Transcript Introduction - CISAT Sharepoint

ISAT 413 - Module III: Building Energy Efficiency
Topic 3:
Refrigeration, Heat Pumps and
Air Conditioning
 Evaporating Cooling
 Vapor-Compression
Refrigeration
 Vapor-Absorption
Refrigeration
 Other Refrigeration
Systems
 The Heat Pump
 Air Conditioning
1
 Evaporative Cooling (Eastop, Example 4.8)
In a cooling tower the water to be cooled enters near the
top of the tower and flows down through packing in
counter flow to an air stream as shown
diagrammatically in Fig. 4.34. Air is induced from the
base of the tower by a fan mounted on the top of the
tower. Using the data given below and neglecting heat
losses and pressure losses, calculate:
(i) the required mass flow rate of air induced by the fan
for the given cooling duty;
(ii) the mass flow rate of make-up water required.
2
Cooling Tower
(Evaporative Cooling)
3
Data:
Water: Mass flow rate, 15 kg/s; temperature of water at
entry, 27oC; temperature of water at exit, 21oC; mean
specific heat, 4.18 kJ/kg.K.
Air: Ambient conditions, 1,01325 bar, 23oC dry bulb (DB),
17oC wet bulb (WB); air at exit, saturated with water vapor
at 25oC; mean specific heat of dry air, 1.005 kJ/kg.K; fan
air power input, 5 kW.
Assumptions: For air at standard atmospheric pressure it
can be assumed that the vapor pressure is given by:
For water vapor at low vapor pressure it is a good
approximation to assume that the enthalpy is equal to the
saturated value at the same dry bulb temperature.
ps bar  PgWB bar  6.748104  t DB  tWB K 
4
(i) the required mass flow rate of air induced by the
fan for the given cooling duty
From T ABLEA - 4 (Cengel pg. 904)
Ps 2  P
o  1 3.169  3.169kPa
sat @ 25 C
0.622Ps 2
0.622 3.169kPa
kg water

 0.02008
P2  Ps 2 101.325 3.169kPa
kg dry air
For theair at intake
2 
Ps1  Pg @ TWB  6.748 10 2  TDB  TWB 
 1.935 6.74810 2  23  17  1.5301kPa
T herefore
0.622Ps1
0.622 1.5301
kg water
1 

 0.009537
P1  Ps1  101.325 1.5301
kg dry air
5
Dry air mass balance: m a1  m a 2  m a .
(1)
Watermass balance: m w  1m a1  m 3  2 m a 2
(2)
 m 3  m w  2  1 m a
Energy balance: m w hw  m a1h1  m a 2 h2  m 3h3
(3)
 m w hw  m a h1  m a h2  m w  2  1 m a h3 (3)
kJ
kJ
Knowing m w  15 kg / s , and hw  h

113
.
1
;
h

88
.
05
w3
f @ 27o C
kg
kg
kJ
h1  ha1  1hg1  1.005 23  0.009537 2544.3  47.38
kg dry air
kJ
h2  ha 2  2 hg 2  1.005 25  0.02008 2547.2  76.27
kg dry air
We can solvefor m a  13.62 kg / s from Equation(3)
T herefore,mass flow rateinduced by thefan
m a  2 m a  1  2 m a  1.02008m a  13.89 kg/s
(ii) and themass flow rateof make- up water required
m makeup  2  1 m a  0 .01054m a  0 .144 kg/s
6
Vapor-Compression Refrigeration
P-h Diagram of an
Ideal Vapor-Compression Refrigeration Cycle
7
Schematic and T-s Diagram for
Ideal Vapor-Compression Refrigeration Cycle
8
Schematic and T-s Diagram for
Actual Vapor-Compression Refrigeration Cycle
9

The performance of refrigerators and heat
pumps is expressed in terms of coefficient of
performance (COP), defined as
10

The standard of comparison for refrigeration
cycles is the reversed Carnot cycle. A
refrigerator or heat pump that operates on the
reversed Carnot cycle is called a Carnot
refrigerator or a Carnot heat pump, and their
COPs are
11

The transfer of heat from lower temperature
regions to higher temperature ones is called
refrigeration. Devices that produce
refrigeration are called refrigerators, and the
cycles on which they operate are called
refrigeration cycles. The working fluids used
in refrigerators are called refrigerants.
Refrigerators used for the purpose of heating
a space by transferring heat from a cooler
medium are called heat pumps.
12

The most widely used refrigeration cycle is the
vapor-compression refrigeration cycle. In an
ideal vapor-compression refrigeration cycle,
the refrigerant enters the compressor as a
saturated vapor and is cooled to the saturated
liquid state in the condenser. It is then
throttled to the evaporator pressure and
vaporizes as it absorbs heat from the
refrigerated space.
13

Very low temperatures can be achieved by
operating two or more vapor-compression
Systems in series, called cascading. The COP
of a refrigeration system also increases as a
result of cascading.
14

