527Elements of Hydroelectric Power Engineering

total electrical power produced. In the Indian scene, currently the ratio of hydro to other sources is about 23:77.

Hydropower has certain unique features that make it highly desirable. The chief among them is that hydropower is clean; with no or very little pollution footprints and is renewable. It is still the largest renewable energy system in use in practically all parts of the world. Some comparative features of hydro and thermal power plants are listed in Table 8.2.

Table 8.2 Comparative features of hydro and thermal power plants

Sl. No

Hydropower Thermal Power

1 Renewable. No (or negligible) fuel cost. Non-renewable source. High fuel cost.

2 Clean and safe energy source. No air pol-lution. No thermal pollution of water.

Heavy pollution and not always safe. Very high safety measures are needed in nuclear thermal plants. Air pollution in fossil-fuel plants. Thermal pollution of water bodies.

3 Ideally suited for peak power. Plant can be run up and synchronised in a few minutes.

Ideally suited for base power. Starting up does take large time.

4 High efficiency, about 90−95% overall energy conversion efficiency.

Efficiency of energy conversion is low with an average of about 35% and a maxi-mum of about 60%

5 High capital cost and very low operating cost. Sometimes, a part of the capital cost of the dam can be shared by making the dam a multipurpose project.

Moderate capital cost and very high oper-ating cost.

6 The locations are geographically fixed leading to large transmission costs.

Can be located in the load centres and transmission costs can be reduced

7 Each hydro plant is site-specific. Standardised modular designs are not possible.

Standardised modular thermal plant units are generally adopted. Leads to consider-able savings in time and cost.

8 Moderate environmental damage at the reservoir, canal and other components.

Heavy environmental damage at the source of the fuel as well as at the power plant and its neighbourhood. Large pollu-tion footprints.

9 Problems related to relocation of project-affected people at the reservoir site may be high.

Smaller problems of relocation of project-affected people.

10 The components are rugged and have long life. A hydro plant having an eco-nomical life of 50 years or more is common. However, sedimentation and related progressive loss of storage capac-ity is a concern.

High depreciation of the components due to the high temperatures and nature of the plant: needs frequent replacements of major components. Economic life of the plant is much smaller than that of a hydro plant.

11 Takes very long time for survey, design, mandatory approvals and construction.

Reasonably quick processing and con-struction time.

320 Hydraulic Machines

NOTEThe above expression for increase in piezometric head could be put in the following form of the familiar Bernoulli equation with the addition of a new term:

Adding 1

2 22

12

gV V( )-ÈÎ ˘̊ to both sides of the equation

p pZ Z

u u

g

v v

gr r2 1

2 122

12

22

12

2 2g g-

ÊËÁ

ˆ¯̃

+ -È

ÎÍ

˘

˚˙ =

--

-È

ÎÍÍ

˘

˚( )

( ) (˙̇˙

pZ

V

g

pZ

V

gH

gV V ueg g

+ +È

ÎÍÍ

˘

˚˙˙

- + +È

ÎÍÍ

˘

˚˙˙

= = - +2

2

2

1

22

12

22

2 2

1

2( ) ( -- - -ÈÎ ˘̊u v vr r1

22

21

2) ( )

pZ

v

g

pZ

v

gu ur r

g g+ +

È

ÎÍÍ

˘

˚˙˙

- + +È

ÎÍÍ

˘

˚˙˙

= -( )22

2

12

1

22

12

2 2

pZ

v

g

u

g

pZ

v

g

u

gr r

g g+ + -

È

ÎÍÍ

˘

˚˙˙

- + + -È

ÎÍÍ

˘

˚˙˙

2 2

2

2 2

12 2 2 2

pZ

v

g

u

g

pZ

v

g

u

gr r

g g+ + -

È

ÎÍÍ

˘

˚˙˙

- + + -È

ÎÍÍ

˘

˚˙˙

2 2

2

2 2

12 2 2 2

= constant

This equation is known as Bernoulli equation in rotating coordinates and applies to ideal incompressible flow.

*EXAMPLE 5.10In a centrifugal pump having a manometric efficiency of 0.7 and having the outer diameter equal to twice the inner diameter, show that the minimum outer diameter D2m of the impeller which will enable it to just start pumping water to a total head of H metres when rotating at a speed of N rpm is given by

DH

Nm281 7

=.

meters.

