Energy Transfer
by Heat, Work
and Mass
1
Objectives
• Introduce the concept of energy transfer by heat.
• Discuss the various forms of work with emphasis on the moving
boundary work.
• Discuss the flow work, the work associated with forcing a fluid into or
out of a control volume.
• Discuss the combination of the internal energy and the flow work (i.e.
Enthalpy).
• Discuss the conservation of mass principle and apply it to various
systems.
• Show that h+ke+pe represents the energy of a flowing fluid per unit of
its mass.
2
Closed
Systems
3
How does energy cross a
closed system boundary?


Energy can cross the
boundaries of a closed Q
system in the form of heat
or work.
If the energy transfer
across the boundaries of a
closed system is due to a
temperature difference, it
is heat; otherwise, it is
work.
System
W
4
So, what do we mean by Heat?
• Heat is defined as the form of
energy (thermal energy) that
is transferred between two
systems (or surroundings) by
virtue of a temperature
difference.
• Heat is energy in transition. It
is recognized as it crosses the
boundary of a system.
5
No heat transfer =Adiabatic
Process



A process during which
there is no heat transfer
is called an adiabatic
process.
The adiabatic process
should not be mixed with
Isothermal process.
A process could be an
adiabatic process while
the temperature of a
system is changing .
6
Symbols of Heat

Q or 1Q2
Definition: The amount of total heat transferred during
a process from state 1 to state 2.
Units: kJ or BTU



q
Heat transfer per unit mass.
Units: kJ/kg or BTU/lbm




Q


Rate of heat transfer
kJ/sec = kW
7
Sign convention
 Heat
transferred to a system is
positive.
 Heat
transferred from a system is
negative.
8
What do we mean by Work?

Work is defined as the energy transfer
associated with force acting through a
distance.
9
Symbols of Work

W or 1W2
The amount of work done by or on a system
during a process from state 1 to state 2.
Units: kJ or BTU



w
Work per unit mass.
Units: kJ/kg or BTU/lbm




W


Power: The work done by unit time.
kJ/sec = kW
10
Sign convention

Work done on a system by an external force
acting in the direction of motion is negative.

Work done by a system against an external
force acting in the opposite direction of the
motion is positive.
11
Characteristics of Heat and
Work




Both of them are boundary phenomena. That
is they are recognized at the boundaries of a
system as they cross them.
Systems possess energy, not heat or work.
Both are associated with a process, not a
state.
Both are path functions (i.e., their
magnitudes depend on the path followed
during a process as well as the end states).
12
What are Path Functions?

Path functions have inexact
differentials designated by the
symbol , eg.  Q and  W.
2
 W  W
1

12


not W 
Point functions have exact
differentials designated by the
symbol , eg.  V

2
1
dV  V2-V1  ΔV
Properties are point functions, but heat and work are
path functions.
13
Types of work

Mechanical Work




Moving boundary work (The most common form)
Shaft work
Spring work
Electric Work

An electric wire crossing the system boundaries,
(We=V I t)



V =voltage
I=current
t =time
14
Two requirements for
Mechanical work to exist
There must be a force on the boundary.
1.

Example: the expansion of a gas into an
evacuated space (is not work since there is no
force).
The boundary must move.
2.

Example: Fixed piston in the presence of
external force (is not work since the boundary
did not move).
15
Computing Moving Boundary Work
Wb   Wb
2
1
  Fds
2
1
F
  Ads
A
2
1
W   PdV
2
1

The area under the process
curve on a P-V diagram
represents the boundary
work.
16
Boundary Work is Path
Dependent
17
Net Work done in a cycle



The cycle shown in the figure
(right) produces a net work
output.
This is because the work done by
the system during the expansion
process (area under path A) is
greater than the work done on the
system during the compression
part of the cycle (area under path
B),
The difference between these two
is the net work done during the
cycle (the colored area).
WA > 0 since V = V1 –V2 > 0
WB < 0 since V = V2 –V1 < 0
Wnet = WA +WB = Area
bounded by the curve
18
Moving Boundary Work for quasiequilibrium process

Strictly speaking the pressure in
the integral is the pressure at the
inner surface of the piston (Pi)

It becomes equal to the pressure
in the cylinder only if the process
is quasi- equilibrium. Thus the
entire gas is at the same pressure
at any given time
W   PdV
2
1
19
Moving Boundary Work for nonquasi- equilibrium process

For non-quasi-equilibrium
process, we still can compute
the work if the pressure at the
inner surface of the piston is
used for Pi.

