Compression of air in a compressor is an adiabatic process. If air at 20^∘ C and 1 atm pressure

Compression of air in a compressor is an adiabatic process....

Key Concepts

Adiabatic Process

An adiabatic process is one in which no heat is exchanged between the system and its surroundings. This means that any change in the internal energy of the gas is solely due to the work done on or by the gas. In thermodynamic analysis, adiabatic processes are often idealized to simplify calculations, using conservation of energy to relate changes in pressure, volume, and temperature.

Specific Heat Ratio

The specific heat ratio, often denoted as ? (gamma), is the ratio of the specific heat at constant pressure (Cp) to the specific heat at constant volume (Cv). This ratio is crucial in determining how a gas behaves during compression or expansion, particularly in adiabatic processes, as it directly influences the relationship between pressure and temperature.

Temperature-Pressure Relationship in Adiabatic Processes

In an adiabatic process involving an ideal gas, the temperature and pressure are related by the equation T2 = T1 (P2/P1)^((?-1)/?). This relationship enables the calculation of the final temperature after a change in pressure, making it a key concept in understanding and quantifying the outcome of adiabatic compression or expansion.

Work Calculation in Adiabatic Compression

The work performed in adiabatic compression is linked to the change in the internal energy of the gas. Since there is no heat transfer, the work done on the gas results in an increase in the gas's internal energy, which can be calculated using thermodynamic relations. This concept is essential for determining the energy or power, such as compressor horsepower, required for compressing a given mass of gas.

Units and Power Conversion in Thermodynamics

In thermodynamic calculations, converting energy into power units (like horsepower) is critical for practical engineering applications. This involves relating the energy per unit time required for a process (work done on the gas) to the commonly used units of power, ensuring that calculated values can be applied in real-world scenarios.

Transcript

0:00 Hi there.

00:01 So for this problem we have an air compressor that takes earth with an initial temperature of 18 celsius degrees and an initial pressure of one atmosphere.

00:20 And the levers compressed earth with a pressure 2 that is equal to 2 .3 atmospheres.

00:32 And the compressor operates with a power that is also given, and that is equal to 230 watts.

00:46 And we need to assume that the compressor operates adiabatically.

00:52 Now, for the first part of this problem, we need to find the temperature of the compressed earth.

00:59 So for part a, we need to find the temperature 2.

01:13 Now, air is mostly diatomic.

01:19 And for that, we are going to use that the constant gap, man, the ratio between the specific heat is 1 .4.

01:33 So we know that the product between the pressure and the volume to gatma is a constant.

01:49 So we know that the pressure 1 times the volume 1 elevated to gatma is equal to the pressure 2 times the volume 2 elevated to gatma.

02:01 Now what we can determine in here is the volume one in terms, the volume two in terms of the volume one.

02:10 Because if we solve in here for the volume two, we will have that that is.

02:17 The volume one times the gatma root of the pressure one over the pressure two.

02:33 And from there we just simply substitute the values that we have.

02:39 The volume one, we don't have it, so we leave it like that.

02:43 And the gamma is equal to 1 .4, and this is equal to one atmospheres.

02:55 Well, let me just...

02:59 Is one atmosphere over the pressure 2, that is 2 .3.

03:05 Atmospheres.

03:08 So from this, using our calculator, we obtain a value of 0 .552 the volume 1.

03:17 Now, we use this because we now want to obtain the temperature 2.

03:23 And for that, we can use the equation for an ideal gas because we know that the temperature 2 is equal to the temperature 1 times the ratio between the pressure 2 over the pressure 1 times the ratio between the volume 2 and the volume 1.

03:45 So what we need to do in here is to simply substitute all of these values.

03:51 Of course, when we substitute the volume in here for b2, you can see that volume 1 cancels with the 1 in the denominator, so we will only have that ratio.

04:06 In there...

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