VPython

Spring 2015
22nd of March

VPython
I followed the tutorial on Youtube and was given the assignment to replicate this 3D model:

This is what my coding came to be:

And then, they asked me to learn to use the # function, that allows the program to ignore certain lines of coding momentarily. Also they asked me to reduce the length of one of the arrows by half, and make it point in the opposite direction.

The reduced arrow is on the bottom sphere, barely visible.

The next assignment looks like this:

This is the way I code mine:

They then wanted me to double the distance of one of the sphere's y-axis, and this is what I coded:

For the final assignment, they wanted to show the function of 'print'. This is what I coded:

Electric Charge and Force

We started the day understanding about electric charges. We use a baloon to find out the relationship. When a baloon is rubbed on a fur, then put on a glass wall, it gets stuck to the wall glass. This is caused by the negative charge being taken out by the fur and leaving it with positive charge, which sticks to the negative charge of the glass wall.

When the baloon is rubbed on a silk, then put on the glass wall, the baloon seems to pushes away from the glass and fall. This is because silk is taking away the negative charge slower than fur, when it actually should have stuck to the glass as well.

Recall from 4a, we learned about forces and free body diagram. First, we are finding the angle by using its relationship with the length of the string. Then, by drawing the free body diagram, we find the sum of forces in x and y directions, then find the force to move the person hanging on the string. Then we replace the angle using what we found earlier.

From 4A, we have learned about universal gravitational force, where G is multiplied by the 2 masses and r of 1&2, divided by the difference of distance squared.

Using that formula, we can solve the given problem below and found the result to force to be 1.35*10^-4 J.

We can see the relationship between force and r as inversely proportional, the bigger the difference between 2 masses, the smaller the Force and vice versa.

When talking about 2 charges, we can also use the same equation by replacing the masses with the charges. We can find the components of forces by finding the angle of the charges. Using that angle and multiply them by the force will give us wach forces and adding them to find the vector force.

Entropy and Cycles

We started the day comparing the two different cycles, auto cycle and diesel cycle. The difference between them is that diesel cycle expands Adiabatic until gas ignites, which makes the pressure constant in the end. It also compresses gas further. Thus, it makes it more efficient, but the max Power is lower because it goes through more cycle?

We now learned about entropy. Entropy is heat divided by temperature. When the situation is adiabatic, entropy becomes what it's called as isoentrophic.  

We then learn briefly about stirling engine. 

It is known to be ecofriendly. There are 4 steps in stirling engine; expansion, transfer, contraction, and transfer. This is called Brayton cycle. From this cycle, we draw the relationship between the pressure and entropy. The shape of the graph becomes parallelogram. We have learned about first law of thermodynamics in the past lecture. Today, we are learning about the second law of thermodynamics, which says energy can be created, but not spontaneously destroyed.

Using this definition, when put into a graph of temperature vs entropy, adiabatic becomes vertical line, and isothermic becomes horizontal line. Whereas Isovolumetric becomes the graph of adiabatic, and isobaric becomes the graph of isothermic on first law.

Stirling engine works with different temperatures, just the same as thermoelectric cooler. The difference is we will be putting a cold temperature on top of the engine, and hot on the bottom to create energy that can move the propeller. The location of hot and cold can also be switched, and it will cause the propeller to spin the other way. 

We learned some new equations about coefficient of ? To find the heat ? First, we wanna find its max coefficient by dividing the temperature of a system with temp of system minus temperature of outside. Using that max coefficient, we can find the heat by multiplying it with the power given as below.

Given a different problem, we find the work the same way as we did just now, except we divide the heat by the max coefficient. We find each work that the system gives out and also the actual work needed. Finding that, we can then calculate the efficiency by dividing the works and multiply them by 100%.

We knew from earlier that heat inside is sum of heat outside and work of the system. Change in entropy of a sytem is zero. Using that, we find the sum of entrphys of a system to find final temperature. Since heat is mc*delta T, we can integrate change in entropy and temperature, and find the equation to be mc*ln(Tf/Ta). Adding the two temperatures, we can put them together in to mc*ln(Tf(square)/Ta*Tb). Now we can solve for final temperature as below.

After finding the final temperature, we can now find the work by subtracting heat in by heat out. 

Thermoelectric cooler & Carnot Engine

We started the day with understanding about thermoelectric cooler. There's two cups on each side of the hypnodisk. Each is to be filled with hot water and iced water. The purpose of this machine is to convert the difference in temperature into electricity. The placement of the water does not matter, it only makes the hypnodisk spins the other way.

when hypnodisk is taken off and replaced with power, it then does the opposite; it converts electricity into two different temperatures as we can see in some of refrigerators.

