Module 02: Energy exchange in Turbo machines Modal 02: Question Number 3 a & 3 b i. Basic Introduction ii. Euler’s turbine equation iii. Alternate form of Euler’s turbine equation iv. Components of energy transfer v. Degree of Reaction vi. Velocity triangles for different values of degree of reaction vii. Utilization factor viii. Relation between degree of reaction and Utilization factor ix. List of Formulas x. Previous Year Question papers

1. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 1
Turbo Machines
18ME54
Course Coordinator
Mr. THANMAY J. S
Assistant Professor
Department of Mechanical Engineering
VVIET Mysore
Module 02: Energy exchange in Turbo machines
Course Learning Objectives
Analyze the energy transfer in Turbo machine with degree of reaction and utilization factor
Course Outcomes
At the end of the course the student will be able to Analyze the energy transfer in Turbo machine

2. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 2
Contents
Modal 02: Question Number 3 a & 3 b
i. Basic Introduction
ii. Euler’s turbine equation
iii. Alternate form of Euler’s turbine equation
iv. Components of energy transfer
v. Degree of Reaction
vi. Velocity triangles for different values of degree of reaction
vii. Utilization factor
viii. Relation between degree of reaction and Utilization factor
ix. List of Formulas
x. Previous Year Question papers

3. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 3
Basic Introduction
Parameter Position in Velocity Triangle
Rotor Speed or tangential speed or peripheral
speed of the shaft 𝑈 =
𝜋𝐷𝑁
60
In velocity triangle is always horizontal
Velocity of fluid (steam, water, air, jet) or Absolute
velocity of fluid (𝑉).
Fluid Angle at inlet or nozzle exit angle (Impulse
turbine), exit angle of guide (fixed) blade is (α1)
with the direction of 𝑈
Absolute Velocity at is to be resolved into two
components
1) Along tangential direction and is called as tangential
component velocity of fluid Vu1 (Tangential or whirl
velocity Vw1) along horizontal direction (along U)
2)Along axial direction in axial turbomachine (Vax1)
is called as axial component, along radial direction in
radial flow turbomachine (Vrd1) is called as radial
component.
Axial and radial direction represented in velocity
triangle in Y direction
Vector difference between absolute velocity of the
fluid and tangential speed of rotor is called as relative
velocity and in velocity diagram this is the line
connecting tip of U and V as given below and arrow
opposes V and Vr follows U.
Direction of Vr is the moving vane angle (vane
(blade)angle, runner vane (blade) angle, moving vane
(blade) angle) and it is denoted by β.
𝑉𝑢1 = 𝑉1 cos𝛼1; 𝑉𝑓1 = 𝑉1 𝑠𝑖𝑛𝛼1; 𝑡𝑎𝑛𝛽1 =
𝑉𝑓1
𝑉𝑢1 − 𝑈1
;𝑉𝑟1 = 𝑉𝑓1 𝑠𝑖𝑛𝛽1
Similarly, outlet Triangle is also represented:
𝑈2 and 𝑉2 are emerging from single point and line
joining tip of 𝑉2 and 𝑈2is relative velocity at outlet
𝑉𝑢2 = 𝑉𝑟2 𝑐𝑜𝑠𝛽2 − 𝑈2; 𝑉𝑓2 = 𝑉𝑟2 𝑠𝑖𝑛𝛽2

4. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 4
Euler’s turbine equation
The Euler turbine equation relates the power added to or removed from the flow, to
characteristics of a rotating blade row. The equation is based on the concepts of conservation
of angular momentum and conservation of energy.
Assumptions:
a) Fluid flow through the turbomachine is steady flow.
b) Mass flow rate is constant and the state of the fluid does not vary with time.
c) Rate of energy transfer at the rotor is constant.
d) Heat and work interactions between the rotor and its surroundings take place at a constant rate.
e) Velocity is uniform over any area normal to the flow.
m = mass flow rate
ω = Angular speed of the rotor
