REGULATION : 2017 ACADEMIC YEAR : 2018-2019 JIT-JEPPIAAR/MECH /Mr.D.Arunkumar & Mr.M.K.Karthik/II rd Yr/SEM 03 /CE8394/FLUID MECHANICS AND MACHINERY/UNIT 1-5/QB+Keys/Ver1.0 3- 1 CE8394 FLUID MECHANICS AND MACHINERY L T P C 4 0 0 4 OBJECTIVES 1. The properties of fluids and concept of control volume are studied 2. The applications of the conservation laws to flow through pipes are studied. 3. To understand the importance of dimensional analysis 4. To understand the importance of various types of flow in pumps. 5. To understand the importance of various types of flow in turbines. UNIT I FLUID PROPERTIES AND FLOW CHARACTERISTICS 12 Units and dimensions- Properties of fluids- mass density, specific weight, specific volume, specific gravity, viscosity, compressibility, vapor pressure, surface tension and capillarity. Flow characteristics – concept of control volume - application of continuity equation, energy equation and momentum equation. UNIT II FLOW THROUGH CIRCULAR CONDUITS 12 Hydraulic and energy gradient - Laminar flow through circular conduits and circular annuli-Boundary layer concepts – types of boundary layer thickness – Darcy Weisbach equation –friction factor- Moody diagram- commercial pipes- minor losses – Flow through pipes in series and parallel. UNITIII DIMENSIONAL ANALYSIS 12 Need for dimensional analysis – methods of dimensional analysis – Similitude –types of similitude - Dimensionless parameters- application of dimensionless parameters – Model analysis. UNIT IV PUMPS 12 Impact of jets - Euler’s equation - Theory of roto-dynamic machines – various efficiencies– velocity components at entry and exit of the rotor- velocity triangles - Centrifugal pumps– working principle - work done by the impeller - performance curves - Reciprocating pump- working principle – Rotary pumps –classification. UNIT V TURBINES 12 Classification of turbines – heads and efficiencies – velocity triangles. Axial, radial and mixed flow turbines. Pelton wheel, Francis turbine and Kaplan turbines- working principles - work done by water on the runner – draft tube. Specific speed - unit quantities – performance curves for turbines – governing of turbines. TOTAL: 60 PERIODS OUTCOMES: Upon completion of this course, the students will be able to CO1 Apply mathematical knowledge to predict the properties and characteristics of a fluid. UNIT 1 CO2 Can analyse and calculate major and minor losses associated with pipe flow in piping networks. UNIT 2 CO3 Can mathematically predict the nature of physical quantities UNIT 3 CO4 Can critically analyse the performance of pumps. UNIT 4 CO5 Can critically analyse the performance of turbines. UNIT 5 TEXT BOOK: 1. Modi P.N. and Seth, S.M. "Hydraulics and Fluid Mechanics", Standard Book House, New Delhi 2013. REFERENCES: 1. Graebel. W.P, "Engineering Fluid Mechanics", Taylor & Francis, Indian Reprint, 2011 2. Kumar K. L., "Engineering Fluid Mechanics", Eurasia Publishing House(p) Ltd., New Delhi 2016 3. Robert W.Fox, Alan T. McDonald, Philip J.Pritchard, “Fluid Mechanics and Machinery”, 2011. 4. Streeter, V. L. and Wylie E. B., "Fluid Mechanics", McGraw Hill Publishing Co. 2010 OUTCOMES: Upon completion of this course, the students will be able to • Apply mathematical knowledge to predict the properties and characteristics of a fluid. • Can analyse and calculate major and minor losses associated with pipe flow in piping networks. • Can mathematically predict the nature of physical quantities • Can critically analyse the performance of pumps • Can critically analyse the performance of turbines.
