Free Essay

Operation of Reciprocating Compressors Used in Vcr Systems

In:

Submitted By Mikejnr
Words 2269
Pages 10
Question
With the aid of a suitable diagram, describe the operation and construction of a reciprocating compressor employed in a VCR system (30 marks).

Introduction
A compressor is the most important and often the costliest component of any vapour compression refrigeration system (VCRS). The function of a compressor in a VCRS is to continuously draw the refrigerant vapour from the evaporator, so that a low pressure and low temperature can be maintained in the evaporator at which the refrigerant can boil extracting heat from the refrigerated space. The compressor then has to raise the pressure of the refrigerant to a level at which it can condense by rejecting heat to the cooling medium in the condenser.
Reciprocating compressors
Description
Reciprocating compressor is the workhorse of the refrigeration and air conditioning industry. It is the most widely used compressor with cooling capacities ranging from a few Watts to hundreds of kilowatts. Modern day reciprocating compressors are high speed (≈ 3000 to 3600 rpm), single acting, single or multi-cylinder (up to 16 cylinders) type.

Construction features and Principle of Operation

Fig. 1

Figure 1 shows the schematic of a reciprocating compressor. Reciprocating compressors consist of a piston moving back and forth in a cylinder, with suction (Sv) and discharge (Dv) valves to achieve suction and compression of the refrigerant vapour. Its construction and working are somewhat similar to a two-stroke engine, as suction and compression of the refrigerant vapour are completed in one revolution of the crank. The suction side of the compressor is connected to the exit of the evaporator, while the discharge side of the compressor is connected to the condenser inlet. The suction (inlet) and the discharge (outlet) valves open and close due to pressure differences between the cylinder and inlet or outlet manifolds respectively. The pressure in the inlet manifold is equal to or slightly less than the evaporator pressure. Similarly the pressure in the outlet manifold is equal to or slightly greater than the condenser pressure. The purpose of the manifolds is to provide stable inlet and outlet pressures for the smooth operation of the valves and also provide a space for mounting the valves.

The valves used are of reed or plate type, which are either floating or clamped. Usually, backstops are provided to limit the valve displacement and springs may be provided for smooth return after opening or closing. The piston speed is decided by valve type. Too high a speed will give excessive vapour velocities that will decrease the volumetric efficiency and the throttling loss will decrease the compression efficiency.

Performance of reciprocating compressors For a given evaporator and condenser pressures, the important performance parameters of a refrigerant compressor are:

a) The mass flow rate (m) of the compressor for a given displacement rate
b) Power consumption of the compressor (WC)
c) Temperature of the refrigerant at compressor exit, Td, and
d) Performance under part load conditions
The mass flow rate decides the refrigeration capacity of the system and for a given compressor inlet condition, it depends on the volumetric efficiency of the compressor. The volumetric efficiency, ηV is defined as the ratio of volumetric flow rate of refrigerant to the maximum possible volumetric flow rate, which is equal to the compressor displacement rate, i.e.,

For a given evaporator and condenser temperatures, one can also use the volumetric refrigeration capacity (kW/m3) to indicate the volumetric efficiency of the compressor. The actual volumetric efficiency (or volumetric capacity) of the compressor depends on the operating conditions and the design of the compressor.
The power consumption (kW) or alternately the power input per unit refrigeration capacity (kW/kW) depends on the compressor efficiency (ηC), efficiency of the mechanical drive (ηmech) and the motor efficiency (ηmotor). For a refrigerant compressor, the power input (WC) is given by: Where Wideal is the power input to an ideal compressor.
The temperature at the exit of the compressor (discharge compressor) depends on the type of refrigerant used and the type of compressor cooling. This parameter has a bearing on the life of the compressor.
The performance of the compressor under part load conditions depends on the type and design of the compressor.

a) Ideal reciprocating compressor:

