Thermodynamics Review Problems for Mechanical Engineering Students | Series 2

josephace135 (69)in #steempress • 6 years ago (edited)

Hi folks!

This is the 2^nd series of my "Thermodynamics Review Problems for Mechanical Engineering Students" and without further ado, allow me to present two review problems for today's series.

Review Problem 1

A 2 kilogram mass of oxygen expands at a constant pressure of 172 kilo-pascal (kPa) in a piston cylinder system from a temperature of 32 degree Celsius to a final temperature of 182 degree Celsius. Determine the work done.

For this review problem, the formula for obtaining the work done at constant pressure is expressed in terms of W = P (change in volume).And since we're not provided with the values of the volume during the initial and final state of the gas, we would be using the relation from the universal gas constant which is mathematically expressed by the formula PV = mRT wherein; P is pressure, V is volume, m is either in terms of mol or in terms of kilograms for SI units or pounds for English units, R is the universal gas constant of a gas and lastly T is the absolute temperature which is expressed in terms of degree Kelvin (K) for SI units or degree Rankine (R) for English units. With that the formula for obtaining the work done by the gas would be simplified to W = mR (change in Temperature).

Before doing so, the very first thing to obtain is the universal gas constant (R) of oxygen. Since the units used are in SI, we would be using the formula, R = 8.314 / MW; for which MW stands for Molecular Weight. By doing so we've obtained the R for oxygen which is equal to 0.25981 kilo-joule per kilogram per unit degree Kelvin. You might noticed that why we used 32 as the molecular weight for the oxygen gas, the explanation to that oxygen gas is diatomic for which its molecular formula is O₂ and since the atomic mass of oxygen is 16 g/mol, it is simply multiplied by 2 to have its molecular weight of 32 g/mol.

Since we've obtained the R for oxygen gas, we can now obtain the work done by the gas during change of state. In the photo below, the computed work done by the oxygen gas is equal to 77.94 kilo-joule (kJ).

_{Note that the change in degree Celsius is equal to the change in degree Kelvin, that's why its cancelled out immediately as shown in the above photo.}

Review Problem 2

While the pressure remains constant at 689.5 kilo-pascal (kPa) the volume of a system of air changes from 0.567 cubic meters (m³) to 0.283 cubic meters (m³). Determine the heat that is added/rejected.

This review problem is somewhat similar to previous problem since when it comes to computation since both questions weren't provided with the initial and final temperatures and is therefore being solve thru simplication of formula with the aid of the universal gas constant formula. In the photo below, the formula for heat which is Q = mc_p (change in Temperature) is simplified Q = kP (change in volume) / (k-1) and for which k is the adiabatic index and for air k = 1.4.

With the above derived formula, we can now obtain the heat that is either added/rejected by the air. In the photo below, the heat was found to be - 685.36 kJ and that signifies that the heat is rejected by the air.