r1 and r2= radius of rotor at inlet and exit
Vu1 and Vu2 = tangential velocity components of fluid ant inlet and exit
U1 and U2 = tangential velocity components of rotor at entry and exit
Angular Momentum
Angular momentum at entry = mass flow rate X Tangential Velocity of fluid X radius of rotor
𝐴𝑛𝑔𝑢𝑙𝑎𝑟 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑎𝑡 𝑒𝑛𝑡𝑟𝑦 = 𝑚 × 𝑉𝑢1 × 𝑟1
Angular momentum at exit = mass flow rate X Tangential Velocity of fluid X radius of rotor
𝐴𝑛𝑔𝑢𝑙𝑎𝑟 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑎𝑡 𝑒𝑥𝑖𝑡 = 𝑚 × 𝑉𝑢2 × 𝑟2
According to Newton’s II law of motion; Torque exerted by the rotor = change of angular
momentum
Change in Angular momentum
Change in Angular momentum = Angular momentum at exit - Angular momentum at entry
𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 = 𝑚 × 𝑉𝑢2 × 𝑟2 − 𝑚 × 𝑉𝑢1 × 𝑟1 = 𝑚(𝑉𝑢2. 𝑟2 − 𝑉𝑢1. 𝑟1)
∴ Torque exerted by the rotor T = 𝑚(𝑉𝑢2. 𝑟2 − 𝑉𝑢1. 𝑟1)

5. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 5
Power Transmitted
Power Transmitted = Angular Velocity × Torque = ω × T
𝑃 = ω × 𝑚(𝑉𝑢2. 𝑟2 − 𝑉𝑢1. 𝑟1) = 𝑚(𝑉𝑢2. 𝑟2. ω − 𝑉𝑢1. 𝑟1. ω)
𝑃 = 𝑚(𝑉𝑢2. 𝑟2. ω − 𝑉𝑢1. 𝑟1. ω)
Tangential Speed or Blade Speed
Tangential Speed (U) = Angular Velocity (ω) × radius(r) ∴ U = ω × r
∴ 𝑃𝑜𝑤𝑒𝑟 𝑇𝑟𝑎𝑛𝑠𝑚𝑖𝑡𝑡𝑒𝑑 𝑃 = 𝑚(𝑉𝑢2. 𝑟2. ω − 𝑉𝑢1. 𝑟1. ω) 𝑤𝑖𝑙𝑙 𝑏𝑒
𝑃 = 𝑚(𝑉𝑢2. 𝑈2 − 𝑉𝑢1. 𝑈1)
𝑻𝒉𝒆 𝑬𝒍𝒖𝒆𝒓′
𝑻𝒖𝒓𝒃𝒊𝒏𝒆 𝑬𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝒊𝒔 𝒈𝒊𝒗𝒆𝒏 𝒂𝒔
Power Absorbing Turbo Machines Power Generation Turbo Machines
Vu2 . U2 > Vu1 . U1 Vu1 . U1 > Vu2 . U2
𝑷 = 𝒎(𝑽𝒖𝟐. 𝑼𝟐 − 𝑽𝒖𝟏. 𝑼𝟏) 𝑷 = 𝒎(𝑽𝒖𝟏. 𝑼𝟏 − 𝑽𝒖𝟐. 𝑼𝟐)
Alternate form of Euler’s turbine equation
Inlet Velocity Triangle Outlet Velocity Triangle
𝑉1
2
= 𝑉𝑓1
2
+ 𝑉𝑢1
2
𝑉𝑓1
2
= 𝑉1
2
− 𝑉𝑢1
2
𝑉2
2
= 𝑉𝑓2
2
+ 𝑉𝑢2
2
𝑉𝑓2
2
= 𝑉2
2
− 𝑉𝑢2
2
𝑉𝑟1
2
= 𝑉𝑓1
2
+ 𝑉𝑟𝑢1
2
𝑏𝑢𝑡 𝑉𝑟𝑢1 = 𝑈1 − 𝑉𝑢1
𝑎𝑙𝑠𝑜 𝑉𝑓1
2
= 𝑉1
2
− 𝑉𝑢1
2
𝑉𝑟1
2
= 𝑉𝑓1
2
+ (𝑈1 − 𝑉𝑢1
2)
𝑉𝑟2
2
= 𝑉𝑓2
2
+ 𝑉𝑟𝑢2
2
𝑏𝑢𝑡 𝑉𝑟𝑢2 = 𝑈2 − 𝑉𝑢2
𝑎𝑙𝑠𝑜 𝑉𝑓2
2
= 𝑉2
2
− 𝑉𝑢2
2
𝑉𝑟2
2
= 𝑉𝑓2
2
+ (𝑈2 − 𝑉𝑢2 )2
∴ 𝑉𝑟1
2
= 𝑉1
2
− 𝑉𝑢1
2
+ (𝑈1 − 𝑉𝑢1
2)
𝑉𝑟1
2
= 𝑉1
2
− 𝑉𝑢12
+ 𝑈1
2
+ 𝑉𝑢12
− 2𝑉𝑢1𝑈1
∴ 𝑉𝑟2
2
= 𝑉2
2
− 𝑉𝑢22
+ (𝑈2 − 𝑉𝑢2 )2
𝑉𝑟2
2
= 𝑉2
2
− 𝑉𝑢2
2
+ 𝑈2
2
+ 𝑉𝑢2
2
− 2𝑉𝑢2𝑈2
∴ 2𝑉𝑢1𝑈1 = 𝑉1
2
+ 𝑈1
2
− 𝑉𝑟1
2
∴ 2𝑉𝑢2𝑈2 = 𝑉2
2
+ 𝑈2
2
− 𝑉𝑟2
2
𝑽𝒖𝟏𝑼𝟏 =
(𝑽𝟏
𝟐
+ 𝑼𝟏
𝟐
− 𝑽𝒓𝟏
𝟐
)
𝟐
𝑽𝒖𝟐𝑼𝟐 =
(𝑽𝟐
𝟐
+ 𝑼𝟐
𝟐
− 𝑽𝒓𝟐
𝟐
)
𝟐
By Euler’s Turbine Equation
Power Absorbing Turbo Machines Power Generation Turbo Machines
𝑃 = (𝑉𝑢2. 𝑈2 − 𝑉𝑢1. 𝑈1) 𝑃 = (𝑉𝑢1. 𝑈1 − 𝑉𝑢2. 𝑈2)
𝑃 =
(𝑉2
2+𝑈2
2−𝑉𝑟2
2)
2
−
(𝑉1
2+𝑈1
2−𝑉𝑟1
2)
2
𝑃 =
(𝑉1
2+𝑈1
2−𝑉𝑟1
2)
2
−
(𝑉2
2+𝑈2
2−𝑉𝑟2
2)
2
𝑷 =
𝟏
𝟐
(𝑽𝟐
𝟐
− 𝑽𝟏
𝟐
) + (𝑼𝟐
𝟐
− 𝑼𝟏
𝟐
) + (𝑽𝒓𝟐
𝟐
− 𝑽𝒓𝟏
𝟐
)
𝑷 =
𝟏
𝟐
(𝑽𝟏
𝟐
− 𝑽𝟐
𝟐
) + (𝑼𝟏
𝟐
− 𝑼𝟐
𝟐
) + (𝑽𝒓𝟏
𝟐
− 𝑽𝒓𝟐
𝟐
)

6. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 6
Components of energy transfer
By Euler’s Turbine Equation or Energy Equation
Power Absorbing Turbo Machines Power Generation Turbo Machines
𝑷 =
𝟏
𝟐
(𝑽𝟐
𝟐
− 𝑽𝟏
𝟐
) + (𝑼𝟐
𝟐
− 𝑼𝟏
𝟐
) + (𝑽𝒓𝟐
𝟐
− 𝑽𝒓𝟏
𝟐
) 𝑷 =
𝟏
𝟐
(𝑽𝟏
𝟐
− 𝑽𝟐
𝟐
) + (𝑼𝟏
𝟐
− 𝑼𝟐
𝟐
) + (𝑽𝒓𝟏
𝟐
− 𝑽𝒓𝟐
𝟐
)
First component: It is the change in the absolute kinetic energy and which causes a change in the dynamic head
or dynamic pressure of the fluid through the machine.
𝟏
𝟐
(𝑽𝟐
𝟐
− 𝑽𝟏
𝟐
)
𝟏
𝟐
(𝑽𝟏
𝟐
− 𝑽𝟐
𝟐
)
Second component: It is the change in the centrifugal energy of the fluid in the motion. This is due to the change
in the radius of rotation of the fluid. This causes a change in the static head or static pressure of the fluid through
the rotor.
𝟏
𝟐
(𝑼𝟐
𝟐
− 𝑼𝟏
𝟐
)
𝟏
𝟐
(𝑼𝟏
𝟐
− 𝑼𝟐
𝟐
)
Third component: It is the change in the relative kinetic energy and which causes a
𝟏
𝟐
(𝑽𝒓𝟐
𝟐
− 𝑽𝒓𝟏
𝟐
)
𝟏
𝟐
(𝑽𝒓𝟏
𝟐
− 𝑽𝒓𝟐
𝟐
)
Degree of Reaction (𝑹)
Degree of Reaction (R) is the ratio of Energy Transfer due to Static Enthalpy change to Total
Energy Transfer due to Total Enthalpy change in a rotor.