An ideal reciprocating compressor is one in which:
i. The clearance volume is zero, i.e., at the end of discharge process, the volume of refrigerant inside the cylinder is zero. ii. No pressure drops during suction and compression iii. Suction, compression and discharge are reversible and adiabatic
Figure 2 shows the schematic of an ideal compression process on pressure-volume and pressure-crank angle (θ) diagrams. As shown in the figures, the cycle of operations consists of:
Process D-A: This is an isobaric suction process, during which the piston moves from the Inner Dead Centre (IDC) to the Outer Dead Centre (ODC). The suction valve remains open during this process and refrigerant at a constant pressure Pe flows into the cylinder.
Process A-B: This is an isentropic compression process. During this process, the piston moves from ODC towards IDC. Both the suction and discharge valves remain closed during the process and the pressure of refrigerant increases from Pe to Pc.
Process B-C: This is an isobaric discharge process. During this process, the suction valve remains closed and the discharge valve opens. Refrigerant at a constant P is expelled from the compressor as the piston moves to IDC. Fig 2 Ideal reciprocating compressor on P-V and P-θ diagrams

Since the clearance volume is zero for an ideal compressor, no gas is left in the compressor at the end of the discharge stroke, as a result the suction process D-A starts as soon as the piston starts moving again towards ODC. The volumetric flow rate of refrigerant at suction conditions is equal to the compressor displacement rate hence, the volumetric efficiency of the ideal compressor is 100 percent.

The mass flow rate of refrigerant of an ideal compressor is given by: Thus for a given refrigeration capacity, the required size of the compressor will be minimum if the compressor behaves as an ideal compressor.

The swept volume,Vsw of the compressor is given by: Where n = Number of cylinders
N = Rotational speed of compressor, revolutions per second
D = Bore of the cylinder, m
L = Stroke length, m

Work input to the ideal compressor: The total work input to the compressor in one cycle is given by:

Wid = W D-A+ W A-B+ W B-C

Where,
W D-A = Work done by the refrigerant on the piston during process D-A
= Area under line D-A on P-V diagram = -Pe.VA

W A-B = Work done by the piston on refrigerant during compression A-B
= Area under the curve A-B on P-V diagram =

WB-C = Work done by the piston on the refrigerant during discharge B-C
= Area under line B-C = Pc.VB

Thus the work input to the ideal compressor per cycle is equal to the area of the cycle on P-V diagram.
The specific work input, wid (kJ/kg) to the ideal compressor is given by:

Where Mr is the mass of refrigerant compressed in one cycle and v is the specific volume of the refrigerant.

The power input to the compressor W c is given by:

The mean effective pressure (mep) for the ideal compressor is given by:

The concept of mean effective pressure is useful for real compressors as the power input to the compressor is a product of mep and the swept volume rate.
Thus the power input to the compressor and its mean effective pressure can be obtained from the above equation if the relation between v and P during the compression process A-B is known. The above equation is valid for both isentropic and non-isentropic compression processes, however, the compression process must be reversible, as the path of the process should be known for the integration to be performed.

For the isentropic process, Pv k= constant, hence the specific work of compression wid can be obtained by integration, and it can be shown to be equal to: In the above equation, k is the index of isentropic compression. If the refrigerant behaves as an ideal gas, then k = γ. In general, the value of k for refrigerants varies from point to point, and if its value is not known, then an approximate value of it can be obtained from the values of pressure and specific volume at the suction and discharge states as The work of compression for the ideal compressor can also be obtained by applying energy balance across the compressor, Fig 3 Since the process is assumed to be reversible and adiabatic and if we assume changes in potential and kinetic energy to be negligible, then from energy balance across the compressor:

The above expression can also be obtained from the thermodynamic relation: The above expression is valid only for reversible, adiabatic compression. Fig 3. Energy balance across a steady flow compressor

b) Ideal compressor with clearance:
In actual compressors, a small clearance is left between the cylinder head and piston to accommodate the valves and to take care of thermal expansion and machining tolerances. As a thumb rule, the clearance C in millimetres is given by:

C = (0.005L + 0.5) mm, where L is stroke length in mm This space along with all other spaces between the closed valves and the piston at the inner dead centre (IDC) is called as Clearance volume, Vc. The ratio of the clearance volume to the swept volume is called as Clearance ratio, ε, i.e.,

The clearance ratio ε depends on the arrangement of the valves in the cylinder and the mean piston velocity. Normally ε is less than 5 percent for well-designed compressors with moderate piston velocities (≈ 3 m/s), however, it can be higher for higher piston speeds.
Due to the presence of the clearance volume, at the end of the discharge stroke, some amount of refrigerant at the discharge pressure Pc will be left in the clearance volume. As a result, suction does not begin as soon as the piston starts moving away from the IDC, since the pressure inside the cylinder is higher than the suction pressure (P c> Pe).

As shown in Fig 4, suction starts only when the pressure inside the cylinder falls to the suction pressure in an ideal compressor with clearance. This implies that even though the compressor swept volume, V SW= VA-VC, the actual volume of the refrigerant that entered the cylinder during suction stroke is VA-VD. As a result, the volumetric efficiency of the compressor with clearance, ηV,cl is less than 100 percent, i.e.,

Fig 4 Ideal reciprocating compressor with clearance

From Fig.4, the clearance volumetric efficiency can be written as:

Substituting the above equation in the expression for clearance volumetric efficiency; we can show that:

Since the mass of refrigerant in the cylinder at points C and D are same, we can express the ratio of cylinder volumes at points D and C in terms of ratio of specific volumes of refrigerant at D and C, i.e., Hence, the clearance volumetric efficiency is given by: If we assume the re-expansion process also to follow the equation Pvk=constant, then:

Hence the clearance volumetric efficiency is given by: The above expression holds good for any reversible compression process with clearance. If the process is not reversible, adiabatic (i.e., non-isentropic) but a reversible polytrophic process with an index of compression and expansion equal to n, then k in the above equation has to be replaced by n, i.e., in general for any reversible compression process;

The above expression shows that ηV,cl ↓ as rp↑ and ε↑ as shown in Fig. 5. It can also be seen that for a given compressor with fixed clearance ratio ε, there is a limiting pressure ratio at which the clearance volumetric efficiency becomes zero. This limiting pressure ratio is obtained from the equation: The mass flow rate of refrigerant compressed with clearance mcl is given by:

Thus the mass flow rate and hence the refrigeration capacity of the system decreases as the volumetric efficiency reduces, in other words, the required size of the compressor increases as the volumetric efficiency decreases.

Fig. 5

Work input to the compressor with clearance:
If we assume that both compression and expansion follow the same equation Pvn = constant (i.e., the index of compression is equal to the index of expansion), then the extra work required to compress the vapour that is left in the clearance volume will be exactly equal to the work output obtained during the re-expansion process. Hence, the clearance for this special case does not impose any penalty on work input to the compressor. The total work input to the compressor during one cycle will then be equal to the area A-B-C-D-A on P-V diagram.

The specific work with and without clearance will be given by the same expression: However, since the mass of refrigerant compressed during one cycle is different with and without clearance, the power input to the compressor will be different with and without clearance. The power input to the compressor and mean effective pressure (mep) with clearance are given by:

Thus the power input to the compressor and mep decrease with clearance due to decrease in mass flow rate with clearance.
If the process is reversible and adiabatic (i.e., n = k), then the power input to the compressor with clearance is given by:

Similar Documents

Free Essay

Maglev

...*3963103* [3963] – 103 T.E. (Petroleum) (Semester – I) Examination, 2011 DRILLING & PRODUCTION OPERATIONS (2003 Course) Time : 3 Hours Max. Marks : 100 Instructions : 1) Question Nos. 1 and 5 are compulsory. Out of the remaining attempt 2 questions from Section I and 2 questions from Section II. 2) Answers to the two Sections should be written in separate books. 3) Neat diagrams must be drawn wherever necessary. 4) Black figures to the right indicate full marks. 5) Use of Logarithmic Tables, Slide Rule, Mollier Charts, Electronic Pocket Calculator and Steam Tables is allowed. 6) Assume suitable data, if necessary. SECTION – I 1. What are different systems on a drilling rig ? Explain any one in detail with suitable diagramme. 18 2. a) Calculate Bottom hole pressure if well depth is 2500 m and mud weight is 1.2 gm/cc. b) Calculate mud weight if mud gradient is 0.87 psi/ft. ′ c) Calculate volume bbl/meter for drill pipe O.D. = 5′ inch and I.D. = 4.276 inch. 2 2 2 10 8 8 16 d) Draw circulation system on a drilling rig. 3. a) Discuss IADC classification of a bit in details. b) Discuss different factors affecting rate of penetration in details. 4. Write short note on : i) Coring ii) Fishing tools iii) BOP iv) Directional well P.T.O. [3963] – 103 -2- *3963103* SECTION – II 5. a) Discuss different types of casings and function of the casings in brief. b) Discuss different types of well completion techniques. 6. a) Discuss primary cementation process with...

Words: 172166 - Pages: 689

Free Essay

Industrial Engineering

...The McGraw-Hill Companies, Inc. All rights reserved. Printed in the United States of America. Except as permitted under the United States Copyright Act of 1976, no part of this publication may be reproduced or distributed in any form or by any means, or stored in a database or retrieval system, without prior written permission of the publisher. This McGraw-Hill Create text may include materials submitted to McGraw-Hill for publication by the instructor of this course. The instructor is solely responsible for the editorial content of such materials. Instructors retain copyright of these additional materials. ISBN-10: 1121789048 ISBN-13: 9781121789043 McGraw-Hill Create™ Review Copy for Instructor Espinoza. Not for distribution. Contents 1. Preface 1 2. Methods, Standards, and Work Design: Introduction 7 Problem-Solving Tools 27 3. Tex 29 4. Operation Analysis 79 5. Manual Work Design 133 6. Workplace, Equipment, and Tool Design 185 7. Work Environment Design 239 8. Design of Cognitive Work 281 9. Workplace and Systems Safety 327 10. Proposed Method Implementation 379 11. Time Study 413 12. Performance Rating and Allowances 447 13. Standard Data and Formulas 485 14. Predetermined Time Systems 507 15. Work Sampling 553 16. Indirect and Expense Labor Standards 585 17. Standards Follow-Up and Uses 611 18. Wage Payment 631 19. Training and Other Management Practices 655 20. Appendix 1: Glossary 685 21. Appendix 2: Helpful Formulas 704 22. Appendix 3: Special Tables 706 23. Index...

Words: 294686 - Pages: 1179

Premium Essay

Manager

...Learning with Cases INTRODUCTION The case study method of teaching used in management education is quite different from most of the methods of teaching used at the school and undergraduate course levels. Unlike traditional lecture-based teaching where student participation in the classroom is minimal, the case method is an active learning method, which requires participation and involvement from the student in the classroom. For students who have been exposed only to the traditional teaching methods, this calls for a major change in their approach to learning. This introduction is intended to provide students with some basic information about the case method, and guidelines about what they must do to gain the maximum benefit from the method. We begin by taking a brief look at what case studies are, and how they are used in the classroom. Then we discuss what the student needs to do to prepare for a class, and what she can expect during the case discussion. We also explain how student performance is evaluated in a case study based course. Finally, we describe the benefits a student of management can expect to gain through the use of the case method. WHAT IS A CASE STUDY? There is no universally accepted definition for a case study, and the case method means different things to different people. Consequently, all case studies are not structured similarly, and variations abound in terms of style, structure and approach. Case material ranges from small caselets (a few paragraphs...

Words: 239776 - Pages: 960