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KV
MOMENT OF MOMENTUM
{Rotodynamic Machines}
KV
Identify the unique vocabulary used in the description and
analysis of fluid flow with an emphasis on Moment of
Momentum and Rotodynamic Machines
Describe and discuss the flow of fluids through
rotodynamic machines and the momentum of the fluids
passing through such devices.
Construct and resolve the velocity vectors associated with
the flow of fluids through rotodynamic devices.
Derive and apply the governing equations associated with
fluid momentum when applied to rotodynamic devices
OBJECTIVES
KV
Consider a two arm sprinkler system as illustrated. The direction
and magnitude of fluid flow through this sprinkler changes direction
as it passes from the inlet to the outlet. The fluid discharging
through the arms of the sprinkler exerts a torque on the sprinkler
head and as a consequence causes it to rotated about the z axis.
Provided that the flow is steady, equation 1 can be applied to the
system.
βˆ‘ π‘Ÿ Γ— π‘ˆ π‘œπ‘’π‘‘π‘š
Β·
π‘œπ‘’π‘‘ βˆ’ βˆ‘ π‘Ÿ Γ— π‘ˆ π‘–π‘›π‘š
Β·
𝑖𝑛 = βˆ‘ π‘Ÿ Γ— 𝐹 (1)
Eq.1 is known as the moment of moment equation
MOMENTUM OF
MOMENTUM
KV
β€’ The fluid enters the control volume of the sprinkler (figure 1)
axially through the stem at section 1.
β€’ The fluid leaves the control volume through each of the two
nozzles at section 2.
β€’ The component of π‘Ÿ Γ— π‘ˆ along the axis of rotation is zero.
β€’ The magnitude of the the axial component of π‘Ÿ Γ— π‘ˆ is π‘Ÿ2 Γ—
π‘ˆπœƒ where r2 is the radius from the axis of rotation to the
nozzle centreline and π‘ˆπœƒ2 is the tangential component of
the velocity of flow exiting the nozzle.
β€’ The fluid velocity measured relative to a fixed control
surface is an absolute velocity, U
β€’ The velocity of the fluid exiting the nozzle is called the
relative velocity, UΞΈ.
β€’ Absolute and relative velocities can be related by a vector
diagram.
Figure 1: Two arm sprinkler (Young, et al. 2012)
KV
For the sprinkler system, it can be stated that;
βˆ‘ π‘Ÿ Γ— π‘ˆ π‘œπ‘’π‘‘π‘š
Β·
π‘œπ‘’π‘‘ βˆ’ βˆ‘ π‘Ÿ Γ— π‘ˆ π‘–π‘›π‘š
Β·
𝑖𝑛 π‘Žπ‘₯π‘–π‘Žπ‘™
= βˆ’π‘Ÿ2 Γ— π‘ˆπœƒ2 π‘š
Β·
(2)
For mass conservation, ṁ is the total mass flowrate through
both nozzles. Given that the direction of rotation is opposite to
that discharging jet, a negative sign is appropriate in the right
hand portion of the equation.
Noting that the mass flow rate ṁ times the velocity U equals
force, then can be stated that
βˆ‘ π‘Ÿ Γ— π‘ˆ π‘š
Β·
π‘Žπ‘₯π‘–π‘Žπ‘™
= βˆ‘ π‘Ÿ Γ— 𝐹 𝑐𝑣 π‘Žπ‘₯π‘–π‘Žπ‘™ = π‘‡π‘ β„Žπ‘Žπ‘“π‘‘ (3)
π‘ˆ2 = π‘Ÿ2πœ”
π‘ˆπœƒ2
π‘ˆπœƒ2
π‘ˆπœƒ
𝑀2
𝑀2
πœ”
π‘‡π‘ β„Žπ‘Žπ‘“π‘‘
Control Volume
Section (2)
Section (1)
Figure 2: Sprinkler control volume and velocities (Young, et al. 2012)
KV
It follows from equations 2 & 3 that
βˆ’π‘Ÿ2 Γ— π‘ˆπœƒ2π‘š
Β·
= π‘‡π‘ β„Žπ‘Žπ‘“π‘‘
(4)
It is obvious that Tshaft is negative thereby the shaft torque
opposes the rotation of the sprinkler. This is true for all turbine
devices.
