Modal 03: Question Number 5 a & 5 b i. Reaction Turbine (Parsons’s turbine) ii. Degree of Reaction for Parsons’s turbine iii. Condition for maximum utilization factor, iv. Reaction staging. v. Numerical Problems. 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 03: Steam Reaction turbine
Course Learning Objectives
Analyze various designs of Reaction turbine and their working principle.
Course Outcomes
Classify, analyze and understand various type of Reaction turbine..

2. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 2
Contents
Modal 03: Question Number 5 a & 5 b
i. Reaction Turbine (Parsons’s turbine)
ii. Degree of Reaction for Parsons’s turbine
iii. Condition for maximum utilization factor,
iv. Reaction staging.
v. Numerical Problems.
Previous Year Question papers

3. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 3
Introduction
Classification of Steam Turbine, based on basic working principle
a) Impulse turbine and
b) Reaction turbine
The machine for which the change in static head in the rotor is zero is known impulse machine. In these
machines, the energy transfer in the rotor takes place only by the change in dynamic head of the fluid.
In reaction turbine energy transfer in the rotor takes place by change in static and dynamic head of the
fluid.
In reaction turbine 𝑉𝑟2 > 𝑉𝑟1 and generally 𝑉f1 = 𝑉f2 (i.e. flow velocity is constant)
Reaction Turbine (Example: Parson’s Turbine)
In the case of reaction turbine, the moving blades of a turbine are shaped in such a way that the steam
