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REFRIGERATIONRefrigeration: The process of cooling a substance and of maintaining it in a temperature below that of the immediate surroundings.Major Uses of Refrigeration:

1. Air Conditioning2. Food Preservation3. Removing of heat from substances in chemical, petroleum and petrochemical plants4. Special applications in the manufacturing and construction industries.

TON OF REFRIGERATIONIt is the heat equivalent to the melting of 1 ton (2000 lb) of water ice at 0C into liquid at 0C in 24 hours.

TR=2000(144 )24

=12,000 BTUhr

TR=211 KJmin

where 144 BTU/lb is the latent heat of fusion of ice

CARNOT REFRIGERATION CYCLE: T QR

TH

Network TL

QA

S

HEAT ADDED (T = C)QA = TH(S1-S4)QA = TH(S)

HEAT REJECTED (T = C)QR = TL (S2-S3)QR = TL (S)(S) = (S2-S3) =(S1-S4)

NET WORKW = QR - QA

W = (TH - TL)(S)COEFFICIENT OF PERFORMANCE

COP=QAW

COP=QAQR−QA

COP=T LTH−T L

where: QA - refrigerating effect or refrigerating capacityQR - heat rejected

3

14

TH

TL

R

1

W - net cycle work

VAPOR COMPRESSION CYCLEComponents:

Compressor Condenser Expansion Valve Evaporator

Processes Compression, 1 to 2 (S = C) Heat Rejection, 2 to 3 (T = C) Expansion, 3 to 4 (S = C) Heat Addition, 4 to 1 (T = C)

P T 2

3 2 3 S = C h = C 4 1 4 1

h S

System: COMPRESSOR (S = C)

W=m(h2−h1 )60

KW

W=kP1V 1'

(k−1)60 [(P2

P1)k−1k

−- 1] KW

P1V 1'=mRT1 System: CONDENSER (P = C) QR = m(h2-h3) KJ/minSystem: EXPANSION VALVE (h = C)

h3 = h4

x4=h4−h f 4

hg 4−h f 4 System: EVAPORATOR (P = C)

QA=m(h1−h4 ) KJ/min

QA=m(h1−h4 )211

Tons of Refrigeration

evaporator

Condenser

expansionvalve

compressor

1

23

4

QA

QR

2

DISPLACEMENT VOLUME (For Reciprocating type of compressor)a. For Single Acting

V D=π LD2 Nn'

4 (60) m3

sec b. For Double Acting without considering piston rod

V D=2π LD2 Nn'

4(60 ) m3

sec c. For Double Acting considering piston rod

V D=π LNn'4(60 )

(2D2 -D2) m3

sec KW PER TON OF REFRIGERATION

KWTon

=211(h2 -h1 )(60 )(h1 -h 4 )

KWTon

COEFFICIENT OF PERFORMANCE

COP=QAW

COP=h1 -h 4

h2 -h1 CUBIC METER/min PER TON OF REFRIGERATION

A. For Single Acting

m3

min-Ton=211π LD2Nn'

4m(h1 -h 4 )

m3

min-Ton B. For double acting without considering piston rod

m3

min-Ton=

211(2) π LD2 Nn'4m (h1 -h 4 )

m3

min-Ton C. For double acting considering piston rod

m3

min-Ton=211π LNn'

4m(h1 -h 4 )(2D2 -d2)

m3

min-Ton where; L - length of stroke, m

D - diameter of bore, md - diameter of piston rod, mN - no. of RPMn' - no. of cylinders

VOLUMETRIC EFFICIENCY

ηv=V 1'

V D x 100%

ηv=[1+C - C(P2

P1)1k ] x 100 %

where: V1' - volume flow rate at intake, m3/secVD - displacement volume, m3/secv - volumetric efficiency

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1 - specific volume, m3/kg v = [1 + C - C(P2/P1)1/k] x 100%EFFICIENCY

A. Compression Efficiency

η

cn=Ideal Work

Indicated Work x 100%

B. Mechanical Efficiency

ηm=Indicated Work

Brake or Shaft Work x 100%

C. Compressor Efficiency

ηc=Ideal WorkBrake or Shaft Work

x 100%

ηc=ηcnηm

EFFECTS ON OPERATING CONDITIONSA. Effects on Increasing the Vaporizing Temperature

P T

h S

1. The refrigerating effect per unit mass increases.2.The mass flow rate per ton decreases3. The volume flow rate per ton decreases.4. The COP increases.5. The work per ton decreases.6. The heat rejected at the condenser per ton decreases.

