CE2016

Ministry of Science and Technology Yangon Technological University Department of Civil Engineering

CE 2016 Fluid Mechanics Problems and Solutions For Second Semester

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Department of Civil Engineering Yangon Technological University

CE 2016 Fluid Mechanics [ Questions and Answers( Second Semester) ]

Chapter 1 Flow Measurement 1. A venturimeter with a 15cm diameter at inlet and 10cm throat is laid with its axis horizontal and is used for measuring the flow of oil of sp.gr. 0.9. The oil-mercury differential manometer shows a gauge difference of 20cm. Assume coefficient of the meter as 0.98. Calculate the discharge.

.

2. A horizontal venturimeter 160mm×80mm used to measure the flow of an oil of sp. gr. 0.8. Determine the deflection of the oil-mercury gauge, if the discharge of the oil is 50liters/sec.

3.(a)Derive the following expression for the actual discharge of a venturimeter.

22

21

21 2

AA

ghAACQ d

−=

Solution See page 1 and 2 (b) Example 1.1 ( page 3 and 4) (c) Example 1.2 ( page 6)

4. A venturimeter has an area ratio of 9 to 1, the larger diameter being 30cm. During the flow, the recorded pressure head in the larger section is 6.5m and that at the throat 4.25m.If the meter coefficient C is 0.99, compute the discharge through the meter.

5. A venturimeter is to be fitted to a 25cm diameter pipe, in which the maximum flow is 7200liters per minute and the pressure head is 6 meter of water. What is the minimum diameter of throat so that there is no negative head in it?

Chapter 2 Flow through Orifices and Mouthpieces 6. A rectangular orifice 1.5m wide and 1.0m deep is discharging water from a tank. If the water level in the tank is 3.0m above the top edge f the orifice, find the discharge through the orifice. Take coefficient of discharge for the orifice as 0.6.

---------------------------------------------------------------------------------------------------------------- 7. Example 2.1 8. Example 2.2

9. (a) Describe some devices for measuring discharge through pipes and for measuring velocity in a closed conduit. Solution devices for measuring discharge through pipes are: - venturimeter - orifice meter - rotameter - nozzle meter -elbow meter device for measuring velocity in a closed conduit is pitot tube. (b)A drowned orifice 1.5m wide and 0.5m deep is provided in one side of a tank. Find the discharge in liters/sec through the orifice, if the difference of water levels on both sides of the orifice be 4m. Take Cd = 0.64.

.

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10. A drowned orifice 1.0m wide has height of water from the bottom and top of the orifice as 2.25m and 2.0m respectively. Find the discharge through the orifice, if the difference of water levels on both sides of the orifice be 37.5cm. Take coefficient of discharge as 0.62.

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11. Water is supplied from a tank into a canal through a rectangular sluice 1m wide and 1.75m high. The water level in the tank is 2m above the top edge of the opening and the canal water level is 30cm below the top edge. If the coefficient of discharge is 0.62for the both free and the submerged portions, calculate the discharge.

Chapter 3 Flow Over Notches and Weirs 12. (a)Find the discharge over a rectangular weir 1.5m long under a head of 400mm by using Banzin’s formula.

(b) Example 3.3 13. (a)A 30m long weir is divided into 10 equal bays by vertical posts each 0.6m wide. Using Francis formula, calculate the discharge over the weir under an effective head of 1m.

(b) Example 3.2

14. Find the discharge in m3/min over a rectangular weir of 300m in length with a head of 1m. Take velocity of approach as 1m/sec and Cd = 0.58.

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15. A weir, 36m long, is divided into 12 equal bays by vertical posts each 60cm wide. Determine the discharge over the weir, if head over the crest is 1.2m and the velocity approach is 2 m/sec.\

Chapter 4 Uniform Flow Through Open Channels 16.(a) What do you understand by the term “most economical section of a channel”?

A channel which gives maximum discharge for a given cross-sectional area and bed slope is called a channel of most economical cross-section. Or in other words, it is a channel, which involves lesser excavation for a designed amount of discharge. A channel of most economical cross-section is, sometimes, also defined as a channel, which has a minimum wetted perimeter; so that there is a minimum resistance to flow and thus resulting in a maximum discharge. (b). A rectangular channel having hydraulic mean depth of 2m discharges water with velocity of 1m/sec. Find the value of Chezy’s constant, if the bed slope of channel is 1 in 8000.

(c) Example 4.3

17.(a) Find the most economical cross section of a rectangular channel to carry 0.3m3/sec of water, when bed slope is 1 in 1000. Assume Chezy’s C=60.

(b) Example 4.1 18.(a) Example 4.2 (b) Example 4.4

Chapter 5 Turbines 19.(a) Describe advantages of impulse turbines and reaction turbines.

The main advantages of impulse turbines are:

• They can be easily adopted to power variation with almost constant efficiency.

• The penstock overpressure and the runner overspeed control are easier.