expands and drops in pressure as it passes through them. As a result of pressure decrease in the moving
blade, a reaction force will be produced. This force will make the blades to rotate.
The combined velocity diagram for an axial flow reaction turbine is as shown in figure below.
From data given in the problem, Vf1=Vf2=Vf.
𝑇ℎ𝑒 𝐸𝑙𝑢𝑒𝑟′
𝑇𝑢𝑟𝑏𝑖𝑛𝑒 𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑖𝑠 𝑔𝑖𝑣𝑒𝑛 𝑎𝑠: 𝑃 = 𝑚(𝑉𝑢1. 𝑈1 ± 𝑉𝑢2. 𝑈2)
𝑓𝑜𝑟 𝑆𝑡𝑒𝑎𝑚 𝑇𝑢𝑟𝑏𝑖𝑛𝑒 𝑈1 = 𝑈2 = 𝑈 ∴ 𝑷 = 𝒎𝑼(𝑽𝒖𝟏 + 𝑽𝒖𝟐)
𝑐𝑜𝑡 𝛽1 =
𝑉𝑢1
𝑉𝑓1
∴ 𝑽𝒖𝟏 = 𝒄𝒐𝒕 𝜷𝟏 × 𝑽𝒇𝟏
𝑆𝑖𝑚𝑖𝑙𝑙𝑎𝑟𝑙𝑦 𝑐𝑜𝑡 𝛽2 =
𝑋2 − 𝑈
𝑉𝑓1
∴ 𝑽𝒖𝟐 = 𝑿𝟐 − 𝑼 = (𝑽𝒖𝟐 + 𝑼) − 𝑼 = 𝒄𝒐𝒕 𝜷𝟐 × 𝑽𝒇𝟐
𝑃 = 𝑚𝑈(𝑉𝑢1 + 𝑉𝑢2) = 𝑚𝑈(𝑉𝑢1 + (𝑋2 − 𝑈)) = 𝑚𝑈(𝑐𝑜𝑡 𝛽1 × 𝑉𝑓1 + 𝑐𝑜𝑡 𝛽2 × 𝑉𝑓2)
𝑷 = 𝒎 𝑼 𝑽𝒇(𝒄𝒐𝒕 𝜷𝟏 + 𝒄𝒐𝒕 𝜷𝟐)𝑎𝑠 𝑽𝒇𝟏 = 𝑽𝒇𝟐 = 𝑽𝒇

4. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 4
Degree of Reaction for Reaction Turbine (Parson’s Turbine)
𝑹 =
1
2
(𝑼𝟏
𝟐
−𝑼𝟐
𝟐)+(𝑽𝒓𝟐
𝟐
−𝑽𝒓𝟏
𝟐)
1
2
(𝑽𝟏
𝟐
−𝑽𝟐
𝟐)+(𝑼𝟏
𝟐
−𝑼𝟐
𝟐)+(𝑽𝒓𝟏
𝟐
−𝑽𝒓𝟐
𝟐)
=
1
2
(𝑽𝒓𝟐
𝟐
−𝑽𝒓𝟏
𝟐)
1
2
(𝑽𝟏
𝟐
−𝑽𝟐
𝟐)+(𝑽𝒓𝟏
𝟐
−𝑽𝒓𝟐
𝟐)
𝑹 =
1
2
(𝑽𝒓𝟐
𝟐
− 𝑽𝒓𝟏
𝟐
)
𝑃
=
(𝑽𝒓𝟐
𝟐
− 𝑽𝒓𝟏
𝟐
)
𝟐 𝑷
𝑆𝑖𝑛 𝛽1 =
𝑉𝑓
𝑉𝑟1
∴ 𝑽𝒓𝟏 =
𝑉𝑓
𝑆𝑖𝑛 𝛽1
= 𝑽𝒇 𝒄𝒐𝒔𝒆𝒄 𝜷𝟏 𝑏𝑒𝑐𝑎𝑢𝑠𝑒
1
𝑆𝑖𝑛 𝛽1
= 𝑐𝑜𝑠𝑒𝑐 𝛽1
𝑆𝑖𝑛 𝛽2 =
𝑉𝑓
𝑉𝑟2
∴ 𝑽𝒓𝟐 =
𝑉𝑓
𝑆𝑖𝑛 𝛽2
= 𝑽𝒇 𝒄𝒐𝒔𝒆𝒄 𝜷𝟐 𝑏𝑒𝑐𝑎𝑢𝑠𝑒
1
𝑆𝑖𝑛 𝛽2
= 𝑐𝑜𝑠𝑒𝑐 𝛽2
(𝑉𝑟1
2
− 𝑉𝑟2
2
) = 𝑉𝑓2 (𝑐𝑜𝑠𝑒𝑐 𝛽12
− 𝑐𝑜𝑠𝑒𝑐 𝛽22)
𝒃𝒚 𝑻𝒓𝒊𝒈𝒏𝒐𝒎𝒆𝒕𝒓𝒚 𝒓𝒆𝒍𝒂𝒕𝒊𝒐𝒏,𝒄𝒐𝒔𝒆𝒄 𝟐
𝜽 = 𝟏 + 𝒄𝒐𝒕 𝟐
𝜽
∴ (𝑽𝒓𝟏
𝟐
− 𝑽𝒓𝟐
𝟐
) = 𝑽𝒇𝟐(𝟏 + 𝒄𝒐𝒕 𝜷𝟏𝟐
− 𝟏 + 𝒄𝒐𝒕 𝜷𝟐𝟐)
(𝑽𝒓𝟏
𝟐
− 𝑽𝒓𝟐
𝟐
) = 𝑽𝒇𝟐(𝒄𝒐𝒕 𝜷𝟏𝟐
− 𝒄𝒐𝒕 𝜷𝟐𝟐)
𝑹 =
(𝑽𝒓𝟐
𝟐
−𝑽𝒓𝟏
𝟐)
𝟐 𝑷
=
𝑽𝒇𝟐(𝒄𝒐𝒕 𝜷𝟐𝟐−𝒄𝒐𝒕 𝜷𝟏𝟐)
𝟐 𝑼 𝑽𝒇(𝒄𝒐𝒕 𝜷𝟏+𝒄𝒐𝒕 𝜷𝟐)
=
𝑽𝒇(𝒄𝒐𝒕 𝜷𝟐𝟐−𝒄𝒐𝒕 𝜷𝟏𝟐)
𝟐 𝑼 (𝒄𝒐𝒕 𝜷𝟏+𝒄𝒐𝒕 𝜷𝟐)
𝑹 =
𝑽𝒇(𝒄𝒐𝒕 𝜷𝟐𝟐−𝒄𝒐𝒕 𝜷𝟏𝟐)
𝟐 𝑼 (𝒄𝒐𝒕 𝜷𝟏+𝒄𝒐𝒕 𝜷𝟐)
=
𝑽𝒇(𝒂𝟐−𝒃𝟐)
𝟐 𝑼 (𝒃+𝒂)
