ACE Engineering Academy … · 2017-03-31 · (ii) Slip and (i ii) coefficient of discharge (Cd): Slip of a reciprocating pump is defined as the difference of the theoretical discharge

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: 2 : Civil Engg. _ ESE MAINS


01. (a)

Sol:

Design traffic,

%r

FDA1)r1(365N

n

Where,

r = rate of growth

n = Design life

D = Lane Distribution factor

F = vehicle damage factor

6

10

10

1003.7

75.01.34411100

3.71365

N

msa

= 9.599 msa ≃ 10 msa

Total thickness required for sub grade

CBR = 5% and traffic ≃ 10 msa is 660

mm(from graph)

(b)

Sol: Width of the rectangular channel b = 4 m

Specific energy in the channel is E = 4 m

Specific energyg2

VyE

2

…… (i)

Discharge Q = AV

= by V

y4

QV

Where y = normal depth

From equation (i)

22

y4g2

Qy4

2

2

gy32

Qy4

Q2 = 2gy32y4

Q2 = (4 – y) 313.92 y2

32 y92.313y68.1255

For maximum discharge 0dy

dQ

0y76.941y36.2511y92.313y68.12552

1 2

32

y = 2.67 m

Maximum discharge

Q = yg32y4

67.281.93267.24Q

= 54.62 m3/sec

(c)

Sol:

(i) NPSH (Net Positive Suction Head): NPSH

available is the theoretical amount of head

that could be lost between suction and point

of minimum pressure without causing

cavitation. NPSH refer to the analysis

of cavitation. The minimum value of NPSH

available that is needed to prevent cavitation

in the pump or turbine i.e., the value of

NPSHA that causes pmin to equal pvap. To

: 3 : Test – 4


avoid cavitation, always operate with

NPSHAvailable NPSHRequired.

Rvap

Lslatm,abs NPSH

γP

ΣhZγ

P

(ii) Slip and (iii) coefficient of discharge (Cd):

Slip of a reciprocating pump is defined as the

difference of the theoretical discharge and

the actual discharge.

i.e. Slip = Theoretical discharge – Actual

discharge

= Qth – Qact

Slip can also be expressed in terms of %age

and given by

100Q

QQSlip%

th

actth

%

100)C1(100Q

Q1 d

th

act

Slip where Cd is known as coefficient of

discharge and is defined as the ratio of the

actual discharge to the theoretical discharge.

th

actd Q

QC

Value of Cd when expressed in percentage is

known as volumetric efficiency of the pump.

Its value ranges between 95-98%. Percentage

slip is of the order of 2% for pumps in good

conditions.

iv) Air Vessels: Air vessels are a closed

container, in which the bottom part is filled

with water & upper half part is filled with

compressed air. These air vessels installed

very near to the suction valve & delivery

valve to avoid the separation in reciprocating

pumps.

An air vessel usually fitted in the discharge

pipe work to dampen out the pressure

variations during discharge. As the discharge

pressure rises the air is compressed in the

vessel, and as the pressure falls the air

expands. The peak pressure energy is thus

stored in the air and returned to the system

when pressure falls.

Purposes of Air vessels:

1) To obtain liquid at uniform discharge.

2) Due to air vessel frictional head and

acceleration head decreases and the

work overcoming friction resistance in

suction and delivery pipe considerably

decreases which results in good amount

of work.



3) Reciprocating pump can run at high

speed without flow separation

(v) Specific speed of pumps: Specific speed is

defined as "the speed of an ideal pump

geometrically similar to the actual pump,

which when running at this speed will raise a

unit of volume in a unit of time through a

unit of head".

In metric units

Specific Speed4/3S H

QN)N(

N = The speed of the pump in revolutions per

minute (rpm)

Q = The flow rate in liters per second (for

either single or double suct)

H = The total dynamic head in meters

In SI units, Q is in cubic meter per

second, N in rpm and H in meters.

02. (a)

(i)

Sol: Given that one road is with one-way traffic

And another road is with two-way traffic.