B. Effects on Increasing the Condensing Temperature

P T

h S

1. The refrigerating effect per unit mass decreases.

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2. The mass flow rate per ton increases.3. the volume flow rate per ton increases.4. The COP decreases.5. The work per ton increases.6. The heat rejected at the condenser per ton increases.C. Effects of Superheating the Suction Vapor

P T

h S

When superheating produces useful cooling:1. The refrigerating effect per unit mass increases.2. The mass flow rate per ton decreases3. The volume flow rate per ton decreases.4. The COP increases.5. The work per ton decreases.When superheating occurs without useful cooling:1. The refrigerating effect per unit mass remains the same.2. The mass flow rate per ton remains the same.3. The volume flow rate per ton increases.4. The COP decreases.5. The work per ton decreases.6. The heat rejected at the condenser per ton increases.

D. Effects of Sub-cooling the Liquid

P T

h S

1. The refrigerating effect per unit mass increases.2. The mass flow rate per ton decreases.3. The volume flow rate per ton decreases.4. The COP increases.

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5. The work per ton decreases.6. The heat rejected at the condenser per ton decreases.

LIQUID-SUCTION HEAT EXCHANGERThe function of the heat exchanger is:1. To ensure that no liquid enters the compressor, and2. To sub-cool the liquid from the condenser to prevent bubbles of vapor from impeding the flow of refrigerant through the expansion valve. QR

W

QA

SAMPLE PROBLEMS

1. An air conditioning system of a high rise building has a capacity of 350 KW of refrigeration, uses R-12. The evaporating and condensing temperature are 0C and 35C, respectively. Determine the following: a) mass of flash gas per kg of refrigerant circulated b) mass of R-12 circulated per second c) volumetric rate of flow under suction conditions d) compression work e) the COP

2. A single cylinder, 6.7 cm x 5.7 cm, R-22 compressor operating at 30 rps indicate a refrigerating capacity of 96.4 KW and a power requirement of 19.4 KW at an evaporating temperature of 5C and a condensing temperature of 35C. Compute: a) the clearance volumetric efficiency if c = 5% b) the actual volumetric efficiency c)the compression efficiency

P1 = 308.6 KPaP2 = 847.7 KPah1 = 351.48 KJ/kgv1 = 0.0554 m3/kgh2 = 368 KJ/kgh3 = h4 = 233.5 KJ/kghf at 0 C = 200 KJ/kghg at 0 C = 351.48 KJ/kg

a) 0.22b) 2.97 kg/secc) 0.1645 m3/secd) 49.06 KWe) 7.14

P1 = 584 KPaP2 = 1355 KPah1 = 407.1 KJ/kgv1 = 40.36 L/kgh2 = 428 KJ/kgh3 = h4 =243.1 KJ/kg

a) 94.91%b) 65.6%c) 63.33%

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condenser

Evaporator

ACTUAL VAPOR COMPRESSION CYCLEAs the refrigerant flows through the system there will be pressure drop in the condenser, evaporator and piping. Heat losses or gains will occur depending on the temperature difference between the refrigerant and the surroundings. Compression will be polytropic with friction and heat transfer instead of isentropic. The actual vapor compression cycle may have some or all of the following items of deviation from the ideal cycle:

Superheating of the vapor in the evaporator Heat gain in the suction line from the surroundings Pressure drop in the suction line due to fluid friction Pressure drop due to wiredrawing at the compressor suction valve. Polytropic compression with friction and heat transfer Pressure drop at the compressor discharge valve. Pressure drop in the delivery line. Heat loss in the delivery line. Pressure drop in the condenser. Sub-cooling of the liquid in the condenser or sub-cooler. Heat gain in the liquid line. Pressure drop in the evaporator.

MULTIPRESSURE SYSTEMSA multi-pressure system is a refrigeration system that has two or more Low - Side pressures. The low-side pressure is the pressure of the refrigerant between the expansion valve and the intake of the compressor. A multi-pressure system is distinguished from the single-pressure system, which has but one low-side pressure.Removal of Flash GasA savings in the power requirement of a refrigeration system results if the flash gas that develops in the throttling process between the condenser and the evaporator is removed and recompressed before complete expansion. The vapor is separated from the liquid by an equipment called the Flash Tank. The separation occurs when the upward speed of the vapor is low enough for the liquid particles to drop back into the tank. Normally, a vapor speed of less than 1 m/sec will provide adequate separation.Inter-coolingInter-cooling between two stages of compression reduces the work of compression per kg of vapor. Inter-cooling in a refrigeration system can be accomplished with a water-cooled heat exchanger or by using refrigerant. The water-cooled intercooler may be satisfactory for two-stage air compression, but for refrigerant compression the water is not usually cold enough. The alternate method uses the liquid refrigerant from the condenser to do the inter-cooling. Discharge gas from the low stage compressor bubbles through the liquid in the intercooler. Refrigerant leaves the intercooler as saturated vapor. Inter-cooling with liquid refrigerant will usually decrease the total power requirement when ammonia is the refrigerant but not when R-12 or R-22 is used.2-Evaporators and 1-Compressor