=
𝑽𝒇(𝒂−𝒃)(𝒂+𝒃)
𝟐 𝑼 (𝒃+𝒂)
=
𝑽𝒇(𝒂−𝒃)
𝟐 𝑼
𝑹 =
𝑽𝒇(𝒄𝒐𝒕 𝜷𝟐 − 𝒄𝒐𝒕 𝜷𝟏)
𝟐 𝑼
𝐹𝑜𝑟 𝑎𝑛 𝑎𝑥𝑖𝑎𝑙 𝑓𝑙𝑜𝑤 50% 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑡𝑢𝑟𝑏𝑖𝑛𝑒, 𝑅 = 0.5
𝑹 =
𝑽𝒇(𝒄𝒐𝒕 𝜷𝟐−𝒄𝒐𝒕 𝜷𝟏)
𝟐 𝑼
=
𝟏
𝟐
≫ 𝑽𝒇(𝒄𝒐𝒕 𝜷𝟐 − 𝒄𝒐𝒕 𝜷𝟏) =
𝟐 𝑼
𝟐
∴ 𝑼 = 𝑽𝒇(𝒄𝒐𝒕 𝜷𝟐 − 𝒄𝒐𝒕 𝜷𝟏)
𝑨𝒍𝒕𝒆𝒓𝒏𝒂𝒕𝒆 𝒎𝒆𝒕𝒉𝒐𝒅: 𝑭𝒓𝒐𝒎 𝒗𝒆𝒍𝒐𝒄𝒊𝒕𝒚 𝒅𝒊𝒂𝒈𝒓𝒂𝒎,
𝑼 = 𝑽𝒖𝟏 − 𝑿𝟏 = (𝑽𝒇𝟏 𝒄𝒐𝒕𝜶𝟏 − 𝑽𝒇𝟏 𝒄𝒐𝒕𝜷𝟏)
𝐹𝑜𝑟 𝑎𝑛 𝑎𝑥𝑖𝑎𝑙 𝑓𝑙𝑜𝑤 50% 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑡𝑢𝑟𝑏𝑖𝑛𝑒,
𝛼1 = 𝛽2 𝑎𝑛𝑑 𝛼2 = 𝛽1 𝑎𝑛𝑑 𝑎𝑙𝑠𝑜 𝑉1 = 𝑉𝑟2 𝑎𝑛𝑑 𝑉2 = 𝑉𝑟1
∴ 𝑖𝑡 𝑐𝑎𝑛 𝑏𝑒 𝑑𝑖𝑟𝑐𝑡𝑙𝑦 𝑝𝑟𝑜𝑣𝑒𝑛 𝑡ℎ𝑎𝑡
𝑼 = 𝑽𝒇(𝒄𝒐𝒕 𝜷𝟐 − 𝒄𝒐𝒕 𝜷𝟏)

5. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 5
𝑭𝒐𝒓 𝒂 𝟓𝟎% 𝒓𝒆𝒂𝒄𝒕𝒊𝒐𝒏 𝒔𝒕𝒆𝒂𝒎 𝒕𝒖𝒓𝒃𝒊𝒏𝒆,𝒔𝒉𝒐𝒘 𝒕𝒉𝒂𝒕 𝜶𝟏 = 𝜷𝟐 𝒂𝒏𝒅 𝜶𝟐 = 𝜷𝟏
𝑅 =
𝑉𝑓(𝑐𝑜𝑡 𝛽2−𝑐𝑜𝑡 𝛽1)
2 𝑈
=
1
2
(
𝑉𝑓
𝑈
) (𝑐𝑜𝑡 𝛽2 − 𝑐𝑜𝑡 𝛽1)
𝑅 =
1
2
(
𝑉𝑓
𝑈
) (𝑐𝑜𝑡 𝛽2 − 𝑐𝑜𝑡 𝛽1) 𝑎𝑑𝑑 𝑎𝑛𝑑 𝑠𝑢𝑏𝑡𝑟𝑎𝑐𝑡 𝒄𝒐𝒕 𝜶𝟏
𝑹 =
𝟏
𝟐
(
𝑽𝒇
𝑼
) (𝒄𝒐𝒕 𝜷𝟐 − 𝒄𝒐𝒕 𝜶𝟏) + (𝒄𝒐𝒕 𝜶𝟏 − 𝒄𝒐𝒕 𝜷𝟏)
𝑤𝑒 𝑐𝑎𝑛 𝑑𝑒𝑟𝑖𝑣𝑒 𝑼 = 𝑽𝒖𝟏 − 𝑿𝟏 = (𝑉𝑓1 𝑐𝑜𝑡𝛼1 − 𝑉𝑓1 𝑐𝑜𝑡𝛽1) = 𝑉𝑓( 𝑐𝑜𝑡𝛼1 − 𝑐𝑜𝑡𝛽1)
𝑜𝑟 ( 𝒄𝒐𝒕𝜶𝟏 − 𝒄𝒐𝒕𝜷𝟏) =
𝑼
𝑽𝒇
𝑇ℎ𝑒𝑛 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑅 𝑐𝑎𝑛 𝑏𝑒 𝑟𝑒 𝑤𝑟𝑖𝑡𝑡𝑒𝑛 𝑎𝑠
𝑅 =
1
2
(
𝑉𝑓
𝑈
) (𝑐𝑜𝑡 𝛽2 − 𝑐𝑜𝑡 𝛼1) + (𝑐𝑜𝑡 𝛼1 − 𝑐𝑜𝑡 𝛽1) =
1
2
(
𝑉𝑓
𝑈
) (𝑐𝑜𝑡 𝛽2 − 𝑐𝑜𝑡 𝛼1 + (
𝑈
𝑉𝑓
))
𝑅 =
1
2
(
𝑉𝑓
𝑈
)𝑐𝑜𝑡 𝛽2 −
1
2
(
𝑉𝑓
𝑈
)𝑐𝑜𝑡 𝛼1 +
1
2
(
𝑉𝑓
𝑈
)(
𝑈
𝑉𝑓
) =
𝟏
𝟐
(
𝑽𝒇
𝑼
) 𝒄𝒐𝒕 𝜷𝟐 −
𝟏
𝟐
(
𝑽𝒇
𝑼
) 𝒄𝒐𝒕 𝜶𝟏 +
𝟏
𝟐
𝑹 =
𝟏
𝟐
(
𝑽𝒇
𝑼
) 𝒄𝒐𝒕 𝜷𝟐 −
𝟏
𝟐
(
𝑽𝒇
𝑼
)𝒄𝒐𝒕 𝜶𝟏 +
𝟏
𝟐
= 𝟎. 𝟓
𝑹 =
𝟏
𝟐
(
𝑽𝒇
𝑼
)𝒄𝒐𝒕 𝜷𝟐 −
𝟏
𝟐
(
𝑽𝒇
𝑼
) 𝒄𝒐𝒕 𝜶𝟏 = 𝟎.𝟓 −
𝟏
𝟐
= 𝟎
𝟏
𝟐
(
𝑽𝒇
𝑼
) (𝒄𝒐𝒕 𝜷𝟐 − 𝒄𝒐𝒕 𝜶𝟏) = 𝟎 ≫ (𝑐𝑜𝑡 𝛽2 − 𝑐𝑜𝑡 𝛼1) = 0 ≫ cot 𝛽2 = cot 𝛼1 ≫
∴ 𝜶𝟏 = 𝜷𝟐
𝑠𝑖𝑚𝑖𝑙𝑙𝑎𝑟𝑙𝑦 𝑤𝑒 𝑐𝑎𝑛 𝑑𝑒𝑟𝑖𝑣𝑒 𝑼 = 𝑿𝟐 − 𝑽𝒖𝟐 = (𝑉𝑓 cot 𝛽2 − 𝑉𝑓 𝑐𝑜𝑡 𝛼2)
∴ 𝑼 = 𝑽𝒇(𝐜𝐨𝐭 𝜷𝟐 − 𝒄𝒐𝒕 𝜶𝟐)
