p = 40kPa Load Case 2 p = 40kPa, 200N force Comosworks p = 20kPa DisplacementStress - Radial Stress - Tang. Displacement Stress - Radial Stress Tang. Displacement: Node X (m) Value (mm) Value (N/m^2) Value (N/m^2) Value (mm) Value (N/m^2) Value (N/m^2) No
Chapter 7: The Energy Equation
Faculty of Engineering and Industrial Sciences
Swinburne University of Technology
Summer 2010
The Energy Equation
Kinetic Energy
(fluid jet)
A fluid can have
several forms of
energy
Potential Energy
(water behind a dam)
Flow
Faculty of Engineering and Industrial Sciences
HES 5320 - SOLID MECHANICS FE ASSIGNMENT - FLAT CIRCULAR PRESS. PLATE - 2009
STUDENT NAMES & No.s : Matthew DeBrincat 6164811 INTRODUCTION :
This assignment combines Practical Lab 1 with the Finite Element An
CHAPTER 05
1. The amount of mass in a system is constant. True or False
A. True
B.False
2.The _ is the control volume expression for the conservation of mass.
YOUR ANSWER: continuity equation
3.If density is known, the volume flowrate can be used to find
Question 1 a) Intensive properties do not depend on the size (extent) of the system but extensive properties do. b) The fluids whose shear stress is proportional to the velocity gradient are called Newtonian fluids. Most common fluids such as water, air,
Chapter 3: Fluid Statics
Faculty of Engineering and Industrial Sciences
Swinburne University of Technology
Summer 2010
HES2340 : Fluid Mechanics 1
1
Fluid Statics
Pressure
Transmission of Fluid Pressure
Ratio A2/A1 is called ideal mechanical advantage,
An
Chapter 5: Control Volume Approach
and Continuity Equation
Faculty of Engineering and Industrial Sciences
Swinburne University of Technology
Summer 2010
Overview
Engineers are often concerned about evaluating the pressure and
velocities at arbitrary locat
Chapter 4: Flowing Fluids and Pressure
Variation
Faculty of Engineering and Industrial Sciences
Swinburne University of Technology
Summer 2010
Streamline
Flow visualization is the visual examination of flow-field features.
A Streamline is a curve that is
Fluid Mechanics 1 Assignment 2 Due Date 5/10/2009 at 5:00 PM
Question 1
I. A velocity field is given by: Velocity fields are plotted V 6 x ^ m/s j below, where the arrows indicate the magnitude and direction of the velocity vectors. The picture that best
A 1. tank is filled with seawater to a depth of 12 ft. If the specific gravity of seawater is 1.03 and the atmospheric pressure at this location is 14.8 psi, the absolute pressure (psi) at the bottom of the tank is most nearly A. 5.4 B. 20.2 C. 26.8 D. 27
QUESTION 2
SUGGESTED SOLUTION:
(a) The magnitude of the shear stress is greater at the moving plate (y = H) since max
shear stress will occur at this point.
(b) Recall
=
= (cfw_(H) + .
Deriving an expression for the y position of zero shear stress implie
15.15: PROBLEM DEFINITION
Situation:
Rectangular, concrete-lined channel
Width is 4m
Slope is .004
Q = 25 m3 / s
Find:
The uniform ow depth ( m).
Assumptions:
n = 0.015
PLAN
Use the Manning equation to solve for d, given that A = d 4 m
SOLUTION
2/3
Q = (1
10.63: PROBLEM DEFINITION
Situation:
Water drains from a tank, passes through a pipe and then jets upward.
D = 1.5 cm, L = 10 m, z = 5 m.
Two 90 elbows in pipe.
Find:
(a) Exit velocity of water ( m/ s).
(b) Height of water jet ( cm).
Assumptions:
The pipe
SUGGESTED SOLUTION:
The control volume selected is shown below .The control volume is stationary.
Free body diagram
ANALYSIS:
Recall the momentum equation
Momentum diagram
Momentum equation in x- and z- directions
In the x- direction
Similarly, in the z d
7.36: PROBLEM DEFINITION
Situation:
Water ows through a pipe with a venturi meter.
D = 30 cm, d = 15 cm.
patm = 100 kPa, H = 5 m.
Find:
Maximum allowable discharge before cavitation.
Assumptions:
= 1.0.