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REGULATION : 2017 ACADEMIC YEAR : 2018-2019
JIT-JEPPIAAR/MECH /Mr.D.Arunkumar & Mr.M.K.Karthik/IIrd Yr/SEM 03 /CE8394/FLUID MECHANICS AND
MACHINERY/UNIT 1-5/QB+Keys/Ver1.0
3- 1
CE8394 FLUID MECHANICS AND MACHINERY L T P C
4 0 0 4
OBJECTIVES
1. The properties of fluids and concept of control volume are studied
2. The applications of the conservation laws to flow through pipes are studied.
3. To understand the importance of dimensional analysis
4. To understand the importance of various types of flow in pumps.
5. To understand the importance of various types of flow in turbines.
UNIT I FLUID PROPERTIES AND FLOW CHARACTERISTICS 12
Units and dimensions- Properties of fluids- mass density, specific weight, specific volume, specific gravity,
viscosity, compressibility, vapor pressure, surface tension and capillarity. Flow characteristics – concept of control
volume - application of continuity equation, energy equation and momentum equation.
UNIT II FLOW THROUGH CIRCULAR CONDUITS 12
Hydraulic and energy gradient - Laminar flow through circular conduits and circular annuli-Boundary layer
JIT-JEPPIAAR/MECH /Mr.D.Arunkumar & Mr.M.K.Karthik/IIrd Yr/SEM 03 /CE8394/FLUID MECHANICS AND
MACHINERY/UNIT 1-5/QB+Keys/Ver1.0
3- 24
θ= 9.0430°
U2= 37.6991m/s (2M)
Vw2=32.5029 m/s (2M)
Discharge,Q=0.2827m3/s (2M)
Work Done =3464008 W (2M)
Efficiency η=60.06% (3M)
2
A single acting reciprocating pump running at 50rpm delivers 0.01 m3/s of water. The
diameter of the piston is 200mm and stroke length 400mm. determine (i) theoretical
discharge (ii) coefficient of discharge (iii) slip and percentage of the pump.
(13M) (APRIL2015) BTL2
Answer: Page 997 – R.K.Bansal
N=50rpm,
Qact=0.01m3/s,
D=0.2m,
L=0.4m
A=πD2/4,
𝑄𝑡ℎ = ALN/60
𝐶𝑑 =𝑄𝑎𝑐𝑡
𝑄𝑡ℎ
𝑠𝑙𝑖𝑝 = 𝑄𝑡ℎ − 𝑄𝑎𝑐𝑡(1M)
A=0.03141m2 (2M)
Theoretical discharge, 𝑄𝑡ℎ = 0.01047m3/s (3M)
Coefficient of discharge, Cd=0.955, Slip = 4.1x10-4 (3M)
%slip = 4.489% (4M)
3
Explain the construction and working of Reciprocating pumps with a neat sketch.(13M)
(APRIL2015) BTL2
Answer: Page 993 – R.K.Bansal
Principle: Reciprocating pump operates on the principle of pushing of liquid by a piston that
executes a reciprocating motion in a closed fitting cylinder. (2M)
Components of reciprocating pumps:-
(i) Piston or plunger: – a piston or plunger that reciprocates in a closely fitted cylinder.
(ii) Crank and Connecting rod: – crank and connecting rod mechanism operated by a power
source. Power source gives rotary motion to crank. With the help of connecting rod we
translate reciprocating motion to piston in the cylinder.
(iii)Suction pipe: – one end of suction pipe remains dip in the liquid and other end attached to
the inlet of the cylinder.
(iv) Delivery pipe: – one end of delivery pipe attached with delivery part and other end at
discharge point.
(v) Suction and Delivery value: – suction and delivery values are provided at the suction end
and delivery end respectively. These values are non-return values. (4M)
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3- 25
(3M) WORKING OF RECIPROCATING PUMP
(i) Operation of reciprocating motion is done by the power source (i.e. electric motor or i.c
engine, etc).
(ii) Power source gives rotary motion to crank; with the help of connecting rod we translate
reciprocating motion to piston in the cylinder (i.e. intermediate link between connecting
rod and piston).