• The turbine enables an easier maintenance.

• Due to the jet the manufacturer of these turbines impose a better solid particle control,

conducting, consequently, to a lower abrasion effect.

The main advantages of reaction turbines are:

• It needs lesser installation space.

• It provides a greater net head and a better protection against downstream high flood

levels.

• It can have greater runner speed.

• It can attain greater efficiencies for high power values.

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(b). Estimate the maximum height of straight conical draft tube of a 18000h.p. Francis

turbine running at 150 r.p.m under a net head of 28m. The turbine is installed at a station

where the effective atmospheric pressure is 10.6m of water. The draft tube must sink at least

0.89 below the tail race.

4/5HPNNs =

= 327

σc = 0.625 (Ns/444)2

= 0.339

Cavitation factor σ = Ha -Hv -Hs

H

Ha -Hv = 10.6 m, H =28 m

= 10.6 -Hs

28

Hs = 1.108 m

Max length of the draft tube = 1.108 + 0.89 = 2 m

20. Powerhouse is equipped with a vertical shaft pelton turbine. The generator is provided

with 6pairs of poles. Design discharge is 1.5 cumecs and net head is 425m . The turbine will

provide 6800hp. Take coefficient of nozzle as 0.95. Determine

(a) the specific speed (b) velocity of jet (c) jet diameter (d) pitch circle diameter of the wheel (e) number of buckets.

(a) p

fN 60=

= 60 x 50/6 = 500 rpm

4/5HPNNs =

= 4/54256800500

= 21.4 Use single jet pelton turbine (b) velocity of jet

2gHv ⋅= vC

= 2x9.81x42595.0 ⋅ = 86.75 m/s (c) jet diameter

jet0.25

0.5

n1

HQ0.54d ⋅⋅=

1

14251.50.54 0.25

0.5

⋅⋅=

= 0.146 m =146 cm (d) pitch circle diameter

NH38D ⋅

=

= 1.57 m (e) Number of buckets

15dD0.5N buc +⋅=

150.1461.570.5Nbuc +⋅=

= 20.4 =21

21. The head available at entrance to the nozzle supplying a pelton wheel is 300m and the

coefficient of velocity for the nozzle is 0.98. The wheel diameter is 1.8m and the nozzle

diameter is 125mm. The buckets deflect the jet through 165°. Assuming the relative velocity

of the jet is reduced by 15%, calculate the theoretical speed in r.p.m for the maximum

hydraulic efficiency. What is the hydraulic efficiency when running at this speed, and what is

the power developed?

Deflection angle =165° = (180- φ )

φ = 15°

k = 0.85

hf

hn

H H'=Vai2/2g H1

For max hydraulic efficiency V/Vai =0.5

Vai = Cv√2gH1

= 75 m/s

V =Vai/2 = 37.5 m/s

V =π DN/60

N =60V/π D = 398 rpm

ηhmax = 1/2 (1+ k Cos φ)

= 91.05 %

mass/sec = m = ρ Q = ρ π/4 d2 Vai

= 920 kg/sec

Inlet diagram:

V Vri

From velocity diagram Vwi = Vai =75 m/s

Vri = Vai- V =37.5 m/s

Vwi = Vai

Outlet diagram:

V Vwo = V -Vro Cos φ

= V - k Vri Cos φ

Vwo = 6.7 m/s

Vro Vao

Power = m V ( Vwi -Vwo)

= 2356350 Watts

Chapter 6 Centrifugal Pump

22. (a)What is a centrifugal pump? On what principle does it work?

Centrifugal pumps are classified as rotodynamic type of pumps in which dynamic pressure is

developed which enables the lifting of liquids from a lower to a higher level. The basic

principle on which a centrifugal works is that when a certain mass of liquid is made to rotate

by an external force, it is thrown away from the central axis of rotation and a centrifugal head

is impressed which enable it to rise to a higher level.

(b). Define specific speed of centrifugal pump.

The specific speed of a centrifugal pump is the speed at which the specific pump must

run to deliver unit quantity against unit head, the efficiency being the same as the actual

pump.

4/3H

QNNs =

where Ns= specific speed

N = rotational speed(rpm)

H = total head

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(c). Describe advantages of centrifugal pump.

The main advantage of a centrifugal pump is that its discharging capacity is very much

greater than a reciprocating pump which can handle relatively small quantity of liquid only.

A centrifugal pump can be operated at very high speeds without any danger of separation and

cavitation . The maintenance cost of a centrifugal pump is low and only periodical check up

is sufficient .

(d) Example 6.1

Chapter 7 Dimensional Analysis

23.(a) What is “hydraulic similitude”? Describe and explain the three important types of

hydraulic similitude.

Solution

To know the complete working and behavior of the prototype, from its model, there should

be a complete similarity between the prototype and its scale model. This similarity is known

as hydraulic similitude.

