thermodynamics 2_chapter 5 part1 (1) - copy

Microsoft PowerPoint - Thermodynamics 2_Chapter 5 part1

(Duty, Honor and Integrity)Chapter 5...Part 1REFRIGERATION CYCLES

(Kitar Penyejuk)

(Thermodynamics II (EMM3503)Chapter 5- REFRIGERATION CYCLES)



Delivered by: Ir Khalid

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Duty, Honor and Integrity

Objectives

Introduce the concepts of refrigerators and heat pumps and the measure of their performance.

Analyze the ideal vapor-compression refrigeration cycle.

Analyze the actual vapor-compression refrigeration cycle.

Review the factors involved in selecting the right refrigerant for an application.

Discuss the operation of refrigeration and heat pump systems.

Evaluate the performance of innovative vapor-compression refrigeration systems.

Analyze gas refrigeration systems.

Introduce the concepts of absorption-refrigeration systems.

Duty, Honor and

IntegrityREFRIGERATORS AND HEAT PUMPS

The Purpose of Refrigeration

The fundamental reason for having a refrigerator is to keep food cold. Cold temperatures help food stay fresh longer. The basic idea behind refrigeration is to slow down the activity of bacteria (which all food contains) so that it takes longer for the bacteria to spoil the food.

The basic mechanism of a refrigerator works like this:

(SIQW Slofl Works)1 .The compressor compresses the refrigerant gas. This raises the refrigerant's pressure and temperature (orange), so the heat-exchanging coils outside the refrigerator allow the refrigerant to dissipate the heat of pressurization.

2. As it cools, the refrigerant condenses into liquid form (purple) and flows through the expansion valve.

3. When it flows through the expansion valve, the liquid refrigerant is allowed to move from a high-pressure zone to a low-pressure zone, so it expands and evaporates (light blue). In evaporating, it absorbs heat, making it cold.

4. The coils inside the refrigerator allow the refrigerant to absorb heat, making the inside of the refrigerator cold. The cycle then repeats.

REFRIGERATORS AND HEAT PUMPS...cont.

A heat pump is a device that uses a small amount of energy to move heat from one location to another. Heat pumps are usually used to pull heat out of the air or ground to heat a home or office building, or they can be switched into reverse to cool a building.

How Heat Pumps Work

Ground-source

System

Blower

Heating

(Coil) (Duty, Honor and Integrity)Closed loop

pipe system

Compressor

Mouse over

button to select

REFRIGERATORS AND HEAT PUMPS...cont.

The transfer of heat from a low-temperature region to a high-temperature one requires special devices called refrigerators.

Refrigerators and heat pumps are essentially the same devices; they differ in their objectives only.

(Cooling effect Work inputHeating effect Work input) (COPRCOPHP) (Duty, Honor and Integrity)Desired output Required input

Desired output Required input

(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.) (Note: The capacity of a refrigeration system that can freeze 1 ton of liquid water at OC in 24 hours is said to 1 ton.1 ton of refrigeration is equivalent to 211 kJ/min)coi'i,]. COP,, - I for fixed values of QL and QH

Duty, Honor and

IntegrityTHE VERSED CARNOT CYCLE

The Carnot cycle is an idealized thermodynamic cycle that describes a perfect heat engine. In the Refrigeration system we need cooling effect, so it has to operate in opposite nature to produce the cooling effect. So we run the carnot cycle reversely in the refrigeration system. So we call the Refrigeration cycle called as REVERSED CARNOT CYCLE.

The reversed Carnot cycle is the most efficient refrigeration cycle operating between TL and TH.

However, it is not a suitable model for refrigeration cycles since processes 2-3 and 4-1 are not practical because

Process 2-3 involves the compression of a liquid-vapor mixture, which requires a compressor that will handle two phases, and process 4-1 involves the expansion of high-moisture-content refrigerant in a turbine.

WARM mediumat T

Condenser

(Turbine)Compressor

Evaporator




(COLD mediumat T,)Schematic of a Carnot refrigerator and T-s diagram of the reversed Carnot cycle.6

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THE REVERSED CARNOT CYCLE...cont

Both COPs increase as the difference between the two temperature decreases, that is, as TL rises or Th falls.

(Turbine) (WARM mediumat T) (Condenser) (Compressor) (Evaporator) (COLD mediumat TV) (COP) (r) (Carnot) (T,) (T,)


THE IDEAL VAPOR-COMPRESSION REFRIGERATION CYCLE

The vapor-compression refrigeration cycle is the ideal model for refrigeration systems. Unlike the reversed Carnot cycle, the refrigerant is vaporized completely before it is compressed and the turbine is replaced with a throttling device.

