The document discusses hydraulic turbines and hydroelectric power plants. It describes how hydraulic turbines convert the kinetic and potential energy of water into mechanical energy using an impulse turbine called a Pelton wheel. The Pelton wheel has buckets that are struck by high-speed jets of water which causes the wheel to spin and power a generator to produce hydroelectricity. The document provides details on the components of a Pelton wheel like the nozzle, runner, and buckets. It also includes diagrams of a typical hydroelectric power plant layout and formulas used in calculating efficiencies and power values.

1. FLUID MECHANICS
AND MACHINERY
TURBINES

2. HYDRAULIC TURBINE
❖ It is a machine which converts the pressure and
kinetic energy of water called hydraulic energy into
mechanical energy.
❖ These are also called as water turbines.
❖ The mechanical energy of turbine is further
converted into electric energy by an electric
generator which is directly coupled to the shaft of
hydraulic turbine.

3. HYDRAULIC TURBINE
❖ The electrical power generated is known as
hydroelectric power.
❖ Hydraulic turbines are efficient.
❖ These have low wear and tear and ease of
maintenance.
❖ The capital cost is high with long gestation period
due to the requirement of constructing the dam
across the river and laying the long pipe lines.

4. CLASSIFICATION OF TURBINE
Turbine
Impulse Turbine Reaction Turbine
Pelton
Wheel
Girard
Turbine
Francis
Turbine
Kaplan
Turbine
Propeller
Turbine
Turgo
Turbine

5. GENERAL LAYOUT OF A HYDRO ELECTRIC POWER PLANT

6. GENERAL LAYOUT OF A HYDRO ELECTRIC POWER PLANT

7. GENERAL LAYOUT OF A HYDRO ELECTRIC POWER PLANT
❖ A dam which is constructed across a river to store water.
❖ It provides necessary potential energy to nozzles of a
turbine.
❖ It acts as water reservoir. Top surface of water in dam is
called head race.
❖ The penstock is a large diameter pipe usually made of
reinforced concrete or steel.
❖ It carries water under pressure from the water stored in a
dam or reservoir to the turbines.

8. GENERAL LAYOUT OF A HYDRO ELECTRIC POWER PLANT
❖ A Water turbine which converts hydraulic energy into
mechanical energy.
❖ A water turbine can be set horizontal or vertical. The
choice is governed by cost, type of turbine, building space
and plant layout etc.
❖ A tail race is a discharge canal into which water is
discharged from the turbine.
❖ Difference of head race and tail race level is called gross
head, Hg.

9. PELTON WHEEL OR PELTON TURBINE
❖ Pelton wheel is a tangential flow impulse turbine.
❖ In Pelton turbines the water strikes the buckets
along the tangent of the runner or wheel.
❖ It is used for high heads more than 100 m of water.
These turbines have been built up to a head of
1600m.

10. CONSTRUCTION AND WORKING OF PELTON WHEEL

11. CONSTRUCTION AND WORKING OF PELTON WHEEL
❖ The main components of a Pelton wheel are
❖ Nozzle and spear assembly
❖ Runner and buckets
❖ Casing
❖ Braking jet
❖ Deflector

12. CONSTRUCTION AND WORKING OF PELTON WHEEL
❖ NOZZLE AND SPEAR ASSEMBLY
❖ The needle spear is provided in the
nozzle to regulate the water flow
through the nozzle.
❖ It provides the smooth flow of water with negligible loss
of energy.
❖ A spear is a conical needle which can be moved in axial
direction by operating the wheel either manually or
automatically.

13. CONSTRUCTION AND WORKING OF PELTON WHEEL
❖ NOZZLE AND SPEAR ASSEMBLY
❖ When the spear is moved in forward
direction into the nozzle, it reduced the
nozzle exit area, hence, the quantity of
water flow striking the buckets is
reduced.
❖ If the spear is moved backwards, it increases the flow rate of water.
❖ The nozzle converts the potential energy of water into kinetic
energy before jet strikes the buckets.

14. CONSTRUCTION AND WORKING OF PELTON WHEEL
❖ RUNNER AND BUCKETS
❖ The turbine rotor called runner is a
circular disc fixed with buckets.
❖ It is provided with cylindrical boss and
keyed to the supporting shaft in small
thrust bearings.
❖ The runner carries cup-shaped buckets more that 15 in number
which are mounted at equidistance around its periphery.
❖ The buckets are either cast integrally with the circular disc or
these are bolted individually to the runner, it helps in easy
replacement of buckets when worn out.

15. CONSTRUCTION AND WORKING OF PELTON WHEEL
❖ RUNNER AND BUCKETS
❖ The turbine rotor called
runner is a circular disc fixed
with buckets.
❖ It is provided with cylindrical boss
and keyed to the supporting shaft in
small thrust bearings.
❖ The runner carries cup-shaped buckets more that 15 in number
which are mounted at equidistance around its periphery.

16. CONSTRUCTION AND WORKING OF PELTON WHEEL
❖ RUNNER AND BUCKETS
❖ The buckets are either cast
integrally with the circular
disc or these are bolted
individually to the runner, it
helps in easy replacement of
buckets when worn out.
❖ Buckets are made of cast iron cast steel, special steels or stainless
steel with inner surface polished to reduce friction losses of water
jet.

17. CONSTRUCTION AND WORKING OF PELTON WHEEL
❖ RUNNER AND BUCKETS
❖ The shape of the buckets is of
double hemispherical cup or
bowl.
❖ Each bowl of the bucket is
separated by a wall called
splitter or a ridge.
❖ The water strikes the bucket at the splitter which splits the water
into two equal streams of the hemispherical bowl.

18. CONSTRUCTION AND WORKING OF PELTON WHEEL
❖ RUNNER AND BUCKETS
❖ The maximum force will be
obtained when the jet is
deflected through 180° into
exact hemispherical bowl.
❖ In practice the jet is deflected through 160° to 170°
❖ It avoids striking the exit jet with the back of the succeeding
bucket, thus exerting a retarding force on it.
❖ This also avoids the splashing of water with a splitter.

19. CONSTRUCTION AND WORKING OF PELTON WHEEL
❖ RUNNER AND BUCKETS
❖ Pelton wheel is provided with
two hemispherical cups since
the splitter splits the jet into
two equal streams, the axial
component of each stream
velocity is equal and opposite
due to which the axial thrust
bn the shaft is negligible
❖ Pelton wheel needs very small thrust bearings.

20. CONSTRUCTION AND WORKING OF PELTON WHEEL
❖ RUNNER AND BUCKETS
❖ An undercut is provided and
surface of spoons is raised so
that water can be deflected
back through the angle of 160 °
to 170° with the vertical
without disturbing the
incoming bucket.

21. CONSTRUCTION AND WORKING OF PELTON WHEEL
❖ BRAKING JET
❖ The turbine is brought to rest, the nozzle
is completely closed by pushing forward
the spear.
❖ The runner continues to rotate due to its
inertia for a considerable period of time
till it comes to rest.
❖ In order to bring the runner to stop in a shortest time, a small
nozzle is provided which issues the water jet and falls on the back
of buckets.
❖ It acts as a hydraulic brake for reducing the speed of runner.

22. CONSTRUCTION AND WORKING OF PELTON WHEEL
❖ DEFLECTOR
❖ A deflector is provided
which is hinged to the
casing to deflect the jet of
water away from striking
the buckets in case the
load on turbine suddenly
reduces.
❖ It prevents the runner of turbine attaining unsafe speeds called
runaway speed.

23. VELOCITY TRIANGLE OF PELTON WHEEL
𝒖𝟏 𝑽𝒓𝟏
𝑽𝒘𝟏 = 𝑽𝟏
𝑨 𝑩 𝑪
𝑬
𝑮
𝑭 𝑯
𝝓
𝑽𝐫𝟐
𝑽𝟐 𝑽𝒇𝟐
𝜷
𝒖𝟐 𝑽𝒘𝟐
𝒖𝟏 & 𝒖𝟐 → 𝑷𝒆𝒓𝒊𝒑𝒉𝒆𝒓𝒂𝒍 𝑽𝒆𝒍𝒐𝒄𝒊𝒕𝒚 𝒐𝒇 𝒕𝒉𝒆
𝒋𝒆𝒕 𝒂𝒕 𝒊𝒏𝒍𝒆𝒕 & 𝒐𝒖𝒕𝒍𝒆𝒕
𝑽𝟏 & 𝑽𝟐 → 𝑨𝒃𝒔𝒐𝒍𝒖𝒕𝒆 𝑽𝒆𝒍𝒐𝒄𝒊𝒕𝒚 𝒐𝒇 𝒕𝒉𝒆
𝒋𝒆𝒕 𝒂𝒕 𝒊𝒏𝒍𝒆𝒕 & 𝒐𝒖𝒕𝒍𝒆𝒕
𝑽𝒓𝟏& 𝑽𝒓𝟐 → 𝑹𝒆𝒍𝒂𝒕𝒊𝒗𝒆 𝑽𝒆𝒍𝒐𝒄𝒊𝒕𝒚 𝒐𝒇 𝒕𝒉𝒆 𝒋𝒆𝒕 𝒂𝒕
𝒊𝒏𝒍𝒆𝒕 & 𝒐𝒖𝒕𝒍𝒆𝒕
𝑽𝒘𝟏& 𝑽𝒘𝟐 → 𝑾𝒉𝒊𝒓𝒍 𝑽𝒆𝒍𝒐𝒄𝒊𝒕𝒚 𝒐𝒇 𝒕𝒉𝒆 𝒋𝒆𝒕 𝒂𝒕
𝒊𝒏𝒍𝒆𝒕 & 𝒐𝒖𝒕𝒍𝒆𝒕
𝜶 & 𝜷 → 𝑮𝒖𝒊𝒅𝒆 𝑩𝒍𝒂𝒅𝒆 𝒂𝒏𝒈𝒍𝒆 𝒂𝒕
𝒊𝒏𝒍𝒆𝒕 & 𝒐𝒖𝒕𝒍𝒆𝒕
𝜽 & 𝝓 → 𝑽𝒂𝒏𝒆 𝒂𝒏𝒈𝒍𝒆 𝒂𝒕 𝒊𝒏𝒍𝒆𝒕 & 𝒐𝒖𝒕𝒍𝒆𝒕
𝝓

