Questions or comments regarding Quiz 1 (two-point control)? Post them here.

22 thoughts on “Quiz 1 Discussion”

To reiterate what we covered today, and what is quite important to solve this problem — treat the curves as approximately linear. Use (if you wish) a ruler to get rise and run.

Also try a “gut feeling” estimate. For example, in Task 3, a lower hysteresis, meaning, a smaller distance between the lower and upper trip points, would mean faster switching. By how much? Well, it’s linear…

I’m sure I’m probably overthinking this, but should the 25% efficiency on the 500W rating of the compressor be accounted for when calculating P(in), or is the efficiency information to be used in task 1 of calculating P(env). I guess in other words, is the 25% efficiency to say that 75% is lost to the environment or that only 25% of the 500W is even making it into the system?

I did not think of it that way. In principle you are right, but for the solution you are overthinking it.

Simply assume that the compressor provides 500W x 0.25 = 125W of cooling power when it runs. The compressor loss — in a large room — would not matter much. The loss that counts, $P_{env}$, is the thermal energy that seeps into the freezer.

The process constants $k_w$ and $k_e$ were specific to the waterbath example. In that example, $k_w = 1/cV$ was related to the water volume and heat capacity, and $k_e$ to the quality of the insulation. Clearly, there are constants with similar interpretation in the freezer problem. When the compressor is off, temperature rises (linear, with good approximation). The two point control is at quasi-equilibrium, so the power put in by the compressor is identical to the power lost to the environment. Since you know the total cycle time (from the diagram), you could calculate $k_e$ and the equivalent of $k_w$. However, these are not asked for and I suggest you do not consider them.

$k_1$ and $k_2$ are simply the slopes of the temperature during on- and off-phases of the cycle. They are not asked for, either. However, these constants could help answer the problems if you consider what influences $k_1$ and $k_2$. In addition, you can read these constants easily from the graph in the quiz for the situation in Task 1. In Task 3, the same conditions as in Task 1 apply with the exception of the smaller hysteresis. Thus, the slopes are unchanged. Conversely, in Task 2 the temperature difference is increased, and the rising slope will be steeper (faster temperature rise due to higher loss), while the falling slope will be less steep (the compressor with constant power struggles more against the higher loss).

In short, $k_w$ and $k_e$ do not apply, and $k_1$ and $k_2$ from the book’s Eq. 1.1 are not needed to solve the problem, but can be used as helper constants.

I think we could easily use k1 and k2 as T'(t) to compute kw and ke using Eq. (5.4). And then in Task 2, use kw and ke obtained in Task 1 get the new slope k1′ and k2′, which can calculate the on-time and the total cycle duration.

The first step is computing $P_{env}$. With this, you can use the on-time to compute $k_w$, because you know that the temperature decrease is governed by thermal capacity of the “stuff” in the freezer and by the power excess from the compressor. Knowing $k_w$, you can obtain $k_e$ from the off-time slope.

But let me emphasize again that this is not needed for the solution.

Could you do another two-point control example in class tomorrow? Or if there isn’t enough time to go over another example could you give us some sort of resource that has more problems like this so we can practice? Thanks

Are you sure you want that? I thought we are done with two-point control, because there is really not much more to it than what we covered. And the exams won’t have two -point controls, either.

I don’t like being so rigid, but with this class size I have no choice. Up to 25, maybe 30, I can collect e-mails and paper. I always tried to consider late work, but with 50, 75, now 140, chasing down missed or late work is practically impossible.

To reiterate what we covered today, and what is quite important to solve this problem — treat the curves as approximately linear. Use (if you wish) a ruler to get rise and run.

Also try a “gut feeling” estimate. For example, in Task 3, a lower hysteresis, meaning, a smaller distance between the lower and upper trip points, would mean faster switching. By how much? Well, it’s linear…

Professor, could we finish the quiz in printed type? For example, I use Word or Latex to complete it and save it in PDF form to submit.

Yes, sure. No reason why not.

