Refrigeration

Topics: Refrigeration, Gas compressor, Heat transfer Pages: 8 (945 words) Published: June 12, 2013
Basic Refrigeration Cycle

Reversed Carnot Cycle

Reversed Carnot Cycle Refrigeration Cycle
Dr. M. Zahurul Haq
Professor Department of Mechanical Engineering Bangladesh University of Engineering & Technology (BUET) Dhaka-1000, Bangladesh zahurul@me.buet.ac.bd http://teacher.buet.ac.bd/zahurul/

ME 415: Refrigeration & Building Mechanical Systems

e720

Coefficient of Performance, COP =

TL QL = Wnet TH − TL
ME 415 (2011) 2 / 14

c Dr. M. Zahurul Haq (BUET)

Refrigeration Cycle Reversed Carnot Cycle

ME 415 (2011)

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c Dr. M. Zahurul Haq (BUET)

Refrigeration Cycle Reversed Carnot Cycle

Basic Refrigeration Cycle

Basic Refrigeration Cycle

Wet Compression in Carnot Cycle vs. Dry Compression

Expansion Process in Carnot Cycle
Carnot cycle demands that the expansion take place isentropically and that the resulting work be used to help drive the compressor. Practical difficulties, however, militate against the expansion engine: the possible work that can be derived from the engine is small fraction that must be supplied to the compressor. practical problems such as lubrication intrude when a fluid of two phases drives the engine. the economics of the power recovery have in past not justified the cost of the expansion engine. A throttling device, such as a valve or other restriction, is almost universally used for this purpose.

During compression, droplets present in liquid are vaporised by the internal heat transfer process which requires finite time. High-speed compressors are susceptible to damage by liquid because of the short time available. In wet compression, the droplets of the liquids may wash the lubricating oil from the walls of the cylinder, accelerating wear. Dry compression takes place with no droplets and is preferable. Liquid refrigerants may be trapped in the head of reciprocating compressor by the rising piston, possibly damaging the valves or the cylinder head.

c Dr. M. Zahurul Haq (BUET)

Refrigeration Cycle

ME 415 (2011)

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c Dr. M. Zahurul Haq (BUET)

Refrigeration Cycle

ME 415 (2011)

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Basic Refrigeration Cycle

Ideal Vapour Compression Refrigeration Cycle

Basic Refrigeration Cycle

Ideal Vapour Compression Refrigeration Cycle

Ideal Vapour Compression Refrigeration Cycle

e722 e723

e494 c Dr. M. Zahurul Haq (BUET) Refrigeration Cycle ME 415 (2011) 5 / 14 c Dr. M. Zahurul Haq (BUET) Refrigeration Cycle ME 415 (2011) 6 / 14

Basic Refrigeration Cycle

Ideal Vapour Compression Refrigeration Cycle

Basic Refrigeration Cycle

Ideal Vapour Compression Refrigeration Cycle

Processes of VC System

QL QH Win COP
e496

= = = =

Q41 = m(h1 − h4 ) Q23 = m(h2 − h3 ) W12 = m(h2 − h1 ) QL /Win

1 → 2: Isentropic compression, Pevap → Pcond 2 → 3: Isobaric heat rejection, QH 3 → 4: Isenthalpic expansion, Pcond → Pevap 4 → 1: Isobaric heat extraction, QL c Dr. M. Zahurul Haq (BUET) Refrigeration Cycle ME 415 (2011) 7 / 14 c Dr. M. Zahurul Haq (BUET) Refrigeration Cycle ME 415 (2011) 8 / 14 e721

Simple vapour compression cycle with pressure & enthalpy values for R134a

Basic Refrigeration Cycle

Ideal Vapour Compression Refrigeration Cycle

Basic Refrigeration Cycle

Ideal Vapour Compression Refrigeration Cycle

Example
A theoretical single stage cycle using R134a as refrigerant operates with a condensing temperature of 30o C and an evaporator temperature of -20o C. The system produces 50 kW of refrigeration effect. Estimate: 1 2

Effect of Evaporator Temperature
R134a: RE = 50 kW, Tcond = 30 C
o

10 9 8 7 6

0.40 0.39 0.38 0.37

Coefficient of performance, COP Refrigerant mass flow rate, m COP

Ref. flow rate COP

0.36 0.35 0.34 0.33 0.32 0.31 0.30

5 4 3 2 1 0

QL Win COP
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= Q41 = m(h1 − h4 ) = 50 kW m = 0.345 Kg/s = W12 = m(h2 − h1 ) = 12.5 kW = QL /Win = 50.0/12.5 = 4.0

-50
e497

-45

-40

-35

-30

-25

-20
o

-15

-10

-5

0

Tevap ( C)
Refrigeration...
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