24. 𝒖𝟏 𝑽𝒓𝟏
𝑽𝒘𝟏 = 𝑽𝟏
𝑨 𝑩 𝑪
𝑬
𝑮
𝑭 𝑯
𝝓
𝑽𝐫𝟐
𝑽𝟐 𝑽𝒇𝟐
𝜷
𝒖𝟐 𝑽𝒘𝟐
FROM INLET VELOCITY TRIANGLE DIAGRAM:
𝑽𝒘𝟏 = 𝑽𝟏
𝑽𝒘𝟏 = 𝒖𝟏 + 𝑽𝒓𝟏
FROM OUTLET VELOCITY TRIANGLE DIAGRAM:
𝒄𝒐𝒔 𝝓 =
𝒖𝟐 + 𝑽𝒘𝟐
𝑽𝒓𝟐
𝒕𝒂𝒏 𝝓 =
𝑽𝒇𝟐
𝒖𝟐 + 𝑽𝒘𝟐
𝒔𝒊𝒏 𝝓 =
𝑽𝒇𝟐
𝑽𝒓𝟐
𝒕𝒂𝒏 𝜷 =
𝑽𝒇𝟐
𝑽𝒘𝟐

25. VELOCITY TRIANGLE OF PELTON WHEEL

26. FORMULA USED
❖ WORK DONE BY JET PER SECOND
𝑾 = 𝝆𝑸 𝑽𝒘𝟏 + 𝑽𝒘𝟐 𝒖
❖ HYDRAULIC EFFICIENCY
𝜼𝒉𝒚𝒅 =
𝟐 𝑽𝒘𝟏 + 𝑽𝒘𝟐 𝒖
𝑽𝟏
𝟐

27. FORMULA USED
❖ OVERALL EFFICIENCY
𝜼𝒐 =
𝑺𝒉𝒂𝒇𝒕 𝑷𝒐𝒘𝒆𝒓
𝑰𝒏𝒑𝒖𝒕 𝑷𝒐𝒘𝒆𝒓
𝜼𝒐 =
𝑺. 𝑷
𝝆𝒈𝑸𝑯

28. FORMULA USED
❖ DISCHARGE OF SINGLE JET
𝒒 =
𝝅
𝟒
× 𝒅𝟐
× 𝑽𝟏
❖ NUMBER OF JET
𝒏 =
𝑸
𝒒

29. FORMULA USED
❖ NUMBER OF BUCKET
𝒁 = 𝟏𝟓 +
𝑫
𝟐𝒅
❖ DIMENSIONS OF BUCKET
𝑨𝒙𝒊𝒂𝒍 𝑾𝒊𝒅𝒕𝒉 𝑩 = 𝟒. 𝟓𝒅
𝑹𝒂𝒅𝒊𝒂𝒍 𝑳𝒆𝒏𝒈𝒕𝒉 𝑳 = 𝟐. 𝟓𝒅
𝑫𝒆𝒑𝒕𝒉 𝒐𝒇 𝑩𝒖𝒄𝒌𝒆𝒕 𝑻 = 𝒅

30. FORMULA USED
❖ KINETIC ENERGY OF JET
𝑲. 𝑬 𝒐𝒇 𝑱𝒆𝒕 =
𝟏
𝟐
𝒎 𝑽𝟏
𝟐
𝑺𝒊𝒏𝒄𝒆 𝒎 = 𝝆𝑨𝑽
𝑻𝒉𝒆𝒓𝒆𝒇𝒐𝒓𝒆 𝑲. 𝑬 𝒐𝒇 𝑱𝒆𝒕 =
𝟏
𝟐
𝝆 × 𝑨 × 𝑽𝟏 × 𝑽𝟏
𝟐
𝑺𝒊𝒏𝒄𝒆 𝑸 = 𝑨𝑽
𝑻𝒉𝒆𝒓𝒆𝒇𝒐𝒓𝒆 𝑲. 𝑬 𝒐𝒇 𝑱𝒆𝒕 =
𝟏
𝟐
𝝆 × 𝑸 × 𝑽𝟏
𝟐

31. FORMULA USED
❖ POWER LOST IN NOZZLE
𝑰𝒏𝒑𝒖𝒕 𝑷𝒐𝒘𝒆𝒓 = 𝑲𝒊𝒏𝒆𝒕𝒊𝒄 𝑬𝒏𝒆𝒓𝒈𝒚 + 𝑷𝒐𝒘𝒆𝒓 𝑳𝒐𝒔𝒕 𝒊𝒏 𝑵𝒐𝒛𝒛𝒍𝒆
❖ POWER LOST IN RUNNER
𝑰𝒏𝒑𝒖𝒕 𝑷𝒐𝒘𝒆𝒓 = 𝑷𝒐𝒘𝒆𝒓 𝒐𝒇 𝑺𝒉𝒂𝒇𝒕 + 𝑷𝒐𝒘𝒆𝒓 𝑳𝒐𝒔𝒕 𝒊𝒏 𝑵𝒐𝒛𝒛𝒍𝒆
+ 𝑷𝒐𝒘𝒆𝒓 𝑳𝒐𝒔𝒕 𝒊𝒏 𝑹𝒖𝒏𝒏𝒆𝒓 +
𝑷𝒐𝒘𝒆𝒓 𝑳𝒐𝒔𝒕 𝑫𝒖𝒆 𝒕𝒐 𝑴𝒆𝒄𝒉𝒂𝒏𝒊𝒄𝒂𝒍 𝑹𝒆𝒔𝒊𝒔𝒕𝒂𝒏𝒄𝒆

32. FORMULA USED
❖ RESULTANT FORCE ON BUCKET
𝑭 = 𝝆𝑸 𝑽𝒘𝟏 + 𝑽𝒘𝟐
❖ TORQUE
𝑻 = 𝑭 ×
𝑫
𝟐

33. FORMULA USED
❖ POWER
𝑷 =
𝟐𝝅𝑵𝑻
𝟔𝟎
❖ SPECIFIC SPEED
𝑵𝒔 =
𝑵 𝑸
𝑯 ൗ
𝟑
𝟒
𝑵𝒔 =
𝑵 𝑷
𝑯 ൗ
𝟓
𝟒

34. PROBLEM 1
A Pelton wheel has a mean bucket speed of 10m/s with a jet of
water flowing at the rate of 0.7 m3/s under a head of 30m. The
buckets deflect the jet through an angle of 160°. Calculate the
power given by the water to the runner and the hydraulic
efficiency of the turbine. Assuming the coefficient of velocity as
0.98
GIVEN:
𝒖 = 𝒖𝟏 = 𝒖𝟐 = 𝟏𝟎 Τ
𝒎
𝒔 𝑸 = 𝟎. 𝟕 ൗ
𝒎𝟑
𝒔 𝑯 = 𝟑𝟎 𝒎
𝝓 = 𝟏𝟖𝟎 − 𝟏𝟔𝟎 = 𝟐𝟎 ° 𝑪𝒗 = 𝟎. 𝟗𝟖

35. SOLUTION:
Velocity of Jet:
𝑽𝟏 = 𝑪𝒗 × 𝟐 × 𝒈 × 𝑯
𝑽𝟏 = 𝟎. 𝟗𝟖 × 𝟐 × 𝟗. 𝟖𝟏 × 𝟑𝟎
𝒖𝟏 𝑽𝒓𝟏
𝑽𝒘𝟏 = 𝑽𝟏
𝑨 𝑩 𝑪
𝑬
𝑮
𝑭 𝑯
𝝓
𝑽𝐫𝟐
𝑽𝟐 𝑽𝒇𝟐
𝜷
𝒖𝟐 𝑽𝒘𝟐
𝑽𝟏 = 𝟐𝟑. 𝟕𝟕 Τ
𝒎
𝒔
From Inlet Velocity Triangle Diagram
𝑽𝒘𝟏 = 𝑽𝟏
𝑽𝒘𝟏 = 𝟐𝟑. 𝟕𝟕 Τ
𝒎
𝒔

36. 𝒖𝟏 𝑽𝒓𝟏
𝑽𝒘𝟏 = 𝑽𝟏
𝑨 𝑩 𝑪
𝑬
𝑮
𝑭 𝑯
𝝓
𝑽𝐫𝟐
𝑽𝟐 𝑽𝒇𝟐
𝜷
𝒖𝟐 𝑽𝒘𝟐
𝒖𝟏 + 𝑽𝒓𝟏 = 𝑽𝟏
𝟏𝟎 + 𝑽𝒓𝟏 = 𝟐𝟑. 𝟕𝟕
𝑽𝒓𝟏 = 𝟏𝟑. 𝟕𝟕 Τ
𝒎
𝒔
From Outlet Velocity Triangle Diagram
𝒄𝒐𝒔 𝝓 =
𝒖𝟐 + 𝑽𝒘𝟐
𝑽𝒓𝟐
𝒄𝒐𝒔 𝟐𝟎 =
𝟏𝟎 + 𝑽𝒘𝟐
𝟏𝟑. 𝟕𝟕
𝑨𝒔𝒔𝒖𝒎𝒆 𝑽𝒓𝟏 = 𝑽𝒓𝟐
𝑽𝒘𝟐 = 𝟐. 𝟗𝟒𝟓 Τ
𝒎
𝒔

