Lec 2: Problem solving,
conservation of mass
1
• For next time:
– Read: § 1-10 to 1-11; 2-1 to 2-4.
• Outline:
– Properties of systems.
– Problem solving.
– Conservation of mass.
• Important points:
– Intensive vs. extensive properties
– Specific volume
– Mass and volume flow rates
2
Review--Property
A property is a characteristic of a system
to which numerical values can be
assigned to describe the system.
Examples are:
• Mass
• Temperature
• Pressure
• Density
3
Review--Intensive Property
Intensive properties are independent of
the size (mass or volume) of the system.
Examples:
•
Density
•
Temperature
4
Review--Extensive Property
Extensive properties are properties which
can be counted and their value for the
whole system is the sum of the value for
subdivisions of the system.
Examples:
•Volume
•Mass
5
TEAMPLAY
• How many properties are needed to define
the state of the air (system) enclosed by a
bicycle tire?
• What are they?
• Make a list of them.
6
Problem Solving
• Steps that will help you think logically
• Steps that will gain you points
• Step 1. Understand the problem
statement
– State the problem in your own words
– State what is given
– State what is to be found
7
Problem Solving
• Step 2. Sketch
– Sketch the physical system (sometimes
a machine) involved.
•Indicate if energy or mass goes in or
out.
•List the given information as “Given”-do not confuse given information
with assumption, the next step.
– Also sketch property diagrams such as
pressure-volume diagrams (to be
covered later)
8
Problem Solving
• Step 3. State assumptions.
– For example, if asked to find the specific
volume of air in Houston at 60 ºF with
no specified pressure, an assumption
might be that the pressure is 14.696
psia.
– In this case the pressure is not given,
but assumed.
9
Problem Solving
• Step 4. Write down physical laws that
apply.
– Examples are conservation of energy
E in  E out  E
– or perhaps the perfect gas law in one of
its many forms
pV  mRT
10
Problem Solving
• Step 5. Find unknown properties
– For example, if you know the
temperature and volume, V, of a
container of water and want to find its
mass, m, you will be able directly to
look up the unknown property specific
volume, v, and then find the mass from
V
m
v
11
Problem Solving
• Step 6. Do the calculation
– neatly begin with the relevant basic
physical relationships
– simplify the relationships
– substitute in given, assumed, and
determined properties
– use horizontal bars, not slashes ( not /)
– write out all units in great detail and
cancel them
– circle or underline important results
12
Problem Solving
• Step 7. Does the result pass the
reasonableness test?
– For example, in calculating your weight
on the moon, if the result is more than
you weigh on earth, doesn’t your
intuition say something is wrong?
13
Problem Solving
• Beware the calculator disease of false
accuracy.
• Answers cannot legitimately cannot be
given to more digits than the least
significant digit in the calculation.
• Assume given information specified in
this course to less than three significant
digits has three significant digits, e.g., 2
kg is 2.00 kg
14
Problem Solving
• Retain excess digits in preliminary steps
and round to the appropriate number of
significant digits in the last step (this is
the way your calculator works).
15
Conservation of mass
Closed systems
• The principle of conservation of mass for
closed systems is used implicitly.
• We simply require that the mass of the
system remain constant at all times.
16
Review
Closed System
“A system is a
region of space or
quantity of matter
we want to study.”
The example
shown here is a
closed system or
fixed quantity of
matter.
Note that dashed
lines enclose and
indicate the
system.
17
Conservation of Mass
Open Systems
• What happens
when the system
is no longer
closed, but
something is
flowing in and out
of it?
• How will that
change our
analysis?
Energy
in
18
Closed and Open Systems
Open systems are also called
“control volumes”
19
There are many examples of
open systems
• Consider an air compressor:
Air in at
14.7 psia
Air out at
120 psia
Work
in
Heat out
20
Or an automobile engine
Fuel in at T
and P
Wout
Air in at T
and P
Qout
Exhaust out at
T and P.
21
Mass flow rates
• Mass enters or leaves a control volume
through
– apertures
– ducts or pipes
22
TEAMPLAY
• Consider flow in a garden hose, crude pipeline,
sewer pipe or air conditioning duct.
• In the middle of a length of hose, pipe, or duct,
sketch the profile of the axial velocity from wall
to middle back to wall.
y
Flow
x
• Ask yourself: What is V at the middle? What is
V at the wall?
23
Velocity profile
•Real system
may have velocity
(or even density)
variations with
cross section.
•We simplify and
use Vavg
24
Mass flow rate
• Uniform--properties do not vary normal to
the flow direction (normal to the velocity
vector). They may vary along the flow
direction.
25
Mass flow rate
• If A is the area of the body surface
across which the mass flows, and if
Vn = Vavg is the normal velocity,
dA
Vn
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Mass flow rate
• Then

d m  ρVn dA

m   ρVn dA
A
• where the use of the “dot” symbol to
indicate a quantity per unit time is
introduced.
27
We can draw a simple schematic
CONTROL SURFACE
2
CONTROL
VOLUME
1
3
28
Make some assumptions
• No generation of mass in control volume
• No destruction of mass in control volume
29
Apply conservation equation
 MASS FLOW 


 RATE INTO 


C.V.


 MASS FLOW 


  RATE OUT OF 


C.V.


 RATE OF CHANGE 


 OF MASS IN THE 


C.V.


30
Look at each term:
 MASS FLOW 


 RATE INTO 


C.V.


 MASS FLOW 


 RATE OUT OF


C.V.


n
i
 m
i 1
k
j
 m
j1
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Accumulation Term
 RATE OF CHANGE 

 dmc.v.
 OF MASS IN THE  
dt


C.V.


32
Put it all together
n
k
dm c.v.
 i  m
j 
m

dt
j1
i 1
 MASS FLOW   MASS FLOW   RATE OF CHANGE 


 
 
 RATE INTO    RATE OUT OF    OF MASS IN THE 






C.V.

C.V.
C.V.

 
 
33
What about the mass in the
control volume?
m c.v. 
ρ
dV

c.v.
We’ll need to know the density variation
across the control volume! If we
assume constant density all over the
control volume, we get
m c.v.  V
34
Integrals introduce some
potentially nasty math!
k
n
d
ρVn dA i    ρVn dA j 
ρ
dV



dt c.v.
i 1 A
j1 A j
i
35
In order to simplify this,
• Assume uniform flow into and out of each
cross section :
V=Vavg
• Assume uniform properties into and out of
each cross section:
 f(A)
36
What does that do to each term?
   ρ Vn dA  ρ Vn  dA
m
A
A
VA
  ρ VA 
m
ν
37
Mass and volumetric flowrates
• Note that the previous equation has
units of mass flow rate or kg/sec (or
lbm/sec).

AV
m  ρVA 
v
• The product AV is the volumetric
flowrate, the flowrate in volume per unit
time (m3/sec or ft3/sec)
38
TEAMPLAY
• Air at 14 ºC (287 K) exits an air
conditioning duct into our room. The duct
opening is 8 inches wide by 2 feet long,
and the velocity is measured to be 4 ft per
second.
• List your givens and your assumptions.
• What is the volumetric flow rate in cubic
feet per minute?
39
The right hand side becomes:
dm c.v.
d
d

ρ dV  (ρ dV)
dt
dt c.v.
dt c.v.


dm c.v.
d
 (ρV )
dt
dt
So we would have to know how the
control volume density and volume
change with time.
40
With most transient problems we’re
not going to worry about the
details of  and V in the control
volume.
We’ll just concern ourselves with what
the change in mass is.
dm c.v.
dt
41