The shaft power αΊ†shaft associated with the shaft torque Tshaft can
be determined
π‘Š
Β·
π‘ β„Žπ‘Žπ‘“π‘‘ = π‘‡π‘ β„Žπ‘Žπ‘“π‘‘ Γ— πœ” = βˆ’π‘Ÿ2π‘ˆπœƒ2π‘š
Β·
πœ”
(5)
Given that π‘Ÿ2π‘ˆπœƒ2 is the speed, U of each nozzle, eq 5 can be
written as
π‘Š
Β·
π‘ β„Žπ‘Žπ‘“π‘‘ = βˆ’π‘ˆ2π‘ˆπœƒ2π‘š
Β·
or written per unit mass π‘Š
Β·
π‘ β„Žπ‘Žπ‘“π‘‘ = βˆ’π‘ˆ2π‘ˆπœƒ2
π‘ˆ2 = π‘Ÿ2πœ”
π‘ˆπœƒ2
π‘ˆπœƒ2
π‘ˆπœƒ
𝑀2
𝑀2
πœ”
π‘‡π‘ β„Žπ‘Žπ‘“π‘‘
Control Volume
Section (2)
Section (1)
Figure 2: Sprinkler control volume and velocities (Young, et al. 2012)
KV
LAWN SPRINKLER
Example Problem
Water enters a rotating lawn sprinkler through its base at the steady rate of 1000 ml
・s-1. The exit area of each of the two nozzles is 30 mm2 and the flow leaving each
nozzle is in the tangential direction. The radius from the axis of rotation to the
centerline of each nozzle is 200 mm. Determine
(a) The resisting torque required to hold the sprinkler head stationary
(b) The resisting torque associated with the sprinkler rotating with a constant
speed of 500 rev・min-1.
(c) The speed of the sprinkler if no resisting torque is applied.
KEITH VAUGH
The fluid enters through the stem of the sprinkler and leaves through each of the
two nozzles as illustrated. The torque can be determined by
π‘‡π‘ β„Žπ‘Žπ‘“π‘‘ = βˆ’π‘Ÿ2π‘ˆπœƒ2π‘š
Β·
where π‘ˆπœƒ2 = π‘ˆ2
SOLUTION
The mass flow rate can be determined
π‘š
Β·
= 𝑉
Β·
𝜌 =
1000π‘šπ‘™/𝑠 Γ— 0.001π‘š3
/𝑙 Γ— 999π‘˜π‘”/π‘š3
1000π‘šπ‘™/𝑙
= 0.999π‘˜π‘”
KEITH VAUGH
The velocity of the water issuing from the nozzle, relative to the nozzle
must be determined. Assuming that the sprinkler head is stationary;
πœ•
πœ•π‘‘
0
∫ πœŒπ‘‘π‘‰ + βˆ‘πœŒπ‘œπ‘’π‘‘π΄π‘œπ‘’π‘‘π‘ˆπ‘œπ‘’π‘‘ βˆ’ βˆ‘πœŒπ‘–π‘›π΄π‘–π‘›π‘ˆπ‘–π‘› = 0
The time rate of changed of the mass of water is zero because the flow
is steady and the control volume is filled with water.
Because there is only one inlet (subscript 1) and two outlets (subscript
2 and 3), the above equation becomes
𝜌2𝐴2π‘ˆ2 + 𝜌3𝐴3π‘ˆ3 βˆ’ 𝜌1𝐴1π‘ˆ1 = 0
Given that outlet 2 and 3 are equal and given that the density of the
fluid does not change;
2 𝐴2π‘ˆ2 βˆ’ 𝐴1π‘ˆ1 = 0
KEITH VAUGH
The volumetric flow rate is 𝑉
Β·
= π΄π‘ˆ
Then the velocity can be determined for the outlet: π‘ˆ2 =
𝑉
Β·
2𝐴2
π‘ˆ2 =
1000π‘šπ‘™/𝑠 Γ— 0.001π‘š3
/𝑙 Γ— 106
π‘šπ‘š2
/π‘š2
1000π‘šπ‘™/𝑙 Γ— 2 Γ— 30π‘šπ‘š2
= 16.7π‘š/𝑠
(a)
The resistance torque can be determined for the insistence when the
sprinkler is not rotating;
π‘‡π‘ β„Žπ‘Žπ‘“π‘‘ = 200 Γ— 10βˆ’3
π‘š Γ— 16.7π‘š/𝑠 Γ— 0.997π‘˜π‘”/𝑠 = βˆ’3.34𝑁 βˆ’ π‘š
KEITH VAUGH
(b)
When the sprinkler rotates at a constant speed, the flow field is unsteady
but cyclical. In such cases the flow field can be regarded as being steady in
the mean.
