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Thermodynamics & Heat Transfer: Refrigeration Cycle - HCMUT 2016, Schemes and Mind Maps of Thermodynamics

A detailed course outline for a thermodynamics & heat transfer course offered at hcmut in 2016. The course covers topics such as the first and second laws of thermodynamics, basic processes of ideal gas, pure substance, vapor power cycle, refrigeration cycle, atmospheric air and air conditioning processes, basics of heat transfer, heat conduction, heat convection, heat radiation, and heat exchanger equipments. The course includes lectures, tests, a midterm exam, a final exam, and a practical experimental component. The course is taught by dr. Tung ha – anh and is divided into 12 weeks, with 4 teaching units per week.

Typology: Schemes and Mind Maps

2020/2021

Uploaded on 03/23/2024

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Download Thermodynamics & Heat Transfer: Refrigeration Cycle - HCMUT 2016 and more Schemes and Mind Maps Thermodynamics in PDF only on Docsity! THERMODYNAMICS & HEAT TRANSFER  Duration: 36 hours during 12 weeks (4 teaching units/week) - Week 1  8 : Thermodynamics + Week 8 : Midterm Exam - Week 9  12 : Heat Transfer Final Exam 20% 40%  Instructor: HÀ ANH TÙNG – Department of Heat and Refrigeration Technology Instructor: Dr. Tung Ha – Anh HCMUT 2/2016 + Week 4 : Test 1 + Week 10 : Test 2 15% 10% - Week 9  13 : Experimental 15% 1 Objectives of the course  to provide students with knowledge and skills required to apply the basic principles of thermodynamics and heat transfer to perform calculations and explain thermal engineering applications: Ex: - Thermal power plants - Refrigeration and air conditioning systems - Dryers, boilers 2/2016 Instructor: Dr. Tung Ha – Anh HCMUT - Internal combustion engines - Heat exchangers, etc. 2 CHAPTER 6 : Refrigeration Cycle 6.1 Introduction 6.2 Basic refrigeration cycle 5 2/2016 Instructor: Dr. Tung Ha – Anh HCMUT Part 1: THERMODYNAMICS 6 6.1 Introduction 2/2016 Instructor: Dr. Tung Ha – Anh HCMUT  is the cycle applied in Air conditioners, Refrigerators, etc. What is Refrigeration? Refrigeration is the process of removing heat from one substance and transferring it to another substance. 7 2/2016 Instructor: Dr. Tung Ha – Anh HCMUT Evaporator refrigerant vapor mixture of liquid and vapor refrigerant B A air 10 2/2016 Instructor: Dr. Tung Ha – Anh HCMUT Compressor low-pressure refrigerant vapor from evaporator C B high-pressure refrigerant vapor to condenser 11 2/2016 Instructor: Dr. Tung Ha – Anh HCMUT Condenser C D refrigerant vapor liquid refrigerant outdoor air 12 2/2016 Instructor: Dr. Tung Ha – Anh HCMUT Refrigerants TABLES  Some types of refrigerant: Fréon 22 (R-22) Ammonia (NH3) Fréon 134A (R-134A) Saturated Tables Superheated Tables Pure substance 15 2/2016 Instructor: Dr. Tung Ha – Anh HCMUT  Energy transfer in refrigeration cycle 1-2: Adiabatic compression wmn 4 2 1 3 q1 2-3: Isobaric condensation 3-4: Constant enthalpy process ( i4 = i3 ) q2 4-1: Isobaric evaporation Compressor Expansion valve Evaporator Condenser 16 2/2016 Instructor: Dr. Tung Ha – Anh HCMUT  Calculation of refrigeration cycle i log p * Example: p3 , T1 given, with R-22 refrigerant State 1: T1 R22 Saturated Table p1 and ii 1 ss 1 State 2: Superheated, having p2 = p3 s2 = s1 R-22 Superheated Table 2i 17 2/2016 Instructor: Dr. Tung Ha – Anh HCMUT  Exercise 20 2/2016 Instructor: Dr. Tung Ha – Anh HCMUT Solution: Having:  Refrigerant: R-22  p3 = 18 bar  t1 = 10oC  Cooling capacity = G*q41 = 10kW 1) Determine enthalpy at 1, 2, 3, 4 - From t1 , p3 known: 21 2/2016 Instructor: Dr. Tung Ha – Anh HCMUT 2) Condenser capacity: 2 3 Condenser   )(321 kWiiGQ  iwhich G (kg/s) is the flow rate of R-22 , determined from the evaporator capacity (cooling capacity): skg ii G / 10 41       )(10 41 32 1 kW ii ii Q    22 2/2016 Instructor: Dr. Tung Ha – Anh HCMUT Trang Thai | Nhiét do Ap suat Entanpi Entropi D6 khé °C bar kJ/kg kJ/(kg.K) % 1. 10 6 709,55 1,7510 - 2. 68,4 18 737,50 1,7510 - 3. 47 18 558,65 1,1941 0 4. 6 6 558,65 1,2230 25,85 3. XAc dinh nang suat lanh Qo (kW), nang sudt ngung ty Qx (kW), c6ng sudt may nén (kW), hé s6 lam lanh. Oy = Gari, —i,)=104,79 kW Ox =Gan Ai, —i,)=124,2 kW N =O, -Q, =19,41 kW Q eé=—=5,4 N 4. Xdc dinh luu lugng nuéec gidi nhiét qua binh ngung G, (kg/h). __ % "¢. At pr n =4,945 kg/s =17.800 kg/h 25 --------------------------------------- END OF CHAPTER 6 26 2/2016 Instructor: Dr. Tung Ha – Anh HCMUT
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