𝐶𝑜𝑚𝑝𝑎𝑟𝑖𝑛𝑔 𝑤𝑖𝑡ℎ 𝑼 = 𝑽𝒖𝟏 − 𝑿𝟏 = (𝑉𝑓1 𝑐𝑜𝑡𝛼1 − 𝑉𝑓1 𝑐𝑜𝑡𝛽1)
𝑖. 𝑒 𝑼 = 𝑽𝒇( 𝒄𝒐𝒕𝜶𝟏 − 𝒄𝒐𝒕𝜷𝟏)
∴ 𝑈 = 𝑉𝑓(cot 𝛽2 − 𝑐𝑜𝑡 𝛼2) = 𝑉𝑓( 𝑐𝑜𝑡𝛼1 − 𝑐𝑜𝑡𝛽1)
𝐜𝐨𝐭 𝜷𝟐 − 𝒄𝒐𝒕 𝜶𝟐 = 𝒄𝒐𝒕𝜶𝟏 − 𝒄𝒐𝒕𝜷𝟏
𝑤𝑒 𝑘𝑛𝑜𝑤 𝑡ℎ𝑎𝑡 𝛼1 = 𝛽2 𝑠𝑜 𝑐𝑜𝑛𝑣𝑒𝑟𝑡𝑖𝑛𝑔 cot 𝛽2 = 𝑐𝑜𝑡 𝛼1
∴ cot 𝛼1 − 𝑐𝑜𝑡 𝛼2 = 𝑐𝑜𝑡𝛼1 − 𝑐𝑜𝑡𝛽1
∴ 𝜶𝟐 = 𝜷𝟏
sin 𝛼2 =
𝑉𝑓
𝑉2
and 𝑠𝑖𝑛 𝛽1 =
𝑉𝑓
𝑉𝑟1
∴ Vf = V2 sin 𝛼2 = 𝑉𝑟1 𝑠𝑖𝑛 𝛽1 𝑏𝑢𝑡 𝛼2 = 𝛽1
∴ 𝐕𝟐 = 𝑽𝒓𝟏 𝑠𝑖𝑚𝑖𝑙𝑙𝑎𝑟𝑙𝑦 𝑽𝟏 = 𝑽𝒓𝟐

6. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 6
Condition for maximum utilization factor,
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) =
𝟏
𝟐
[(𝑽𝟏
𝟐
− 𝑽𝟐
𝟐
) + (𝑽𝒓𝟏
𝟐
− 𝑽𝒓𝟐
𝟐
)]
𝟏
𝟐
[(𝑽𝟏
𝟐
− 𝑽𝟐
𝟐
) + (𝑽𝒓𝟏
𝟐
− 𝑽𝒓𝟐
𝟐
)] +
𝟏
𝟐
(𝑽𝟐
𝟐
)
∴ 𝑼𝒕𝒊𝒍𝒊𝒛𝒂𝒕𝒊𝒐𝒏 𝑭𝒂𝒄𝒕𝒐𝒓 (𝝐) =
𝑷
𝑷 +
𝟏
𝟐
(𝑽𝟐
𝟐
)
Reaction staging
Reaction staging the expansion of steam and enthalpy drop occurs both in fixed and moving blades.
Due to the effect of continuous expansion during flow over the moving blades, the relative velocity of
steam increases i.e., Vr2>Vr1.
Parsons stages of steam turbines have a degree of reaction (R) equal to 0.5. One stator blade-ring
followed by a rotor blade-ring together make up the one stage. The blades are designed so that the
passages between the blades act like nozzles in both the stator and rotor. The expansion of steam takes
place in both the rings.
The areas of flow have to be increased continuously to accommodate the increased volume flow rates.
The velocity triangles at the inlet and outlet of every rotor blade ring become symmetrical. The usual
practice is to have the same geometry (α1, α2, β1, β2) of the blades with continuously increasing heights.