Or
The degree of reaction is also defined as the ratio of energy transfer due to the change in static
pressure in the rotor to total energy transfer due to the change in total pressure in the rotor.
𝑹 =
𝑺𝒕𝒂𝒕𝒊𝒄 𝒉𝒆𝒂𝒅
𝑻𝒐𝒕𝒂𝒍 𝒉𝒆𝒂𝒅
=
𝑺𝒕𝒂𝒕𝒊𝒄 𝒆𝒏𝒕𝒉𝒂𝒍𝒑𝒚 𝒄𝒉𝒂𝒏𝒈𝒆
𝑻𝒐𝒕𝒂𝒍 𝒆𝒏𝒕𝒉𝒂𝒍𝒑𝒚 𝒄𝒉𝒂𝒏𝒈𝒆
=
𝜟𝒉
𝜟𝒉𝟎
𝜟𝒉 =
𝟏
𝟐
(𝑼𝟏
𝟐
− 𝑼𝟐
𝟐
) + (𝑽𝒓𝟏
𝟐
− 𝑽𝒓𝟐
𝟐
)
"𝑬" 𝒐𝒓 "𝒆" 𝒐𝒓 "𝑷" = 𝜟𝒉𝟎 =
𝟏
𝟐
(𝑽𝟏
𝟐
− 𝑽𝟐
𝟐
) + (𝑼𝟏
𝟐
− 𝑼𝟐
𝟐
) + (𝑽𝒓𝟏
𝟐
− 𝑽𝒓𝟐
𝟐
)
∴ 𝑅 =
𝛥ℎ
𝛥ℎ0
=
𝟏
𝟐
(𝑼𝟏
𝟐
− 𝑼𝟐
𝟐
) + (𝑽𝒓𝟏
𝟐
− 𝑽𝒓𝟐
𝟐
)
𝟏
𝟐
(𝑽𝟏
𝟐
− 𝑽𝟐
𝟐) + (𝑼𝟏
𝟐
− 𝑼𝟐
𝟐) + (𝑽𝒓𝟏
𝟐
− 𝑽𝒓𝟐
𝟐)
𝑹 =
(𝑼𝟏
𝟐
− 𝑼𝟐
𝟐
) + (𝑽𝒓𝟏
𝟐
− 𝑽𝒓𝟐
𝟐
)
(𝑽𝟏
𝟐
− 𝑽𝟐
𝟐
) + (𝑼𝟏
𝟐
− 𝑼𝟐
𝟐
) + (𝑽𝒓𝟏
𝟐
− 𝑽𝒓𝟐
𝟐
)
𝑹 =
𝑷 −
𝟏
𝟐
(𝑽𝟏
𝟐
− 𝑽𝟐
𝟐
)
𝑷
𝑤ℎ𝑒𝑟𝑒 𝑷 =
𝟏
𝟐
(𝑽𝟏
𝟐
− 𝑽𝟐
𝟐
) + (𝑼𝟏
𝟐
− 𝑼𝟐
𝟐
) + (𝑽𝒓𝟏
𝟐
− 𝑽𝒓𝟐
𝟐
)

7. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 7
Velocity triangles for different values of degree of reaction [ R=0, R=0.5, R=1]
Case 01) When R =0
(i.e., Impulse type Vr1 = Vr2, and hence β1= β2)
Energy transfer occurs purely due to the change in
absolute kinetic energy. Zero degree of reaction is the
characteristics of Impulse machine i.e Vr1 = Vr2.
Here energy transfer is purely due to change in dynamic
pressure. (U1=U2)
𝑹 =
(𝑼𝟏
𝟐
− 𝑼𝟐
𝟐
) + (𝑽𝒓𝟏
𝟐
− 𝑽𝒓𝟐
𝟐
)
(𝑽𝟏
𝟐
− 𝑽𝟐
𝟐
) + (𝑼𝟏
𝟐
− 𝑼𝟐
𝟐
) + (𝑽𝒓𝟏
𝟐
− 𝑽𝒓𝟐
𝟐
)
𝑹 =
(𝟎) + (𝟎)
(𝑽𝟏
𝟐
− 𝑽𝟐
𝟐
) + (𝟎) + (𝟎)
= 0
Case 02) When R = 0.5
(i.e., 50% Reaction axial flow)
This implies U1= U2, V1 = Vr2, V2=Vr1 and Vf1 = Vf2.
For symmetric Velocity α2 = β1 and α1= β2. Energy
transfer occurs initially by impulse action and then by
reaction.
𝑹 =
(𝑼𝟏
𝟐
− 𝑼𝟐
𝟐
) + (𝑽𝒓𝟏
𝟐
− 𝑽𝒓𝟐
𝟐
)
(𝑽𝟏
𝟐
− 𝑽𝟐
𝟐
) + (𝑼𝟏
𝟐
− 𝑼𝟐
𝟐
) + (𝑽𝒓𝟏
𝟐
− 𝑽𝒓𝟐
𝟐
)
𝑹 =
(𝟎) + (𝑽𝒓𝟏
𝟐
− 𝑽𝒓𝟐
𝟐
)
(𝑽𝟏
𝟐
− 𝑽𝟐
𝟐
) + (𝟎) + (𝑽𝒓𝟏
𝟐
− 𝑽𝒓𝟐
𝟐
)
=
1
2
Case 03) When R = 1
(i.e., 100% Fully reaction)
In this case V1 = V2, U1 = U2 and V2>Vr1
Energy transfer occurs purely due to change in relative
Kinetic Energy of fluid. The rotor acts both as the nozzle
and as the energy transfer device, so energy transfer is
purely due to change in static pressure
𝑹 =
(𝑼𝟏
𝟐
− 𝑼𝟐
𝟐
) + (𝑽𝒓𝟏
𝟐
− 𝑽𝒓𝟐
𝟐
)
(𝑽𝟏
𝟐
− 𝑽𝟐
𝟐
) + (𝑼𝟏
𝟐
− 𝑼𝟐
𝟐
) + (𝑽𝒓𝟏
𝟐
− 𝑽𝒓𝟐
𝟐
)
𝑹 =
(𝟎) + (𝑽𝒓𝟏
𝟐
− 𝑽𝒓𝟐
𝟐
)
(𝟎) + (𝟎) + (𝑽𝒓𝟏
𝟐
− 𝑽𝒓𝟐
𝟐
)
= 1

8. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 8
Utilization factor (𝝐)
The utilization factor is the ratio of the ideal (Euler) work output to the energy available for conversion
into work. Under ideal conditions, it should be possible to utilize all of the kinetic energy of the fluid at
the rotor inlet and also the increase in kinetic energy obtained in the rotor due to static pressure drop
(i.e., the reaction effect).
