1. No category

Fluid Systems - Model Paper 2

Model Question Paper 2
Fluid Systems
Prepared By: Kumar Ankur
Note: Attempt any five questions including Q. No. 1 which is compulsory.
Q 1. Define unit quantities?
Ans. Three types of unit quantities are:1. Unit speed:- it is defined as the speed of a turbine working under a unit head(i.e., under a
head of 1m). It is denoted by β€˜β€²π‘π‘’β€² .
π‘‘β„Žπ‘’ 𝑒π‘₯π‘π‘Ÿπ‘’π‘ π‘ π‘–π‘œπ‘› π‘“π‘œπ‘Ÿ 𝑒𝑛𝑖𝑑 𝑠𝑝𝑒𝑒𝑑 (𝑁𝑒 )𝑖𝑠 π‘œπ‘π‘‘π‘Žπ‘–π‘›π‘’π‘‘ π‘Žπ‘  ∢
𝑙𝑒𝑑 𝑁 = 𝑠𝑝𝑒𝑒𝑑 π‘œπ‘“ π‘Ž π‘‘π‘’π‘Ÿπ‘π‘–π‘›π‘’ π‘’π‘›π‘‘π‘’π‘Ÿ π‘Ž β„Žπ‘’π‘Žπ‘‘ 𝐻,
𝐻 = β„Žπ‘’π‘Žπ‘‘ π‘’π‘›π‘‘π‘’π‘Ÿ π‘€β„Žπ‘–π‘β„Ž π‘Ž π‘‘π‘’π‘Ÿπ‘π‘–π‘›π‘’ 𝑖𝑠 π‘€π‘œπ‘Ÿπ‘˜π‘–π‘›π‘”,
𝑒 = π‘‘π‘Žπ‘›π‘”π‘’π‘›π‘‘π‘–π‘Žπ‘™ π‘£π‘’π‘™π‘œπ‘π‘–π‘‘π‘¦
π‘’βˆπ‘‰
π‘€β„Žπ‘’π‘Ÿπ‘’ 𝑉 ∝ √𝐻
… … (1)
∝ √𝐻
π‘‘π‘Žπ‘›π‘”π‘’π‘›π‘‘π‘–π‘Žπ‘™ π‘£π‘’π‘™π‘œπ‘π‘–π‘‘π‘¦(𝑒)𝑖𝑠 𝑔𝑖𝑣𝑒𝑛 𝑏𝑦
𝑒=
πœ‹π·π‘
60
π‘€β„Žπ‘’π‘Ÿπ‘’ 𝐷 = π‘‘π‘–π‘Žπ‘šπ‘’π‘‘π‘’π‘Ÿ π‘œπ‘“ π‘‘π‘’π‘Ÿπ‘π‘–π‘›π‘’
π‘“π‘œπ‘Ÿ π‘Ž π‘‘π‘’π‘Ÿπ‘π‘–π‘›π‘’ , π‘‘β„Žπ‘’ π‘‘π‘–π‘Žπ‘šπ‘’π‘‘π‘’π‘Ÿ (𝐷)𝑖𝑠 π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘.
∴
𝑒 ∝ 𝑁 π‘œπ‘Ÿ 𝑁 ∝ 𝑒 π‘œπ‘Ÿ 𝑁 ∝ √𝐻
∴
𝑁 = 𝐾1 √𝐻
π‘€β„Žπ‘’π‘Ÿπ‘’ 𝐾1 𝑖𝑠 π‘Ž π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘ π‘œπ‘“ π‘π‘œπ‘Ÿπ‘π‘Ÿπ‘œπ‘‘π‘–π‘œπ‘›π‘Žπ‘™π‘–π‘‘π‘¦.