π‘ˆπœƒ = π‘Š2 βˆ’ π‘ˆ2 where π‘Š = 16π‘š/𝑠 and π‘ˆ2 = π‘Ÿ2πœ”
Applying the axial component of the moment of momentum equation
π‘ˆπœƒ = 16.7π‘š/𝑠 βˆ’ π‘Ÿ2πœ” = 16.7π‘š/𝑠 βˆ’ 200 Γ— 10βˆ’3
π‘š Γ— 500π‘Ÿπ‘’π‘£/π‘šπ‘–π‘› Γ—
2πœ‹
60
π‘ˆπœƒ = 6.2π‘š/𝑠
The torque can now be determined for this case;
π‘‡π‘ β„Žπ‘Žπ‘“π‘‘ = 200 Γ— 10βˆ’3
Γ— 6.2π‘š/𝑠 Γ— 0.999π‘˜π‘”/𝑠 = βˆ’1.24𝑁 βˆ’ π‘š
It can be observed from this result that the resisting torque associated with
the head rotating is less that the torque required to hold it stationary.
KEITH VAUGH
(c)
When no resisting torque is applied, a maximum constant speed of rotation
will occur.
π‘‡π‘ β„Žπ‘Žπ‘“π‘‘ = βˆ’π‘Ÿ2 π‘Š2 βˆ’ π‘Ÿ2πœ” π‘š
Β·
therefore 0 = βˆ’π‘Ÿ2 π‘Š2 βˆ’ π‘Ÿ2πœ” π‘š
Β·
When the mass flow rate is constant
πœ” =
π‘Š2
π‘Ÿ2
=
16.7π‘š/𝑠
200 Γ— 10βˆ’3π‘š
= 83.5π‘Ÿπ‘Žπ‘‘/𝑠
πœ” =
83.5π‘Ÿπ‘Žπ‘‘/𝑠 Γ— 60𝑠/π‘šπ‘–π‘›
2πœ‹π‘Ÿπ‘Žπ‘‘/π‘Ÿπ‘’π‘£
= 797π‘Ÿπ‘π‘š
The torque can now be determined for this case;
π‘‡π‘ β„Žπ‘Žπ‘“π‘‘ = 200 Γ— 10βˆ’3
Γ— 6.2π‘š/𝑠 Γ— 0.999π‘˜π‘”/𝑠 = βˆ’1.24𝑁 βˆ’ π‘š
It can be observed from this result that the resisting torque associated with
the head rotating is less that the torque required to hold it stationary.
KEITH VAUGH
(d)
Investigate varying Ο‰ between 0 and 1000 rpm and plot the results against
the torque.