I’m sure I’m probably overthinking this, but should the 25% efficiency on the 500W rating of the compressor be accounted for when calculating P(in), or is the efficiency information to be used in task 1 of calculating P(env). I guess in other words, is the 25% efficiency to say that 75% is lost to the environment or that only 25% of the 500W is even making it into the system?

I did not think of it that way. In principle you are right, but for the solution you are overthinking it.

Simply assume that the compressor provides 500W x 0.25 = 125W of cooling power when it runs. The compressor loss — in a large room — would not matter much. The loss that counts, $P_{env}$, is the thermal energy that seeps into the freezer.

How can we relate K1 and K2 to Kw and Ke? Do we need to do this to solve the problem?

The process constants $k_w$ and $k_e$ were specific to the waterbath example. In that example, $k_w = 1/cV$ was related to the water volume and heat capacity, and $k_e$ to the quality of the insulation. Clearly, there are constants with similar interpretation in the freezer problem. When the compressor is off, temperature rises (linear, with good approximation). The two point control is at quasi-equilibrium, so the power put in by the compressor is identical to the power lost to the environment. Since you know the total cycle time (from the diagram), you could calculate $k_e$ and the equivalent of $k_w$. However, these are not asked for and I suggest you do not consider them.

$k_1$ and $k_2$ are simply the slopes of the temperature during on- and off-phases of the cycle. They are not asked for, either. However, these constants could help answer the problems if you consider what influences $k_1$ and $k_2$. In addition, you can read these constants easily from the graph in the quiz for the situation in Task 1. In Task 3, the same conditions as in Task 1 apply with the exception of the smaller hysteresis. Thus, the slopes are unchanged. Conversely, in Task 2 the temperature difference is increased, and the rising slope will be steeper (faster temperature rise due to higher loss), while the falling slope will be less steep (the compressor with constant power struggles more against the higher loss).

In short, $k_w$ and $k_e$ do not apply, and $k_1$ and $k_2$ from the book’s Eq. 1.1 are not needed to solve the problem, but can be used as helper constants.

If we don’t obtain k_e and k_w, how can we compute P_env?

I thought there was a BIIIG HINT in the text above…

Remember, over many cycles you have a quasi-equilibrium.

I think we could easily use k1 and k2 as T'(t) to compute kw and ke using Eq. (5.4). And then in Task 2, use kw and ke obtained in Task 1 get the new slope k1′ and k2′, which can calculate the on-time and the total cycle duration.

The first step is computing $P_{env}$. With this, you can use the on-time to compute $k_w$, because you know that the temperature decrease is governed by thermal capacity of the “stuff” in the freezer and by the power excess from the compressor. Knowing $k_w$, you can obtain $k_e$ from the off-time slope.

But let me emphasize again that this is not needed for the solution.

Could you do another two-point control example in class tomorrow? Or if there isn’t enough time to go over another example could you give us some sort of resource that has more problems like this so we can practice? Thanks

Are you sure you want that? I thought we are done with two-point control, because there is really not much more to it than what we covered. And the exams won’t have two -point controls, either.

Hi Proffessor,

The Tenv for Task 1 signifies the lab temperature (+20 degree C) or the freezer temperature (-20 degree C)

Lab temperature. So that a 40 degree difference is responsible for the slow warming inside the freezer while the compressor is off.

Hi Professor,

How do we need to submit the Quiz? Also, is there a time deadline for the submission of the Quiz?

You need to use the eLC drop box for this course. The deadline is today, end of class.

Once you have submitted, you will have access to the solution. Get the quiz policy document from our class web page.

Deadline for second submission (graded) is Thursday, class time.

However, we couldn’t submit eLC drop box now. We don’t have the folder for this quiz.

I can already see 115 submissions at this time.

Professor, we can’t see the folder. We are Graduate students. That might be the reason.

I’m assuming our self-graded quiz that is due tomorrow will also be submitted through eLC?

Yes… please.

I don’t like being so rigid, but with this class size I have no choice. Up to 25, maybe 30, I can collect e-mails and paper. I always tried to consider late work, but with 50, 75, now 140, chasing down missed or late work is practically impossible.