Air enters the turbocharger compressor (see Fig. P8.51) of...

Air enters the turbocharger compressor (see Fig. P8.51) of an automotive engine at $100 \mathrm{kPa}, 30^{\circ} \mathrm{C}$ and exits at $200 \mathrm{kPa}$. The air is cooled by $50^{\circ} \mathrm{C}$ in an intercooler before entering the engine. The isentropic efficiency of the compressor is $75 \%$. Determine the temperature of the air entering the engine and the irreversibility of the compressioncooling process.

Key Concepts

Compressor Process

The compressor process in thermodynamics involves increasing the pressure of a gas by reducing its volume, typically resulting in a rise in temperature. In devices such as turbocharger compressors, air is drawn in at ambient conditions and compressed to a higher pressure. The process often deviates from ideal behavior due to various irreversibilities, making the analysis of real performance crucial for engine efficiency and design.

Intercooling

Intercooling refers to the process where a compressed gas is cooled between successive stages of compression. By lowering the temperature of the gas before further compression, intercooling reduces the work required in subsequent stages and limits the overall thermal stress on the system. This temperature reduction can also improve efficiency and performance by increasing the density of the air entering the engine.

Isentropic Efficiency

Isentropic efficiency is a measure of how effectively a real compressor approximates an ideal, reversible (isentropic) process. It is defined as the ratio of the minimum work (or enthalpy change) required under ideal conditions to the actual work (or enthalpy change) observed in the process. This metric helps quantify the losses due to irreversibilities like friction, turbulence, and heat losses, which are inevitable in practical applications.

Irreversibility

Irreversibility in a thermodynamic process denotes the losses that occur due to non-ideal factors such as friction, unrestrained expansion, heat transfer through a finite temperature difference, and other dissipative effects. These losses lead to an increase in entropy and a decrease in the usable energy available for work, making the quantification of irreversibility essential for assessing the efficiency and performance of engineering systems.

Transcript

00:01 In this question it is given that in a turbocharger compression air enters at a pressure of 100 kioskal and the temperature is 30 degrees centigrade the exit pressure is 200k kioskl now before the air enters into the engine air is cooled by 50 degrees centigrade in an intercooler the compressures isentropic efficiency is 75 25 % we are required to calculate the temperature of the air entering the engine and the irreversibility of the compression cooling process.

00:46 So let's see how to solve this question.

00:49 First of all, let's apply the formula to calculate exit temperature.

00:53 So we can write t2s upon t1 and this will be equals to p2 upon p1 to the power k minus 1 upon k.

01:05 Now substitute all the values so we can write p2 s upon t1 that means 30 degree centigrade or we can write 30 plus 273 so by adding 273 we have converted this temperature into kelvin now this will be equals to p2 upon p1 that means final pressure is 200 upon initial pressure is 100 to the power k that means ratio of specific heats and for air its value is 1 .4 minus 1 divided by 1 .4.

01:44 So when we further calculate, we get exit temperature t2s is equals to 369 .36 kelvin.

01:55 Now let's apply the formula to calculate isentropic work of the compression, that is, ws is equal to cp into t1 minus t2s.

02:12 Now substitute all the values, so we will have ws is equal to cp, that means the specific heat of air at constant pressure and its value is 1 .004 kilojure per kilogram kelvin, multiplied by t1, that means 303.

02:31 Minus t2s that means 369 .36.

02:37 So when we further calculate we get minus 66 .62 kilojoule per kilogram.

02:47 And now let's find the actual work of the compression.

02:51 So we have w actual is equals to w isentropic divided by efficiency isentropic.

03:00 Now substitute all the values so we can we can write w actual is equal to minus 66 .62 divided by efficiency isentropic that means 0 .75 and when we further calculate we get minus 88 .82 kilojoule per kilogram and now let's apply the formula to calculate final temperature corresponding to actual work so we can write w.

03:36 Actual is equal to tp into t1 minus t2 now substitute all the values so we can write minus 88 .82 and this will be equals to 1 .004 multiplied by t1 that means 303 minus t2 so when we further calculate we get t2 is equals to 3991 .46 kelvin now let let's find the final temperature of air entering to the engine...

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