(ii)

Sol:

Before the imposition of the ban on private

cars:

The weighted spacing of vehicles

= 7 × 0.8 + 12 × 0.18 + 10 × 0.02

= 5.6 + 2.16 + 0.2 = 7.96 m

Jamming concentration before,

96.7

1000K j

= 126 veh/hr

The vs-K curve and Q-K curve before the

imposition of the ban are given in figure

4

vKQ sfj

(max)

4

50126 = 1575 veh/hour/lane

After the imposition of the ban on private

cars

Conflicts: Number of points

4

4

1

2

Total 11

one-way

two-

way

: 5 : Test – 4


The number of persons traveling by car

before the ban = 1200 × 0.80 × 1.5

= 1440 persons/hour

The number of persons now switching over

to buses

= 50% of 1440

= 720 persons/hour

No. of extra buses needed

30

720 = 24/hour

No. of buses originally present

100

21200 = 24/hour

Total No. of buses after the ban = 48/hour

No. of commercial vehicles

= 1200 × 0.18= 216/hour

Total flow = 48 + 216

= 264 vehicles/hour

Weighted spacing of vehicles now:

264

4810

264

21612

= 9.8 + 1.82 = 11.62, say 12 m

Jamming concentration now

12

1000 = 83 veh/km/lane

The vs-K and Q-K curves after the imposition

of the ban are given in figure

4

vKQ sfj

(max)

4

5083 = 1038 veh/hour/lane

(b) (i)

Sol:

Coefficient velocity Cv = 0.98

Coefficient of contraction Cc = 0.62

v1 = 2.006 m/sec

g = 9.8 m/sec2

Coefficient of contraction

Cc =openingofHeight

ctavenacontraatDepth

Height of opening

62.0

8.0

C

8.0

c

29032.1 m

Equating specific energy at upstream and

down stream sections

8.0g2

V2

g2V 2

22

1

8.0

81.92V

281.92

006.2 22

2

8.081.92

V205098.2

22

V2 = 5.25 m/sec

Discharge Q = CdA2V2

= y2 0.98 v2

= 0.98 0.8 5.25

= 4.1164 m2/sec



: 7 : Test – 4


(ii)

Sol:

Uniform flow depth yo = 1.5 m

Let computing critical depth of trapezoidal

section

Uniform flow depth is 1.5 m,

Bed slope So = 1 10–4

Area 2m5.345.1

22620

A

Perimeter P = b + 2y 2m1

= 20 + 2 1.5 41

= 26.708 m

Discharge 2/13/2 3Rn1

AQ

R = hydraulic radius =PA

2/1o

3/2 SRn1

AQ

43/2

101708.265.34

2.01

5.34

Q = 2.046 m3/sec

Critical depth equationTA

gQ 32

Where A = area at critical depth

T = top width at critical depth

cccc yy220y

2y42020

A

T = 20 + 4yc

Critical depthTA

gQ 32

c

322

y420

y2y20

81.9

046.2

32ccc )y20y20()y420(4267.0

yc = 0.1018 m

yc < yo mild slope

y > yo > yc M1 profile

03. (a)

Sol:

Given

Discharge = 270 m3/sec

Constant width of spillway = 15 m

U/S level of water above datum = 58 m

D/s level of water above datum = 26 m

3m 3m

y

b=20 m2

1

2yc 2yc

yc

b=20 m2

1y2

E

y1

Z

(1) (2) (3)

g2

V 21 g2

V 22

26m

58 m



Let the basin invert be at a height ‘Z’ above

the datum

Further let y1, y2 and V1 , V2 be the depth of

flow and velocities of flow before and after

the hydraulic jump respectively.

Then11

1 y

18

y15

270V

222 y

18

y15

270V

If q is the discharge per component width of

the spillway

msec//m1815

270q 3

Applying Bernoulli’s equation between the

section behind the spillway and

where the hydraulic jump begin’s assuming

that no energy is dissipated in the flow

down the spillway, we get

g2

VyZ58

21

1 (i)

and Z + y2 = 26 m

Z = 26 y2 (ii)

Further hydraulic jump in a rectangular

channel is

2121

21 yyyy

g

q2

2121

2

yyyy81.9

182

66.055 = y1y2 (y1 + y2) (iii)

From equation (i) and (ii)

21

2

12 qy2

qyy2658

21

2

12 y81.92

18yy2658

21

12 y

513.16yy32

32y

513.16yy

21

12 (iv)

From equation (iii) and (iv)

32

y

513.16yy32

y

513.16yy055.66

21

1121

11

By solving above equation

y1 = 0.6325

From equation (iii)

