Thermodynamics Lecture
Series
First Law of
Thermodynamics &
Energy Balance – Control
Mass, Open System
Applied Sciences Education Research Group
(ASERG)
Faculty of Applied Sciences
Universiti Teknologi MARA
email: drjjlanita@hotmail.com
http://www5.uitm.edu.my/faculties/fsg/drjj1.html
Quotes
Quotes
"One who learns by finding out has
sevenfold the skill of the one who learned
by being told.“ - Arthur Gutterman
"The roots of education are bitter, but
the fruit is sweet." -Aristotle
Research Findings
Research Findings – Retention % of
Learning After 3 days period
Read only – 10%
See only 30%
Hear only – 20%
See + hear – 50%
Say only – 70%
Say & do simultaneously - 90%
CHAPTER
4
The First Law of
Thermodynamics
Introduction
Objectives:
1. State the conservation of energy principle.
2. Write an energy balance for a general system
undergoing any process.
3. Write the unit-mass basis and unit-time basis
(or rate-form basis) energy balance for a general
system undergoing any process.
4. Identify the energies causing the system to
change.
Introduction
Objectives:
5. Identify the energy changes within the system.
6. Write the energy balance in terms of all the
energies causing the change and the energy
changes within the system.
7. Write a unit-mass basis and unit-time basis (or
rate-form basis) energy balance in terms of all
the energies causing the change and the energy
changes within the system.
Introduction
Objectives:
8. State the conditions for stationary, closed
system and rewrite the energy balance and the
unit-mass basis energy balance for stationaryclosed systems.
9. Apply the energy conservation principle for a
stationary, closed system undergoing an
adiabatic process and discuss its physical
interpretation.
Introduction
Objectives:
10.Apply the energy conservation principle for a
stationary, closed system undergoing an
isochoric, isothermal, cyclic and isobaric
process and discuss its physical interpretation.
11.Give the meaning for specific heat and state its
significance in determining internal energy and
enthalpy change for ideal gases, liquids and
solids.
12.Use the energy balance for problem solving.
Energy Transfer -Heat Transfer
P = 100 kPa
T = 99.6 C
Oven
qin
Nasi
Lem
Lemak 20C
200C
SODA
5C
qout
Sat. Vapor
H2O:
Sat. Liq.
qin
3-1
Qin
25C
What happens to
the properties of
the system after the
energy transfer?
Example: A steam power cycle.
Steam Power Plant
Combustion
Products
Steam
Turbine
Required Qin
Fuel
input
Heat
Exchanger
Win
Pump
Wout
Qout
Air
Mechanical Energy
to Generator
Cooling Water


Desired
output
System Boundary
for Thermodynamic
Analysis

The net work output is W net ,out  W out  W in , kW
Energy Transfer – Work Done
Mechanical work:
Piston moves up
Boundary work is
done by system
Electrical work is done on system
Wpw,in ,kJ
H2O:
Sat. liquid
H2O:
Super
Vapor
No heat transfer
T increases
after some time

i
W e  Vi
We,in = Vit/100, kJ
Voltage, V
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
FIGURE 4-46
Pipe or duct flow may
involve more than one
form of work at the
same time.
4-8
First Law – Energy Transfer
Can it change?
How? Why?
System’s initial total
energy is
E1= U1+KE1+PE1 or
e1= u1+ke1+pe1, kJ/kg
System
Total energy
E1
System in thermal
equilibrium
First Law – Energy Transfer
Movable boundary
position gone up
System
expands
E1= U1+KE1+PE1
System,
E1
System
A change has taken
place.
First Law – Energy Transfer
Movable boundary
position gone up
System
expands
E1= U1+KE1+PE1
Initial
System
System,
E1
Final
System
System’s final energy is E2=U2+KE2+PE2
A change has taken place
First Law – Energy Transfer
How to relate
changes to
the cause

Q in , kW

qin, or Qin
qout, or, Qout
Q out , kW
Properties will change
indicating change of
state
System
E1, P1, T1, V1
To
E2, P2, T2, V2
Heat as a cause
(agent) of change
First Law – Energy Transfer
How to relate
changes to
the cause
Properties will change
indicating change of

state
W in , kW
System
E1, P1, T1, V1
To
E2, P2, T2, V2
Work as a cause
(agent) of change
Win, in, kJ/kg
Wout, in, kJ/kg

W out , kW
First Law – Energy Transfer
How to relate
changes to
the cause
Mass
in