We knew from the heat formula about heat capacity, but we did not go deeper into it. So today, we learned about the understanding of heat capacity. To find heat capacity when heat and change in temperature is given, we just divide them. When molar is involve, we just multiply them by the inverse of mole.

In isochoric, we knew from the past lecture that internal energy is 3/2 PV or 3/2 NKT. Since the volume is constant, means that there is no work. This leaves change in energy equals to heat. When we set these equations equal to each other, we found heat capacity to be 3/2R, which we also know this to be noble gasses from chemistry class.

We know that Cp=Cv+R from ...

We can replace R into Cp-Cv in ideal gas law. Given that PV= P*delta V + V*delta P. By combining these two equations, we found their relationship as below.

By deriving it again with the ideal gas law, can now find that by adding V*delta P/P* delta V and Cp/Cv, they become zero.

By separating pressure and volume on each side of the equation and integrating them. We found the relationship of them to be inverse as below.

Work, as we knew, to be efficient has to be heat in minus heat out. We knew that power is work divided by heat in. By collaborating these equations, we found the efficiency equation to be 1- heat in/heat out. Given some numbers as below, we calculated the efficiency using the equation that we have derived.

The relationship between heat in and heat out is proportional to temperature in and temperature out.

In a car engine, to make it efficient, we would want the radiator to be cooled when the engine heats up.

Lastly, we learned about engine. There are four process in working engine, which are intake, compression, power, and exhaust. During intake, the intake valves open, and the piston moves down and intake valves close. During compression, all valves are closed, piston moves up, and the gas is ignited. During power, there's explosion, piston moves down to increase volume. During exhaust, piston moves back up to create pressure, and exhaust valves open.

From those four processes, we make a graph of Pressure vs. Volume and connect them. The graph looks like a parallelogram. The enclosed graph shows the amount of work done in the engine.

There are three ways to increase the work, which is shown below.

Work

Spring 2015, Prof. Mason

5 March 2015

We started the day with an experiment. By connecting a syringe to a isolated flask, we want to find the pressure on the syringe when temperature increases. When the flask is placed in the heated water in the beaker, the syringe end is pushed up by the expanded gas and increasing the volume in the syringe. This shows the work done in the syringe.

Next, we want to know the relationship between work and pressure. Work that we have known from our past physics class was W= Int of F.dx and Pressure was F/A. By combining these two equations together, we found the Work to be int. of PA*dx.

Thus, we know that the graph of Force vs. distance will give us Work as well as the graph of Pressure vs. Volume by calculating the Area of the graph.

We have known from our past physic classes about Newton's first law; Energy can't be destroyed nor created, it can only transform to a different type of Energy. We chose to explain this law by using the example of skidding tire. When tire is skidding, we can hear and see the energy in a form of sound and smoke. That's because energy from the skidding tire transform.

From this law, we found the equation to be E=Q-W. From this equation, we derive them to find work, heat, pressure, kinetic energy, and relationships between them. The velocity in x direction is x/delta t of x and in y direction is y/delta t of y. To find velocity total we use 3D-Pythagorean theorem. Given than Vx square, Vy square, and Vz square equal to each other, we know by the Pythagorean theorem, total v square equals to 3 Vx square. We know that when the molecule bounces back on the x direction, it will be 2 delta t of x.

We found the pressure to be NmVx(square)/x(square), but we know that Vx (square) is Vt(square)/3 and x(square) is volume. By replacing this we get MVt(square) in the numerator, which is the same as 2KE; thus, we can replace this and found the relationship between pressure and kinetic energy.

Next we want to find the relationship between Temperature and root mean square velocity. We know that PV=NKbT. From there, we found T= 2/3 K/Kb. By replacing the K into its rquation that we know, we found the Root mean square velocity.










Given that the system is isothermal (temperature is constant), we then know that change in internal energy is also zero.

Given that the system is adiabatic (Heat is constant), we then set change in heat to be zero, and find the internal energy to be equal to negative work. From earlier, we have found change in internal energy to be 3/2 NKbT, and Work is int. Of PdV. Replacing the variables with this, we found the relationship between temperature and volume, which Temperature is -2/3 of Volume in ratio.

Last experiment, we are using the fire. We want to change volume without changing the pressure. By pressing the platform very quickly, we are going to avoid change in pressure as seen in the video below. The temperature increase very rapidly when doing this. First, we collect the data before doing the experiment. Then, we make some calculations to predict the final temperature. The ignition point of cotton is about 210*C or 483K. We found the final temperature to be 2860.8K. Thus, the cotton did light up as shown in the video as well.

