Transcript First Law – Control Mass - Universiti Teknologi MARA

Thermodynamics Lecture
Series
Second Law – Quality of
Energy-Part1
Applied Sciences Education Research Group
(ASERG)
Faculty of Applied Sciences
Universiti Teknologi MARA
email: [email protected]
http://www5.uitm.edu.my/faculties/fsg/drjj1.html
Quotes
• It does not matter how slowly you go, so
long as you do not stop.
--Confucius
• To be wronged is nothing unless you
continue to remember it.
--Confucius
Symbols
•
•
•
•
•
•
•
•
Q
q
W

V

E


W

Q



V

m
Introduction - Objectives
Objectives:
1. Explain the need for the second law of
thermodynamics on real processes.
2. State the general and the specific statements of
the second law of thermodynamics.
3. State the meaning of reservoirs and working
fluids.
4. List down the characteristics of heat engines.
Introduction - Objectives
Objectives:
5. State the difference between thermodynamic
heat engines and mechanical heat engines.
6. Sketch an energy-flow diagram indicating the
flow of energy and label all the energies and all
the reservoirs for a steam power plant.
7. Sketch a schematic diagram for a steam power
plant and label all the energies, flow of energies
and all the reservoirs.
Introduction - Objectives
Objectives:
8. State the desired output and required input for
a steam power plant.
9. State the meaning of engines’ performance and
obtain the performance of a steam power plant
in terms of the heat exchange.
10.State the Kelvin-Planck statement on steam
power plant.
Introduction - Objectives
Objectives:
11.Sketch an energy-flow diagram indicating the
flow of energy and label all the energies and all
the reservoirs for a refrigerator.
12.Sketch a schematic diagram for a refrigerator
and label all the energies, flow of energies and
all the reservoirs.
13.State the desired output and required input for
a refrigerator.
Introduction - Objectives
Objectives:
14.Obtain the performance of a refrigerator in
terms of the heat exchange.
15.Sketch an energy-flow diagram indicating the
flow of energy and label all the energies and all
the reservoirs for a heat pump.
16.Sketch a schematic diagram for a a heat pump
and label all the energies, flow of energies and
all the reservoirs.
Introduction - Objectives
Objectives:
17.State the desired output and required input for
a a heat pump.
18.Obtain the performance of a a heat pump in
terms of the heat exchange.
19.State the Clausius statement for refrigerators
and heat pumps.
20.Solve problems related to steam power plants,
refrigerators and heat pumps.
Review - First Law
• All processes must obey energy conservation
• Processes which do not obey energy conservation
cannot happen.
• Processes which do not obey mass conservation
cannot happen
Piston-cylinders, rigid tanks
Turbines, compressors,
Nozzles, heat exchangers
Review - First Law
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
Review - First Law
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
Review - First Law
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
Review - First Law
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
Review - First Law
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, Vsys= 0, m3;
msys= 0 or min = mout , kg  m sys  0 , kg/s




m in  mout  0 or m in  mout , kg/s
Review - First Law
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
qin + in + qin = qout+ out+ qout, kJ/kg
Review - First Law
Mass & Energy Balance–SteadyFlow: Single Stream
Mass balance




 m sys  0. So, min  mout  m, kg/s

Energy balance





 E sys  0. So, E in  E out , kJ/s


Qin  Qout  W in  W out  m(out  in ), kW
qin – qout+ in – out = qout – qin, kJ/kg
= hout - hin + keout– kein + peout - pein, kJ/kg
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
-2 inlets and 2 exits
In, 1
Exit, 2
Exit, 4
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
Heat Exchanger
3
1
Case 1
2
Energy balance
0  4 - 3  2 - 1
where  h  ke  pe
4
Mass balance




m1  m2 , m3  m4 , kg/s
First Law of Thermodynamics
Heat Exchanger
3
Case 2
Mass balance
1
2

Qout
4



m1  m2  m, kg/s
Energy balance

Q net ,in  0  m(2 - 1 )
where  h  ke  pe
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 h1  h2  ke1  ke2  pe1  pe2 

 m3 h4  h3  ke4  ke3  pe4  pe3 
where ke  pe  0
1
2
4
First Law of Thermodynamics
Heat Exchanger
Purpose: Remove or add heat