E mass,in  (m  ) in , kW
 in , kJ / kg
Properties will change
indicating change of
state
System
E1, P1, T1, V1
To
  
 m  ,k W

 out
E2, P2, T2, V2
Mass
out
 out , kJ / kg
Mass transfer as a
cause (agent) of
change
First Law – Energy Transfer
How to relate
changes to
the cause
Mass
in
Qin
Qout
Properties will change
indicating change of
state
System
E1, P1, T1, V1
To
E2, P2, T2, V2
Dynamic Energies
as causes (agents)
of change
Win
Wout
Mass
out
First Law-Conservation of Energy Principle
Energy must be conserved in any process.
Energy cannot be created nor destroyed. It
can only change forms. Total Energy
before a process must equal total energy
after process
In any
E=U+KE+PE = U+PE
z =h
process,
every bit
of energy
z =h/2
should be
accounte
z =0
d for!!
Known as Conservation of Energy Principle
First Law-Conservation of Energy Principle
Energy must be conserved in any process.
Energy cannot be created nor destroyed. It
can only change forms. Total Energy
before a process must equal total energy
after process
In any
z =h
E=U+KE+PE=U+0+PE
process,
every bit
of energy
z =h/2
E=U+KE+PE
should be
accounte
z =0
d for!!
Known as Conservation of Energy Principle
First Law-Conservation of Energy Principle
Energy must be conserved in any process.
Energy cannot be created nor destroyed. It
can only change forms. Total Energy
before a process must equal total energy
after process
2
2
In any

ke




f
i
z =h
process,
pe  z f  zi
every bit
of energy
z =h/2
E=U+KE+PE
u  T
should be
accounte
E=U+KE+0
z =0
d for!!
Known as Conservation of Energy Principle
First Law Energy Balance
Energy
Energy
Change of
Entering - Leaving = system’s
a system
a system
energy
Energy Balance
Amount of energy causing
change must be equal to amount
of energy change of system
First Law of Thermodynamics
Energy
Energy
Change of
Entering - Leaving = system’s
a system
a system
energy
Energy Balance
Ein – Eout = Esys, kJ or
ein – eout = esys, kJ/kg or



E in  E out  E sys , kW
First Law of Thermodynamics
How to relate
changes to
the cause
Mass in
Qin
Qout
Properties will change
indicating change of
state
System
E1, P1, T1, V1
To
E2, P2, T2, V2
Dynamic Energies
as causes (agents)
of change
Win
Wout
Mass
out
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
FIGURE 4–7
The energy change of
a system during a
process is equal to the
net work and heat
transfer between the
system and its
surroundings.
4-1
First Law – Interaction Energies
Energy Balance – The
Agent
Ein = Qin+Win+Emass,in ,kJ
ein = qin+ in+ qin, kJ/kg




E in  Q in  W in  E mass ,in ; kW
For Closed system: Emass,in = 0, kJ, qin= 0, kJ/kg

E mass ,in  0 , kW
First Law - Interaction Energies
Energy Balance – The
Agent
E out = Q out +W out +Emass,out ,kJ
eout = qout+ out+ qout, kJ/kg




E in  Qout  W out  E mass ,out ; kW
For Closed system: Emass,out = 0, kJ, qout= 0, kJ/kg

E mass ,out  0, kW
First Law - System’s Energy
Energy Balance – The
Change WIthin
Energy change within the system, Esys = E2-E1
is the sum of
Internal energy change, U = U2 – U1
kinetic energy change, KE = KE2 – KE1
potential energy change, PE = PE2 – PE1
First Law – Energy Change
Energy Balance – The
Change WIthin
Esys = U+KE+PE, kJ
esys = u+ke+pe, kJ/kg




 E sys  U   KE   PE , kW
For Stationary system: KE=PE = 0, kJ
ke=pe=0, kJ/kg


 KE   PE  0
First Law – General Energy Balance
Energy
Energy
Change of
Entering - Leaving = system’s
a system
a system
energy
Energy Balance
Ein – Eout = Esys, kJ or
ein – eout = esys, kJ/kg or



E in  E out  E sys , kW
First Law – General Energy Balance
Energy Balance –General
system
Qin + Win + Emass,in – Qout – Wout - Emass,out
= U+ KE + PE, kJ
qin + in + qin – qout – out – qout
= u+ ke + pe, kJ/kg






Q in  W in  E mass,in  Q out  W out  E mass,out



 U   KE   PE, kW
First Law – Stationary System
Energy Balance –Stationary
system
Qin – Qout+ Win – Wout+ Emass,in - Emass,out = U+0+0
qin – qout+ in – out+ qin - qout = u + 0 + 0, kJ/kg






Q in  Q out  W in  W out  E mass,in  E mass,out

 U  0  0
First Law – Closed System
Energy Balance –Closed
system
Qin – Qout+ Win – Wout+ 0 - 0
= U+ KE + PE, kJ
qin – qout+ in – out + 0 – 0
= u+ ke + pe, kJ/kg




