UNIT 1
Classification of fluid machinery
Q1. Write a short note on “applications of turbomachines”.
Answer: -
Some of the major applications of turbomachines are given below.
Q2. On what principle the energy transfer takes place between the fluid and rotor?
Answer: - Consider a rotor rotating at an angular speed of Ω rad/s
Consider points one and two on the rotor at radii r1 and r2. As in that the fluid enters at point 1 and leaves at point 2.
Let, m = mass flow rate of fluid in kg per second
V1 = absolute velocity of fluid at point 1 in meters per second
V2 = absolute velocity of fluid at point 2 in meters per second
V1 = Vf1 + VR1 + Vw1
V2 = Vf2 + VR2 + Vw2
The energy transferred between the fluid and machine is only due to change in momentum caused by tangential component of velocity.
Therefore, the rate of change of momentum between points 1 and 2
= m Vw1 - m Vw2
The momentum at point 1 = m Vw1 r1
The momentum at point 2 = m Vw2 r2
This rate of change is momentum represent torque produced on the rotor.
Torque, T = rate of change of angular momentum
= m Vw1 r1 - m Vw2 r2
= m (Vw1 r1 - Vw2 r2 )
Q3. Define momentum of a fluid and Newton’s second law of motion.
Answer: - The momentum of fluid is defined as the product of mass and its velocity. therefore,
momentum M = Mass, m X Velocity, C --------- (1)
According to Newton's second law of motion the rate of change of momentum is directly proportional to external force acting on the body.
F α dM/dt
F = gc X dM/dt ---------(2)
Where, gc is the constant of proportionality
From (1) and (2) and product rule of differentiation, we can substitute and rewrite (2) as
F = d (m.C)/dt = [m.dC/dt + C.dm/dt]
In hydrodynamic machines, dm/dt = 0, since there is no change in mass with respect to time (Steady flow).
F = m.dC/dt = m.a ----------(3)
Where, acceleration a = dC/dt.
Q4. Derive the impulse momentum equation.
Answer: - The momentum of fluid is defined as the product of mass and its velocity. therefore,
momentum M = Mass, m X Velocity, C --------- (1)
According to Newton's second law of motion the rate of change of momentum is directly proportional to external force acting on the body.
F α dM/dt
F = gc X dM/dt ---------(2)
Where, gc is the constant of proportionality
From (1) and (2) and product rule of differentiation, we can substitute and rewrite (2) as
F = d (m.C)/dt = [m.dC/dt + C.dm/dt]
In hydrodynamic machines, dm/dt = 0, since there is no change in mass with respect to time.
F = m.dC/dt = m.a ----------(3)
Where, acceleration a = dC/dt.
Q5. State the basic classification of fluid machines.
Answer: - The fluid machines can be broadly classified into 2 main categories, namely power consuming and power producing machines.
In case of power producing devices, the energy of the working fluid is converted into mechanical power. Example: - reciprocating engines and turbines. While in power absorbing devices, the mechanical power supplied at the input shaft is utilized and transfer to the working fluid either in the form of pressure energy or kinetic energy or both. Example: - Pumps, fans, compressors, etc.
Both these types of machines use liquid as working fluid in case of hydraulic machines and gases or vapour in case of thermal machines.
Q6. Discuss the detailed classification of Turbo machines with examples.
Answer: -
Classification of turbomachines
1 Based on working medium
A) Using air or gases or vapour. These are called thermal or pneumatic turbomachines.
B) Using liquids. These are hydro turbomachines.
2 According to energy conversion
A) Power producing devices which convert energy of fluid into mechanical work
B) Power absorbing devices which convert mechanical energy into pressure or kinetic energy
3 According to direction of flow
a) Radial flow. There is substantial change in radius during fluid flow. Thus, there is change in energy due to change in radius. These maybe inward or outward flow machines. Example: - Francis turbine.
b) Axial flow. They have no significant change of radius during fluid flow at entry and exit.
Example: - axial flow pumps and turbines, windmills
c) Mixed flow. They have mixed fluid flow that is both axial and radial flow.
d) Tangential flow Example: - impulse hydraulic turbine
4 According to action of fluid on the rotor vanes
a) Impulse machines which work on principle of impulse. A machine in which there is no change in static head or static pressure in the rotor is called impulse machine. The degree of reaction of impulse machines is 0. Therefore, in these types of machines, velocity at inlet is equal to velocity at outlet and the energy transfer is purely due to change in absolute kinetic energy of the fluid passing over the rotor. In other words, energy transfer in these machines is purely due to dynamic action of fluid. Example: - Pelton Wheel.
b) Reaction machines which work on principle of reaction. In a pure reaction machine, the degree of reaction R is equal to 1. In such a machine the energy transfer is by virtue of change of static pressure in the router and eventually V1 is equal to V2 . However, the absolute velocity at inlet and outlet may not be equal due to which the degree of reaction turbine is usually less than one. Reaction machines with any degree of reaction must have an enclosed rotor be'cause the fluid cannot expand freely in any direction. A simple example of a reaction machine would be a lawn sprinkler. In this the water enters the nozzles under high pressure or high head. High pressure is converted into velocity head in nozzle. The high velocity jet leaves the rotor in tangential direction. The change in momentum of the fluid in nozzle which is free to rotate gives rise to reactive force hence it is reactive type of simple machine.
Q7. Derive the fundamental equation for energy transfer between fluid and rotor.
Answer: - Let, m = mass flow rate of fluid in kg per second
V1 = absolute velocity of fluid at point 1 in meters per second
V2 = absolute velocity of fluid at point 2 in meters per second
V1 = Vf1 + VR1 + Vw1
V2 = Vf2 + VR2 + Vw2
The energy transferred between the fluid and machine is only due to change in momentum caused by tangential component of velocity.
Therefore, the rate of change of momentum between points 1 and 2
= m Vw1 - m Vw2
The momentum at point 1 = m Vw1 r1
The momentum at point 2 = m Vw2 r2
This rate of change is momentum represent torque produced on the rotor.
Torque, T = rate of change of angular momentum
= m Vw1 r1 - m Vw2 r2
= m (Vw1 r1 - Vw2 r2 )
Q8. What are the principal components of Turbo machines?
Answer: -
Q9. Classify the turbomachines according to direction of flow in detail.
Answer: -
Classification of turbomachines according to direction of flow
a) Radial flow. There is substantial change in radius during fluid flow. Thus, there is change in energy due to change in radius. These maybe inward or outward flow machines. Example: - Francis turbine.
b) Axial flow. They have no significant change of radius during fluid flow at entry and exit.
Example: - axial flow pumps and turbines, windmills
c) Mixed flow. They have mixed fluid flow that is both axial and radial flow.
d) Tangential flow Example: - impulse hydraulic turbine
Q10. Explain the working of a simple Reaction machine with example.
Answer: - In a pure reaction machine, the degree of reaction R is equal to 1. In such a machine the energy transfer is by virtue of change of static pressure in the router and eventually V1 is equal to V2 . However, the absolute velocity at inlet and outlet may not be equal due to which the degree of reaction turbine is usually less than one. Reaction machines with any degree of reaction must have an enclosed rotor because the fluid cannot expand freely in any direction. A simple example of a reaction machine would be a lawn sprinkler. In this the water enters the nozzles under high pressure or high head. High pressure is converted into velocity head in nozzle. The high velocity jet leaves the rotor in tangential direction. The change in momentum of the fluid in nozzle which is free to rotate gives rise to reactive force hence it is reactive type of simple machine.