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Geek Retreat: Heat Transfer or Why Engineering is Easy

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Old Nov 8, 2002 | 09:03 PM
  #1  
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Geek Retreat: Heat Transfer or Why Engineering is Easy

Geeks Only:

Alright so the consensus is a Max engine and tranny of weight 525 pounds. Included is 9 liters of coolant, use water.

Maxima pulls into an unheated garage of 20' x 20' x 9' = 3600 ft3. Garage has insulation on the walls and ceiling. Why, don't ask, none of your business. Garage is at lets say 50 degrees F.

Max's engine is 180 F, it has been running for about 20 minutes.

The 400 pounds of Aluminum in the engine cools eventually down to the temperature of the garage. The 125 pounds of Iron and 20 pounds of coolant does the same. Combined they give off, if they cool to 50F, 15,500 Btu's approximately.

What does the temperature of the garage rise to. Would you believe in a perfect world, something in the range of 270F.

Hey I didn't make the rules in this crazy world. Just remember the engine would take along time to lose that amount of heat and as it did the temperature of the air in the garage would increase. Therefore the engine would not cool to 50 F so my calculations are screwed. Then this guy named what, Newton or something shows up and decides we need calculus so we can calculate **** thats in a transitional nature of this manner. Somebody who has way too much time on there hands and a big brain could actually calculate this.

Hey it took me an hour just to do the calculation, and yeah its right, about right anyways, if there was no heat loss in the garage walls and if the engine cooled immediately down to the 50F.

But who cares anyways, this all came to pass because I saw that my contractor who is rebuilding my home after a flood threw some insulation over the top of the garage which is an unheated space and it got me to thinking what temperature would my garage get to in the winter or fall if I pull the car into an insulated garage.

If your still here then maybe you can shed some further insight into this mass of nothingness.....
Old Nov 8, 2002 | 09:11 PM
  #2  
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cliff notes? I didn't take calculus in school, how about a simplier explaination for us idiots. Does this mean that a fully insulated room would try to match the temp of the engine in the max? If so that is one hot room
Old Nov 8, 2002 | 09:32 PM
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i think you would need a little more chemistry than calculus ... but of course this is a very intense calculation and i don't want to bother with it and i wouldn't know what i'm doing ...
Old Nov 8, 2002 | 10:02 PM
  #4  
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Originally posted by Conrad283
i think you would need a little more chemistry than calculus ... but of course this is a very intense calculation and i don't want to bother with it and i wouldn't know what i'm doing ...
Not chemistry, maybe common sense. I'd hate to see what his calculations would show in the 3-car garages that some of the folks im my area have...
Old Nov 8, 2002 | 11:10 PM
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Re: Geek Retreat: Heat Transfer or Why Engineering is Easy

Originally posted by rp95se
Geeks Only:

Alright so the consensus is a Max engine and tranny of weight 525 pounds. Included is 9 liters of coolant, use water.

Maxima pulls into an unheated garage of 20' x 20' x 9' = 3600 ft3. Garage has insulation on the walls and ceiling. Why, don't ask, none of your business. Garage is at lets say 50 degrees F.

Max's engine is 180 F, it has been running for about 20 minutes.

The 400 pounds of Aluminum in the engine cools eventually down to the temperature of the garage. The 125 pounds of Iron and 20 pounds of coolant does the same. Combined they give off, if they cool to 50F, 15,500 Btu's approximately.

What does the temperature of the garage rise to. Would you believe in a perfect world, something in the range of 270F.

Hey I didn't make the rules in this crazy world. Just remember the engine would take along time to lose that amount of heat and as it did the temperature of the air in the garage would increase. Therefore the engine would not cool to 50 F so my calculations are screwed. Then this guy named what, Newton or something shows up and decides we need calculus so we can calculate **** thats in a transitional nature of this manner. Somebody who has way too much time on there hands and a big brain could actually calculate this.

Hey it took me an hour just to do the calculation, and yeah its right, about right anyways, if there was no heat loss in the garage walls and if the engine cooled immediately down to the 50F.

But who cares anyways, this all came to pass because I saw that my contractor who is rebuilding my home after a flood threw some insulation over the top of the garage which is an unheated space and it got me to thinking what temperature would my garage get to in the winter or fall if I pull the car into an insulated garage.

