The temperature of the air in a building can be maintained...

Name: Problem 148, Chapter 8: Thermodynamics: An Engineering Approach
Uploaded: 2021-05-25T06:36:01-08:00
Duration: 2 min 56 s
Channel: Ajay Singhal
Description: The temperature of the air in a building can be maintained at a desirable level during winter by using different methods of heating. Compare heating this air in a heat exchanger unit with condensing steam to heating it with an electric-resistance heater. Perform a second-law analysis to determine the heating method that generates the least entropy and thus causes the least exergy destruction.

The temperature of the air in a building can be maintained at a desirable level during winter by using different methods of heating. Compare heating this air in a heat exchanger unit with condensing steam to heating it with an electric-resistance heater. Perform a second-law analysis to determine the heating method that generates the least entropy and thus causes the least exergy destruction.

Key Concepts

Second Law of Thermodynamics

This law governs the directionality of natural processes and quantifies the generation of entropy in any energy conversion system. It is fundamental in analyzing the irreversibility of processes, thereby influencing how energy resources are effectively utilized and how efficient various heating methods are.

Entropy Generation

Entropy generation is a measure of irreversibility in a thermodynamic process. Lower entropy generation indicates a process that is closer to the ideal reversible case, which is important for assessing and optimizing the efficiency of systems such as heating units.

Exergy Destruction

Exergy destruction represents the loss of useful work potential due to irreversibilities in a process. By quantifying exergy destruction, engineers can compare different systems and identify those that utilize energy more effectively by minimizing inherent losses.

Exergy Analysis

Exergy analysis involves evaluating the quality of energy and the potential for doing work within a system. It extends beyond the simple energy balance by accounting for the second law of thermodynamics, proving crucial in comparing alternative heating methods based on their thermodynamic efficiency.

Heat Exchanger

A heat exchanger is a device designed to transfer thermal energy between fluids. In thermodynamic systems, the performance of heat exchangers is evaluated based on how efficiently they transfer heat with minimal entropy generation, which directly impacts overall system efficiency.

Phase Change and Latent Heat Effects

In systems where condensing steam is used, phase change plays a critical role as it involves the release of latent heat. This process can be more efficient compared to direct conversion methods by exploiting the large energy change during phase transitions, thereby reducing entropy generation and exergy destruction.

Transcript

00:01 In this question, we have th equals to 35 degrees centigrade that is 308 kelvin and we have tl equals to 18 degree centigrade that is 291 kelvin.

00:10 So for the part a, we have to calculate how much heat flow out of the air conditioner to the outdoor.

00:16 So we have to calculate qh and it will be given by ql plus work done w.

00:22 And we know that for the refrigerator coefficient of performance k it is given by ql divided by w and it is also given by t l divided by t h minus t l okay and from these two equations we can write that w it will be equal to it w it will be equals to t h minus t l divided t l multiplied by q l okay so substituting this value of w here so we will get that q h this will be equals to q l plus t h minus t l divided by t l by qh ql.

01:01 So from hereafter solving further we will get that this will be equals to this will be equals to t h t h multiplied by ql.

01:16 So substituting values so we get qh that is equals to t h which is 308 kelvin multiplied by ql which is equals to 1 .00 t l.

01:30 X divided by t l okay 1 .00 and t l which is equal to 291 calvin so from here after solving we get q h equals to 1 .06 jule so this is the answer for the part a of the problem okay now moving to the next part b in which we have to calculate by approximately how much does the entropy of the room decreases so delta s this will be given by q by temperature t so so, q, this is equals to the decrease in entropy, so minus 1 .00 jule and temperature t which is equals to 291 kelvin for the room.

02:10 So from here we get delta s it is equals to minus 3 .24 multiplied by 10 to the power minus 3 jule per kelvin...

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