Unformatted text preview: III. Control Volume Relations for Fluid Analysis
From consideration of hydrostatics, we now move to problems involving fluid
flow with the addition of effects due to fluid motion, e.g. inertia and convective
mass, momentum, and energy terms.
We will present the analysis based on a control volume (not differential element)
formulation, e.g. similar to that used in thermodynamics for the first law.
Basic Conservation Laws:
Each of the following basic conservation laws is presented in its most
fundamental, fixed mass form. We will subsequently develop an equivalent
expression for each law that includes the effects of the flow of mass, momentum,
and energy (as appropriate) across a control volume boundary. These transformed
equations will be the basis for the control volume analyses developed in this
chapter.
Conservation of Mass:
Defining m as the mass of a fixed mass system, the mass for a control volume
V is given by m sys = ∫ ρ dV sys The basic equation for conservation of mass is then expressed as
The time rate of change of mass for
the control volume is zero since at
this point we are still working with
a fixed mass system. dm =0
dt sys
Linear Momentum: Defining P sys as the linear momentum of a fixed mass, the linear momentum
of a fixed mass control volume is given by:  v v III1  eText Main Menu  Textbook Table of Contents  Study Guide P sys = m V = ∫ V ρ dV sys where V is the local fluid velocity and d V is a differential volume element
in the control volume.
The basic linear momentum equation is then written as d (m V)
dP = ∑ F = dt dt sys sys Moment of Momentum:
Defining H as the moment of momentum for a fixed mass, the moment of
momentum for a fixed mass control volume is given by H sys = ∫ r × V ρ dV sys where r is the moment arm from an inertial coordinate system to the
differential control volume of interest. The basic equation is then written as dH ∑ M sys = ∑ r × F = dt sys Energy:
Defining E sys as the total energy of an element of fixed mass, the energy of a
fixed mass control volume is given by E sys = ∫ e ρ dV sys  v v III2  eText Main Menu  Textbook Table of Contents  Study Guide where e is the total energy per unit mass ( includes kinetic, potential, and
internal energy ) of the differential control volume element of interest.
The basic equation is then written as &
& dE
Q −W = (Note: written on a rate basis) d t sys It is again noted that each of the conservation relations as previously written
applies only to fixed, constant mass systems.
However, since most fluid problems of importance are for open systems, we
must transform each of these relations to an equivalent expression for a control
volume which includes the effect of mass entering and/or leaving the system.
This is accomplished with the Reynolds transport theorem.
Reynolds Transport Theorem
We define a general, extensive property ( an extensive property depends on the
size or extent of the system) B sys where B sys = ∫ β ρd V sys Bsys could be total mass, total energy, total momentum, etc., of a system.
and B sys per unit mass is defined as
Thus, β β= or β is the intensive equivalent of B sys dB
dm . Applying a general control volume formulation to the time rate of change of
B sys , we obtain the following (see text for detailed development):  v v III3  eText Main Menu  Textbook Table of Contents  Study Guide dB
=
d t sys ∂
β ρ dV
∂ t cv ∫ ↓ ∫β + e i ρi Vi d Ai Ai ↓ Rate of
change of
B in c.v. ↓ ∫β − Ae ↓ System rate
of change
of B ρe Ve d Ae ↓ Rate of B
leaving c.v. Rate of B
entering c.v. ↓ transient term convective terms where B is any conserved quantity, e.g. mass, linear momentum, moment of
momentum, or energy.
We will now apply this theorem to each of the basic conservation equations to
develop their equivalent open system, control volume forms.
Conservation of mass
For conservation of mass, we have that
B=m β =1 and From the previous statement of conservation of mass and these definitions,
Reynolds transport theorem becomes
∂
ρ dV
∂ t cv ∫ ∫ + ρe Ve d Ae − Ae ∫ ρ i Vi d Ai =0 Ai or
∂
ρ dV
∂ t cv ∫ + ∫ ρe Ve d Ae − Ae ↓ ∫ ↓ Rate of mass
leaving c.v., ↓ ↓ = 0 for steadystate =0 Ai ↓ Rate of change
of mass in c.v., ρ i Vi d Ai Rate of mass
entering c.v., ↓ &
me &
mi  v v III4  eText Main Menu  Textbook Table of Contents  Study Guide This can be simplified to
dm
&
& + ∑ me − ∑ mi = 0
d t cv Note that the exit and inlet velocities Ve and Vi are the local components of
fluid velocities at the exit and inlet boundaries relative to an observer
standing on the boundary. Therefore, if the boundary is moving, the velocity
is measured relative to the boundary motion. The location and orientation of a
coordinate system for the problem are not considered in determining these
velocities.
