Chapter 5 Dimensional Analysis and Similarity
5.1 Solution: For kerosene at 20C, take = 804 kg/m3 and = 0.00192 kg/ms. The only unknown in the transition Reynolds number is the fluid velocity: Re tr 2300 =
Vd (804)V(0.05) = , solve for Vtr = 0.11 m/s 0.0
Lecture 8: MEC516 FLUID MECHANICS I
2012
1/3
Hydrostatic Forces on Submerged Plane Surfaces (continued)
Inclined Surfaces
Lets consider a submerged inclined plane surface with the xy-plane portrays the inclined plane.
p = pa
F
hCG
p = h + pa
The pressure
Lecture 13: MEC516 FLUID MECHANICS I
2012
1/3
Conservation of Mass
In this case, the property is mass.
msyst
dB Bsyst
1
Hence: Bsyst m syst , then
dm m syst
msyst
The mass of the system is fixed and remains constant. Hence,
d Bsyst
dt
dm syst
dt
0 (i.
Lecture 14: MEC516 FLUID MECHANICS I
2012
1/2
Examples: (continued)
Continuity equations for a CV with multiple inlets and outlets:
for steady flow assuming V A and 1-D flow:
or
in Vin Ain out Vout Aout
in Qin out Qout
in
or
out
in
in
m in m out
out
o
Lecture 12: MEC516 FLUID MECHANICS I
2012
1/3
Volume and Mass Rate of Flow
Suppose that we have a surface A through which the fluid passes over a time interval of dt.
n
unit outward normal n
V
V
dA
A
V dt
dA
dA
n
V
dA cos
Parallelepiped slanted at
an a
Lecture 11: MEC516 FLUID MECHANICS I
2012
1/2
Chapter 3: Integral Relations for a Control Volume
Flow Concepts and Kinematics:
1.
Lagrangian
- closed system
- control mass (fixed mass)
vs.
Eulerian Approach
- open system
- control volume (fixed space/poin
Lecture 10: MEC516 FLUID MECHANICS I
2012
1/5
Buoyancy
Archimedes laws of buoyancy:
1. A body immersed in a fluid experiences a vertical buoyant force equal to the weight of the fluid
it displaces.
FB fluid displaced
where displaced body if the body is fu
Print Name:
SOLUTIO NS
Section # Student ID:
RYERSON UNIVERSITY
Department of Mechanical and Industrial Engineering
MEC Slé/BME 516
Fluids Mechanics
Common Midterm Test
Friday, Oct 17, 2014
4:10 to 6:00pm
Examiners: Drs. Alan Fung, David Naylor, and Toora
Lecture 22: MEC516 FLUID MECHANICS I
2012
1/3
The Differential Equation of Linear Momentum (continued)
Because CV is so small (infinitesimal CV), Eq. (3.37) becomes simply:
VdAout Vout VdAin Vin
dF t V d
out
in
dF t V d M xdx M x M y dy M y M z dz
Lecture 23: MEC516 FLUID MECHANICS I
2012
1/3
The Differential Equation of Linear Momentum (continued)
Finally, substituting the external forces into Eq. (3) yields the full momentum equation in vector
notation as:
dV
g p ij
dt
(4.32)
dV V
V
V
V
where
i
Lecture 21: MEC516 FLUID MECHANICS I
2012
1/2
The Differential Equation of Mass Conservation
Consider an elemental CV in a fluid flow
y
dy
m x u dydz
m x dx u u dx dydz
x
x
dz
dx
Fig. 4.1
z
Similar in y-direction: m y v dxdz , m y dy v v dy dxdz
y
Similar
Lecture 24: MEC516 FLUID MECHANICS I
2012
1/3
An Example on differential equations of motion
A continuous stream of Newtonian fluid flows steadily in a very long, horizontal, straight circular pipe
of radius R. Assume incompressible, constant-property flo
Revisit the force polygon diagram for the example problem:
I apologize for the confusion yesterday when I tried to present to you the force polygon diagram
for the example problem. When I was drawing the force polygon diagram, I kept thinking about
the li
Lecture 15: MEC516 FLUID MECHANICS I
2012
1/3
The Control Volume Linear-Momentum Equation
Recalling the Reynolds transport theorem:
dBsyst
d
d Vr n dA
A
dt
dt cv
cs
which is the general control volume conservation equation.
(3.15)
The amount of proper
Lecture 9: MEC516 FLUID MECHANICS I
2012
1/3
Example
A concrete gate is used to keep water level of a river upstream.