Thus the Energy avilable for conversion of Work is
𝑷𝑨𝒗𝒂𝒊𝒍𝒂𝒃𝒍𝒆 =
𝟏
𝟐
[(𝑽𝟏
𝟐
) + (𝑼𝟏
𝟐
− 𝑼𝟐
𝟐
) + (𝑽𝒓𝟏
𝟐
− 𝑽𝒓𝟐
𝟐
)]
Total work output is
"E" 𝒐𝒓 "e" 𝒐𝒓 P 𝒐𝒓 "𝜟𝒉𝟎" =
𝟏
𝟐
[(𝑽𝟏
𝟐
− 𝑽𝟐
𝟐
) + (𝑼𝟏
𝟐
− 𝑼𝟐
𝟐
) + (𝑽𝒓𝟏
𝟐
− 𝑽𝒓𝟐
𝟐
)]
∴ 𝑈𝑡𝑖𝑙𝑖𝑧𝑎𝑡𝑖𝑜𝑛 𝐹𝑎𝑐𝑡𝑜𝑟 (𝜖) =
𝑃
𝑃𝐴𝑣𝑎𝑖𝑙𝑎𝑏𝑙𝑒
=
1
2
[(𝑉1
2
− 𝑉2
2
) + (𝑈1
2
− 𝑈2
2
) + (𝑉𝑟1
2
− 𝑉𝑟2
2
)]
1
2
[(𝑉1
2) + (𝑈1
2
− 𝑈2
2) + (𝑉𝑟1
2
− 𝑉𝑟2
2)]
∴ 𝑼𝒕𝒊𝒍𝒊𝒛𝒂𝒕𝒊𝒐𝒏 𝑭𝒂𝒄𝒕𝒐𝒓 (𝝐) =
(𝑽𝟏
𝟐
− 𝑽𝟐
𝟐
) + (𝑼𝟏
𝟐
− 𝑼𝟐
𝟐
) + (𝑽𝒓𝟏
𝟐
− 𝑽𝒓𝟐
𝟐
)
(𝑽𝟏
𝟐
) + (𝑼𝟏
𝟐
− 𝑼𝟐
𝟐
) + (𝑽𝒓𝟏
𝟐
− 𝑽𝒓𝟐
𝟐
)
𝒐𝒓 𝑼𝒕𝒊𝒍𝒊𝒛𝒂𝒕𝒊𝒐𝒏 𝑭𝒂𝒄𝒕𝒐𝒓 (𝝐) =
𝟏
𝟐
[(𝑽𝟏
𝟐
− 𝑽𝟐
𝟐
) + (𝑼𝟏
𝟐
− 𝑼𝟐
𝟐
) + (𝑽𝒓𝟏
𝟐
− 𝑽𝒓𝟐
𝟐
)]
𝟏
𝟐
[(𝑽𝟏
𝟐
− 𝑽𝟐
𝟐) + (𝑼𝟏
𝟐
− 𝑼𝟐
𝟐) + (𝑽𝒓𝟏
𝟐
− 𝑽𝒓𝟐
𝟐)] +
𝟏
𝟐
(𝑽𝟐
𝟐)
∴ 𝑼𝒕𝒊𝒍𝒊𝒛𝒂𝒕𝒊𝒐𝒏 𝑭𝒂𝒄𝒕𝒐𝒓 (𝝐) =
𝑷
𝑷 +
𝟏
𝟐
(𝑽𝟐
𝟐)
Relation between degree of reaction and Utilization factor
(𝝐) =
(𝑽𝟏
𝟐
−𝑽𝟐
𝟐
)
(𝑽𝟏
𝟐
−𝑹 𝑽𝟐
𝟐
)
𝑊𝑒 𝑘𝑛𝑜𝑤 𝑡ℎ𝑎𝑡 𝐷𝑒𝑔𝑟𝑒𝑒 𝑜𝑓 𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑹 =
(𝑼𝟏
𝟐
−𝑼𝟐
𝟐
)+(𝑽𝒓𝟏
𝟐
−𝑽𝒓𝟐
𝟐
)
(𝑽𝟏
𝟐
−𝑽𝟐
𝟐
)+(𝑼𝟏
𝟐
−𝑼𝟐
𝟐
)+(𝑽𝒓𝟏
𝟐
−𝑽𝒓𝟐
𝟐
)
By cross multiplication
𝑅(𝑉1
2
− 𝑉2
2
) + 𝑅 [(𝑈1
2
− 𝑈2
2
) + (𝑉𝑟1
2
− 𝑉𝑟2
2
)] = (𝑈1
2
− 𝑈2
2
) + (𝑉𝑟1
2
− 𝑉𝑟2
2
)
𝑅(𝑉1
2
− 𝑉2
2
) = (1 − 𝑅) [(𝑈1
2
− 𝑈2
2
) + (𝑉𝑟1
2
− 𝑉𝑟2
2
)]
𝑹
(𝟏−𝑹)
(𝑽𝟏
𝟐
− 𝑽𝟐
𝟐
) = (𝑼𝟏
𝟐
− 𝑼𝟐
𝟐
) + (𝑽𝒓𝟏
𝟐
− 𝑽𝒓𝟐
𝟐
)
𝑊𝑒 𝑎𝑙𝑠𝑜 𝐾𝑛𝑜𝑤 𝑡ℎ𝑎𝑡 𝑈𝑡𝑖𝑙𝑖𝑧𝑎𝑡𝑖𝑜𝑛 𝐹𝑎𝑐𝑡𝑜𝑟 (𝝐) =
(𝑽𝟏
𝟐
−𝑽𝟐
𝟐
)+(𝑼𝟏
𝟐
−𝑼𝟐
𝟐
)+(𝑽𝒓𝟏
𝟐
−𝑽𝒓𝟐
𝟐
)
(𝑽𝟏
𝟐
)+(𝑼𝟏
𝟐
−𝑼𝟐
𝟐
)+(𝑽𝒓𝟏
𝟐
−𝑽𝒓𝟐
𝟐
)
(𝜖) =
(𝑉1
2
−𝑉2
2
)+
𝑅
(1−𝑅)
(𝑉1
2
−𝑉2
2
)
(𝑉1
2
)+
𝑅
(1−𝑅)
(𝑉1
2
−𝑉2
2
)
(𝜖) =
(1 − 𝑅)(𝑉1
2
− 𝑉2
2
) + 𝑅 (𝑉1
2
− 𝑉2
2
)
(1 − 𝑅)(𝑉1
2
) + 𝑅 (𝑉1
2
− 𝑉2
2
)
(𝝐) =
(𝑽𝟏
𝟐
− 𝑽𝟐
𝟐
)
(𝑽𝟏
𝟐
− 𝑹 𝑽𝟐
𝟐
)

9. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 9
List of Formulas
Basic Trigonometry Cosine Rule
Euler’s Turbine Equation
Power Absorbing Turbo Machines Power Generation Turbo Machines
𝑃 = (𝑉𝑢2. 𝑈2 − 𝑉𝑢1. 𝑈1) 𝑃 = (𝑉𝑢1. 𝑈1 − 𝑉𝑢2. 𝑈2)
𝑷 =
𝟏
𝟐
[(𝑽𝟐
𝟐
− 𝑽𝟏
𝟐
) + (𝑼𝟐
𝟐
− 𝑼𝟏