1- (WARM) (eiivfrtmincnL)2Isentropic compression in a compressor

2- 3Constant-pressure heat rejection in a condenser

3- 4Throttling in an expansion device

4- 1Constant-pressure heat absorption in an evaporator

(y Expansion) (v valve) (Compressor) (Evaporutoi) (COLD refrigeratedspace) (This is the most widely used cycle for refrigerators, A-C systems, and heat pumps.) (s)Condctisei

Schematic and T-s diagram for the ideal vapor- compression refrigeration cycle.

Duty, Honor and

Integrity3 ideal vapor-compression refrigeration cycle involves an irreversible (throttling) process to make it a more realistic model for the actual systems.

(/?, = hg @ P[ and h3 = hf @ for the ideal case)Replacing the expansion valve by a turbine is not practical since the added benefits cannot justify the added cost and complexity.

(Steady-flowenergy balance-") (Capillary tubeKitchen air 25C) (Condensercoils) (Compressor) (The P-h diagram of an ideal vapor- compression refrigeration cycle.)y

r.

ACTUAL VAPOR-COMPRESSION REFRIGERATION CYCLE

(Duty, Honor and Integrity) (WARMenvironment) (Qn) (Condenser) (y Expansion valve Compressor) (Evaporator ^ Ql) (Schematic and T-s diagram for the actual vapor-compression refrigeration cycle.) (DIFFERENCESNon-isentropic compressionSuperheated vapor at evaporator exitSubcooled liquid at condenser exitPressure drops in condenser and evaporator)An actual vapor-compression refrigeration cycle differs from the ideal one in several ways, owing mostly to the irreversibilities that occur in various components, mainly due to fluid friction (causes pressure drops) and heat transfer to or from the surroundings. The COP decreases as a result of irreversibilities.

COLD refrigeratedspace

11-2 A steady-flow Carnot refrigeration cycle with refrigerant-13 4a as the working fluid is considered, The coefficient of performance, the amount of heat absorbed from the refrigerated space, and the net work input

are to be determined.

Assumptions 1 Steady operating conditions exist, 2 Kinetic and potential energy changes are negligible.

Analysis (a) Noting that TH = 30C = 303 K and TL = T^m mkPa = -15.60C = 257.4 K. the COP of this

Camot refrigerator is detenmned from

(= 3.64)(303 K)/(257.4 K) 1

(p) From the refrigerant tables (Table A-l 1)

'h ~ "smart - 266.66 kJ/kg

c -93.58 kJ/kg

Thus

q = /, ha =266.66 -93,58 = 173.08 kJ/kg

(r 257.4 K) ((173.08 kJ/kg) = 147*03 kJ/kg) (303 K)and

(ii') (net) (= qH - qL = 173,08 - 147.03 = 26.05 kJ/kg)(c) The net work input is deteimmed from

(Duty, Honor and Integrity)11-3 A steady-flow Carnot refrigeration cycle with refrigerant-134a as the working fluid is considered. Hie coefficient of performance, the quality at the beginning of the he at-absorption process, and the net work input are to be determined.

(Analysis (a) Noting that TK= .^0^iIPa = 21.55C = 294.6 K and TL = rat@02mp3 = -10.09C = 262.9 K.1 1) (COIVc =) (= 8.31) (1(294.6 ^(262.9 K1-1(6) Process 4-1 is isentropic, and thus(*| + W,, 0SMKl = 0.30799+ (0.05)(0.6137) = 0.33868 fcJ-kg-K) (Xi =) (V0.2 MP;i) (0.33868-0.154570.7S316) (= 0.2351)Assmitptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

(c) Remembering that on a T-s diagram the area enclosed represents the net work, and s3 = jE u Q = 0.92177 kJ 'kg'K.



liW* = [TH - Tl {s3 - r4 : = [21.5 5 - (-10.09)](0.9 2177 - 0.3 3 3 68} ti/kg K = 18.4 ? kJ, kg

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11-11 A commercial refrigerator with refrigerant-134a as the working fluid is considered. Hie qtvality of the refrigerant at the evaporator inlet, the refrigeration load the COP of the refrigerator, and the theoretical maximum refrigeration load for the same power input to the compressor are to be determined.

Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

Water

(4 2=C) (Conti enser) (Expansion) (ah) (Compressor) (Evaporator) (MPa) (55) (C) (bOkPa) (34C)1SC

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(60 kPa -34*C) (Evaporator) (11-11 A commercial refrigerator with refrigerant-134a as the working fluid is considered. The quality of the refrigerant at the evaporator inlet, the refrigeration load, the COP of the refrigerator, and the theoretical maximum refrigeration load for the same power input to the compressor are to be determined.Assumptions 1 Steady operating conditions exist. 1 Kinetic and potential energy' changes are negligible.Analysis (a) From refrigerant-134a tables (Tables A-11 through A-13)= 60kPa jIi=- 34-Cj'26C) (P2 = 1200 kPa ] T2 = 65C) (^ = 230.03 kJ/kg \h2 =295.1(5 kJ/kg) (Water18C) (42 C) (J>3 = 1200kPa Ti = 42 C) (h3 =111.23 kJ/kg) (hA =h2 =111.23 kJ/kg P4 = 60 kPa h4= 111.23 kJ/kg) (=0.4795) (Condenser) (r Expansion valve) (Compressor) (1.2 MPa65C)Using saturated liquid enthalpy at the given temperature, for water we have (Table A-4)

^ = &f@iarc = 75.47 kJ/kg =;26ac = l'^S.94 kJ/kg

(b) The mass flow rate of the refrigerant may be determined from an energy' balance on the compressor

mK{h2 -h2) =- ig)

(295.16 -111.23)k J/kg = (0.25 kg/s)(l 08.94- 75.4 7)kJ/kg >mR - 0.0455 kg/s

Tilt waste heat transferred from the refrigerant, the compressor power input, and the refrigeration load are Qh =mi(fct-Ai>=(004551^3X295.16-1 !L23)tJ/kg = S.367 kW

fTb= (0 0455bg/sX295 i6-230.03)kJ'kg-Q.45kW = 2 5BkW

Ql=367-2.513=5.85 kW

(tr) The COP of the refrigerator is determined from its definition

(-2.33) (COP-)Ql _ 5-35 2513

(d) The reversible COP of the refrigerator for the same temperature limits is

COP i-i

Th i Tl -1 (18 + 273) /(30 + 273) -1

Then, the maximum refrigeration load becomes

Qlu* = COP^&m = (5.063X2.513 kW) =12.72 kW

compression refrigeration cycle. The mass now rate of the refrigerant, the condenser pressure, and the

COP of the refrigerator are to be determined

Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible

Analysis (it) (e) From the refrigerant-134a tables (Tables A-l 1 through A-13)

P* = 120kPa

(h4 = 86.83 kJ/kg)x& - 0.30

(11-23 A refrigerator uses refrigerant-134a as the working Stud and operates on the ideal vapor-)



Condenser

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(h? = 36.83kJ/kg) (P, =671.8) (jr3 = 0 (sat. liq.)) (Expansion) (Pn = P-) (valve) (Compressor)

R = 671.8kPa

(h2 = 298.87 kJ/kg) (Pj = P4 = 120 kPa) (a-! = l(sat.vap.)) (hi =23(5.97 kJ-'Kg) (120 kPa) (a-0.3) (Evaporator)t2 = eoc

The mass flow rate of the refrigerant is

(0.45 kW) (= 0.00727 kg/s) (h2 -hi) ((29S.S7 - 236.97)kJ/kg)determined from

(c) The refrigeration load and the COP are

Ql =m(h1 -h4)

= (0.0727 kg# s)(23 6 .97 - 86.8 3)kJ/kg = 1.091 kW

(COP =) (Ql) (1091kW) (2.43) (0.45 kW)


SELECTING THE RIGHT REFRIGERANT

Several refrigerants may be used in refrigeration systems such as chlorofluorocarbons (CFCs), ammonia, hydrocarbons (propane, ethane, ethylene, etc.), carbon dioxide, air (in the air-conditioning of aircraft), and even water (in applications above the freezing point).

R-11, R-12, R-22, R-134a, and R-502 account for over 90 percent of the market.

The industrial and heavy-commercial sectors use ammonia due to low cost, higher COPs, higher heat transfer coefficients and no effect to ozone layer. The major drawback is toxicity which makes it unsuitable for domestic used.

R-11 is used in large-capacity water chillers serving A-C systems in buildings.

R-134a (replaced R-12, which damages ozone layer) is used in domestic refrigerators and freezers, as well as automotive air conditioners.

R-22 is used in window air conditioners, heat pumps, air conditioners of 17 commercial buildings, and large industrial refrigeration systems, and offers

Duty, Honor and

IntegritySELECTING THE RIGHT REFRIGERANT.. .cont.

R-22 is used in window air conditioners, heat pumps, air conditioners of commercial buildings, and large industrial refrigeration systems, and offers strong competition to ammonia.

R-502 (a blend of R-115 and R-22) is the dominant refrigerant used in commercial refrigeration systems such as those in supermarkets.

CFCs allow more ultraviolet radiation into the earths atmosphere by destroying the protective ozone layer and thus contributing to the greenhouse effect that causes global warming. Fully halogenated CFCs (such as R-11, R-12, and R- 115) do the most damage to the ozone layer. Refrigerants that are friendly to the ozone layer have been developed.

