Lecture 14 – 16 on
Momentum Equation and its
Application
SEMESTER 1, 2024
Dr J Lem
Dept of Mining
PNGUOT
MOMENTUM EQUATION
• We have seen that moving fluids exert forces. For example, a lift force
on an aircraft is exerted by air moving over the wing or a jet of water
from a hose exerts a force on whatever it hits.
• In solid mechanics, the momentum of a particle or object is defined as
the product of its mass m and its velocity v, i.e. momentum = mv.
• The particle of a fluid stream will posses momentum and whenever the
velocity of the stream is changed in magnitude or direction, there will be
a corresponding change in the momentum of the fluid particle.
One Dimensional Flow
• To determine the rate of change of momentum for a fluid, we will
consider a stream tube and assume the flow is steady (& non-uniform):
• The continuity of mass flow across the control volume may be
expressed as;
Mass of fluid entering at AB = mass of fluid leaving at CD
i.e.,
𝝆𝟏 𝒗𝟏 𝑨𝟏 =𝝆𝟐 𝒗𝟐 𝑨𝟐 =𝒎ሶ
• That is no accumulation of fluid and 𝑚ሶ is the fluid mass of flow per
unit time
Momentum Equation
• The rate at which momentum exits the control volume across CD may be
expressed as;
ሶ 𝟐
𝝆𝟐 𝒗𝟐 𝑨𝟐 𝒙 𝒗𝟐 = 𝒎𝒗
Since 𝑚ሶ = 𝜌2 𝑣2 𝐴2
• Similarly the rate at which momentum enters the control volume across AB may be
expressed as;
ሶ 𝟏
𝝆𝟏 𝒗𝟏 𝑨𝟏 𝒙 𝒗𝟏 = 𝒎𝒗
ሶ 𝟐 − 𝒗𝟏 ) =
• The rate of change of momentum across the control volume = 𝒎(𝒗
𝒎𝒂𝒔𝒔 𝒇𝒍𝒐𝒘 𝒓𝒂𝒕𝒆 𝒙 𝒄𝒉𝒂𝒏𝒈𝒆 𝒐𝒇 𝒗𝒆𝒍𝒐𝒄𝒊𝒕𝒚
• According to Newton’s 2nd Law, this change of momentum per unit time will be
caused by a force, F
• That is,
F=
F = Q𝝆(𝒗𝟐 − 𝒗𝟏 )
𝑚ሶ = 𝑄𝜌
F = Q𝝆(𝒗𝟐 − 𝒗𝟏
This is the Resultant Force acting in the
direction of the flow of fluid. This analysis
assumes one-dimensional flow (i.e v1 and
v2 in the same direction)
Momentum Equation Two- Dimensional Flow along a Streamline
• Consider 2-dimensional system shown:
• Since both momentum and force are vector
quantities, they can be resolving into components
in the x and y directions
• Force in the x-direction;
𝐹𝑋
= 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑐ℎ𝑎𝑛𝑔𝑒 𝑜𝑓 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑖𝑛 𝑥 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛
= mass flow rate x change of velocity in x direction
= 𝑚ሶ 𝑉2 𝐶𝑜𝑠∅ − 𝑉1 𝐶𝑜𝑠𝜃
= 𝒎ሶ 𝑽𝒙𝟐 − 𝑽𝒙𝟏
• Similarly in the y-direction
𝐹𝑦
= 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑐ℎ𝑎𝑛𝑔𝑒 𝑜𝑓 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑖𝑛 𝑦 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛
= mass flow rate y change of velocity in y- direction
= 𝑚ሶ 𝑉2 𝐶𝑜𝑠∅ − 𝑉1 𝐶𝑜𝑠𝜃
= 𝒎ሶ 𝑽𝒚𝟐 − 𝑽𝒚𝟏
Momentum Equation Two- Dimensional Flow along a Streamline
• These components force can be combined to give the
resultant force, F
𝑭=
𝑭𝑿 𝟐 + 𝑭𝒀 𝟐
• The direction of the resultant force can then be determined by;
𝜶 = 𝐭𝐚𝐧−𝟏
𝑭𝒀
𝑭𝑿
• For a three-dimensional (x, y, z) system we then have an extra
force to calculate and resolve in the z direction.
• This is considered in exactly the same way.