SolutionFor a centrifugal pump to just start pumping, if there are no losses, the centrifugal head must be equal to the actual head. Hence, under ideal conditions, the total head

H is related to the speed as Hu u

g=

-( )22

12

2. However, for actual pumps to account

for losses in the suction and delivery pipes, the manometer head is used to represent total head as H = Hm and in the pump, the manometer efficiency is used to account for losses in the pump as

( )u u

g

Hm

ma

22

12

2

-≥

h

Guided TourMore than 290 illustrations and diagrams are present to enhance the concepts.

Most chapters have a set of References for additional reading.

Useful Notes and Tables given at appropriate places to provide additional information

119Reaction Turbines-1: Francis Turbine

2. Simple Elbow-TypeIn this kind, a divergent pipe is bent at 90° to create an elbow. The divergence of the flow is restricted to 10°, and the 90° bend helps in restricting the vertical distance covered by the draft tube. The overall geometry is not the best, hydraulically, and hence the overall efficiency is average being 50−60% (Fig. 2.34). This type of draft tube is seldom used these days.

3. Elbow-type with Varying Cross-Sectional Shape and SizeThis is a modification of the simple elbow type and has the cross-sectional form varying from circular at the top of the draft tube to a rectangular diverging conduit at the lower end. The change of shape allows the flow to be expanded considerably and gradually. Figure 2.35 shows a simple elbow-type draft tube in which the vertical circular pipe is gradually enlarged into a rectangular cross section in the horizontal portion.

Figure 2.36 shows a well-designed, high-efficiency, elbow-type draft tube of varying cross-sectional shape. The change of the cross-section happens right at the top of the tube in a three-dimensional manner. The bottom rectangular section may

Fig. 2.34 Simple elbow-type draft tube

Tailwater

Fig. 2.35 Elbow-type draft tube with varying cross section

Hs

Turbine

CircularCrossSection

RectangularCross Section

Datum

Tailrace

y

Fig. 2.36 A well-designed elbow-type draft tube of varying cross section

D

Normal Tailwater Level

Lowest Permissible Level

Turbine

TurbineOutletLevel

Hs

A

B

C

E F G

H

A B C D

E F G H

84 Hydraulic Machines

Fig. 2.11(a) Definition sketch

Vu1

u1

a1b1

V1

or 1 Vf 1

b1

a1

Fig. 2.11(b) b1 > 90° , Velocity triangles of slow Francis

Outlet

V Vf f1 2=

u2

b2a2 = 90°

or 2

Vu1

V uu1 1–

b1 b'1

Vf 1or 1

V1

Inlet

u1

a1

Fig. 2.11(c) b1 c 90° ,Velocity triangles of normal Francis

a1

u2

b2

or 2

a2 = 90°

OutletInlet

b1 = 90°

V1

u V1 1= u

or f1 1= V

V =2 2Vf

Fig. 2.11(d) b1 < 90° , velocity triangles of fast Francis

u1

Vu1

a1

V1 Vf 1

or 1

b1V V2 2= f

u2

b2a2 = 90°

or2

OutletInlet

37Basic Principles

GuideVane

ScrollCase

From Penstock

Tailrace

Draft TubeScroll CaseRunner

Vane

ShroundRingh

GuideVane

ScrollCase

Main Shaft

Pivot

(a)

Fig. 1.23(a) Schematic sketch of a Francis turbine

BladesHub

WaterFlow

Turbine

Shaft

Generator

Rotor

Stator

GuideVanes

(b)

Fig. 1.23(b) Schematic sketch of a Kaplan turbine and generator set.

Development of electric generators and fast hydroturbines as prime movers of these generators led to the obsolescence of slow water wheels. Very good descriptions of earlier water wheels along with relevant illustrations are available on the Internet, for example at http://en.wikipedia.org/wiki/Water_wheel#Greco-Roman_world.

A modern hydraulic reaction turbine consists essentially of a rotating element, called runner, which is acted upon by water. The runner will have a series of blades. Appropriate casings (scroll casing) and guide mechanisms (guide vanes) are provided to have efficient entry of water to the rotor. The blades of the rotor are so shaped that they have efficient interaction with flowing water all the way from the inlet up to the exit. Figure 1.23(a) shows the basic layout and components of a reaction turbine known as Francis turbine. Water from a penstock enters the scroll casing, gets guided by the guide vanes and passes through the passage between the blades of the runner. While doing so the pressure of the water in the rotor changes. This type of turbine develops torque by reacting to the pressure of the fluid in a manner analogous to the rotation of a lawn sprinkler. The water that exits the runner passes through a divergent pipe known as draft tube and finally reaches the tailrace.