𝐼𝑓 𝐻 = 1, 𝑁 = 𝑁𝑒
𝑠𝑒𝑏𝑠𝑖𝑑𝑒𝑑𝑖𝑛𝑔 π‘‘β„Žπ‘’π‘ π‘’ π‘£π‘Žπ‘™π‘’π‘’π‘  𝑖𝑛 π‘’π‘ž(2) … 𝑀𝑒 𝑔𝑒𝑑
𝑁𝑒 = 𝐾1 √𝐻 = 𝐾1
𝑠𝑒𝑏𝑠𝑖𝑑𝑒𝑑𝑖𝑛𝑔 π‘‘β„Žπ‘’ π‘£π‘Žπ‘™π‘’π‘’ π‘œπ‘“ 𝑁𝑒 𝑖𝑛 π‘’π‘ž(2) …
𝑁 = 𝑁𝑒 √𝐻
π‘“π‘Ÿπ‘œπ‘š. . (1)
2. Unit discharge:-It is defined as the discharge passing through a turbine , which is
working under a unit head(i.e., 1m). it is denoted by symbol ′𝑄𝑒′
𝑙𝑒𝑑 𝐻 = β„Žπ‘’π‘Žπ‘‘ π‘œπ‘“ π‘€π‘Žπ‘‘π‘’π‘Ÿ π‘œπ‘› π‘‘β„Žπ‘’ π‘‘π‘’π‘Ÿπ‘π‘–π‘›π‘’ ,
𝑄 = π‘‘π‘–π‘ π‘β„Žπ‘Ÿπ‘”π‘’ π‘π‘Žπ‘ π‘ π‘–π‘›π‘” π‘‘β„Žπ‘Ÿπ‘œπ‘’π‘”β„Ž π‘‘π‘’π‘Ÿπ‘π‘–π‘›π‘’ π‘€β„Žπ‘’π‘› β„Žπ‘’π‘Žπ‘‘ 𝑖𝑠 𝐻 π‘œπ‘› π‘‘β„Žπ‘’ π‘‘π‘’π‘Ÿπ‘π‘–π‘›π‘’
π‘Ž = π‘Žπ‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘“π‘™π‘œπ‘€ π‘œπ‘“ π‘€π‘Žπ‘‘π‘’π‘Ÿ
π‘‘β„Žπ‘’ π‘‘π‘–π‘ π‘β„Žπ‘Žπ‘Ÿπ‘”π‘’ π‘π‘Žπ‘ π‘ π‘–π‘›π‘” π‘‘β„Žπ‘Ÿπ‘œπ‘’π‘”β„Ž π‘Ž 𝑔𝑖𝑣𝑒𝑛 π‘’π‘›π‘‘π‘’π‘Ÿ π‘Ž β„Žπ‘’π‘Žπ‘‘
β€²
𝐻 β€² 𝑖𝑠 𝑔𝑖𝑣𝑒𝑛 𝑏𝑦,
𝑄 = π‘Žπ‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘“π‘™π‘œπ‘€ × π‘£π‘’π‘™π‘œπ‘π‘–π‘‘π‘¦
𝑏𝑒𝑑 π‘“π‘œπ‘Ÿ π‘Ž π‘‘π‘’π‘Ÿπ‘π‘–π‘›π‘’, π‘Žπ‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘“π‘™π‘œπ‘€ 𝑖𝑠 π‘π‘›π‘ π‘‘π‘Žπ‘›π‘‘ π‘Žπ‘›π‘‘ π‘£π‘’π‘™π‘œπ‘π‘–π‘‘π‘¦ 𝑖𝑠 π‘π‘Ÿπ‘œπ‘π‘œπ‘Ÿπ‘‘π‘–π‘œπ‘›π‘Žπ‘™ π‘‘π‘œβˆšπ».
∴
𝑄 ∝ π‘£π‘’π‘™π‘œπ‘π‘–π‘‘π‘¦ ∝ √𝐻
π‘œπ‘Ÿ
𝑄 = 𝐾2 √𝐻
… . . (1)
π‘€β„Žπ‘’π‘Ÿπ‘’ 𝐾2 𝑖𝑠 π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘‘ π‘œπ‘“ π‘π‘Ÿπ‘œπ‘π‘œπ‘Ÿπ‘‘π‘–π‘œπ‘›π‘Žπ‘™π‘–π‘‘π‘¦.