We can not speak of pressure
of a system during non-quasiequilibrium process.
W   Pi dV
2
1
20
Computing the work for
some special processes
 Constant
volume
 Constant pressure
 Isothermal
 Polytropic
21
Constant Volume Process
P
1
2
Wb   PdV  0
1
2
V
22
Constant Pressure Process
P 1
2
W
V
Wb   PdV  P  dV  PV2  V1 
2
1
2
1
23
Example 3-6: Boundary Work during a ConstantPressure Process
A floating frictionless piston-cylinder device contains 10
lbm of water vapor at 60 psia and 320oF. Heat is now
transferred to the steam until the temperature reaches
400oF.
Determine the work done by the steam during this
process. (Answer: 96.4 Btu)
24
Isothermal Process
Wb   PdV
But for an
2
1
ideal gas
P 1
mRT Cons
P

V
V
mRT
Wb  
dV
V
Wb  mRT lnV2 / V1 
2
1
2
V
Wb  PV lnV2 / V1 
25
Example 3-7: Boundary Work during an
Isothermal Process
A piston-cylinder device initially contains 0.4 m3 of air at
100 kPa and 80oC. The air is now compressed to 0.1 m3
in such a way that the temperature inside the cylinder
remains constant.
Determine the work done during this process.
(Answer: -55.45 kJ)
26
Polytropic Process
For General Gases, P and V are
often related as
PV  Cons tant
n
Const
P
Vn
 P2V2  P1V1
,
n

1
 1 n
dV

2
2
Wb  1 PdV  1 Const n  


V
V
2
 PV ln  , n  1
V 

 1
27
Lets go through that step-wise
PV  C
n
C
n
P  n  CV
V
2
2
Wb   PdV   CV dV
1
n
1
 n 1
V
Wb  C
 n 1
2
1
V
V
 C
  n 1
 n 1
2
 n 1
1



28
Wb
V
V
 C
  n 1
 n 1
2
 n 1
1
  CV2  CV1
n
n

V2
V1

 n 1
But…
C
P2  n
V2
And….
C
P1  n
V1
Wb  P2V2  P1V1
1 n
So…
Polytropic Process
n 1
29
Wb  P2V2  P1V1
1 n
So far, we have not assumed an ideal
gas in this derivation, if we do, then….
PV  mRT
mRT2  mRT1
Wb 
1 n
mRT2  T1 
Wb 
1 n
Ideal gas
undergoing a
Polytropic
Process
30
Computing the work of a Gas
expanding against a Spring
Fspring
x
Patm
Patm
x0
Wpiston
x
Wpiston
P
P1
Spring just
touching piston
Pa A p
W piston
Spring exerting
force on piston
  x  Ap
Pi Ap
 x   / Ap
31
Computing the Spring work
alone
x
Patm
Wpiston
x
2
1
P
W
No Spring exerting
force on piston
Spring exerting
force on piston
V1
V2
V2
Volume
32
Computing the Spring work
alone
Wspring  Fdx
F  kx
x
1
Wspring   Fdx  x kxdx
2
1
1
2
2
Wspring  k( x2  x1 )
2
33
Example 3-9: Expansion of a Gas against a Spring
A piston-cylinder device
contains 0.05 m3 of gas
initially at 200 kPa. At
this state, a linear spring
that has a spring
constant of 150 kN/m is
touching the piston but
exerting no force on it.
Now heat is transferred
To the gas, causing the piston to rise and to compress
the spring until the volume inside the cylinder doubles.
If the cross-sectional area of the piston is 0.25 m2,
determine (a) the final pressure inside the cylinder, (b)
the total work done by the gas, and c) the fraction of
this work done against the spring to compress it.
34
Computing the Shaft work
T  Fr
W sh  Fs
T
F
r
s   2 r  n
W sh  2 nT
35
Example 3-8: Power transmission by the shaft
of a car
Determine the power transmission through the shaft of a car
when the torque applied is 200 Nm and the shaft rotate at a
rate of 4000rpm.
W sh  2 nT
(Answer: 83.8 kW)
36
Open System
or
Control Volume
37
How does energy cross an
open system boundary?