The same set of velocity triangles and analysis hold good for a few of the rotor rings in succession.
When the increase in the height of blades becomes limited after a few rings, the mean diameter of rotor
rings can be increased so that another set of velocity triangles and analysis can hold good for another
series of rotor rings.
For Parson’s (axial flow 50% reaction) turbine, α1=β2 and α2=β1 and also V1=Vr2 and V2=Vr1, then
the velocity triangles are symmetric.
Work done or Power developed is:
𝑷 = 𝑼(∆𝑽𝒖) = 𝑈(𝑉𝑢1 + 𝑉𝑢2) = 𝑼(𝑽𝒖𝟏 + 𝑿𝟐 − 𝑼)
𝑽𝒖𝟏 = 𝑽𝟏 𝒄𝒐𝒔 𝜶𝟏
𝑋2 − 𝑈 = 𝑉𝑟2 𝑐𝑜𝑠 𝛽2 − 𝑈
𝑏𝑢𝑡 𝜶𝟏 = 𝜷𝟐 𝒂𝒏𝒅 𝑽𝟏 = 𝑽𝒓𝟐
∴ 𝑿𝟐 − 𝑼 = 𝑽𝟏 𝒄𝒐𝒔 𝜶𝟏 − 𝑼
∴ 𝑃 = 𝑈(𝑉𝑢1 + 𝑋2 − 𝑈) = 𝑈(𝑉1 𝑐𝑜𝑠 𝛼1 + 𝑉1 𝑐𝑜𝑠 𝛼1 − 𝑈)
𝑷 = 𝟐𝑼𝑽𝟏 𝒄𝒐𝒔 𝜶𝟏 − 𝑼𝟐

7. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 7
𝑷 = 𝟐𝑼𝑽𝟏 𝒄𝒐𝒔 𝜶𝟏 − 𝑼𝟐
𝑃 =
𝑉12
𝑉12
(2𝑈𝑉1 𝑐𝑜𝑠 𝛼1 − 𝑈2) = 𝑉12
(
2𝑈𝑉1 𝑐𝑜𝑠 𝛼1
𝑉12
−
𝑈2
𝑉12
) = 𝑉12(2𝜑 𝑐𝑜𝑠 𝛼1 − 𝜑2)
𝑃 = 𝑉12(2𝜑 𝑐𝑜𝑠 𝛼1 − 𝜑2) 𝑤ℎ𝑒𝑟𝑒 𝜑 =
𝑈
𝑉1
𝑤𝑒 𝑘𝑛𝑜𝑤 𝑡ℎ𝑎𝑡 𝑃𝑎𝑣𝑖𝑙𝑎𝑏𝑙𝑒 =
1
2
[(𝑉1
2
) − (𝑉𝑟1
2
− 𝑉𝑟2
2
)]
𝑃𝑎𝑣𝑖𝑙𝑎𝑏𝑙𝑒 =
1
2
(𝑉1
2
) −
1
2
(𝑉𝑟1
2
) +
1
2
(𝑉𝑟2
2
) 𝑏𝑢𝑡 𝑉1 = 𝑉𝑟2
∴ 𝑃𝑎𝑣𝑖𝑙𝑎𝑏𝑙𝑒 =
1
2
(𝑉1
2
) −
1
2
(𝑉𝑟1
2
) +
1
2
(𝑉1
2
) ≫ 𝑉1
2
−
1
2
(𝑉𝑟1
2
)
𝑏𝑦 𝐶𝑜𝑠𝑖𝑛𝑒 𝑟𝑢𝑙𝑒 𝑉𝑟1
2
= 𝑉1
2
+ 𝑈2
− 2𝑈𝑉1 𝑐𝑜𝑠 𝛼1
𝑃𝑎𝑣𝑖𝑙𝑎𝑏𝑙𝑒 = 𝑉1
2
−
1
2
(𝑉1
2
+ 𝑈2
− 2𝑈𝑉1 𝑐𝑜𝑠 𝛼1)
𝑃𝑎𝑣𝑖𝑙𝑎𝑏𝑙𝑒 = 𝑉1
2
−
1
2
𝑉1
2
−
1
2
𝑈2
+
1
2
2𝑈𝑉1 𝑐𝑜𝑠 𝛼1
𝑃𝑎𝑣𝑖𝑙𝑎𝑏𝑙𝑒 =
1
2
𝑉1
2
−
1
2
𝑈2
+
1
2
