Thermodynamics Review Problems for Mechanical Engineering Students | Series 12

in #steemstem6 years ago (edited)

This is the 12th series of my "Thermodynamics Review Problems for Mechanical Engineering Students". If you've missed the previous series you may try scrolling this blog and head over to the "Curriculum". This series features two problems relating to Polytropic Process for which the first one requires us to compute for the heat that has been removed during compression of air and the last one requires us to compute for the power that is necessary to compress the air; wherein it is with respect to their operating conditions as provided by the problems. Without much ado, let’s start solving these review problems.

Review Problem 1

Air is compressed polytropically from 101 kPa and 23°C and delivered to a tank at 1500 kPa and 175°C. Determine per kilogram of air the heat removed during compression.

Solution

In obtaining the heat (Q) of the aforementioned review problem, the first thing to obtain is the polytropic index (n) which can be obtain since the initial and final temperatures (T) and pressures (P) are provided. These two thermodynamic properties share a relation in a polytropic process for which the polytropic index plays an important role. Using the relation of temperature and pressure and taking the natural logarithm of both sides, we’ve found out that the polytropic index of air in this scenario is equal to 1.181. Looking into the computation, when dealing with temperatures (T) and pressures (P), these two thermodynamic properties need to be expressed in absolute to follow the STP[2] (Standard Temperature and Pressure).



Since the computation of heat under polytropic conditions, the polytropic heat capacity plays a very vital role on that one. Polytropic heat capacity (cn) uses three important thermodynamic parameters namely: specific heat capacity of the substance at constant volume (cv); adiabatic index (k) and polytropic index (n). With those three thermodynamic parameters, only the specific heat capacity at constant volume has units which is kJ per kg per °K, for air cv = 0.7186 kJ per kg per °K ; whereas the adiabatic and polytropic indices are dimensionless. In the computation below, we’ve found out that the polytropic heat capacity is equal to - 0.8695 kJ per kg per °K, well looking at the denominator (1 – n) it is expected to be negative since polytropic index must always be greater than one (n > 1) in order that the thermodynamic process be identified as a polytropic.[3]


Finally, we can now obtain the heat that is being removed during the compression of air from the thermodynamic scenario that the problem had stated – since we now have the polytropic heat capacity of air. In the computation below, we’ve found out that the heat that is being removed during compression is equal to (-) 132.16 kJ per kg. The negative sign indicates that is the heat is being removed or rejected.

Review Problem 2

One kilogram per second of air initially at 101 kPa and 300°K is compressed polytropically according to the process PV1.3 = C. Calculate the power necessary to compress the air to 1380 kPa.

Solution

The power necessary to compress the air as described by the review problem is the work that is being done during compression. For this scenario wherein we are provided with the initial temperature (T1) and both the initial and final pressure (P1 and P2); the easiest way to compute for the work is to use the formula for which the work of compression is expressed as the quotient of the product of mass flow rate (ṁ), gas constant (R) and the change in temperature (∆T) to the difference of 1 to its polytropic index (n)as shown in photo below.



And before we compute for the power, we need to obtain first the final temperature (T2), which can be obtain using the temperature (T) and pressure (P) relations at polytropic process; using the relation, we’ve found out that the final temperature (T2) is equal to 548.50°K.


Finally, we can now compute for the power that is necessary to compress the air. And using the formula above, we’ve found out that the power is equal to 237.73 kW. In the computation below, that negative sign indicates the work is being done to the surroundings during the compression of air.

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Curriculum

Series 1
Series 2
Series 3
Series 4
Series 5
Series 6
Series 7
Series 8
Series 9
Series 10
Series 11

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Reference

  1. PIPE - PROBLEM SET #2 by Alcorcon Engineering Review Center
  2. https://en.wikipedia.org/wiki/Standard_conditions_for_temperature_and_pressure.
  3. https://en.wikipedia.org/wiki/Polytropic_process.
  4. Sta. Maria, H. [1990] 2012. Thermodynamics 1. Mandaluyong City: National Bookstore Inc..

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Computations and screenshots are made by the author.

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Special thanks to @jbeguna04 for designing the GIF photos.

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