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A pplied Thermodynamics. 6. REFRIGERATION Definition Refrigeration is the process of providing and maintaining temperature of the system below that of the surrounding atmosphere. - PowerPoint PPT Presentation
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1
Applied Thermodynamics
2
6. REFRIGERATION Definition Refrigeration is the process of providing and
maintaining temperature of the system below that of the surrounding atmosphere.The refrigeration effect can be accomplished by non – cyclic processes, making use of substances at temperature well below the temperature of the surroundings – e.g., ice, snow, dry ice (solid CO2) etc.
However, of greater importance are cyclic refrigeration systems, wherein the cooling substance (called refrigerant) is not consumed and discarded, but used again and again in a thermodynamic cycle.
3
REFRIGERATION
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A ton of refrigeration is defined as the quantity of heat required to be removed to produce one ton (1000kg) of ice within 24 hours when the initial condition of water is 00C.
Consider a refrigerator of T tons capacity, Refrigeration capacity = 3.5 kJ/sHeat removed from refrigerator = Refrigeration effect
=R.E. kJ/sPower of the compressor =work/kg of refrigerant x mass flow rate
kJ/s 5.3360024
3351000ionrefrigerat of xxTon
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Reversed Heat Engine Cycle A reversed heat engine is a potential refrigerating machine. It receives heat from a low temperature region at T2, discharge heat to a high temperature region at T1, and requires a net inflow of work. Removal of heat from a low temperature region reduces the temperature of that region below the temperature of the surroundings, thus producing refrigeration.
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According to First Law Q2 – Q1 = -W i.e., Q1 = Q2 + W Such a device is called a Refrigerator or Heat Pump,
depending on whether the focus is on heat received from the low temperature region Q2 or the heat discharged to the high temperature region Q1. Q2 is known as the refrigeration effect.The performance of a refrigerator/heat pump is measured by means of its coefficient of performance (COP). COP of a refrigeration/heat pump is defined as
The working fluid in a refrigeration cycle is called a Refrigerant.
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Important application of Refrigeration1. Ice plants2. Food processing units and transportation, including
dairies3. Industrial air – conditioning4. Comfort air – conditioning5. Chemical and related industries.6. Hospitals.7. Laboratories.8. Domestic applications
Basic processes (operations) in a Refrigeration CycleSince a refrigeration cycle is essentially a reversed heat engine cycle, the working substance (refrigerant) will undergo the following basic operations.
1. Compression - resulting in increase in pressure and temperature.
2. Heat rejection at high temperature.3. Expansion – resulting in reduction in pressure and
temperature and4. Heat addition at low temperature – during which heat
is transferred from the body to be cooled to the refrigerant.
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Vapour Compression Refrigeration CycleIn this, the refrigerant used is a vapour (e.g., ammonia, Freon-22, Freon-11, Freon -12 etc). The refrigerant undergoes the following operations in a cyclic manner.
1. Compression in a compressor (Usually reciprocating), with work input.
2. Condensation of the vapour into liquid in a condenser, wherein heat is rejected to a cooling medium (air, water) at high pressure and temperature.
3. Expansion of the liquid refrigerant in a suitable device (engine, expansion valve, capillary etc). There may or may not be work output. The liquid may evaporate partially.
4. Evaporation of the mixture of liquid and vapour in an evaporator where heat is added to the refrigerant from the substance to be cooled, producing the necessary refrigeration effect.
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4-1: Reversible adiabatic (isentropic) compression, with work
input WC.1-2: Condensation at constant pressure and temperature with
heat Q1 rejected to some cooling medium.2-3: Reversible adiabatic expansion, with work output WE.
3-4: Evaporation at constant pressure and temperature wherein heat Q2 is absorbed from the substance to be cooled.
