(9/10/14) Heat Engine

In this class, we learned about the basics of how a heat engine (and the wonders of the 20th century)

Candle in Cylinder Activity:

Our prediction of what would occur to the candle

During the beginning portion of lecture, we were asked to predict what would occur if we were to place a lit candle inside a graduated cylinder.
Our prediction states that we believed that the candle would go down, since it takes longer for the oxygen to reach to the graduated cylinder

The video of what actually occurred

It was through our surprised that the candle quickly let out. This occurred because the fire was cut off from its plentiful environment of oxygen, and burned out what little it could have.
Additionally, this actually took two attempts to do, as the first time the candle ran out faster than we could place it in to the bottom of the cylinder.

Prediction part 2

After understanding what made the candle light out so quickly, we then were asked a similar question. "What would occur to the fire of the candle, if you were to place an additional cylinder to be placed upon the candle"
My original thought was that since you were decreasing the amount of oxygen inside the cylinder even further, that the candle will then diminish, at around the same speed.

Video of Experiment part 2

Once again, we were proven to be incorrect. Now that the candle has a way for oxygen to reach into the cylinder, it was able to last longer than it was during the first portion of the experiment.
We can see this kind of effect happening in our daily lives via a chimney, as a chimney is basically a funnel for oxygen to enter into the house, in which the fire can use allow itself to continue burning






Candle in a closed container activity:

In this activity, we took a candle inside a closed container, and were asked what was to happen to the candle while the closed container were to fall a few feet.
Our prediction states that the candle's light should get dimmer, due to the amount of gravity that it is being pushed on about.

A video of a video of the experiment

Watching the video, we came to the conclusion that our prediction was indeed correct, but the reasoning was completely off. The fact is that while it was falling, convection current was no longer occurring, which is what gives the candle is straight-like figure. What took over instead was diffusion, which forces the candle to become dimmer, as the energy was dispersed evenly

Additionally, if a candle was to be taken to space and lit, we also learned that a similar effect were to occur, where diffusion becomes the primary effect, and the candle would be a arc-shaped, rather than the common fire shape that we know it as.








State Variable and Ideal Gas Law Activity:
We then were sent to do the next activity on the computer, answering the six questions that were given to us

Answer to question 1

The first question was about isobaric (constant pressure) process, in which we were to find a relationship between volume and temperature.

We answered stating that since pressure is constant, then, by the ideal gas law, we can then state that volume and temperature (which is supposed to be notated by T), is directly proportional to each other.

Answer to question 2

The second question related to isochoric (constant volume) process, once again finding a relationship between pressure and temperature

Using the ideal gas law, and know that volume is constant, we found out that pressure and temperature is again directly proportional to each other.

Answer to question 3

The third question talks about isothermal (constant temperature) process, and asks us to find a relationship between pressure and volume

Using the Ideal Gas Law as a base once again, we find that pressure and volume are actually inversely proportional to each other, when pressure increases, volume decreases, and vice versa.

Answer to question 4

Question 4 gives us a scenario in which we have 1 moles of a gas at 500K and pressure of 42kPa. They want to know the pressure in the gas when the temperature is lowered to 300\1.8K
There are two ways to approaching this question. One was by using the ideal gas law, and given the final temperature, amount of moles, the constant R and the pressure, we can easily find the volume.
The second way was by understanding that since moles, the constant R, and pressure are all constant, we can easily find the volume by reducing the ideal gas law to V1/T1 = V2/T2, and solve for V2, with V1, T1 and T2 all given to us.
The answer we came up was around 25 m^3, which is the same answer as the advisor obtains as well

Answers to question 5 and question 6

Both question 5 and question 6 are a variation of what question 4 was asking of us, instead of asking for volume, asked for pressure.
Like question 4, we can easily solve this question by understand which variables of the ideal gas law is not constant, in this case, pressure and temperature, and solving for them, given the initial pressure, initial temperature and final temperature.
Question 5 we obtain around 126kPa
Question 6 is a two-part in which we obtained 124kPa and 248kPa respectively

Working Rubber Band Activity:
In this activity we were asked what would happen to rubber band as it is heated up.
We predicted as a group, that what we knew about thermal expansion of a substance, that if you were to heat up a rubber band, in turn the rubber band will expand.
However, within the lecture, we learned that not to be the case that a rubber band will indeed contract upon itself due to the chemical structure of rubber.

We then used this idea of contracting rubber to create a makeshift mass-lifting engine thought idea.

A version of a workable cycle

We were asked to create a cycle the can repeatedly lift cans to the packing conveyor using rubber.
We stated that one cycle of lifting a can requires the following steps:
1) Put can on rubber
2) Heat up rubber
3) Unload
4) Cool down rubber band
We can point out that the heat is entered into the rubber during the second step of the cycle, that the heat energy is going along positively, and then removed at the last step, and the rubber band is once again elongating to its original position.

Additionally, if we were to take this idea, and place it into a factory during a hot day, we would observe that our makeshift engine would not work, as there would be no way to remove the heat, which is a necessary step for this model engine to work.

We also have to note that not all 100% of the heat energy used in this experiment goes through the process of our makeshift engine. We should note that a portion of the energy is going to heating up, and cooling down the rubber itself, which we cannot gain back in terms of mechanical energy.

Defining Efficiency Activity:

Our next portion of the lab goes to understanding engine efficiency, notated by η
we can state that η is equal to W/QH , where W is the work and QH is our hot reservoir. We can then state that W = QH-QC  where Qc is the cold reservoir, and with a simple amount of math, obtain the term η = 1-Qc/QH 

Its to note that if we were to obtain an engine with no waste heat, Qc would equal to zero, QH would equal to infinity, and η would equal to 1

Analyzing the Cycles:

Our processes of finding the work and heat (incomplete)

In our last experiment, we wree to mathmatically find the work, heat, and internal energy of a cycle in each of their respective parts.
By creating a crude graph, we found that this cycle creates a square-like graph (as shown by the photo) which each point being a different step of the cycle.
Additionally, we found work to be done on the gas at step A and step C, while by the gas at steps B and step D. while heat energy transferred to the gas from a reservoir at steps AB and BC, and vice versa at steps CD and DA

Finding work and heat (completed) 

Afterwards, we were to evaluate Eint at each ponits using the equation Eint = (3/2)PV
We came with the following answers for this portion of the lab
1) 12240 J
2) 15300 J
3) 11,350 J
4) 9,480 J
We then were able to similarily find ΔEint, work and heat energy respectively (all numbers are rounded up to make overall math simplier)

ΔE int (A) (15300-12240) = 3100 J
ΔE int (B) (11350-15300) = -4000 J
ΔE int (C)  (9480-11350) = -1900 J

ΔE int (D (12240-9480) = 2800 J

Wa = 2000 J

Wb = 0 J

Wc = -1600 J

Wd = 0J

Qa = 5100 J

Qb = -4000 J

Qc = -3500 J

Qd = 2800 J

We know that since the work is being done on its surroundings, the gas is therefore expanding, and work becomes positive, and vice versa for work negative, as it is contracting and work is being done on the gas. 

We were then asked to find the net work, the sum of all the work (Wnet = Wa + Wb + Wc + Wd)

Wnet = 2000 + 0 -1600 + 0 = 400 J

Lastly, we were asked to use the rectangular graph, and use it to find the net work (finding the area enclosed in the rectangle) and see whether or not they are around the same. 

Since area of a rectangle is length x width

A = (0.10-0.08) x (1.02x10^5- 0.079x10^5) = 400 J
Its to note that due to the rounding, the numbers tend to be exactly the same, in which might not have been the case, if we took their precise numbers.