If your still here then maybe you can shed some further insight into this mass of nothingness.....
Ummm you obviously screwed something in your calculation up. You CANNOT physicaly heat the garage to a temperture of 270F, when the max temperature of the engine (your only heat source) is 180F that just doesnt make sense.
See im thinking you took the calculated BTUs and dumped all of it instantly into your theoretical garage, you have to calculate the release of heat energy over time and take into account that the insulation isnt perfect and some heat will escape into the atmosphere.

Also you should be using Celsius when doing these calculations, its a whole lot easier.

See im also thinking you used Newton's thermal law to determine how hot your garage would get. (correct me if im wrong) Keep in mind Newton's law is very limited and doesnt help all that much especially in a dynamic enviroment, also to use it properly you would have to take real measurements of the engine's drop in temperature over time (Delta T). Say for instance at the zero minute your engine is 180 F, after cooling for 20 minutes the engine is now 100 F. Subtract 180 F from 100 F this would give you 80 F as your delta T or change in temperature. Or better yet measure the increase of the temp of the room and do the same, either way IMO the only way to solve this problem in our lifetimes is to take some real data.

Back to the example engine, with delta T = 80 F and the time being 20 minutes you can calculate the heat coefficent of the engine/car and determine how much heat was dumped into the room over time and ultimitly how hot the room would get. But my advice would be to just drive the car around for 20 minutes and park it in your garage and measure the temp increase, this requires no calculations and no headaches; just a thermometer.
Old Nov 8, 2002 | 11:48 PM
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Re: Geek Retreat: Heat Transfer or Why Engineering is Easy

Originally posted by rp95se
...
What does the temperature of the garage rise to. Would you believe in a perfect world, something in the range of 270F.
...
If your still here then maybe you can shed some further insight into this mass of nothingness.....
What you are up against is the first law of thermodynamics which essentially tells us heat energy only flows from a hot source to a colder source. This means that once the temperature inside the garage reaches the temperature of the engine the transfer of energy stops. So in short the final temperature of the air cannot ever exceed the starting temperature of the engine or put differently, even in a perfect world, the temperature of the garage cannot exceed 180F. Common sense tells us though that air will not reach 180F and luckily we don't need Calculus to prove it. All we need to do is set the energy absorbed by the Air equal to the energy released by engine and solve for the final temperature as follows:
Code:
Assuming the garage is 6m x 6m x 3m we have 108,000 litres or 4821 moles of air.
I'm going to assume our garage is a constant pressure system (otherwise you'd 
blow the door off as the heated air expanded) so the heat capacity of the air (C_air)
is roughly 139,809 J/C. Similarly, using your numbers for the composition of the engine, I 
get a heat capacity for the engine (C_eng) of about 225,662 J/C. And so the fun begins,

The energy of the air is given by:

  E_air = C_air * (T_airfinal - T_airinitial)

  where T_airinitial = 10 C

and the energy of the engine is given by:

  E_eng = C_eng * (T_enginitial -T_engfinal)

  where T_enginitial = 82 C

Note that by the first law of thermodynamics the final temperature of the air 
will equal the final temperature of the engine. We will then call this temperature T_final 
giving us:

  E_air = C_air * (T_final - 10)
  E_eng = C_eng * (82 - T_final)

Now set the two equations equal and solve for T_final:

    C_air *(T_final - 10) = C_eng *(82 - T_final)

->  -10*C_air + C_air*T_final = -C_eng*T_final + 82*C_eng

->  10*C_air + 82*C_eng = (C_air + C_eng)*T_final

    10*C_air + 82*C_eng 
->  ------------------- = T_final
      (C_air + C_eng) 

And subbing in the numbers gives us:

  T_final ~ 54 C
or about 129 F assuming I didn't royaly screw up somewhere (what are these pounds and F things anyway....).

Better than 270F but still pretty warm. Of course we are still ignoring a few vital things. For instance, in addition to heating up the air we have to heat up the other 2700 lbs of Maxima sitting in the garage. This will soak up a good amount of the heat bringing our final temperature even lower. That said, assuming a perfectly insulated garage I put the final temp around 70-80F (but this is just a guess).

