Small Scale Nuclear Reactors
Discussion of thermodynamic systems & processes
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Lesson 46: Engine of Nature
There was a young man named CarnotInstructional Objectives
Whose logic was able to show
For a work source proficient
There's none so efficient
As an engine that simply won't go. - David L. Goodstein, Physics undergraduate (1958)
- Be able to state the first law of thermodynamics and use it in solving problems.
- Be able to calculate the work done by a gas during various quasi-static processes and sketch the processes on a PV diagram.
- Be able to give the definition of the efficiency of a heat engine.
- Be able to describe a Carnot engine.
- Be able to use the expression for the efficiency of a Carnot engine.
Since the graphs for the adiabatic and the isothermal systems are the same, shouldn't the pictures of the pistons regarding pressure and volume also be the same as the adiabatic one?
ReplyDeleteI'm not 100% sure, but I don't think so... because in the adiabatic system the decrease in volume causes an increase in temperature and the temperature for a isothermal system is constant.
DeleteAlso, for the iso-volumetric picture, since volume is constant, does that mean that the volume is independent of any change in pressure? If that's the case, would the piston scenario work with an iso-volumetric system?
ReplyDeleteYes the difference will be the influx or outflow of heat (Q)
DeleteYes the difference will be the influx or outflow of heat (Q)
DeleteOn this worksheet you gave me today im not sure how to figure out b) If the plant generates useful energy at the rate of 100 megawatts while operating with the efficeny found in part a (7.37%) at what rate is heat given off the the surrounding
ReplyDeleteZach Makki
Deficiency is like percent where if the efficiencies 7% then 93% of the energy input gets wasted as heat
Deletethe equation for the efficiency is e= Watts(used)/Q(in). First you have to find Q(in), by rearranging that equation. The efficiency is what you got in part a) and they give us the watts used. You would then use this Q(in) along with the equation they gave in the video ( e= 1- Q(out)/ Q(in) ). Since you have efficiency and Q(in), you can solve for your unknown, Q(out).
DeleteDon't you have to find the rate for part b? Or is that just the same as find the Q in Joules? I don't know if this is right but I thought since we need to find a rate, you could divide the Qin by time and get a rate for that and then use it to find the rate of Qout.
DeleteYes. Essentially you are using part one to figure out what the output would be. Does time rate factor to in efficiency? That's really the question. Anyone else want to weigh in on this?
DeleteI was just watching the video you posted and the equation when the rearranged it said Q/T. I think I just mistook it for time when I watched it instead of temperature for some reason. That makes more sense.
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ReplyDeleteAlso, I like the little poem.
Deletehttp://www.learner.org/vod/vod_window.html?pid=616
ReplyDeleteOn the 1986 AP set, what equation or combination of equations did you use that involved mass? I'm not sure where mass would fit in, but I might also be missing something obvious.
ReplyDeleteMaybe e=mc^2? Am I on the right track, did anyone else use that?
DeleteI have a question on the 2001 worksheet that we worked on in class on Monday... for part c, I think the process from 2 to 3 is isobaric by process of elimination. But for the explanation, would it be because the volume and temperature go up, the pressure stays constant?
ReplyDelete