Another type of reaction turbine, known as Kaplan turbine consists essentially of a propeller acted upon by water under pressure. Figure 1.23(b) shows schematically the arrangements of such a turbine along with the generator.

The reaction turbines are very versatile in their capability to handle a wide range of discharges and heads and thus nearly 80% of water turbines in the world are of reaction type. The details of the reaction turbines are discussed in sufficient detail in chapters 2 and 3. The impulse turbines are described in detail in chapter 4.

(Image from Wikipedia: The copyright holder of this work has released this work into the public domain.)

275Impulse Turbines—Pelton Turbine, and Cross-flow and Banki Turbines

hh = 2 1 11

2C KKv e e

ee

( )( )

- +-

ÊËÁ

ˆ¯̃

= 2 1 11

22 2C KK

Cv ve ee

ee( )

( )- +

-ÊËÁ

ˆ¯̃

=

But e = u

V

K gH

C gH

K

Cu

v

u

v1

2

2= =

Hence, hh = 2 22CK

CC Kv

u

vv u= .

RefeRences

4.1 Anagnostopoulos, JS, Flow Modeling and Runner Design Optimization in Turgo Water Turbines, World Academy of Science, Engineering and Technology, 28, 2007.

• http//www.waset.org/journals/waset/v28/v28-38.pdf4.2 Dritina, P and Sallaberger, M, Hydraulic turbines − basic principles and State of

Art Computational Fluid Dynamics Applications Proc. of Inst. of Mech. Engineers, J. of Mech. Engg. Science, Vol. 213, Part–C−1999. pp. 85–102

• www.mecanica.eafit.edu.co/.../mg/...turbine/hydraulicturbines-drtina.pdf 4.3 ESHA, Guide on How to Develop a small Hydropower Plant–European Small

Hydropower Association–2004. • http//www.esha.be/fileadmin/esha_files/documents/publications/GUIDES/

GUIDE_SHP/GUIDE_SHP_EN.pdf • http //bit.ly/rmZA104.4 Keck, H, et al., Commissioning and Operation Experience with the World’s largest

Pelton turbines−Bieudron, • http//www.andritz.com/hydro-media-media-center-large-hydro-bieudron_

worlds_largest_pelton_en_1_.pdf4.5 USBR, USA, Mechanical Governors for Hydroelectric Units, Facilities, Instruc-

tions, Standards and Techniques,Vol. 2−3, USBR, USA, 2002 • www.usbr.gov/power/data/fist/fist2_3/vol2-3.pdf

Review Questions

4.1. Differentiate between the following with the help of neat sketches: (a) Net head in a Pelton turbine (b) Net head in a reaction turbine 4.2. Give at least two major functional differences between a Pelton turbine and a

reaction turbine. 4.3. Sketch the velocity triangles for the inlet and outlet of the buckets of a Pelton turbine

and label all the salient velocities and angles. Indicate clearly the velocity of whirl at inlet and outlet. Include a list of description of the symbols used in the diagram.

23Basic Principles

(a) (i) Stationary Plate: Consider the normal and tangential directions and a control volume as in Fig. 1.14 . Let Rn = Normal reaction on the control volume. Let b = Inclination of the jet axis with the normal to the plate.

Since the pressure is atmospheric on all the control volume surfaces, by linear momentum equation:

0 – Rn = rQ(0 – V cos b) where V = Velocity of the jet, and b = Inclination of the jet to the normal of the plate.

Rn = rQV cos b and the normal force on the plate is equal and opposite to R. Thus, Fn = rQV cos b in the inward normal direction as indicated in Fig. 1.14.

Here, b = 30°, Area A = p4

× (0.06)2 = 0.002827 m2 and V = 10 m/s.