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Reversed Carnot Cycle as a Refrigeration Cycle
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Q1 = area under 2 – 3 = Tmax (s2 – s3)
Q2 = area under 4 – 1 = Tmin (s1 – s4)
= Tmin (s2 – s3), s1 = s2 & s3 = s4
Wnet = WC – WE = Q1 – Q2
= (Tmax-Tmin) (s2-s3)
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These are the maximum values for any refrigerator or heat pump operating between two fixed temperatures Tmax and Tmin. In other words, no refrigerator/ heat pump has a COP greater than that of a Carnot refrigerator/heat pump, operating between the same maximum and minimum temperatures.
When the refrigerator/heat pump operates on a cycle other than a Carnot cycle, the heat rejection (condensation) and heat addition (evaporation) process may not be isothermal. Then the COPs are given by
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Where Tcond = average temperature during condensation. Tevap = average temperature during evaporation.It can be seen that the closer the temperatures Tcond and Tevap, the higher the COP.
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In practice, an expansion engine is not used in a vapour compression refrigeration unit.
This is because; the power output of such an engine is too small to justify its cost. Instead, some kind of expansion device – like a throttling valve or a capillary tube – is used to reduce the pressure and temperature of the refrigerant.
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• The most convenient property diagram.
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Process 1-2 or 1’–2’: Reversible adiabatic compression. Process 1–2, starting with saturated vapour (state 1) and ending in the superheated region (state 2) is called Dry compression. Process 1’-2’, starting with wet vapour (state 1’) and ending as saturated vapour (state 2’) is called wet compression. Dry compression is always preferred to wet compression. .
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Process 2-3 (or 2’–3): Reversible constant pressure heat rejection, at the end of which the refrigerant is in saturated liquid state. 2–2’ is desuperheating, and 2’-3 is condensation.
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Process 3-4: Adiabatic throttling process, for which enthalpy before is equal to enthalpy after throttling. This process is adiabatic but not isentropic. Since it is irreversible, it cannot be shown on a property diagram. States 3 and 4 are equilibrium points and are simply joined by a dotted line following a constant enthalpy line.
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Process 3-4: Adiabatic throttling process, for which enthalpy before is equal to enthalpy after throttling. This process is adiabatic but not isentropic. Since it is irreversible, it cannot be shown on a property diagram. States 3 and 4 are equilibrium points and are simply joined by a dotted line following a constant enthalpy line.
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Analysis:The compressor, the condenser and the evaporator can be treated as steady–flow devices, governed by the Steady Flow Energy Equation. Application of S.F.E.E. to these devices results in:
Compressor: Process 1–2 isentropic Q1-2 = 0 W1-2 = - ∆h
W1-2 = - (h2 - h1)
Compressor work WC = (h2-h1) kJ/kg, on a unit mass basis.
If mr is the mass flow rate of the refrigerant in kg/sec, then the power input in the compressor is given by
Power input = mr (h2 – h1) kW
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Condenser: Process 2–3: reversible constant pressure process W2-3 = 0
Q2-3 = ∆h = (h3 – h2) kJ/kgThis is negative i.e., heat rejected.Heat rejected per unit mass of the refrigerant isQ1 = (h2 - h3) kJ/kg.Rate of heat rejection Q1 = mr (h2 - h3) kJ/sec
23
Evaporator: Process 4 – 1: reversible constant pressure process , W4-1 = 0
Q4-1 = ∆h = (h1 - h4) kJ/kg Heat received by unit mass of the refrigerant= heat received from the substance being cooled= Q2 = (h1 – h4) kJ/kg of refrigerant.Rate of heat removed = Refrigerating effect = Q2 = m r(h1 - h4) kJ/secRefrigerating Effect in terms of refrigeration
24
Expansion: for process 3 – 4, h3 = h4
but, it is not a constant enthalpy process.Note: Values of enthalpy h1, h2, h3 & h4 can be obtained
from property Tables or Property Charts (Diagrams).
25
Actual Vapour Compression Refrigeration Cycle:A constant amount of superheating of the vapour before it enters the compressor is recommended. This is to ensure that no liquid refrigerant droplets enter the compressor. Further, a small degree of sub cooling (under cooling) of the liquid refrigerant at the condenser exit is desirable, in order to reduce the mass of vapour formed during expansion. Excessive formation of vapour bubbles may obstruct the flow of liquid refrigerant through the expansion valve.
