Unit -1
Introduction to Turbo Machinery & Impact of J
Q1) Define & classify turbo machines.
Ans.
The principle component of a turbo machine are :
Classification of Turbo machines
1. Based on energy transfer
2. Based on fluid flowing in turbo machine
3. Based on direction of flow through the impeller or vanes or blades, with reference to the axis of shaft rotation
4. Based on condition of fluid in turbo machine
5. Based on position of rotating shaft
Q2) State different losses associated with turbo machines.
Ans. Losses in Turbo machines
The two types of losses are as following
2. External losses:
Q3) Compare turbomachines with positive displacement machines.
Ans.
Turbo machines | Positive displacement machines |
1. It creates thermodynamics and dynamic action between rotating element and following fluid, energy transfer takes place if pressure and momentum changes. | 1. It creates thermodynamics and mechanical action between moving member and static fluid, energy transfer takes place with displacement of fluid. |
2. It involves a study flow of fluid and rotating motion of mechanical element. | 2. It involves a unsteady flow of fluid and reciprocating motion. |
3. They operate at high rotational speed. | 3. They operate at low speed. |
4. Change of phase during fluid flow causes serious problems in Turbo machine. | 4. Change of phase during fluid flow causes less problems in positive displacement machines. |
5. Efficiency is usually less. | 5. Efficiency is higher. |
6. It is simple in design. | 6. It is complex in design. |
7. Due to rotatory motion vibration problems are less. | 7. Due to reciprocating motion vibration problems are more. |
Q4) Explain fundamental equations which govern turbo machines.
Ans. Euler’s Equation
dM1 = (Vn1 da1) Vu1 R1
dM2 = (Vn2 da2) Vu2 R2
Thus the total moments of momentum are
M1 =
M2 =
The fluid torque is the net effect given by
T = M1 + M2
= ………………. 1
It is assumed that VuR is a constant across each surface and it is noted thatis the mass flow rate m.
Then equation (1) becomes
T = m(Vu1 R1 - Vu2 R2) ………………………………………….. 2
The rate of doing work is T and since R is the rotor peripheral velocity u
at radius R, equation (2) can be transformed to give work done per unit
mass:
gH = u1 Vu1 - u2Vu2
This is one form of the Euler equation.
Application of the First Law of Thermodynamics
Work output of the turbo machine = Total fluid energy at inlet - Total fluid energy
at outlet
= E1 – E2
On the basis of unit mass flow rate,
W = e1 – e2
= -
Where W, is the “specific work” of a turbo machine. It is the work interaction of 1 kg of fluid while flowing over the rotor of the turbo machine.
W = -
Application of the Second Law of Thermodynamics
Q5) State different applications of turbo machines.
Ans. Applications of Turbo machines
Q6) State & derive impulse moment principle& state its applications.
Ans. It is based on law of conservation of momentum or on the momentum principle, which states that the net force acting on a fluid mass is equal to the change in momentum of flow per unit time in that direction.
The force acting on a fluid mass m is given by the Newtons second law of motion.
F = m x a
Where a is the acceleration acting in the same direction as force F
But a =
F = m x
= m is constant can be taken inside the differential
Above equation is known as the momentum principle.
It can be written as F dt = d(mv)
Which is known as the momentum equation & states that the impulse of force F acting on a fluid of mass m in a short interval of time dt is equal to the change of momentum d(mv) in the direction of force.
Applications of the Impulse - Momentum Equation
1. For any problems involving F, v, t:
The impulse Momentum equation may be used for any problems involving the variables force F , velocity v, and time t. The IM equation is not directly helpful for determining acceleration, a, or displacement, s.
2. Helpful for impulsive forces:
The IM equation is most helpful for problems involving impulsive forces. Impulsive forces are relatively large forces that act over relatively short periods of time for example during impact. If one knows the velocities, and hence momenta, of a particle before and after the action of an impulse, then one can easily determine the impulse. If the time of impulse is known, then one can calculate the average force that act during the impulse.
3. For problems involving graph of F vs. t :
Some problems give a graph of Force vs. time. The area under this curve is impulse.