(iii)When crank moves from inner dead centre to outer dead centre vacuum will create in the
cylinder. When piston moves outer dead centre to inner dead centre and piston force the water at outlet or delivery value (4M)
4
A centrifugal pump has an impeller 500mm diameter running at 400rpm. The discharge at
the inlet is entirely radial. The velocity of the flow at inlet is 1m/s. the vanes are curved
backwards at outlet at 30 to the wheel tangent. If the discharge of the pump is 0.14m3/s.
calculate the impeller power and torque on the shaft.
(13M) (APRIL2011) BTL2
Answer: Page 955 – R.K.Bansal
D2=0.5m,
R2=0.25m,
N=400rpm,
α=90°,
Vf2=1m/s,
ᴓ=30°,
Q=0.14m3/s
𝑈2 = πD2N
60
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3- 26
𝑡𝑎𝑛𝜃 = Vf2
(u2 − Vw2)
𝑇 = ρQ
(u2. Vw2)
𝑃 = 2πNT
60
U2=10.4719 m/s (3M)
Vw2=8.7399 m/s (3M)
Torque, T=305.8965 N-m (3M)
Power, P=12.8133 kW (4M)
5
Explain the performance characteristics curves of centrifugal pump (NOVEMBER2015)
(13M) BTL2
Answer: Page 978 – R.K.Bansal
The curves which are plotted from the series of a number of tests on the centrifugal pump are known as characteristics or performance curves. It refers to the graphical representation of variation in head, power and efficiency of pump drawn to a common base line of flow rate. (5M)
The following four types are the characteristic curves used for centrifugal pumps.
(i) Main characteristic curves,
(2M) (ii) Operating characteristic curves,
(2M) (iii) Constant efficiency curves
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3- 27
(2M) (iv) Constant head and constant discharge curves.
(2M)
6
With neat sketch describe about the various components of centrifugal pump and its
working principle. (13M) (NOVEMBER 2016) BTL2
Answer: Page 945 – R.K.Bansal
Working principle of a centrifugal pump remains the same, based on the impeller and suction.
Working of the centrifugal motor is pretty simple. The rotation of the impeller creates a
very low pressure at its inlet, called the eye of the impeller. The fluid gets carried along
the impeller towards the casing. (3M)
Some of the most common components found in centrifugal pumps are:
(i) Pump main housing.
(ii) Impeller.
(iii) Impeller seal.
(iv) Impeller bearings.
(v) Motor.
(vi) Coupling.
(vii) Shaft-drive. (2M)
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(5M)
Vane pumps use vanes (flat blades) that slide in and out as they rotate, moving the fluid
from the inlet to the outlet and flinging it out at speed. Impeller pumps use a wheel with
curved blades called an impeller, which is a bit like a multi-bladed propeller fitted snugly
in the middle of a closed pipe. (3M)
PART * C
1
The internal and external diameters of the impellers of a centrifugal pump are 300mm and 600mm resp. The pump is running at 1000rpm. The vane angles of the impeller at inlet and outlet are 20 and 30 resp. the water enters the impeller radially and velocity of flow is constant. Determine the work done by the impeller per unit weight of water. (15M) (NOVEMBER2012) BTL2
Answer: Page 951 – R.K.Bansal
D1=0.2m,
D2=0.4m,
N=1200rpm,
θ=20°,
ᴓ=30°,
α=90°,
Vw1=0, Vf1=Vf2.
𝑈1 = πD1N
60
𝑡𝑎𝑛𝜃 = Vf1
𝑢𝑙
𝑈2 = πD2N
60
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𝑡𝑎𝑛𝜃 = Vf2
(u2 − Vw2)
𝑊𝐷 = ρg
(u2.Vw2) (2M)
U1=12.5663 m/s (2M)
U2=25.1327m/s (2M)
Vf1=4.5737m/s (2M)
Vf2=4.5737m/s (2M)
Vw2 =17.6444m/s (2M)
Work Done = 45.9839 W (3M)
2
Discuss the working of lobe pumps with a neat sketch. (15M) (NOVEMBER 2012) BTL2
Answer: Page 1066 – R.K.Bansal
Lobe Pump:
(i) Lobe pumps are similar to external gear pumps in operation in that fluid flows around
the interior of the casing.
