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Project Lead The Way is slowly getting bigger every year in our school. This coming year we'll have an Inventor course offered, and I'm sure more courses are coming in the future. PLTW is in no way "worthless". Even if your school's program may not be up to your standards, I guarantee at least a few people will benefit from it.
Thumbs up to PLTW :) |
Sample Principles Of Engineering Homework Problem from PLTW Class
OK well since I'm bored now i pulled my P.O.E. notebook out from this past school year and got out one of the homework sheets we have, this one being on thermodynamics. The following was the second of 8 questions, which get much harder as they go on, this is a simple single equations question, which my brother just went through in his junior year of college for mechanical engineering at Clarkson University in the class, Fundamentals of heat and mass transfer. The problem is copied as follows:
2.) A wall is made of an outer brick (R=0.3ft[^2]*hr*[degree]F/BTU), a piece of plywood (R=1), 6” of insulation (R=25), an air gap (R=4), and drywall (R=0.5). Calculate the coefficient of thermal conductivity. now i am wondering if anyone that says that P.L.T.W. is not worth the time, can answer that question on their own without looking to the web for help. ~Mike P.S.--> Also this problem is expected to be done in a time frame of 3-5 minuets at max. |
Our county is supposed to start that up next year at one of our HS's... though not mine.
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its been great hearing all your comments, keep them coming!
~Mike |
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that is probablly one of the simpelest questions we got for a homework, if im feeling up to it, maybe in a little while ill put the equation and answer to Q.2 and question 3 which uses that answer to find something else...
~Mike P.S.~Resistivity is = to 1/Conductivity LOL |
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good job adam thats how that question would be done... i will add my work, with answer to Q2 later along with q3 which will show how much you do learn from just one day in POE.
~Mike |
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Q2 answewr with Q3
Here is the work and answer for Question 2, along followed by Q3.
Q2 Answer: R=Coefficient of Resistivity U=Coefficient of Conductivity R=1/U R(total)=R1+R2+R3.... R(total)=(0.3ft[^2]*hr*[degree]F/BTU)+1+25+4+0.5 R(total)=30.8ft[^2]*hr*[degree]F/BTU U(total)=1/30.8ft[^2]*hr*[degree]F/BTU U(total)=.032 BTU/ ft[^2]*hr*[degree]F Question #3 3.) Use the answer to problem 2 to find the heat load on an air conditioner if a house is built of walls like these. The house is 20 feet by 40 feet with 8 foot tall walls, and you may consider the roof has the same construction as the walls (for the purpose of making this problem simpler). Neglect any windows and doors. That question is a 2 part problem, found everyday while trying to find the load on an air conditioner. This type of problem was taught in class over a 3 day span, after learning about thermal conductivity. ~Mike |
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OK lets put this back on track, anyone who has heard of or has interacted with the Project Lead The Way pre-engineering program for high school, please answer the following questions:
The questions i have are: -who's heard of it? -who has it in their school? -what classes have you taken (if any)? -also what experience have you gotten from it? -What would you say to others thinking about taking any of the classes? Anyone can chime in if they want to... ~Mike Project Lead The Way Homepage |
Oops :yikes:
Ment for a nother topic... Musta been the late night time that threw me off :D |
Matt, I think you wanted to reply to another topic...
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