Thermodynamics Lecture
Series
First Law – Control
Volume
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
First Law - Quotes
Quote
“Education is not the piling on of learning, information, data, facts, skills,
or abilities--that's training or instruction--but is rather a making visible
what is hidden as a seed... To be educated, a person doesn't have to know
much or be informed, but he or she does have to have been exposed
vulnerably to the transformative events of an engaged human life... One
of the greatest problems of our time is that many are schooled but few are
educated.”
Author:Thomas Moore
Source:he Education of the Heart by Thomas Moore
Introduction - Objectives
Objectives:
1. State the conservation of mass principle.
2. State the meaning of steady-flow process and
the implications on a system’s properties.
3. Write the unit-mass basis and unit-time basis
(or rate-form basis) energy balance for a general
steady-flow process.
4. Write the unit-time basis (or rate-form basis)
mass balance for a general steady-flow process.
Introduction - Objectives
Objectives:
4. State the assumptions for steady-flow devices
such as nozzles, diffusers, turbines,
compressors, throttle valves, heat exchangers
and mixing chambers.
5. State the purpose for each of the steady-flow
device noted above.
6. Write the unit-mass basis and unit-time basis
(or rate-form basis) energy balance for each of
the steady-flow devices noted above
Introduction - Objectives
Objectives:
7. Write the unit-time basis (or rate-form basis)
mass balance for each of the steady-flow
device.
8. Use the energy and mass balance to solve
problems related to each of the steady-flow
device.
First Law - General
Control Volume
Open System
Steady-flow
Devices
First Law - General
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 – 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 – 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 – Energy Balance
Energy Balance –General
system
Qin – Qout+ Win – Wout+ Emass,in - Emass,out
= U+ KE + PE, kJ
qin – qout+ win – wout + qin – qout
= u+ ke + pe, kJ/kg






Q in  Q out  W in  W out  E mass ,in  E mass ,out



  U   KE   PE , kW
First Law – Stationary Closed
Energy Balance –Stationary
Closed system
Qin – Qout+ Win – Wout+ 0 – 0 = U+ 0 + 0, kJ
qin – qout+ win – wout + 0 – 0 = u+ 0 + 0, kJ/kg





Qin  Q out  W in  W out  0  0  U  0  0, kW
First Law – Mass Balance
Mass
Mass
Change of
Entering - Leaving = system’s
a system
a system
mass
Mass Balance
min – mout = msys, kg or



m in  m out  msys , kg / s
First Law - Mass flow rates
Mass Balance – Mass &
Volume Flow Rate
The volume of
the cylinder is
V  A  m , m 3
The velocity of
the mass is



,m / s
t
The volume flow rate is
3
V

m
m

A  ,
t t
t 3
s


 m
V  m  A,
s
Then, mass flow rate is

A kg
m
,
 s

First Law – Steady - flow
Energy Balance – Control
Volume Steady-Flow
Steady-flow is a flow where all properties
within boundary of the system
remains constant with time
First Law – single stream CV
How to relate
changes to
the cause
Mass in,
State 1
Properties will change
indicating change of
state

E mass ,in  m1 1
System
E1, P1, T1, V1
To
Qin
E1, P1, T1, V1




W in

W out

Mass out,
State 2

E mass ,out  m2 2
Q out
Single Stream CV
First Law – Steady - flow
Energy Balance – Control
Volume Steady-Flow
Steady-flow is a flow where all properties within
boundary of the system remains constant with time

Esys= 0, kJ; esys= 0 , kJ/kg,  E sys  0, kJ/s
Vsys= 0, m3; msys= 0

 m sys  0 , kg/s
First Law – Energy Balance CV
Mass & Energy Balance–
Steady-Flow CV
Mass balance



 msys  0, So, min  mout or

m

in



m
out
, kg/s

Energy balance  E sys  0, So, E in  E out kJ/s


Qin  Win 



 
 m   Qout  W out 

in

 
 m  , kW

out
q  w  in  qout  wout  out , kW
First Law – Energy balance CV
Mass & Energy Balance–SteadyFlow: Single Stream
Mass balance