26
Both the superheating at the evaporator outlet and the subcooling at the condenser outlet contribute to an increase in the refrigerating effect. However, the load on the condenser also increases. There will be an increase in the compressor discharge temperature. Since the compressor input more or less remains unchanged, the COP of the cycle appears to increase due to this superheating/subcooling.
However, for a fixed temperature of the refrigerated space, the evaporation temperature must be lowered (i.e., Tevap is reduced). Further, for a fixed temperature of the cooling medium, the condensation temperature must be raised (i.e., Tcond will be higher). Hence COP will reduce.
27
Refrigerants and desirable properties:
The most commonly used refrigerants are a group of
halogenated hydrocarbons, marketed under various
proprietary names of freon, genetron, arcton etc.
Among them Freon–22 (Mono-chloro Difluoro Methane),
Freon–11 (Tri-chloro – mono-fluoro methane) & Freon–12
(Dichloro Difluoro methane) are extensively used.
Ammonia is another commonly used refrigerant. Other
refrigerants include CO2, SO2, Methyl chloride, Methylene
chloride, Ethyl chloride etc.
28
Desirable properties of a good refrigerant:Thermodynamic properties1. Low boiling point2. Low freezing point3. Positive gauge pressure in condenser and evaporator,
but not very high4. High latent heat of vaporization Chemical properties5. Non–toxic6. Non–inflammable & non–explosive7. Non–corrosive8. Chemically stable9. No effect on quality of stored products
29
Desirable properties of a good refrigerant: Physical properties.1. Low specific volume of vapour2. Low specific heat3. High thermal conductivity4. Low viscosity
Other properties5. Ease of leakage detection6. Cost7. Ease of handling
30
Ammonia is a good refrigerant with the highest refrigerating effect per unit mass.It is relatively cheap. But it is toxic and corrosive. Leakage can be easily detected because if its pungent odour. Freons are Non–toxic & non–inflammable. Leakage cannot be detected easily as they are odour less and colour less. Some coloured additives are sometimes mixed with Freons to facilitate detection of leakage.
31
Gas (Air) Cycle Refrigeration:Refrigeration can also be accomplished by means of
a gas cycle, the most common being the one using air as a refrigerant. In such a cycle, a throttle valve cannot be used for expansion of the working fluid. During the throttling process, enthalpy at the beginning is equal to enthalpy at the end. For an ideal gas, (all gases including air are assumed to be ideal), enthalpy is a function of temperature only. Hence, during throttling temperature at the beginning will be equal to temperature at the end.
32
Gas (Air) Cycle Refrigeration contin…..:Since there is no cooling of air during expansion,
refrigeration is not possible. In place of a throttle valve, an expander is used. Work output obtained from the expander can be utilized for compression, thus decreasing the net work input. In a gas refrigeration cycle, the refrigerant (gas/air) remains in a gaseous state throughout the cycle. Since there is no phase change, the terms ‘condenser’ and ‘evaporator’ are not appropriate. The device in which heat is rejected at a higher temperature can be called a cooler, while the device in which heat is absorbed at a lower temperature is called the ‘refrigerator’.
33
Reversed Carnot CycleA reversed Carnot cycle
using air as the working substance can be a Refrigeration cycle, through it is not practicable.1 – 2: isentropic compression.2 – 3: heat rejection at constant temperature.3 – 4: Expansion4 – 1: heat addition at constant temperature (refrigeration)
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35
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Heat rejected during process 2 – 3= Q1 = Tmax (s2-s3)= Tmax (s1-s4)Heat received during process 4 – 1= Q2 = Tmin (s1-s4)Wnet = WC - WE = Q1 - Q2 (First Law)= (Tmax - Tmin) (s1-s4)
These COPs are the maximum possible COPs for given maximum and minimum temperatures.