-3.94
-3.38
-2.81
-2.25
-1.69
-1.13
-0.56
0.00
0.56
0 100 200 300 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500 1600
Torque
N-m
Rotational Speed (rpm)
Torque Vs. Rotation Speed
KV
Ο‰
Runner
blade
u0
u1β†’ Absolute velocity vector. Velocity of the
flow at the inlet. It’s direction is governed
by the guide vane angle Ξ±
v1β†’ Velocity of blade at inlet
ur1β†’ The relative velocity vector of the jet at
the inlet
uf1β†’ The velocity of flow in the radial
direction
uw1β†’The component of the velocity of the jet
u1 in horizontal/tangential direction
Ξ±1β†’ Guide vane angle, or the angle between
the velocity vector of the blade and the
direction of the jet
Ξ²1β†’ Blade angle at outlet. Angle between the
relative velocity vector and the direction
of motion of the blade at the inlet
KV
Ο‰
Runner
blade
u0
u2β†’ Absolute velocity vector water leaving
the runner. It’s direction is governed
by the guide vane angle Ξ± and if it is in
the radial direction, will be equal to uf2
v2β†’ Velocity of blade at exit
ur2β†’ The relative velocity vector of the water
leaving the impeller
uf2β†’ The velocity of flow in the radial
direction
uw2β†’The component of the velocity for exit
u2 in horizontal/tangential direction
Ξ±2β†’ Guide vane angle, or the angle between
the velocity vector of the blade and the
direction of the u2 exiting
Ξ²2β†’ Blade angle at outlet, Angle between the
relative velocity vector and the direction
of motion of the blade at the outlet
𝜢2
uf2
Ξ²2
u0
KV
uf1
uf1
u1
uw1
ur1
uf1
u1
uf1 ur1
Velocity triangles Use trigonometry to solve for each vectors
uw1
Two triangles within
uw1
uw1
uw1 - v1
v1 - uw1
KV
Ο‘
Opposite
Adjacent
π‘ π‘–π‘›πœƒ =
𝑂𝑃𝑃
π»π‘Œπ‘ƒ
π‘π‘œπ‘ πœƒ =
𝐴𝐷𝐽
π»π‘Œπ‘ƒ
π‘‘π‘Žπ‘›πœƒ =
𝑂𝑃𝑃
𝐴𝐷𝐽
Also remember that
π‘‘π‘Žπ‘›πœƒ =
π‘ π‘–π‘›πœƒ
π‘π‘œπ‘ πœƒ
π‘π‘œπ‘‘πœƒ =
π‘π‘œπ‘ πœƒ
π‘ π‘–π‘›πœƒ
π‘ π‘’π‘πœƒ =
1
π‘π‘œπ‘ πœƒ
π‘π‘ π‘πœƒ =
1
π‘ π‘–π‘›πœƒ
Recall from Trigonometry in Mathematics
KV
π‘ƒπ‘œπ‘€π‘’π‘Ÿ = 𝜌 Γ— 𝑔 Γ— 𝑉
Β·
Γ— 𝐻
π‘˜π‘”
π‘š3
Γ—
π‘š
𝑠2
Γ—
π‘š3
𝑠
Γ— π‘š
π‘˜π‘”
π‘š3
Γ—
π‘š5
𝑠3
π‘˜π‘” Γ—
π‘š2
𝑠3
π‘˜π‘” Γ— π‘š
𝑠2
Γ—
π‘š
𝑠
𝑁 Γ—
π‘š
𝑠
=
𝐽
𝑠
= π‘Šπ‘Žπ‘‘π‘‘π‘ 

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  • 2. KV Identify the unique vocabulary used in the description and analysis of fluid flow with an emphasis on Moment of Momentum and Rotodynamic Machines Describe and discuss the flow of fluids through rotodynamic machines and the momentum of the fluids passing through such devices. Construct and resolve the velocity vectors associated with the flow of fluids through rotodynamic devices. Derive and apply the governing equations associated with fluid momentum when applied to rotodynamic devices OBJECTIVES
  • 3. KV Consider a two arm sprinkler system as illustrated. The direction and magnitude of fluid flow through this sprinkler changes direction as it passes from the inlet to the outlet. The fluid discharging through the arms of the sprinkler exerts a torque on the sprinkler head and as a consequence causes it to rotated about the z axis. Provided that the flow is steady, equation 1 can be applied to the system. βˆ‘ π‘Ÿ Γ— π‘ˆ π‘œπ‘’π‘‘π‘š Β· π‘œπ‘’π‘‘ βˆ’ βˆ‘ π‘Ÿ Γ— π‘ˆ π‘–π‘›π‘š Β· 𝑖𝑛 = βˆ‘ π‘Ÿ Γ— 𝐹 (1) Eq.1 is known as the moment of moment equation MOMENTUM OF MOMENTUM