The three important types of hydraulic similitude are:

- Geometric similitude

- Kinematics similitude

- Dynamic similitude

Geometric similitude

The model and the prototype are identical in shape, but differ only in size. (The ratios of

all the corresponding linear dimensions are equal.)

Kinematics similitude

The model and the prototype have identical motions. (The ratios of the velocities at

corresponding points are equal.)

Dynamic similitude

The model and the prototype have identical forces. (The ratios of the corresponding forces

acting at corresponding points are equal.)

(b). A velocity at a point on a spillway model of a dam is 1.5m/sec for a prototype of model

ratio1:10. What is the velocity at the corresponding point in the prototype?

Velocity in the model v = 1.5 m/s

Model ratio, 1/s = 1/10

s = 10

Velocity in the prototype V = v √s

V =1.5 x √10

= 4.74 m/sec

24. Prove that the discharge over a spillway is given by the relation.

⎥⎥⎦

⎤

⎢⎢⎣

⎡=

DH

VgD

fVDQ ,2

Where V = velocity of flow D = Depth of throat H = Head of water g = Acceleration due to gravity

Q= f (V,D,H,G)

f1 (Q,V,D,H,G) = C

Choose V and D as repeating variables

n = 5, m =2, n-m =3

π1= Va1 Db1 Q = (LT-1)a1 (L)b1 (L3T-1)

π2 = Va2 Db2 H = (LT-1)a2 (L)b2 (L)

π3 = Va3 Db3 g = (LT-1)a3 (L)b3 (LT-2)

M0L0T 0 = (LT-1)a1 (L)b1 (L3T-1)

0 = -a1-1 ; a1 = -1

0 = a1+b1+3 ; b1= -2

π1 = V -1 D -2 Q

π1 = Q

VD2

M0L0T 0 =(LT-1)a2 (L)b2 (L)

0 = -a2

a2+b2+1 =0 ; b2 = -1

π2 = V0 D-1 H = H/D

M0L0T 0 = (LT-1)a3 (L)b3 (LT-2)

0 = -a3-2 ; a3 = -2

0 = a3+b3+1 ; b3 =1

π3 = V-2 D g

= VgD

f2 ( 0),,2 =VgD

DH

VDQ

),(2 DH

VgD

fVD

Q=

),(2DH

VgD

fVDQ =

25. A V-notch weir is a vertical plate with a notch angle ф cut into the top of it and placed across an open channel. The liquid in the channel is backed up and forced to flow through the notch. The discharge Q is some function of the elevation H of upstream liquid surface above the bottom of the notch. In addition it depends upon gravity and upon the velocity of approach Vo to the weir. Determine the form of the form of discharge equation:

⎥⎥⎦

⎤

⎢⎢⎣

⎡= φ,2

5

gHV

fHgQ o

Q = f (H, g , Vo , φ )

f1 ( Q, H ,g ,Vo, φ ) = C

Choose g and H as repeating variables

n = 5; n-m =3 ; m=2

π1 = Ha1 gb1 Q = (L)a1 (LT-2)b1 L3 T-1

π2 =Ha2 gb2 Vo = (L)a2 (LT-2)b2 LT-1

π3 = φ

(M)o (L)o (T)o = (L)a1 (LT-2)b1 L3 T-1

a1+b1+3 =0 ⇒ a1 = -5/2

-2b1-1 =0 ⇒ b1 = -1/2

π1 = H-5/2 g-1/2 Q = Q

√g H 5/2

(M)o (L)o (T)o = (L)a2 (LT-2)b2 L T-1

a2 = -1/2

b2 = -1/2

π2 =H-1/2 g-1/2 Vo= Vo

√gH

12/52 ),,( CgHV

HgQf o =φ

),(2/5 φgHVf

HgQ o=

),(2/5 φgHVfHgQ o=

26. (a)The Reynolds number (Re) is a function of density, viscosity, and velocity of a fluid and a characteristic length. Establish the Reynolds number relation by dimensional analysis. Density = ρ = M L-3

Viscosity = µ =M L-1 T-1 Velocity = V = L T-1 Length = L =L Re = f (ρ ,µ, V, L) Re = k ρa µb Vc Ld M0 L0 T0 = ( Ma L-3a )( Mb L-b T-b )( Lc T-c )( Ld ) M => 0 = a + b --------------------eq 1 L => 0 = –3a –b + c + d --------------------eq 2 T => 0 = –b –c --------------------eq 3 From eq 1 => a = -b From eq 2 => c = -b From eq 3 => d = -b Re = k ρ-b µb V-b L-b

b

VLk ⎟⎟

⎠

⎞⎜⎜⎝

⎛=

ρµRe

b

VLk−

⎟⎟⎠

⎞⎜⎜⎝

⎛=

µρRe

Assume k = 1 and b = -1

µρVL

=Re

(b)

27. (a)

(b)

28.

29.

30.

31.