Q7) Derive expression for force exerted by the jet on the flat plate in the direction of jet, when the jet strikes the vertical plate at its center when the plate is stationary.
Ans. Consider a jet of water coming out from the nozzle, strikes a flat vertical plate as shown in Fig
V = velocity of the jet
d = diameter of the jet
a = area of the cross section of the jet
=
The jet after striking the plate will move along the plate.
But the plate is at right angles to the jet.
Hence the jet after striking will get deflected through 90°.
Hence the component of the velocity of jet, in the direction of jet, after striking will be zero.
The force exerted by the jet on the plate in the direction of jet,
=rate of change of momentum in the direction of force
=
=
=
= (Mass/sec) x (velocity of Jet before striking - velocity of Jet after striking)
=
=
Q8) Derive expression for force exerted by the jet on the flat plate in the direction of jet, when the jet strikes the inclined plate at its center when the plate is stationary.
Ans. Let Jet of water, coming out from the nozzle, strikes and inclined flat plate as shown in Fig.
V = Velocity of jet in the direction of
= Angle between the jet and plate
a = area of cross section of the jet
Then mass of water per second striking the plate =
If the plate is smooth and if it is assumed that there is no loss of energy due to impact of the jet, then Jet will move over the plate after striking with a velocity equal to initial velocity.
The force exerted by the jet on the plate in the direction normal to the plate.
= mass of Jet striking per second × [ initial velocity of Jet before striking in the direction of - final velocity of Jet after striking in the direction of ]
= θ
= component of perpendicular to flow
= θ) = θ = θ θ
=
=component of perpendicular to flow.
= θ) = θ = θ θ
Q9) Derive expression for force exerted by the jet on the curved plate in the direction of jet, when the jet strikes at its center when the plate is stationary.
Ans. Let a Jet of water strikes a fixed curved plate at the centre as shown in Fig.
The Jet after striking the plate, comes out with the same velocity. If the plate is smooth and there is no loss of energy due to impact of the jet, in the tangential direction of the curved plate.
The velocity at outlet of the plate can be resolved into two components, one in the direction of jet and other perpendicular to the direction of the jet.
Component of velocity in the direction of jet
= θ
Component of velocity perpendicular to the jet
= θ
Force exerted by the jet in the direction of jet,
= × [
= Initial Velocity in the direction of jet=
= Final velocity in the direction of jet= θ
= θ = θ
= θ
Similarly,
=initial velocity in the direction of
=final velocity in the direction of θ
θ] = θ
Negative sign means that forces acting in the downward direction. In this case the angle of deflection of the jet = (180°- θ)
Q10) Derive expression for force exerted by the jet on the hinged flat plate.
Ans.
Consider a jet of water striking a vertical plate at the centre which is hinged at O.
Due to the force exerted by the jet on the plate, the plate swing through some angle about the hinge as shown in fig.
Let x= distance of the centre of Jet from hinge O
Θ= Angle of Swing about hinge
W=weight of plate acting at C.G. of the plate
The dotted line shows the position of the plate, before the jet strikes the plate. The point A on the plate will be of A' after the jet strikes the plate.
The distance OA= OA'=x
Let the weight of the plate is acting at A'. When the plate is in equilibrium after the jet strikes the plate. The moment of all the forces about the hinge must be zero.
Two force are acting on the plate
1. Force due to Jet of water, normal to the plate
Where = angle between jet and plate
= (90-)
2. Weight of the plate, W
Moment of force about hinge
=
Moment of weight W About hinge
= W × OA’ sin θ
= W× θ
For equilibrium of the plate
Q11) Derive expression for work done per second in case of radially mounted curved vane & jet strikes tangentially.
Ans. For a radial curved vane, the radius of the vane at inlet and outlet is different& hence the tangential velocities of the radial vane at the inlet and outlet are not equal.
Consider a series of radial curved vanes mounted on a wheel as shown in fig.