Properties:
Water (10 C) , Table A.5: pv = 2340 Pa,
4.4: PROBLEM DEFINITION
Situation:
The valve in a system is gradually opened to have a constant rate of increase in
discharge.
Find:
Describe the ow at points A and B.
SOLUTION
A: Unsteady, uniform.
B: Non-uniform, unsteady.
4
4.16: PROBLEM DEFINITION
Sit
SWINBURNE UNIVERSITY OF TECHNOLOGY (SARAWAK CAMPUS)
FACULTY OF ENGINEERING AND INDUSTRIAL SCIENCE
HES2340 Fluid Mechanics 1
Semester 1, 2012
Assignment 1: Basic Concepts of Fluid Flow
By
Stephen, P. Y. Bong (4209168)
Lecturer: Professor Alexander Gorin
Du
1.1: PROBLEM DEFINITION
Find: List three common units for each variable:
a. Volume ow rate (Q), mass ow rate (m), and pressure (p).
b. Force, energy, power.
c. Viscosity, surface tension.
PLAN
Use Table F.1 to nd common units
SOLUTION
a. Volume ow rate, m
Fluid mechanics 1 Assignment 2
Question 1
A velocity field is given by:
Velocity fields are plotted
V 6 x m/s
j
below, where the arrows indicate the magnitude and direction of the velocity
vectors. The picture that best describes the field is:
I.
y
0
0
0
5.4: PROBLEM DEFINITION
Situation:
Liquid ows through a pipe at constant velocity.
Find:
Flow rate is (a) halved, (b) doubled, (c) quadrupled if pipe diameter is doubled
but velocity remains the same.
SOLUTION
Use discharge equation, Q = AV .
Since the di
3.4: PROBLEM DEFINITION
Situation:
A Crosby gage tester is applied to calibrate a pressure gage.
Indicated pressure on the gage is p = 200 kPa.
W = 140 N, D = 0.03 m.
Find:
Percent error in gage reading.
PLAN
1. Calculate the pressure that the gage should
2.5: PROBLEM DEFINITION
Situation:
Carbon dioxide.
Find:
Density and specic weight of CO2 .
Properties:
From Table A.2, RCO2 = 189 J/kgK.
p = 300 kPa, T = 60 C.
PLAN
1. First, apply the ideal gas law to nd density.
2. Then, calculate specic weight using =
6.7: PROBLEM DEFINITION
Situation:
A water jet is lling a tank.
m = 20 kg, V = 20 L.
d = 30 mm, v = 20 m/s.
Find:
Force on the bottom of the tank (N).
Force acting on the stop block (N).
Assumptions:
Steady ow.
Properties:
Water (15 C), Table A.5: = 999 k
7.13: PROBLEM DEFINITION
Situation:
The velocity distribution in a pipe with turbulent ow is given by
n
V
y
=
Vmax
r0
Find:
Derive a formula for as a function of n.
Find for n = 1/7.
SOLUTION
Flow rate equation
V
=
Vmax
Q=
Z
y
r0
n
=
r0 r
r0
n
n
r
= 1
r0
10.6: PROBLEM DEFINITION
Situation:
Air is owing from a large tank to ambient through a horizontal pipe.
Pipe is 1" Schedule 40. D = 1.049 in = 0.0266 m.
V = 10 m/ s. f = 0.015, L = 50 m.
Sketch:
Find:
Pressure in the tank (Pa).
Assumptions:
Air has const
CHAPTER 03
1. Write Newton's second law of motion.
YOUR ANSWER: F = ma
2.The lines that are tangent to the velocity vectors throughout the flow field are called
steady flow lines. True or False
A. True
B. False
3.Streamwise acceleration is the product of
Chapter 10: Flow in Conduits
Faculty of Engineering and Industrial Sciences
Swinburne University of Technology
Summer 2010
Pressure Drop (Loss)
Pressure Drop (Loss)
are the sum of
+
Major Losses
due to
pipe friction
Minor Losses
due to
Flow separation and
Chapter 15: Open Channel Flow
Faculty of Engineering and Industrial Sciences
Swinburne University of Technology
Summer 2010
Objectives
FLOW IN OPEN CHANNELS
Understand
State of Flow
Flow Classification
Hydraulic jumps
Flow regimes
Best hydraulic cross
sec