The energy content of a control volume
can be changed by mass flow as well as
heat and work interactions
38
Mass Flow Rate


The amount of mass flowing
through a cross section per unit
time is called the mass flow rate.
It is given as
m  Vm A




 = density of fluid, Kg/m3
Vm = mean velocity normal to A
A= cross sectional area of pipe.
Usually we drop the subscript m
m  VA
The unit is Kg/s
39
Volume Flow Rate


Volume flow rate is the
volume of fluid flowing
through a cross section per
unit of time
It is given as Q or V  Vm A

Vm = mean velocity normal to A
A= cross sectional area of
pipe.

Usually we drop the subscript m

Q or V  VA
The unit is m3/s
40
Conservation of mass
principle
Total mass
entering the
system
_
Total mass
leaving the
system
=
Net change in
mass within
the system
min  mout  msystem
On a rate basis,
dmsys
m in  m out 
dt
mi=50 kg
m= mi-me
=20 kg
me=30 kg
41
Conservation of mass
principle (continued)
For many inlets and
exits,
m1=2 kg
m2=3 kg
 min   mout
 ( m2  m1 )sys
m3=2+3 =5 kg
On a rate basis,
 in   m out  dmsys / dt
m
42
For steady process, no
accumulation of mass in CV
 in   m out  dmsys / dt
m
 in   m out
m
0
m 1  m 2
 1V1 A1  2V2 A2
mi
Control
volume
mcv=const.
For single stream,
me=mi
43
Flow rate for Steady compressible
flow ( = not const.)
Although, 1V1 A1   2V2 A2
in a steady flow process,
 Volume Flow Rates are not
Necessarily the same

V1 A1  V2 A2

This is because
1   2
44
Special Case: Flow rate for Steady
Incompressible Flow ( =constant)
 in   m out
m
 ( VA )in   ( VA )out
 ( VA )in   ( VA )out
For single stream
V1 A1  V2 A2
45
Flow Work (Flow Energy)



Some people call it
“flow energy”
Flow work is defined
as the work required
to push a unit mass of
fluid into or out of a
control volume.
It is equal to Pv
46
Flow Work derivation
The force applied on the fluid by
the imaginary piston is
F  PA
W flow  FL
W flow  PAL  PV
(KJ )
Wf per unit mass is
 w flow  Pv ( KJ / Kg )
47
Total Energy of a Flowing Fluid
Recall that The total energy of a system is
e  u  V / 2  gz
2
For control volume, the fluid entering or leaving possesses
an additional form of energy (the flow work or the flow
energy)
e  Pv  u V / 2  gz
2
e  h  V / 2  gz  
2
The total energy of a flowing fluid
48
Observations

Using h instead of u to represent the energy
of a flowing fluid, results in

Taking care of the energy required to push
the fluid in or out of the control volume.

In fact, this is the main reason behind
introducing the enthalpy.
49
The Total Energy of a Flowing
and non-Flowing Fluid
50
Energy Transport by Mass
The total energy of a flowing fluid of mass m
with uniform properties is
Emass  m
(Amount of energy transport)
 The rate of energy transport by a fluid with a
mass flow rate of m
 is
 mass  m

E

(Rate of energy transport)
 When the KE and PE of a flow stream are
negligible, then

Emass  mh
E mass  m h
51
Example 3-14: Energy transport by mass
Steam is leaving a 4-L pressure cooker whose operating
pressure is 150 Kpa. The liquid in the cooker
decreased 0.6 L in 40 min. The cross sectional area
of the exit is 8 mm2. Determine:




The mass flow rate of steam
The exit velocity
The total and flow energies of the
steam per unit mass.
The rate at which the energy is
leaving the cooker by steam.
steam
Sat. vapor,
x=1
150 kPA
Sat. liquid,
x=0
52
Solution: Energy transport by mass

Assumptions:





The flow exiting the cooker is
steady.
KE and PE are negligible.
Saturation conditions exist all
times. Thus:
(1) The steam leaving the cooker
is saturated vapor (x=1) at the
operating pressure.
(2) The liquid remaining is
saturated liquid (x-0) at the
operating pressure.
steam
Sat. vapor,
x=1
150 kPA
Sat. liquid,
x=0
53