  • 4. KV β€’ The fluid enters the control volume of the sprinkler (figure 1) axially through the stem at section 1. β€’ The fluid leaves the control volume through each of the two nozzles at section 2. β€’ The component of π‘Ÿ Γ— π‘ˆ along the axis of rotation is zero. β€’ The magnitude of the the axial component of π‘Ÿ Γ— π‘ˆ is π‘Ÿ2 Γ— π‘ˆπœƒ where r2 is the radius from the axis of rotation to the nozzle centreline and π‘ˆπœƒ2 is the tangential component of the velocity of flow exiting the nozzle. β€’ The fluid velocity measured relative to a fixed control surface is an absolute velocity, U β€’ The velocity of the fluid exiting the nozzle is called the relative velocity, UΞΈ. β€’ Absolute and relative velocities can be related by a vector diagram. Figure 1: Two arm sprinkler (Young, et al. 2012)
  • 5. KV For the sprinkler system, it can be stated that; βˆ‘ π‘Ÿ Γ— π‘ˆ π‘œπ‘’π‘‘π‘š Β· π‘œπ‘’π‘‘ βˆ’ βˆ‘ π‘Ÿ Γ— π‘ˆ π‘–π‘›π‘š Β· 𝑖𝑛 π‘Žπ‘₯π‘–π‘Žπ‘™ = βˆ’π‘Ÿ2 Γ— π‘ˆπœƒ2 π‘š Β· (2) For mass conservation, ṁ is the total mass flowrate through both nozzles. Given that the direction of rotation is opposite to that discharging jet, a negative sign is appropriate in the right hand portion of the equation. Noting that the mass flow rate ṁ times the velocity U equals force, then can be stated that βˆ‘ π‘Ÿ Γ— π‘ˆ π‘š Β· π‘Žπ‘₯π‘–π‘Žπ‘™ = βˆ‘ π‘Ÿ Γ— 𝐹 𝑐𝑣 π‘Žπ‘₯π‘–π‘Žπ‘™ = π‘‡π‘ β„Žπ‘Žπ‘“π‘‘ (3) π‘ˆ2 = π‘Ÿ2πœ” π‘ˆπœƒ2 π‘ˆπœƒ2 π‘ˆπœƒ 𝑀2 𝑀2 πœ” π‘‡π‘ β„Žπ‘Žπ‘“π‘‘ Control Volume Section (2) Section (1) Figure 2: Sprinkler control volume and velocities (Young, et al. 2012)
  • 6. KV It follows from equations 2 & 3 that βˆ’π‘Ÿ2 Γ— π‘ˆπœƒ2π‘š Β· = π‘‡π‘ β„Žπ‘Žπ‘“π‘‘ (4) It is obvious that Tshaft is negative thereby the shaft torque opposes the rotation of the sprinkler. This is true for all turbine devices. The shaft power αΊ†shaft associated with the shaft torque Tshaft can be determined π‘Š Β· π‘ β„Žπ‘Žπ‘“π‘‘ = π‘‡π‘ β„Žπ‘Žπ‘“π‘‘ Γ— πœ” = βˆ’π‘Ÿ2π‘ˆπœƒ2π‘š Β· πœ” (5) Given that π‘Ÿ2π‘ˆπœƒ2 is the speed, U of each nozzle, eq 5 can be written as π‘Š Β· π‘ β„Žπ‘Žπ‘“π‘‘ = βˆ’π‘ˆ2π‘ˆπœƒ2π‘š Β· or written per unit mass π‘Š Β· π‘ β„Žπ‘Žπ‘“π‘‘ = βˆ’π‘ˆ2π‘ˆπœƒ2 π‘ˆ2 = π‘Ÿ2πœ” π‘ˆπœƒ2 π‘ˆπœƒ2 π‘ˆπœƒ 𝑀2 𝑀2 πœ” π‘‡π‘ β„Žπ‘Žπ‘“π‘‘ Control Volume Section (2) Section (1) Figure 2: Sprinkler control volume and velocities (Young, et al. 2012)
  • 7. KV LAWN SPRINKLER Example Problem Water enters a rotating lawn sprinkler through its base at the steady rate of 1000 ml ・s-1. The exit area of each of the two nozzles is 30 mm2 and the flow leaving each nozzle is in the tangential direction. The radius from the axis of rotation to the centerline of each nozzle is 200 mm. Determine (a) The resisting torque required to hold the sprinkler head stationary (b) The resisting torque associated with the sprinkler rotating with a constant speed of 500 rev・min-1. (c) The speed of the sprinkler if no resisting torque is applied.