37
Reversed Brayton Cycle.A reversed Brayton cycle with air as the working substance is a more practical refrigeration cycle.1 – 2: isentropic compression2 – 3: constant pressure heat rejection3 – 4: isentropic expansion4 - 1: constant pressure heat addition.
38
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On a unit mass basis,Compressor work input = WC = h2 - h1 = Cp (T2 - T1)Expansion work output = WE = h3 - h4 = Cp (T3 - T4)Heat rejected at constant pressure = Q1 = h2 - h3 = Cp (T2 - T3)Heat received at constant pressure = Q2 = h1 - h4 = Cp (T1 - T4)
40
On a unit mass basis,Compressor work input = WC = h2 - h1 = Cp (T2 - T1)Expansion work output = WE = h3 - h4 = Cp (T3 - T4)Heat rejected at constant pressure = Q1 = h2 - h3 = Cp (T2 - T3)Heat received at constant pressure = Q2 = h1 - h4 = Cp (T1 - T4)
41
For the isentropic process 1 – 2,
For the isentropic process 3-4,
42
The COP of a gas cycle refrigeration system is low. The power required per unit capacity is high. Its main application is in aircrafts and missiles, where a vapour compression refrigeration system becomes heavy and bulky. Another application of gas cycle refrigeration is in the liquefaction of gases.
Shown below is a schematic flow diagram of an open cycle air refrigeration system.
43
A small amount of compressed air is blend from the main compressor of a turbojet or a supercharged aircraft engine, and is cooled by rejecting heat to large amounts of cooler ambient air.
The cooled compressed air expands in an expander, and as a result cools further.
The cool air enters the cabin. The output of the expander is used to run a blower which sucks the ambient air in.
In addition to cooling, replacement of stale air in the cabin is possible. At high altitudes the pressurization of the cabin is also possible. Because of this consideration air cycle refrigeration is extensively used in aircrafts.
44
Vapour Absorption RefrigerationA vapour absorption refrigeration system uses a refrigerant as well as an absorbent which can be a liquid or solid.
Possibly the best known combination is ammonia as the refrigerant and water as the absorbent.
A vapour absorption refrigeration system does not have a compressor.
The compressor is replaced by a combination of generator, an absorber and a pump.
45
46
Working:The generator acts as a reservoir for the solution of ammonia in water.
Heat from an external source QG is supplied to the solution, leading to evaporation of ammonia and water.
The mixture of ammonia vapour and water vapour rises through the analyzer, where most of the water vapour condenses, gets separated from NH3 and drops back into the generator.
The analyzer is a direct–contact heat exchanger consisting of a series of trays mounted above the generator.
The strong solution of NH3 from the absorber flows down over the trays, comes into contact with and cools the rising vapours. Since the saturation temperature of water is higher than that of NH3 at a given pressure, water vapour will condense first.
47
As the vapour passes upward through the analyzer, it is cooled and enriched by ammonia.
The ammonia vapour leaving the analyzer may still contain traces of water vapour.
If allowed to flow through the condenser and expansion valve, the water vapour will freeze and block the expansion valve.
Traces of water vapour are separated from ammonia vapour in the rectifier.
The rectifier is a water cooled heat exchanger, wherein all of the remaining water vapour and some ammonia vapour condense and return to the generator through the drip line.
48
The net result is that pure ammonia vapour flows into the condenser and condenses to form saturated or slightly under cooled liquid.
The refrigerant then expands through the valve, resulting in a drop in its pressure and temperature.
The cold refrigerant then flows through the evaporator, extracting heat from the substance to be cooled.
Saturated or slightly saturated ammonia vapour from evaporator flows into the absorber.
The weak solution of ammonia (with low concentration of ammonia in water) coming from the generator is sprayed into the absorber.
The ammonia vapour comes into contact with the weak solution, and gets readily absorbed, releasing the latent heat of condensation.
49
This heat QA taken away by cooling water, thereby maintaining the temperature in the absorber constant.
The resulting strong NH3 solution is pumped to the generator, where heat Qa is supplied to it from an external source.