(ii) As the lobes come out of mesh, they create expanding volume on the inlet side of the
pump. Liquid flows into the cavity and is trapped by the lobes as they rotate.
(iii) Lobe pumps are used in a variety of industries including pulp and paper, chemical,
food, beverage, pharmaceutical, and biotechnology. Rotary pumps can handle solids
(e.g., cherries and olives), slurries, pastes, and a variety of liquids. (8M)
(7M)
3 The impeller of a centrifugal pump having external and internal diameter 450mm and
225mm respectively. Width at outlet is 45mm and running at 1250rpm, works against a
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3- 30
head of 50m. The velocity of flow through the impeller is constant and equal to 2.8 m/s. The
vanes are set back at an angle of 40° at the outlet. Determine (i) inlet vane angle, (ii) work
done by the impeller on water per second, (iii) manometric efficiency. (15M) (NOVEMBER
2014) BTL2
Answer: Page 990 – R.K.Bansal
D1=0.45m,
D2=.225m,
N=1250rpm,
θ=40°,
Vf1= Vf2= 2.8m/s (2M)
𝑈1 = πD1N
60
𝑡𝑎𝑛𝜃 = Vf1
𝑢𝑙
𝑈2 = πD2N
60
𝑊𝐷 = ρQ
(u2.Vw2) (2M)
U1= 8.5663 m/s
U2=10.7m/s (2M)
Vw2 =17.6444m/s (2M)
Work Done = 136.9839 W (3M)
Manometric efficiency = 63.7% (4M)
UNIT V TURBINES
Classification of turbines – heads and efficiencies – velocity triangles. Axial, radial and mixed flow
turbines. Pelton wheel, Francis turbine and Kaplan turbines- working principles - work done by
water on the runner – draft tube. Specific speed - unit quantities – performance curves for
turbines – governing of turbines.
PART * A
Q.No. Questions
1.
Give an example for low head, medium head and high head turbine. BTL1
Low head turbine – Kaplan turbine
Medium head turbine – Modern Francis turbine
High head turbine – Pelton wheel.
2
What is impulse turbine? Give example. BTL2
In impulse turbine all the energy converted into kinetic energy. From these the turbine will
develop high kinetic energy power. This turbine is called impulse turbine.
Example: Pelton turbine
3 What is reaction turbine? Give example. BTL2
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In a reaction turbine, the runner utilizes both potential and kinetic energies. Here portion of
potential energy is converted into kinetic energy before entering into the turbine.
Example: Francis and Kaplan turbine.
4
What is axial flow turbine? BTL2
The water flows parallel to the axis of the turbine shaft is called axial flow turbine
Example: Kaplan turbine.
5
What is mixed flow turbine? BTL2
In mixed flow water enters the blades radially and comes out axially, parallel to the turbine shaft.
Example: Modern Francis turbine.
6
What is the function of spear and nozzle? BTL2
The nozzle is used to convert whole hydraulic energy into kinetic energy. Thus the nozzle
delivers high speed jet. To regulate the water flow through the nozzle and to obtain a good jet of
water spear or nozzle is arranged.
7
Define gross head and net or effective head. BTL1
Gross Head:
The gross head is the difference between the water level at the reservoir and the level at the
tailstock.
Effective Head:
The head available at the inlet of the turbine.
8 What is hydraulic efficiency? BTL1
It is defined as the ratio of power developed by the runner to the power supplied by the water jet.
9
Define mechanical efficiency. BTL1
It is defined as the ratio of power available at the turbine shaft to the power developed by the
turbine runner.