The post jump depth

32y

513.16yy

21

12

m09.9y2

We know Z + y2 = 26 m

Z + 9.09 = 26

Z = 26 –9.09 = 16.91

Pre jump depth y1 = 0.6325 m

Post jump depth y2 = 9.09 m

Invert basin level = 16.91 m

: 9 : Test – 4




(b) (i)

Sol:

In order to predict the behaviour of a water

turbine working under varying conditions of

head, speed, and power, there is need of the

concept of unit quantities. The unit quantities

give the speed, discharge and power for a

particular turbine under a head of one meter

assuming the same efficiency. The following

are the three unit quantities.

Unit speed

Unit discharge

Unit power

1) Unit Speed (Nu): The speed of the turbine,

working under unit head (say 1m) is known

as unit speed of the turbine.

The tangential velocity is given by,

60

DNu or

D

u60N

If H = 1; then HNN u

Where, H = head of water, under which the

turbine is working; N = speed of turbine

under a head; H; u = tangential velocity; Nu =

speed of the turbine under a unit head.

2) Unit discharge (Qu): The discharge of the

turbine working under a unit head (say 1m) is

known as unit discharge.

HKgH2aQ 3

If H = 1; Then Q =Qu ;

33u K1KQ

Or, HQu

Thus,H

aQu

3) Unit Power (Pu): The power developed by a

turbine, working under a unit head (say 1m)

is known as unit power of the turbine

Power developed by a trubine is given as

QHP and gH2V

H)gH2a(P

P = K2H3/2

If H = 1; then, P = Pu

22/3

2u K1KP

P = Pu H3/2

Thus,2/3u H

PP

If a turbine is working under different heads,

the behaviour of the turbine can be easily

known from the unit quantities.

2

2

1

1u

H

N

H

NN

2

3

2

2

2

3

1

1u

H

P

H

PP

2

2

1

1u

H

Q

H

QQ

: 11 : Test – 4


Where H1, H2 are the heads under which a

turbine works; N1,N2 are the corresponding

speeds; Q1,Q2 are the discharges, and P1,P2

are the power developed by the turbine.

(ii)

Sol:

Given data:

Power of the prototype (larger turbine)

turbine (Pp) =7500kW

Head on the prototype Kaplan turbine

(Hp) =10m

Speed of the prototype turbine (Np) =100rpm

Scale Ratio= 1:10

Head on the model turbine (Hm) =4m

Efficiency of both turbine equal= 0.85

Nm, Qm, Pm, Nsm =?

Head coefficient = N.D H

Discharge coefficient = Q D2. H

Power coefficient =PD2. 2

3

H

p

m

pp

mm

H

H

D.N

D.N

10

4

10

1

100

Nm

Nm = 632.46 rpm (Speed of the model

turbine)

2

3

m

p

2

m

p

m

p

H

H.

D

D

P

P

5.126

4

10

1

10

Pm

105.7

Pm =18.97kW

model =mm H.gQ

Pm

0.85 = 4Qm81.91000970,18

Qm = 0.569m3/sec

Nsm =4/5)4(

97.1846.632 = 486.96



04. (a)

(i)

Sol:

SL No Aspects Impulse Turbine Reaction turbine

1 Conversion of fluid

energy

The available fluid

energy is converted

into K.E by a nozzle.

The energy of the fluid is partly

transformed into K.E before it (fluid) enters

the runner of the turbine.

2 Changes in pressure

and velocity

The pressure remains

same (atmospheric)

throughout the action

of water on the runner

After entering the runner with an excess

pressure, water undergoes changes both in

velocity and pressure while passing

through the runner.

3 Admittance of water

over the wheel

Water may be allowed

to enter a part or whole

of the wheel

circumference.

Water is admitted over the circumference

of the wheel

4 Water-tight casing Not Required Necessary

5 Extent to which the

water fills the

wheel/turbine

The wheel/turbine does

not run full and air has

a free access to the

buckets.

Water completely fills all the passages

between the blades and while flowing

between inlet and outlet sections does work

on the vanes.

6 Installation of Unit Always installed above

the tail race. “No drafttube is used”.

Unit may be installed above or below the

tail race-“use of a draft tube is necessary”

7 Relative velocity of

water

Either remaining

constant or reduces

slightly due to friction.