Also, the result of Ve ⋅ dA e and Vi ⋅ dA i is the product of the normal
velocity component times the flow area at the exit or inlet, e.g. Ve,n dA e Vi,n dA i and Special Case: For incompressible flow with a uniform velocity over the flow
area, the previous integral expressions simplify to:
&
m= ∫ ρ V d A = ρ AV cs Conservation of Mass Example
Water at a velocity of 7 m/s exits a
stationary nozzle with D = 4 cm and is
o
directed toward a turning vane with θ = 40 ,
Assume steadystate.
Determine:
a. Velocity and flow rate entering the c.v.
b. Velocity and flow rate leaving the c.v.  v v III5  eText Main Menu  Textbook Table of Contents  Study Guide &
a. Find V1 and m1 Recall that the mass flow velocity is the normal component of velocity
measured relative to the inlet or exit area.
Thus, relative to the nozzle, V(nozzle) = 7 m/s and since there is no relative
motion of point 1 relative to the nozzle, we also have V1 = 7 m/s ans.
From the previous equation:
&
m= ∫ ρ V d A = ρ AV = 998 kg/m3*7 m/s*π*0.042/4 cs &
m1 = 8.78 kg/s ans.
&
b. Find V2 and m2 Determine the flow rate first.
Since the flow is steady state and no mass accumulates on the vane:
&
&
&
m1 = m2 , m2 = 8.78 kg/s ans.
&
m2 = 8.78 kg/s = ρ A V)2 Now: Since ρ and A are constant, V2 = 7 m/s ans.
Key Point: For steady flow of a constant area, incompressible stream, the
flow velocity and total mass flow are the same at the inlet and exit, even
though the direction changes.
or alternatively:
Rubber Hose Concept: For steady flow of an incompressible fluid, the
flow stream can be considered as a rubber hose and if it enters a c.v. at a
velocity of V, it exits at a velocity V, even if it is redirected.  v v III6  eText Main Menu  Textbook Table of Contents  Study Guide Problem Extension:
Let the turning vane (and c.v.) now
move to the right at a steady velocity of
2 m/s (other values remain the same);
perform the same calculations.
Therefore:
Given: Uc = 2 m/s VJ = 7 m/s For an observer standing at the c.v. inlet (point 1)
V1 = VJ – Uc = 7 – 2 = 5 m/s
&
m1 = ρ1 V1 A1 = 998 kg/m3*5 m/s*π*0.042/4 = 6.271 kg/s Note: The inlet velocity used to specify the mass flow rate is again measured
relative to the inlet boundary but now is given by VJ – Uc .
Exit:
&
&
m1 = m2 = 6.271 kg/s Again, since ρ and A are constant, V2 = 5 m/s. Again, the exit flow is most easily specified by conservation of mass concepts.
Note: The coordinate system could either have been placed on the moving cart or
have been left off the cart with no change in the results.
Key Point: The location of the coordinate system does not affect the calculation
of mass flow rate which is calculated relative to the flow boundary. It could have
been placed at Georgia Tech with no change in the results.
Review material and work examples in the text on conservation of mass.  v v III7  eText Main Menu  Textbook Table of Contents  Study Guide Linear Momentum
For linear momentum, we have that B = P = m V and β = V
From the previous statement of linear momentum and these definitions, Reynolds
transport theorem becomes ∑F = d ( mV ) ∂
=
∫ V ρ dV
∂ t cv
dt sys or ∑F ∂
∫ V ρ dV
∂ t cv = ↓ + &
∫V dm e − ↓ = the rate of
change of
momentum
in the c.v. = body + point +
distributed, e.g.
(pressure) forces ∫V ρ e Ve ⋅ d Ae Ae Ae ↓ = the ∑ of the
external forces
acting on the c.v. + − ∫V ρ V ⋅d A
i i Ai &
∫V d m i Ai ↓ = 0 for
steadystate = the rate of
momentum
leaving the
c.v. = the rate of
momentum
entering the c.v. and where V is the vector momentum velocity relative to an inertial reference
frame.