Discharge Q = AV = 0.002827 × 10 = 0.02827 m3/s. Normal force on the plate, Fn = 998 × 0.2827 × 10 × cos 30° = 244.4 N Since there is no friction, there is no tangential component of the force on the

plate. Further, since u = 0, work done on the plate is zero. (ii) Moving Plate: (a) When u = 5 m/s equivalent relative motion with respect to a stationary plate is

to be considered. Relative velocity vr = (V – u) = (10 – 5) = 5 m/s Relative discharge Qr = Avr = 0.002827 × 5 = 0.01414 m3/s Normal force on the plate, Fn = rQr, vr cos q = 998 × 0.01414 × 5 × cos 30°

= 61.09 N Since there is no friction, there is no tangential component of the force on the

plate. Also, since u = 5 m/s, work done on the plate per second is = P = Fnu cos q P = 61.09 × 5 × cos30° = 264.5 W (b) When the plate is moving in opposite direction to that of the jet, u = – 5 m/s Relative velocity vr = (V + u) = (10 + 5) = 15 m/s Relative discharge Qr = Avr = 0.02827 × 15 = 0.0424 m3/s Normal force on the plate Fn = rQrvr cos b = 998 × 0.0424 × 15 × cos 30°

= 549.8 N Since there is no friction, there is no tangential component of the force on the

plate.

*EXAMPLE 1.3A two-dimensional free jet of water of thickness B and discharge q per unit width of the jet strikes a stationary, frictionless, plate at angle b to the normal to the plate. Calculate (a) the force on the plate, (b) the ratio of discharges in the two streams that move on the plate on either side of the impact zone, and (c) the force on the plate when b = 15°, B = 10 cm and q = 1.5 m3/s.

Solutionb is the angle made by the axis of the jet with the normal to the plate. (a) Consider a unit width of the jet. In the direction normal to the plate let Rn be the

reaction of the plate on the control volume enclosing the flow of water, (Fig. 1.15).

In each chapter, the theoretical

derivations and technical

descriptions are followed by a

set of carefully chosen and

graded Illustrative Examples

256 Hydraulic Machines

Euler head He = 1

gu(V1 – u) (1 + K cos bl2)

= 1

9 81.× 39.27 (75.186 – 39.27) (1 + 0.95 cos 15°)

= 275.7 m

Hydraulic efficiency h0 = H

He = =

275 7

3000 919

..

Overall efficiency h0 = (hmhh) = 0.919 × 0.95 = 0.873

Shaft power P = h0 g QH = 0.873 × 9.79 × 2.362 × 300 = 6057 kW

Specific speed Ns = N P

H5 4 5 4

300 6057

30018 7

/ /( ).= =

**EXAMPLE 4.8A double-jet Pelton wheel has specific speed of 26 and is required to deliver 10 MW of power. The turbine is supplied through a pipeline from a reservoir whose level is 400 m above the nozzles. Allowing 5% for friction loss in the pipe, calculate (a) speed in rpm, (b) diameter of the jets and (c) pitch diameter of the wheel, [Assume Ku = 0.46, Cv = 0.98 and h0 = 0.85].

SolutionNet available head H = 400 (1 – 0.05 ) = 380 m

Specific speed Ns = N P

H5 4/= 26

NOTEThat for multijet Pelton turbines the specific speed is calculated on power produced per jet.

Power per jet P = 10,000/2 = 5000 kW

Ns = 26 = N 5000

380 5 4( ) /

Operating speed N = 616.9 rpm

u = Ku 2gH = 0.46 × 2 9 81 380¥ ¥. = 39.72 m/s

Since u = p DN

60, the pitch diameter D =

60 60 39 72

616 91 23

¥=

¥¥

=u

Np p.

.. m

Total power developed = Pt = g h0 Qt H

1000 = 9.79 × 0.85 × Qt × 380

Total discharge Qt = 3.162 m3/s

387Centrifugal Pumps

Review Questions

5.1 What are manometric head and manometric efficiency of a centrifugal pump? 5.2 With the aid of neat sketches, describe three kinds of casings adopted in centrifugal

pumps. 5.3 Explain different types of energy losses encountered in centrifugal pumps. 5.4 Explain different types of efficiencies used in connection with centrifugal pumps. 5.5 Derive Euler’s equation for a centrifugal pump and explain the physical signifi-

cance of each of the terms in the equation. 5.6 How are pumps classified? List the further subclassifications of centrifugal pumps. 5.7 Explain with appropriate sketches the components of a typical centrifugal pump. 5.8 Draw the inlet and outlet velocity diagram of a centrifugal pump with radial inlet

velocity. Label the various velocities and angles involved. 5.9 Describe the need for priming of centrifugal pumps. How is it achieved? 5.10 Distinguish between Euler head and manometric head of a centrifugal pump. 5.11 Describe a multistage centrifugal pump. 5.12 Explain why the head-discharge curve of an ideal centrifugal pump is linear and

that of an actual pump is non-linear. 5.13 Explain the head-discharge characteristics of a centrifugal pump. Discuss the slip

factor. 5.14 Discuss the main and operating characteristics of a centrifugal pump. What is the

importance of constant efficiency curves?

