Jntu Anan Civil 2 2 HHM Set 4
Jntu Anan Civil 2 2 HHM Set 4
Jntu Anan Civil 2 2 HHM Set 4
S.26
Code No: 9A01402/R09
April/May - 2012
Set-4
Solutions
Time: 3 Hours
Max. Marks: 70
Answer any FIVE Questions
All Questions carry equal marks
---
1.
2.
3.
4.
5.
6.
7.
8.
(a)
(b)
Derive an expression for the discharge through a channel by Chezys formula. (Unit-I, Topic No. 1.2)
Determine the dimensions of the most economical trapezoidal earth lined channel (Mannings n = 0.020) to
carry 14 m3/s at a slope of 4 in 10,000. (Unit-I, Topic No. 1.2)
(a) What are alternate depths? What are conjugate depths? Can a sequent depth be called a conjugate depth?
(Unit-II, Topic No. 2.2)
(b) The initial and sequent depths of a hydraulic jump in a horizontal rectangular channel are 0.6 m and 3.6 m
respectively. What is the initial Froude number? What is the energy loss? (Unit-II, Topic No. 2.2)
(a) For models governed by viscous forces, obtain the scaling ratios for velocity, discharge energy and power.
(Unit-III, Topic No. 3.1)
(b) The time period T of water surface waves is known to depend on the wave length , depth of flow D, density
of the fluid , acceleration due to gravity g and surface tension . Obtain the dimensionless form of the
functional relationship. (Unit-III, Topic No. 3.1)
(a) Obtain an expression for the force exerted by a jet of water on a fixed vertical plate in the direction of the jet.
(Unit-IV, Topic No. 4.1)
(b) A jet of water of diameter 50 mm moving with a velocity of 20 m/s strikes a fixed plate in such a way that the
angle between the jet and the plate is 60. Find the force exerted by a jet on the plate,
(i) In the direction normal to the plate and
(ii) In the direction of the jet. (Unit-IV, Topic No. 4.1)
A Kaplan turbine develops 60,000 kW of power under a head of 25 m with overall efficiency of 90%. Taking the value
of flow ratio = 0.5, speed ratio = 1.6, the hub diameter as 0.35 times the diameter of the runner. Find,
(i) The diameter of the runner
(ii) The speed of the turbine
(iii) The specific speed of the turbine. (Unit-V, Topic No. 5.4)
(a) Tests were conducted on a Francis turbine of 0.8 m diameter under a head of 9 m. The turbine developed 115 kW
running at 240 r.p.m and consuming 1.2 m3/sec. If the same turbine is operated under a head of 16 m, predict its
new speed, discharge and power. (Unit-V, Topic No. 5.4)
(b) What do you mean by cavitation? What are the physical indicators for the presence of cavitation in turbine?
(Unit-VI, Topic No. 6.4)
(a) Draw a typical layout and explain the working of centrifugal pump. Also indicate various components.
(Unit-VII, Topic No. 7.1)
(b)
(a)
(b)
Explain the method of selection of centrifugal pumps with the aid of characteristic curves. (Unit-VII, Topic No. 7.4)
Describe different structures associated with the intake to penstock. Discuss their uses. (Unit-VIII, Topic No. 8.1)
Write short notes on,
(i) Forebay
(ii) Intake structure
(iii) Penstock. (Unit-VIII, Topic No. 8.1)
( JNTU-Anantapur)
S.27
(a)
1
AR 2 / 3 S 10 / 2
n
14 =
4
1
2.4721 y e 2 ( y e / 2) 2 / 3
0.020
10,000
Answer :
1/ 2
2 23
14 = 1.557 ye y e
Answer :
83
8.992 = y e
Given that,
ye = 2.279 m
Mannings, n = 0.020
Bottom Width
Discharge, Q = 14 m3/s
Be = 0.4721 2.279
Slope, m = 4 in 10,000
= 1.076 m
Q2.
(a)
Answer :
Alternate Depths
1/ 6
1 4
=
0.020 10,000
= 13.572
Condition for most economical trapezoidal section
is,
R=
ye
2
Be + 2mye = 2 m + 1. ye
2
m=2
Be + 4ye = 2 5 ye
Be = (2 5 4) y e
Be = 0.4721ye
A = (Be + mye)ye
= (0.4721 + 2.0)ye2
= 2.4721 ye2
( JNTU-Anantapur)
S.28
(b)
Answer :
Q3.