Oh, and Conrad283, this is more a physics problem than chemistry
Old Nov 9, 2002 | 12:38 AM
  #7  
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Re: Re: Geek Retreat: Heat Transfer or Why Engineering is Easy

Originally posted by Myrv
What you are up against is the first law of thermodynamics which essentially tells us heat energy only flows from a hot source to a colder source. This means that once the temperature inside the garage reaches the temperature of the engine the transfer of energy stops. So in short the final temperature of the air cannot ever exceed the starting temperature of the engine or put differently, even in a perfect world, the temperature of the garage cannot exceed 180F. Common sense tells us though that air will not reach 180F and luckily we don't need Calculus to prove it. All we need to do is set the energy absorbed by the Air equal to the energy released by engine and solve for the final temperature as follows:
Code:
Assuming the garage is 6m x 6m x 3m we have 108,000 litres or 4821 moles of air.
I'm going to assume our garage is a constant pressure system (otherwise you'd 
blow the door off as the heated air expanded) so the heat capacity of the air (C_air)
is roughly 139,809 J/C. Similarly, using your numbers for the composition of the engine, I 
get a heat capacity for the engine (C_eng) of about 225,662 J/C. And so the fun begins,

The energy of the air is given by:

  E_air = C_air * (T_airfinal - T_airinitial)

  where T_airinitial = 10 C

and the energy of the engine is given by:

  E_eng = C_eng * (T_enginitial -T_engfinal)

  where T_enginitial = 82 C

Note that by the first law of thermodynamics the final temperature of the air 
will equal the final temperature of the engine. We will then call this temperature T_final 
giving us:

  E_air = C_air * (T_final - 10)
  E_eng = C_eng * (82 - T_final)

Now set the two equations equal and solve for T_final:

    C_air *(T_final - 10) = C_eng *(82 - T_final)

->  -10*C_air + C_air*T_final = -C_eng*T_final + 82*C_eng

->  10*C_air + 82*C_eng = (C_air + C_eng)*T_final

    10*C_air + 82*C_eng 
->  ------------------- = T_final
      (C_air + C_eng) 

And subbing in the numbers gives us:

  T_final ~ 54 C
or about 129 F assuming I didn't royaly screw up somewhere (what are these pounds and F things anyway....).

Better than 270F but still pretty warm. Of course we are still ignoring a few vital things. For instance, in addition to heating up the air we have to heat up the other 2700 lbs of Maxima sitting in the garage. This will soak up a good amount of the heat bringing our final temperature even lower. That said, assuming a perfectly insulated garage I put the final temp around 70-80F (but this is just a guess).

Oh, and Conrad283, this is more a physics problem than chemistry
Good points, I agree. Kinda reminds me of the Calculus Physics class that I slept thru in college. But in the real world I would just skip all the calculations and just measure the garage temp with a thermometer. Its the most practical and simple test.

Also your calculations are not all done because we would need to take into account the heat soak of the Maxima's body absorbing the heat from the engine. Also there are losses from radiant heat escaping the garage into the atmosphere. Without all that info its hard to put an actual number on the final rise in temp. But it would be interesting to see how close your calculations are when compared to the real measured rise in temperature in the garage. All in all a final temp increase of 20~30 F seems resonable to me.
Old Nov 9, 2002 | 04:55 AM
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Myrv, thanks for shedding some light on the subject. Unquestionably I knew that the garage would not heat up over the temperature of the engine however, I did not memorize the laws of thermo, or maybe I just do not understand them as well as I should.

I don't know why my brain does not have the capacity to take the two equations and equate them. I always just go the easy route to mcdeltaT.

Besides being a physics dude what practical field are you in now, I presently am testing pharmaceutical process control systems manufacturing systems.
Old Nov 9, 2002 | 06:44 AM
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The above few guys are on the right track (engineer checking in ), assuming of course that the insulation is perfect and that the garage is 100% air tight.

In the real world of course, neither of those two things are even remotely true, and the hot air is going to rise and escape out the garage door (even when closed) while cooler air seeps in, constantly cooling the engine to the point where you're back at the outside air temperature.
Old Nov 9, 2002 | 11:41 AM
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Originally posted by rp95se
Besides being a physics dude what practical field are you in now, I presently am testing pharmaceutical process control systems manufacturing systems.
My official title is Optical Systems Designer although I'm not really involved in the actual design per se (of components at least). What I currently do is write simulation programs that model optical systems used for telecommunications. In other words, I write software that figures out what laser pulses will look like once they've travelled through 500+ km of fiber optics, filters, and amplifiers.
Old Nov 9, 2002 | 11:47 AM
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Originally posted by Myrv
My official title is Optical Systems Designer although I'm not really involved in the actual design per se (of components at least). What I currently do is write simulation programs that model optical systems used for telecommunications. In other words, I write software that figures out what laser pulses will look like once they've travelled through 500+ km of fiber optics, filters, and amplifiers.
Computer simulated Fourier analysis of optical signals? When I worked at Lucent Technologies they were looking for a scientist for that kinda work. They had to analyize the change of an optical signal as it went thru a sample piece of defective fiber optic cable. By doing this they could partly determine what was wrong with the fiber sample and make improvements in its manufacturing process.
Old Nov 9, 2002 | 12:46 PM
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I just started college and i'm an E.E. major... haha this stuff is startin to scare me now
Old Nov 9, 2002 | 12:59 PM
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Originally posted by nadir_s
I just started college and i'm an E.E. major... haha this stuff is startin to scare me now
This shouldn't scare you. Just wait
Old Nov 9, 2002 | 01:36 PM
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Re: Re: Geek Retreat: Heat Transfer or Why Engineering is Easy