٨
Summary
The total force exerted on the fluid in
=
a control volume in a given direction
Rate of change of momentum in
the given direction of fluid passing
through the control volume
𝑭𝑻 = 𝒎ሶ (𝒗𝒐𝒖𝒕 − 𝒗𝒊𝒏 )
𝑭𝑻 = ρ𝑸 (𝒗𝒐𝒖𝒕 − 𝒗𝒊𝒏 )
• The value of F is positive in the direction in which v is assumed to be
positive
Summary
• This resultant force is made up of three components;
1.
F1 =FR = Force exerted in the given direction on the fluid by
any solid body touching the control volume
2.
F2 =FB = Force exerted in the given direction on the fluid by
body force (e.g. gravity)
3.
F3 =FP = Force exerted in the given direction on the fluid by
fluid pressure outside the control volume
• So we say that the total force, FT, is given by the sum of these forces:
𝑭𝑻 = 𝑭𝑹 + 𝑭𝑩 + 𝑭𝑷 = 𝒎ሶ (𝒗𝒐𝒖𝒕 − 𝒗𝒊𝒏 )
• The force exerted by the fluid
on the solid body touching
the control volume is equal
and opposite to FR .
• So the reaction force, R, is
given by:
R = -FR
• Because fluid changes direction, a Force (so large in the
case of water supply pipes) will act in the bend. If the
bend is not fixed, it will move and eventually break at the
joints
• One must know how much force a support (thrust block)
must withstand
STEPS IN ANALYSIS
ONE-DIMENSIONAL FLOW
• Steps 1 & 2
Step 3: Calculate the
total Force
Step 4: Calculate the
pressure forces
Step 5: Calculate the
body Force
Step 6: Calculate the
reaction force that
nozzle acts on the fluid
A. Force due to deflection of a Jet by a curved vane
❖ Both velocity and momentum are vector quantities
❖ Even if the magnitude of the velocity remains unchanged, a changed in direction
of a stream of fluid will give rise to a change of momentum.
❖ If the stream is deflected by a curved vane (entering and leaving tangentially
without impact). A force will be exerted between the fluid and the surface of the
vane to cause this change in momentum.
❖ It is usually convenient to calculate the components of this force parallel and
perpendicular to the direction of the incoming stream
❖ The resultant can be combined to give the magnitude of the resultant force which the
vane exerts on the fluid, and equal and opposite reaction of the fluid on the vane
Worked Example
A jet of water from a nozzle is deflected through an angle, θ=60o
from its original direction by a curved vane which enters
tangentially without shock with mean velocity of 30 m/s and
leaves with mean velocity of 25 m/s. If the discharge from the
nozzle is 0.8 kg/s.
Calculate the magnitude and direction of the resultant
force on the vane if the vane is stationary
Solution
F = ρQ (V2 – V1) = F1 + F2 + F3
F1 =-R (reaction force)
F2 = 0 (body force)
F3 = Pressure force = constant pressure for
free jet = 0
So,
ρQ (V2 – V1) = 𝑚ሶ (V2 – V1) = – R
R = 𝑚ሶ (V1 – V2)
Given,
V1 = 30m/s, V2 = 25m/s, 𝑚ሶ = 0.8kg/s
Θ = 60° ρ = 1000kg/m3,
𝑚ሶ = ρQ so Q = 0.8*1000 = 8 x 10-4m3/s
To calculate the resultant force, R, let us determine the component forces in the x- and y-direction first:
In x-direction:
Rx = 𝑚ሶ (V1 – V2X) but V2X = V2Cos60 = 12.5m/s
So,
Rx = 0.8(30-12.5) = 14 N
And in the y-direction:
Ry = 𝑚V
ሶ 2y but V2y = V2Sin60 = 21.7m/s
So,
Ry = 0.8(21.7) = 17.4N
Then the resultant force R is;
𝑅 = 𝑅𝑋 2 + 𝑅𝑌 2
𝑅 = (14)2 + (17.4)2
𝑹 = 22.3N
𝑅𝑦
Direction, θ = tan -1 (𝑅𝑋)
17.4
θ = tan -1 ( 14 )
B. Force exerted on a bend and closed conduit
A force will act on the bend due to:
• The fluid changes its direction
• If the pipe tapers, there is a change in velocity
magnitude.
• Pressure forces.
Why do we want to know the forces here?