Key Point: Thus, the momentum velocity has magnitude and direction and is
measured relative to the reference frame (coordinate system) being used for the
&
&
problem. The velocities in the mass flow terms mi and me are scalars, as noted
previously, and are measured relative to the inlet or exit boundary.
Always clearly define a coordinate system and use it to specify the value of all
inlet and exit momentum velocities when working linear momentum problems.  v v III8  eText Main Menu  Textbook Table of Contents  Study Guide i For the 'x' direction, the previous equation becomes ∑F ∂
∫ V ρ dV
∂ t cv x = x + ∫V x ,e &
d me − Ae ∫V x ,i &
d mi Ai Note that the above equation is also valid for control volumes moving at
constant velocity with the coordinate system placed on the moving control
volume. This is because an inertial coordinate system is a nonaccelerating
coordinate system which is still valid for a c.s. moving at constant velocity.
Example:
A water jet 4 cm in diameter with a velocity
of 7 m/s is directed to a stationary turning
o
vane with θ = 40 . Determine the force F
necessary to hold the vane stationary. Governing equation: ∑F = x ∂
∫ V ρ dV
∂ t cv x + ∫V x ,e &
d me ∫V − Ae x ,i &
d mi Ai Since the flow is steady and the c.v. is stationary, the time rate of change of
momentum within the c.v. is zero. Also with uniform velocity at each inlet and
exit and a constant flow rate, the momentum equation becomes
&
&
− Fb = me Ve − mi Vi Note that the braking force, Fb, is written as negative since it is assumed to be
in the negative x direction relative to positive x for the coordinate system.  v v III9  eText Main Menu  Textbook Table of Contents  Study Guide From the previous example for conservation of mass, we can again write
&
m= = 998 kg/m3*7 m/s*π*0.042/4 ∫ ρ V d A = ρ AV cs &
m1 = 8.78 kg/s and V1 = 7 m/s and for the exit:
&
m2 = 8.78 kg/s and V2 = 7 m/s inclined 40û above the horizontal. Substituting in the momentum equation, we obtain
o Fb = 8.78 kg/s * 7 m/s *cos 40  8.78 kg/s * 7 m/s
and Fb =  14.4 kg m/s 2 or Fb = 14.4 N ← ans. Note: Since our final answer is positive, our original assumption of the
applied force being to the left was correct. Had we assumed that the
applied force was to the right, our answer would be negative, meaning
that the direction of the applied force is opposite to what was assumed.
Modify Problem:
Now consider the same problem but with
the cart moving to the right with a velocity
Uc = 2 m/s. Again solve for the value of
braking force Fb necessary to maintain a
constant cart velocity of 2 m/s.
Note: The coordinate system for the
problem has now been placed on the
moving cart.  v v III10  eText Main Menu  Textbook Table of Contents  Study Guide The transient term in the momentum equation is still zero. With the coordinate
system on the cart, the momentum of the cart relative to the coordinate system
is still zero. The fluid stream is still moving relative to the coordinate system,
however, the flow is steady with constant velocity and the time rate of change
of momentum of the fluid stream is therefore also zero. Thus
The momentum equation has the same form as for the previous problem
(However the value of individual terms will be different.)
&
&
− Fb = me Ve − mi Vi
&
&
m1 = ρ1 V1 A1 = 998 kg/m3*5 m/s*π*0.042/4 = 6.271 kg/s = m2 Now we must determine the momentum velocity at the inlet and exit. With the
coordinate system on the moving control volume, the values of momentum
velocity are
V1 = VJ – Uc = 7 – 2 = 5 m/s and V2 = 5 m/s inclined 40 o The momentum equation ( x  direction ) now becomes
o Fb = 6.271 kg/s * 5 m/s *cos 40  6.271 kg/s * 5 m/s
and Fb =  7.34 kg m/s 2 or Fb = 7.34 N ← ans. Question:
What would happen to the braking force Fb if the turning
o
o
angle had been > 90 , e.g., 130 ? Can you explain based on your
understanding of change in momentum for the fluid stream?
