the control surface)

Fx = equation (3) – Equation (4)

2o

2o U

6c5U

18c19

U�¸¹·

¨©§U

184cU2

oU�

Therefore it is thrust per unit length 184cU2

oU

By substituting all the known value into the above equation, we get

Total Thrust N133.0184)06.0()1.0)(1000( 2

20

(Q10)

Since water is discharged equally through each nozzle, therefore

(a) when rotating freely at an angular velocity of Z

Q1 = Q2 = 0.5 x10-4 m3/s.

Vr1 = Vr2 = 4/)005.0(

10x5.02

4

S

�

=2.54 m/s

Since Vt1 and Vt2 are in OPPOSITE direction.

Torque on fluid > @1t12t2 VrVrQ �U

Torque on Sprinkler by fluid

- > @1t12t2 VrVrQ �U

External torque on Sprinkler (friction) + Torque on Sprinkler by fluid = 0 (1) External torque on Sprinkler + > @ 0VrVrQ 1t12t2 �U where Vt1 and Vt2 are the absolute velocities of flow relative to the rotating axis.

Vt1=Vr1cos T-Zr1

Vt2= Vr2cos T-Zr2

45o

30o

150 mm 200 mm

Z

2 1

21

Since there is no friction involved

� � � �> @ 0rcosVrrcosVrQ 111r1222r2 Z�T�Z�TU [0.15 {(2.54 cos 45-Z (0.15)} +0.2{(2.54 cos 30-Z (0.2)}] =0

0.0625Z = 0.71 Z =11.36 rad/s or 108.4 rpm

(b) When stationary (Z=0), the torque due to each nozzle is For nozzle 1: Torque on fluid (T1)=UQ1Vr1cos30 x 0.2 =1000x0.5x10-4x2.54 cos30x0.2

=0.022 Nm For nozzle 2: Torque on fluid (T2) =UQ2Vr2cos45 x 0.15 =1000x0.5x10-4x2.54 cos45x0.15

=0.0135 Nm

Total torque on fluid at nozzles 1 and 2 are: 0.022 +0.0135 =0.0355 Nm Total torque on Sprinkler by fluid= -0.0355 External torque on Sprinkler (friction) + Torque on sprinkler by fluid = 0

Torque on Sprinkler � 0.0355 =0

Torque on Sprinkler = 0.0355