Due to continuous drop in pressure during

flow through the blade, the relative velocity

increases.

8 Flow regulation By means of a

needle valve fitted

into the nozzle.

Possible without

losses

By means of a guide-vane assemble.

Always accompanied by loss

: 13 : Test – 4




(ii)

Sol: Given Data:

H = 250m

= 0.45

Cv=0.98

m=18

0 =0.76

f = 50Hz

Pair of poles=6

D=? ; d=? ; Po=?

Synchronous speed (N) =60f/P = 60*50/6

= 500rpm

Speed ratio ():

=gH2

U

0.45 =25081.9260

500D

D = 1.2 m

Jet Ratio (m)= D/d = 18

d= D/18=1.2/18=0.06667m=66.67mm

Discharge through the nozzle

(Q) = Cv A nozzle

VJet=0.98.*((22/7)/4)*(0.06667)2

* 25081.92

= 16.8m3/sec

Output power of the pelton turbine

= 0. g Q H

= 0.76*1000*9.81*16.8* 250=31.31MW

(b)

Sol:

The mid-point of speed class (vi) and qi are

tabulated below, along with qivi and ki = qi/vi

Speed

Range

Volume

(qi)

Speed

mid-

point

(vi)

qivi

Ki =

qi/vi

2-5 1 3.5 3.5 0.29

6-9 4 7.5 30.0 0.54

10-13 0 11.5 0 0

14-17 7 15.5 108.5 0.45

18-21 20 19.5 390.0 1.02

22-25 44 23.5 1034.0 1.87

26-29 80 27.5 2200.0 2.91

30-33 82 31.5 2583.5 2.60

34-37 79 35.5 2804.5 2.23

38-41 49 39.5 1935.5 1.24

42-45 36 43.5 1566.0 0.83

46-49 26 47.5 1235.0 0.55

50-53 9 51.5 493.5 0.17

54-57 10 55.5 555.0 0.18

58-61 3 58.5 178.5 0.05

450 15087.0 14.93

qi = Q = 450 veh/hour

viqi = 15087.0

: 15 : Test – 4


ki = k = 14.93

450

15087

Q

vqv ii

t = 33.53 km/hr

93.14

450

k

Qvs 30.1 km/hr

s

2s

st vvv

s2 = (33.53 – 30.1) 30.1 = 103.5

05. (a)

Sol:

(i) For plain terrain:

v = 80 × 0.278 kmph = 22.24 m/s

From equation

gR

)V75.0(e

2

max

2

e1319.0)21581.9(

)24.2275.0(e

Therefore, provide, e = emax and check the

value of fmax

From equation

Available max

2

egR

vf

max

2

f164.007.0)21581.9(

)24.22(

Therefore, determine maximum allowable

speed when e = emax and f = fmax values are

taken

s/m54.21)15.007.0(21581.9VL

= 77.48 kmph

(ii) For rolling terrain:

v = 80 × 0.278 kmph = 22.24 m/s

From equation

max

2

e094.0)30081.9(

)24.2275.0(e

Therefore, provide, e = emax = 0.07 and check

the value of fmax

From equation

Available max

2

egR

vf

max

2

f098.007.0)30081.9(

)24.22(

Provide e = 0.07 for the given speed

(iii) For hilly areas:

v = 100 × 0.278 kmph = 27.8 m/s

From equation

max

2

e080.0)13781.9(

)8.2775.0(e

Where, emax = 0.1

Provide super elevation rate = 0.08 × 100

= 8%

(b) (i) (1)

Sol:

Base load: It is the minimum level of

electricity demand required over a period of

24 hours. It is needed to provide power to

components that keep running at all times

(also referred as continuous load).



Refrigerator and HVAC (Heating ventilation

systems) are examples of base demand.

Peak load: It is the time of high demand.

These peaking demands are often for only

shorter durations. In mathematical terms,

peak demand could be understood as the

difference between the base demand and the

highest demand.

Examples of household loads: microwave

oven, toaster and television are examples of

peak demand,

Power plants are also categorised as base

load and peak load power plants.

Base Load Power plants:

Plants that are running continuously over

extended periods of time are said to be base

load power plant. The power from these

plants is used to cater the base demand of the

grid. A power plant may run as a base load

power plant due to various factors (long

starting time requirement, fuel requirements,

etc.).