𝑖𝑓
𝐻 = 1, 𝑄 = 𝑄𝑒
𝑠𝑒𝑏𝑠𝑒𝑑𝑒𝑑𝑖𝑛𝑔 π‘‘β„Žπ‘’π‘ π‘’ π‘£π‘Žπ‘™π‘’π‘’π‘  𝑖𝑛 π‘’π‘ž(1), 𝑀𝑒 𝑔𝑒𝑑
𝑄𝑒 = 𝐾2 √1.0 = 𝐾2
𝑠𝑒𝑏𝑠𝑖𝑑𝑒𝑑𝑖𝑛𝑔 π‘‘β„Žπ‘’ π‘£π‘Žπ‘™π‘’π‘’ π‘œπ‘“ π‘˜2 𝑖𝑛 π‘’π‘ž(1), 𝑀𝑒 𝑔𝑒𝑑
𝑄 = 𝑄𝑒 √𝐻 π‘œπ‘Ÿ 𝑄𝑒 =
𝑄
√𝐻
3. Unit power:- IT is defined as the power developed by a turbine , working under a unit
head (i.e., under a head or 1m). it is denoted by symbol ′𝑃𝑒′ .
𝑙𝑒𝑑 𝐻 = β„Žπ‘’π‘Žπ‘‘ π‘œπ‘“ π‘€π‘Žπ‘‘π‘’π‘Ÿ π‘œπ‘› π‘‘β„Žπ‘’ π‘‘π‘’π‘Ÿπ‘π‘–π‘›π‘’ ,
𝑃 = π‘π‘œπ‘€π‘’π‘Ÿ π‘‘π‘’π‘£π‘’π‘™π‘œπ‘π‘’π‘‘ 𝑏𝑦 π‘‘β„Žπ‘’ π‘‘π‘’π‘Ÿπ‘π‘–π‘›π‘’ π‘’π‘›π‘‘π‘’π‘Ÿ π‘Ž β„Žπ‘’π‘Žπ‘‘ π‘œπ‘“ 𝐻,
𝑄 = π‘‘π‘–π‘ π‘β„Žπ‘Žπ‘Ÿπ‘”π‘’ π‘‘β„Žπ‘Ÿπ‘œπ‘’π‘”β„Ž π‘‘π‘’π‘Ÿπ‘π‘–π‘›π‘’ π‘’π‘›π‘‘π‘’π‘Ÿ π‘Ž β„Žπ‘’π‘Žπ‘‘ 𝐻
π‘‘β„Žπ‘’ π‘œπ‘£π‘’π‘Ÿπ‘Žπ‘™π‘™ 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦(ƞ)𝑖𝑠 𝑔𝑖𝑣𝑒𝑛 π‘Žπ‘ 
∴
ƞ=
∴
π‘π‘œπ‘€π‘’π‘Ÿ π‘‘π‘’π‘£π‘’π‘™π‘œπ‘π‘’π‘‘
𝑃
=
𝜌×𝑔×𝑄×𝐻
π‘€π‘Žπ‘‘π‘’π‘Ÿ π‘π‘œπ‘€π‘’π‘Ÿ
1000
𝜌×𝑔×𝑄×𝐻
𝑃 =ƞ×
1000
βˆπ‘„×𝐻
∝ √𝐻 × π»
3
∝ 𝐻2
3
∴
𝑃 = 𝐾3 𝐻 2
π‘€β„Žπ‘’π‘Ÿπ‘’πΎ3 𝑖𝑠 π‘Ž π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘ π‘œπ‘“ π‘π‘Ÿπ‘œπ‘π‘œπ‘Ÿπ‘‘π‘–π‘œπ‘›π‘Žπ‘™π‘–π‘‘π‘¦.
π‘€β„Žπ‘’π‘›
𝐻 = 1π‘š 𝑃 = 𝑃𝑒
3
∴ 𝑃𝑒 = 𝐾3 (1)2 = π‘˜3
𝑠𝑒𝑏𝑠𝑖𝑑𝑒𝑑𝑖𝑛𝑔 π‘‘β„Žπ‘’ π‘£π‘Žπ‘™π‘’π‘’ π‘œπ‘“ 𝐾3 𝑖𝑛 π‘’π‘ž(1)
3
𝑃 = 𝑃𝑒 𝐻 2
∴ 𝑃𝑒 =
𝑃
3
𝐻2
Q 2.what is cavitation ? write down the effects and precaution of cavitation?