𝟐
) + (𝑽𝒓𝟐
𝟐
− 𝑽𝒓𝟏
𝟐
)] 𝑷 =
𝟏
𝟐
[(𝑽𝟏
𝟐
− 𝑽𝟐
𝟐
) + (𝑼𝟏
𝟐
− 𝑼𝟐
𝟐
) + (𝑽𝒓𝟏
𝟐
− 𝑽𝒓𝟐
𝟐
)]
𝟏
𝟐
(𝑽𝟐
𝟐
− 𝑽𝟏
𝟐
)
𝟏
𝟐
(𝑽𝟏
𝟐
− 𝑽𝟐
𝟐
)
Centrifugal Energy
𝟏
𝟐
(𝑼𝟐
𝟐
− 𝑼𝟏
𝟐
)
𝟏
𝟐
(𝑼𝟏
𝟐
− 𝑼𝟐
𝟐
)
Relative Kinetic Energy
𝟏
𝟐
(𝑽𝒓𝟐
𝟐
− 𝑽𝒓𝟏
𝟐
)
𝟏
𝟐
(𝑽𝒓𝟏
𝟐
− 𝑽𝒓𝟐
𝟐
)
Degree of Reaction (R)
𝑹 =
𝜟𝒉
𝜟𝒉𝟎
=
𝑷 −
𝟏
𝟐
(𝑽𝟏
𝟐
− 𝑽𝟐
𝟐
)
𝑷
𝑹 =
(𝑼𝟏
𝟐
− 𝑼𝟐
𝟐
) + (𝑽𝒓𝟏
𝟐
− 𝑽𝒓𝟐
𝟐
)
(𝑽𝟏
𝟐
− 𝑽𝟐
𝟐
) + (𝑼𝟏
𝟐
− 𝑼𝟐
𝟐
) + (𝑽𝒓𝟏
𝟐
− 𝑽𝒓𝟐
𝟐
)
Different values of degree of reaction [ R=0, R=0.5, R=1]
When R =0 Vr1 = Vr2, U1=U2 and β1= β2
When R = 0.5 = 50% U1= U2, V1 = Vr2, V2=Vr1 and Vf1 = Vf2
When R = 1 = 100% V1 = V2, U1 = U2 and V2>Vr1
𝑼𝒕𝒊𝒍𝒊𝒛𝒂𝒕𝒊𝒐𝒏 𝑭𝒂𝒄𝒕𝒐𝒓 (𝝐)
(𝝐) =
(𝑽𝟏
𝟐
− 𝑽𝟐
𝟐
) + (𝑼𝟏
𝟐
− 𝑼𝟐
𝟐
) + (𝑽𝒓𝟏
𝟐
− 𝑽𝒓𝟐
𝟐
)
(𝑽𝟏
𝟐
) + (𝑼𝟏
𝟐
− 𝑼𝟐
𝟐
) + (𝑽𝒓𝟏
𝟐
− 𝑽𝒓𝟐
𝟐
)
(𝝐) =
𝑷
𝑷 +
𝟏
𝟐
(𝑽𝟐
𝟐)
Relation between degree of reaction and Utilization factor
(𝝐) =
(𝑽𝟏
𝟐
− 𝑽𝟐
𝟐
)
(𝑽𝟏
𝟐
− 𝑹 𝑽𝟐
𝟐
)

10. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 10
Previous Year Question papers
18ME54 Model Question Paper -1 with effect from 2020-21(CBCS Scheme)
Module – 2
3 a) Define Degree of reaction. Obtain an expression for Utilization factor in terms of degree of
reaction and absolute velocities.
6
Ans ∴ 𝑼𝒕𝒊𝒍𝒊𝒛𝒂𝒕𝒊𝒐𝒏 𝑭𝒂𝒄𝒕𝒐𝒓 (𝝐) =
𝑬
𝑬+
𝟏
𝟐
(𝑽𝟐
𝟐
)
Page Number 8
b) For 50% degree of reaction axial flow turbomachine, inlet fluid velocity is 230 m/s, out
angle of inlet guide blade is 30°, inlet rotor angle is 60° and outlet rotor angle is 25°. Find
the utilization factor, axial thrust and power output per unit mass flow.
6
Ans
Case 02) When R = 0.5
(i.e., 50% Reaction axial flow)
This implies
U1= U2,
V1 = Vr2,
V2=Vr1 and
Vf1 = Vf2.
For symmetric Velocity
α2 = β1 and α1= β2.
Given Data: V1 = 230 m/s; α1=30°; β1= 60°; β2= 25°;
Although R = 50% α2 ≠ β1 and α1≠ β2 as Vf1 ≠ Vf2.
utilization factor (↋) =?
F(axial) =?
P =?