5.15 Describe the salient features of a mixed-flow pump.

5.16 Explain the chief characteristics of an axial-flow pump.

5.17 Sketch typical performance characteristics curves of an axial flow pump. Use per-centages of best efficiency, best efficiency head and best efficiency Power in the y-axis and percentage best efficiency flow in the x-axis.

5.18 Compare the operating characteristics of radial-flow, axial-flow and mixed-flow pumps.

5.19 Explain with a sketch the variation of peak efficiency of a centrifugal pump as a function of specific speed and peak discharge.

5.20 Draw a sketch showing the variation of peak efficiency of large pumps with spe-cific speed. Mark clearly the zones of operation of radial flow, mixed flow and axial flow pumps.

5.21 Distinguish between NPSHR and NPSHA.

5.22 Explain how the operating point of a pump–pipeline system is determined when

(a) a single pump is working

(b) series duplex pumps are working

(c) parallel duplex pumps are working.

5.23 Write short notes on:

(a) Double suction pump

(b) Booster pump

(c) Priming of pumps

(d) Nondimensional specific speed of centrifugal pumps

551Elements of Hydroelectric Power Engineering

= 0.2 × 33.6 × 24 × 3600 = 0.5806 Mm3

Total pondage required = S1 + S2 = 0.2592 + 0.5806 = 0.8398 Mm3

(a) Maximum allowed water-level fluctuation in the pond = 0.8 m

Area of pond = Volume of pond

Permitted change in water level= 0 8398

0 8

.

. = 1.0498 Mm2

Average head during the week = 8.0 + 0 80

2

. = 8.40 M

(b) Peak power generated = P = hg QH = 0.72 × 9.79 × 33.6 × 8.40 = 1990 kW = 1.99 MW

Weekly energy output = 1.99 × 5 × 20 × 0.60 = 119.4 MWh

***EXAMPLE 8.4A run-of-river plant supplies power to a variable load as indicated below:

Time 0−4 hours

4−8 hours

8–12 hours

12–16 hours

16–20 hours

20−24 hours

Load in MW

2.1 3.2 7.4 8.2 6.2 4.1

(a) Draw the load curve and determine (i) average load, and (ii) daily load factor. (b) If the net head is 8.0 m and overall efficiency is 82%, determine the average flow required and the pondage required to meet daily load fluctuation.

Solution

Average load = (2.1 + 3.2 + 7.4 + 8.2 + 6.2 + 4.1) = 31 2

6

.= 5.2 MW

Daily load factor = Average load in a day

Peak load in a day= 5 2

8 2

.

.= 0.634 = 63.4%

The load curve is plotted from given data and is shown in Fig. 8.12.

Fig. 8.12 Load curve

0–4 h 4–8 h 8–12 h 12–16 h 16–20 h 20–24 hTime Interval

0.0

1.0

2.0

3.0

4.0

5.0

6.0

7.0

8.0

9.0

Average Load

5.2 MW

Peak LoadL

oad

(MW

)

3.2

6.2

4.1

8.27.4

2.1

Review Questions to test the student’s subjective grasp on the topics

Numerical problems

for practice

Objective-type questions for quick revision and recapitulation of concepts

144 Hydraulic Machines

2.23 What are the main features in classifying the runners of a Francis turbine as slow, medium and fast?

Problems

1. [All Problems and Objective Questions have been graded in three levels: − Simple, Medium and Difficult. The markings for these are: Simple = *, Medium = **, Difficult = ***]

Velocity Triangles P2.1 * At what angle should the wicket gates of a Francis turbine be set to extract

8000 kW of power from a flow of 30 m3/s when running at a speed of 200 rpm? The diameter of the runner at inlet is 3.0 m and the breadth of the openings at inlet is 0.9 m. The flow can be assumed to leave the runner radially and the blade angle is obtuse. [Ans: a1 = 22.58°]

P2.2 * An inward-flow reaction turbine has inlet and outlet diameters of 1.2 m and 0.6 m respectively. The breadth at the inlet is 0.25 m and at the outlet, it is 0.35 m. At a speed of rotation of 250 rpm, the relative velocity at the entrance is 3.5 m/s and it is radial. Calculate the (a) absolute velocity at the entrance and its inclination to the tangent of the runner, (b) discharge, and (c) the velocity of flow at the outlet.