Q in  Q out  W in  W out  0  0



 U   KE   PE, kW
First Law – Stationary & Closed
Energy Balance –Stationary
Closed system
Qin – Qout+ Win – Wout+ 0 - 0 = U + 0 + 0
qin – qout+ in – out+ 0 - 0 = u + 0 + 0, kJ/kg





Q in  Qout  W in  W out  0  0  U  0  0
First Law – Historical Perspective
Energy Balance –Stationary
Closed system-History
(Qin – Qout) + (Win - Wout) = (Qin – Qout) - (Wout - Win)
= Qnet,in – Wnet,out
=Q–W
q –   qnet,in – net,out = (qin –qout) – (out – in)
First Law – Adiabatic Process
Energy Balance Stationary
Closed system - Special
Adiabatic: 0 – 0 + Win – Wout+ 0 - 0 = U + 0 + 0
0 – 0 + in – out+ 0 - 0 = u + 0 + 0, kJ/kg
in = elec + pw + b,compress and out = b,expand kJ/kg
First Law – Adiabatic Process
Energy Balance Closed
Stationary system - Special
Adiabatic: in – out = u + 0 + 0, kJ/kg
Spontaneous Expansion: pistoncylinder device
System
System
in = 0, and out = b,expand kJ/kg
Work is expansion work: 0 - out = -b,expand = u < 0
u2 < u1. Final u is smaller than initial u, T drops
First Law – Adiabatic Process
Energy Balance Closed
Stationary system - Special
Adiabatic: in – out - 0 = u+0+0, kJ/kg
Compression: piston-cylinder device
out = 0, and in = b,compress kJ/kg
System
System
Work is compression work: in = b,compress = u > 0
u2 > u1. Final u is bigger than initial u; T increases
First Law – Cyclic Process
Energy Balance Closed
Stationary system - Special
Cyclic: Qin – Qout+ Win – Wout+ 0 - 0 = Esys = 0
qin – qout+ in – out+ 0 - 0 = esys = 0, kJ/kg
qin - qout = out – in
or qnet,in = net,out
Expansion: qin - 0 =out = b,expand
All heat absorbed is used to do expansion work
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
FIGURE 3-22
The net work
done during a
cycle is the
difference
between the
work done by
the
system and the
work done on
the system.
Work done - Cyclic process
Total work is area of A minus area
of B. Total work is shaded area
Input power