IN, 1
OUT, 2
 msys  0. So, min  mout , kg/s

Energy balance





 E sys  0. So, E in  E out , kJ/s

Qin  Qout  W in  W out
  
  
  m     m   , kW

 out 
 in
qin – qout+ win – wout = qout – qin, kJ/kg
First Law - Single Stream
Mass & Energy Balance–SteadyFlow: Single Stream

Energy balance




Mass balance m1  m 2  m, kg/s



Qin  Qout  W in  W out  mout  in , kW




Qin  Qout  W in  W out


 mh2  ke2  pe2   mh1  ke1  pe1 , kW

 mh2  h1  ke2  ke1  pe2  pe1 , kW

 mh  ke  pe, kW
First Law Energy balance CV
Mass & Energy Balance–SteadyFlow: Single Stream



Mass balance: m1  m 2  m, kg/s
Energy balance:
qin – qout+ win – wout = qout – qin, kJ/kg
= h2 – h1 + ke2 – ke1 + pe2 – pe1, kJ/kg
= h + ke + pe, kJ/kg
First Law – Energy Balance CV
Mass & Energy Balance–SteadyFlow: Single Stream
Energy balance:
qin – qout+ win – wout = h + ke + pe, kJ/kg
 2 2
where
 
kJ
ke  kefinal  keinitial 
and
2
1
2000
,
kg
pe  pefinal  peinitial  g y 2  y1  , kJ
1000
kg
First Law of – Ideal Gas
Mass & Energy Balance–SteadyFlow CV: Ideal Gases
Use Ideal Gas Equation of State for real
gases that behave like ideal gases. Criteria:
Pgas << Pcrit and/or Tgas >> Tcrit. P = RT
Where  is the specific volume,m3/kg, R is gas
constant, kJ/kgK, T is absolute temperature in Kelvin
For known P and T, use to determine
, and hence the mass flow rate.
First Law of – Ideal Gas
Mass & Energy Balance–SteadyFlow CV: Ideal Gases
Use of property table for real gases that
behave like ideal gases.
Knowing T, read value for h and viceversa. If T or h not found, do interpolation
First Law - Nozzles
Nozzles
Purpose: Increase velocity Effect: Pressure drops



Mass balance m1  m 2  m, kg/s In
Out
State 1
Energy balance





State 2
A2 << A1
Qin  Qout  W in  W out  mout  in , kW





Qin  Qout  W in  W out  mh  ke  pe, kW

0  0  0  0  mh  ke  0 , kW
First Law - Nozzles
Nozzles
Purpose: Increase velocity Effect: Pressure drops




1 A1  2 A2

, kg/s
m1  m 2
Mass balance:

1
2
Energy balance: 0  0  0  0  mh  ke  0 , kW
0  h  ke, kJ/kg 0  h2  h1  ke2  ke1 , kJ/kg
2  2
1   2
h2  h1  ke1  ke2 
, kJ/kg
2000
First Law - Nozzles
Nozzles
Purpose: Increase velocity Effect: Pressure drops
2  2
1   2
, kJ/kg
Energy balance: h2  h1  ke1  ke2 
2000
Air:Use energy balance to find h2 and use table A17 (& interpolation technique) to determine T2.
Air: Use table A-17 to find h1
In
(& interpolation technique)
State 1
for a given T1.
A2 << A1
Out
State 2
First Law - Diffusers
Diffusers
Purpose: Increase Pressure Effect: Velocity drops
2  2
1   2
, kJ/kg
Energy balance: h2  h1  ke1  ke2 
2000
Air:Use energy balance to find h2 and use table A17 (& interpolation technique) to determine T2.
Air: Use table A-17 to
find h1 (& interpolation
technique) for a given T1.
In
State 1
A2 >> A1
Out
State 2
First Law - Turbines
Turbines
Purpose: Produce Work Effect: Pressure

Drops


1 A1  2 A2

, kg/s
Mass balance: m1  m 2
1
2
Energy balance:
qin – qout+ win - wout = h2 – h1+ ke2 – ke1+ pe2 – pe1





Qin  Qout  W in  W out  mh  ke  pe, kW

In

0  0  0  W out  mh2  h1  0  0 , kW
– wout = h2 – h1, kJ/kg
wout = h1 – h2, kJ/kg
Out
First Law - Compressors
Compressors
Purpose: Increase Pressure Sacrifice:
 Work
 supplied