Originally posted by Myrv
What you are up against is the first law of thermodynamics which essentially tells us heat energy only flows from a hot source to a colder source. This means that once the temperature inside the garage reaches the temperature of the engine the transfer of energy stops. So in short the final temperature of the air cannot ever exceed the starting temperature of the engine or put differently, even in a perfect world, the temperature of the garage cannot exceed 180F. Common sense tells us though that air will not reach 180F and luckily we don't need Calculus to prove it. All we need to do is set the energy absorbed by the Air equal to the energy released by engine and solve for the final temperature as follows:
Code:
Assuming the garage is 6m x 6m x 3m we have 108,000 litres or 4821 moles of air.
I'm going to assume our garage is a constant pressure system (otherwise you'd 
blow the door off as the heated air expanded) so the heat capacity of the air (C_air)
is roughly 139,809 J/C. Similarly, using your numbers for the composition of the engine, I 
get a heat capacity for the engine (C_eng) of about 225,662 J/C. And so the fun begins,

The energy of the air is given by:

  E_air = C_air * (T_airfinal - T_airinitial)

  where T_airinitial = 10 C

and the energy of the engine is given by:

  E_eng = C_eng * (T_enginitial -T_engfinal)

  where T_enginitial = 82 C

Note that by the first law of thermodynamics the final temperature of the air 
will equal the final temperature of the engine. We will then call this temperature T_final 
giving us:

  E_air = C_air * (T_final - 10)
  E_eng = C_eng * (82 - T_final)

Now set the two equations equal and solve for T_final:

    C_air *(T_final - 10) = C_eng *(82 - T_final)

->  -10*C_air + C_air*T_final = -C_eng*T_final + 82*C_eng

->  10*C_air + 82*C_eng = (C_air + C_eng)*T_final

    10*C_air + 82*C_eng 
->  ------------------- = T_final
      (C_air + C_eng) 

And subbing in the numbers gives us:

  T_final ~ 54 C
or about 129 F assuming I didn't royaly screw up somewhere (what are these pounds and F things anyway....).

Better than 270F but still pretty warm. Of course we are still ignoring a few vital things. For instance, in addition to heating up the air we have to heat up the other 2700 lbs of Maxima sitting in the garage. This will soak up a good amount of the heat bringing our final temperature even lower. That said, assuming a perfectly insulated garage I put the final temp around 70-80F (but this is just a guess).

Oh, and Conrad283, this is more a physics problem than chemistry
future M.E. checkin in....ummmm your calculations are very thorough...i'm assuming you are using the equation q=mc(delta T)? but ignored the masses of "air?" Also, there is no such thing as a mole of air, for the fact that air is made up of various gases...but the assumption of course can be made...boy this stuff is makin my head hurt
Old Nov 9, 2002 | 02:14 PM
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Originally posted by Myrv
My official title is Optical Systems Designer although I'm not really involved in the actual design per se (of components at least). What I currently do is write simulation programs that model optical systems used for telecommunications. In other words, I write software that figures out what laser pulses will look like once they've travelled through 500+ km of fiber optics, filters, and amplifiers.
Would that make you a EE with specialization in computer engineering?

Just curious.
-hype
Old Nov 9, 2002 | 02:31 PM
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I could've contributed to the solution of this heat transfer problem 8 years ago, but since I started working as an engineer I haven't used 90% of what they teach in college, plus I've gotten lazy. Good thing I passed the EIT my senior year.

I like how the airtight garage has perfectly insulated walls and door, and the effects of heating the wall and door materials are ignored. Likewise, the mass of the concrete floor would act as a heatsink. How about making the garage floor a frictionless plane? Then it would be exactly like all those ideal situations we encountered in engineering homework.

