TBA

1. KV
MOMENT OF MOMENTUM
{Rotodynamic Machines}

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.
∑ 𝑟 × 𝑈 𝑜𝑢𝑡𝑚
·
𝑜𝑢𝑡 − ∑ 𝑟 × 𝑈 𝑖𝑛𝑚
·
𝑖𝑛 = ∑ 𝑟 × 𝐹 (1)
Eq.1 is known as the moment of moment equation
MOMENTUM OF
MOMENTUM

4. KV
• The fluid enters the control volume of the sprinkler (figure 1)
axially through the stem at section 1.
• The fluid leaves the control volume through each of the two
nozzles at section 2.
• The component of 𝑟 × 𝑈 along the axis of rotation is zero.
• The magnitude of the the axial component of 𝑟 × 𝑈 is 𝑟2 ×
𝑈𝜃 where r2 is the radius from the axis of rotation to the
nozzle centreline and 𝑈𝜃2 is the tangential component of
the velocity of flow exiting the nozzle.
• The fluid velocity measured relative to a fixed control
surface is an absolute velocity, U
• The velocity of the fluid exiting the nozzle is called the
relative velocity, Uθ.
• 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;
∑ 𝑟 × 𝑈 𝑜𝑢𝑡𝑚
·
𝑜𝑢𝑡 − ∑ 𝑟 × 𝑈 𝑖𝑛𝑚
·
𝑖𝑛 𝑎𝑥𝑖𝑎𝑙
= −𝑟2 × 𝑈𝜃2 𝑚
·
(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
∑ 𝑟 × 𝑈 𝑚
·
𝑎𝑥𝑖𝑎𝑙
= ∑ 𝑟 × 𝐹 𝑐𝑣 𝑎𝑥𝑖𝑎𝑙 = 𝑇𝑠ℎ𝑎𝑓𝑡 (3)
𝑈2 = 𝑟2𝜔
𝑈𝜃2
𝑈𝜃2
𝑈𝜃
𝑤2
𝑤2
𝜔
𝑇𝑠ℎ𝑎𝑓𝑡
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
−𝑟2 × 𝑈𝜃2𝑚
·
= 𝑇𝑠ℎ𝑎𝑓𝑡
(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 Ẇshaft associated with the shaft torque Tshaft can
be determined
𝑊
·
𝑠ℎ𝑎𝑓𝑡 = 𝑇𝑠ℎ𝑎𝑓𝑡 × 𝜔 = −𝑟2𝑈𝜃2𝑚
·
𝜔
(5)
Given that 𝑟2𝑈𝜃2 is the speed, U of each nozzle, eq 5 can be
written as
𝑊
·
𝑠ℎ𝑎𝑓𝑡 = −𝑈2𝑈𝜃2𝑚
·
or written per unit mass 𝑊
·
𝑠ℎ𝑎𝑓𝑡 = −𝑈2𝑈𝜃2
𝑈2 = 𝑟2𝜔
𝑈𝜃2
𝑈𝜃2
𝑈𝜃
𝑤2
𝑤2
𝜔
𝑇𝑠ℎ𝑎𝑓𝑡
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
𝑇𝑠ℎ𝑎𝑓𝑡 = −𝑟2𝑈𝜃2𝑚
·
where 𝑈𝜃2 = 𝑈2
SOLUTION
The mass flow rate can be determined
𝑚
·
= 𝑉
·
𝜌 =
1000𝑚𝑙/𝑠 × 0.001𝑚3
/𝑙 × 999𝑘𝑔/𝑚3
1000𝑚𝑙/𝑙
= 0.999𝑘𝑔

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;
𝜕
𝜕𝑡
0
∫ 𝜌𝑑𝑉 + ∑𝜌𝑜𝑢𝑡𝐴𝑜𝑢𝑡𝑈𝑜𝑢𝑡 − ∑𝜌𝑖𝑛𝐴𝑖𝑛𝑈𝑖𝑛 = 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
𝜌2𝐴2𝑈2 + 𝜌3𝐴3𝑈3 − 𝜌1𝐴1𝑈1 = 0
Given that outlet 2 and 3 are equal and given that the density of the
fluid does not change;
2 𝐴2𝑈2 − 𝐴1𝑈1 = 0

10. KEITH VAUGH
The volumetric flow rate is 𝑉
·
= 𝐴𝑈
Then the velocity can be determined for the outlet: 𝑈2 =
𝑉
·
2𝐴2
𝑈2 =
1000𝑚𝑙/𝑠 × 0.001𝑚3
/𝑙 × 106
𝑚𝑚2
/𝑚2
1000𝑚𝑙/𝑙 × 2 × 30𝑚𝑚2
= 16.7𝑚/𝑠
(a)
The resistance torque can be determined for the insistence when the
sprinkler is not rotating;
𝑇𝑠ℎ𝑎𝑓𝑡 = 200 × 10−3
𝑚 × 16.7𝑚/𝑠 × 0.997𝑘𝑔/𝑠 = −3.34𝑁 − 𝑚