  • 8. KEITH VAUGH The fluid enters through the stem of the sprinkler and leaves through each of the two nozzles as illustrated. The torque can be determined by π‘‡π‘ β„Žπ‘Žπ‘“π‘‘ = βˆ’π‘Ÿ2π‘ˆπœƒ2π‘š Β· where π‘ˆπœƒ2 = π‘ˆ2 SOLUTION The mass flow rate can be determined π‘š Β· = 𝑉 Β· 𝜌 = 1000π‘šπ‘™/𝑠 Γ— 0.001π‘š3 /𝑙 Γ— 999π‘˜π‘”/π‘š3 1000π‘šπ‘™/𝑙 = 0.999π‘˜π‘”
  • 9. KEITH VAUGH The velocity of the water issuing from the nozzle, relative to the nozzle must be determined. Assuming that the sprinkler head is stationary; πœ• πœ•π‘‘ 0 ∫ πœŒπ‘‘π‘‰ + βˆ‘πœŒπ‘œπ‘’π‘‘π΄π‘œπ‘’π‘‘π‘ˆπ‘œπ‘’π‘‘ βˆ’ βˆ‘πœŒπ‘–π‘›π΄π‘–π‘›π‘ˆπ‘–π‘› = 0 The time rate of changed of the mass of water is zero because the flow is steady and the control volume is filled with water. Because there is only one inlet (subscript 1) and two outlets (subscript 2 and 3), the above equation becomes 𝜌2𝐴2π‘ˆ2 + 𝜌3𝐴3π‘ˆ3 βˆ’ 𝜌1𝐴1π‘ˆ1 = 0 Given that outlet 2 and 3 are equal and given that the density of the fluid does not change; 2 𝐴2π‘ˆ2 βˆ’ 𝐴1π‘ˆ1 = 0
  • 10. KEITH VAUGH The volumetric flow rate is 𝑉 Β· = π΄π‘ˆ Then the velocity can be determined for the outlet: π‘ˆ2 = 𝑉 Β· 2𝐴2 π‘ˆ2 = 1000π‘šπ‘™/𝑠 Γ— 0.001π‘š3 /𝑙 Γ— 106 π‘šπ‘š2 /π‘š2 1000π‘šπ‘™/𝑙 Γ— 2 Γ— 30π‘šπ‘š2 = 16.7π‘š/𝑠 (a) The resistance torque can be determined for the insistence when the sprinkler is not rotating; π‘‡π‘ β„Žπ‘Žπ‘“π‘‘ = 200 Γ— 10βˆ’3 π‘š Γ— 16.7π‘š/𝑠 Γ— 0.997π‘˜π‘”/𝑠 = βˆ’3.34𝑁 βˆ’ π‘š
  • 11. KEITH VAUGH (b) When the sprinkler rotates at a constant speed, the flow field is unsteady but cyclical. In such cases the flow field can be regarded as being steady in the mean. π‘ˆπœƒ = π‘Š2 βˆ’ π‘ˆ2 where π‘Š = 16π‘š/𝑠 and π‘ˆ2 = π‘Ÿ2πœ” Applying the axial component of the moment of momentum equation π‘ˆπœƒ = 16.7π‘š/𝑠 βˆ’ π‘Ÿ2πœ” = 16.7π‘š/𝑠 βˆ’ 200 Γ— 10βˆ’3 π‘š Γ— 500π‘Ÿπ‘’π‘£/π‘šπ‘–π‘› Γ— 2πœ‹ 60 π‘ˆπœƒ = 6.2π‘š/𝑠 The torque can now be determined for this case; π‘‡π‘ β„Žπ‘Žπ‘“π‘‘ = 200 Γ— 10βˆ’3 Γ— 6.2π‘š/𝑠 Γ— 0.999π‘˜π‘”/𝑠 = βˆ’1.24𝑁 βˆ’ π‘š It can be observed from this result that the resisting torque associated with the head rotating is less that the torque required to hold it stationary.