Aren't you also ignoring the rest of the car? That's a significant amount of metal, plastic, and glass to which the engine heat would spread.
Old Nov 9, 2002 | 04:10 PM
  #17  
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another MechE here...

so basically if you consider everything the engine cools to ambiant temp and thats that. : )

**sidenote**
i love the frictionless massless pulleys. they rock. where can i get some?
Old Nov 9, 2002 | 04:24 PM
  #18  
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Originally posted by 98SEBlackMax
Computer simulated Fourier analysis of optical signals?
Partly. I use the split-step fourier method for optical propagation down the fibre as well as straight forward fourier analysis of the signals for processing.

Originally posted by 98SEBlackMax
When I worked at Lucent Technologies they were looking for a scientist for that kinda work.
Are they still looking? I'm overworked and severely underpaided but after seeing 65,000 of my colleagues lose their jobs (in my company alone) I guess I should be happy I still have one. Unfortunately too many of my friends are saying "when I used to work at....".

Originally posted by xHypex
Would that make you a EE with specialization in computer engineering?
Actually I hold two separate degrees, one in Honours Physics and one in Honours Computer Science (both from the University of Waterloo). I'm currently finishing up my Masters in Physics (if I can ever find the time to finish writing my damn thesis....).

Originally posted by
Gotrice4

future M.E. checkin in....ummmm your calculations are very thorough...i'm assuming you are using the equation q=mc(delta T)? but ignored the masses of "air?" Also, there is no such thing as a mole of air, for the fact that air is made up of various gases...but the assumption of course can be made...
Actually for the air I used q = n C_p (delta T) but it amounts to pretty much the same thing. And yes, I realize there is no such thing as a mol of air but I made the approximation for simplicities sake (in fact I cheated even more and just assumed the air in the garage was at STP (which it isn't) and divided by 22.4 L to get n). This is just a back of the envelop calculation remember


Originally posted by
ru4real

Aren't you also ignoring the rest of the car? That's a significant amount of metal, plastic, and glass to which the engine heat would spread.
Yes, and I mentioned this at the end of my original post. My calculation is ignoring the other 2700 lbs of car and whatever else may be in the garage as well as assuming the walls and floor of the garage are perfect insulators. Rumour has it that the 5th floor of my schools physics building contains all the frictionless planes, massless pulleys, perfect springs, and point masses they talk about in the books (of course, one should note that the building only had 4 floors....).

Anyway, if anyone is looking for a Computer Scientist with a physics background and good taste in cars, drop me a line
Old Nov 10, 2002 | 12:00 AM
  #19  
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Originally posted by vinnieloo
another MechE here...

so basically if you consider everything the engine cools to ambiant temp and thats that. : )

**sidenote**
i love the frictionless massless pulleys. they rock. where can i get some?
That and the massless rope. I need some of that. BTW I'm an ME student too.
Old Nov 10, 2002 | 09:23 AM
  #20  
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I am pleased that my heat transfer calculation has sparked interest.

Just to let the ee's, meche's know who are in engineering school, all that massless stuff above, yeah it's basically the direction one's engineering career is likely to take. Do not plan on going in a straight line they don't exist. I have had 5 jobs now in 15 years. Passed my PE test this year after the second try, busted my *** to do so too. Yet Myrv still had to help me out with the heat transfer calc. hmm.

Myrv, interesting stuff that optical analysis and fourier series analysis. I would tell all the ee's w/comp sci that those are great career paths for potential employment. The meche's and cheme's have to deal with the real world, more macro than micro. Although I am presently enjoying both, I get to play on computers all day, mix drugs up and filter them in tanks, and then figure out why the damn things won't pass the quality tests.

Anyways, on the subject of my garage and the heat transfer calc., it was just something that popped into my head during the flood rebuild, and yeah I new I couldn't exceed the temperature of the engine....

It is interesting the number of engineers that are on this board, especially students.
Old Nov 10, 2002 | 09:52 AM
  #21  
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Future Computer Engineer here!!

Myrv: How'd you find UofWaterloo? I'm at UofT and there a few other Engs from UofT on this board as well.

The Optical stuff that you are talking about seems interesting, but is only usually for hte top of the top students? Or can any student who does reasonably well (above 70% avg) be able to do some research in Optics (in relation to networking), VLSI, etc...?

Cause that's the reason I'm putting up with 1st year (STUPID CIV101!!), so I can do cool stuff like you guys.
Old Nov 10, 2002 | 09:53 AM
  #22  
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Oh..and we have a saying for Engineering...
"Your Mark in Computer/Elec Engineering is inversely propionate to the the amount of sleep you get."
Old Nov 10, 2002 | 12:49 PM
  #23  
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Originally posted by Myrv
Partly. I use the split-step fourier method for optical propagation down the fibre as well as straight forward fourier analysis of the signals for processing.

