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In this report we will discuss about the important parameters of an AC that works on a refrigeration cycle. In this report we will find the temperature of AC at all its working stages , its enthalpy at each stage. We will also determine the COP of an AC that decides its performance and also has been used as the measure to compare two products. In this report we will discuss the analysis of these parameters of the AC installed in my house.
It will help us to get more knowledge of working about my AC.
The AC unit under discussion is an LG 1.5 Ton 5 Star Dual Inverter Split AC (Copper KS-Q18HNZD White), with indoor dimensions of 998 mm x 330 mm x 210 mm and outdoor dimensions of 770 mm x 545 mm x 288 mm. This model is notable for its energy efficiency and inverter technology, designed to provide effective cooling with minimal energy consumption.
In this section of report we will discuss the data obtained about compressor form the compressor’s came with air conditioner.
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It is a rotary compressor based refrigerant..This compressor uses R-134a as refrigerant. The data obtained from compressor’s data regarding pressure is as follows:
These are four main components of the refrigeration cycle: the compressor, condenser, expansion unit, and evaporator. Refrigerant stays trapped in the refrigerant system between these four elements. The refrigerant begins as a cool vapor and goes to the part one: the compressor.
The compressor is commonly known as the refrigeration cycle engine (Turns, Stephen (2006) , p : 152) it uses the most power out of the components of the HVAC device and drives the coolant through the machine. The cold, ionized refrigerant is converted into a very hot and elevated-pressure vapor when being concentrated.
The role of the condenser is to cool the refrigerant such that it transforms, or condenses, from a gas to a liquid. (The Ideal Vapor-Compression Cycle Archived 2007-02-26 , p : 246) It occurs as moist outside air is packed with soft, ionized refrigerant over the condenser wire. It enables heat to be moved from the refrigerant to the colder outside air, where the surplus heat is released into the air. The coolant now enters the expanding system as a hot liquid with a strong pressure. The expanding state is responsible for slowing down the coolant stress such that it will steam (evaporate) in the evaporator quite effectively.
Because the refrigerant becomes a liquid combination of fluid and gas (vapor), it continues flowing into the evaporator. The evaporator is responsible for refrigerating the air that goes through space through heating (evaporating) the refrigerant that runs via it. (Fundamentals of Engineering Thermodynamics , p : 348) This occurs because, when cold refrigerant passes through the evaporator wire, warm air is forced over the evaporator. Heat move from the air to the refrigerant, which immediately cools the air until it is vented into the vacuum.
By the help of the refrigerant property table ('The Basic Vapor Compression Cycle and Component” , 2006 , p : 543), we get
T2’ = T3 = 36C = 36 +273 = 309 K
T1 = T4 = -7C = 7 + 273 = 266 K
So, Temperature at compressor entrance T1= 266 K
Temperature at compressor exit T2= 309 K
Temperature at condenser exit T3= 309 K
Temperature at expansion valve exit T4 = 266 K
By the use of refrigerant enthalpy table ('The Basic Vapor Compression Cycle and Component” , 2006 , p : 543) , we get
Enthalpy at entrance of evaporator hf3 = h4 =70.55kJ/kg
Enthalpy at exit of expansion valve hf3 = h4 =70.55kJ/kg
Enthalpy at vapor coming out from evaporator hf1 = hf4 = 293.62kJ/kg
Enthalpy at the exit of compressor h2 = 201.8 kJ/jkg
Enthalpy at the entrance of condenser h2 = 201.8 kJ/jkg
Enthalpy at the entrance of compressor h1 = 184.5 kJ/kg
Specific heat of the refrigerant cp = 0.64 kJ/jkg K
We can calculate that enthalpy of superheated vapour at point 2 as
h2s = h2 + cp (T2 - T2' )
=201.8 + 0.64 (317-309) = 206.92 kJ/kg.
The efficiency coefficient (COP) ( Cengel, Yunus A. and Michael A. Boles (2008) , p : 635) is a output metric that shows us how powerful a heat pump or air conditioner is when transmitting heat and how much electrical power it absorbs. Note, heating systems and air conditioners transfer heat from a low temperature environment and transport this 'uphill' heat to a high temperature region, so according to the rules of thermodynamics (heat rules), which claim that heat moves automatically from anything hot to anything cold. The 'uphill' heat transfer from a low temperature environment to a high temperature environment thus needs effort, and the COP determines how effectively a heat pump or air conditioner does this function by asking us how much power is needed to work it efficiently.
h2 = 201.8 kJ/jkg
h1 = 184.5 kJ/kg
hf3 = h4 =70.55kJ/kg
= 5.1
This detailed analysis of the AC's performance based on its refrigeration cycle reveals the intricate balance between temperature, pressure, and enthalpy changes. The calculated COP offers insight into the unit's efficiency, providing a comparative measure of its operational effectiveness. Understanding these parameters not only demystifies the AC's functionality but also aids in optimizing its performance for energy conservation and enhanced cooling efficiency.
Analysis of Air Conditioning Performance Based on Refrigeration Cycle. (2024, Feb 17). Retrieved from https://studymoose.com/document/analysis-of-air-conditioning-performance-based-on-refrigeration-cycle
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