# Thermal Cycles - Rankine Cycle with Reheat - Download Now PowerPoint

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```					     Thermodynamics

First Law Efficiencies
Efficiency(Performance) =

"What You Get (useful )"
"What you pay for "

1
Thermodynamics
  Wnetout / Qin
Note for an heat engine:

Qin                           Wnet out

Wnet  Qout  Qin
Qout         or Wnet  Qin  Qout
Qin  Qout      Qout
             1
Qin          Qin
2
Thermodynamics
Example:

70 kW (238,840 btu/h)

Qin                    Wnet out
400,000 btu/h

Qout = Qin – W net out
Qout
  Wnetout / Qin          Qout = 400,000 –
  238,840/400,000        238,840

= .597 or 60%              = 161,160 btu/h          3
Thermodynamics
We may be able to convert some of Qout to additional
work if the quality (temperature) of the energy is high
enough.
Qin                                 Wnet out

Qout

Wnet out

Called “Bottoming Cycle” on a power
Qout       plant.

4
Thermodynamics
Let’s Talk Refrigeration Cycle

Qout          (yes, Ingley likes air conditioning stuff)
Useful _ effect   
Qin
COPR                   

Re quired _ input Wnet

Condenser                                 or

Win

Qin
Compressor       COPR       
Qout  Qin

Not a typical efficiency!! Can be
Throttling Device    Evaporator            greater than 1. More on refrigeration
cycles later.
Q’in
5
Thermodynamics
Heat Pump Cycle                             Useful _ effect   
Qout
COPHP                   
Q’out                                               
Re quired _ input Wnet

or

Qout
W’in   COPHP          COPR  1
Condenser                          Qout  Qin

Compressor

What are the sources and sinks for
Throttling Device    Evaporator       these two cycles? (thermal reservoirs)

Q’in
6
Thermodynamics
All of these efficiencies we have talked about can be applied to ideal cycles,
perfect cycles or actual cycles with assumptions. Let’s talk about “perfect”
processes for a minute.
Reversible and Irreversible Processes
Reversible Process-a process that can be reversed without leaving any trace
on the surroundings.
• Do not exist in the real world.
• They are theoretical limits for the corresponding irreversible (real) process.
• We learn how to make real processes more efficient by analyzing reversible
theoretical cycles.
Irreversibilities:
friction
unrestrained expansion of gases
heat transfer                                                          7
Thermodynamics
Carnot Principle
Carnot Cycle: the Carnot Heat Engine
• The efficiency of this heat engine would be the best we could expect to
achieve with a heat engine.
• Reversing the Carnot heat engine provides us with the Carnot refrigeration
cycle.
Remember from an earlier slide:

Qin  Qout      Qout
             1
Qin          Qin

8
Thermodynamics
Let’s look at heat transfer out as being the heat transferred to a low temperature
sink (reservoir) and we’ll look at the heat transfer in as being the heat transferred
from a high temperature reservoir:

Qin  Qout      Qout      QL
             1       1
Qin          Qin       QH
If we now look at the discussion on absolute temperatures in your text, we see
that for the Carnot Cycle QL=TL and QH = TH. Therefore,

TL TH  TL
Carnot    1    
TH   TH

Note: must use absolute temperatures!!!                                   9
Thermodynamics
For the refrigeration cycle and heat pump cycle where we use COP in lieu of
efficiency:

1        TL  1
Carnot    COPR               
QH
1
TH
1   TH  TL
QL      TL

1       1     TH
Carnot    COPHP                     
QL      TL TH  TL
1       1
QH       TH
10
Thermodynamics
Let’s look at an example for a heat engine working between a reservoir at 800
deg F and a reservoir at 85 deg F

TL TH  TL
Carnot    1    
TH   TH

TH = 800 + 460 = 1260 R
TL = 85 + 460 = 545 R

TL TH  TL 1260  545
Carnot    1                      0.57 or57 %
TH   TH       1260

Remember: this is the best efficiency we can expect operating between these
two reservoir temperatures.
11
Thermodynamics
Let’s look at an example for a refrigeration device that is cooling a building with
air at 75 F and rejecting heat to outside air at 95 F:

TH = 95+460 = 555R
TL = 75 + 460 = 535R

1          TL
1        535
COPR                                 26 .75
 1 TH  TL 555  535
QH      TH
1
QL      TL

In the real world, we are lucky to get a COP = 4.0
Note: Some engineers use the refrigerant evaporator temperature and
condenser temperature to calculate COP.

12

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