L 3 2 4 2 5 ft
hinge
l = 3 ft
h1 = 3 ft
h
SG = 1
h2 = 4 ft
4 ft
SG = 1
Gate with 5-ft wide (into the page), i.e. b = 5 ft
SGg = 2.5
To wh
MEC516 Fluid Mechanics I
2012
An example on buoyancy
King Hiero of Syracuse wanted to know whether his new golden crown was really made of pure
gold (SGgold = 19.3). Archimedes discovered the buoyancy laws when he attempted to find the
answer. There were
Lecture 4: MEC516 FLUID MECHANICS I
2012
1/2
Chapter 2: Pressure Distribution in a Fluid
Hydrostatics or fluid statics is a study of fluid at rest or static.
Lets consider a differential volume of fluid or a fluid parcel.
Under the gravity, the fluid parc
Lecture 9: MEC516 FLUID MECHANICS I
2012
1/3
Example
A concrete gate is used to keep water level of a river upstream.
L 3 2 4 2 5 ft
hinge
l = 3 ft
h1 = 3 ft
h
SG = 1
h2 = 4 ft
4 ft
SG = 1
Gate with 5-ft wide (into the page), i.e. b = 5 ft
SGg = 2.5
To wh
Lecture 6: MEC516 FLUID MECHANICS I
2012
1/2
Pressure Variation in an Incompressible Fluid with a constant or
For a fluid, which is homogeneous and incompressible, its or is constant.
Integrating Eq. (I) with respect to a reference point:
p
z
z
dp z z 0
Lecture 3: MEC516 FLUID MECHANICS I
2012
1/3
Dimensions & Units
A dimension is the measure by which a physical variable is expressed quantitatively.
e.g.: length is a dimension [associated with] distance, displacement, height, depth, etc. as variables
A u
Lecture 7: MEC516 FLUID MECHANICS I
2012
1/2
Manometers (continued)
Examples
1) Inclined manometer:
a
R
A
p B p a 1 h1 y 2 y R sin (abs)
or p B 1 h1 y 2 y R sin (gage)
Substituting for y gives:
a
a
p B 1 h1 R 2 R sin (gage)
A
A
In terms of head:
a
a
h
Lecture 18: MEC516 FLUID MECHANICS I
2012
1/2
Bernoullis Equation
V2
g z (mechanical energy), to a streamline in a
2
steady, frictionless and incompressible flow, the mechanical energy will remain constant at any point on
the streamline because there is
Lecture 5: MEC516 FLUID MECHANICS I
2012
1/2
Basic Equation of Fluid Statics
Lets consider a small volume (element) of fluid in a static fluid.
Note: Static fluid has no shear stress ( = 0), because there is no motion.
z
d = dxdydz
dx
g
g
dz
y
dy
x
Net fo
Lecture 19: MEC516 FLUID MECHANICS I
2012
1/2
Reversibility, Irreversibility, and Losses (continued)
When the shaft work is extracted from fluid by a turbine, hs ht (turbine head). Eqn. (13) becomes
p
p
V2
V2
z
zc
c
2g
2g
in
When the shaft work is put
Lecture 17: MEC516 FLUID MECHANICS I
2012
1/2
The Energy Equation
The 1st law of thermodynamics or energy balance for a system (fixed mass) changing from state 1 to
state 2 over time t is
Q W E 2 E1
(1)
where: Q is the heat added to the system over the ti
Lecture 16: MEC516 FLUID MECHANICS I
2012
1/6
The Control Volume Linear-Momentum Equation (continued)
Note:
If the velocities are not uniform over the inlet and outlet surfaces,
2
V
V n dA Vav A n . Hence, in
Ain
order to be equal, we use the momentu
RYERSON UNIVERSITY
Department of Mechanical and Industrial Engineering
MEC/BME516
Fluid Mechanics
Additional Suggested Problems from Chapter 5
5.43 Non-dimensionalize the thermal energy partial differential equation (4.75) and its boundary
conditions (4.6
2.8 Buoyancy
(Archimedes Principle)
MEC516 / BME516
Fluid Mechanics I, F2015
Chapter 2G Notes
Buoyancy force is due to displacement of fluid
Always acts upward
Objects can "float" due to the buoyancy force, i.e.,
boats, balloons, etc.)
Copyright Prof. R.
Lecture 20: MEC516 FLUID MECHANICS I
2012
1/3
An example of using energy equation
The overall loss coefficient in the system (excluding turbine loss) is estimated to be K = 1.8, based on
the velocity Vp, which includes frictional, entrance and exit losses
1.8 Suppose that bending stress 0' in a beam depends upon bending moment M and
beam area moment of inertia I and is proportional to the beam half-thickness y. Suppose
also that, for the paiticular case M = 2900 in-lbf, y = 1.5 in, and I = 0.4 m4, the pred
Error: This equation
should be set eqaul
to zero.
=0
Typographical error. This should be
V_1=V_2(1-sin beta). But rest of solution
looks ok.
This is only the normal
force. They should really
have computed the shear
force too.
Note: There is no
torque. The