linear momentum - fkm.utm.mysyahruls/resources/skmm2313/7-linear-momentum.pdf · chapter 7 –...

Chapter 7 – Continuity Equation and Linear Momentum

LINEAR MOMENTUM Derivation of the Linear Momentum Equation Newton’s second law of motion for a system is Time rate of change of

the linear momentum of the system

Sum of the external forces acting on the

system =

Momentum is mass times velocity. Then, the Newton’s second law becomes ;

∑∫ = syssys

FVdVDtD ρ

Furthermore, for a system and the contents of a coincident control volume that is fixed and non-deforming control volume, the Reynolds transport theorem (with b=velocity) allows us to conclude that ; Net rate of flow

of linear momentum through the

control surface

= + Time rate of change of the

linear momentum of the contents of

the control volume

Time rate of change of the

linear momentum of

the system

1


It can be written as ;

∫ ∫∫ ⋅+∂∂

=cv cssys

dAnVVVdVt

VdVDtD ˆρρρ

For a control volume that is fixed and non-deforming, the appropriate mathematical statement of Newton’s second law of motion is ;

∑∫ ∫ =⋅+∂∂

volumecontrol theof contentsˆ FdAnVVVdV

t cv cs

ρρ

We call above equation is the linear momentum equation.

2


Forces Due to Fluids in Motion Newton’s second law of motion is

maF = In fluid flow problems, we use mass flow rate (kg/s) to determine “mass” that involve in the motion.

vt

mtV

mmaF ∆⋅∆

=∆∆⋅==

Mass flowrate can be written as ;

Qm ρ=& For fluid, Newton’s second law of motion is ;

vQvmvt

mmaF ∆=∆=∆⋅

∆== ρ&

3


Linear momentum idea is usually used for water jet and vane problems. Because velocities has magnitude and direction, forces act on vane can be determine as ;

22yxR

yy

xx

FFF

vQFvQF

+=

∆=∆=

ρρ

4


Force on x-direction

112 )( QvvvQvQRF xxxxx ρρρ =−=∆== Force on y-direction

212 )( QvvvQvQRF yyyyy ρρρ =−=∆==

5


TUTORIAL FOR LINEAR MOMENTUM QUESTION 1

Figure 1

Water flows through the 20º reducing bend shown in Figure 1 at rate of 0.025m3/s. The flow is frictionless, gravitational effects are negligible, and the pressure at section (1) is 150kPa. Determine the x and y components of force required to hold the bend in place.

1


QUESTION 2

Figure 2

Determine the magnitude and direction of the anchoring force needed to hold the horizontal elbow and nozzle combination shown in Figure 2 in place. Atmospheric pressure is 100kPa. The gage pressure at section (1) is 100kPa. At section (2), the water exits to the atmosphere.

2


QUESTION 3

Figure 3

Water flows as two free jets from the tee attached to the pipe shown in Figure 3. The exit speed is 15m/s. If viscous effects and gravity are negligible, determine the x and y components of the force that the pipe exerts on the tee.

3


QUESTION 4

Figure 5

A free jet o fluid strikes a wedge as shown in Figure 5. Of the total flow, a portion is deflected 30º, the reminder is not deflected. The horizontal and vertical components of force needed to hold the wedge stationary are FH and FV, respectively. Gravity is negligible, and the fluid speed remains constant. Determine the force ratio, FH/FV.

4


QUESTION 5

Figure 5

A converging elbow as shown in Figure 5 turns water through an angle of 135º in a vertical plane. The flow cross section diameter is 400mm at the elbow inlet, section (1), and 200mm at the elbow outlet, section (2). The elbow flow passage volume is 0.2m3 between sections (1) and (2). The water volume flowrate is 0.4m3/s and the elbow inlet and outlet pressures are 150kPa and 90kPa. The elbow mass is 12kg. Calculate the horizontal (x-direction) and vertical (z-direction) anchoring forces required to hold the elbow in place.

5


QUESTION 6

Figure 6

Water flows from a large tank into a dish as shown in Figure 6. a) If at the instant shown the tank and the water in it

weigh, W1 in kg, what is the tension, T1, in the cable supporting the tank?

b) If at the instant shown the dish and the water in it weigh W2 in kg, what is the force, F2, needed to support the dish?

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linear momentum - fkm.utm.mysyahruls/resources/skmm2313/7-linear-momentum.pdf · chapter 7 –...

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