37. Work done by Jet per Second:
𝑾 = 𝝆 × 𝑸 × 𝑽𝒘𝟏 + 𝑽𝒘𝟐 × 𝒖
𝑾 = 𝟏𝟎𝟎𝟎 × 𝟎. 𝟕 × 𝟐𝟑. 𝟕𝟕 + 𝟐. 𝟗𝟒𝟓 × 𝟏𝟎
𝑾 = 𝟏. 𝟖𝟕 × 𝟏𝟎𝟓
𝑾
Hydraulic Efficiency:
𝜼𝒉𝒚𝒅 =
𝟐 × 𝑽𝒘𝟏 + 𝑽𝒘𝟐 × 𝒖
𝑽𝟏
𝟐
=
𝟐 × 𝟐𝟑. 𝟕𝟕 + 𝟐. 𝟗𝟒𝟓 × 𝟏𝟎
𝟐𝟑. 𝟕𝟕𝟐
𝜼𝒉𝒚𝒅 = 𝟎. 𝟗𝟒𝟓𝟔 𝜼𝒉𝒚𝒅 = 𝟗𝟒. 𝟓𝟔 %

38. PROBLEM 2
A Pelton wheel is to be designed for the following specifications :
Shaft power =11,772 KW ; head = 380 m; speed = 750 rpm,
Overall efficiency=86%. Jet diameter is not to exceed one-sixth of the
wheel diameter. Determine the
i) Wheel diameter ii) Number of jets required
iii) Diameter of the jet.
GIVEN:
𝑷𝒔 = 𝟏𝟏𝟕𝟕𝟐 × 𝟏𝟎𝟑
𝑾 𝑯 = 𝟑𝟖𝟎 𝒎 𝑵 = 𝟕𝟓𝟎 𝒓𝒑𝒎
𝜼𝒐 = 𝟖𝟔%
𝒅 =
𝟏
𝟔
𝑫

39. SOLUTION:
Velocity of Jet:
𝑽𝟏 = 𝑪𝒗 × 𝟐 × 𝒈 × 𝑯
𝑽𝟏 = 𝟎. 𝟗𝟖 × 𝟐 × 𝟗. 𝟖𝟏 × 𝟑𝟖𝟎
𝑽𝟏 = 𝟖𝟒. 𝟔𝟏 Τ
𝒎
𝒔
Velocity of Wheel:
𝒖 = 𝑲𝒖 × 𝟐 × 𝒈 × 𝑯
𝒖 = 𝟎. 𝟒𝟓 × 𝟐 × 𝟗. 𝟖𝟏 × 𝟑𝟖𝟎
𝒖 = 𝟑𝟖. 𝟖𝟓 Τ
𝒎
𝒔
𝑨𝒔𝒔𝒖𝒎𝒆 𝑲𝒖 = 𝟎. 𝟒𝟓

40. Wheel Diameter:
𝒖 =
𝝅 × 𝑫 × 𝑵
𝟔𝟎
𝟑𝟖. 𝟖𝟓 =
𝝅 × 𝑫 × 𝟕𝟓𝟎
𝟔𝟎
𝑫 = 𝟎. 𝟗𝟖𝟗𝟑 𝒎
Jet Diameter:
𝒅 =
𝟏
𝟔
× 𝑫 =
𝟏 × 𝟎. 𝟗𝟖𝟗𝟑
𝟔
𝒅 = 𝟎. 𝟏𝟔𝟒𝟗 𝒎

41. Overall Efficiency:
𝜼𝒐 =
𝑷𝒔
𝑷𝒘
=
𝑷𝒔
𝝆 × 𝒈 × 𝑸 × 𝑯
𝑸 = 𝟑. 𝟔𝟕 ൗ
𝒎𝟑
𝒔
𝟎. 𝟖𝟔 =
𝟏𝟏. 𝟕𝟕𝟐 × 𝟏𝟎𝟑
𝟏𝟎𝟎𝟎 × 𝟗. 𝟖𝟏 × 𝑸 × 𝟑𝟖𝟎
Jet Discharge:
𝒒 = 𝑨 × 𝑽𝟏 =
𝝅
𝟒
× 𝒅𝟐
× 𝑽𝟏 =
𝝅
𝟒
× 𝟎. 𝟏𝟔𝟒𝟗𝟐
× 𝟖𝟒. 𝟔𝟏
𝒒 = 𝟏. 𝟖𝟎𝟔 ൗ
𝒎𝟑
𝒔

42. No of Jet:
𝒏 =
𝑸
𝒒
=
𝟑. 𝟔𝟕
𝟏. 𝟖𝟎𝟔
𝒏 = 𝟐. 𝟎𝟑
𝑱𝒆𝒕
𝒏 ≈ 𝟑

43. PROBLEM 3
A Pelton turbine is required to develop 9000 kW when working
under a head of 300m the impeller may rotate at 500 rpm.
Assuming a jet ratio of 10 and overall efficiency of 85% calculate
(i) Quantity of water required (ii) Diameter of the wheel (iii)
Number of jets (iv) Number and size of the bucket vanes on the
runner
GIVEN:
𝑷𝒔 = 𝟗𝟎𝟎𝟎 × 𝟏𝟎𝟑
𝑾 𝑯 = 𝟑𝟎𝟎 𝒎 𝑵 = 𝟓𝟎𝟎 𝒓𝒑𝒎
𝜼𝒐 = 𝟖𝟓%
𝒅 =
𝟏
𝟏𝟎
𝑫

44. SOLUTION:
Velocity of Jet:
𝑽𝟏 = 𝑪𝒗 × 𝟐 × 𝒈 × 𝑯
𝑽𝟏 = 𝟎. 𝟗𝟖 × 𝟐 × 𝟗. 𝟖𝟏 × 𝟑𝟎𝟎
𝑽𝟏 = 𝟕𝟓. 𝟏𝟖 Τ
𝒎
𝒔
Velocity of Wheel:
𝒖 = 𝑲𝒖 × 𝟐 × 𝒈 × 𝑯
𝒖 = 𝟎. 𝟒𝟓 × 𝟐 × 𝟗. 𝟖𝟏 × 𝟑𝟎𝟎
𝒖 = 𝟑𝟒. 𝟓𝟐 Τ
𝒎
𝒔
𝑨𝒔𝒔𝒖𝒎𝒆 𝑲𝒖 = 𝟎. 𝟒𝟓

45. Wheel Diameter:
𝒖 =
𝝅 × 𝑫 × 𝑵
𝟔𝟎
𝟑𝟒. 𝟓𝟐 =
𝝅 × 𝑫 × 𝟓𝟎𝟎
𝟔𝟎
𝑫 = 𝟏. 𝟑𝟏𝟖 𝒎
Jet Diameter:
𝒅 =
𝟏
𝟏𝟎
× 𝑫 =
𝟏 × 𝟏. 𝟑𝟏𝟖
𝟏𝟎
𝒅 = 𝟎. 𝟏𝟑𝟏𝟖 𝒎

46. Overall Efficiency:
𝜼𝒐 =
𝑷𝒔
𝑷𝒘
=
𝑷𝒔
𝝆 × 𝒈 × 𝑸 × 𝑯
𝑸 = 𝟑. 𝟓𝟗𝟕 ൗ
𝒎𝟑
𝒔
𝟎. 𝟖𝟓 =
𝟗𝟎𝟎𝟎 × 𝟏𝟎𝟑
𝟏𝟎𝟎𝟎 × 𝟗. 𝟖𝟏 × 𝑸 × 𝟑𝟎𝟎
Jet Discharge:
𝒒 = 𝑨 × 𝑽𝟏 =
𝝅
𝟒
× 𝒅𝟐
× 𝑽𝟏 =
𝝅
𝟒
× 𝟎. 𝟏𝟑𝟏𝟖𝟐
× 𝟕𝟓. 𝟏𝟖
𝒒 = 𝟏. 𝟎𝟐𝟓 ൗ
𝒎𝟑
𝒔

47. No of Jet:
𝒏 =
𝑸
𝒒
=
𝟑. 𝟓𝟗𝟕
𝟏. 𝟎𝟐𝟓
𝒏 = 𝟑. 𝟓 𝑱𝒆𝒕
𝒏 ≈ 𝟒
Size of Bucket:
𝑩 = 𝟒. 𝟓 𝒅 = 𝟒. 𝟓 × 𝟎. 𝟏𝟑𝟏𝟖
Axial Width:
𝑩 = 𝟎. 𝟓𝟗𝟑𝟏 𝒎

48. 𝑳 = 𝟐. 𝟓 𝒅 = 𝟐. 𝟓 × 𝟎. 𝟏𝟑𝟏𝟖
Radial Length:
𝑳 = 𝟎. 𝟑𝟐𝟗𝟓 𝒎
𝑻 = 𝒅 = 𝟎. 𝟏𝟑𝟏𝟖
Depth of Bucket:
𝑻 = 𝟎. 𝟏𝟑𝟏𝟖 𝒎
Number of Bucket:
𝒁 = 𝟏𝟓 +
𝑫
𝟐𝒅
= 𝟏𝟓 +
𝟏. 𝟑𝟏𝟖
𝟐 × 𝟎. 𝟏𝟑𝟏𝟖
𝒁 = 𝟐𝟎 𝑩𝒖𝒄𝒌𝒆𝒕𝒔

