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2
REFRIGERATION Refrigeration: 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 Conditioning 2. Food Preservation 3. Removing of heat from substances in chemical, petroleum and petrochemical plants 4. Special applications in the manufacturing and construction industries. TON OF REFRIGERATION
It is the heat equivalent to the melting of 1 ton (2000 lb) of water ice at 0C into liquid at
0C in 24 hours.
min
KJ 211TR
hr
BTU 12,000
24
2000(144)TR
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 WORK W = QR - QA
W = (TH - TL)(S) COEFFICIENT OF PERFORMANCE
LH
L
AR
A
A
TT
TCOP
QCOP
W
QCOP
3
1 4
TH
TL
R
2
where: QA - refrigerating effect or refrigerating capacity QR - heat rejected W - net cycle work VAPOR COMPRESSION CYCLE Components:
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)
11'1
k
1k
1
21'1
12
mRTVP
KW 1 -P
P
1)60(k
VkPW
KW 60
)hm(hW
System: CONDENSER (P = C) QR = m(h2-h3) KJ/min System: EXPANSION VALVE (h = C) h3 = h4
evaporator
Condenser
expansion
valve
compressor
1
2 3
4
QA
QR
3
44
44
4
fg
f
hh
hhx
System: EVAPORATOR (P = C)
ionRefrigerat of Tons
211
)hm(hQ
KJ/min )hm(hQ
41
A
41A
DISPLACEMENT VOLUME (For Reciprocating type of compressor) a. For Single Acting
sec
m
4(60)
Nn'LDV
32
D
π
b. For Double Acting without considering piston rod
sec
m
4(60)
Nn'LDV
32
D
2π
c. For Double Acting considering piston rod
sec
m D-2D
4(60)
LNn'V
3
22
D
π
KW PER TON OF REFRIGERATION
Ton
KW
)h-(60)(hTon
KW
41
)h-211(h12
COEFFICIENT OF PERFORMANCE
12
41
A
h-h
h-hCOP
W
QCOP
CUBIC METER/min PER TON OF REFRIGERATION A. For Single Acting
Ton-min
m
)h-4m(h
Nn'LD
Ton-min
m 3
41
23 211π
B. For double acting without considering piston rod
Ton-min
m
)h-4m(h
Nn'LD
Ton-min
m 3
41
23 211(2)π
C. For double acting considering piston rod
Ton-min
m d-2D
)h-4m(h
LNn'
Ton-min
m 3
22
41
3 211π
where; L - length of stroke, m D - diameter of bore, m d - diameter of piston rod, m N - no. of RPM n' - no. of cylinders VOLUMETRIC EFFICIENCY
4
%xηk
100 P
PC - C1
100% xV
Vη
1
1
2
v
D
1'
v
where: V1' - volume flow rate at intake, m3/sec VD - displacement volume, m3/sec
v - volumetric efficiency
1 - specific volume, m3/kg
v = [1 + C - C(P2/P1)1/k] x 100% EFFICIENCY A. Compression Efficiency
100% xWork Indicated
Work Idealη
cn
B. Mechanical Efficiency
100% xWorkShaft or Brake
Work Indicatedη
m
C. Compressor Efficiency
mη
cnη
c
c
η
100% xWorkShaft or Brake
Work Idealη
EFFECTS ON OPERATING CONDITIONS A. 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 decreases 3. 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.
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B. Effects on Increasing the Condensing Temperature P T h S 1. The refrigerating effect per unit mass decreases. 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 decreases 3. 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.
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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. 5. The work per ton decreases. 6. The heat rejected at the condenser per ton decreases. LIQUID-SUCTION HEAT EXCHANGER The function of the heat exchanger is: 1. To ensure that no liquid enters the compressor, and 2. 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
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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 ACTUAL VAPOR COMPRESSION CYCLE As 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.
P1 = 308.6 KPa P2 = 847.7 KPa h1 = 351.48 KJ/kg v1 = 0.0554 m3/kg h2 = 368 KJ/kg h3 = h4 = 233.5 KJ/kg
hf at 0 C = 200 KJ/kg
hg at 0 C = 351.48 KJ/kg
a) 0.22 b) 2.97 kg/sec c) 0.1645 m3/sec d) 49.06 KW e) 7.14
P1 = 584 KPa P2 = 1355 KPa h1 = 407.1 KJ/kg v1 = 40.36 L/kg h2 = 428 KJ/kg
h3 = h4 =243.1 KJ/kg
a) 94.91% b) 65.6%
c) 63.33%
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condenser
Evaporator
Evaporator
MULTIPRESSURE SYSTEMS A 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 Gas A 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-cooling Inter-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
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2-Compressors and 1-Evaporator
2-Compressors and 2-Evaporators
HP Compressor
LP Compressor
Condenser
Flash Tank and Intercooler
Evaporator
condenser
HP compressor
HP Evaporator
LP Compressor LP Evaporator
Flash Tank and Intercooler
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SAMPLE PROBLEMS 1. 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/kg from 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 with inter-cooling and removal of flash gas. The condensing temperature
is 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 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
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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
LP Compressor
Open Type
Cascade Condenser
Evaporator
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SAMPLE PROBLEMS 1. 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/kg 2. 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
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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
1P
P
k1
VkPW
k
1k
1
211
c
Expander Work
1P
P
k1
VkPW
k
1k
3
433
E
Net Work W = WC - WE
Coefficient of Performance
Work Net
Effect ingRefrigeratCOP
1
2 3
4
P
V
T
S
1
2
3
4
S = C S = C
P = C
P = C
15
t1
m
tf
m
t2
m
SAMPLE PROBLEMS 1. 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 load Q1 - heat to cool product from t2 to tf Q2 - heat to freeze Q3 - 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
Tons 211
)]t - (t C h )t - (t [C m Q
2fpbLf1pa
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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
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