Pressure

reducing valve

7

Evaporator

2-Compressors and 1-Evaporator

2-Compressors and 2-Evaporators

HP Compressor

LP Compressor

Condenser

Flash Tank andIntercooler

Evaporator

condenser

HP compressor

HP Evaporator

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SAMPLE PROBLEMS1. Calculate the power required by a system of one compressor serving two evaporators.

One evaporator carries a load of 35 KW at 10C and the other at a load of 70 KW at -5C. A backpressure valve reduces the pressure in the 10C evaporator to that of the -5C evaporator. The condensing temperature is 37C. The refrigerant is ammonia. What is the COP.

h3 = h4 = h7 = hf at 37C = 375.9 KJ/kg h5 = h6 = hg at 10C = 1471.6 KJ/kg h8 = hg at -5C = 1456.2 KJ/kg

by energy balance at 10C evaporator m4 = m5 = m6 = 35/(h5 - h4) = 0.0319 kg/sec

by energy balance at -5C evaporator m7 = m8 = 70/(h8 - h7) = 0.0648 kg/sec

by mass and energy balance at the mixing point as shown on figure above m6h6 + m8h8 = m1h1

h1 = 1461.3 KJ/kgfrom chart, at S1 = S2 to P2

h2 = 1665 KJ/kg W = m1(h2 - h1) = 19.7 KW COP = 35 + 70 19.7 COP = 5.33

2. Calculate the power required in an ammonia system that serves a 210 KW evaporator at -20C. The system uses two-stage compression withinter-cooling and removal of flash gas. The condensing temperatureis 32C.

For minimum work and with perfect inter-cooling, the intermediate pressure P2 is equal to P2 = (P1P4)1/2

m7 = m8 = m1 = m2 Q = m1(h1 - h8)

m1 = 0.172 kg/sec

LP CompressorLP Evaporator

Flash Tank andIntercooler

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by mass and energy balance about the intercooler; m2h2 + m6h6 = m7h7 + m3h3

m3 = 0.208 kg/sec WLP = m1(h2 - h1) = 21.6 KW WHP = m3(h4 - h3) = 31.1 KW W = WHP + WLP

CASCADE SYSTEM

A cascade system combines two vapor compression units, with the condenser of the low temperature system discharging its heat to the evaporator of the high temperature system. This system can furnish refrigeration for about -100 C. There are two types of a cascade system, the closed cascade condenser and the direct contact heat exchanger. In the closed cascade condenser fluids in the low-pressure and high-pressure may be different, but in the direct contact heat exchanger the same fluid is used throughout the system.

CLOSED CASCADE CONDENSER

DIRECT CONTACT CONDENSER

Cascade Condenser

HP Compressor

LP Compressor

Condenser

Evaporator

Condenser

HP Compressor

Open TypeCascade Condenser

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SAMPLE PROBLEMS1. A cascade refrigerating system uses R-22 in the low-temperature unit and R-12 in the high- temperature unit. The system develops 28 KW of refrigeration at -40C. the R-12 system operates at -10C evaporating and 38C condensing temperature. There is a 10C overlap of temperatures in the cascade condenser. Calculate: a) the mass flow rate of R-22 (0.1485 kg/sec) b) the mass flow rate of R-12 (0.305 kg/sec) c) the power required by the R-22 compressor (5.7 KW) d) the power required by the R-12compressor (7.6 KW) h1 = 388.6 KJ/kg h5 = 347.1 KJ/kg h2 = 427 KJ/kg h6 = 372 KJ/kg h3 = h4 = 200 KJ/kg h7 = h8 = 236.5 KJ/kg2. In a certain refrigeration system for low temperature application, a two stage operation is desirable which employs ammonia system that serves a 30 ton evaporator at -30 C. the system uses a direct contact cascade condenser, and the condenser temperature is 40 C. find the following: a) Sketch the schematic diagram of the system and draw the process on the Ph diagram b) the cascade condenser pressure in KPa for minimum work c) the mass flow rate in the high pressure and low pressure loops in kg/sec d) the total work in KW