Ans. Cavitation is defined the phenomenon of formation of vapour bubbles of a flowing liquid
in a region where the pressure of the liquid falls below its vapour pressure and sudden collapsing
of these vapour bubbles in a region of higher pressure . when the vapour bubbles collapse, a very
high pressure is created. The metallic surfaces , above which these vapour bubbles collapse, is
subjected to these high pressures, which cause pitting action on the surface . thus cavities are
formed on the metallic surface and also considerable noise and vibrations are produced.
Effects :1. The metallic surfaces are damged and cavities are formed on the surfaces.
2. Due to sudden collapse of vapour bubble, considerable noise and vibrations are produced.
3. The efficiency of a turbine decreases due to cavitation .
Precaution:-
1. The pressure of the flowing liquid in any part of the hydraulic system should not be
allowed to fall below its vapour pressure. If the flowing liquid is water, then the absolute
pressure head should not be below 2.5m of water.
2. The special materials or coatings such as aluminium-bronze and stainless steel, which are
cavitation resistant materials, should bs used.
Q 3.. A one –fifth scale model of a pump was tested in a laboratory at 1000r.p.m. the head
developed and the power input at the best efficiency point were found to be 8m and 30kW
respectively. If the prototype pump has to work against a head of 25m, determine its working
speed, the power required to drive it and the ratio of the flow rates handled by the two pumps.
Ans. One-fifthe model means that the ratio of linear dimensions of a model and its prototype is
equal to 1/5.
𝑠𝑝𝑒𝑒𝑑 π‘œπ‘“ π‘šπ‘œπ‘‘π‘’π‘™,
π‘π‘š = 1000π‘Ÿ. 𝑝. π‘š.
β„Žπ‘’π‘Žπ‘‘ π‘œπ‘“ π‘šπ‘œπ‘‘π‘’π‘™,
π»π‘š = 8π‘š
π‘π‘œπ‘€π‘’π‘Ÿ π‘œπ‘“ π‘šπ‘œπ‘‘π‘’π‘™,
π‘ƒπ‘š = 30π‘˜π‘Š
β„Žπ‘’π‘Žπ‘‘ π‘œπ‘“ π‘π‘Ÿπ‘œπ‘‘π‘œπ‘‘π‘¦π‘π‘’,
𝑙𝑒𝑑
𝐻𝑝 = 25π‘š
𝑁𝑝 = 𝑠𝑝𝑒𝑒𝑑 π‘œπ‘“ π‘π‘Ÿπ‘œπ‘‘π‘œπ‘‘π‘¦π‘π‘’
𝑃𝑝 = π‘π‘œπ‘€π‘’π‘Ÿ π‘œπ‘“ π‘π‘Ÿπ‘œπ‘‘π‘œπ‘‘π‘¦π‘π‘’
𝑄𝑝 = π‘“π‘™π‘œπ‘€ π‘Ÿπ‘Žπ‘‘π‘’ π‘œπ‘“π‘π‘Ÿπ‘œπ‘‘π‘œπ‘‘π‘¦π‘π‘’
π‘„π‘š = π‘“π‘™π‘œπ‘€ π‘Ÿπ‘Žπ‘‘π‘’ π‘œπ‘“ π‘šπ‘œπ‘‘π‘’π‘™
(𝑖) 𝑠𝑝𝑒𝑒𝑑 π‘œπ‘“ π‘π‘Ÿπ‘œπ‘‘π‘œπ‘‘π‘¦π‘π‘’
√𝐻
√𝐻
( ) =( )
𝐷𝑁 π‘š
𝐷𝑁 𝑝
π‘œπ‘Ÿ
βˆšπ»π‘š
βˆšπ»π‘
=
π·π‘š π‘π‘š 𝐷𝑝 𝑁𝑝
π‘œπ‘Ÿ
𝑁𝑃 =
=
βˆšπ»π‘ƒ
βˆšπ»π‘š
×
π·π‘š
× π‘π‘€
𝐷𝑝
1
× × 1000
√8 5
√25
= 353.5 π‘Ÿ. 𝑝. π‘š. π‘Žπ‘›π‘ .
