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Lec 2: Problem solving, conservation of mass

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• 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

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Review--Property

• Mass

• Temperature

• Pressure

• Density

A property is a characteristic of a system to which numerical values can be assigned to describe the system.

Examples are:

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Review--Intensive Property

Intensive properties are independent of the size (mass or volume) of the system.

Examples:

• Density

• Temperature

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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

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TEAMPLAYTEAMPLAY

• 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.

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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

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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)

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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.

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Problem Solving

• Step 4. Write down physical laws that apply.– Examples are conservation of energy

– or perhaps the perfect gas law in one of its many forms

EEE outin

mRTpV

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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

Vm

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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

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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?

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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

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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).

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Conservation of massClosed 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.

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ReviewClosed 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.

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Conservation of MassOpen 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

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Closed and Open Systems

Open systems are also called “control volumes”

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There are many examples of open systems

• Consider an air compressor:

Work in

Air in at 14.7 psia

Air out at 120 psia

Heat out

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Or an automobile engine

Fuel in at T and P

Air in at T and P

Wout Qout

Exhaust out at T and P.

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Mass flow rates

• Mass enters or leaves a control volume through– apertures– ducts or pipes

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TEAMPLAYTEAMPLAY• 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.

• Ask yourself: What is V at the middle? What is V at the wall?

y

x

Flow

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Velocity profile

•Real system may have velocity (or even density) variations with cross section.

•We simplify and use Vavg

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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.

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Mass flow rate

dA

Vn

• If A is the area of the body surface across which the mass flows, and if Vn = Vavg is the normal velocity,

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Mass flow rate

• Then

• where the use of the “dot” symbol to indicate a quantity per unit time is introduced.

dAρmd nV

dAρmA

nV

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We can draw a simple schematic

CONTROLVOLUME

1

2

3

CONTROL SURFACE

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Make some assumptions

• No generation of mass in control volume

• No destruction of mass in control volume

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Apply conservation equation

C.V.

INTO RATE

FLOW MASS

C.V.

THE IN MASS OF

CHANGE OF RATE

C.V.

OF OUT RATE

FLOW MASS

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Look at each term:

C.V.

INTO RATE

FLOW MASS

n

1iim

C.V.

OF OUT RATE

FLOW MASS

k

1jjm

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Accumulation Term

dt

dm

C.V.

THE IN MASS OF

CHANGE OF RATEc.v.

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Put it all together

n

1iim

k

1jjm

dt

dmc.v.

C.V.

INTO RATE

FLOW MASS

C.V.

OF OUT RATE

FLOW MASS

C.V.

THE IN MASS OF

CHANGE OF RATE

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What about the mass in the control volume?

c.v.

c.v. dVρm

We’ll need to know the density variation across the control volume! If we assume constant density all over the control volume, we get

Vmc.v.

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Integrals introduce some potentially nasty math!

n

1i A

in

i

dAρV

k

1j A

jn

j

dAρV c.v.

dVρdt

d

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• Assume uniform flow into and out of each cross section : V=Vavg

• Assume uniform properties into and out of each cross section: f(A)

In order to simplify this,

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What does that do to each term?

dAρmA

nV A

dAρ nV

Aρm Vν

AV

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Mass and volumetric flowrates

• Note that the previous equation has units of mass flow rate or kg/sec (or lbm/sec).

• The product AV is the volumetric flowrate, the flowrate in volume per unit time (m3/sec or ft3/sec)

v

Aρm

VVA

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TEAMPLAYTEAMPLAY

• 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?

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The right hand side becomes:

c.v.

dVρdt

ddt

dmc.v. )dV(ρdt

d

c.v.

)ρV(dt

d

dt

dmc.v.

So we would have to know how the control volume density and volume change with time.

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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.

dt

dmc.v.