[Ans: a1 = 12.56°, Q = 3.299 m3/s, Vf 2 = 5.0 m/s]

P2.3 * A reaction turbine works under a net head of 12.0 m. The guide-vane angle and the inlet-blade angle are 20° and 60° respectively. Assuming radial discharge at the outlet and constant flow velocity, determine the hydraulic efficiency of the turbine. [Ans: hh = 94.8%]

P2.4 * An inward flow reaction turbine has an outer diameter of 1.2 m and an inner diameter of 0.6 m. The breadth at the inlet is 0.20 m and it is 0.3 m at the outlet. Water enters the turbine with a flow velocity of 5.0 m/s. (a) Calcu-late the discharge and velocity of flow at the outlet. Assume vane thickness coefficient = 0.95 at both inlet and outlet. (b) If the speed of the turbine is 300 rpm, calculate the blade angle at the outlet. Assume the discharge to be radial at the outlet.

[Ans: (a) Q = 3.5814 m3/s, Vf2= 6.67 m/s, (b) b2 = 19.486°]

P2.5 * For a Francis turbine with radial entry and radial exit, the ratio of diameter at inlet to the diameter at outlet is Dr . Show that the exit blade angle b2 is related to the inlet guide vane angle a1 as

tan b2 = Dr tan a1.

P2.6 * An inward-flow reaction turbine has a runner of 1.2 m outer diameter and 0.6 m inner diameter. The blades occupy 5% of the peripheral area and the widths of the blades are 25 cm and 30 cm at the inlet and outlet respectively. If a discharge of 3.0 m3/s enters radially, determine the flow velocities at the inlet and outlet of the runner. What is the inlet-vane angle required to have a speed of 300 rpm? [Ans: Vf1 = 3.35 m/s, Vf 2 = 5.58 m/s, a1 = 10.078°]

468 Hydraulic Machines

Objective-type Questions

O6.1 * In positive-displacement pumps, the slip can sometimes be negative when the actual discharge is greater than the theoretical discharge. This happens in

(a) small suction pipes coupled with low delivery head

(b) small suction pipes coupled with medium delivery head

(c) long suction pipes coupled with low delivery head

(d) long suction pipes coupled with medium delivery head

O6.2 * In a reciprocating pump without air vessels, the friction head in the delivery pipe is maximum at crank angle q =

(a) 0° (b) 90° (c) 60° (d) 180°

O6.3 * In a reciprocating pump without air vessels, the acceleration head in the suction pipe is maximum at the crank angle value of q =

(a) 0° (b) 90° (c) 60° (d) 180°

O6.4 ** In a single-acting reciprocating pump, without air vessels, the average veloc-ity in delivery pipe is given by Vd =

(a) A

Ar

d

ÊËÁ

ˆ¯̃

w (b) A

A

r

d

ÊËÁ

ˆ¯̃

wp

(c) A

A

r

d

ÊËÁ

ˆ¯̃

pw2

(d) A

A

r

d

ÊËÁ

ˆ¯̃

wp2

where w = Angular velocity of the crank, r = Radius of crank, A = Area of cylinder, Ad = Area of delivery pipe.

O6.5 ** In a single reciprocating pump without air vessels, the ratio of the time-averaged friction head to the instantaneous maximum friction head in delivery

pipe is given by h

hfdav

fdm

=

(a) 1/2 (b) 1/3 (c) 2/3 (d) 4/3

O6.6 * The indicator diagram of a reciprocating pump is a plot of (a) work done vs stroke length (b) acceleration head vs stroke length (c) pressure head vs stroke length (d) crank speed vs power developed

O6.7 ** A double-stoke reciprocating pump has its stroke length reduced by half and the speed is doubled. This would cause the discharge to

(a) decrease by 25% (b) remain unaltered (c) decrease by 50% (d) increase by 100%

O6.8 ** Figure 6.19 shows an indicator diagram of a single-acting reciprocating pump. This indictor diagram belongs to

(a) a reciprocating pump with two air vessels, one each in suction pipe and delivery pipe at some distance from the cylinder

(b) a reciprocating pump with one air vessel (c) a reciprocating pump with two ideal air vessels (d) a reciprocating pump with no air vessels