# 1.6 Mechanisms of heat transfer  (Page 15/27)

 Page 15 / 27

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 $\text{Δ}T$ for the 120-day heating season to be $15.0\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}.$ We found in the preceding problem that $P=126\text{Δ}T\phantom{\rule{0.2em}{0ex}}\text{W}·\text{°}\text{C}$ as baseline energy use. So the total heat loss during this period is $Q=\left(126\phantom{\rule{0.2em}{0ex}}\text{J/s}·\text{°}\text{C}\right)\left(15.0\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}\right)\left(120\phantom{\rule{0.2em}{0ex}}\text{days}\right)\left(86.4\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{s/day}\right)=1960\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{6}\phantom{\rule{0.2em}{0ex}}\text{J}$ . 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 $3695\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}$ 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 $–15\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}$ to $40\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}$ but it cannot expand? For steel, the Young’s modulus $Y=210\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{9}\phantom{\rule{0.2em}{0ex}}{\text{N/m}}^{2}$ from Stress, Strain, and Elastic Modulus . (Ignore the change in area resulting from the expansion.)

$\frac{F}{A}=\left(210\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{9}\phantom{\rule{0.2em}{0ex}}\text{Pa}\right)\left(12\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-6}\text{/}\text{°}\text{C}\right)\left(40\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}-\left(-15\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}\right)\right)=1.4\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{8}\phantom{\rule{0.2em}{0ex}}{\text{N/m}}^{2}$ .

A brass rod $\left(Y=90\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{9}\phantom{\rule{0.2em}{0ex}}{\text{N/m}}^{2}\right),$ with a diameter of 0.800 cm and a length of 1.20 m when the temperature is $25\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}$ , 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 $1\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}$ 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 $\left(1\phantom{\rule{0.2em}{0ex}}\text{watt}=1\phantom{\rule{0.2em}{0ex}}\text{joule/second}$ or $1\phantom{\rule{0.2em}{0ex}}\text{W}=1\phantom{\rule{0.2em}{0ex}}\text{J/s}$ and
$1\phantom{\rule{0.2em}{0ex}}\text{MW}=1\phantom{\rule{0.2em}{0ex}}\text{megawatt}\right).$ (a) Calculate the rate of temperature increase in degrees Celsius per second $\left(\text{°}\text{C/s}\right)$ if the mass of the reactor core is $1.60\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{5}\phantom{\rule{0.2em}{0ex}}\text{kg}$ and it has an average specific heat of $0.3349\phantom{\rule{0.2em}{0ex}}\text{kJ/kg}·\text{°}\text{C}$ . (b) How long would it take to obtain a temperature increase of $2000\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}$ , 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.)

The properties of a system during a reversible constant pressure non-flow process at P= 1.6bar, changes from constant volume of 0.3m³/kg at 20°C to a volume of 0.55m³/kg at 260°C. its constant pressure process is 3.205KJ/kg°C Determine: 1. Heat added, Work done, Change in Internal Energy and Change in Enthalpy
please how do dey get 5/9 in the conversion of Celsius and Fahrenheit
what is copper loss
this is the energy dissipated(usually in the form of heat energy) in conductors such as wires and coils due to the flow of current against the resistance of the material used in winding the coil.
Henry
it is the work done in moving a charge to a point from infinity against electric field
what is the weight of the earth in space
As w=mg where m is mass and g is gravitational force... Now if we consider the earth is in gravitational pull of sun we have to use the value of "g" of sun, so we can find the weight of eaeth in sun with reference to sun...
Prince
g is not gravitacional forcé, is acceleration of gravity of earth and is assumed constante. the "sun g" can not be constant and you should use Newton gravity forcé. by the way its not the "weight" the physical quantity that matters, is the mass
Jorge
Yeah got it... Earth and moon have specific value of g... But in case of sun ☀ it is just a huge sphere of gas...
Prince
Thats why it can't have a constant value of g ....
Prince
not true. you must know Newton gravity Law . even a cloud of gas it has mass thats al matters. and the distsnce from the center of mass of the cloud and the center of the mass of the earth
Jorge
please why is the first law of thermodynamics greater than the second
define electric image.obtain expression for electric intensity at any point on earthed conducting infinite plane due to a point charge Q placed at a distance D from it.
explain the lack of symmetry in the field of the parallel capacitor
pls. explain the lack of symmetry in the field of the parallel capacitor
Phoebe
does your app come with video lessons?
What is vector
Vector is a quantity having a direction as well as magnitude
Damilare
tell me about charging and discharging of capacitors
a big and a small metal spheres are connected by a wire, which of this has the maximum electric potential on the surface.
3 capacitors 2nf,3nf,4nf are connected in parallel... what is the equivalent capacitance...and what is the potential difference across each capacitor if the EMF is 500v
equivalent capacitance is 9nf nd pd across each capacitor is 500v
santanu
four effect of heat on substances
why we can find a electric mirror image only in a infinite conducting....why not in finite conducting plate..?
because you can't fit the boundary conditions.
Jorge
what is the dimensions for VISCOUNSITY (U)
Branda