Refrigerant 22 enters the compressor of an ideal vaporcompression refrigeration system as satura

Refrigerant 22 enters the compressor of an ideal...

Key Concepts

Vapor?Compression Refrigeration Cycle

This cycle is a common refrigeration method that involves the circulation of a refrigerant through a series of processes: evaporation, compression, condensation, and expansion. It is designed to extract heat from a low?temperature source and reject it to a higher temperature sink, using a compressor to raise the refrigerant pressure and temperature. Understanding this cycle is fundamental for analyzing system performance metrics such as compressor power and coefficient of performance.

Refrigerant Thermodynamic Properties

Thermodynamic properties of refrigerants, such as temperature, pressure, specific volume, and enthalpy at various states (saturated vapor or liquid), are essential for accurately determining state conditions throughout the refrigeration cycle. These properties are typically obtained from refrigerant property tables or computational tools, and they enable calculations of heat transfer, work, and efficiency.

Phase Change and Saturation States

Recognizing the saturation state, whether the refrigerant is a saturated vapor or liquid, is crucial because the properties significantly differ at phase change conditions. Saturated conditions indicate that the refrigerant is at the brink of changing phase, which affects volume, pressure, and energy interactions and is critical for determining mass flow rates and energy balances.

Mass Flow Rate Calculation

The mass flow rate is often derived from the given volumetric flow rate using the density (or specific volume) of the refrigerant at a given state. This step is key because many energy and work calculations in the cycle, such as compressor power, are based on the mass flow rate rather than the volumetric flow rate.

Power Calculation in Compressors

Calculating the compressor power involves determining the work input required to compress the refrigerant from inlet to outlet conditions. This is typically performed by evaluating the specific work (work per unit mass) associated with the compression process and then multiplying it by the mass flow rate. Accurate power calculations are essential for designing and evaluating the energy efficiency and performance of the system.

Coefficient of Performance (COP)

The coefficient of performance for a refrigeration cycle is defined as the ratio of the useful refrigeration effect (cooling capacity) to the work input (typically the compressor work). It is a critical figure of merit in refrigeration systems, providing a measure of efficiency; a higher COP indicates a more efficient system.

Unit Conversion for Refrigerating Capacity

In many refrigeration applications, capacity is expressed in tons, where one ton of refrigeration is a standardized unit equivalent to a specific amount of cooling power. Converting between units such as kW and tons allows for consistent comparison and evaluation of refrigeration systems, making it essential to understand the conversion factors involved.

Transcript

00:01 At refrigeration, heat, pop and air conditioning system for this id before compression revolution cycle want to determine the compressor power due to the refrigeration capacity as well as the coefficient of performance, alright.

00:21 So we have this.

00:26 So this is point one, point two, and let's call here point two point three point four okay and to solve this problem we just applied the governing principles for solving problems in refrigeration compression refusional cycle rights okay this is this okay this is the entropy axis, inflageo per kilogram per kelvin, and that's the temperature axis.

01:27 We are given that at the evaporator, sorry, at the condensate 6 degrees celsius, and that is 10 bar, 10 bar.

01:39 Okay, at the evaporator, the temperature is minus 45 degree.

01:48 Degree the refrigerant is arrow 22 that's revision 22 right given the the odometric flow rates this is called it is equal to 20 cubic meter per minutes all right so at at at 5 degrees celsius 1 .45 degrees celsius, h .1, is equal to 2095, a kilo -kilio per kilogram.

02:59 Of course, the saturated vapor line is equal to also s1, is equal to 1 .0, 1 ,2.

03:12 One two six kilo per kilogram per kelvin same unit as we have here okay so all the entire besides kloge about kilogram entropy's like per kilogram so at ten banner hg is saturated vipolein to that project that point alone such as well as a good line, i give you equal to h2.

03:54 Check it is to give you to 7 .28 kilojou per kilogram.

04:06 That's a point among the saturated liquid line, is h3.

04:15 It's going to give you equal to 73 .3 0 kilo per kilogram okay if you check sg is equal to s a point 2 is equal to 0 .8 952 kilo kilo per kilogram per kelvin at that point along the resulted ripple line because in that timber if you is this let's go to sr3, sorry, along the subter liquid line, okay? give you 0 .2748, alright.

05:24 So you can look for smg difference in entropy, put up points along the, considering the subcited paper line and liquid line, are the difference, okay? 3 that pressure, 0 .6 .204 per kilo per kilogram per kilogram.

05:58 You can get also the entropy is going to give you 183.

06:09 Okay.

06:10 0 .99 kilo per kilogram.

06:14 Okay.

06:15 Of course, we know that s2 prime is equal to s at 1 right so from this you can look for this this is a 2 prime is equal to sorry is equal to s prime minus as along the saturated liquid line as entropy there okay difference in entropy consider saturated with a pipeline and liquid line with that particular point along that pressure so if you plug in all the values in there you arrive at one point one eight okay nine two nine two one point one eight nine two okay then we can go ahead now and do other things okay we know that it's flow right is equal to row density area velocity okay okay which is also equal to area time velocity over specific volume okay right so we check this we given this in you can change it of cubic meter pair our q, remember that we're related to our monometric flow rate to be equal to q, it's equal to a, b.

08:37 If we do that, converting it to cubic meter per second is going to give us 1 times, over 1 times 60, 10 .60, which will give us 0 .0056 cubic meter, a second okay all right and if we apply this apply this okay okay annoying this checking for this one we're gonna give us at point one you're gonna give us 0 .2564 cubic meter 0 .2564 okay cubic meter 2 .564 okay so if you check this on the table, you have that.

10:01 So if you plug it now into that equation at the start, you'll have mass...

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