The jet of water strikes the vanes and the wheel start rotating at a constant angular speed
Radius of wheel at the inlet of vane
Radius of wheel at the outlet of vane
=Angular speed of the wheel
Then,
And
The mass of water striking per second for a series of vane =
Where a= Area of jet
Velocity of jet
Momentum of water striking the vanes in the tangential direction per second at inlet= Mass of water per second Component of V in the tangential direction
=
Similarly momentum of water at outlet per second =Component of in the
tangential direction
=(-cos)
=
Now angular momentum per second in the inlet
=Momentum at inlet Radius at outlet
=
Angular momentum per second at outlet
=momentum at outlet Radius at outlet
= -
Torque exerted by the water on the wheel
T= Rate of change of angular momentum
=[Initial angular momentum per second – Final angular momentum per second]
=-(-
=+
Work done per second on the wheel
=Torque Angular velocity
=T
=+
=+
=+
If angle in fig. is on obtuse angle then work done per second will be given as
-
General expression for the work done per second on the wheel
if the discharge is radial at outlet then & work done becomes as
Q12) Derive expression for work done per second in case of flat plates mounted curved vane & jet strikes tangentially.
Ans.
The force exerted by a jet of water on a single moving plate is not practically feasible. This case is only a theoretical one.
In actual practice a large number of plates are mounted on the circumference of a wheel at a fixed distance apart as shown in fig.
The jet strikes a plate. Due to the force exerted by the jet on the plate wheel starts moving at a constant speed.
Let V=Velocity of jet
d=distance of jet
a=cross sectional area of jet =
u=Velocity of vane
In this case mass of water coming out from the nozzle per second is always in contact with the plates, when all the plates are considered. Hence mass of water per second striking the series of plate =
Also the jet strikes the plate with a velocity (V-u)
After striking, the jet moves tangential to the plate and hence the velocity component in the direction of motion of plate is equal to zero.
The force exerted by the jet in the direction of motion of plate.
=Mass per second [initial velocity – Final velocity]
=[(V-u) - 0]
= (V-u)
Work done by the jet on the series of plate per second
=Force Distance per second in the direction of force
=
=[V-u]u
Kinetic energy of jet per second
=
Efficiency =
Q13) Draw & explain inlet & outlet velocity triangles for moving unsymmetrical curved vanes mounted radially and jet strikes tangentially.
Ans.
Fig. shows a jet of water strikes a moving curved plate (vane) tangentially at one of its tips.
Let
=Velocity of the jet of inlet
=Velocity of the vane at inlet
=Relative velocity of jet and vane of inlet
= Guide blade angle
(Angle between the direction of jet and direction of motion of the plate)
Angle made by the relative velocity with direction of motion at inlet (vane angle at inlet )
=The components of velocity of jet in the direction of motion and perpendicular to direction of motion of the vane respectively.
= Whirl velocity at inlet
= Flow velocity at inlet
=Velocity of jet leaving the vane or velocity of the jet at the outlet of vane
= Velocity of the vane at outlet
=Relative velocity of the jet with respect to the vane
=Angle made by the velocity with direction of motion of the vane at outlet
=Vane angle at outlet
=Whirl velocity of outlet
=Velocity of the flow at outlet
The triangles ABD and EGF are called the velocity triangles of inlet and outlet
Velocity triangle at inlet
Then BD= Velocity of the flow at inlet=
AD=Velocity of the whirl at inlet-
Angle BCD = Vane angle of the inlet
2. Velocity triangle of outlet
If vane surface is assumed to be very smooth, the loss of energy due to friction will be zero. The water will be gliding over the surface of the vane with relative velocity equal to and will come out of the vane with the relative velocity . Which means . And also relative velocity at outlet and EG=. From G draw a line GF in the direction of vane at outlet & equal to u2 , the velocity of vane at outlet. Join EF. Then EF represents the absolute velocity of the jet at outlet in magnitude & direction. From E draw vertical line EH to meet the line GF produced at H. Then
EH=Velocity of flow of outlet =
FH=Velocity of whirl at outlet =
Angle of vane of outlet
=Angle made by with the direction of motion of vane at outlet.
If the vane is smooth and is having velocity in the direction of motion at inlet and outlet equal then we have = u = velocity of vane in the direction of motion