Reversed cycle machines Dr. Tamás Szakács

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Reversed cycle machines
Dr. Tamás Szakács
college senior lecturer
University of Óbuda
Budapest Bécsi út 96/B
H-1034 Budapest, Hungary
[email protected]
Content of the presentation
•
•
•
•
•
Introduction
Reverse cycle machines
Design of the inner cycle
Design of the heating/cooling properties
Examples
Introduction
Reversed cycle machines
• Machine is producing work from introduced
heat. E.g. the idealized Carnot cycle, or the
realized Otto, Diesel or Joule cycles.
• Reversed cycle transforms heat opposite the
natural
direction,
but
requires
work.
For example: Refrigerator, heat-pump.
Introduction
Importance of reversed machines
• Increasing energy consumption of EU countries.
• Peak-loads in summer (originated from air
conditioning).
• Reversed machines can help reducing energy
consumprion even in increasing demands.
Reversed cycle machines
• Coolong (refrigerating)
• Air-conditioning
• Heating
• Drying
Heating by reverse cycle air
conditioning
Reverse cycle air conditioning extracts heat from a
natural heat source (e.g. soil, or the outside air,
even on mid-winter nights) and transfers it inside.
A refrigerant is passed through an external coil,
absorbing the heat. This refrigerant is then
compressed by a compressor into a fan coil unit (or
‘condenser’) inside the home, releasing its heat into
the room.
Heating by reverse cycle air
conditioning
Advantages:
• One of the most economical forms of heating
• Able to provide both heating and cooling
• Have no exposed elements or flames
• Lifetime of up to 20 years
• Filter and dehumidify air
• Can utilize district- and waste heat, heat from
COG, and renewable heat sources.
Reverse cycle air conditioning
Types:
•
•
•
•
•
Portable
Window/wall units
Split systems
Multi-split systems
Ducted systems
Mollier diagram with approximation range and
rated refrigeration cycle
Solar assisted air conditioning
Principle of an absorption chiller
Key Issues for Renewable Heat in Europe (K4RES-H)
Solar Assisted Cooling – WP3, Task 3.5 Contract EIE/04/204/S07.38607
Keep Cool
Solar Cooling
www.energyagency.at
Principle of an absorption chiller
Keep Cool
Solar Cooling
www.energyagency.at
Principle of an absorption chiller
Principle of an adsorption chiller
Key Issues for Renewable Heat in Europe (K4RES-H)
Solar Assisted Cooling – WP3, Task 3.5 Contract EIE/04/204/S07.38607
Principle of an adsorption chiller
Solid desiccant cooling with rotating wheels
Solid desiccant cooling with rotating wheels
Key Issues for Renewable Heat in Europe (K4RES-H)
Solar Assisted Cooling – WP3, Task 3.5 Contract EIE/04/204/S07.38607
Overview of the most common solar assisted air conditioning technologies
Overview of the most common solar assisted air conditioning technologies
Mollier h-x-Diagramm für
feuchte Luft für p=1 bar
Idealized Carnot cycles
Ambient
temperature
S
Cooler
Single purpose S
heat-pump
Multipurpose S
heat-pump
Electric S
powerplant
Calculation example
Example 1
How much is the cooling capacity of a cooler
made of an adiabatic compression and expansion
pistion cycle with ammonia refrigerant?
The temperature of the condensation is 30C, the
evaporation -10C degrees. The compressor is
supplied with dry saturated gas. (x2=1). The
condensation is performed till x3= 0.
3
2
1
4
240
275
1355
1542
h3  275kJ kg
h1  1355kJ kg
h4  240kJ kg
h2  1542kJ kg
Transferred heat in the evaporator by 1 kg
cooling agent:
kJ
q0  h1  h4  1355 240  1115
kg
Mass flow of the circulated cooling agent:
Q0  3600 100 3600
kg
K

 322,9
q0
1115
h
Work required to power the compressor:
kJ
wk  h2  h1  1542 1355 187
kg
Work done by the expansion piston
kJ
we  h3  h4  275 240  35
kg
Work required to keep the cycle up:
w  wk  we  187  35  152
W
COP:
kJ
kg
K  w 322,9 152

 13,6 kWh
3600
3600
q0 1115
 
 7,3
w 152
Supplied volume flow of the compressor:
Vk  K  v1  322,9  0,4  129,2 m3 h
Calculation example
Example 2
The temperature of a m=18000kg air t1=80°C,
moisture content x1=0,03kg/kg. Make a x4 = 0,01
kg/kg dry air by cooling, while keep the outlet
temperature same as t1. How much water has to be
drained? How much heat has to be subtracted for
the cooling and added for the re-heating?
Properties of point 1
t1 = 80 oC
h1 = 160 kJ/kg
x1 = 0,03 kg/kg.
Properties of point 2
t2 = 14 oC
h2 = 141 kJ/kg
x2 = x1
Properties of point 3
t3 = t2 = 14 oC
h3 = 39 kJ/kg
x3 = 0,01
=1
Properties of point 4
t4 = t1 = 80 oC
h4 = 108 kJ/kg
x4 = x3 = 0,01 kg/kg.
Water removed
GH 2O 
m  x2  x3   18000 0,03  0,01  360kg
Heat removed by cooling
Qh 
m  h1  h2   18000 160 41  2,142106 kJ
Heat added by re-warming
Qm 
m  G  h
H 2O
4
 h3   18000 360  108 39
Qm  1,217106 kJ
Thank you for the attention
[email protected]
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