𝑹 =
(𝑽𝒓𝟏
𝟐
− 𝑽𝒓𝟐
𝟐
)
(𝑽𝟏
𝟐
− 𝑽𝟐
𝟐
) + (𝑽𝒓𝟏
𝟐
− 𝑽𝒓𝟐
𝟐
)
=
1
2
= 0.5
From Inlet Velocity Triangle
𝑉𝑓1 = 𝑉1 sin 𝛼1 𝑉𝑟1 =
𝑉𝑓1
𝑠𝑖𝑛𝛽1
𝑈1 = 𝑉1 cos 𝛼1 − 𝑉𝑟1cos 𝛽1 ;
𝑉𝑓1 = 230 sin 30 = 115 𝑚/𝑠 𝑉𝑟1 =
115
𝑠𝑖𝑛60
= 132.8 𝑚/𝑠
𝑈1 = 230 cos30 − 𝑉𝑟1 cos60 ;
𝑈1 = 132.8 𝑚/𝑠;
tan 𝛼1 =
𝑉𝑓1
𝑉𝑢1
; 𝑉𝑢1 =
𝑉𝑓1
tanα 1
𝑉𝑢1 =
𝑉𝑓1
tanα1
=
115
tan 30
= 199.18 𝑚/𝑠
𝑽𝟏 = 𝟐𝟑𝟎 𝒎/𝒔 𝑽𝒓𝟏 = 𝟏𝟑𝟐. 𝟖 𝒎/𝒔
For 50% Reaction V2=Vr1 = 132.8
(𝝐) =
(𝑽𝟏
𝟐
− 𝑽𝟐
𝟐
)
(𝑽𝟏
𝟐
− 𝑹 𝑽𝟐
𝟐
)
=
(𝟐𝟑𝟎 𝟐
− 𝟏𝟑𝟐. 𝟖𝟐
)
(𝟐𝟑𝟎 𝟐
− (𝟎. 𝟓)𝟏𝟑𝟐.𝟖𝟐
)
=
𝟑𝟓, 𝟐𝟔𝟒. 𝟏𝟔
𝟒𝟒, 𝟎𝟖𝟐. 𝟎𝟖
= 𝟎. 𝟕𝟗𝟗𝟗
𝑉𝑓1 = 𝑉1 sin 𝛼1 = 115𝑚/𝑠
𝑉𝑓2 = 𝑉𝑟2 sin 𝛽2 = 97.202
𝑭 (𝒂𝒙𝒊𝒂𝒍) = 𝒎(𝑽𝒇𝟏 − 𝑽𝒇𝟐);
𝐹 (𝑎𝑥𝑖𝑎𝑙)
𝑚
= (115 − 97.202) = 𝟏𝟕. 𝟕𝟗𝟕 𝑵 − 𝒎/𝒔
𝑷𝒐𝒘𝒆𝒓 (𝑷) = 𝒎𝑼(𝑽𝒖𝟏 − 𝑽𝒖𝟐);
𝑷
𝒎
= (𝑽𝟏 𝒄𝒐𝒔𝜶𝟏 − 𝑽𝒓𝟐 𝒄𝒐𝒔𝜷𝟐 − 𝑼)
𝑷 = 𝟏𝟑𝟐. 𝟖 × (𝟏𝟗𝟗. 𝟏𝟖𝟓 − 𝟕𝟓. 𝟔𝟓) = 𝟏𝟔, 𝟒𝟎𝟒. 𝟔𝟕𝟖 = 𝟏𝟔. 𝟒𝟎𝟒 𝒌𝑾

11. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 11
Previous Year Question papers
18ME54 Model Question Paper -2 with effect from 2020-21(CBCS Scheme)
Module – 2
3 a) With the help of velocity triangles at inlet and outlet, derive an alternate form of Euler’s
turbine equation.
b) Show that maximum utilization factor of an axial flow turbine with degree of reaction(
𝟏
𝟑
),
the relationship of blade speed U to absolute velocity at rotor inlet V1(speed ratio) is given
by∅ =
𝑼
𝑽𝟏
=
𝟏
𝟒
𝒄𝒐𝒔 𝜶𝟏, where α1 is the nozzle angle with respect to tangential direction at inlet.
𝑹 =
(𝑼𝟏
𝟐
−𝑼𝟐
𝟐
)+(𝑽𝒓𝟏
𝟐
−𝑽𝒓𝟐
𝟐
)
(𝑽𝟏
𝟐
−𝑽𝟐
𝟐
)+(𝑼𝟏
𝟐
−𝑼𝟐
𝟐
)+(𝑽𝒓𝟏
𝟐
−𝑽𝒓𝟐
𝟐
)
𝒇𝒐𝒓 𝒂𝒙𝒊𝒂𝒍 𝒇𝒍𝒐𝒘 (𝑼𝟏 = 𝑼𝟐) 𝒂𝒏𝒅 𝑽𝒇𝟏 = 𝑽𝒇𝟐 = 𝑽𝒇 = 𝑽𝟐
∴ 𝑹 =
(𝑽𝒓𝟏
𝟐
−𝑽𝒓𝟐
𝟐
)
(𝑽𝟏
𝟐
−𝑽𝟐
𝟐
)+(𝑽𝒓𝟏
𝟐
−𝑽𝒓𝟐
𝟐
)
=
1
3
𝑠𝑖𝑛𝛼1 =
𝑽𝟐
𝑽𝟏
≫ 𝑽𝟐 = 𝑽𝟏 𝑠𝑖𝑛𝛼1
𝑽𝟐
𝟐
= 𝑽𝟏
𝟐
𝒔𝒊𝒏𝜶𝟏𝟐
𝑽𝒓𝟐
𝟐
= 𝑽𝟐
𝟐
+ 𝑼𝟐
𝑽𝒓𝟐
𝟐
= 𝑽𝟏
𝟐
𝒔𝒊𝒏𝜶𝟏𝟐
+ 𝑼𝟐
𝐴𝑝𝑝𝑙𝑦𝑖𝑛𝑔 𝐶𝑜𝑠𝑖𝑛𝑒 𝑅𝑢𝑙𝑒 𝑡𝑜 𝑂𝐴𝐵
𝑽𝒓𝟏
𝟐
= 𝑽𝟏
𝟐
+ 𝑼𝟐
