Thermodynamics Review Problems for Mechanical Engineering Students | Series 4

Hi folks!

This is the 4^th series of my "Thermodynamics Review Problems for Mechanical Engineering Students" and once again I'll be showing two review problems wherein the solution is explained and the computation is presented in a form of screenshots. And without further ado, lets get started for this series' set of problems.

Review Problem 1

A rigid tank contains a perfect gas with R = 2.08 kilojoule per kilogram per degree Kelvinand k = 1.67. Calculate the final temperature in degree Celsius if it is initially at 30 degree Celsiusand heat added is 15 kilojoule per kilogram.

Solution

For this review problem, we are provided with the keyword rigid tank which enables us to identify that this perfect gas is undergoing a constant volume or isometric thermodynamic process. For gases undergoing isometric thermodynamic process, heat added/rejected is equal to the change in internal energy (∆U). And for change in internal energy, specific heat capacity at constant volume is used (c_v). And with that since we are provided with the gas constant (R) of the perfect gas and its adiabatic index (k) as well. The first thing to obtain is the specific heat capacity of the perfect gas at constant volume (c_v), for which it is equal to 3.10 kilojoule per kilogram per degree Kelvin.

With that in mind, we can now proceed in obtaining the final temperature (T₂) and this can be done by using the relation in isometric thermodynamic processes that the heat added/rejected is always equal to the change in internal energy since the work done in that process is always equal to zero due to the fact that the volume is held constant. And with that, final temperature (T₂) is equal to 34.84 degree Celsius.

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Review Problem 2

Air is heated from 27 degree Celsius to 327 degree Celsius. How much does the specific internal energy of the air changed as a result of heating?

Solution

For this review problem, the difficulty is the same to the previous problem since we are provided with all the parameters to solve for the thermodynamic property that is being asked. The substance that is being heated which is AIR and then we are provided with both the initial and final temperatures. And for internal energy the thermodynamic property for specific heat capacity should be at constant volume (c_v) for which air's c_v is equal to 0.718 kilojoule per kilogram per degree Kelvin. And with we obtained the change in specific internal energy of the air which is equal to ∆U = 215.4 kilojoule per kilogram for which the computation is shown below. Additionally, I'd like to repeat that the change in degree Celsius is just equal to the change in degree Kelvin (∆ degree K = ∆ degree C); that's why the those two are directly cancelled out in the computation below .

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Curriculum

• Series 1
• Series 2
• Series 3

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_{Computations and screenshots are made by the author, @josephace135.Special thanks to @jbeguna04 for designing the GIF photos.}

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Posted from my blog with SteemPress : https://geuseppedeacenet.000webhostapp.com/2018/07/thermodynamics-review-problems-for-mechanical-engineering-students-series-4

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