(a)
Answer :
Given that,
Initial depth, y1 = 0.6 m
Reynolds number, Re =
( y 2 y1 ) 3
Energy loss =
4 y1 y 2
.V .L
... (1)
... (2)
(3.6 0.6) 3
4 0.6 3.6
m .Vm .Lm p .V p .L p
=
m
E L = 3.125 m
y2 1
1 + 1 + 8 F12
=
y1 2
Lm
Lp
Scale Ratios
3.6 1
1 + 1 + 8 F12
=
0.6 2
6=
12 + 1 =
(i)
1
[1 + 1 + 8F12 ]
2
Vr =
r
r .Lr
(or)
Vr =
r
Lr
1 + 8 F12
13 = 1 + 8 F12
Vm
Vp
[Q From (2)]
F12 =
F1 =
168
= 21
8
(ii)
21
F1 = 4.58
Qm
Qp
Qr =
r
.L2r
r .Lr
r .Lr
r
( JNTU-Anantapur)
Em
Ep
Ist Term
1 = T a g b c
= (Force Distance)r
= (. L3. V2)r
r2
= r. Lr3 2 2
r .Lr
Er =
(iv)
2r .Lr
r
c =0
a + b 3 c=0
pr
pp
a = 1/2
On substituting a, b, c values, we get,
= (Force Velocity)r
1 =
= (. L2 V3)r
= r. Lr2 .
pr =
(b)
S.29
2 = D a gb c
3r .L3r
2 = D/
IIIrd Term
3r
2r .Lr
3 = a gb c
[M0 L0 T0] = [MT2] [L]a [LT2]b [ML3]c
b= 1
T = f(, D, g, )
a + b 3c = 0
a= 2
Variables
[T]
[L]
[L]
[ML3]
[LT2]
[MT2]
[ML1 T1]
II ndTerm
3r
Answer :
T g
Q4.
D
g
= f , 2
g
(a)
Answer :
( JNTU-Anantapur)
S.30
(b)
Given that,
Power, P = 60, 000 kW
Head, H = 25 m
overall efficiency, 0 = 90% = 0.90
Flow ratio, F.R = 0.5
Speed ratio, S.R = 1.6
Hub diameter = 0.35 runner diameter
Flow ratio,
V f1
2 gH
V f1
2 9.81 25
d2
4
Fn = 692.800 N
Force in the Direction of Jet
Fx = AV2 sin2
= 1000 (0.002) (20)2 sin2 (60)
= 1000 0.002 400 0.750
Fx = 600 N
= 0.5
Speed ratio,
u1
2 gH
u1
2 9.81 25
(i)
= 1.6
= 1.6
u1 = 35.435 m/s
Diameter of boss, Db = 0.35 D0
Runner Diameter
0 =
w.P =
0.90 =
0.90 =
= 0.5
V f1 = 11.073 m/s
(0.050) 2
4
= 0.785 0.003
= 0.002 m2
Force exerted by jet in the direction normal to plate
can be calculated as,
Fn = AV2 sin
= 1000 (0.002) (20)2 sin 60
= (1000) (0.002) (400) (0.866)
Q5.
Answer :
Q=
P
w.P
g Q H
1000
60,000
g Q H
1000
60,000 1000
g Q H
60,000 1000
g 0.90 H
60,000 1000
1000 9.81 0.90 25
= 271.831 m3/s
( JNTU-Anantapur)
2
( D0 Db2 ) V f1
4
2
[ D0 (0.35 D0 ) 2 ] 11.073
4
Speed, N2 =
2
2
271.831 = [ D0 0.122 D0 ) 11.073
4
D02 = 35.602
Q2 =
D0 = 5.966 m
Speed of the Turbine
35.435 =
D0 N
60
5.966 N
60
Power, P2 =
Q1 H 2
H1
1.2 16
9
P1 H 23 2
H13 2
N P
NS =
H5 4
=
113.43 60,000
(a)
(b)
Answer :
Given that,
Head on turbine, H1 = 9 m
Power developed, P1 = 115 kW
Speed, N1 = 240 r.p.m
Discharge, Q1 = 1.2 m3/sec
115 (16)3 2
(9)3 2
P2 = 272.6 kW
(255 / 4 )
= 497.025 r.p.m
Q6.