The weak solution leaving the generator and the pressurized strong solution going to the generator flow through a heat exchanger.
In this heat exchanger, the strong solution is preheated while the weak solution is pre-cooled, reducing both Qa, the heat to be supplied in the generator and QA the heat to be removed in the absorber.
50
The combination of the generator and absorber is equivalent to a heat engine, which does the job of the compressor, namely, receiving from the evaporator at low pressure, comparatively low temperature ammonia vapour and delivering high pressure, higher temperature ammonia vapour to the condenser. This is shown in the diagram.
51
Advantages of vapour absorption Refrigeration System:1. No moving parts (in some cases, there is a small pump) Less
wear and tear. Less maintenance cost.2. Low grade fuels can be used. Waste heat can be used.3. System not affected by variation of loads.4. No electricity required.5. No chance of leakage.Disadvantages6. Low COP7. Bulky8. Higher cost initially Vapour absorption refrigeration systems using solar energy as
the heat source to the generator, hold a lot of promise for the future, in the areas of food preservation and comfort cooling.
52
Steam jet refrigeration: It operates on the principle of reducing the boiling point
of water below 100OC by reducing the pressure on the surface of water below the atmospheric pressure. This low pressure or high vacuum is maintained by throttling the steam through jet or nozzles.
53
Working: The system consists of an evaporator, one or more booster ejectors, a surface type of barometric condenser and two stage ejector air pump.
54
Boiler supplies high pressure steam to the nozzle inlet where steam expands.
The warm water returning from the A/C plant is sprayed through the nozzle in the flash chamber to ensure maximum surface area for evaporation.
The water vapor leaving the flash chamber mixes with the high velocity steam from the nozzle and gets further compressed in thermo compressor.
The K.E is converted into pressure energy and the mass discharged into the condenser returns as condensate to the boiler.
About 1% evaporation of water in evaporator is sufficient to reduce the temperature of water to 6OC. the chilled water is circulated to the A/C plant, which returns as warm water into the flash chamber.
55
The water lost due to evaporation in the flash chamber and any loss of cold water is compensated by make-up water.
Air dissolved in feed water is released in the condenser of the system and covers the condensing surfaces along with other gases and increases the condenser pressure.
This air is removed by using small two stage air ejector to maintain high vacuum.
In order to maintain temperature in the evaporator below 0OC it is necessary to add antifreeze or brine.
56
Limitations:1. It requires very high vacuum and cannot be used if
temperature in the evaporator is below 0OC.2. Size of the compressor should be larger to handle
larger volume.3. The compression ratio used in the thermo compressor
is limited to 8 only.4. Heat removed in the condenser is almost double as
compared to vapor compression system.
57
Advantages:1. Due to no moving parts it is noiseless.2. Flexible in operation as cooling capacity can be quickly
changed.3. Weight/Ton of Ref is low and plant life is more.4. Used in cold water processing of rubber mills, chemical
and food processing plants, breweries, refineries etc.5. Safer in operation and absolutely no hazard from
leakage.6. Cheaper in operation and easy to maintain.7. Useful in comfort air conditioning, but not suitable if
water temperature is below 4OC.
58
Prob 1. A cold storage is to be maintained at -5°C (268k) while the surroundings are at 35°C. the heat leakage from the surroundings into the cold storage is estimated to be 29kW. The actual C.O.P of the refrigeration plant is one third of an ideal plant working between the same temperatures. Find the power required to drive the plant. VTU Jan 2007.
7.6682308
268
TTT ideal C.O.P
cycle.carnot on based C.O.Pbut nothing isplant ideal theof C.O.P
2685T 30835T-:Solution
21
2
1 2
kCkC -
59
233.27.631
..31 ..
x
POidealCPOCActual
Q2 = The heat removed from low temperature reservoir (cold storage) must be equal to heat leakage from surroundings to the cold storage(which is 29kw)
kW 12.98required 233.229
.. W
..