Are they still looking? I'm overworked and severely underpaided but after seeing 65,000 of my colleagues lose their jobs (in my company alone) I guess I should be happy I still have one. Unfortunately too many of my friends are saying "when I used to work at....".
Sorry bro that division of Lucent was sold off over a year and a half ago to JDS Uniphase. I left there December of 2000 and from what Ive heard from my former co-workers is that they layed off just about everyone there. They had like over 300 people working there on 4 shifts, now its like 50 people and they only run 2 shifts. The telecom industry is down the drain, heck you should be lucky to still have a job. I got laid off last September and I was working at a job were I was overworked and underpaid from day one.

Ehh I should of just listened to my dad and became a professional auto mechanic, like he wanted me to be; because people are always gonna have cars that need to be worked on. Its ironic nowadays because im doing repair and mod work for people's cars on the side for some extra cash to get by with.
Old Nov 18, 2002 | 10:45 AM
  #24  
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Originally posted by mAdD MAX
Myrv: How'd you find UofWaterloo? I'm at UofT and there a few other Engs from UofT on this board as well.
Waterloo was fun. Because of the heavy co-op population everything is done on a 4 month term cycle. Even in the summer about 80% of student population is still around doing courses. Made my summer jobs as a lab assistant a lot more fun.

As for the physics program, you couldn't ask for a better group of friends. Maybe we all just shared the same masochistic tendencies but you could always find someone to help you with the lastest quantum assignment. The profs were great as well, always willing to answer questions in class or engage in a friendly game of darts. The compsci program on the other hand was not nearly as fun. It was a lot more competitive and the profs were not as easy to approach. A good program nonetheless but I'm disappointed that in later years they've dropped much of the math requirement.

The Optical stuff that you are talking about seems interesting, but is only usually for hte top of the top students? Or can any student who does reasonably well (above 70% avg) be able to do some research in Optics (in relation to networking), VLSI, etc...?

Cause that's the reason I'm putting up with 1st year (STUPID CIV101!!), so I can do cool stuff like you guys.
Well, to put it one way I'm arguably the "dumbest" person in my group if you just look at the academic record. Three quarters of the people in my group have PhDs. I'm only finishing my masters but I make up for it by having a second bachelors degree in compsci. I can run circles around most of my group with respect to code but some of the theory that comes across my desk I just shake my head at and try to not look like a complete fool when asked what I think. That said, the field isn't completely reserved for the best of the best (I'm enough proof of that).

The key is getting to know the right people (isn't that always the case?). If you're really interested in the field I would advise you to find a prof in your department working in it and either volunteer or if you lucky get a job as a lab assistant with them. Getting a few terms in the lab (or office if theory is your thing) is invaluable. I can't stress that enough. I know two guys from Waterloo that had less than stellar marks but got a couple of terms in with a world reknown laser specialist and now one is working for an imaging company (and has personally been on the space shuttle twice installing hardware before the launch) and the other is still slugging away in telecom (which is remarkable given the magnitude of the current telecom downturn).

So no, you don't have to be the class star to get into optics, but you do need to put a little effort into getting some experience under your belt. Hope that helps.

Oh and one last thing, because you mentioned it as one of your interests, in my opinion Optical Networking (Optical Switching, Wavelength switching, Add/Drop provisioning, etc) will probably be the next big thing in telecom so it's definitely not a bad area to go into. Anyway, the best of luck to you in your future studies.
Old Nov 18, 2002 | 12:29 PM
  #25  
mAdD MAX's Avatar
Senior Member
 
Joined: Oct 2000
Posts: 1,677
Thanks a lot for your in-depth reply. I just have one more question: how do you end up having time to be a lab assistant or volunteer with the prof AND keep up in school??

Also how do you get to know the Profs if they don't teach you? In my lectures, the Prof is usually the first one to leave!
Old Nov 18, 2002 | 02:25 PM
  #26  
Mizeree_X's Avatar
Getting back to his roots
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Joined: Mar 2001
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ChemE checking in!

You can assume a mole of air, because at 1 atm and low T, all gases are essentially ideal, so taking air as having an average molecular weight of 29 g/mol would be perfectly reasonable.

And my M.S. thesis progress (or complete lack thereof) has now pushed my graduation time back to the summer (at best).

Sorry to chime in so late
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