  • 12. KEITH VAUGH (c) When no resisting torque is applied, a maximum constant speed of rotation will occur. π‘‡π‘ β„Žπ‘Žπ‘“π‘‘ = βˆ’π‘Ÿ2 π‘Š2 βˆ’ π‘Ÿ2πœ” π‘š Β· therefore 0 = βˆ’π‘Ÿ2 π‘Š2 βˆ’ π‘Ÿ2πœ” π‘š Β· When the mass flow rate is constant πœ” = π‘Š2 π‘Ÿ2 = 16.7π‘š/𝑠 200 Γ— 10βˆ’3π‘š = 83.5π‘Ÿπ‘Žπ‘‘/𝑠 πœ” = 83.5π‘Ÿπ‘Žπ‘‘/𝑠 Γ— 60𝑠/π‘šπ‘–π‘› 2πœ‹π‘Ÿπ‘Žπ‘‘/π‘Ÿπ‘’π‘£ = 797π‘Ÿπ‘π‘š The torque can now be determined for this case; π‘‡π‘ β„Žπ‘Žπ‘“π‘‘ = 200 Γ— 10βˆ’3 Γ— 6.2π‘š/𝑠 Γ— 0.999π‘˜π‘”/𝑠 = βˆ’1.24𝑁 βˆ’ π‘š It can be observed from this result that the resisting torque associated with the head rotating is less that the torque required to hold it stationary.
  • 13. KEITH VAUGH (d) Investigate varying Ο‰ between 0 and 1000 rpm and plot the results against the torque. -3.94 -3.38 -2.81 -2.25 -1.69 -1.13 -0.56 0.00 0.56 0 100 200 300 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500 1600 Torque N-m Rotational Speed (rpm) Torque Vs. Rotation Speed
  • 14. KV Ο‰ Runner blade u0 u1β†’ Absolute velocity vector. Velocity of the flow at the inlet. It’s direction is governed by the guide vane angle Ξ± v1β†’ Velocity of blade at inlet ur1β†’ The relative velocity vector of the jet at the inlet uf1β†’ The velocity of flow in the radial direction uw1β†’The component of the velocity of the jet u1 in horizontal/tangential direction Ξ±1β†’ Guide vane angle, or the angle between the velocity vector of the blade and the direction of the jet Ξ²1β†’ Blade angle at outlet. Angle between the relative velocity vector and the direction of motion of the blade at the inlet
  • 15. KV Ο‰ Runner blade u0 u2β†’ Absolute velocity vector water leaving the runner. It’s direction is governed by the guide vane angle Ξ± and if it is in the radial direction, will be equal to uf2 v2β†’ Velocity of blade at exit ur2β†’ The relative velocity vector of the water leaving the impeller uf2β†’ The velocity of flow in the radial direction uw2β†’The component of the velocity for exit u2 in horizontal/tangential direction Ξ±2β†’ Guide vane angle, or the angle between the velocity vector of the blade and the direction of the u2 exiting Ξ²2β†’ Blade angle at outlet, Angle between the relative velocity vector and the direction of motion of the blade at the outlet 𝜢2 uf2 Ξ²2 u0
  • 16. KV uf1 uf1 u1 uw1 ur1 uf1 u1 uf1 ur1 Velocity triangles Use trigonometry to solve for each vectors uw1 Two triangles within uw1 uw1 uw1 - v1 v1 - uw1
  • 17. KV Ο‘ Opposite Adjacent π‘ π‘–π‘›πœƒ = 𝑂𝑃𝑃 π»π‘Œπ‘ƒ π‘π‘œπ‘ πœƒ = 𝐴𝐷𝐽 π»π‘Œπ‘ƒ π‘‘π‘Žπ‘›πœƒ = 𝑂𝑃𝑃 𝐴𝐷𝐽 Also remember that π‘‘π‘Žπ‘›πœƒ = π‘ π‘–π‘›πœƒ π‘π‘œπ‘ πœƒ π‘π‘œπ‘‘πœƒ = π‘π‘œπ‘ πœƒ π‘ π‘–π‘›πœƒ π‘ π‘’π‘πœƒ = 1 π‘π‘œπ‘ πœƒ π‘π‘ π‘πœƒ = 1 π‘ π‘–π‘›πœƒ Recall from Trigonometry in Mathematics
  • 18. KV π‘ƒπ‘œπ‘€π‘’π‘Ÿ = 𝜌 Γ— 𝑔 Γ— 𝑉 Β· Γ— 𝐻 π‘˜π‘” π‘š3 Γ— π‘š 𝑠2 Γ— π‘š3 𝑠 Γ— π‘š π‘˜π‘” π‘š3 Γ— π‘š5 𝑠3 π‘˜π‘” Γ— π‘š2 𝑠3 π‘˜π‘” Γ— π‘š 𝑠2 Γ— π‘š 𝑠 𝑁 Γ— π‘š 𝑠 = 𝐽 𝑠 = π‘Šπ‘Žπ‘‘π‘‘π‘ 

Editor's Notes

