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Many decisions are made on the basis of the payback period: the time it will take through savings to equal the capital cost of an investment. Acceptable payback times depend upon the business or philosophy one has. (For some industries, a payback period is as small as 2 years.) Suppose you wish to install the extra insulation in the preceding problem. If energy cost $\text{\$}1.00$ per million joules and the insulation was $4.00 per square meter, then calculate the simple payback time. Take the average $\text{\Delta}T$ for the 120-day heating season to be $15.0\phantom{\rule{0.2em}{0ex}}\text{\xb0}\text{C}.$
We found in the preceding problem that $P=126\text{\Delta}T\phantom{\rule{0.2em}{0ex}}\text{W}\xb7\text{\xb0}\text{C}$ as baseline energy use. So the total heat loss during this period is $Q=\left(126\phantom{\rule{0.2em}{0ex}}\text{J/s}\xb7\text{\xb0}\text{C}\right)\left(15.0\phantom{\rule{0.2em}{0ex}}\text{\xb0}\text{C}\right)\left(120\phantom{\rule{0.2em}{0ex}}\text{days}\right)\left(86.4\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{s/day}\right)=1960\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{6}\phantom{\rule{0.2em}{0ex}}\text{J}$ . At the cost of $1/MJ, the cost is $1960. From an earlier problem, the savings is 12% or $235/y. We need $150\phantom{\rule{0.2em}{0ex}}{\text{m}}^{2}$ of insulation in the attic. At $\$4\text{/}{\text{m}}^{2}$ , this is a $500 cost. So the payback period is $\$600\text{/}\left(\$235\text{/}\text{y}\right)=2.6\phantom{\rule{0.2em}{0ex}}\text{years}$ (excluding labor costs).
In 1701, the Danish astronomer Ole Rømer proposed a temperature scale with two fixed points, freezing water at 7.5 degrees, and boiling water at 60.0 degrees. What is the boiling point of oxygen, 90.2 K, on the Rømer scale?
What is the percent error of thinking the melting point of tungsten is $3695\phantom{\rule{0.2em}{0ex}}\text{\xb0}\text{C}$ instead of the correct value of 3695 K?
$7.39\%$
An engineer wants to design a structure in which the difference in length between a steel beam and an aluminum beam remains at 0.500 m regardless of temperature, for ordinary temperatures. What must the lengths of the beams be?
How much stress is created in a steel beam if its temperature changes from $\mathrm{\u201315}\phantom{\rule{0.2em}{0ex}}\text{\xb0}\text{C}$ to $40\phantom{\rule{0.2em}{0ex}}\text{\xb0}\text{C}$ but it cannot expand? For steel, the Young’s modulus $Y=210\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{9}\phantom{\rule{0.2em}{0ex}}{\text{N/m}}^{2}$ from Stress, Strain, and Elastic Modulus . (Ignore the change in area resulting from the expansion.)
$\frac{F}{A}=\left(210\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{9}\phantom{\rule{0.2em}{0ex}}\text{Pa}\right)\left(12\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-6}}\text{/}\text{\xb0}\text{C}\right)\left(40\phantom{\rule{0.2em}{0ex}}\text{\xb0}\text{C}-\left(\mathrm{-15}\phantom{\rule{0.2em}{0ex}}\text{\xb0}\text{C}\right)\right)=1.4\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{8}\phantom{\rule{0.2em}{0ex}}{\text{N/m}}^{2}$ .
A brass rod $\left(Y=90\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{9}\phantom{\rule{0.2em}{0ex}}{\text{N/m}}^{2}\right),$ with a diameter of 0.800 cm and a length of 1.20 m when the temperature is $25\phantom{\rule{0.2em}{0ex}}\text{\xb0}\text{C}$ , is fixed at both ends. At what temperature is the force in it at 36,000 N?
A mercury thermometer still in use for meteorology has a bulb with a volume of $0.780\phantom{\rule{0.2em}{0ex}}{\text{cm}}^{3}$ and a tube for the mercury to expand into of inside diameter 0.130 mm. (a) Neglecting the thermal expansion of the glass, what is the spacing between marks $1\phantom{\rule{0.2em}{0ex}}\text{\xb0}\text{C}$ apart? (b) If the thermometer is made of ordinary glass (not a good idea), what is the spacing?
a. $1.06$ cm; b. $1.11$ cm
Even when shut down after a period of normal use, a large commercial nuclear reactor transfers thermal energy at the rate of 150 MW by the radioactive decay of fission products. This heat transfer causes a rapid increase in temperature if the cooling system fails
$(1\phantom{\rule{0.2em}{0ex}}\text{watt}=1\phantom{\rule{0.2em}{0ex}}\text{joule/second}$ or
$1\phantom{\rule{0.2em}{0ex}}\text{W}=1\phantom{\rule{0.2em}{0ex}}\text{J/s}$ and
$1\phantom{\rule{0.2em}{0ex}}\text{MW}=1\phantom{\rule{0.2em}{0ex}}\text{megawatt}).$ (a) Calculate the rate of temperature increase in degrees Celsius per second
$\left(\text{\xb0}\text{C/s}\right)$ if the mass of the reactor core is
$1.60\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{5}\phantom{\rule{0.2em}{0ex}}\text{kg}$ and it has an average specific heat of
$0.3349\phantom{\rule{0.2em}{0ex}}\text{kJ/kg}\xb7\text{\xb0}\text{C}$ . (b) How long would it take to obtain a temperature increase of
$2000\phantom{\rule{0.2em}{0ex}}\text{\xb0}\text{C}$ , which could cause some metals holding the radioactive materials to melt? (The initial rate of temperature increase would be greater than that calculated here because the heat transfer is concentrated in a smaller mass. Later, however, the temperature increase would slow down because the 500,000-kg steel containment vessel would also begin to heat up.)
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