(𝑖𝑖)π‘π‘œπ‘€π‘’π‘Ÿ π‘‘π‘’π‘£π‘’π‘™π‘œπ‘π‘’π‘‘ 𝑏𝑦 π‘π‘Ÿπ‘œπ‘‘π‘œπ‘‘π‘¦π‘π‘’
𝑃
𝑃
( 5 3) = ( 5 3)
𝐷 𝑁 π‘š
𝐷 𝑁 𝑝
π‘œπ‘Ÿ
π‘œπ‘Ÿ
𝑃𝑝
π‘ƒπ‘š
=
5 3
π·π‘š
π‘π‘š 𝐷𝑝5 𝑁𝑝3
𝐷𝑝 5
𝑁𝑝 3
353.5 3
𝑃𝑝 = π‘ƒπ‘š × ( ) × ( ) = 30 × 55 × (
)
π·π‘š
π‘π‘š
1000
= 30 × 3125 × 0.04419 = 4143π‘˜π‘Š. π‘Žπ‘›π‘ .
(𝑖𝑖𝑖)π‘Ÿπ‘Žπ‘‘π‘–π‘œ π‘œπ‘“ π‘‘β„Žπ‘’ π‘“π‘™π‘œπ‘€ π‘Ÿπ‘Žπ‘‘π‘’π‘  π‘œπ‘“ π‘‘π‘€π‘œ π‘π‘’π‘šπ‘π‘ (𝑖. 𝑒. , π‘šπ‘œπ‘‘π‘’π‘™ π‘Žπ‘›π‘‘ π‘π‘Ÿπ‘œπ‘‘π‘œπ‘‘π‘¦π‘π‘’)
𝑄
𝑄
( 3 ) =( 3 )
𝐷 𝑁 π‘š
𝐷 𝑁 𝑝
π‘œπ‘Ÿ
π‘œπ‘Ÿ
𝑄𝑝
π‘„π‘š
=
3
π·π‘š
π‘π‘š 𝐷𝑝3 𝑁𝑝
𝑄𝑝
𝐷𝑝3 𝑁𝑝
𝐷𝑝 3 𝑁𝑃
𝐷𝑝 5
353.5
= 3
=( ) ×
= 5×
(∴
= )
π‘„π‘š π·π‘š π‘π‘š
π·π‘š
π‘π‘š
1000
π·π‘š 1
= 44.1875 π‘Žπ‘›π‘ .
Q 4.
Q 5. Derive the expression for specific speed of a turbine?
Q 6.Explain Hydroelectric power plants with neat sketch.
Sol. Introduction The purpose of a hydroelectric power plant is to harness power from water
flowing under pressure. As such it incorporates a number of water driven prime-movers known
as water turbines. Water flowing under pressure has two forms of energy kinetic and potential.
The kinetic energy depends on the mass of water flowing and its velocity while the potential
energy exists as result of the difference in water level between two points which is known as
"head". The water or hydraulic turbine, as it is sometimes named, converts the kinetic and
potential energies possessed by water into mechanical power.
Head and flow rate or discharge Head is the difference in elevation between two levels of
water. The head of a hydroelectric power plant is entirely dependent on the topographical
conditions. Head can be characterized as: gross head, and net or effective head.
Gross head Is defined as the difference in elevation between the head race level at the intake and
the tail race level at the discharge side, naturally, both the elevations have to be measured
simultaneously. The gross head may vary as both the elevations of water do not remain the same
at all times. It is essential to known the maximum and minimum as well as the normal values of
the gross head. The normal value would be that for which the plant works most of the time. In
rainy season the flood may raise the elevation of tail race, thus, reducing the gross head. On the
other hand at the time of draught the same may be increased. Net or effective head Is the head
obtained by subtracting from gross head all losses in carrying water from the head race to the
entrance of the turbine. The losses are due to friction occurring in tunnels, canals and penstocks
which lead the water into the turbine. Net or effective head is, therefore, the true pressure
difference between the entrance to the turbine casing and the tail race water elevation.
Flow rate or discharge of water It is the quantities of water used by the water turbine in unit
time and is generally measured in (m3/s) or ( l/s).
Essential components of hydroelectric power plant.