Two important parameters that need to be considered in the selection of a refrigerant are the temperatures of the two media (the refrigerated space and the environment) with which the refrigerant exchanges heat.

(Duty, Honor and Integrity) (A heat pump heats your home in the winter...) (and cools your home in the summer.)HEAT PUMP SYSTEMS

A heat pump is a machine or device that moves heat from one location (the 'source') at a lower temperature to another location (the 'sink' or 'heat sink') at a higher temperature using mechanical work or a high-temperature heat source. The difference between a heat pump and a normal air conditioner is that a heat pump can be used to heat a home as well as cool it. Heat pumps are a unique kind of heating system, because they can do the work of both a furnace and an air conditioner.



http://home.howstuffworks.com/home-improvement/heating-and-cooling/heat-pump1.htm

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(Duty, Honor andIntegrity HEAT PUMP SYSTEMS...cont) (HEAT PUMP OPERATIONHEATIMS MODEOuKlooreuilRq ve rsj ng valvepressure [itjutid Lowjjjresstirfc li^Likl-vyjjurLow-pressure vapor 111 H igh-pressure yqporA heat pump can be used to heat a house in winter and to cool it in summer.) (HEAT PUMP OPERATIONCOO! IKC MODE Out(l(Wr eoiIReversing valve)The most common energy source for heat pumps is atmospheric air (air-to- air systems).

Water-source systems usually use well water and ground-source (geothermal) heat pumps use earth as the energy source. They typically have higher COPs but are more complex and more expensive to install.

Both the capacity and the efficiency of a heat pump fall significantly at low temperatures. Therefore, most air-source heat pumps require a supplementary heating system such as electric resistance heaters or a gas furnace.

Heat pumps are most competitive in areas that have a large cooling load during the cooling season and a relatively small heating load during the heating season. In these areas, the heat pump can meet the entire cooling and heating needs of residential or commercial buildings. 20


HEAT PUMP SYSTEMS...cont.

(refrigerators) (area lo a warme-r area.) (Coefficient of Performance =) (General define on) (- -) (-G,) (H) (H) (H) (Carnot caseT) (eal coefficient l^eflormanoe)Heat Pumps are usually characterized by a coefficient of performance which is the number of units of energy delivered to the hot reservoir per unit work input.

11-3-1 An actual heat pump cycle with R-134a as the refrigerant is considered. The isentropic efficiency of the compressor, the rate of heat supplied to the heated room, the COP of the heat pump, and the COP and the rate of heat supplied to the heated room if this heat pump operated on the ideal vapor-compression cycle between the same pressure limits are to be determined.

Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) The properties of refrigerant-134a are (Tables A-ll through A-13)

(COP64)

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(790) (W) (The) (ideal) (apor) (compression) (cycle analysis of me) (cycle) (IS) (as) (follows) (PS) (200) (kfa) (kJ/kg) (200) (kPa) (9377) (td/kg) (K) (P) (800) (MPa) (273) (25) (kJ/kg) (95) (47) (kJ'kg) (273) (25) (95) (47) (COP) (18) (273) (2 s) (244) (46) (Qe) ((0) (01Ekg/s)(273) (25) (95) (47)kJ/kg) (20) (kW)

11-32 A hear pump operating on the ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The COP and the rate of heat supplied to the evaporator are to be determined.

Assumptions 1 Steady op dating conditions exist. 2 Kinetic and potential energy changes are negligible.

Analysis In an ideal vapor-compression refrigeration cycle, the compression process is isentropic. the refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the; condenser as saturated liquid at the condenser pressure. From the refrigerant tables (Tables A-11. A-12. and A-13),

(Px = 280 kPa sat. vapor) (P2 = 1200 kPa) (jg 2S0 kPa _ 249. '2 kJ kg si -s:s@2mm = 0.93210kJ kg-K h, =280.00 kike) (P3 = 1200 kPa) (^3 - hf @ 1200 IcPa 117.17 kJ/kg) (sat. liquid hA = h3 = 117.77 kJ kg (throttling)The mass flow rate of the refrigerant is determined from) (W-m = mih-, - /q )> m = ) (W r r 1+) (20 kJ i) ( = 0.0605 kWs) (h1-h1(280.00 249.72) kJ.'kgThen the rate of heat supplied to the evaporator isql =m{hx -h4) = (0.6605 kg/s)(249.72-117.77) kJ kg = 87.15 kW The COP of the heat pump is determined from its definition.qH h2-h3280.00-117.77) (COPjjp =) (h2-hx 280.00-249.72) (5.36)

thermodynamics 2_chapter 5 part1 (1) - copy

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