10
What is volumetric efficiency? BTL1
It is defined as the volume of water actually striking the buckets to the total water supplied by the
jet.
11
Define overall efficiency. BTL1
It is defined as the ratio of power available at the turbine shaft to the power available from the
water jet.
12
Differentiate between Kaplan turbine and propeller turbine.BTL1
The difference between the Propeller and Kaplan turbines is that the Propeller turbine has fixed
runner blades while the Kaplan turbine has adjustable runner blades.
It is a pure axial flow turbine uses basic aerofoil theory.
13
List down the main components of pelton wheel. BTL1
➢ Nozzle and flow regulating arrangements
➢ Runner and buckets
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➢ Casing, and
➢ Breaking jet.
14
What do you mean by Net Positive Suction Head (NPSH)? (NOVEMBER2015) BTL1
NPSH can be defined as two parts:
NPSH Available (NPSHA): The absolute pressure at the suction port of the pump.
NPSH Require (NPSHR): The minimum pressure required at the suction port of the pump to
keep the pump from cavitating.
15
Define negative slip in reciprocating pump. BTL1
The difference between theoretical discharge and actual discharge is called the slip of the pump.
Negative slip occurs when delivery pipe is short suction pipe is long and pump is running at
high speed.
PART * B
1
A Pelton wheel, working under a head of 500 m develops 13 MW when running at a speed
of 430 rpm. If the efficiency of the wheel is 85%, determine the rate of flow through the
turbine, the diameter of the wheel and the diameter of the nozzle. Take speed ratio as 0.46
and coefficient of velocity for the nozzle as 0.98. (13M) (APRIL2014) BTL2
Answer: Page: 870 – R.K.Bansal
H=500m,
P=13MW,
N=430rpm
η= 0.85
Kv=0.98
N=0.46
𝑢 = πDN
60
η = p
ρgQH
𝑢 = 𝐾𝑣. √2gH (1M)
U = 45.56 m/s (4M)
Rate of flow =3.11m3/s (4M)
Diameter = 2.02 m (4M)
2
A Pelton wheel works under a gross head of 510 m. One third of gross head is lost in
friction in the penstock. The rate of flow through the nozzle is 2.2 m3/sec. The angel of
deflection of jet is 165°. Find the (i) power given by water to the runner (ii) hydraulic
efficiency of Pelton wheel. Take CV = 1.0 and speed ratio = 0.45. (13M)(APRIL2017) BTL1
Answer: Page: 863 – R.K.Bansal
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3- 33
Vw1 = 𝑉1 = 𝐶𝑣. √2gH
𝑢 = 𝐾𝑢. √2gH
Vr1 = 𝑉1 − 𝑢
𝑊𝐷 = ρQ
(Vw1 + Vw2)𝑢
η = 2(Vw1+Vw2)𝑢
v2 (1M)
Vw1= 81.67 m/s (2M)
U = 36.75 m/s (2M)
Vr1= 44.92 m/s (2M)
Work done/s = 713986.3 Nm/s (3M)
η= 97.3% (3M)
3
A 137 mm diameter jet of water issuing from a nozzle impinges on the buckets of a Pelton
wheel and the jet is deflected through an angle of 165 by the buckets. The head available
at the nozzle is 400m. Find: (a) Force exerted on the buckets and (b) Power developed.