Review and work examples for linear momentum with fixed and nonaccelerating (moving at constant velocity) control volumes.
Accelerating Control Volume
The previous formulation applies only to an inertial coordinate system, i.e.,
fixed or moving at constant velocity (nonaccelerating).  v v III11  eText Main Menu  Textbook Table of Contents  Study Guide We will now consider problems with accelerating control volumes. For these
problems we will again place the coordinate system on the accelerating control
volume, thus making it a noninertial coordinate system.
For coordinate systems placed on an accelerating control volume, we must
account for the acceleration of the c.s. by correcting the momentum equation
for this acceleration. This is accomplished by including the term as shown
below: ∑F − ∫ a cv cv d mcv ∂
∫ V ρ dV
∂ t cv = ↓ + &
∫V dm e − Ae &
∫V d m i Ai integral sum of
the local c.v. (c.s.)
acceleration * the c.v. mass
The added term accounts for the acceleration of the control volume and allows
the problem to be worked with the coordinate system placed on the accelerating
c.v.
Note: Thus, all vector (momentum) velocities are then measured relative to an
observer (coordinate system) on the accelerating control volume. For example,
the velocity of a rocket as seen by an observer (c.s.) standing on the rocket is
zero and the time rate of change of momentum is zero in this reference frame
even if the rocket is accelerating.
Accelerating Control Volume Example
o A turning vane with θ = 60 accelerates
from rest due to a jet of water
(VJ = 35 m/s, AJ = 0.003 m2 ). Assuming
the mass of the cart mc, is 75 kg and
neglecting drag and friction effects, find:
a. Cart acceleration at t = 0.
b. Uc as a f(t)  v v III12  eText Main Menu  Textbook Table of Contents  Study Guide Starting with the general equation shown above, we can make the following
assumptions:
1. ∑ Fx = 0, no friction or body forces.
2. The jet has uniform velocity and constant properties.
3. The entire cart accelerates uniformly over the entire control volume.
4. Neglect the relative momentum change of the jet stream that is within the
control volume.
With these assumptions, the governing equation simplifies to
&
&
− ac mc = me Vx ,e − mi Vx ,i We thus have terms that account for the acceleration of the control volume, for
the exit momentum, and for the inlet momentum (both of which change with
time.)
Mass flow:
As with the previous example for a moving control volume, the mass flow terms
are given by:
&
&
&
mi = me = m = ρ AJ (VJ – Uc) Note that since the cart accelerates, Uc is not a constant but rather changes with
time.
Momentum velocities:
Ux,i= VJ  Uc Ux,e = (VJ  Uc ) cos θ Substituting, we now obtain
2  ac mc = ρ AJ (VJ – Uc) cos θ  ρ AJ (VJ – Uc)
Solving for the cart acceleration, we obtain ρ AJ (1 − cos θ ) (VJ − Uc ) 2 ac = mc  v v III13  eText Main Menu  Textbook Table of Contents  Study Guide Substituting for the given values at t = 0, i.e., Uc = 0, we obtain
2 ac (t = 0) = 24.45 m/s = 2.49 g’s
Note: The acceleration at any other time can be obtained once the cart velocity
Uc at that time is known.
To determine the equation for cart velocity as a function of time, the equation
for the acceleration must be written in terms of Uc (t) and integrated.
1
V
dUc ρ AJ ( − cosθ )( J − Uc )
=
dt
mc 2 Separating variables, we obtain
Uc ( t ) ∫ 0 dUc (VJ − U c )2 = t ρ AJ (1 − cosθ ) 0 mc ∫ dt Completing the integration and rearranging the terms, we obtain a final
expression of the form Uc
V bt
=J
VJ 1+ VJ b t where b= ρ AJ (1 − cosθ ) Substituting for known values, we obtain VJ b = 0.699 s mc
1 Thus the final equation for Uc is give by 0.699 t
Uc
=
VJ 1+0.699 t  v v III14  eText Main Menu  Textbook Table of Contents  Study Guide The final results are now given as shown below:
t
(s)
0
2
5
10
15
∞ Uc/VJ Uc
(m/s)
0.0
20.0
27.2
30.6
31.9
35 0.0
0.583
0.757
0.875
0.912
1.0 ac
(m/s2)
24.45
4.49
1.22
0.39
0.192
0.0 Uc vs t
35
30
25
20
15
10
5
0
0 5 10 15 t(s) Note that the limiting case occurs when the cart velocity reaches the jet
velocity. At this point, the jet can impart no more momentum to the cart, the
acceleration is now zero, and the terminal velocity has been reached.