LP CompressorEvaporator

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AIR CYCLE REFRIGERATION

Closed or Dense Air System

Open Air System

The air cycle refrigeration system in which a gaseous refrigerant is used throughout the cycle. The compression is accomplished by a reciprocating or a centrifugal compressor as in the vapor-compression system, but condensation and evaporation is replaced by a sensible cooling and heating of the gas. An air cooler is used in place of a condenser and a refrigerator in place of an evaporator. The expansion valve is replaced by an expansion engine or turbine.The air cycle refrigeration system is ideally suited for use in air craft, because it is lighter in weight and requires less space than the vapor-compression cycle. One disadvantage of the air cycle is that it is not as efficient as the vapor-compression cycle. The air cycle refrigeration may be designed an operated either as an open or a closed system as shown above. In the closed or dense-air-system, the air refrigerant is contained within the piping or component parts of the system at all times and with the refrigerator usually maintained at pressures above atmospheric level. In the open system, the refrigerator is an actual space to be cooled with the air expanded to atmospheric pressure, circulated through the cold room and then compressed to the cooler pressure.

Cooler

Expander Compressor

Refrigerator

Cooler

Refrigerator

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IDEAL AIR REFRIGERATION CYCLE Processes: 1 to 2: Isentropic Compression 2 to 3: Constant Pressure Heat Rejection 3 to 4: Isentropic Expansion 4 to 1: Constant Pressure Heat Absorption

Refrigerating Effect QA = mCp(T1-T4)Heat Rejected

QR = mCp(T3-T2)Compressor Work

W c=kP1V 1

1−k [( P2

P1)k−1k −1]

Expander Work

W E=kP3V 3

1−k [( P4

P3)k−1k −1]

Net Work

W = WC - WE

Coefficient of Performance

COP=Refrigerating Effect

Net Work

1

23

4

P

V

T

S

1

2

3

4

S = CS = C

P = C

P = C

13

t1

m tf

m t2

m

SAMPLE PROBLEMS1. An air cycle refrigeration system operating on a closed cycle is required to produced 50 KW of refrigeration with a cooler pressure of 1550 KPa and a refrigerator pressure of 448 KPa. Leaving air temperature are 25C for cooler and 5C for the refrigerator. Assuming a theoretical cycle with isentropic compression and expansion, no clearance and no losses. Determine; a) the mass flow rate (0.720 kg/sec) b) the compressor displacement (0.1283 cu.m./sec) c) the expander displacement (0.0964 cu.m./sec d) the COP (2.35)

PRODUCT LOAD

Cpa Cpb

Q1 Q2 Q3

Q = Q1 + Q2 + Q3

Q - product loadQ1 - heat to cool product from t2 to tf

Q2 - heat to freezeQ3 - heat to cool product from tf to final storage temperature t2

Q1 = m Cpa (t1 - tf) KJ/min Q2 = mhL KJ/min Q3 = m Cpb (tf - t2) where: m - mass rate in kg/min t1 - entering temperature in C tf - freezing temperature in C t2 - storage temperature in C Cpa - specific heat above freezing, KJ/kg-C or KJ/kg-K Cpb - specific heat below freezing, KJ/kg-C or KJ/kg-K hL - latent heat of freezing of product, KJ/kg Q = m [Cpa (t1 - tf) + hL + Cpb (tf - t2)] KJ/min

SAMPLE PROBLEMS 1. Compute the heat to be removed from 110 kg of lean beef if it were to be cooled from 20C to 4C, after which it is frozen and cooled to -18C. Specific heat of beef above freezing is given as 3.23 KJ/kg-C, and below freezing is

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Q=m [Cpa( t1 - tf )+ h L+ Cpb( t f - t2 )]211

Tons

1.68 KJ/kg-C. Freezing point of beef is -2.2C, and latent heat of fusion is 233 KJ/kg. Given: m = 110 kg Cpa = 3.23 KJ/kg-C t1 = 20C Cpb = 1.68 KJ/kg-C t2 = 4C hL = 233 KJ/kg tf =-2.2C t3 = -18C Q = m[Cpa(t1 - t2) + Cpa(t2 - tf) + hL + Cpb(tf - t3)] Q = 36 438 KJ

PREPARED BY: ENGR. YURI G. MELLIZA

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