2𝑈𝑉1 𝑐𝑜𝑠 𝛼1 ≫
1
2
[𝑉1
2
− 𝑈2
+ 2𝑈𝑉1 𝑐𝑜𝑠 𝛼1]
𝑡𝑎𝑘𝑖𝑛𝑔 𝑉1
2
𝑐𝑜𝑚𝑚𝑜𝑛 𝑜𝑢𝑡𝑠𝑖𝑑𝑒 𝑃𝑎𝑣𝑖𝑙𝑎𝑏𝑙𝑒 =
1
2
𝑉1
2
[1 −
𝑈2
𝑉1
2 +
2𝑈𝑉1
𝑉1
2 𝑐𝑜𝑠 𝛼1]
𝑃𝑎𝑣𝑖𝑙𝑎𝑏𝑙𝑒 =
1
2
𝑉1
2
[1 − 𝜑2
+ 2𝜑 𝑐𝑜𝑠 𝛼1]
𝑜𝑟 𝑷𝒂𝒗𝒊𝒍𝒂𝒃𝒍𝒆 =
𝑽𝟏
𝟐
𝟐
[𝟏 + 𝟐𝝋 𝒄𝒐𝒔 𝜶𝟏 − 𝝋𝟐]
𝜼𝒃 =
𝑷
𝑷𝒂𝒗𝒊𝒍𝒂𝒃𝒍𝒆
=
𝑉12(2𝜑 𝑐𝑜𝑠 𝛼1 − 𝜑2)
𝑉1
2
2
[1 + 2𝜑 𝑐𝑜𝑠 𝛼1 − 𝜑2]
𝜂𝑏 =
2(2𝜑 𝑐𝑜𝑠 𝛼1−𝜑2)
[1+2𝜑 𝑐𝑜𝑠 𝛼1−𝜑2]
≫
(4𝜑 𝑐𝑜𝑠 𝛼1−2𝜑2+2−2)
[1+2𝜑 𝑐𝑜𝑠 𝛼1−𝜑2]
≫
2(1+2𝜑 𝑐𝑜𝑠 𝛼1−𝜑2)−2
[1+2𝜑 𝑐𝑜𝑠 𝛼1−𝜑2]
𝜼𝒃 =
2(1 + 2𝜑 𝑐𝑜𝑠 𝛼1 − 𝜑2) − 2
[1 + 2𝜑 𝑐𝑜𝑠 𝛼1 − 𝜑2]
=
𝟐(𝟐𝝋 𝒄𝒐𝒔 𝜶𝟏 − 𝝋𝟐)
[𝟏 + 𝟐𝝋 𝒄𝒐𝒔 𝜶𝟏 − 𝝋𝟐]
𝜂𝑏 = 2 −
2
[1+2𝜑 𝑐𝑜𝑠 𝛼1−𝜑2]
≫ 𝟐 − 𝟐[𝟏 + 𝟐𝝋 𝒄𝒐𝒔 𝜶𝟏 − 𝝋𝟐]
−𝟏
𝐹𝑜𝑟 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑏𝑙𝑎𝑑𝑒 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦
𝒅𝜼𝒃
𝒅𝝋
= 𝟎
𝑑
𝑑𝜑
(2 − 2[1 + 2𝜑 𝑐𝑜𝑠 𝛼1 − 𝜑2]−1) = 0
2[1 + 2𝜑 𝑐𝑜𝑠 𝛼1 − 𝜑2]−2[2𝜑 𝑐𝑜𝑠 𝛼1 − 2𝜑] = 0
𝑖𝑓 [𝟐𝝋 𝒄𝒐𝒔 𝜶𝟏 − 𝟐𝝋] = 𝟎 𝑡ℎ𝑒𝑛 𝝋 = 𝒄𝒐𝒔 𝜶𝟏
𝜂𝑏 =
2(2𝜑 𝑐𝑜𝑠 𝛼1 − 𝜑2)
[1 + 2𝜑 𝑐𝑜𝑠 𝛼1 − 𝜑2]
=
2(2 𝑐𝑜𝑠 𝛼12
− 𝑐𝑜𝑠 𝛼12)
[1 + 2𝑐𝑜𝑠 𝛼12 − 𝑐𝑜𝑠 𝛼12]
𝜼𝒃 =
𝟐 𝒄𝒐𝒔 𝜶𝟏𝟐
𝟏 + 𝒄𝒐𝒔 𝜶𝟏𝟐

8. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 8
Previous Year Question Papers
VTU Modal Question Paper 01
6 c)
In Parson turbine running at 1500 rpm, the available enthalpy drop for the expansion is 63
kJ/kg. If the mean diameter of the rotor is 100 cm, find the number of moving rows required.
Assume the efficiency of the stage is 0.8, blade outlet angle is 20° and speed ratio 0.7.
