The only known sources of helium are the atmosphere (mole...

The only known sources of helium are the atmosphere (mole fraction approximately $5 \times 10^{-6}$ ) and natural gas. A large unit is being constructed to separate $100 \mathrm{m}^{3} / \mathrm{s}$ of natural gas, assumed to be 0.001 He mole fraction and 0.999 $\mathrm{CH}_{4}$. The gas enters the unit at $150 \mathrm{kPa}, 10^{\circ} \mathrm{C}.$ Pure helium exits at $100 \mathrm{kPa}, 20^{\circ} \mathrm{C},$ and pure methane exits at $150 \mathrm{kPa}, 30^{\circ} \mathrm{C}$. Any heat transfer is with the surroundings at $20^{\circ} \mathrm{C}$. Is an electrical power input of $3000 \mathrm{kW}$ sufficient to drive this unit?

Key Concepts

Thermodynamic Separation Limits

This concept refers to the fundamental minimum energy required to separate a mixture into its components, as dictated by the second law of thermodynamics. It involves calculating the minimum work by assessing the change in Gibbs free energy or the entropy increase associated with mixing. The theoretical limit provides a benchmark that any practical separation process cannot surpass, emphasizing that any real process will require additional energy due to irreversibilities.

Gibbs Free Energy of Mixing

Gibbs free energy of mixing quantifies the energy change when two or more components form a homogeneous mixture. In the context of separation, recovering separate pure components implies overcoming this energy benefit. The difference in chemical potentials on removing the mixing effect is critical in determining the minimum theoretical work required for the separation process.

Entropy and Irreversibility in Separation Processes

The concept of entropy is central to understanding the irreversibility associated with the mixing and separation of gases. When a mixture is separated, the entropy reduction must be compensated by energy input. Real-world processes always involve additional energy consumption above the theoretical minimum due to unavoidable inefficiencies and irreversibilities, which is crucial for process design and evaluation.

Process Operating Conditions and Energy Balances

This concept involves analyzing how operating parameters, such as pressure, temperature, and flow rates, affect the energy inputs and outputs in separation units. Evaluating these conditions using energy and mass balance equations helps in determining whether the available energy (or power input) is sufficient to drive the process, considering both ideal and non-ideal behaviors.

Transcript

00:01 Hi, everybody.

00:01 So this question is asking us is an electrical power input of 300, 3 ,000 k watts sufficient to drive this unit.

00:13 And so we need to find n1, which is the key, the pressure of volume of 1 of r over t1.

00:25 So it's going to be 150 times 100.

00:31 Times the standard rate arm.

00:37 So times temperature to 83 .2.

00:41 And we get 6 .3703k mole per second.

00:49 Okay.

00:51 And your nt2 is going to be 0 .001 times your 6 .3703 .0.

01:02 And it's going to be 0 .00637 k mole per second.

01:15 So let me, there we go.

01:19 And your n3 is going to be y, c, h4 of your n1.

01:27 So it's going to be 0 .999 times 6 .3703.

01:35 Equals 6 .3639 k mole per second.

01:45 And now for your q value, we have n -h -e minus h -i, the intern, exit, plus the warp.

01:57 And we got n2 times c -p -h -e times 2 -t -2 -t minus t1 plus n3 times cpch times t3 minus t1 plus the work.

02:21 And so before we solve for the q, we need to find the c -p -h -e, helium.

02:32 So that's going to be 4 .003 times 5 .193, which is 20 .787 kilojoules per kilomol times kelvin.

02:54 And now we have cpch.

02:58 And that is 11 .043 times 2 .254 equals 36 .1609 kilojou per kilmole times kelvin.

03:21 Okay.

03:23 And now we can plug that back into the k, the q function.

03:27 So for that, which is this bad boy here okay okay so 0 .00637 and let me think that seven looking nice seven times 20 .787 times 20 minus 10 plus 639 times 20 minus 10 plus 6 .3639 times 36 .1609 times 30 minus 10 minus 3 ,000.

04:16 Okay.

04:19 And let me like get that 3 ,000 looking normal.

04:23 There we go.

04:24 So 3 ,000.

04:27 Okay.

04:28 And this is going to equal to 1 ,6003 .804 kilowatts.

04:36 Okay.

04:38 And now we can find the entropy.

04:43 So the entropy is going to be, and i'm just going to write it out with numbers...