Examples of base load power plants are:

Nuclear power plant

Coal (Thermal)power plant

Hydroelectric plant

Geothermal plant

Biogas plant

Biomass plant

Solar thermal with storage

Ocean thermal energy conversion

Peak Load Power plants:

To cater the demand peaks, peak load power

plants are used. They are started up whenever

there is a spike in demand and stopped when

the demand recedes.

Examples of peak load power plants are:

Gas plant

Solar power plants

Wind turbines

Diesel generators

(i) (2)

Sol:

Firm Power: The net amount of power

which is continuously available from a plant

without any break on firm or guaranteed

basis.

Secondary Power: The excess power

available over the firm power during the off

peak hours or during monsoon season.

Firm (primary) power: Minimum Power

that can be produced by a plant with no risk.

For a single hydroelectric plant, it

corresponds to the minimum availability of

: 17 : Test – 4


storage based. Firm energy is marketed with

high price.

Secondary (Surplus) power: All the power

available in excess of firm power. Secondary

power cannot be relied upon. Its rate is

usually less than that of firm power. It can

be generated ~9 to 14 hours/day.

Firm (primary) power: It depends upon

whether storage is available or not for the

plant since a plant without storage like run-

of-river plants would be produced power as

per the minimum stream flow. For a plant

with storage, the minimum power produced

is likely to be more since some of the stored

water would also be used for power

generation when there is low flow or no flow

in the river.

Secondary (surplus) Power: This is the

power produced by a hydropower plant over

and above the firm power.

(ii)

Sol:

Given data:

Discharge (Q) = 11m3/sec

Head (H) = 16m

Efficiency of the plant () =0.75

Peak load operation time = 5 hours per day

Firm capacity of the plant =?

Case-I: Without pondage (peak load

operation)

Output Power of the plant

= Efficiency × Hydro power

= 0.75 × 1000 × 9.81 × 11 × 16

= 1.3 MW

Firm capacity of the plant (Energy generated

for specific duration)

= 1.3 × 5 = 6.5 MWH per day

or

= 6.5 × 365× 24 =56,940 MWH per annum

Case-II: With pondage (peak load

operation)

Output Power of the plant = Different

efficiencies (Plant efficiency × Evaporation

loss factor) × Hydro power

= 0.75 × 0.9 × 1000 × 9.81 × 11 × 16

= 1.17 MW

Firm capacity of the plant (Energy generated

for specific duration)

= 1.17 × 5= 5.85 MWH per day

(or)

= 5.85 × 365 × 24 =51,246 MWH per annum



06. (a)

Sol: Prototype Model

bp = 1200 m bm = ?

QP = 2830 m3/sec

nP = 0.025 nm = ?

Vertical scale ratio (Lv)r =40

1

Horizontal scale ratio (Lh)r =800

1

Depth in prototype

Q = b × y × V

2830 = 1200 × y × 0.9

yp = 2.1225 m

40

1

y

y

p

m

ym = 0.053 m

In rivers gravitational force is predominant.

Froude’s law is same in both prototype an

model.

mpgy

V

gy

V

rv

p

m

p

m

p

m Ly

y

gy

gy

V

V

∵ Acceleration due to gravity is constant in

both prototype and model

Mean velocity in prototype 9.0Vp m/sec

Vm= prv VL

9.040

1

Vm = 0.1423 m/sec

Manning’s constant n =VSR 2/13/2

Manning’s constant

3/2P

p

m

3/2m

p

m

R

V

V

R

n

n

Assume both prototype and model has

same bed slope

1423.0

9.0L 3/2

rv

3245.640

13/2

5407.0n

n

p

m

nm= 0.0135

Model width prhm b)(Lb

1200800

1 m5.1

Reynolds number in model

VbVbRe

The kinematic viscosity did not change

because the fluid is same in both model &

prototype

mm bV

6101.1

5.11423.0Re

Re = 194.045 103 > 500

The flow will be turbulent

: 19 : Test – 4


(b) (i)

Sol:

These curves are obtained by running the

centrifugal pump at the design speed, which

is also the driving speed of the motor. The

design discharge, power and head are

obtained from the corresponding curves,

where the efficiency is maximum as shown

below figure.