Heat Processes (Isobaric, Isothermic, Isochoric, Adiabatic)

We started the class with understanding from past lectures about a system. System includes Temperature, Pressure, and Volume. Recall from last lecture, we learned about isothermal where temperature is constant, and adiabatic where heat does not change. Today we learn 2 more, which are Isobaric where pressure is constant, and Isovolumetric or Isochoric where Volume is constant.

We are given 6 questions from a website to see the relationship between Pressure, Volume, and Temperature at the 3 different situations that we discussed earlier. When the system is isobaric, we found the Volume and Temperature to be directly proportional by using the gas law formula that we have learned before. When it's isochoric, the Presure and Temperature is also directly proportional. When it's isothermal, we found the Pressure and Volume to be inversely proportional; thus, it gives a curve on the graph. 

After finding these relationships, we are given some knowns and find the unknown by using the relationships. For isobaric, we plug in the knowns and find the final volume to be 25.1 dm(cubic). For isothoric, we found the final pressure to be 126 kPa, and for isothermal, we found the initial pressure to be 124 kPa when final volume is 20 dm(cubic), abd 248 kPa when the final volume is 40 dm(cubic).

We are then given 4 different graphs and we need to predict which graphs shows the situations that we talked about earlier. For the first graph where the line is vertical, we predict it to be isochoric, since the volume is constant as pressure goes up. For the horizontal line graph, we predict it to be isobaric since pressure is constant as volume increases. For the third and forth graph is kind of tricky since the difference between them is only the length of the line. We predict the longer graph to be isothermal since it only depends on molar mass and Ideal gas constant; thus, we believe isothermal to have faster rate, whereas the adiabatic depends on mass and specific heat as well as those two in isothermal. Thus, it has a slower rate.

Next, we discuss about rubber band. When rubber band is heated, at first we thought that they will stretch and elongate. After seeing the youtube experiment, we found that the rubber band actually shrinks. This happens because rubber band is polymer. When it's heated, it curls up and shrink.

After understanding that, we now can use it in real life application. Since rubber band curls up when it gets heated, we can use it to transfer an object from lower to higher location. We are going to apply it to moving a can from a lower conveyer belt to higher conveyer belt. The steps are as below. When the engine moved by the conveyer belt there is a positive work done, but no heat in or out. When the rubber band is shrinking, there is heat going in, as well as work done on it, but when the rubber band cools down. It is the opposite for the hear and work.

Now, we imagine a situation where the outside temperature is fairly cold. Is it possible for the heat to fully goes into the rubber and have the work done? The answer is no, because when the rubber band cools down, there is heat loss, which means the heat can't be efficiently become work.

For the reason above, we take the heat loss into an account. Thus, work is equal to heat gain - heat loss. To calculate energy efficiency, we also learned a new formula, which e=1-Qh/Qc.

We are given another problem. A 1 mole of air is put into an iced water. When we transform this into a graph, the cycle looks like a rectangle. From there, we decide where the work and the heat are. Feom point 1 to 2, there is positive work done and heat in because the volume increase, but pressure is constant. From point 2 to 3, there is heat out but no work done because the volume does not change. From point 3 to 4, there is negative work done and heat out as well because the volume decrease, but the pressure does not change. From point 4 back to 1, volume is constant but pressure increases; therefore, there is only heat in, but no work done.

We are then given that internal energy is 3/2PV. We plugged in all the pressure and volume from each point an find the internal energy. From there, we want to find the change in internal energy by finding the difference between one point to the next point. After finding that, we make a table with heat and work done. We know from before that W=P*delta V. We use that equation to find work where the work is done. Then we'll find the heat by adding the change in internal energy and work.

By adding all the work done in the system, we find the net work on that cycle. When we calculate the area of the enclosed graph, it turns out to be exactly the same as the net work. Thus, we can calcualate the work both ways.

For last experiment. We use the syringe again thats connected to the flask. After running this experiment, we got the data as shown in the table below. Then we calculate the internal energy the same way as we did before as shown below.

Using point 3 where the temperature is closest to atmosphere temperature, we then calculate the mole using the ideal gas law with the known at that point. Then we found the mass by dividing the mole found by the molar mass of ? After finding the differences in internal energy of each point to the next point, we calculate the heat using the formula we have known (mc*delta t). Then we subtract internal energy by the heat to find the work. Finally, we recorded the data in the table as we did below.