Transcript Notes 11 - CEProfs

Lec 11: Flow work, energy
transport by mass, first law of
thermodynamics
1
• For next time:
– Read: § 5-2 to 5-3
– HW 6 due Wednesday, October 8, 2003
• Outline:
– Flow work
– Conservation of energy
– Open and closed systems
• Important points:
– Know how to calculate start and end states
– Know how to find the path between states
– Know when to apply the various forms of the
conservation of energy equation
2
Flow work
• For open systems, obviously work must
be done to move the fluid into and out
of the control volume.
• It is a form of boundary work.
L
Flow
“piston” of fluid having m, P, V. A
is crossectional area of pipe.
3
Flow work
• The force on the fluid element F=PA
W   Fds  F ds  FL  PAL
• But AL is V, so W=PV or
W
V
 w  P  Pv
m
m
4
Energy of fluid in closed system
• Energy of a simple, compressible system
is
e  u  pe  ke
5
Energy of fluid in open system
• Now there is a fourth term: the flow
energy
θ  Pv  u  pe  ke
• Where θ  Pv  e
• Pv + u is just h, so
θ  h  pe  ke
6
Energy transport by mass
• Now that we know the energy per unit
mass =h+ke+pe for an open system, the
energy transported in and out with the
mass is just the product of the mass and
the energy:
θ
E mass  mθ or E mass  m
7
Conservation of Energy
First Law of Thermodynamics
• The net change (increase or decrease) in
the total energy (Esystem) of a system
during a process is equal to the difference
between the total energy transferred in or
entering (Ein) and the total energy
transferred out or leaving the system
during that process.
E system  E in  E out
8
Conservation of Energy
First Law of Thermodynamics
• For a simple, compressible system,
E system  U  KE  PE
• and a change in E is
E system  U  KE  PE
9
The kinetic energy is given by:
KE 
1
2
2
m(V f  V i )
2
To what does this kinetic energy term
apply ? Explain to the person across
the table.
10
Energy has units of force times
distance. The energy change
in accelerating a mass of 10 kg
from Vi= 0 to Vf = 10 m/s is?
1
2
2
KE 
m( V f  V i )
2
1
m2
1N
 (10 kg  100 2 ) 
 500 N  m
m
2
s
kg  2
s
J
kJ
 500 N  m 

 0.5 kJ
N  m 1,000 J
11
NOTE THE CONVERSION TO GET
FROM m2/s2 to kJ/kg
2
m
kJ
1000 2 1
s
kg
REMEMBER IT! YOU WILL
NEED IT.
12
TEAMPLAY
How much energy in ft lbf must a shot
putter exert in throwing a 16 lbm shot put
if it is released at 30 ft/s?
How does this compare to a 60 W light
bulb burning for 10 sec?
13
Gravity is another force acting on
our system. It shows up in the
potential energy change.
PE  mg(z f  z i )
Work can be done by a change in
elevation of the system.
14
TEAMPLAY
Let’s say we have a 10 kg mass A that we
drop 100 m. We also have a device that will
convert all the potential energy of A into
kinetic energy of an object B. If the mass of
object B is 1 kg and it is initially at rest,
what would be B’s final velocity from
absorbing the potential from a 100 m drop
of A? Assume that object B travels
horizontally.
15
Internal energy…..
Internal energy is the energy a molecule
possesses, mostly as a result of.
Translation
Vibration
Rotation
All these are forms of kinetic energy. We will
neglect other forms of molecular energy
which exist on the atomic level.
16
Conservation of Energy
First Law of Thermodynamics
• Energy can be transferred into (or out of)
the system in three ways:
• Heat interactions (called heat transfer)
• Work interactions
• Mass flow (carrying energy with the mass).
E in  E out  (Qin  Q out )  ( Win  Wout )  ( E mass,in  E mass,out )
 E system
17
Conservation of Energy
First Law of Thermodynamics
(Q in  Q out )  ( Win  Wout )  ( E mass,in  E mass,out )  E system
Viewed in this way, E on the right hand side
is a property of the system. However, the
left side has Q and W which depend on the
path, in general. This leads us to
sometimes write
E in  E out  dE system
18
Conservation of Energy
First Law of Thermodynamics
• Adiabatic means simply that there is no
heat transfer, or Q=0.
• Consider a closed, adiabatic system.
0
0
(Q in  Q out )  ( Win  Wout )  ( E mass,in  E mass,out )  E system
adiabatic
closed
19
Conservation of Energy
First Law of Thermodynamics
• Now we have
Win  Wout  E system
• or in differential form
Win  Wout  dE system
20
Conservation of Energy
First Law of Thermodynamics
• So for an adiabatic closed system the
work is equal to a path independent
quantity, which means adiabatic work is
independent of path.
• This is another form of the first law;
• “For all adiabatic processes between two
specified states of a closed system, the
work is the same, regardless of the
nature of the closed system and the
details of the process.”
21
TEAMPLAY
Is an adiabatic process the same as a
constant temperature process?
Explain to the teammate next to you
why or why not - try to think of some
examples.
22
Conservation of Energy
First Law of Thermodynamics
• Let us replace Qin- Qout by a single Q that
represents the sum of all heat interactions
during the process.
• Remember--if heat is transferred in it is
positive--if transferred out it is negative.
23
Conservation of Energy
First Law of Thermodynamics
• Let us replace Win- Wout by a single -W
that represents the sum of all work
interactions during the process.
• Remember--if work is done on the
system it is negative. If done by the
system, it is positive.
24
Conservation of Energy
First Law of Thermodynamics
• The first law is now simply
Q - W  E
• Note that heat transferred to the system
(+) or work done on the system (-) both
will raise the energy of the system.
25
So let’s put it all together again:
Q  W  U  KE  PE
This is the algebraic statement of the
first law for closed systems of all
kinds, not just adiabatic.
26
Conservation of Energy
First Law of Thermodynamics
• Stationary means not moving--so PE
and KE are zero and the first law
becomes
Q  W  U
27
TEAMPLAY
• Solve problem 5-15.
28
Energy analysis of cycles
4
3
For the cycle,
E1- E1 = 0, or
Ecycle  0
1
2
E cycle  Q cycle  Wcycle  0
29
For cycles, we can write:
Qcycle = Wcycle,
Qcycle and Wcycle represent net amounts
which can also be represented as:
Q  W
cycle
cycle
30
TEAMPLAY
• A closed system undergoes a cycle consisting
of two processes. During the first process,
40 Btu of heat is transferred to the system
while the system does 60 Btu of work.
During the second process, 45 Btu of work is
done on the system.
– Determine the heat transfer during the
second process.
– Determine the net work and net heat
transfer of the cycle.
31
TEAMPLAY
• Solve problem 5-47
32