A large air separation plant takes in ambient air (79 % N2, 21 % O2 by mole ) at 100 kPa and 20

A large air separation plant takes in ambient air (79 % N2,...

A large air separation plant takes in ambient air $\left(79 \% \mathrm{N}_{2}, 21 \% \mathrm{O}_{2} \text { by mole }\right)$ at $100 \mathrm{kPa}$ and $20^{\circ} \mathrm{C}$ at a rate of $25 \mathrm{kg} / \mathrm{s}$. It discharges a stream of pure $\mathrm{O}_{2}$ gas at $200 \mathrm{kPa}$ and $100^{\circ} \mathrm{C}$ and a stream of pure $\mathrm{N}_{2}$ gas at $100 \mathrm{kPa}$ and $20^{\circ} \mathrm{C}$. The plant operates on an electrical power input of 2000 $\mathrm{kW},$ shown in Fig. $\mathrm{P} 12.68 .$ Calculate the net rate of entropy change for the process.

Key Concepts

Second Law of Thermodynamics

The second law establishes that the net change in entropy for an isolated system must be greater than or equal to zero, providing the basis for understanding the irreversibility of processes and the necessity of work input in separation processes.

Control Volume Analysis

This approach examines a fixed region in space through which mass and energy can flow. It is essential for assessing system performance by accounting for inputs, outputs, and internal generation of entropy.

Ideal Gas Behavior

For many practical applications, gases are assumed to behave ideally, meaning their thermodynamic properties (such as entropy, enthalpy, and internal energy) can be related simply to variables like temperature and pressure. This assumption simplifies calculations in processes involving air, oxygen, and nitrogen.

Entropy

Entropy is a thermodynamic property that quantifies the degree of disorder or randomness in a system. It is fundamental in determining the direction of natural processes and assessing the irreversibility of a process.

Entropy Balance

An entropy balance involves accounting for the entropy entering and leaving a control volume along with any entropy generated due to irreversibilities. In open systems, this requires analysis of both mass flows and energy interactions.

Transcript

00:01 Hi, everybody.

00:02 So we are asked, calculate the net rate of entropy change in the process.

00:07 Okay, so we have to convert mass fraction, mole fraction to mass fraction, okay? so we have the n, cn2 for the mass fraction, so it was 0 .79.

00:21 So it's 28 .013, okay.

00:27 And we're convertinging the mole fraction into mass fraction, okay.

00:30 And this bad boy is multiple.

00:36 And so we have 0 .79 times 28 .013 plus 0 .29.

00:49 Okay.

00:50 This gives us 0 .76707.

00:55 Okay.

00:55 And now we got to do the 02 .2.

00:57 And the o2 is 0 .21 times 31 .999.

01:05 Okay.

01:07 And that's just the mass.

01:08 So 0 .79 times 28 .0 .13 plus 0 .21 plus 0 .29 times 31 .9 times 31 .99.

01:22 Amen.

01:23 It doesn't really matter.

01:25 And so many 0 .23.

01:27 And now we need um now we're finding the m of 0 2 which is going to be c o 2 which is it's 0 .2329 times 25 which we're using this number here times 25 and that is and that is can give us 5 .82 5 kilograms per second and now we have m and 2 which is going to be cn2 of m1 which is the 0 .7707 times 25 which is 19 .167 kilograms per second okay and now what we can do is find the q value, okay? so we have the q value which equals m -o -2 times h -2 minus h1 plus m -n -2 times h3 minus h1 plus the work equals m -o2, actually we're gonna make this less confusing.

03:03 Okay, there we go, guys.

03:09 M .o2 times the c .o .2 2 times t2 minus t1, okay, plus mn2, cpn2, times t3 minus t1, okay, plus mn2, c -p -n2, times t3 minus t1, plus the w.

03:38 And this is equal to 5 .8 .225 times 0 .9222 times 100 minus 20 plus 19 .1765 times 1 .042 times 20 plus negative 2 .2...

Please give Ace some feedback