49. PROBLEM 4
A Pelton wheel supplied water from reservoir under a gross head of 112m
and the friction losses in pen stock amounts to 20m of head. The water from
pen stock is discharged through a single nozzle of diameter of 100mm at the
rate of 0.30m3/s. Mechanical losses due to friction amounts to 4.3kW of
power and the shaft power available is 208kW. Determine velocity of jet,
water power at inlet to runner, power losses in nozzles, power lost in runner
due to hydraulic resistance.
GIVEN:
𝒉𝒈 = 𝟏𝟏𝟐 𝒎 𝒉𝒇 = 𝟐𝟎 𝒎 𝒅 = 𝟏𝟎𝟎 × 𝟏𝟎−𝟑
𝒎
𝑸 = 𝒒 = 𝟎. 𝟑 ൗ
𝒎𝟑
𝒔 𝑴𝑳 = 𝟒. 𝟑 × 𝟏𝟎𝟑
𝑾 𝑷𝒔 = 𝟐𝟎𝟖 × 𝟏𝟎𝟑
𝑾

50. SOLUTION:
Velocity of Jet:
𝑽𝟏 = 𝑪𝒗 × 𝟐 × 𝒈 × 𝑯 𝑽𝟏 = 𝟎. 𝟗𝟖 × 𝟐 × 𝟗. 𝟖𝟏 × 𝟗𝟐
𝑽𝟏 = 𝟒𝟏. 𝟔𝟑𝟔 Τ
𝒎
𝒔
Velocity of Wheel:
𝒖 = 𝑲𝒖 × 𝟐 × 𝒈 × 𝑯 𝒖 = 𝟎. 𝟒𝟓 × 𝟐 × 𝟗. 𝟖𝟏 × 𝟗𝟐
𝒖 = 𝟏𝟗. 𝟏𝟏𝟖 Τ
𝒎
𝒔
Total Head:
𝑯 = 𝒉𝒈 − 𝒉𝒇 𝑯 = 𝟏𝟏𝟐 − 𝟐𝟎
𝑯= 𝟗𝟐 𝒎

51. Power at Base of Nozzle:
𝑷 = 𝝆 × 𝒈 × 𝑸 × 𝑯
𝑷 = 𝟏𝟎𝟎𝟎 × 𝟗. 𝟖𝟏 × 𝟎. 𝟑 × 𝟗𝟐
𝑷 = 𝟐. 𝟕𝟎𝟕𝟓 × 𝟏𝟎𝟓 𝑾
Kinetic Energy of Jet:
𝑲. 𝑬 𝒐𝒇 𝑱𝒆𝒕 =
𝟏
𝟐
𝒎 𝑽𝟏
𝟐 𝑺𝒊𝒏𝒄𝒆 𝒎 = 𝝆𝑨𝑽
𝑲. 𝑬 𝒐𝒇 𝑱𝒆𝒕 =
𝟏
𝟐
𝝆 × 𝑨 × 𝑽𝟏 × 𝑽𝟏
𝟐 𝑺𝒊𝒏𝒄𝒆 𝑸 = 𝑨𝑽

52. 𝑲. 𝑬 𝒐𝒇 𝑱𝒆𝒕 =
𝟏
𝟐
𝝆 × 𝑸 × 𝑽𝟏
𝟐
𝑲. 𝑬 𝒐𝒇 𝑱𝒆𝒕 =
𝟏
𝟐
× 𝟏𝟎𝟎𝟎 × 𝟎. 𝟑 × 𝟒𝟏. 𝟔𝟑𝟔𝟐
𝑲. 𝑬 𝒐𝒇 𝑱𝒆𝒕 = 𝟐. 𝟔 × 𝟏𝟎𝟓 𝑾
Power Loss in Nozzle:
𝑰𝒏𝒑𝒖𝒕 𝑷𝒐𝒘𝒆𝒓 = 𝑲𝒊𝒏𝒆𝒕𝒊𝒄 𝑬𝒏𝒆𝒓𝒈𝒚 + 𝑷𝒐𝒘𝒆𝒓 𝑳𝒐𝒔𝒕 𝒊𝒏 𝑵𝒐𝒛𝒛𝒍𝒆
𝟐. 𝟕𝟎𝟕𝟓 × 𝟏𝟎𝟓
= 𝟐. 𝟔 × 𝟏𝟎𝟓
+ 𝑷𝒐𝒘𝒆𝒓 𝑳𝒐𝒔𝒕 𝒊𝒏 𝑵𝒐𝒛𝒛𝒍𝒆
𝑷𝒐𝒘𝒆𝒓 𝑳𝒐𝒔𝒕 𝒊𝒏 𝑵𝒐𝒛𝒛𝒍𝒆 = 𝟎. 𝟏𝟎𝟕 × 𝟏𝟎𝟓 𝑾

53. Power Loss in Runner:
𝑰𝒏𝒑𝒖𝒕 𝑷𝒐𝒘𝒆𝒓 = 𝑷𝒐𝒘𝒆𝒓 𝒐𝒇 𝑺𝒉𝒂𝒇𝒕 + 𝑷𝒐𝒘𝒆𝒓 𝑳𝒐𝒔𝒕 𝒊𝒏 𝑵𝒐𝒛𝒛𝒍𝒆
+ 𝑷𝒐𝒘𝒆𝒓 𝑳𝒐𝒔𝒕 𝒊𝒏 𝑹𝒖𝒏𝒏𝒆𝒓
+𝑷𝒐𝒘𝒆𝒓 𝑳𝒐𝒔𝒕 𝑫𝒖𝒆 𝒕𝒐 𝑴𝒆𝒄𝒉𝒂𝒏𝒊𝒄𝒂𝒍 𝑹𝒆𝒔𝒊𝒔𝒕𝒂𝒏𝒄𝒆
𝑷𝒐𝒘𝒆𝒓 𝑳𝒐𝒔𝒕 𝒊𝒏 𝑹𝒖𝒏𝒏𝒆𝒓 = 𝟎. 𝟒𝟕𝟕𝟓 × 𝟏𝟎𝟓 𝑾
𝟐. 𝟕𝟎𝟕𝟓 × 𝟏𝟎𝟓
= 𝟐. 𝟎𝟖 × 𝟏𝟎𝟓
+ 𝟎. 𝟏𝟎𝟕 × 𝟏𝟎𝟓
+𝑷𝒐𝒘𝒆𝒓 𝑳𝒐𝒔𝒕 𝒊𝒏 𝑹𝒖𝒏𝒏𝒆𝒓 + 𝟒. 𝟑 × 𝟏𝟎𝟑

54. PROBLEM 5
A Pelton turbine having 1.6m bucket diameter develops a
power of 3600kW at 400rpm, under a net head of 275m. If the
overall efficiency is 88%, and the coefficient of velocity is 0.97,
find speed ratio, discharge, diameter of the nozzle and specific
speed.
GIVEN:
𝑫 = 𝟏. 𝟔 𝒎 𝑵 = 𝟒𝟎𝟎 𝒓𝒑𝒎 𝑷𝒔 = 𝟑𝟔𝟎𝟎 × 𝟏𝟎𝟑
𝑾
𝑯 = 𝟐𝟕𝟓 𝒎 𝑪𝒗 = 𝟎. 𝟗𝟕 𝜼𝒐 = 𝟖𝟖%

55. SOLUTION:
Velocity of Jet:
𝑽𝟏 = 𝑪𝒗 × 𝟐 × 𝒈 × 𝑯
𝑽𝟏 = 𝟎. 𝟗𝟕 × 𝟐 × 𝟗. 𝟖𝟏 × 𝟐𝟕𝟓
𝑽𝟏 = 𝟕𝟏. 𝟐𝟓 Τ
𝒎
𝒔
Velocity of Wheel:
𝒖 =
𝝅 × 𝑫 × 𝑵
𝟔𝟎
=
𝝅 × 𝟏. 𝟔 × 𝟒𝟎𝟎
𝟔𝟎
𝒖 = 𝟑𝟑. 𝟓𝟏 Τ
𝒎
𝒔

56. Speed Ratio
𝒖 = 𝑲𝒖 × 𝟐 × 𝒈 × 𝑯
𝟑𝟑. 𝟓𝟏 = 𝑲𝒖 × 𝟐 × 𝟗. 𝟖𝟏 × 𝟐𝟕𝟓
𝑲𝒖 = 𝟎. 𝟒𝟓𝟔𝟐 𝒏𝒐 𝒖𝒏𝒊𝒕
Overall Efficiency:
𝜼𝒐 =
𝑷𝒔
𝑷𝒘
=
𝑷𝒔
𝝆 × 𝒈 × 𝑸 × 𝑯
𝑸 = 𝟏. 𝟓𝟏𝟔 ൗ
𝒎𝟑
𝒔
𝟎. 𝟖𝟖 =
𝟑𝟔𝟎𝟎 × 𝟏𝟎𝟑
𝟏𝟎𝟎𝟎 × 𝟗. 𝟖𝟏 × 𝑸 × 𝟐𝟕𝟓

57. Diameter of Nozzle:
𝒒 = 𝑨 × 𝑽𝟏
𝟏. 𝟓𝟏𝟔 =
𝝅
𝟒
× 𝒅𝟐
× 𝟕𝟏. 𝟐𝟓
𝒅 = 𝟎. 𝟏𝟔𝟒𝟓 𝒎
𝑨𝒔𝒔𝒖𝒎𝒆 𝒔𝒊𝒏𝒈𝒍𝒆 𝒋𝒆𝒕 𝑸 = 𝒒
Specific Speed :
𝑵𝒔 =
𝑵 𝑷
𝑯 ൗ
𝟓
𝟒
=
𝟒𝟎𝟎 𝟑𝟔𝟎𝟎
𝟐𝟕𝟓 ൗ
𝟓
𝟒
Note: Power should be
substituted in kW
𝑵𝒔 = 𝟐𝟏. 𝟒𝟑 𝑵𝒐 𝑼𝒏𝒊𝒕

