CONTINUITY EQUATION - Universiti Teknologi Malaysiasyahruls/resources/SKMM2313/7-Linear.pdf ·...

Chapter 7 – Continuity Equation and Linear Momentum

CONTINUITY EQUATION

Derivation of the Continuity Equation A system is defined as a collection of unchanging contents, so the conservation of mass principle for a system is simply stated as ;

Time rate of change of the system mass = 0

Or

0=Dt

DM sys

where the system mass, Msys, is more generally expressed as ;

∫ ⋅=sys

sys VdM ρ

and the integration is over the volume of the system. In words, we can say that the system mass is equal to the sum of all the density-volume element products for the contents of the system.

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The Reynolds transport theorem states that ;

∫ ∫∫ ⋅⋅+∂∂

=⋅sys cscv

dAnVVdt

VdDtD ˆρρρ

∫∂∂

cv

Vdt

ρ shows the time rate of change of the

mass of the contents of the control volume.

∫ ⋅cs

dAnV ˆρ shows the net rate of mass flow

through the control surface. When a flow is steady, all field properties including density remain constant with time, and the time rate of change of the mass of the contents of the control volume is zero. That is ;

0=∂∂∫cv

Vdt

ρ

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∫ ⋅cs

dAnV ˆρ is mass flowrate through dA

“ + ” is for flow out of the control volume “ - ” is for flow into the control volume It can be shown as ;

∑ ∑∫ −=⋅ inoutcs

mmdAnV &&ˆρ

An often-used expression for mass flowrate, , through a section of control surface having area A is ;

m&

AVQm ρρ ==&

The average value of velocity, V , defined as ;

VA

dAnVV A =

⋅=∫

ρ

ρ ˆ

If the velocity is considered uniformly distributed (one-dimensional flow) over the section A, the

VV =

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Fixed, Non-deforming Control Volume In many applications of fluid mechanics, an appropriate control volume to use is fixed and non-deforming. It will give the value of the time rate of change of the mass of the contents of the control volume is equal to zero.

0=∂∂∫cv

Vdt

ρ

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Moving, Non-deforming Control Volume It is sometimes necessary to use a non-deforming control volume attached to a moving reference frame. For example, the exhaust stacks of a ship at sea and the gasoline tank of an automobile passing by. When a moving control volume is used, the fluid velocity relative to the moving control volume (relative velocity) is an important flow field variable. The relative velocity, W, is the fluid velocity seen by an observer moving with the control volume. The control volume velocity, Vcv, is the velocity of the control volume as seen from a fixed coordinate system. The absolute velocity, V, is the fluid velocity seen by a stationary observer in a fixed coordinate system. Their relation is ;

V = W + Vcv

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We can get the control volume expression for conservation of mass (the continuity equation) for a moving, non-deforming control volume, namely ;

∫∫ =⋅+∂∂

cscv

dAnWVdt

0ˆρρ

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CONTINUITY EQUATION

Derivation of the Continuity Equation The Reynolds transport theorem states that ;

∫ ∫ ∫+∂∂

==system cv cs

system dAnVbVdbt

VdDtD

DtDB

ˆρρρ

)mass()parameter changeable( ×=⋅= mbB

to complete above mention Reynolds Transport Theorem, we found that, if B is mass, the b will be equal to one (1). B = mass b = 1 We found that the time rate of change of the system is

zero, so we can say that ;

0==Dt

DmDt

DB systemsystem

1


∫∂∂

cv

Vdt

ρ shows the time rate of change of the

mass of the contents of the control volume.

∫ ⋅cs

dAnV ˆρ shows the net rate of mass flow

through the control surface. When a flow is steady, all field properties including density remain constant with time, and the time rate of change of the mass of the contents of the control volume is zero. That is ;

0=∂∂∫cv

Vdt

ρ

2


Example:

Figure 1 Air at standard conditions enters the compressor shown in Figure 1 at rate of 0.3m3/s. It leaves the tank through a 3cm diameter pipe with a density of 1.80kg/m3 and a uniform speed of 210m/s. a) Determine the rate at which the mass of air in the

tank in increasing or decreasing. b) Determine the average time rate of change of air

density within the tank.

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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

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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.

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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&

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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

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Force on x-direction

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

212 )( QvvvQvQRF yyyyy ρρρ =−=∆==

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TUTORIAL FOR LINEAR MOMENTUM QUESTION 1

Figure 3

Water flows steadily through the control volume shown in Figure 3. The volumetric flowrate across section (3) is 2m3/s and the mass flowrate across section (2) is 3 kg/s. Determine the numerical value for the following quantities :

(a) ∫ ⋅1

ˆdAngVρ

(b) ∫ ⋅1

ˆdAnVVρ

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QUESTION 2

Figure 2

Two rivers merge to form a larger river as shown in Figure 2. At a location downstream from the junction (before the two streams completely merge), the non-uniform velocity profile is as shown. Determine the value of V.

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QUESTION 3

Figure 1

Air flows steadily between two cross sections in a long straight section of 0.1m inside diameter pipe. The static temperature and pressure at each section are indicated in Figure 1. If the average air velocity at section (1) is 205m/s, determine the average air velocity at section (2).

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Figure 4

Various types of attachments can be used with the shop vacuum such as nozzle and brush. The flowrate is 1.5m3/s. a) Determine the average velocity through the nozzle

entrance, Vn b) Assume the air enters the brush attachment in a

radial direction all around the brush with a velocity profile that varies linearly from 0 to Vb along the length of the bristles as shown in Figure 4. Determine the value of Vb.

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QUESTION 5

Figure 5

The pump shown in Figure 5 produces a steady flow of 100m3/s through the nozzle. Determine the nozzle exit diameter, D2, if the exit velocity is to be V2=30m/s.

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QUESTION 6

Figure 6

An evaporative cooling tower as shown in Figure 6 is used to cool water from 110ºC to 80ºC. When water enters the tower at a rate of 35kg/s. Dry air (no water vapor) flows into the tower at a rate of 18kg/s. If the rate of wet air flow out of the tower is 20kg/s. Determine the rate of water evaporation and the rate of cooled water flow.

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The wind blows through a 2.5m x 3m garage door opening with a speed of 5m/s as shown in the figure. Determine the average speed, V, of the air through the two 90cm x 120cm openings in the windows.

QUESTION 2 Water flows along the centerline of a 50mm diameter pipe with an average velocity of 10m/s and out radially between two large circular disks are parallel and spaced 10mm apart. Determine the average velocity of the water at a radius of 300mm in space between the disks.

QUESTION 3 Oil having a specific gravity of 0.9 is pumped as illustrated with a water jet pump. The water volume flowrate is 2m3/s. The water and oil mixture has a average specific gravity of 0.95. Calculate the rate, in m3/s, at which the pump moves oil.

QUESTION 4

As shown in figure, at the entrance to a 1m wide channel the velocity distribution is uniform with a velocity V. Further downstream the velocity profile is given by u=4y-2y2. Determine the value of V.

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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.

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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.

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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.

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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.

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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.

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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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CONTINUITY EQUATION - Universiti Teknologi Malaysiasyahruls/resources/SKMM2313/7-Linear.pdf ·...

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