Review the text example on accelerating control volumes. Moment of Momentum (angular momentum)
For moment of momentum we have that B = H = r × (m V ) and β = r ×V From the previous equation for moment of momentum and these definitions,
Reynolds transport theorem becomes  v v III15  eText Main Menu  Textbook Table of Contents  Study Guide 20 ∑M ∂
∫ r × V ρ dV
∂ t cv = ↓ + &
∫ r ×V d m Ae ↓ = the ∑ of all
external
moments
acting
on the c.v. − e &
∫ r ×V d m i Ai ↓ = the rate of
change of moment of momentum
in the c.v. = 0
for steady state ↓ = the rate of
moment of
momentum
leaving
the c.v. = the rate of
moment of
momentum
entering
the c.v. For the special case of steadystate, steadyflow and uniform properties at any
exit or inlet, the equation becomes
&
&
∑ M = ∑ me r ×Ve − ∑ mi r ×Vi For moment of momentum problems, we must be careful to correctly evaluate
the moment of all applied forces and all inlet and exit momentum flows, with
particular attention to the signs.
Moment of Momentum Example:
A small lawn sprinkler operates as indicated.
The inlet flow rate is 9.98 kg/min with an inlet
pressure of 30 kPa. The two exit jets direct
o
flow at an angle of 40 above the horizontal. 160 mm For these conditions, determine the following:
a. Jet velocity relative to the nozzle.
b. Torque required to hold the arm stationary. D J = 5 mm c. Friction torque if the arm is rotating at 35 rpm.
d. Maximum rotational speed if we neglect
friction.  v v III16  eText Main Menu  Textbook Table of Contents  Study Guide a. R = 160 mm, DJ = 5 mm, Therefore, for each of the two jets:
3 3 QJ = 0.5* 9.98 kg/min/998 kg/m = 0.005 m /min
2 5 2 AJ = π π 0.0025 = 1.963*10 m
3 5 2 VJ = 0.005 m /min / 1.963*10 m /60 s/min VJ = 4.24 m/s relative to the nozzle exit ans.
b. Torque required to hold the arm stationary.
First develop the governing equations and analysis for the general case of the
arm rotating. V cos θ
J rω With the coordinate system at the
center of rotation of the arm, a
general velocity diagram for the case
when the arm is rotating is shown in
the adjacent schematic. R
+
ω o Taking the moment about the center of rotation, the moment of the inlet flow is
zero since the moment arm is zero for the inlet flow.
The basic equation then becomes
&
T0 = 2 me R (VJ cos α − R ω ) Note that the net momentum velocity is the difference between the tangential
component of the jet exit velocity and the rotational speed of the arm. Also note
that the direction of positive moments was taken as the same as for VJ and
opposite to the direction of rotation.
For a stationary arm R ω = 0. We thus obtain for the stationary torque  v v III17  eText Main Menu  Textbook Table of Contents  Study Guide To = 2 ρ QJ R VJ cos α
To = 2 * 998 kg
m3 1min
m
.005
m * 4.24 cos 4.160o
3
min 60
s
m To = 0.0864 N m clockwise. ans. A resisting torque of 0.0864 N m must be applied in the clockwise direction to
keep the arm from rotating in the counterclockwise direction.
c. At ω = 30 rpm, calculate the friction torque Tf
ω = 30 To = 2 * 998 rev
rad 1min
rad
2π
=π
min
rev 60
s kg
m3 1min
m
rad 0.005
0.16m 4.24 cos 40o − .16m * π
3
min 60
s
s
m ans.
Note; The resisting torque decreases as the speed increases.
d. Find the maximum rotational speed.