29
2
2
2
PowerPOCActual
QWQPOCActual
kWQ
60
2. A refrigeration machine of 6 tones capacity working on Bell coleman cycle has an upper limit pressure of 5.2 bar. The pressure and temperature at the start of the compression are 1 bar and 18°C respectively. The cooled compressed air enters the expander at 41°C, assuming both expansion and compression to be adiabatic with an index of 1.4.
Calculate:-Co-efficient of performance.Quantity of air circulated per minute.Piston displacement of compressor and expanderBore of compression and expansion cylinder when the
unit runs at 240 rpm and is double acting with stroke length =200 mm , Power required to drive the unit
61
bar2.5P 41T
1bar P 18T-:Solution
23
11
C
C
kgkJ
TTTTWork
/57196314291466005.1
Cinput 4312p
21kJ/s6x3.5 tons6capacity Re
67.157
95.42
inputWork Re..
nfrigeratio
effectngriferatioPOC
62
13.2kg/min0.22x60air/min of Mass12.54kW0.22 x 57
air/sec of Massair x of /kgrequired
/22.095.42
21
R.Eapacity Reair/sec of
workdonePower
skg
cngriferatioMass
min/42.7Vexpander ofnt displaceme
min/42.7101
196287.02.13min/11V compressor ofnt displaceme
min/11101
291287.02.13
34
32
4
44
31
32
1
11
mPiston
mx
xxP
mRTV
mPiston
mx
xxP
mRTV
63
31.3cm0.313mcylinder expander of
2402.04
242.7
42
38cm0.38mcylinder compressor of
2402.04
211
42
1
21
224
1
21
211
diameterd
xxd
LNdV
diameterd
xxd
LNdVBut
64
Problem3 An air refrigerator system operating on Bell Column cycle, takes in air from cold room at 268 K and compresses it from 1 bar to 5.5 bar the index of compression being 1.25. the compressed air is cooled to 300 K. the ambient temperature is 200C. Air expands in expander where the index of expansion is 1.35.
Calculate:C.O.P of the systemQuantity of air circulated per minute for production
of 1500 kg of ice per day at 0°C from water at 20ºC.Capacity of the plant.
65
K8.376
5.5268 25.1125.1
1
1
212
PPTT
Solution
K83.1925.5
130035.1
135.11
3
434
PPTT
kgkJ
TTCnn
p
/2.1562688.376005.14.1
14.1125.1
25.1
11
W 12C
66
kgkJ
TTCnn
p
/69.11883.192300005.14.1
14.1135.1
35.1
11
W 43E
25.37
54.75..
/54.75)83.192268(005.1)(./5.3769.1182.156
41
workREPOC
skJTTCERkgkJWWNetwork
p
EC
skg
heatLatentHeat
/0173.024x3600
1500ecproduced/s ice of Mass
g418.74kJ/k3354.187(20)
_)020(Cice of kgextracted/ pw
67
skg
Mass
tons
heatActual
/096.0 54.7526.7
efection RefrigeratCapacityion Refrigeratrate flow
02.25.3
7.267.26kJ/scapacityion Refrigerator
173418.74x0.0secextracted/
68
Problem 4An air refrigeration system is to be designed according to the
following specificationsPressure of air at compressor inlet=101kPaPressure of work at compressor outlet=404kPaPressure loss in the inter cooler=12kPaPressure loss in the cold chamber=3kPaTemperature of air at compressor inlet=7°Temperature of air at turbine inlet=27° Isentropic efficiency of compressor =85%Isentropic efficiency of turbine =85%DetermineC.O.P of cyclePower required to produce 1 ton of refrigerationMass flow rate of air required for 1 ton of refrigeration
69
85.0;85.0 27T
101kPaP 7T-:Solution
CT3
11
C
C
K4.395101404266
'Hence ,isentropic is 2-1 Pr
4.114.1
1
1
212
PPTTocess
kPaxPPPPPkPaxPPPPP
kT
TTTTTT
C
39240497.097.0 03.010410103.103.1 03.0
2.418'88.0
2664.395'or '
23232
14114
2
1212
12
70
202.3K392104300
PPTT ,isentropic is 4-3 Process
1.411.4
γ1γ
3
434
g152.96kJ/k266]-.21.005x[418
'Cair of work/kgompressor kJ/kg47.05216.53]-1.005x[266
Cair of effect/kg Re53.216]3.205300[85.0300'
' '
12p
41p
4
433443
43
TTC
TTnfrigeratiokxT
TTTTTTTT
TE
71
kgkJ
TTT
/06.729.8096.152
WWair W of Input/kgNet work
kJ/kg9.84216.53]-1.005x[300
'Cair W of work/kgurbine
TCnet
43pT
73.006.7273.46..