1 A1  2 A2

, kg/s
Mass balance: m1  m 2
1
2
Energy balance:
qin – qout+ win - wout = h2 – h1+ ke2 – ke1+ pe2 – pe1





Qin  Qout  W in  W out  mh  ke  pe, kW Out


0  0  0  W in  mh2  h1  0  0 , kW
win
= h2 – h1, kJ/kg
In
First Law – Increase Pressure
Pressure Increasing Devices
Compressors: Increase Pressure of gas to high P
Fans: Increase Pressure of gas slightly to move air
Pumps: Increase Pressure of liquid to high P
Sacrifice: Work supplied
First Law - Throttle
Throttle
First Law - Throttle
Throttle
Purpose: Reduce Pressure
Effect: Temp drops



1 A1  2 A2

, kg/s
m1  m 2
Mass balance:
1
2
Energy balance:
qin – qout+ win - wout = h2 – h1+ ke2 – ke1+ pe2 – pe1





Qin  Qout  W in  W out  mh  ke  pe, kW
0  0  0  0  h2  h1  0  0 , kJ/kg
h2  h1 , kJ/kg isenthalpic process In
Out
First Law – Mixing Chamber
Mixing Chamber
Purpose: Mixing Mass balance:



m1  m2  m3 , kg/s
Energy balance: qnet,in – wnet,out = q3 – q1 - q2, kJ/kg


(m1  m 2 )h3  ke3  pe3 


 m 2 h2  ke2  pe2   m1h1  ke1  pe1 , kW
ke3  ke2  ke1  ke3  0; pe3  pe2  pe1  pe3  0


Then, m2 ( h3  h2 )  m1 (h1  h3 ), kW
1
3
2
First Law – Heat Exchanger
Heat
Exchanger
Boundary has
2 inlets and
2 exits
First Law – Heat Exchanger
Heat
Exchanger
Boundary has
1 inlet and
1 exit
First Law of Thermodynamics
In, 3
Heat Exchanger
-no mixing
-1 inlet and 1 exit
In, 1
Exit, 2
Exit, 4
First Law of Thermodynamics
In, 3
Heat Exchanger
-no mixing
-2 inlets and 2 exits
In, 1
Exit, 2
Exit, 4
First Law of Thermodynamics
3
Heat Exchanger
Case 1
1
2
4
0  4 - 3  2 - 1




m1  m2 , m3  m4 , kg/s
First Law of Thermodynamics
3
Heat Exchanger
Case 2
1
2




m1  m2 , kg/s
Qout
4
m3  m4 , kg/s
Qnet ,in  0  2 - 1
First Law of Thermodynamics
Heat Exchanger
Energy
balance: Case 1
Purpose: Remove or add heat

0  0  0  0  m 4 h4  ke4  pe4  - m3 h3  ke3  pe3 


 m 2 h2  ke2  pe2   m1 h1  ke1  pe1 , kW
3
Mass




m1  m2 , m3  m4 , kg/s
balance:

m1 h2  h1  ke2  ke1  pe2  pe1 

 m3 h4  h3  ke4  ke3  pe4  pe3 
where ke  pe  0
1
2
4
First Law of Thermodynamics
Heat Exchanger
Mass balance:
Purpose: Remove or add heat




m1  m2 , m3  m4 , kg/s
Energy balance: Case 2




Q in  Q out  0  0  m2 h2  m1 h1 , kW


 Qout  0  0  m2 h2  h1 , kW

where
3
h2  h1  
1
2
m3 h3  h4 

m1
Qout4
First Law of Thermodynamics
Heat Exchanger
Mass balance:
Purpose: Remove or add heat




m1  m2 , m3  m4 , kg/s
Energy balance: Case 2




Q in  Q out  0  0  m2 h2  m1 h1 , kW

Qin

Qin  0  0  0  m2 h2  h1 , kW

where
3
h2  h1  
1
2
m3 h3  h4 

m1
4
First Law of Thermodynamics
Concluding remarks - Quotes
“The illiterate of the 21st century will not be those who cannot read and
write, but those who cannot learn, unlearn, and relearn.”
Author: Alvin Toffler
Source: Lessons from the Art of Juggling; How to Achieve Your Full
Potential in Business, Learning and Life by Michael
Gelb
and Tony Buzan