3 Engineering Example 2
3.1 Heat flow in an insulated metal plate
Introduction
Thermal insulation is important in many domestic (e.g. central heating) and industrial (e.g cooling and heating) situations. Although many real situations involve heat flow in more than one dimension, we consider only a one dimensional case here. The flow of heat is determined by temperature and thermal conductivity. It is possible to model the amount of heat $Q$ (J) crossing point $x$ in one dimension (the heat flow in the $x$ direction) from temperature ${T}_{2}$ (K) to temperature ${T}_{1}$ (K) (in which ${T}_{2}>{T}_{1}$ ) in time $t$ s by
$\phantom{\rule{2em}{0ex}}\frac{Q}{t}=\lambda A\left(\frac{{T}_{2}{T}_{1}}{x}\right)$ ,
where $\lambda $ is the thermal conductivity in ${\text{Wm}}^{1}\text{K}$ .
Problem in words
Suppose that the upper and lower sides of a metal plate connecting two containers are insulated and one end is maintained at a temperature ${T}_{2}$ (K) (see Figure 7).
The plate is assumed to be infinite in the direction perpendicular to the sheet of paper.
Figure 7:
 Find a formula for $T$ .
 If $\lambda =205$ ( ${\text{Wm}}^{1}{\text{K}}^{1}$ ), ${T}_{1}=300$ (K), $A=0.004{\text{m}}^{2}$ ), $x=0.5$ (m), calculate the value of ${T}_{2}$ required to achieve a heat flow of $100{\text{Js}}^{1}$ .
Mathematical statement of the problem
 Given $\frac{Q}{t}=\lambda A\left(\frac{{T}_{2}{T}_{1}}{x}\right)$ express ${T}_{2}$ as the subject of the formula.

In the formula found in part
 substitute $\lambda =205,\phantom{\rule{1em}{0ex}}{T}_{1}=300,\phantom{\rule{1em}{0ex}}A=0.004$ , $x=0.5$ and $\frac{Q}{t}=100$ to find ${T}_{2}$ .
Mathematical analysis

$\frac{Q}{t}=\lambda A\left(\frac{{T}_{2}{T}_{1}}{x}\right)$
Divide both sides by $\lambda A$
$\phantom{\rule{2em}{0ex}}\frac{Q}{t\lambda A}=\frac{{T}_{2}{T}_{1}}{x}$
Multiply both sides by $x$
$\phantom{\rule{2em}{0ex}}\frac{Qx}{t\lambda A}={T}_{2}{T}_{1}$
Add ${T}_{1}$ to both sides
$\phantom{\rule{2em}{0ex}}\frac{Qx}{t\lambda A}+{T}_{1}={T}_{2}$
which is equivalent to
$\phantom{\rule{2em}{0ex}}{T}_{2}=\frac{Qx}{t\lambda A}+{T}_{1}$

Substitute
$\phantom{\rule{1em}{0ex}}\lambda =205,\phantom{\rule{1em}{0ex}}{T}_{1}=300,\phantom{\rule{1em}{0ex}}A=0.004,\phantom{\rule{1em}{0ex}}x=0.5$
and
$\frac{Q}{t}=100$
to find
${T}_{2}$
:
$\phantom{\rule{2em}{0ex}}{T}_{2}=\frac{100\times 0.5}{205\times 0.004}+300\approx 60.9+300=360.9$
So the temperature in container 2 is $361\text{K}$ to 3 sig.fig.
Interpretation
The formula $\phantom{\rule{1em}{0ex}}{T}_{2}=\frac{Qx}{t\lambda A}+{T}_{1}$ can be used to find a value for ${T}_{2}$ that would achieve any desired heat flow. In the example given ${T}_{2}$ would need to be about $361$ K ( $\approx 7{8}^{\circ}\text{C}$ ) to produce a heat flow of $100{\text{Js}}^{1}$ .
Exercises
 The formula for the volume of a cylinder is $V=\pi {r}^{2}h$ . Find $V$ when $r=5$ cm and $h=15$ cm.