WorkREPOC
C.O.Pcapacityion Refrigerat
ionrefergerat of per tons required
Power
72
kWxxmassofairWPower
skg
net 42.5075.006.72sec/
/075.047.505.3
REcapacityion Refrigeratair of Mass
3.5kJ/s ton1capacityion Refrigerat
73
Problem5: 20 tons of ice is produced from water at 200C to
ice at -60C in a day of 24 hours, when the temperature range in the compressor is from -150C to 250C. The condition of the vapour is dry at the end of compression.
Assuming relative C.O.P as 80%, calculate the power required to drive the compressor.
Take Cpice=2.1kJ/kg, Latent heat of ice=335k/kg
74
Temp ºC
Liquid
Vapour
Enthalpy hf Entropy Sf Enthalpy hg Entropy Sg
25 100.04 0.347 1319.2 4.4852
-15 -54.55 -2.1338 1304.99 5.0585
75
KT
TLnTTLnC
To
b
b
b
bp
15.312
7019.0303
56.06853.0ss
aat Entropy bat Entropy b'.'point at vapour ofcondition thefind
ga'
gb'
kgkJHorkkgkJH
TTChH
kgkJh
TTChH
a
c
ccPLfcc
ga
bbpgbb
/55.2119.18374.204Hw/61.123575.5919.183HR.E
kJ/kg 59.57525)-1.003(30-64.59
)(
/19.183H /kg204.74kJ303)-50.56(312.1100.62
)('
b
a
''
'a
'
76
431.346 x 2.133520 x 4.187
)]6(0[Cheat
)020(Cice of kgextracted/Heat 7.130.8x8.913C.O.P ctual
0.8C.O.P Re
913.898.12218.1096..
pice
pw
Latent
Alative
workREPOC
kWPowerActual
Actual
Actual
skgx
Mass
147.13
99.84 work/sec
work/sececextrated/sheat ActualC.O.P Actual
99.84kJ/s 31431.34x0.2secextracted/heat
/231.0360024
20x1000ecproduced/s ice of
77
skJ
capacityf
nfrigeratioworkREPOC
/424.0123.61
52.5
RE.Referon of Mass
52.5kJ/kg15x3.5 15tonscapacity Re
73.555.2161.123..
9.152kW421.55x0.42 freon/s of xMasswork/kgrequired
Power
78
Problem6: A food storage locker requires a refrigeration system
of 12 tons capacity at an evaporator temperature of -80C and a condenser temperature of 300C. The refrigerant freon-12 is sub cooled to 250C before entering the expansion valve and the vapour is superheated to -20C before entering the compressor. The compression of the refrigerant is reversible adiabatic. A double action compressor with stroke equal to 1.5 times the bore is to be used operating at 900 rpm.
DetermineCOPTheoretical piston displacement/minMass of refrigerant to be circulated/minTheoretical bore and stroke of the compressor.Take liquid specific heat of refrigerant as 1.23 kJ/kg K and the
specific heat of vapour refrigerant is 0.732 kJ/kg K.