58. PROBLEM 6
A Pelton wheel which is receiving water from a penstock with a gross
head of 510m. One - third of Gross head is lost in the penstock. The rate
of flow through the nozzle fitted at the end of the penstock is 2.2
m3/sec. The angle of deflection of the jet is 165°. Determine (1) The
power given by the water to the runner (2) Hydraulic efficiency of the
Pelton wheel. Take Cv=1 and speed ratio =0.45
GIVEN:
𝒉𝒈 = 𝟓𝟏𝟎 𝒎 𝒉𝒇 =
𝟏
𝟑
𝒉𝒈 𝑸 = 𝒒 = 𝟐. 𝟐 ൗ
𝒎𝟑
𝒔
𝝓 = 𝟏𝟖𝟎 − 𝟏𝟔𝟓 = 𝟏𝟓° 𝑪𝒗 = 𝟏 𝑲𝒖 = 𝟎. 𝟒𝟓

59. SOLUTION:
Velocity of Jet:
𝑽𝟏 = 𝑪𝒗 × 𝟐 × 𝒈 × 𝑯 𝑽𝟏 = 𝟏 × 𝟐 × 𝟗. 𝟖𝟏 × 𝟑𝟒𝟎
𝑽𝟏 = 𝟖𝟏. 𝟔𝟕𝟒𝟗 Τ
𝒎
𝒔
Velocity of Wheel:
𝒖 = 𝑲𝒖 × 𝟐 × 𝒈 × 𝑯 𝒖 = 𝟎. 𝟒𝟓 × 𝟐 × 𝟗. 𝟖𝟏 × 𝟑𝟒𝟎
𝒖 = 𝟑𝟔. 𝟕𝟓𝟑𝟕 Τ
𝒎
𝒔
Total Head:
𝑯 = 𝒉𝒈 − 𝒉𝒇
𝑯 = 𝟓𝟏𝟎 −
𝟏
𝟑
× 𝟓𝟏𝟎
𝑯= 𝟑𝟒𝟎 𝒎

60. 𝒖𝟏 𝑽𝒓𝟏
𝑽𝒘𝟏 = 𝑽𝟏
𝑨 𝑩 𝑪
𝑬
𝑮
𝑭 𝑯
𝝓
𝑽𝐫𝟐
𝑽𝟐 𝑽𝒇𝟐
𝜷
𝒖𝟐 𝑽𝒘𝟐
From Inlet Velocity Triangle Diagram
𝑽𝒘𝟏 = 𝑽𝟏
𝑽𝒘𝟏 = 𝟖𝟏. 𝟔𝟕𝟒𝟗 Τ
𝒎
𝒔
𝒖𝟏 + 𝑽𝒓𝟏 = 𝑽𝟏
𝟑𝟔. 𝟕𝟓𝟑𝟕 + 𝑽𝒓𝟏 = 𝟖𝟏. 𝟔𝟕𝟒𝟗
𝑽𝒓𝟏 = 𝟒𝟒. 𝟗𝟐𝟏𝟐 Τ
𝒎
𝒔

61. 𝒖𝟏 𝑽𝒓𝟏
𝑽𝒘𝟏 = 𝑽𝟏
𝑨 𝑩 𝑪
𝑬
𝑮
𝑭 𝑯
𝝓
𝑽𝐫𝟐
𝑽𝟐 𝑽𝒇𝟐
𝜷
𝒖𝟐 𝑽𝒘𝟐
From Outlet Velocity Triangle Diagram
𝒄𝒐𝒔 𝝓 =
𝒖𝟐 + 𝑽𝒘𝟐
𝑽𝒓𝟐
𝒄𝒐𝒔 𝟏𝟓 =
𝟑𝟔. 𝟕𝟓𝟑𝟕 + 𝑽𝒘𝟐
𝟒𝟒. 𝟗𝟐𝟏𝟐
𝑨𝒔𝒔𝒖𝒎𝒆 𝑽𝒓𝟏 = 𝑽𝒓𝟐
𝑽𝒘𝟐 = 𝟔. 𝟔𝟑𝟔𝟖 Τ
𝒎
𝒔

62. Power given by Water to Runner
𝑷 = 𝝆 × 𝑸 × 𝑽𝒘𝟏 + 𝑽𝒘𝟐 × 𝒖
𝑷 = 𝟏𝟎𝟎𝟎 × 𝟐. 𝟐 × 𝟖𝟏. 𝟔𝟕𝟒𝟗 + 𝟔. 𝟔𝟑𝟔𝟖 × 𝟑𝟔. 𝟕𝟓𝟑𝟕
𝑷 = 𝟕𝟏. 𝟒𝟎𝟕 × 𝟏𝟎𝟓
𝑾
Hydraulic Efficiency:
𝜼𝒉𝒚𝒅 =
𝟐 × 𝑽𝒘𝟏 + 𝑽𝒘𝟐 × 𝒖
𝑽𝟏
𝟐
=
𝟐 × 𝟖𝟏. 𝟔𝟕𝟒𝟗 + 𝟔. 𝟔𝟑𝟔𝟖 × 𝟑𝟔. 𝟕𝟓𝟑𝟕
𝟖𝟏. 𝟔𝟕𝟒𝟗𝟐
𝜼𝒉𝒚𝒅 = 𝟎. 𝟗𝟕𝟑𝟗 𝜼𝒉𝒚𝒅 = 𝟗𝟕. 𝟑𝟕 %

63. PROBLEM 7
A Pelton wheel is having a mean bucket diameter of 1 m and is
running at 1000 rpm. The net head on the Pelton wheel is 700
m. If the side clearance angle is 15°and the discharge through
the nozzle is 0.1m3/sec, find the i) power available at the
nozzle and ii) hydraulic efficiency of the turbine. Take CV=1.
GIVEN:
𝑫 = 𝟏 𝒎 𝑵 = 𝟏𝟎𝟎𝟎 𝒓𝒑𝒎 𝑯 = 𝟕𝟎𝟎 𝒎
𝝓 = 𝟏𝟓 ° 𝒒 = 𝟎. 𝟏 ൗ
𝒎𝟑
𝒔 𝑪𝒗 = 𝟏

64. SOLUTION:
Velocity of Jet:
𝑽𝟏 = 𝑪𝒗 × 𝟐 × 𝒈 × 𝑯
𝑽𝟏 = 𝟏 × 𝟐 × 𝟗. 𝟖𝟏 × 𝟕𝟎𝟎
𝑽𝟏 = 𝟏𝟏𝟕. 𝟏𝟗𝟐 Τ
𝒎
𝒔
Velocity of Wheel:
𝒖 =
𝝅 × 𝑫 × 𝑵
𝟔𝟎
𝒖 =
𝝅 × 𝟏 × 𝟏𝟎𝟎𝟎
𝟔𝟎
𝒖 = 𝟓𝟐. 𝟑𝟓𝟗 Τ
𝒎
𝒔

65. 𝒖𝟏 𝑽𝒓𝟏
𝑽𝒘𝟏 = 𝑽𝟏
𝑨 𝑩 𝑪
𝑬
𝑮
𝑭 𝑯
𝝓
𝑽𝐫𝟐
𝑽𝟐 𝑽𝒇𝟐
𝜷
𝒖𝟐 𝑽𝒘𝟐
From Inlet Velocity Triangle Diagram
𝑽𝒘𝟏 = 𝑽𝟏
𝑽𝒘𝟏 = 𝟏𝟏𝟕. 𝟏𝟗𝟐 Τ
𝒎
𝒔
𝒖𝟏 + 𝑽𝒓𝟏 = 𝑽𝟏
𝟓𝟐. 𝟑𝟓𝟗 + 𝑽𝒓𝟏 = 𝟏𝟏𝟕. 𝟏𝟗𝟐
𝑽𝒓𝟏 = 𝟔𝟒. 𝟖𝟑𝟑 Τ
𝒎
𝒔

66. 𝒖𝟏 𝑽𝒓𝟏
𝑽𝒘𝟏 = 𝑽𝟏
𝑨 𝑩 𝑪
𝑬
𝑮
𝑭 𝑯
𝝓
𝑽𝐫𝟐
𝑽𝟐 𝑽𝒇𝟐
𝜷
𝒖𝟐 𝑽𝒘𝟐
From Outlet Velocity Triangle Diagram
𝒄𝒐𝒔 𝝓 =
𝒖𝟐 + 𝑽𝒘𝟐
𝑽𝒓𝟐
𝒄𝒐𝒔 𝟏𝟓 =
𝟓𝟐. 𝟑𝟓𝟗 + 𝑽𝒘𝟐
𝟔𝟒. 𝟖𝟑𝟑
𝑨𝒔𝒔𝒖𝒎𝒆 𝑽𝒓𝟏 = 𝑽𝒓𝟐
𝑽𝒘𝟐 = 𝟏𝟎. 𝟐𝟔𝟒 Τ
𝒎
𝒔

67. Power available at Nozzle:
𝑷 = 𝝆 × 𝑸 × 𝑽𝒘𝟏 + 𝑽𝒘𝟐 × 𝒖
𝑷 = 𝟏𝟎𝟎𝟎 × 𝟎. 𝟏 × 𝟏𝟏𝟕. 𝟏𝟗𝟐 + 𝟏𝟎. 𝟐𝟔𝟒 × 𝟓𝟐. 𝟑𝟓𝟗
𝑷 = 𝟔. 𝟔𝟕𝟑𝟓 × 𝟏𝟎𝟓
𝑾
Hydraulic Efficiency:
𝜼𝒉𝒚𝒅 =
𝟐 × 𝑽𝒘𝟏 + 𝑽𝒘𝟐 × 𝒖
𝑽𝟏
𝟐
=
𝟐 × 𝟏𝟏𝟕. 𝟏𝟗𝟐 + 𝟏𝟎. 𝟐𝟔𝟒 × 𝟓𝟐. 𝟑𝟓𝟗
𝟏𝟏𝟕. 𝟏𝟗𝟐𝟐
𝜼𝒉𝒚𝒅 = 𝟎. 𝟗𝟕𝟏𝟖 𝜼𝒉𝒚𝒅 = 𝟗𝟕. 𝟏𝟖 %

