# The tutor examines the idea that internal combustion engines are more efficient in cold weather.

An upper limit for efficiency of an internal combustion engine is

eff = (Tcombust – Tsurrounding)/Tcombust

where

Tcombust is the temp of the combustion cylinder

Tcombust, Tsurrounding both in degrees Kelvin (Celsius + 273).

Let’s imagine a diesel engine, whose average internal cylinder temperature might be around 1600°C. Then at outdoor temp of 25°C (293K) we have

eff = (1873 – 298)/1873 = 84.1%

Likewise, at outdoor temp -25°C we have

eff = (1873 – 248)/1873 = 86.8%

The 2.7% increase in efficiency at -25°C vs 25°C may be noticeable to an operator.

Source:

www.ncert.nic.in

White, J. Edmund. Physical Chemistry: College Outline Series. New York: Harcourt Brace Jovanovich, 1987.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

# The tutor continues about a toy repair, with some reinforcement ideas.

While we were repairing the toy a few nights ago (see my previous post), my father-in-law suggested that, after the first repair cured, a second application should be considered around the outside. Such reinforcement, he commented, would give the repair its best chance of holding.

I considered his counsel from an engineering point of view: how much extra strength could we anticipate from application of J-B Weld around the outside of the repair site?

Let’s imagine the shear force to be straight forward. The strength of a reinforcement can be, generally, proportional to its left-right length multiplied by its height, then by the square of its forward length. Assuming the J-B Weld works as an integral piece after drying, I imagined an outside application along each side plane. The application would be about 20 times the height of the original shear, then its same forward length, but only about 1/30 of its left-right length. Compared with the first repair reuniting the two sheared surfaces, the reinforcement strength per side might be 20(1/30) or 2/3. Both sides together could offer reinforcement strength of 2(2/3)=4/3 or 1.33 times the strength of the original repair, more than doubling its shear resistance.

With these numbers in mind, I took my father-in-law’s advice and made the reinforcement application about 24 hours ago. The repair should be ready right now.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

# The tutor wonders about a pending toy repair.

My younger son has a toy he really loves that broke. Specifically, a pot metal part sheared off. It’s a toy out of production, so can’t be replaced.

Looking at the breakage, a repair seemed unlikely to work. I went to the hardware store and explained the situation. The man handed me J-B Weld: “If anything could work, this will.”

The J-B Weld label suggests a tensile strength of 3960 psi. The tensile strength of pot metal might be around 40 000 psi. In the toy, the strain on the metal part is not tensile, but rather shear. The shear strength of pot metal is around 75% of its tensile; if the same for the repair, it might be around 3000 psi.

I estimate the surface of the repair to be about 1/64, or 0.015625, inch2, suggesting a repair strength of 3000(0.015625) = 47 pounds.

The pot metal part stretches an elastic. Stretching the same elastic even further by a hanging mass, I’ve determined that the toy’s pulling force is less than 2.5 lbs. With estimated strength of 47 pounds, the repair should hold.

J-B Weld needs 24 hours to cure; the repair has had about 36. I guess we’ll know soon if it works. I’ll keep you posted:)

Source:

wikipedia

sciencedaily.com

jbweld.com

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

# Tutoring high school math, unit conversions are often visited.  The math tutor discusses the conversion of engine displacements.

Back when I was a kid, I had a 50cc (cc:cubic centimetre, also written as cm3) minibike. My friends soon got an 80cc one. That’s when the numbers “50cc” or “80cc” became important to me: their 80cc bike was not just a minibike. It could go 60mph (we still talked in miles back then); it always had more power than you needed. That was the difference between a “50” and an “80”.

Through many moves, new friends who didn’t ride motorbikes, jobs, then university, I became completely removed from the motorcycle culture. In high school I heard about old muscle cars; a few of my friends even had them. The numbers were “289”, “350”,or “400”. I was told those numbers also described engine size, but they meant cubic inches rather than cubic centimetres. At the same time, more of my friends had smaller cars whose engines were measured in litres: 1.6L, 2.6L, or 3.8L, for instance.

The inevitable question in such a context: what is a 350 cubic inch engine in litres? Or in cubic cm? How can we arrive at a common unit so as to compare the sizes of those engines?

We can do so with the help of this diagram:

It’s a fact that, for a box structure, the volume (aka, displacement) is given by

Volume=length*width*height

Therefore, the volume of the cubic inch pictured above, in cubic centimetres, would be

Volume=(2.54cm)*(2.54cm)*(2.54cm)=16.387064cm^3

Therefore, rounded to the nearest hundredth,
1inch^3=16.39cm^3

It follows that a 350inch3 engine is 350*16.39=5736.5cm^3, or 5736.5cc’s, if you prefer that way of saying it.

Another fact is that

1cm^3=1cc=1mL

Furthermore,

1L=1000mL

Therefore, your 5736.5cc engine is also 5736.5mL, which is 5.7365L.

My old minibike of 50cc? Well, I guess it was 50/16.39=3.05inch^3. For its size, though, it really got around:)

We’ve hit a cool, damp stretch of days here, but we needed it. Wherever you are, hope your summer is turning out well, too.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.