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.
𝑈𝜃 = 𝑊2 − 𝑈2 where 𝑊 = 16𝑚/𝑠 and 𝑈2 = 𝑟2𝜔
Applying the axial component of the moment of momentum equation
𝑈𝜃 = 16.7𝑚/𝑠 − 𝑟2𝜔 = 16.7𝑚/𝑠 − 200 × 10−3
𝑚 × 500𝑟𝑒𝑣/𝑚𝑖𝑛 ×
2𝜋
60
𝑈𝜃 = 6.2𝑚/𝑠
The torque can now be determined for this case;
𝑇𝑠ℎ𝑎𝑓𝑡 = 200 × 10−3
× 6.2𝑚/𝑠 × 0.999𝑘𝑔/𝑠 = −1.24𝑁 − 𝑚
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.
𝑇𝑠ℎ𝑎𝑓𝑡 = −𝑟2 𝑊2 − 𝑟2𝜔 𝑚
·
therefore 0 = −𝑟2 𝑊2 − 𝑟2𝜔 𝑚
·
When the mass flow rate is constant
𝜔 =
𝑊2
𝑟2
=
16.7𝑚/𝑠
200 × 10−3𝑚
= 83.5𝑟𝑎𝑑/𝑠
𝜔 =
83.5𝑟𝑎𝑑/𝑠 × 60𝑠/𝑚𝑖𝑛
2𝜋𝑟𝑎𝑑/𝑟𝑒𝑣
= 797𝑟𝑝𝑚
The torque can now be determined for this case;
𝑇𝑠ℎ𝑎𝑓𝑡 = 200 × 10−3
× 6.2𝑚/𝑠 × 0.999𝑘𝑔/𝑠 = −1.24𝑁 − 𝑚
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 ω 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
ω
Runner
blade
u0
u1→ Absolute velocity vector. Velocity of the
flow at the inlet. It’s direction is governed
by the guide vane angle α
v1→ Velocity of blade at inlet
ur1→ The relative velocity vector of the jet at
the inlet
uf1→ The velocity of flow in the radial
direction
uw1→The component of the velocity of the jet
u1 in horizontal/tangential direction
α1→ Guide vane angle, or the angle between
the velocity vector of the blade and the
direction of the jet
β1→ Blade angle at outlet. Angle between the
relative velocity vector and the direction
of motion of the blade at the inlet

15. KV
ω
Runner
blade
u0
u2→ Absolute velocity vector water leaving
the runner. It’s direction is governed
by the guide vane angle α and if it is in
the radial direction, will be equal to uf2
v2→ Velocity of blade at exit
ur2→ The relative velocity vector of the water
leaving the impeller
uf2→ The velocity of flow in the radial
direction
uw2→The component of the velocity for exit
u2 in horizontal/tangential direction
α2→ Guide vane angle, or the angle between
the velocity vector of the blade and the
direction of the u2 exiting
β2→ Blade angle at outlet, Angle between the
relative velocity vector and the direction
of motion of the blade at the outlet
𝜶2
uf2
β2
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
ϑ
Opposite
Adjacent
𝑠𝑖𝑛𝜃 =
𝑂𝑃𝑃
𝐻𝑌𝑃
𝑐𝑜𝑠𝜃 =
𝐴𝐷𝐽
𝐻𝑌𝑃
𝑡𝑎𝑛𝜃 =
𝑂𝑃𝑃
𝐴𝐷𝐽
Also remember that
𝑡𝑎𝑛𝜃 =
𝑠𝑖𝑛𝜃
𝑐𝑜𝑠𝜃
𝑐𝑜𝑡𝜃 =
𝑐𝑜𝑠𝜃
𝑠𝑖𝑛𝜃
𝑠𝑒𝑐𝜃 =
1
𝑐𝑜𝑠𝜃
𝑐𝑠𝑐𝜃 =
1
𝑠𝑖𝑛𝜃
Recall from Trigonometry in Mathematics

18. KV
𝑃𝑜𝑤𝑒𝑟 = 𝜌 × 𝑔 × 𝑉
·
× 𝐻
𝑘𝑔
𝑚3
×
𝑚
𝑠2
×
𝑚3
𝑠
× 𝑚
𝑘𝑔
𝑚3
×
𝑚5
𝑠3
𝑘𝑔 ×
𝑚2
𝑠3
𝑘𝑔 × 𝑚
𝑠2
×
𝑚
𝑠
𝑁 ×
𝑚
𝑠
=
𝐽
𝑠
= 𝑊𝑎𝑡𝑡𝑠