− 𝟐𝑼𝑽𝟏𝒄𝒐𝒔𝜶𝟏
𝐴𝑝𝑝𝑙𝑦𝑖𝑛𝑔 𝑽𝟐
𝟐
, 𝑽𝒓𝟐
𝟐
𝒂𝒏𝒅 𝑽𝒓𝟏
𝟐
𝒕𝒐 𝑹
𝑹 =
(𝑽𝒓𝟏
𝟐
−𝑽𝒓𝟐
𝟐
)
(𝑽𝟏
𝟐
−𝑽𝟐
𝟐
)+(𝑽𝒓𝟏
𝟐
−𝑽𝒓𝟐
𝟐
)
=
1
3
(𝑽𝟏
𝟐
− 𝑽𝟐
𝟐
) + (𝑽𝒓𝟏
𝟐
− 𝑽𝒓𝟐
𝟐
) = 𝟑 × (𝑽𝒓𝟏
𝟐
− 𝑽𝒓𝟐
𝟐
)
(𝑽𝟏
𝟐
− 𝑽𝟐
𝟐
) = 𝟑 × (𝑽𝒓𝟏
𝟐
− 𝑽𝒓𝟐
𝟐
) − (𝑽𝒓𝟏
𝟐
− 𝑽𝒓𝟐
𝟐
)
∴ (𝑽𝟏
𝟐
− 𝑽𝟐
𝟐
) = 𝟐 × (𝑽𝒓𝟏
𝟐
− 𝑽𝒓𝟐
𝟐
)
𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑛𝑔 𝑽𝟐
𝟐
, 𝑽𝒓𝟐
𝟐
𝒂𝒏𝒅 𝑽𝒓𝟏
𝟐
(𝑽𝟏
𝟐
− 𝑽𝟏
𝟐
𝒔𝒊𝒏𝜶𝟏𝟐) = 𝟐 × ((𝑽𝟏
𝟐
+ 𝑼𝟐
− 𝟐𝑼𝑽𝟏𝒄𝒐𝒔𝜶𝟏) − (𝑽𝟏
𝟐
𝒔𝒊𝒏𝜶𝟏𝟐
+ 𝑼𝟐
𝟐
))
(𝑽𝟏
𝟐
− 𝑽𝟏
𝟐
𝒔𝒊𝒏𝜶𝟏𝟐) = 𝟐(𝑽𝟏
𝟐
+ 𝑼𝟐
− 𝟐𝑼𝑽𝟏𝒄𝒐𝒔𝜶𝟏 − 𝑽𝟏
𝟐
𝒔𝒊𝒏𝜶𝟏𝟐
− 𝑼𝟐
𝟐
)
(𝑽𝟏
𝟐
− 𝑽𝟏
𝟐
𝒔𝒊𝒏𝜶𝟏𝟐) = 𝟐(𝑽𝟏
𝟐
− 𝟐𝑼𝑽𝟏𝒄𝒐𝒔𝜶𝟏 − 𝑽𝟏
𝟐
𝒔𝒊𝒏𝜶𝟏𝟐
)
𝑽𝟏
𝟐
− 𝑽𝟏
𝟐
𝒔𝒊𝒏𝜶𝟏𝟐
= 𝟐𝑽𝟏
𝟐
− 𝟒𝑼𝑽𝟏𝒄𝒐𝒔𝜶𝟏 − 𝟐𝑽𝟏
𝟐
𝒔𝒊𝒏𝜶𝟏𝟐
𝑽𝟏
𝟐
− 𝟐𝑽𝟏
𝟐
− 𝑽𝟏
𝟐
𝒔𝒊𝒏𝜶𝟏𝟐
+ 𝟐𝑽𝟏
𝟐
𝒔𝒊𝒏𝜶𝟏𝟐
= −𝟒𝑼𝑽𝟏𝒄𝒐𝒔𝜶𝟏
−𝑽𝟏
𝟐
+ 𝑽𝟏
𝟐
𝒔𝒊𝒏𝜶𝟏𝟐
= −𝟒𝑼𝑽𝟏𝒄𝒐𝒔𝜶𝟏
𝑽𝟏
𝟐
𝒔𝒊𝒏𝜶𝟏𝟐
− 𝑽𝟏
𝟐
= 𝟒𝑼𝑽𝟏𝒄𝒐𝒔𝜶𝟏
𝑽𝟏
𝟐
𝒄𝒐𝒔𝜶𝟏𝟐
= 𝟒𝑼𝑽𝟏𝒄𝒐𝒔𝜶𝟏
𝟏
𝟒
𝒄𝒐𝒔𝜶𝟏𝟐
𝒄𝒐𝒔𝜶𝟏
=
𝑼𝑽𝟏
𝑽𝟏
𝟐 ≫=
𝑼
𝑽𝟏
=
𝟏
𝟒
𝒄𝒐𝒔𝜶𝟏
𝑺𝒕𝒖𝒅𝒆𝒏𝒕 𝑨𝒄𝒕𝒊𝒗𝒊𝒕𝒚 𝒊𝒇 𝑹 =
𝟏
𝟒
𝒕𝒉𝒆𝒏 𝒔𝒉𝒐𝒘 ≫=
𝑼
𝑽𝟏
=
𝟐
𝟑
𝒄𝒐𝒔𝜶𝟏

12. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 12
3 c) The mean rotor blade speed of an axial flow turbine stage with a degree of reaction of 50% is
210 m/s. The steam emerges from the nozzle inclined at 28º to the wheel plane, with an axial
velocity component which is equal to the blade speed. Assuming symmetric inlet and outlet
velocity triangles, find the rotor blade angles and the utilization factor. Find also the degree of
reaction to make the utilization a maximum, if the axial velocity and the blade speed as well as
the nozzle remain the same as above.
𝑹 = 𝟎.𝟓, 𝑼 = 𝟐𝟏𝟎
𝒎
𝒔
, 𝜶𝟏 = 𝟐𝟖° 𝑽𝒇𝟏 = 𝑼 = 𝟐𝟏𝟎
𝒎
𝒔
,
𝑹 = 𝟎. 𝟓 𝒕𝒉𝒆𝒏 𝑽𝒓𝟏 = 𝑽𝟐 & 𝑽𝒓𝟐 = 𝑽𝟏
Case 02) When R = 0.5
(i.e., 50% Reaction axial flow)
This implies
U1= U2,
V1 = Vr2,
V2=Vr1 and
Vf1 = Vf2.
For symmetric Velocity
α2 = β1 and α1= β2.