If
$R=5{p}^{2}$
, find
$R$
when
 $p=10$ ,
 $p=16$ .

For the following formulae, find
$y$
at the given values of
$x$
.
 $y=3x+2$ , $x=1$ , $x=0$ , $x=1$ .
 $y=4x+7$ , $x=2$ , $x=0$ , $x=1$ .
 $y={x}^{2}$ , $x=2$ , $x=1$ , $x=0$ , $x=1$ , $x=2$ .
 If $P=\frac{3}{QR}$ find $P$ if $Q=15$ and $R=0.300$ .
 If $y=\sqrt{\frac{x}{z}}$ find $y$ if $x=13.200$ and $z=15.600$ .
 Evaluate $M=\frac{\pi}{2r+s}$ when $r=23.700$ and $s=0.2$ .

To convert a length measured in metres to one measured in centimetres, the length in metres is multiplied by 100. Convert the following lengths to cm.
 5 m,
 0.5 m,
 56.2 m.

To convert an area measured in
${\text{m}}^{2}$
to one measured in
${\text{cm}}^{2}$
, the area in
${\text{m}}^{2}$
is multiplied by
$1{0}^{4}$
. Convert the following areas to
${\text{cm}}^{2}$
.
 $5{\text{m}}^{2}$ ,
 $0.33{\text{m}}^{2}$ ,
 $6.2{\text{m}}^{2}$ .

To convert a volume measured in
${\text{m}}^{3}$
to one measured in
${\text{cm}}^{3}$
, the volume in
${\text{m}}^{3}$
is multiplied by
$1{0}^{6}$
. Convert the following volumes to
${\text{cm}}^{3}$
.
 $15{\text{m}}^{3}$ ,
 $0.25{\text{m}}^{3}$ ,
 $8.2{\text{m}}^{3}$ .
 If $\eta =\frac{4{Q}_{P}}{\pi {d}^{2}Ln}$ evaluate $\eta $ when ${Q}_{P}=0.0003$ , $d=0.05$ , $L=0.1$ and $n=2$ .

The moment of inertia of an object is a measure of its resistance to rotation. It depends upon both the mass of the object and the distribution of mass about the axis of rotation. It can be shown that the moment of inertia,
$J$
, of a solid disc rotating about an axis through its centre and perpendicular to the plane of the disc, is given by the formula
$\phantom{\rule{2em}{0ex}}J=\frac{1}{2}M{a}^{2}$
where $M$ is the mass of the disc and $a$ is its radius. Find the moment of inertia of a disc of mass 12 kg and diameter 10 m. The SI unit of moment of inertia is $\text{kg}\phantom{\rule{0.3em}{0ex}}{\text{m}}^{2}$ .

Transpose the given formulae to make the given variable the subject.
 $y=3x7$ , for $x$ ,
 $8y+3x=4$ , for $x$ ,
 $8x+3y=4$ for $y$ ,
 $132x7y=0$ for $x$ .

Transpose the formula
$PV=RT$
for
 $V$ ,
 $P$ ,
 $R$ ,
 $T$ .

Transpose
$v=\sqrt{x+2y}$
,
 for $x$ ,
 for $y$ .
 Transpose $8u+4v3w=17$ for each of $u$ , $v$ and $w$ .

When a ball is dropped from rest onto a horizontal surface it will bounce before eventually coming to rest after a time
$T$
where
$\phantom{\rule{2em}{0ex}}T=\frac{2v}{g}\left(\frac{1}{1e}\right)$
where $v$ is the speed immediately after the first impact, and $g$ is a constant called the acceleration due to gravity. Transpose this formula to make $e$ , the coefficient of restitution, the subject.
 Transpose $q={A}_{1}\sqrt{\frac{2gh}{{\left({A}_{1}\u2215{A}_{2}\right)}^{2}1}}$ for ${A}_{2}$ .