  1. To appreciate energy conversion such as hydro, wave, tidal and wind power a detailed knowledge of fluid mechanics is essential. During the course of this lecture, a brief summary of the basic physical properties of fluids is provided and the conservation laws of mass and energy for an ideal (or inviscid) fluid are derived. The application of the conservation laws to situations of practical interest are also explored to illustrate how useful information about the flow can be derived. Finally, the effect of viscosity on the motion of a fluid around an immersed body (such as a turbine blade) and how the flow determines the forces acting on the body of interest.
  2. Consider an inward flow Francis turbine. The total head available is H and the fluid velocity entering is u0. The velocity leaving the guide vanes is u1 and is related to u0 by the continuity equation; u0A0 = uf1A1 However, uf1= u1sinΟ‘, therefore u0A0 = u1 A1 sinΟ‘
  3. Consider an inward flow Francis turbine. The total head available is H and the fluid velocity entering is u0. The velocity leaving the guide vanes is u1 and is related to u0 by the continuity equation; u0A0 = uf1A1 However, uf1= u1sinΟ‘, therefore u0A0 = u1 A1 sinΟ‘
  4. Consider an inward flow Francis turbine. The total head available is H and the fluid velocity entering is u0. The velocity leaving the guide vanes is u1 and is related to u0 by the continuity equation; u0A0 = uf1A1 However, uf1= u1sinΟ‘, therefore u0A0 = u1 A1 sinΟ‘
  5. Consider an inward flow Francis turbine. The total head available is H and the fluid velocity entering is u0. The velocity leaving the guide vanes is u1 and is related to u0 by the continuity equation; u0A0 = uf1A1 However, uf1= u1sinΟ‘, therefore u0A0 = u1 A1 sinΟ‘
  6. Consider an inward flow Francis turbine. The total head available is H and the fluid velocity entering is u0. The velocity leaving the guide vanes is u1 and is related to u0 by the continuity equation; u0A0 = uf1A1 However, uf1= u1sinΟ‘, therefore u0A0 = u1 A1 sinΟ‘
  7. Consider an inward flow Francis turbine. The total head available is H and the fluid velocity entering is u0. The velocity leaving the guide vanes is u1 and is related to u0 by the continuity equation; u0A0 = uf1A1 However, uf1= u1sinΟ‘, therefore u0A0 = u1 A1 sinΟ‘
  8. Consider an inward flow Francis turbine. The total head available is H and the fluid velocity entering is u0. The velocity leaving the guide vanes is u1 and is related to u0 by the continuity equation; u0A0 = uf1A1 However, uf1= u1sinΟ‘, therefore u0A0 = u1 A1 sinΟ‘
  9. Consider an inward flow Francis turbine. The total head available is H and the fluid velocity entering is u0. The velocity leaving the guide vanes is u1 and is related to u0 by the continuity equation; u0A0 = uf1A1 However, uf1= u1sinΟ‘, therefore u0A0 = u1 A1 sinΟ‘
  10. Consider an inward flow Francis turbine. The total head available is H and the fluid velocity entering is u0. The velocity leaving the guide vanes is u1 and is related to u0 by the continuity equation; u0A0 = uf1A1 However, uf1= u1sinΟ‘, therefore u0A0 = u1 A1 sinΟ‘