tan 𝛼1 =
𝑉𝑓1
𝑉𝑢1
≫ 𝑉𝑢1 =
𝑉𝑓1
tan 𝛼1
=
210
𝑡𝑎𝑛28
= 394.95 𝑉𝑟1 =
𝑉𝑓1
𝑠𝑖𝑛𝛽1
tan 𝛽1 =
𝑉𝑓1
𝑉𝑢1 − 𝑈
=
210
394.95 − 210
=
210
184.95
𝑉𝑟1 =
115
𝑠𝑖𝑛60
= 132.8 𝑚/𝑠
∴ tan 𝛽1 = 1.13544 ≫ 𝛽1 = tan−1(1.13544)
𝛽1 = 48.62°
𝜶𝟏 = 𝜷𝟐 & 𝜶𝟐 = 𝜷𝟏
𝜶𝟏 = 𝜷𝟐 = 𝟐𝟖° & 𝜶𝟐 = 𝜷𝟏 = 𝟒𝟖. 𝟔𝟐°
(𝝐) =
(𝑽𝟏
𝟐
− 𝑽𝟐
𝟐
)
(𝑽𝟏
𝟐
− 𝑹 𝑽𝟐
𝟐
)
sin 𝛼1 =
𝑉𝑓1
𝑉1
≫ 𝑉1 =
𝑉𝑓1
sin 𝛼1
=
210
𝑠𝑖𝑛28
= 447.31
𝑉1 = 447.31 𝑎𝑛𝑑 𝑉2 = 279.87 & 𝑅 = 0.5
sin 𝛽1 =
𝑉𝑓1
𝑉𝑟1
≫ 𝑉𝑟1 =
𝑉𝑓1
sin𝛽1
=
210
𝑠𝑖𝑛48.62
= 279.87
(𝝐) =
(𝑽𝟏
𝟐
− 𝑽𝟐
𝟐
)
(𝑽𝟏
𝟐
− 𝑹 𝑽𝟐
𝟐
)
=
(𝟒𝟒𝟕. 𝟑𝟏𝟐
− 𝟐𝟕𝟗. 𝟖𝟕𝟐
)
(𝟒𝟒𝟕. 𝟑𝟏𝟐
− (𝟎. 𝟓)𝟐𝟕𝟗. 𝟖𝟕𝟐
)
(𝝐) = 𝟎. 𝟕𝟓𝟕
𝑓𝑜𝑟 𝑅 = 50%; 𝑉2 = 𝑉𝑟1 = 279.87

13. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 13
Previous Year Question papers
Model Question Paper (CBCS) with effect from 2015-16)
Module – 2
3 a) Derive an alternate form of Euler Turbine equation. 8
b) In an axial flow turbine, the discharge blade angles are 20° each for both the stator and the
rotor. The steam speed from the nozzle exit is 140m/s. The ratio of
𝑽𝒂
𝑼
= 𝟎. 𝟕 at the entry
and 0.76 at the exit of the rotor blade. Find the rotor inlet blade angle and the power
developed by the blade ring for a mass flow rate of 2.6kg/s.
8
axial flow turbine 𝑈1 = 𝑈2 = 𝑈
the discharge blade angle is 20ºeach, for both the stator and the rotor.
i.e., ∝1= 20º; 𝛽2 = 20º;
The steam speed at the exit of the fixed blade is 140m/s i.e., 𝑉1 = 140m/s
𝑽𝒂𝟏
𝑼
= 𝟎. 𝟕𝑎𝑡 𝑡ℎ𝑒 𝑒𝑛𝑡𝑟𝑦 𝑎𝑛𝑑
𝑽𝒂𝟐
𝑼
= 𝟎.𝟕𝟔 𝑎𝑡 𝑡ℎ𝑒 𝑒𝑥𝑖𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑟𝑜𝑡𝑜𝑟 𝑏𝑙𝑎𝑑𝑒
Find the rotor inlet blade angle
power developed by the blade ring
mass flow rate of 2.6kg/s.
𝑃 = (𝑉𝑢1. 𝑈1 − 𝑉𝑢2. 𝑈2)
𝑷 = (𝑽𝒖𝟏 − 𝑽𝒖𝟐)𝑼
cos𝛼1 =
𝑉𝑢1
𝑉1
≫ 𝑉𝑢1 = 𝑉1 cos𝛼1
𝑉𝑢1 = 140 cos20 ≫ 𝑽𝒖𝟏 = 𝟏𝟑𝟏. 𝟓𝟓
sin 𝛼1 =
𝑉𝑓1
𝑉1
≫ 𝑉𝑓1 = 𝑉1 sin 𝛼1
𝑉𝑓1 = 140 sin 20 ≫ 𝑽𝒇𝟏 = 𝟒𝟕. 𝟖𝟖
𝑽𝒂𝟏
𝑼
=
𝑽𝒇𝟏
𝑼
=
𝟒𝟕.𝟖𝟖
𝑼
= 𝟎.𝟕
47.88
0.7
= 𝑼 = 𝟔𝟖.𝟒𝟎
𝑽𝒂𝟐
𝑼
=
𝑽𝒇𝟐
𝑼
= 𝟎.𝟕𝟔
𝑽𝒇𝟐 = 𝟎. 𝟕𝟔 × 𝑼 ≫ 𝑽𝒇𝟐 = 𝟓𝟏.𝟗𝟖
tan𝛽1 =
𝑉𝑓1
𝑉𝑢1 − 𝑈
=
47.88
131.55 − 68.40
𝛽1 = 37.16°
sin𝛽2 =
𝑉𝑓2
𝑉𝑟2
≫ 𝑉𝑟2 =
𝑉𝑓2
sin 𝛽2
=
51.98
sin20
𝑽𝒓𝟐 = 𝟏𝟓𝟏. 𝟗𝟕
cos 𝛽2 =
(𝑉𝑢2 + 𝑈)
𝑉𝑟2
= (𝑉𝑢2 + 𝑈) = cos 𝛽2 𝑉𝑟2
𝑉𝑢2 = 𝑉𝑟2cos 𝛽2 − 𝑈 𝑉𝑢2 = 151.97cos20 − 68.40 ≫ 𝑽𝒖𝟐 = 𝟕𝟒. 𝟒𝟎
𝑷 = (𝑽𝒖𝟏 − 𝑽𝒖𝟐)𝑼 𝑷 = (𝟏𝟑𝟏.𝟓𝟓 − 𝟕𝟒.𝟒𝟎)𝟔𝟖.𝟒𝟎
𝑷 = 𝟑𝟗𝟎𝟗.𝟎𝟔 𝑱
𝑷
𝒎
=
𝟑𝟗𝟎𝟗.𝟎𝟔
𝟐.𝟔
= 𝟏𝟓𝟎𝟑.𝟒𝟖 𝑱/𝒌𝒈

14. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 14
Previous Year Question papers
Model Question Paper (CBCS) with effect from 2015-16)
Module – 2
3 a) In a certain turbomachine, the blade speed at exit is twice that at inlet (u2=2u1), the meridian
component of fluid velocity at inlet is equal to that at exit and the blade angle at inlet is 450. Show
that the energy transfer per unit mass and degree of reaction is given 8
b) At a stage of 50% reaction axial flow turbine running at 3000 rpm, the mean blade diameter is 68.5
cm. If the maximum utilization factor for the stage is 0.915, Calculate (a) the inlet and outlet absolute
velocities and (b) the power output. Also, find the power developed for a steam flow rate of 15 kg/s.
8
Case 02) When R = 0.5
(i.e., 50% Reaction axial flow)
This implies
U1= U2,
V1 = Vr2,
V2=Vr1 and
Vf1 = Vf2.
For symmetric Velocity
α2 = β1 and α1= β2.