Make
$x$
the subject of
 $y=\frac{r+x}{1rx}$ ,
 $y=\sqrt{\frac{x1}{x+1}}$ .

In the design of orifice plate flowmeters, the volumetric flowrate, Q (
${\text{m}}^{3}$
${\text{s}}^{1}$
), is given by
$\phantom{\rule{2em}{0ex}}Q={C}_{d}{A}_{o}\sqrt{\frac{2g\Delta h}{1{A}_{o}^{2}\u2215{A}_{p}^{2}}}$
where ${C}_{d}$ is a dimensionless discharge coefficient, $\Delta h$ (m) is the head difference across the orifice plate and ${A}_{o}$ ( ${\text{m}}^{2}$ ) is the area of the orifice and ${A}_{p}$ ( ${\text{m}}^{2}$ ) is the area of the pipe.
 Rearrange the equation to solve for the area of the orifice, ${A}_{o}$ , in terms of the other variables.

A volumetric flowrate of
$100{\text{cm}}^{3}{\text{s}}^{1}$
passes through a
$10$
cm inside diameter pipe. Assuming a discharge coefficient of
$0.6$
, calculate the required orifice diameter, so that the head difference across the orifice plate is
$200$
mm.
[Hint: be very careful with the units!]
 $1178.1{\text{cm}}^{3}$

 $500$ ,
 $1280$

 $1,2,5$ ,
 $15,7,3$ ,
 $5,3,1,0$ ,
 $P=0.667$
 $y=0.920$
 $M=0.067$

 $500\text{cm}$ ,
 $50\text{cm}$ ,
 $5620\text{cm}$ .

 $50000{\text{cm}}^{2}$ ,
 $3300{\text{cm}}^{2}$ ,
 $62000{\text{cm}}^{2}$ .

 $15000000{\text{cm}}^{3}$ ,
 $250000{\text{cm}}^{3}$ ,
 $8200000{\text{cm}}^{3}$ .
 $\eta =0.764$ .
 $150\text{kg}\phantom{\rule{0.3em}{0ex}}{\text{m}}^{2}$

 $x=\frac{y+7}{3}$ ,
 $x=\frac{48y}{3}$ ,
 $y=\frac{48x}{3}$ ,
 $x=\frac{137y}{2}$

 $V=\frac{RT}{P}$ ,
 $P=\frac{RT}{V}$ ,
 $R=\frac{PV}{T}$ ,
 $T=\frac{PV}{R}$

 $x={v}^{2}2y$ ,
 $y=\frac{{v}^{2}x}{2}$
 $u=\frac{174v+3w}{8}$ , $v=\frac{178u+3w}{4}$ , $w=\frac{8u+4v17}{3}$
 $e=1\frac{2v}{gT}$
 ${A}_{2}=\pm \sqrt{\frac{{A}_{1}^{2}{q}^{2}}{2{A}_{1}^{2}gh+{q}^{2}}}$

 $x=\frac{yr}{1+yr}$ ,
 $x=\frac{1+{y}^{2}}{1{y}^{2}}$

 ${A}_{0}=\frac{Q{A}_{p}}{\sqrt{{Q}^{2}+2g\Delta h{A}_{p}^{2}{C}_{d}^{2}}}$

$Q=100{\text{cm}}^{3}{\text{s}}^{1}=1{0}^{4}{\text{m}}^{2}{\text{s}}^{1}$
${A}_{p}=\pi \frac{0.{1}^{2}}{4}=0.007854$
${\text{m}}^{2}$
${C}_{d}=0.6$
$\Delta h=0.2$
m
$g=9.81{\text{ms}}^{2}$
Substituting in answer
 gives ${A}_{o}=8.4132\times 1{0}^{5}{\text{m}}^{2}$ so diameter $=\sqrt{\frac{4{A}_{o}}{\pi}}=0.01035\text{m}=1.035\text{cm}$