Storage reservoir The water available from a catchment area is stored in a reservoir, so that it
can be utilized to run the turbines for producing electric power according to the
requirement through out the year. The storage reservoir may be natural or artificial.
Dam with its control works Dam is a structure erected on suitable site to provide for the
storage of water and to create head. Dam may be built to make an artificial reservoir from
a valley or it may be erected in a river to control the flowing water. Structures and
appliances to control the supply of water from the storage reservoir through the dam, are
known as control works or head works. The principal elements of control works are: a.
Gates and valves. b. Structures necessary for their operation. c. Devices for the protection
of gates and hydraulic machines, which consist of: i. Trashracks: They are made up of a
row of rectangular cross sectional structural steel bars placed across the intake opening is
an inclined position. They are used to obstruct debris from going into the intake. ii.
Debris cleaning device fitted on the trashrack. iii. Heating element against ice troubles.
Waterways with their control works.
Is a passage through which the water is carried from the storage reservoir to the power house. It
may consist of tunnels, canals, forebays and pipes ( i.e., penstocks) as shown in figure below.
The control works for the tunnels, canals, forebays and pipes may be different types of gates in
additional to these, surge tank which is reservoir fitted at some opening made on a long pipe line
to receive the rejected flow when the pipeline is suddenly closed by a valve at its steep end. The
surge tank, therefore, controls the pressure variations resulting from the rapid changes in pipeline
flow thus eliminating water hammer effects.
Power house Is a building to house the turbines, generators and other accessories for operating
the machines.
Tail race Is a waterway to conduct the water discharged from the turbines to a suitable point
where it can be safely disposed of or stored to be pumped back into the original reservoir.
Generation and transmission of electric power It consists of electrical generating machines,
transformers, switching equipments and transmission lines.
Q 7.Explain the construction of reaction turbine
Francis turbine
Main components
β€’ Penstock
Penstock is a waterway to carry water from the reservoir to the turbine casing. Trashracks are
provided at the inlet of penstock in order to obstruct the debris entering in it.
β€’ Casing
The water from penstocks enter the casing which is of spiral shape. In order to distribute the
water around the guide ring evenly, the area of cross section of the casing goes on decreasing
gradually. The casing is usually made of concrete, cast steel or plate steel.
β€’ Guide vanes
The stationary guide vanes are fixed on stationary circular wheel which surrounds the runner.
The guide vanes allow the water to strike the vanes fixed on the runner without shock at the
inlet. This fixed guide vanes are followed by adjustable guide vanes. The cross sectional area
between the adjustable vanes can be varied for flow control at part load.
β€’ Runner
It is circular wheel on which a series of radial curved vanes are fixed. The water passes into
the rotor where it moves radially through the rotor vanes and leaves the rotor blades at a
smaller diameter. Later, the water turns through 90ointo the draft tube.
β€’ Draft tube
β€’ The pressure at the exit of the rotor of a reaction turbine is generally less than the
atmospheric pressure. The water at exit can not be directly discharged to the tail race. A
tube or pipe of gradually increasing area is used for discharging the water from the
turbine exit to the tail race. In other words, the draft tube is a tube of increasing cross
sectional area which converts the kinetic energy of water at the turbine exit into pressure
energy.
Q 8. Describe the classificationof centrifugal pump.
Sol. Centrifugal pumps can be classified according to :
a) Working head β€’ Low lift centrifugal pumps (up to 15 m).
β€’ Medium lift centrifugal pumps (15-40 m).
β€’ High lift centrifugal pumps (above 40 m).
b) Type of casing β€’ Volute pump.
β€’ Turbine pump or diffusion pump
c) Number of impeller β€’ Single stage centrifugal pump.
β€’ Multi stage centrifugal pump.
d) Number of entrances to the impeller β€’ Single entry.
β€’ Double entry.
e) Disposition of shaft β€’ Horizontal.
β€’ Vertical.
f) Liquid handled.
g) Specific speed.
h) Non-dimensional factor ks.
The specific speed Nsis a dimensional quantity, but a ksis a non-dimensional quantity.
Where Q= flow rate. (m3/s)
N= speed. (rpm)
V= velocity of water ( m/s) ,
H= total head.

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