68. PROBLEM 8
A Pelton wheel is supplied with water under a head of 35
m at the rate of 40.5 kl/min. The bucket deflects the jet
through an angle of 160° and the mean bucket speed is
13m/s. Calculate the power and hydraulic efficiency of the
turbine.
GIVEN:
𝑯 = 𝟑𝟓 𝒎
𝑸 = 𝟒𝟎. 𝟓 ൗ
𝒌𝒍
𝒎𝒊𝒏 = 𝟒𝟎. 𝟓 × 𝟏𝟎𝟑 ൗ
𝒍
𝒎𝒊𝒏 = 𝟎. 𝟔𝟕𝟓 ൗ
𝒎𝟑
𝒔
𝒖 = 𝟏𝟑 Τ
𝒎
𝒔
𝝓 = 𝟏𝟖𝟎 − 𝟏𝟔𝟎 = 𝟐𝟎 °

69. SOLUTION:
Velocity of Jet:
𝑽𝟏 = 𝑪𝒗 × 𝟐 × 𝒈 × 𝑯
𝑽𝟏 = 𝟎. 𝟗𝟖 × 𝟐 × 𝟗. 𝟖𝟏 × 𝟑𝟓
𝒖𝟏 𝑽𝒓𝟏
𝑽𝒘𝟏 = 𝑽𝟏
𝑨 𝑩 𝑪
𝑬
𝑮
𝑭 𝑯
𝝓
𝑽𝐫𝟐
𝑽𝟐 𝑽𝒇𝟐
𝜷
𝒖𝟐 𝑽𝒘𝟐
𝑽𝟏 = 𝟐𝟓. 𝟔𝟖 Τ
𝒎
𝒔
From Inlet Velocity Triangle Diagram
𝑽𝒘𝟏 = 𝑽𝟏
𝑽𝒘𝟏 = 𝟐𝟓. 𝟔𝟖 Τ
𝒎
𝒔

70. 𝒖𝟏 𝑽𝒓𝟏
𝑽𝒘𝟏 = 𝑽𝟏
𝑨 𝑩 𝑪
𝑬
𝑮
𝑭 𝑯
𝝓
𝑽𝐫𝟐
𝑽𝟐 𝑽𝒇𝟐
𝜷
𝒖𝟐 𝑽𝒘𝟐
𝒖𝟏 + 𝑽𝒓𝟏 = 𝑽𝟏
𝟏𝟑 + 𝑽𝒓𝟏 = 𝟐𝟓. 𝟔𝟖
𝑽𝒓𝟏 = 𝟏𝟐. 𝟔𝟖 Τ
𝒎
𝒔
From Outlet Velocity Triangle Diagram
𝒄𝒐𝒔 𝝓 =
𝒖𝟐 + 𝑽𝒘𝟐
𝑽𝒓𝟐
𝒄𝒐𝒔 𝟐𝟎 =
𝟏𝟑 + 𝑽𝒘𝟐
𝟏𝟐. 𝟔𝟖
𝑨𝒔𝒔𝒖𝒎𝒆 𝑽𝒓𝟏 = 𝑽𝒓𝟐
𝑽𝒘𝟐 = −𝟏. 𝟎𝟖 Τ
𝒎
𝒔

71. 𝒖𝟏 𝑽𝒓𝟏
𝑽𝒘𝟏 = 𝑽𝟏
𝑨 𝑩 𝑪
𝑬
𝑮
𝑭
𝑯
𝝓
𝑽𝐫𝟐
𝑽𝟐 𝑽𝒇𝟐
𝜷
𝒖𝟐 𝑽𝒘𝟐
From Outlet Velocity Triangle Diagram
𝒄𝒐𝒔 𝝓 =
𝒖𝟐 − 𝑽𝒘𝟐
𝑽𝒓𝟐
𝒄𝒐𝒔 𝟐𝟎 =
𝟏𝟑 − 𝑽𝒘𝟐
𝟏𝟐. 𝟔𝟖
𝑨𝒔𝒔𝒖𝒎𝒆 𝑽𝒓𝟏 = 𝑽𝒓𝟐
𝑽𝒘𝟐 = 𝟏. 𝟎𝟖 Τ
𝒎
𝒔
𝟏𝟖𝟎 − 𝜷

72. Work done by Jet per Second:
𝑾 = 𝝆 × 𝑸 × 𝑽𝒘𝟏 + 𝑽𝒘𝟐 × 𝒖
𝑾 = 𝟏𝟎𝟎𝟎 × 𝟎. 𝟔𝟕𝟓 × 𝟐𝟓. 𝟔𝟖 − 𝟏. 𝟎𝟖 × 𝟏𝟑
𝑾 = 𝟐. 𝟏𝟓 × 𝟏𝟎𝟓
𝑾
Hydraulic Efficiency:
𝜼𝒉𝒚𝒅 =
𝟐 × 𝑽𝒘𝟏 + 𝑽𝒘𝟐 × 𝒖
𝑽𝟏
𝟐
=
𝟐 × 𝟐𝟓. 𝟔𝟖 − 𝟏. 𝟎𝟖 × 𝟏𝟑
𝟐𝟓. 𝟔𝟖𝟐
𝜼𝒉𝒚𝒅 = 𝟎. 𝟗𝟔𝟗𝟖 𝜼𝒉𝒚𝒅 = 𝟗𝟔. 𝟗𝟖 %

73. PROBLEM 9
A single jet Pelton wheel runs at 300 rpm under a head of 510
m. The jet diameter is 200 mm and its deflection inside the
bucket is 165°. Assuming that its relative velocity is reduced
by 15% due to friction, determine (i) water power (ii)
resultant force on bucket and (iii) overall efficiency
GIVEN:
𝑵 = 𝟑𝟎𝟎 𝒓𝒑𝒎
𝝓 = 𝟏𝟖𝟎 − 𝟏𝟔𝟓 = 𝟏𝟓°
𝒅 = 𝟐𝟎𝟎 × 𝟏𝟎−𝟑
𝒎
𝑯 = 𝟓𝟏𝟎 𝒎
𝑽𝒓𝟐 = 𝟎. 𝟖𝟓 𝑽𝒓𝟏

74. SOLUTION:
Velocity of Jet:
𝑽𝟏 = 𝑪𝒗 × 𝟐 × 𝒈 × 𝑯
𝑽𝟏 = 𝟎. 𝟗𝟖 × 𝟐 × 𝟗. 𝟖𝟏 × 𝟓𝟏𝟎
𝑽𝟏 = 𝟗𝟖. 𝟎𝟑 Τ
𝒎
𝒔
Velocity of Wheel:
𝒖 = 𝑲𝒖 × 𝟐 × 𝒈 × 𝑯 𝒖 = 𝟎. 𝟒𝟓 × 𝟐 × 𝟗. 𝟖𝟏 × 𝟓𝟏𝟎
𝒖 = 𝟒𝟓. 𝟎𝟏𝟒 Τ
𝒎
𝒔

75. Discharge:
𝒒 = 𝑨 × 𝑽 =
𝝅
𝟒
× 𝒅𝟐
× 𝑽
𝒖𝟏 𝑽𝒓𝟏
𝑽𝒘𝟏 = 𝑽𝟏
𝑨 𝑩 𝑪
𝑬
𝑮
𝑭 𝑯
𝝓
𝑽𝐫𝟐
𝑽𝟐 𝑽𝒇𝟐
𝜷
𝒖𝟐 𝑽𝒘𝟐
𝒒= 𝟑. 𝟎𝟕𝟗 ൗ
𝒎𝟑
𝒔
From Inlet Velocity Triangle Diagram
𝑽𝒘𝟏 = 𝑽𝟏
𝑽𝒘𝟏 = 𝟗𝟖. 𝟎𝟑 Τ
𝒎
𝒔
𝒒 =
𝝅
𝟒
× 𝟎. 𝟐𝟐
× 𝟗𝟖. 𝟎𝟑

76. 𝒖𝟏 𝑽𝒓𝟏
𝑽𝒘𝟏 = 𝑽𝟏
𝑨 𝑩 𝑪
𝑬
𝑮
𝑭 𝑯
𝝓
𝑽𝐫𝟐
𝑽𝟐 𝑽𝒇𝟐
𝜷
𝒖𝟐 𝑽𝒘𝟐
𝒖𝟏 + 𝑽𝒓𝟏 = 𝑽𝟏
𝟒𝟓. 𝟎𝟏𝟒 + 𝑽𝒓𝟏 = 𝟗𝟖. 𝟎𝟑
𝑽𝒓𝟏 = 𝟓𝟑. 𝟎𝟏𝟔 Τ
𝒎
𝒔
𝑽𝒓𝟐 = 𝟎. 𝟖𝟓 𝑽𝒓𝟏
𝑽𝒓𝟐 = 𝟎. 𝟖𝟓 × 𝟓𝟑. 𝟎𝟏𝟔
𝑽𝒓𝟐 = 𝟒𝟓. 𝟎𝟔𝟑 Τ
𝒎
𝒔