Assume Cv as 0.97, speed ratio as 0.46 and reduction in velocity while passing through the
buckets as 15%. (13M)(APRIL2010) BTL1
Answer: Page: 869 – R.K.Bansal
Vw1 = 𝑉1 = 𝐶𝑣. √2gH
𝑢1 = 𝐾𝑢. √2gH
Vr1 = 𝑉1 − 𝑢1
𝑃 = Fx . u
1000
Fx = ρav1(Vw1 − Vw2)𝑢 (1M)
Vw1= 85.93 m/s (2M)
U1 = 40.75 m/s (2M)
Vr1= 45.18 m/s (2M)
Vr2= 38.40 m/s (2M)
Fx = 104206 N (2M)
P = 4246.4 kW (2M)
4
A Pelton turbine is required to develop 9000 KW when working under a head of 300 m the
impeller may rotate at 500 rpm. Assuming a jet ratio of 10 and an overall efficiency of 85%
calculate (i) Quantity of water required, (ii) Diameter of the wheel, (iii) No of jets, (iv) No
and size of the bucket vanes on the runner. (13M)(APRIL2011) BTL1
Answer: Page: 940 – R.K.Bansal
𝑉1 = 𝐶𝑣. √2gH
𝑢 = 𝐾𝑢. √2gH
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3- 34
𝑢 = πDN
60
Vr1 = 𝑉1 − 𝑢
η = 𝑃
ρgQH
m = 𝐷
d (1M)
V1= 75.56 m/s (2M)
U = 34.52 m/s (2M)
Diameter = 0.1318 m (2M)
Discharge=3.59 m3/s (2M)
No. of jets = 3.48 (2M)
No. of bucket = 15+0.5 m = 20
Width = 0.659 mm
Depth = 0.158 mm (2M)
5
A pelton wheel turbine develops 3000kW power under a head of 300m. The overall
efficiency of the turbine is 83%. If the speed ratio = 0.46, Cv = 0.98 and specific speed is
16.5, then find diameter of the turbine and diameter of the jet. (13M)(APRIL2016) BTL1
Answer: Page: 924 – R.K.Bansal
𝑉1 = 𝐶𝑣. √2gH
𝑢 = 𝐾𝑢. √2gH
𝑢 = πDN
60
Po = 𝑃
ρgQH1000
Ns = 𝑁
𝐻24
V1= 75.12 m/s (2M)
U = 34.95 m/s (2M)
Diameter = 0.142 m (2M)
Discharge=1.23 m3/s (3M)
N = 375 rpm (4M)
PART * C
1
A hub diameter of a Kaplan turbine, working under a head of 12m, is 0.35 times the
diameter of the runner. The turbine is running at 100rpm. If the vane angle of the runner
at outlet is 15deg. And flow ratio 0.6, find (i) diameter of the runner, (ii) diameter of the
boss, and (iii) Discharge through the runner. Take the velocity of whirl at outlet as zero.
(15M)(APRIL2012) BTL1
Answer: Page: 909 – R.K.Bansal
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3- 35
H =12 m
N = 100 rpm
𝐹𝑙𝑜𝑤 𝑟𝑎𝑡𝑖𝑜 = 𝑉𝑓1/√2gH = 0.6
tan 𝜙 = 𝑉𝑓1
𝑢2
U2 = 34.33 m/s (3M)
Diameter, Db = 2.3 m (4M)
Flow velocity,Vf1= 9.2 m/s (4M)
Discharge Q = 271.7 m3/s (4M)
2
A reaction turbine works at 450 r.p.m. under a head of 120 m. Its diameter at inlet is 1.2 m
and the flow area is 0.4 m2. The angles made by absolute and relative velocities at inlet are
20° and 60° respectively with the tangential velocity. Determine: (i)the volume rate of flow,
(ii) the power developed, and (iii) the hydraulic efficiency. (15M)(APRIL2016) BTL2
Answer: Page: 523 – R.K.Bansal
tan 𝜃 = 𝑉𝑓1
𝑉𝑤1 − 𝑢1
tan 𝜃 = 𝑉𝑓1
𝑉𝑤1
𝑢1 = πDN
60
𝑄 = 𝜋𝐷𝐵. 𝑉𝑓1
W = ρQ(Vw1. u1)
ηH = Vw1.u1
gH (1M)
u1 = = 28.27 m/s (2M)
Vf1 = 0.364 Vw1 (3M)
Vw1 = 35.79 m/s (3M)
Discharge = 5.211 m3/s (3M)
Hydraulic efficiency ,ɳ = 85.95% (3M)
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