Q in  Q out  0  0  m2 h2  m1 h1 , kW

3

 Q out  0  0  mh2  h1 , kW

where

m1  m2  m, kg/s
Energy balance: Case 2

Mass balance:
h2  h1  
m3 h3  h4 

m1
1
2
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
Second Law
0 -qout+ 0 - 0 = -u = u1 - u2, kJ/kg
• First Law
involves
quantity or
amount of
energy to be
conserved
in processes
Qout
Qout
Tsys,initial=40C
Tsys,final=25C
•OK for
this cup
Tsurr=25C
This is a natural process!!!
Q flows from high T to low T
medium until thermal
equilibrium is reached
Second Law
qin – 0 + 0 - 0 = u = u2 - u1, kJ/kg
Qin
Qin
Tsys,initial=25C
Tsys,final=40C
Tsurr=25C
This is NOT a natural process!!!
Q does not flow from low T to
high T medium. Never will the
coffee return to its initial state.
Second Law
qin – 0 + 0 - 0 = u = u2 - u1, kJ/kg
Qin
Qin
Tsys,initial=25C
Tsys,final=40C
Tsurr=25C
This is NOT a natural process!!!
Q does not flow from low T to
high T medium. Never will the
coffee return to its initial state.
Second Law
• First Law
involves
quantity or
amount of
energy to be
conserved
in processes
qin – 0 + 0 - 0 = u =u2-u1 kJ/kg
Qin
Qin
Tsys,initial=25C
Tsys,final=40C
Tsurr=25C
This is a NOT a natural process!!!
But is the
Q does not flow from low T to
process in this
high T medium. Never will
cup possible??
equilibrium be reached
Second Law
• First Law is not sufficient to determine if a
process can or cannot proceed
Second Law
• First Law is not sufficient to determine if a
process can or cannot proceed
•Introduce the second law of thermodynamics –
processes occur in its natural direction.
Heat (thermal energy) flows from high
temperature medium to low temperature
medium.
Energy has quality & quality is higher with
higher temperature. More work can be
done.
Second Law
Considerations:
•Work can be converted to heat directly &
totally.
•Heat cannot be converted to work directly &
totally.
Requires a special device – heat engine.
Second Law
Heat Engine Characteristics:
•Receive heat from a high T source.
•Convert part of the heat into work.
•Reject excess heat into a low T sink.
•Operates in a cycle.
Second Law
Heat Engines
•Thermodynamics heat engines – external
combustion: steam power plants
Combustion outside system
•Mechanical heat engines – internal
combustion: jets, cars, motorcycles
Combustion inside system
Performance = Desired output / Required input
Second Law
Working fluid:
Water
qin - qout = out - in
High T Res., TH
Furnace
qin = qH
qin = net,out + qout
Purpose:
Produce work,
Wout, out
Steam Power Plant
net,out = qin - qout
qout = qL
Low T Res., TL
Water from river
An Energy-Flow diagram for a SPP
net,out
Second Law
Working fluid:
Water
High T Res., TH
Furnace
qin = qH
Boiler
Pump
Turbin
e
in
out
Condenser
qin - qout = out - in
qout = qL
Low T Res., TL
Water from river
qin - qout = net,out
A Schematic diagram for a Steam Power Plant
Second Law
Thermal Efficiency for steam power plants
desired output  net ,out


requiredinput
qin

 net ,out
q in
q in qout
qin  qout



q in
q in
qin
qout
qL
1 
1 
qin
qH
Second Law
Thermal Efficiency for steam power plants

desiredoutput W net ,out



requiredinput
Q in





W net ,out
Q in  Q out
Q in Q out


   


Q in
Q in
Q in
Q in

1 
Q out

Q in
1 

QL

QH
Second Law
Kelvin Planck Statement for steam power plants
• It is impossible for engines operating in a cycle
to receive heat from a single reservoir and
convert all of the heat into work.
• Heat engines cannot be 100% efficient.
Second Law
Working fluid:
Ref-134a
High T Res., TH, Kitchen
room / Outside house
qout – qin = in - out
qout = qH
Refrigerator/ Air Cond
net,in = qout - qin
net,in = qH - qL
qin = qL
Low Temperature
Res., TL, Inside
fridge or house
net,in
Purpose:
Maintain space
at low T by
Removing qL
An Energy-Flow diagram for a Refrigerator/Air
Second Law
Working fluid:
Refrigerant-134a
High T Res., TH
Kitchen/Outside house
qout = qH
Condenser
Com
pressor
Throttle Valve
in
Evaporator
qin = qL
Low T Res., TL
Ref. Space/Room
A Schematic diagram for a Refrigerator/Air Cond.
Second Law
Coefficient of Performance for a Refrigerator
qin
desiredoutput
COPR 