• This force can be very large in the case of water
supply pipes.
• If the bend is not fixed it will move and eventually
break at the joints.
• We need to know how much force a support (thrust
block) must withstand.
Worked Problem
1
A pipe bend tapers from a diameter of d1 of 500mm
at inlet to a diameter d2 of 250mm at outlet and turns
the flow trough an angle θ of 45o.
Measurements of pressure at inlet and outlet show that
P1 = 40 kPa and P2 = 23 kPa. If the pipe is conveying oil
of density, ρ = 850 kg/m3
Calculate the magnitude and direction of the
resultant force on the bend when the oil is flowing
at the rate of 0.45m3/s.
Note: The bend is in horizontal plane
2
Solution
F = ρQ (V2 – V1) = F1 + F2 + F3
F1 =-R (reaction force)
F2 = 0 (body force)
So,
ρQ (V2 – V1) = F3 - R
R = F3 + ρQ(V1 – V2) =resultant force
Given,
d1 = 500mm, d2 = 250mm
Θ = 45° ρ = 850kg/m3,
F3 = Pressure forces
𝑚ሶ = ρQ = 850 x 0.45 m3/s = 382.5kg/s
1
1
1
1
𝐴1 = 4 𝜋𝑑1 2 = 4 𝜋(0.5)2 = 0.196m2
𝐴2 = 4 𝜋𝑑2 2 = 4 𝜋(0.25)2 = 0.0491m2
Using continuity equation, let us determine inlet velocity, V1 and outlet velocity, V2
𝑄
0.45
Q = A1V1 = A2V2 =0.45 ; V1 =𝐴 = 0.196 = 2.3 m/s
𝑄
0.45
V2 =𝐴 = 0.0491 = 9.2 m/s
2
1
To calculate the resultant force, R, let us first calculate the component forces in the x- and y-directions
In the x-direction:
Rx = F3X + ρQ (V1 – V2X) , but F3X = P1A1 – P2A2 Cosθ (by taking right direction of flow as positive)
Rx = P1A1 – P2A2 Cosθ+ ρQ(V1 – V2X) but V2X = V2Cos45° = 9.2 x Cos45° =4.8m/s
and P1A1 = 40000 x 0.196 = 7840 N
P2A2 Cosθ = 23000 x 0.0491 x Cos 45° = 593 N
ρQ(V1 – V2X) = 382.5(2.3 – 4.8) = -1836N
So,
Rx = 7840 - 593 – 1836 = 5152 N
Similarly in the y-direction:
Ry = F3y + ρQ V2y) , but F3Xy= P2A2 Sinθ and V2y = V2 Sin45° = 9.2 x Sin45° =7.8m/s
Ry = P2A2 Sinθ+ ρQV2y but
P2A2 Sin45 = 23000 x 0.0491 x Sin 45° = 961 N
ρQ V2X = 382.5 x 7.8 = 2983.5N
So,
Rx = 961 + 2983.5 = 3944.5 N
Then the resultant force R is;
𝑅 = 𝑅𝑋 2 + 𝑅𝑌 2
𝑅 = (5152)2 + (3944.5)2
𝑅 = 6489 N
Direction, θ = tan -1 (
θ = tan -1 (
3944.5
)
5152
𝑅𝑦
)
𝑅𝑋
C. Reaction on a Jet
❖ Whenever the momentum of a stream of fluid increased in a given direction in passing from one section to another,
there must be a net force acting on the fluid in that direction.
❖ By Newton’s third law, there will be an equal and opposite force exerted by the fluid on the system.
Typical example:
❖ Force on the nozzle at the outlet of a pipe. Anything holding the nozzle (e.g. a fireman) must be strong enough to
withstand these forces.
❖ Or the reactive force exerted when the fluid is discharge in the form of high-velocity jet. (aircraft, rocket motors…)
Worked Example
A jet of water of diameter d = 50 mm issues with velocity of 4.9 m/s
from a hole in the vertical side of an open tank which kept filled with
water to a height of 1.5 m above the center of the hole. Calculate the
reaction of the jet on the tank and its contents when:
1. It is stationary.
2. Its moving with a velocity u = 1.2m/s in the
opposite direction to the jet while the velocity
of the jet relative to the tank remains
unchanged.
3. In the latter case, what would be the work
done per second.