10
𝑵 = 𝟏𝟓𝟎𝟎 𝒓𝒑𝒎,𝑫 = 𝟏𝒎,
𝜼𝟎 = 𝜼𝑺 × 𝜼𝒃
𝜼𝑺 = 𝟎.𝟖,
𝜷𝟐 = 𝟐𝟎°
𝑩𝒍𝒂𝒅𝒆 𝑺𝒑𝒆𝒆𝒅 𝑹𝒂𝒕𝒊𝒐 =
𝑼
𝑽𝟏
= 𝟎. 𝟕
𝜶𝟏 = 𝜷𝟐 𝒂𝒏𝒅 𝜶𝟐 = 𝜷𝟏
𝑽𝟏 = 𝑽𝒓𝟐 𝒂𝒏𝒅 𝐕𝟐 = 𝑽𝒓𝟏
𝑼 =
𝝅𝑫𝑵
𝟔𝟎
=
𝝅 × 𝟏 × 𝟏𝟓𝟎𝟎
𝟔𝟎
=∴ 𝑼 = 𝟕𝟖.𝟓𝟑 𝒎/𝒔
𝑼
𝑽𝟏
= 𝟎.𝟕 ∴ 𝑽𝟏 = 𝟏𝟏𝟐.𝟏𝟖 𝒎/𝒔
𝑷 = 𝑼(∆𝑽𝒖)
(∆𝑽𝒖) = 𝑽𝟏 𝒄𝒐𝒔𝜶𝟏 + (𝑽𝒓𝟐 𝒄𝒐𝒔𝜷𝟐 − 𝑼)
(∆𝑽𝒖) = 𝑽𝟏 𝒄𝒐𝒔𝜶𝟏 + (𝑽𝟏 𝒄𝒐𝒔𝜶𝟏 − 𝑼)
(∆𝑽𝒖) = 𝟐𝑽𝟏 𝒄𝒐𝒔𝜶𝟏 − 𝑼
𝑷 = 𝑼(∆𝑽𝒖) = 𝑼(𝟐𝑽𝟏 𝒄𝒐𝒔𝜶𝟏 − 𝑼)
𝑷 = 𝟕𝟖.𝟓𝟑(𝟐 × 𝟏𝟏𝟐.𝟏𝟖 𝒄𝒐𝒔𝟐𝟎 − 𝟕𝟖.𝟓𝟑)
𝜼𝑺 =
𝑷
∆𝒉°𝑺𝒕𝒂𝒈𝒆
= 𝟎. 𝟖 ∴ ∆𝒉° =
𝑷
𝟎.𝟖
= 𝟏𝟐.𝟗𝟖 𝒌𝑱/𝒌𝒈
𝑷 = 𝟏𝟎𝟑𝟗𝟎.𝟑𝟏𝟖𝟏 𝑾 = 𝟏𝟎.𝟑𝟖 𝒌𝑾 ∆𝒉°𝑨𝒗𝒂𝒊𝒍𝒂𝒃𝒍𝒆
∆𝒉°𝑺𝒕𝒂𝒈𝒆
=
𝟔𝟑
𝟏𝟐.𝟗𝟖
= 𝟒.𝟖
≈ 𝟓 𝒓𝒐𝒘𝒔 𝒐𝒇 𝒎𝒐𝒗𝒊𝒏𝒈 𝒃𝒍𝒂𝒅𝒆𝒔 𝒂𝒓𝒆 𝒓𝒆𝒒𝒖𝒊𝒓𝒆𝒅
VTU Question Paper Aug./Sept. 2020
6 b)
The following particulars refer to a Parson’s reaction turbine : Mean diameter of the blade
ring = 90 cm, Speed = 3000 rpm, Inlet absolute velocity = 350 m/s, Blade outlet angle = 20°,
Steam flow rate = 7.2 kg/s. Calculate (i) Blade inlet angle (ii) Tangential force (iii) Power
developed.
10
𝑵 = 𝟑𝟎𝟎𝟎 𝒓𝒑𝒎,𝑫 = 𝟎.𝟗𝒎,
𝑽𝟏 = 𝑽𝒓𝟐 = 𝟑𝟓𝟎 𝒎/𝒔
𝒎 = 𝟕. 𝟖
𝜷𝟐 = 𝜶𝟏 = 𝟐𝟎°
𝑵𝒐𝒕𝒆 𝑷𝒂𝒓𝒔𝒐𝒏 𝑻𝒖𝒓𝒃𝒊𝒏𝒆
𝜶𝟏 = 𝜷𝟐 𝒂𝒏𝒅 𝜶𝟐 = 𝜷𝟏
𝑽𝟏 = 𝑽𝒓𝟐 𝒂𝒏𝒅 𝐕𝟐 = 𝑽𝒓𝟏
𝑼 =
𝝅𝑫𝑵
𝟔𝟎
=
𝝅 × 𝟎. 𝟗 × 𝟑𝟎𝟎𝟎
𝟔𝟎
=∴ 𝑼 = 𝟏𝟒𝟏.𝟑𝟕 𝒎/𝒔 sin 𝛼1 =
𝑉𝑓1
𝑉1
≫ 𝑉𝑓1 = 𝑉1 × 𝑠𝑖𝑛𝛼1 = 119.70 𝑚/𝑠
cos 𝛽2 =
𝑋2
𝑉𝑟2
≫ 𝑋2 = 𝑉𝑟2 × cos 𝛽2 = 328.89 𝑚/𝑠 cos 𝛼1 =
𝑉𝑢1
𝑉1
≫ 𝑉𝑢1 = 𝑉1 × 𝑐𝑜𝑠𝛼1 = 328.89 𝑚/𝑠
tan 𝛽1 =
𝑉𝑓1
𝑉𝑢1 − 𝑈
≫ 𝛽1 = 𝑡𝑎𝑛−1
(
𝑉𝑓1
𝑉𝑢1 − 𝑈
) 𝛽1 = 𝑡𝑎𝑛−1
(
119.70
328.89 − 141.37
) = 32.55° = 𝛼2
𝑉𝑢2 = 𝑋2 − 𝑈 = 328.89 − 141.37 = 187.53 𝑚/𝑠 𝑷 = 𝑼(∆𝑽𝒖) = 𝑼(𝟐𝑽𝟏 𝒄𝒐𝒔𝜶𝟏 − 𝑼) = 𝟕𝟑𝟎𝟎𝟓.𝟓𝟔 𝑾
𝑷 = 𝒎 × 𝟕𝟑𝟎𝟎𝟓.𝟓𝟔 𝑾 = 𝟕𝟑.𝟎 𝒌𝑾, 𝑼 = 𝟏𝟒𝟏.𝟑𝟕 𝒎/𝒔,𝜷𝟏 = 𝟑𝟐.𝟓𝟓°

9. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 9
VTU Question Paper Dec. 2019/Jan. 2020
6 b)
In a Parson’s turbine, the axial velocity of flow of steam is 0.5 times the mean blade speed.