(ii)

Sol:

Given Data:

D2=2D1

N= 1200 rpm

H=75m

Vf1=Vf2=3m/s

D2=0.6m,

Therefore D1=D2/2 =0.6/2=0.3m

B2=0.05m

0 =?

W.D/sec=?

mano =?

Tangential Velocity of the pump wheel

(Impeller)at inlet (U1):

60

NDU 1

1

60

12003.0 = 18.85 m/s

Outlet Diagram

1

1

u

Vftan 159.0

85.18

0.3

= 9.04 (inlet vane angle)

U2 =60

12006.0

60

ND2

= 37.7 m/sec

H

Q

PInput power

Max eff

Head

Q optimum

(N = constant)



Assumption: It seems that the impeller vane

bend backward exit angle typing or print

error, in order to get the solution, it may be

assumed exit vane angle as 300

tan30o=2

2

w2

f

VU

V

= 3.0 / (37.7-Vw2)

Vw2 = 32.5m/s

Discharge through the impeller (Q):

Q=D2.B2.2fV = .(0.6).(0.05).(3.0)=0.283m3/sec

Work done per second by the impeller

= Q.Vw2. U2

= 1000 × 0.283 × 32.5 ×37.7

= 346.75 kN-m/sec

2wmano uV

gHmn

2

= ( 9.81 × 75) / (32.5 × 37.7) = 0.6

= 60%

07. (a)

Sol:

(i) Bulk density of compacted mix (without

paraffin coating):

339.24.6844.1195

4.1195G compbulk

Unit weight of compacted bituminous mix

= 2.339 ×1 g/cc = 2.339 g/cc

ii) Bulk density of compacted mix specimen

coated with paraffin:

the bulk density of paraffin or wax coated

specimen

wax

waxwaterwaxairwax

.spec

G

WWW

W

33.2

9.0

5.453.6789.1240

4.1195

(Wwax = 1240.9 – 1195.4 = 45.5 g)

iii) Combined specific gravity of aggregates:

bulk specific gravity of combined aggregates

(Gbulk-agg)

3

3

2

2

1

1

321

G

p

G

p

G

pppp

On substtiuon, Gbulk-agg=

69.2

5.7

71.2

0.38

61.2

5.545.70.385.54

78.202.1488.20

100

= 2.653

iv) Maximum Theoretical Density,

02.1

5.5

69.2

1.7

71.2

1.35

61.2

5.51100

G t

= 2.438

v) Percentage of voids present in the compacted

mix:

100G

GG%V

t

mtv

100438.2

339.2438.2

= 4.07%

vi) Voids in mineral aggregates:

Vb% = (percent f bitumen)b

compbulk

G

G

: 21 : Test – 4


%61.1202.1

339.25.5

VMA (%) = Vv + Vb = 4.07 +12.61= 16.69%

vi) Voids filled with bitumen:

Voids filled with bitumen,

VMA

%V100(%)VFB b

VFB (%) = %56.7569.16

61.12100

(b)

Sol:

By Karman momentum integral equation

xU2

o

thicknessMomentum

0

2o dy

U

u1

U

u

xU

Put y/ = , dy = d and the limits of

are 0 and 1

4322)(fU

u

du = U df()

Substituting,

1

0

2o d)(f1)((f

dx

dU

But

1

0d)(f1)((f

1

0

4343 d)221)(22(

1

0 8754

7643542

d)22

24422442(

1

0

8765432 d)4449242(

=1

09

9

8

84

7

74

6

64

5

59

4

42

3

34

2

22

= 37/315

Therefore,dx

dU

315

37 2o

…..(i)

From the boundary conditions for a laminar

boundary layer

00y

o d

)(dfU

y

u

Since 4322)(f

2]462[d

(df0

32

0

U2

o ………….(ii)

Equating equation (i) and (ii)

dx

dU

315

37U2 2

dxU37

630d

CxU37

630

2

2

Since at x = 0, = 0, C = 0

)/Ux(

x05.34x

U37

1260 22

xRe

835.5

x

where

pUx

Rex



Documents

ACE Engineering Academy … · 2017-03-31 · (ii) Slip and (i ii) coefficient of discharge (Cd): Slip of a reciprocating pump is defined as the difference of the theoretical discharge