Another way of improving the performance of
a vapor-compression refrigeration system is
by using multistage compression with
regenerative cooling. A refrigerator with a
single compressor can provide refrigeration at
several temperatures by throttling the
refrigerant in stages. The vapor-compression
refrigeration cycle can also be used to liquefy
gases after some modifications
Q:
Cycles? Components? Working Fluids?
15
Working Fluids – Refrigerants
The designation of CFCs (ChloroFluoroCarbons)
is by a two or three digit number:
• The digit on the right represents the number of
fluorine (F) atoms;
• The middle digit represents one more than the
number of hydrogen (H) atoms;
• The digit on the left represents one less than the
number of carbon (C) atoms and omitted when
there is only one carbon atom.
CHClF2 (R22) or called HCFC 22
16
Some Common CFC, HCFC Refrigerants:
CCl3F (R11), CCl2F2 (R12), and CHClF2 (R22), HFC 134a,
CCl2FCClF2 (R113) and CClF2CClF2 (R114)
R502 which is an azeotropic mixture (acts as a single
substance and cannot be separated into its
components by distillation.) of R22 and R115
The commonly accepted designation for refrigerants
other than halocarbons (compounds containing carbon,
fluorine and chlorine in varying proportions, known as
CFCs) is to put the digit 7 in front of the relative
molecular mass (e.g. NH3 (R717), CO2 ( R744).)
(Notice that researches are now underway to develop
acceptable alternatives.)
17
Vapor-Absorption Refrigeration
Ammonia Absorption Refrigeration Cycle
18

Another form of refrigeration that becomes
economically attractive when there is a source of
inexpensive heat energy at a temperature of 100 to
2000C is absorption refrigeration, where the
refrigerant is absorbed by a transport medium and
compressed in liquid form. The most widely used
absorption refrigeration system is the ammoniawater system, where ammonia serves as the
refrigerant and water as the transport medium. The
work input to the pump is usually very small, and
the COP of absorption refrigeration systems is
defined as
19

The maximum COP an absorption refrigeration
system can have is determined by assuming
totally reversible conditions, which yields
where T0, TL, and Ts are the absolute
temperatures of the environment, refrigerated
space, and heat source, respectively.
20
 Other Refrigeration Systems
Schematic of Simple
Thermoelectric Power
Generator
21
10-13
A Thermoelectric Refrigerator

(Fig. 10-28)
22

A refrigeration effect can also be achieved
without using any moving parts by simply
passing a small current through a closed
circuit made up of two dissimilar materials.
This effect is called the Peltier effect, and a
refrigerator that works on this principle is
called a thermoelectric refrigerator.
23
The Heat Pump
Refrigerator and Heat Pump Objectives
The objective of
a refrigerator is
to remove heat
(QL) from the
cold medium;
the objective of a
heat pump is to
supply heat (QH)
to a warm
medium.
24
Heat Pump
Heats a House in Winter
and Cools it in Summer
25
 Air Conditioning (Eastop, Example 4.11)
An air conditioning plant is designed to maintain a
room at 20oC, percentage saturation 50%, which an
air supply to the room of 1.8 kg/s at 14oC,
percentage saturation 60%. The design outside air
conditions are 27oC,percentage saturation 70%. The
plant consists of a mixing chamber for re-circulated
and fresh air, a cooling coil supplied with chilled
water, heating coil, and a supply fan. The ratio of recirculated air to fresh air is 3; the cooling coil has
an apparatus dew point of 5oC, and the refrigeration
unit supplying the chilled water has an overall
coefficient of performance 2. Neglecting all losses
and fan and pump work, calculate:
26
(i) the total air conditioning load for the room;
(ii) the required total energy input;
(iii) the required energy input if the energy to the
heating coil is supplied from the refrigeration plant
condenser cooling water.
27
(i) the total air conditioning load for the room;
h
kJ
 29.23
kg
h
kJ
 38.8
kg
1 T1 14o C ,60% 


2 T2  20o C ,50% 


Total load on room
kg
kJ
 m h2  h1   1.8 38.8  29.23
 17.2kW
s
kg
28
(ii) the required total energy input;
kJ
h
 68.0
o

3T3  27 C,70% 
kg


3h2  h3
kJ
h4 
 46.1
(Mixing 2 and 3 into4)
4
kg
h
5T5 8o C.,5 1 


kJ
Mixing 4 and A into5
 22.7
kg
Coolingload from the chilledwater is
 h4  h5   1.8 kg 46.1  22.7  kJ  42.1kW
m
s
kg
 h1  h5   W comp  32.93kW
Total energyinputis m
29
(iii) the required energy input if the energy to the
heating coil is supplied from the refrigeration plant
condenser cooling water.
Heat rejectedto coolingwater is
42.1 



Wcooling load  Wcomp   42.1 
  63.15kW
2 

Total energyinput woul
d be just thepowerinput tothe
compressor, i.e, 21.05kW if the energy tothe heatingcoil
is suppliedfrom the refrigeration plantcondensercooling
water.
30