W
W b,in  , kW
t
f
b   Pd
i
f
Wb   PdV
i
3-4
Output power

W
W b ,out  , kW
t
First Law – Cyclic Process
Energy Balance Closed
Stationary system - Special
Cyclic: Qin – Qout+ Win – Wout+ 0 - 0 = Esys = 0
qin – qout+ in – out+ 0 - 0 = esys = 0, kJ/kg
qin - qout = out – in
Compression: 0 - in = -b,compress = qin –
qout
All compression work is removed as heat
= - qout
First Law – Isochoric Process
Energy Balance Closed
Stationary system - Special
Isochoric : Q – Q + W – W + 0 - 0 = U
in
out
in
out
Rigid Tank
qin – qout+ in – out+ 0 - 0 = u, kJ/kg
Since, in - out = others + 0 = elec + pw+ 0 + 0
Then, qin - qout + elec + pw+ 0 + 0 = u
First Law - Isobaric Process
Energy Balance Closed
Stationary system - Special
Isobaric : Qin – Qout+ Win – Wout+ 0 - 0 = U
Piston-Cyl
qin – qout+ in – out+ 0 - 0 = u, kJ/kg
For expansion, in - out = elec + pw - b,expand
qin – qout+ elec + pw + 0 - 0 = b,expand + u, kJ/kg
First Law - Isobaric Process
Energy Balance Closed
Stationary system - Special
Isobaric
qin – qout+ in – out+ 0 - 0 = u, kJ/kg
expansion
qin – qout+ elec + pw + 0 - 0 = b,expand + u, kJ/kg
q in  qout   elec   pw 
final
2
initial
1
 Pd  u  P  d  u
q in  qout   elec   pw  P2 2  P1 1  u2  u1
First Law - Isobaric Process
Energy Balance Closed
Stationary system - Special
Isobaric expansion
q in  qout   elec   pw 
final
2
initial
1
 Pd  u  P  d  u
q in  qout   elec   pw  P2 2  P1 1  u2  u1
q in  qout   elec   pw  P2 2  u 2  P1 1  u1 
q in  qout   elec   pw  h2  h1  h
First Law - Closed System
Energy Balance Closed
Stationary system
qin – qout+ in – out+ 0 - 0 = u, kJ/kg
For pure substances, use property table and
mathematical manipulations to determine u2 and
u1. Then u = u2 – u1.
And h2 and h1, . Then h = h2 – h1.
First Law –Specific Heat
Energy Balance Closed
Stationary system C  ,
Cp,
Specific heats to find U and H
Specific Heat Capacity
C  at constant volume, Cp, at constant pressure
Amount of heat necessary to increase temperature of a
unit mass by 1K or 1degree Celcius
First Law –Specific Heat
Energy Balance Closed
Stationary system C  ,
Cp,
For Ideal Gases: Estimate internal energy change and
enthalpy change
Assume smooth change of C with T, & approximate to be
linear over small T (approx. a few hundred degrees)
2
u   CV dT  CV ,avg T2  T1   CV ,avg T
1
2
h   C P dT  C P ,avg T2  T1   C P ,avg T
1
kJ
,
kg  K
First Law –Specific Heat
Energy Balance Closed
Stationary system C  ,
Cp,
For Ideal Gases : Estimate internal energy change and
enthalpy change
Assume smooth change of C with T, & approximate to be
linear over small T (approx. a few hundred degrees)
C , avg is determined using interpolation technique & use
Table A-2b
First Law –Specific Heat
Energy Balance Closed
Stationary system C  ,
Cp,
For solids & Liquids: May consider as incompressible or
constant volume
Cv = Cp = C , kJ/kg K ,
Enthalpy h = u + P,
Hence the enthalpy
change,
u = Cav (T2 - T1), kJ/kg
So, dh =du + dP +Pd, kJ/kg
h = u + P + P, kJ/kg
h = Cav T + P + P, kJ/kg
First Law –Specific Heat
Energy Balance Closed
Stationary system C  ,
Cp,
For solids & liquids: u = Cav (T2 - T1), kJ/kg
Cv = Cp = C , kJ/kg K ,
For solids: Enthalpy, h = Cav T + P + 0, kJ/kg
Since, P = 0, Then, h  Cav T, kJ/kg
First Law –Specific Heat
Energy Balance Closed
Stationary system C  ,
Cp,
For Liquids: u = Cav (T2 - T1), kJ/kg
Enthalpy, h = Cav T + P + 0, kJ/kg
Heaters where P = 0,
h = u  Cav T, kJ/kg
Pumps where T = 0,
h = P, kJ/kg
Or written as h2 - h1= (P2 - P1)
First Law –Example – Prob 4-20
Initial: V = 5 L
phase sat. liq. P = 150 kPa,
H2O
P1= P2
t = 45 min x 60s/min
t = 45 x 60s
Current, i = 8 A,
i
Wpw,in ,kJ
Voltage, V
Wpw,in = 300 kJ,
Final phase is sat. liq.-vapor mix at 150 kPa. Quality of
steam is 0.5. Since mg = m/2, hence x = mg/m = m/2m
=0.5.
First Law –Example – Prob 4-20
Initial: V = 5 L
phase sat. liq. P = 150 kPa,
Wpw,in ,kJ
Voltage, V
Wpw,in = 300 kJ,
H2O
t = 45 x 60s
Current, i = 8 A,
i
P1= P2
t = 45 min x 60s/min
1= f@150 kPa, = 0.0010528 m3/kg
x2 = 0.5
h1= hf@150 kPa, = 467.1 kJ/kg
2= [f + x2vfg]@150 kPa, = 0.579938 m3/kg
h2= [hf + xhfg]@150 kPa =1580 kJ/kg
First Law –Example – Prob 4-20
Energy balance,
Ein - Eout = Esys
0 + Win + 0 – 0 – Wout - 0 = U+ 0 + 0, kJ
2
Wpw,in + We,in= U + Wout
where Wb ,out 
 PdV
1
Then We ,in  U 2  U1  P2V2  P1V1  W pw,in
We,in  U 2  P2V2  (U1  P1V1 )  W pw,in
We,in  H 2  H1  W pw,in  H  W pw,in
First Law –Example – Prob 4-20
Energy balance,
Ein - Eout = Esys
Since We ,in  H 2  H1  W pw,in  H  W pw,in
Vit
Electrical work done is
 mh  W pw,in , kJ
1000
1000 (mh  W pw,in )
V
, kJ
Voltage source is
it
First Law –Example – Prob 4-20
Energy balance,
Ein - Eout = Esys
Voltage source is
5 L *10-3 m3 / L
1000 [(
)(1580.36  467.11) kJ / kg  300 kJ ]
3
0.0010528 m /kg
V
(8 A)( 45 min x60 s / min)
Note that the unit kJ/s = Volts-Ampere or VA
230.9 kJ
AV
V 
 230.9
 230.9 V
As
A