Reconsider Prob. 10-18E. How much error is caused in the...

Key Concepts

Pump Work

Pump work refers to the energy required to move the working fluid through the cycle. Even though the power consumed by a pump may be relatively small compared to the overall energy transfers, it still constitutes energy input that must be subtracted from the net work output. Neglecting pump work can lead to an overestimation of the system’s net work and, consequently, its thermal efficiency.

Error Analysis

Error analysis in this context involves assessing the impact of omitting a component of energy consumption—such as pump work—on the calculated thermal efficiency. By comparing the efficiency calculated with and without the pump work, one can quantify the error introduced by this simplification. This process is essential to validate the assumptions made in thermal system modeling and to ensure that the efficiency figures reflect all energy inputs and losses.

Thermal Efficiency

Thermal efficiency is a measure of how well a system converts input heat energy into useful work. It is typically defined as the ratio of the net work output of a system to the heat input. In cycle analysis, accurately calculating thermal efficiency requires accounting for all relevant energy transfers, including losses and auxiliary work inputs.

Transcript

00:01 And this problem, we're asked to revisit the previous problem.

00:06 And let me pull this up.

00:08 And see how much error is caused in the thermal efficiency if the power required by the pump were completely neglected.

00:17 Okay.

00:18 So we have in previous problem, i flip back to that.

00:22 Again, we have an isotropic pump.

00:27 And so, you know, the power that the pump, you know, puts in is this amount, right? and so you can see that, you know, the net power is, you know, this amount, specific power anyway.

00:43 And so this is, so you can see that this is like 1 % of the, you know, the power, the net power that's out.

00:50 So if we, you know, if we took this, if we neglected this, we would just have this would be, you know, roughly 140, ptu per pound mass.

01:04 So we shouldn't get much of a difference and we'll see that that is indeed the case.

01:11 We'll kind of proceed as before.

01:13 So everything up to this point is exactly the same.

01:18 And again, i calculated this, but we're not going to use it.

01:24 I think, yeah, this is also the same.

01:28 And so what we're going to do here is just we can figure out the network that's going out.

01:35 From just to change in enthalpy from here to here.

01:42 And that will give us a mass flow rate.

01:45 Now, we can see that we have a little bit different mass flow rate.

01:52 And it is slightly less because we're basically, to get the same power, we need the, and we're neglecting some of the work that we're actually doing.

02:05 So we'll need less for mass flow rate.

02:08 Now here we calculated the net, work or the specific work in a little different way, right? so here, um, the specific work, we calculated it from here.

02:21 Over here, we did it by taking the difference in these energies or these heat transfers.

02:27 So either way, you know, you can see how that, you know, if you were to plug q in in here, you would get that that was, uh, let's see, h3 minus h2 and then minus q out, which is h4 minus plus h1.

02:45 So you can see here what we're doing is we're basically saying that h1 and h2 cancel out.

02:53 So roughly that these are have the same enthalpy.

02:56 And so that's why we get h3 minus h4 here.

03:00 And then we can figure out the heat transfer in.

03:04 We can see that it's a little bit different from over there because again we have a different mass flow rate and so i think the anthopies between here and here are the same as in the previous problem.

03:20 Let me see here...

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