requiredinput  net ,in
COPR 

qin
 net ,in
qin

qout  qin
1
1
qout
qin

qin
qin

qout
1
qin
Divide top and
bottom by qin

1
qH
1
qL
Second Law
Coefficient of Performance for a Refrigerator

desiredoutput
COPR 

requiredinput

COPR 


Q in


W net ,in
1

Q out

Q in

Q in

W net ,in
Q in


Q out  Q in
1
 

Q in
Q out

Q in

Q in

1
1

QH

QL
1
Second Law
Working fluid:
Ref-134a
High Temperature
Res., TH, Inside house
qout = net,in + qin
qout = qH
Heat Pump
net,in = qout - qin
qin = qL
net,in = qH - qL
Low Temperature
Res., TL, Outside
house
An Energy-Flow diagram for a Heat Pump
Purpose:
Maintain space
at high T by
supplying qH
net,in
Second Law
Working fluid:
Refrigerant-134a
High T Res., TH
Inside house
qout = qH
Condenser
Com
pressor
Throttle Valve
Evaporator
qin = qL
Low T Res., TL
Outside the house
A Schematic diagram for a Heat Pump
in
Second Law
Coefficient of Performance for a Heat Pump
COPHP
COPHP
qout
desired output


requiredinput  net ,in
qout
qout


 net ,in qout  qin

1
qout
qin

qout
qout
1
1


qin
qL
1
1
qout
qH
Second Law
Clausius Statement on Refrigerators/Heat Pump
• It is impossible to construct a device operating
in a cycle and produces no effect other than the
transfer of heat from a low T to a high T
medium.
• Must do external work to the device to make it
function. Hence more energy removed to the
surrounding.
Second Law – Energy Degrade
What is the maximum performance of
real engines if it can never achieve
100%??
Factors of irreversibilities
• less heat can be converted to work
– Friction between 2 moving surfaces
– Processes happen too fast
– Non-isothermal heat transfer
Second Law – Dream Engine
Carnot Cycle
• Isothermal expansion
Slow adding of Q resulting in work done by
system (system expand)
Qin – Wout = U = 0. So, Qin = Wout . Pressure
drops.
• Adiabatic expansion
0 – Wout = U. Final U smaller than initial U.
T & P drops.
Second Law – Dream Engine
Carnot Cycle
• Isothermal compression
Work done on the system
Slow rejection of Q
- Qout + Win = U = 0. So, Qout = Win .
Pressure increases.
• Adiabatic compression
0 + Win = U. Final U higher than initial U.
T & P increases.
Second Law – Dream Engine
Carnot Cycle
P, kPa
P -  diagram for a Carnot (ideal) power plant
qin
1
2
4
3
qout
, m3/kg
Second Law – Dream Engine
Reverse Carnot Cycle
P, kPa
P -  diagram for a Carnot (ideal) refrigerator
1
qout
4
2
3
qi
n
, m3/kg
Second Law – Dream Engine
Carnot Principles
• For heat engines in contact with the same hot
and cold reservoir
All reversible engines have the same
performance.
Real engines will have lower performance
than the ideal engines.
 qH

q
 L

TH (K)



 rev TL (K)
Second Law
Working fluid:
Not a factor
P1: 1 = 2 = 3
High T Res., TH
Furnace
qin = qH
real
Steam Power Plants
P2: real < rev
net,out
qout = qL
rev
Low T Res., TL
Water from river
qL
1
qH
TL (K)
1
TH (K)
An Energy-Flow diagram for a Carnot SPPs
Second Law
Working fluid:
Not a factor
COPHP
High T Res., TH,
Kitchen room / Outside
house
1

qL
1
qH
qout = qH
Rev. Fridge/ Heat Pump
COPHPrev 
1
 qL
1  
 qH
COPHPrev 
1
T
1 L
TH


rev
qin = qL
Low Temperature
Res., TL, Inside
fridge or house
COPR 
1
qH
1
qL
net,in
1
COPRrev 
 qH 

  1
 qL rev
1
COPR rev 
TH
1
TL
An Energy-Flow diagram for Carnot Fridge/Heat Pump