The outlet angle of the blade is 20° diameter of the blade ring is 1.3m and rotational speed
3000rpm. Determine inlet blade angles, power developed for steam flow of 65kg/sec and
isentropic enthalpy drop, if the stage efficiency is 80%.
10
𝑽𝒇 = 𝟎. 𝟓 𝑼
𝑫 = 𝟏.𝟑 𝒎
𝜷𝟐 = 𝜶𝟏 = 𝟐𝟎°
𝑵 = 𝟑𝟎𝟎𝟎 𝒓𝒑𝒎
𝒎 = 𝟔𝟓 𝒌𝒈/𝒔
∆𝒉°𝑺𝒕𝒂𝒈𝒆 =?
𝜼𝑺 = 𝟎.𝟖𝟎
𝑵𝒐𝒕𝒆 𝑷𝒂𝒓𝒔𝒐𝒏 𝑻𝒖𝒓𝒃𝒊𝒏𝒆
𝜶𝟏 = 𝜷𝟐 𝒂𝒏𝒅 𝜶𝟐 = 𝜷𝟏
𝑽𝟏 = 𝑽𝒓𝟐 𝒂𝒏𝒅 𝐕𝟐 = 𝑽𝒓𝟏
𝑼 =
𝝅𝑫𝑵
𝟔𝟎
=
𝝅 × 𝟏. 𝟑 × 𝟑𝟎𝟎𝟎
𝟔𝟎
=∴ 𝑼 = 𝟐𝟎𝟒.𝟐𝟎 𝒎/𝒔 sin 𝛼1 =
𝑉𝑓
𝑉1
=
0.5 𝑈
𝑉1
≫ 𝑉1 =
0.5𝑈
𝑠𝑖𝑛𝛼1
= 298.52 𝑚/𝑠
cos 𝛼1 =
𝑉𝑢1
𝑉1
≫ 𝑉𝑢1 = 𝑉1 × 𝑐𝑜𝑠𝛼1 = 280.51 𝑚/𝑠
tan 𝛽1 =
𝑉𝑓
𝑉𝑢1 − 𝑈
≫ 𝛽1 = 𝑡𝑎𝑛−1
(
0.5𝑈
𝑉𝑢1 − 𝑈
) = 53°
𝑐𝑜𝑠𝛽1 =
𝑋1
𝑉𝑟1
≫ 𝑉𝑟1 =
𝑉𝑢1 − 𝑈
𝑐𝑜𝑠𝛽1
≫ 126.81 𝑚/𝑠
𝑉2 = 𝑉𝑟1 = 126.81 𝑚/𝑠
cos 𝛼2 =
𝑉𝑢2
𝑉2
≫ 𝑉𝑢2 = 𝑉2 × cos 𝛼2 = 76.31 𝑚/𝑠 𝜼𝑺 =
𝑷
∆𝒉°𝑺𝒕𝒂𝒈𝒆
= 𝟎.𝟖 ≫ ∆𝒉°𝑺𝒕𝒂𝒈𝒆 =
𝑷
𝜼𝑺
∆𝒉°𝑺𝒕𝒂𝒈𝒆 = 𝟗𝟏.𝟎𝟖 𝒌𝑱/𝒌𝒈
𝑷 = 𝑼(∆𝑽𝒖) = 𝑼(𝟐𝑽𝟏 𝒄𝒐𝒔𝜶𝟏 − 𝑼) = 𝟕𝟐𝟖𝟔𝟓.𝟓𝟏 𝑾 𝑷 = 𝒎 × 𝟕𝟐𝟖𝟔𝟓.𝟓𝟏 𝑾 == 𝟒𝟕𝟑𝟔.𝟐𝟓 𝒌𝑾
Other Relevant Questions & Problems:
 A Parson's turbine is running at 1200 rpm. The mean rotor diameter is 1m.
Blade outlet angle is 23°, speed ratio is 0.75. Stage efficiency is 0.8.
Find enthalpy drop in this stage VTU Question Paper Dec. Dec.2018/Jan. 2019
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