77. 𝒖𝟏 𝑽𝒓𝟏
𝑽𝒘𝟏 = 𝑽𝟏
𝑨 𝑩 𝑪
𝑬
𝑮
𝑭 𝑯
𝝓
𝑽𝐫𝟐
𝑽𝟐 𝑽𝒇𝟐
𝜷
𝒖𝟐 𝑽𝒘𝟐
𝒄𝒐𝒔 𝝓 =
𝒖𝟐 + 𝑽𝒘𝟐
𝑽𝒓𝟐
𝒄𝒐𝒔 𝟏𝟓 =
𝟒𝟓. 𝟎𝟏𝟒 + 𝑽𝒘𝟐
𝟒𝟓. 𝟎𝟔𝟑
𝑽𝒘𝟐 = −𝟏. 𝟒𝟖𝟔 Τ
𝒎
𝒔
From Outlet Velocity Triangle Diagram

78. 𝒖𝟏 𝑽𝒓𝟏
𝑽𝒘𝟏 = 𝑽𝟏
𝑨 𝑩 𝑪
𝑬
𝑮
𝑭
𝑯
𝝓
𝑽𝐫𝟐
𝑽𝟐 𝑽𝒇𝟐
𝜷
𝒖𝟐 𝑽𝒘𝟐
𝒄𝒐𝒔 𝝓 =
𝒖𝟐 − 𝑽𝒘𝟐
𝑽𝒓𝟐
𝒄𝒐𝒔 𝟏𝟓 =
𝟒𝟓. 𝟎𝟏𝟒 − 𝑽𝒘𝟐
𝟒𝟓. 𝟎𝟔𝟑
𝑽𝒘𝟐 = 𝟏. 𝟒𝟖𝟔 Τ
𝒎
𝒔
𝟏𝟖𝟎 − 𝜷

79. Resultant Force:
𝑭 = 𝝆 × 𝑸 × 𝑽𝒘𝟏 − 𝑽𝒘𝟐
𝑭 = 𝟏𝟎𝟎𝟎 × 𝟑. 𝟎𝟕𝟗 × 𝟗𝟖. 𝟎𝟑 − 𝟏. 𝟒𝟖𝟔
𝑭 = 𝟐𝟗𝟕. 𝟐𝟓𝟖 × 𝟏𝟎𝟑
𝑵
Hydraulic Efficiency:
𝜼𝒉𝒚𝒅 =
𝟐 × 𝑽𝒘𝟏 + 𝑽𝒘𝟐 × 𝒖
𝑽𝟏
𝟐
=
𝟐 × 𝟗𝟖. 𝟎𝟑 − 𝟏. 𝟒𝟖𝟔 × 𝟒𝟓. 𝟎𝟏𝟒
𝟗𝟖. 𝟎𝟑𝟐
𝜼𝒉𝒚𝒅 = 𝟎. 𝟗𝟎𝟒𝟒 𝜼𝒉𝒚𝒅 = 𝟗𝟎. 𝟒𝟒 %

80. Shaft Power:
𝑷𝒔 =
𝟐 × 𝝅 × 𝑵 × 𝑻
𝟔𝟎
𝑻 = 𝑭 ×
𝑫
𝟐
𝒖 =
𝝅 × 𝑫 × 𝑵
𝟔𝟎
𝟒𝟓. 𝟎𝟏𝟒 =
𝝅 × 𝑫 × 𝟑𝟎𝟎
𝟔𝟎
𝑫 = 𝟐. 𝟖𝟔𝟓 𝒎

81. 𝑻 = 𝑭 ×
𝑫
𝟐
𝑻 = 𝟐𝟗𝟕. 𝟐𝟓𝟖 × 𝟏𝟎𝟑
×
𝟐. 𝟖𝟔𝟓
𝟐
𝑻 = 𝟒𝟐𝟓. 𝟖𝟐 × 𝟏𝟎𝟑 𝑵 − 𝒎
𝑷𝒔 =
𝟐 × 𝝅 × 𝟑𝟎𝟎 × 𝟒𝟐𝟓. 𝟖𝟐 × 𝟏𝟎𝟑
𝟔𝟎
𝑷𝒔 = 𝟏𝟑. 𝟑𝟕𝟕 × 𝟏𝟎𝟔 𝑾

82. Water Power:
𝑷𝒘 = 𝝆 × 𝒈 × 𝑸 × 𝑯
𝑷𝒘 = 𝟏𝟎𝟎𝟎 × 𝟗. 𝟖𝟏 × 𝟑. 𝟎𝟕𝟗 × 𝟓𝟏𝟎
𝑷𝒘 = 𝟏𝟓. 𝟒𝟎𝟒𝟓 × 𝟏𝟎𝟔
𝑾
Overall Efficiency:
𝜼𝒐 =
𝑷𝒔
𝑷𝒘
𝜼𝒐 =
𝟏𝟑. 𝟑𝟕𝟕 × 𝟏𝟎𝟔
𝟏𝟓. 𝟒𝟎𝟒𝟓 × 𝟏𝟎𝟔
𝜼𝒐 = 𝟎. 𝟖𝟔𝟖𝟒 𝜼𝒐 = 𝟖𝟔. 𝟖𝟒 %

83. PROBLEM 10
Consider an impulse wheel with a pitch diameter of 2.75m and
a bucket angle of 170°. If the velocity is 58m/s, the jet
diameter is 100mm, and the rotational speed is 320rpm, find
the force on the buckets, the torque on the runner, and the
power transferred to the runner. Assume 𝑽𝒓𝟐 = 𝟎. 𝟗𝑽𝒓𝟏.
GIVEN:
𝑫 = 𝟐. 𝟕𝟓 𝒎
𝒅 = 𝟏𝟎𝟎 × 𝟏𝟎−𝟑
𝒎
𝑽𝟏 = 𝟓𝟖 Τ
𝒎
𝒔
𝝓 = 𝟏𝟖𝟎 − 𝟏𝟕𝟎 = 𝟏𝟎°
𝑵 = 𝟑𝟐𝟎 𝒓𝒑𝒎 𝑽𝒓𝟐 = 𝟎. 𝟗 𝑽𝒓𝟏

84. SOLUTION:
Velocity of Wheel:
𝒖 =
𝝅 × 𝑫 × 𝑵
𝟔𝟎
𝒖 =
𝝅 × 𝟐. 𝟕𝟓 × 𝟑𝟐𝟎
𝟔𝟎
𝒖 = 𝟒𝟔. 𝟎𝟕𝟔 Τ
𝒎
𝒔
Velocity of Jet:
𝑽𝟏 = 𝑪𝒗 × 𝟐 × 𝒈 × 𝑯
𝟓𝟖 = 𝟎. 𝟗𝟖 × 𝟐 × 𝟗. 𝟖𝟏 × 𝑯
𝑯 = 𝟏𝟕𝟖. 𝟓𝟐𝟕 𝒎

85. Discharge:
𝒒 = 𝑨 × 𝑽 =
𝝅
𝟒
× 𝒅𝟐
× 𝑽
𝒖𝟏 𝑽𝒓𝟏
𝑽𝒘𝟏 = 𝑽𝟏
𝑨 𝑩 𝑪
𝑬
𝑮
𝑭 𝑯
𝝓
𝑽𝐫𝟐
𝑽𝟐 𝑽𝒇𝟐
𝜷
𝒖𝟐 𝑽𝒘𝟐
𝒒= 𝟎. 𝟒𝟓𝟓 ൗ
𝒎𝟑
𝒔
From Inlet Velocity Triangle Diagram
𝑽𝒘𝟏 = 𝑽𝟏
𝑽𝒘𝟏 = 𝟓𝟖 Τ
𝒎
𝒔
𝒒 =
𝝅
𝟒
× 𝟎. 𝟏𝟐
× 𝟓𝟖

86. 𝒖𝟏 𝑽𝒓𝟏
𝑽𝒘𝟏 = 𝑽𝟏
𝑨 𝑩 𝑪
𝑬
𝑮
𝑭 𝑯
𝝓
𝑽𝐫𝟐
𝑽𝟐 𝑽𝒇𝟐
𝜷
𝒖𝟐 𝑽𝒘𝟐
𝒖𝟏 + 𝑽𝒓𝟏 = 𝑽𝟏
𝟒𝟔. 𝟎𝟕𝟔 + 𝑽𝒓𝟏 = 𝟓𝟖
𝑽𝒓𝟏 = 𝟏𝟏. 𝟗𝟐𝟒 Τ
𝒎
𝒔
𝑽𝒓𝟐 = 𝟎. 𝟗 𝑽𝒓𝟏
𝑽𝒓𝟐 = 𝟎. 𝟗 × 𝟏𝟏. 𝟗𝟐𝟒
𝑽𝒓𝟐 = 𝟏𝟎. 𝟕𝟑𝟏𝟔 Τ
𝒎
𝒔

87. 𝒖𝟏 𝑽𝒓𝟏
𝑽𝒘𝟏 = 𝑽𝟏
𝑨 𝑩 𝑪
𝑬
𝑮
𝑭 𝑯
𝝓
𝑽𝐫𝟐
𝑽𝟐 𝑽𝒇𝟐
𝜷
𝒖𝟐 𝑽𝒘𝟐
𝒄𝒐𝒔 𝝓 =
𝒖𝟐 + 𝑽𝒘𝟐
𝑽𝒓𝟐
𝒄𝒐𝒔 𝟏𝟎 =
𝟒𝟔. 𝟎𝟕𝟔 + 𝑽𝒘𝟐
𝟏𝟎. 𝟕𝟑𝟏𝟔
𝑽𝒘𝟐 = −𝟑𝟓. 𝟓𝟎 Τ
𝒎
𝒔
From Outlet Velocity Triangle Diagram

