Thermodynamics Review Problem | Law of Conservation of Energy | SI Units | Series 2

Hi folks!

In my previous post, I did discussed an overview about the Law of Conservation of Energy which was accompanied by a review problem wherein the computations are explained and shown. And most importantly, the parameters were all elaborated. So today, allow me to continue presenting review problems relating to the one of the fundamental laws in the field of Thermodynamics. The same as the previous post, the unit of measurements are based on SI.

Without further ado, here's the review problem.

Air flows steadily at the rate of 0.50 kilograms per second (kg/sec) through an air compressor, entering at a speed of 7 meters per second (m/s), a pressure of 100 kilopascal (kPa) and a specific volume of 0.95 cubic meters per kilogram (m³/kg), and leaving at a speed of 5 meters per second (m/s), pressure of 700 kilopascal (kPa) and a specific volume of 0.19 cubic meters per kilogram (m³/kg). The internal energy of the air leaving is 90 kilojoules per kilogram (kJ/kg) greater than that of the air entering. Cooling water in the compressor jackets absorbs heat from the air at the rate of 58 kilowatts (kW). Compute the work in kilowatts (kW).

SOLUTION

In the above mentioned review problem, the only thermodynamic property that aren't provided is for the potential energy and it is due to the fact the compressor is just in a stationary position thus there exists no change in potential energy. And with that we can directly omit the potential energy (PE) from the general law of conservation of energy formula, as shown below.

wherein:

• Q stands for heat
• PE stands for potential energy
• KE stands for kinetic energy
• U stands for internal energy
• Wf stands for flow work
• W stands for work

Since the work is what the review problem is asking for, we will simplify the equation as shown below.

wherein:

• delta denotes change

Since we now have a simplified formula, we need to obtain the needed parameters such as the change in kinetic energy, internal energy and the flow work.

Starting with the change in kinetic energy (KE), the computation is shown below.

_{Thus, the change in kinetic energy is equal to - 0.006 kW, well, that is expected since there was a decrease in the speed of the air in the air compressor.}

Since the change in internal energy (U) is provided in the above problem, the only thing we need to do for this one is to multiply the change in internal energy to the mass flow rate of the air, so that we can arrive to an answer expressed in kilowatt (kW) which is a unit of power and power can be simply expressed as the ratio of energy with respect to time, and the computation is shown below.

_{The change in internal energy is equal to 45 kW.}

And then, for the change in flow work (W_f), the computation is shown below.

_{The change in the flow work is equal to 19 kW.}

Lastly, since the cooling water in the compressor jackets absorbs the heat that is being released by the air, in other words it is the heat that is being rejected by the compressor that that is being cooled by way of cooling water from the compressor jackets. So,

_{Thus, Q (heat) is equal to - 58 kW which means heat that is being rejected.}

Finally, we can compute for the work in kilowatts (kW) being made by the air compressor, and the computation is shown below.

_{Thus, we arrive of - 122 kW of work, which means that it is the work done by the air compressor to the surrounding.}

---

References

Sta. Maria, H. (1990). Thermodynamics 1. Mandaluyong City, Philippines: National Book Store.

---

Well, that's it and I hope you learnt something from this review problem that I've solved and presented.

Thank you for spending your precious time reading this blog of mine.

Much love and respect.

Ace | @josephace135

---

_{All the computations are made using the Equation feature of the MS Word application. All the screenshots are made possible using the Snipping Tool application.}