79
Temp ºC
EnthalpyEntropy
hf hg Sf Sg
30 64.59 199.62 0.24 0.6853
-8 25.75 184.2 0.1142 0.7002
Solution:From tables the properties of Freon 12 are
80
265271Ln 733.07002.0
303TLn 732.06853.0
SS
aat entropy bat
/235.1,/732.0
b
'ga'
'gb'
a
ap
b
bp
PLp
TTLnC
TTLnC
Entropy
kgKkJCkgKkJC
KTb 22.317
kJ/kg02.210 )30322.318(732.062.991
)(kJ/kg59.188
)265271(732.02.841
)(
''
''
bbpgbb
aapgaa
TTChH
TTChH
81
kgkJHorkkgkJH
TTChH
a
c
ccPLgcc
/43.2159.18802.210Hw/181.13041.5859.188HR.E
kJ/kg41.85298)-1.235(303-64.59
)(
b
a
''
07.643.2118.130..
workREPOC
a
a
a
ga
TPV
TPV
kgm
skg
capacityf
'
'
3ga' /0441995.0V C,8-at tablesFrom
n19.35kg/mi0.322x60
/322.0130.1812x3.5
RE.Retrefrigeran of Mass
82
0.0203m
9004
5.1287462.0
)5.1( 4
2min/0.87462m5219.35x0.04
xVmass Vnt displacemepiston
0452.00441995.0265271
2
2
3
a
'
'
xdxxd
dLLNdV
lTheoretica
xTxVT
Va
gaaa
11.08cm 1.5x7.381.5dL
7.38cm 0.0738m d
83
Temp Pressure Vf Vg hf hg Sf Sg
°C bar m3/kg kJ/kg kJ/kgK
-10 226 0.7x10-3 0.08 190 345 0.95 1.5
30 7.5 0.77x10-3 0.02 220 220 1.10 1.45
Problem7: A vapour compression refrigeration system of 5kW cooling
capacity operates between -10ºC and 30ºC. The enthalpy of refrigerant vapour after compression is 370kJ/kg. Find the COP, refrigerating effect, mass flow rate of the refrigerant and the compressor power. The extract of the refrigerant property table is given below
84
kgkJorkkgkJ
givenkgkJH
kgkJh
kgkJhH
b
ga
fcc
/25345370HHw/125220345HHR.E
)(/370
/345H
/220
ab
ca
'a
Solution:Assume the condition before compression as dry saturated vapour
525
125.. workREPOC
85
1kW25x0.04-
trefrigeran of mass x .
/04.0125
5
RE.Retrefrigeran of Mass
kJ/sor 5kW capacity ion Refrigerat
kgworkworkCompressor
skg
capacityf
86
Problem8: A vapour compression refrigerator uses methyl chloride and works in the pressure rang of 1.19 bar and 5.67 bar. At the beginning of compression, the refrigerant is 0.96 dry and at the end of isentropic compression, its temperature is 55ºC. The refrigerant liquid leaving the condenser is saturated. If the mass flow of refrigerant is 1.8kg/min,DetermineCOPThe rise in temperature of cooling water if the water flow rate is16 kg/min. the properties of methyl chloride is given below
87
Temp ºC
Pressurebar
EnthalpyEntropy
hf Hfg hg Sf Sg
30 1.19 64.59 135.03 199.62 0.24 0.6853
-10 5.67 26.87 156.31 183.19 0.108 0.7019
kgkJ
hhxhhxhHCT
xSolution
Take
fagaafafgaafaa
b
a
/196.43830.1)-0.96(455.2430.1
)()(55
96.0
K 0.75kJ/kg as chloride methylheat super ofheat specific
88
kgkJHorkkgkJH
kgkJh
TTChH
a
c
fc
bbpgbb
/8.60196.438499Hw/669.3375.100196.438HR.E
/5.100HkJ/kg49925)-(5575.05.764
)(
b
a
c
''
55.58.60
669.337.. workREPOC
watercoolingby gain heat condenser in the vapour by thelost
Heat
CriseeTemperaturriseetemperaturx
riseetemperaturxCmhmTTCm pwfgbrbbpr
7.10 x 187.4 16
100.5)-1.8(476.525)-0.75(55 x 8.1
)( ''
89
90