The maximum rotational speed occurs when the opposing torque is zero and all
the moment of momentum goes to the angular rotation. For this case,
VJ cos θ – Rω = 0
V cosθ
ω= J
=
R 4.2 rad
= 193.8 rpm 4 m / s • cos 40
rad
s
= 20.3
= 193.8 rpm
0.16 m
s Review material and examples on moment of momentum.  v v III18  eText Main Menu  Textbook Table of Contents  Study Guide ans. Energy Equation (Extended Bernoulli Equation)
For energy, we have that B = E = ∫eρd V 1
2 β = e = u + V2 + g z and cv From the previous statement of conservation of energy and these definitions,
Reynolds transport theorem becomes:
&
& d E = ∂ e ρ dV +
Q −W =
∫
d t sys ∂ t cv ∫ ee ρe Ve ⋅ d Ae − A ei ρi Vi ⋅ d Ai
∫
A
e i After extensive algebra and simplification (see text for detailed development),
we obtain:
2 P1 − P2
ρg = 2 V2 − V1
2g ↓ Pressure
drop due to
acceleration
of the fluid Z2 − Z1 ↓ Pressure drop
from 1 – 2,
in the flow
direction + + hf,1− 2 ↓
Pressure
drop due to
elevation
change − ↓ hp ↓ Pressure
drop due to
frictional
head loss Pressure
drop due to
mechanical
work on fluid Note: this formulation must be written in the flow direction from 1  2 to be
consistent with the sign of the mechanical work term and so that hf,12 is always
a positive term. Also note the following: ❑ ❑ The points 1 and 2 must be specific points along the flow path
Each term has units of linear dimension, e.g., ft or meters, and z2 – z1 is
positive for z2 above z1
The term hf,12 is always positive when written in the flow direction and for
internal, pipe flow includes pipe or duct friction losses and fitting or piping
component (valves, elbows, etc.) losses, III19  v v ❑  eText Main Menu  Textbook Table of Contents  Study Guide ❑ The term hp is positive for pumps and fans ( i.e., pumps increase the
pressure in the flow direction) and negative for turbines (turbines decrease
the pressure in the flow direction) ❑ For pumps: hp = ws
g where ws = the useful work per unit mass to the fluid Therefore: w s = g hp and &
&
Wf = mw s = ρ Q g hp where &
Wf = the useful power delivered to the fluid and &
W
&
Wp = f where ηp Example ηp is the pump efficiency 2 Water flows at 30 ft/s through a
1000 ft length of 2 in diameter pipe.
The inlet pressure is 250 psig and the
exit is 100 ft higher than the inlet. 2 Assuming that the frictional loss is
2
given by 18 V /2g, 100 ft
1 Determine the exit pressure. 250 psig Given: V1 = V2 = 30 ft/s, L = 1000 ft, Z2 – Z1 = 100 ft, P1 = 250 psig
Also, since there is no mechanical work in the process, the energy equation
simplifies to  v v III20  eText Main Menu  Textbook Table of Contents  Study Guide P − P2
1
= Z2 − Z1
ρg + hf
2 2 2 30 ft / s
P − P2
1
= 100 ft + 18
ρg
64.4 ft / s 2 = 351.8 ft 3 P1 – P2 = 62.4 lbf/ft 351.8 ft = 21,949 psf = 152.4 psi
P2 = 250 – 152.4 = 97.6 psig ans.
Problem Extension
A pump driven by an electric motor is now added to the system. The motor
delivers 10.5 hp. The flow rate and inlet pressure remain constant and the pump
efficiency is 71.4 %, determine the new exit pressure.
2 2 3 Q = AV = π π (1/12) ft * 30 ft/s = 0.6545 ft /s
Wf = ηp Wp= ρ Q g hp
hp = 0.714 * 10.5 hp * 550ft − lbf / s / hp
62.4 lbm / ft 3 * 0.6545ft 3 / s = 101ft The pump adds a head increase equal to 101 ft to the system and the exit pressure
should increase.
Substituting in the energy equation, we obtain
2 2 2 30 ft / s
P − P2
1
= 100 ft + 18
ρg
64.4 ft / s 2 − 101 ft = 250.8 ft 3 P1 – P2 = 62.4 lbf/ft 250.8 ft = 15,650 psf = 108.7 psi
P2 = 250 – 108.7 = 141.3 psig ans.
Review examples for the use of the energy equation  v v III21  eText Main Menu  Textbook Table of Contents  Study Guide ...
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This note was uploaded on 10/27/2009 for the course MAE 101a taught by Professor Sakar during the Spring '08 term at UCSD.
 Spring '08
 Sakar

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