240 16
Q2 = 1.6 m 3 sec
N = 113.43 r.p.m
(iii)
H1
Discharge,
u1 =
N1 H 2
N 2 = 320 r.p.m
(ii)
S.31
Answer :
Cavitation
For answer refer Unit-VI, Q21, Topic: Cavitation and
its Causes.
The occurrence of cavitation in turbines can be on
the suction surfaces of the runner blades. These blades due
to their dynamic action creates low pressure zones in a region
where the static pressure is less.
Generally the cavitation occurs and starts increasing,
if the head is low and the velocity is high at a region and the
elevations z of the turbine is kept too high above the tail
race.
( JNTU-Anantapur)
S.32
Turbines with horizontal shafts delivers low pressure at the upper part of the runner, which is the important part in
case of large machines. Cavitation occurs in this region of the turbine with horizontal shafts.
Generally cavitation mostly occur in the reaction turbines at the outlet of the runner or at inlet of the draft tube. The
pressure at the outlet of runner and inlet of draft tube is substantially reduced. Thus giving rise to the
occurrence of cavitation in these zones. The material at these zones is eaten gradually resulting in the poor efficiency of the
turbine.
Q7.
(a)
Draw a typical layout and explain the working of centrifugal pump. Also indicate various
components.
April/May-12, Set-4, Q7(a)
Answer :
For answer refer Unit-VII, Q3.
(b)
Explain the method of selection of centrifugal pumps with the aid of characteristic curves.
April/May-12, Set-4, Q7(b)
Answer :
For answer refer Unit-VII, Q23.
Q8.
(a)
Describe different structures associated with the intake to penstock. Discuss their uses.
April/May-12, Set-4, Q8(a)
Answer :
Following are the various structures associated with the intake to penstock,
1.
1.
Canal
2.
Forebay
3.
Surge tank.
Canal
Canals are artificial channel constructed in trapezoidal shape on ground surface. Canals are used as conveyance
system, which carry water from reservoir to the power house.
2.
Forebay
For answer refer April/May-12, Set-4, Q8(b)(i)
3.
Surge Tank
For answer refer Unit-V, Q2, Topic: Surge Tank.
River
Intake
Dam
Forebay
Penstock
Surge
tank
River
Canal
Power house
Tail Race
Figure: Showing the various Structures Associated with Intake to Penstock
( JNTU-Anantapur)
S.33
Forebay
Answer :
(i)
Forebay
Forebay are enlarged water body provided infront of the penstock. These are used when the length of penstock is
short and the head of water level is low. In the upstream side of the power house, if the load is suddenly reduced, then
forebay provide a small balancing storage in order to store the water rejected by the turbine temporarily and increase the
demand of load. At the same time the water in the penstock gets accelerated.
(ii)
Intake Structures
Intake structures are provided at the opening of canals or tunnels, in order to carry the water to power house. These
intake structure control the flow of water and stop the entry of boulders, logs of wood etc., into the conveyance passage.
Based on the type of power plant, the intake structure can be classified as,
(iii)
1.
Run-of-river intake
2.
Canal intake
3.
Dam intake
4.
Shaft intake.
Penstock
Penstock is a pipe which convey the water from reservoir to the turbine house. Penstock should be laid above the
hydraulic gradient at any point and should follow the shortest route. The principle for designing penstock is similar to the
principle of pressure vessels and tanks, along with that hammer pressure causes due to governor control is also taken into
consideration. As per ASME code, the thickness of penstock is given as,
t=
PR
+ 0.15
S (0.6 P )
Where,
t = Thickness of penstock in cm
P= Pressure (kg/m2)
R= Internal radius (cm)
S = Design stress (kg/cm2)
= Joint efficiency factor.
If the distance between forebay and power house is long, then a single penstock are used for carrying water.
Whereas, for shorter distance separate penstock need to be provided.
( JNTU-Anantapur)