88. 𝒖𝟏 𝑽𝒓𝟏
𝑽𝒘𝟏 = 𝑽𝟏
𝑨 𝑩 𝑪
𝑬
𝑮
𝑭
𝑯
𝝓
𝑽𝐫𝟐
𝑽𝟐 𝑽𝒇𝟐
𝜷
𝒖𝟐 𝑽𝒘𝟐
𝒄𝒐𝒔 𝝓 =
𝒖𝟐 − 𝑽𝒘𝟐
𝑽𝒓𝟐
𝒄𝒐𝒔 𝟏𝟎 =
𝟒𝟔. 𝟎𝟕𝟔 − 𝑽𝒘𝟐
𝟏𝟎. 𝟕𝟑𝟏𝟔
𝑽𝒘𝟐 = 𝟑𝟓. 𝟓𝟎 Τ
𝒎
𝒔
𝟏𝟖𝟎 − 𝜷

89. Resultant Force:
𝑭 = 𝝆 × 𝑸 × 𝑽𝒘𝟏 − 𝑽𝒘𝟐
𝑭 = 𝟏𝟎𝟎𝟎 × 𝟎. 𝟒𝟓𝟓 × 𝟓𝟖 − 𝟑𝟓. 𝟓𝟎
𝑭 = 𝟏𝟎. 𝟐𝟑 × 𝟏𝟎𝟑
𝑵
Hydraulic Efficiency:
𝜼𝒉𝒚𝒅 =
𝟐 × 𝑽𝒘𝟏 + 𝑽𝒘𝟐 × 𝒖
𝑽𝟏
𝟐
=
𝟐 × 𝟓𝟖 − 𝟑𝟓. 𝟓𝟎 × 𝟒𝟔. 𝟎𝟕𝟔
𝟓𝟖𝟐
𝜼𝒉𝒚𝒅 = 𝟎. 𝟔𝟏𝟔𝟑 𝜼𝒉𝒚𝒅 = 𝟔𝟏. 𝟔𝟑 %

90. Shaft Power:
𝑷𝒔 =
𝟐 × 𝝅 × 𝑵 × 𝑻
𝟔𝟎
𝑻 = 𝑭 ×
𝑫
𝟐
𝑻 = 𝟏𝟎. 𝟐𝟑 × 𝟏𝟎𝟑
×
𝟐. 𝟕𝟓
𝟐
𝑻 = 𝟏𝟒. 𝟎𝟔𝟔 × 𝟏𝟎𝟑 𝑵 − 𝒎

91. 𝑷𝒔 =
𝟐 × 𝝅 × 𝟑𝟐𝟎 × 𝟏𝟒. 𝟎𝟔𝟔 × 𝟏𝟎𝟑
𝟔𝟎
𝑷𝒔 = 𝟒. 𝟕𝟏 × 𝟏𝟎𝟓 𝑾
Water Power:
𝑷𝒘 = 𝝆 × 𝒈 × 𝑸 × 𝑯
𝑷𝒘 = 𝟏𝟎𝟎𝟎 × 𝟗. 𝟖𝟏 × 𝟎. 𝟒𝟓𝟓 × 𝟏𝟕𝟖. 𝟓𝟐𝟕
𝑷𝒘 = 𝟕. 𝟗𝟔𝟖 × 𝟏𝟎𝟓
𝑾

92. Overall Efficiency:
𝜼𝒐 =
𝑷𝒔
𝑷𝒘
𝜼𝒐 =
𝟒. 𝟕𝟏 × 𝟏𝟎𝟓
𝟕. 𝟗𝟔𝟖 × 𝟏𝟎𝟓
𝜼𝒐 = 𝟎. 𝟓𝟗𝟏𝟕
𝜼𝒐 = 𝟓𝟗. 𝟏𝟕 %

93. PROBLEM 11
The nozzle of a Pelton wheel gives a jet of 9cm diameter and velocity
75m/s. Coefficient of velocity is 0.978. The pitch circle diameter is 1.5m
and the deflection angle of the buckets is 170°. The wheel velocity is
0.46 times the jet velocity. Estimate the speed of the Pelton wheel
turbine in rpm, theoretical power developed and also the efficiency of
the turbine.
GIVEN:
𝒅 = 𝟗 × 𝟏𝟎−𝟐
𝒎
𝑫 = 𝟏. 𝟓 𝒎
𝑽𝟏 = 𝟕𝟓 Τ
𝒎
𝒔
𝝓 = 𝟏𝟖𝟎 − 𝟏𝟕𝟎 = 𝟏𝟎°
𝒖 = 𝟎. 𝟒𝟔 𝑽𝟏 = 𝟎. 𝟒𝟔 × 𝟕𝟓 = 𝟑𝟒. 𝟓 Τ
𝒎
𝒔

94. SOLUTION:
Speed of Turbine:
𝒖 =
𝝅 × 𝑫 × 𝑵
𝟔𝟎
𝟑𝟒. 𝟓 =
𝝅 × 𝟏. 𝟓 × 𝑵
𝟔𝟎
𝑵 = 𝟒𝟑𝟗. 𝟐𝟔 𝒓𝒑𝒎
Velocity of Jet:
𝑽𝟏 = 𝑪𝒗 × 𝟐 × 𝒈 × 𝑯
𝟕𝟓 = 𝟎. 𝟗𝟕𝟖 × 𝟐 × 𝟗. 𝟖𝟏 × 𝑯
𝑯 = 𝟐𝟗𝟗. 𝟕𝟒 𝒎

95. Discharge:
𝒒 = 𝑨 × 𝑽 =
𝝅
𝟒
× 𝒅𝟐
× 𝑽
𝒖𝟏 𝑽𝒓𝟏
𝑽𝒘𝟏 = 𝑽𝟏
𝑨 𝑩 𝑪
𝑬
𝑮
𝑭 𝑯
𝝓
𝑽𝐫𝟐
𝑽𝟐 𝑽𝒇𝟐
𝜷
𝒖𝟐 𝑽𝒘𝟐
𝒒= 𝟎. 𝟒𝟕𝟕 ൗ
𝒎𝟑
𝒔
From Inlet Velocity Triangle Diagram
𝑽𝒘𝟏 = 𝑽𝟏
𝑽𝒘𝟏 = 𝟕𝟓 Τ
𝒎
𝒔
𝒒 =
𝝅
𝟒
× 𝟎. 𝟎𝟗𝟐
× 𝟕𝟓

96. 𝒖𝟏 𝑽𝒓𝟏
𝑽𝒘𝟏 = 𝑽𝟏
𝑨 𝑩 𝑪
𝑬
𝑮
𝑭 𝑯
𝝓
𝑽𝐫𝟐
𝑽𝟐 𝑽𝒇𝟐
𝜷
𝒖𝟐 𝑽𝒘𝟐
𝒖𝟏 + 𝑽𝒓𝟏 = 𝑽𝟏
𝟑𝟒. 𝟓 + 𝑽𝒓𝟏 = 𝟕𝟓
𝑽𝒓𝟏 = 𝟒𝟎. 𝟓 Τ
𝒎
𝒔
𝑽𝒓𝟐 = 𝑽𝒓𝟏
𝑽𝒓𝟐 = 𝟒𝟎. 𝟓
𝑽𝒓𝟐 = 𝟒𝟎. 𝟓 Τ
𝒎
𝒔

97. 𝒖𝟏 𝑽𝒓𝟏
𝑽𝒘𝟏 = 𝑽𝟏
𝑨 𝑩 𝑪
𝑬
𝑮
𝑭 𝑯
𝝓
𝑽𝐫𝟐
𝑽𝟐 𝑽𝒇𝟐
𝜷
𝒖𝟐 𝑽𝒘𝟐
𝒄𝒐𝒔 𝝓 =
𝒖𝟐 + 𝑽𝒘𝟐
𝑽𝒓𝟐
𝒄𝒐𝒔 𝟏𝟎 =
𝟑𝟒. 𝟓 + 𝑽𝒘𝟐
𝟒𝟎. 𝟓
𝑽𝒘𝟐 = 𝟓. 𝟑𝟖𝟒𝟕 Τ
𝒎
𝒔
From Outlet Velocity Triangle Diagram

98. Theoretical Power:
𝑷 = 𝝆 × 𝑸 × 𝑽𝒘𝟏 + 𝑽𝒘𝟐 𝒖
𝑷 = 𝟏𝟎𝟎𝟎 × 𝟎. 𝟒𝟕𝟕 × 𝟕𝟓 + 𝟓. 𝟑𝟖𝟒𝟕 × 𝟑𝟒. 𝟓
𝑷 = 𝟏. 𝟑𝟐𝟐𝟖𝟓 × 𝟏𝟎𝟔
𝑾
Overall Efficiency:
𝜼𝒐 =
𝑷𝒔
𝑷𝒘
=
𝑷𝒔
𝝆 × 𝒈 × 𝑸 × 𝑯
=
𝟏. 𝟑𝟐𝟐𝟖𝟓 × 𝟏𝟎𝟔
𝟏𝟎𝟎𝟎 × 𝟗. 𝟖𝟏 × 𝟎. 𝟒𝟕𝟕 × 𝟐𝟗𝟗. 𝟕𝟒
𝜼𝒐 = 𝟎. 𝟗𝟒𝟑𝟏 𝜼𝒐 = 𝟗𝟒. 𝟑𝟏 %

99. Overall Efficiency:
𝜼𝒐 =
𝑷𝒔
𝑷𝒘
𝜼𝒐 =
𝑷𝒔
𝝆 × 𝒈 × 𝑸 × 𝑯
𝜼𝒐 =
𝟗𝟎. 𝟔𝟑𝟐 × 𝟏𝟎𝟑
𝟏𝟎𝟎𝟎 × 𝟗. 𝟖𝟏 × 𝟎. 𝟎𝟔𝟔 × 𝟏𝟓𝟎
𝜼𝒐 = 𝟎. 𝟗𝟐𝟐𝟎 𝜼𝒐 = 𝟗𝟐. 𝟐𝟎 %

100. THANK YOU