Question 15.3: Saturated steam at 0.276 MPa flows inside a steel pipe havin......

In the case of the bare pipe there are three thermal resistances to evaluate

R_{1}=R_{\mathrm{convection\ \mathrm{inside}}}=1/h_{i}A_{i}

R_{2}\equiv R_{\mathrm{convection~outside}}\equiv1/h_{o}A_{o}

R_{3}=R_{\mathrm{conduction}}=\ln(r_{o}/r_{i})/2\pi k L

For conditions of this problem, these resistances have the values

R_{1}=1/[(5680\,\mathrm{W/m^{2}}\cdot\mathrm{K})(\pi)(0.0209\,\mathrm{m})(1\,\mathrm{m})]

=\;0.00268\ \mathrm{K/W}\qquad\left(0.00141\frac{\mathrm{h}\ \mathrm{R}}{\mathrm{Btu}}\right)

R_{2}=1/[(22.7\,\mathrm{W/m}^{2}\cdot{K})(\pi)(0.0267\,\mathrm{m})(1\,\mathrm{m})]

=\;0.525\;\mathrm{K/W}\qquad\left(0.277\frac{\mathrm{h\;R}}{\mathrm{Btu}}\right)

and

R_{3}={\frac{\ln(2.67/2.09)}{2\pi(42.9\;\mathrm{W/m}\cdot\mathrm{K})(1\,\mathrm{m})}}

=\;0.00091\;{\mathrm{K/W}}\qquad\ \;\left(0.0048\;{\frac{\mathrm{hR}}{\mathrm{Btu}}}\right)

The inside temperature is that of 0.276 MPa saturated steam, 404 K or 267 F. The heat-transfer rate per meter of pipe may now be calculated as

q={\frac{\Delta T}{\sum R}}={\frac{404-294\,\mathrm{K}}{0.528\,\mathrm{KW}}}

\begin{array}{r l}{=\ 208\,\mathrm{W}}&{{}\quad\left(710{\frac{\mathrm{Btu}}{\mathrm{h}}}\right)}\end{array}

In the case of an insulated pipe, the total thermal resistance would include R_{1} and R_{3} evaluated above, plus additional resistances to account for the insulation. For the insulation,

R_{4}=\frac{\ln(10.27/2.67)}{2\pi(0.0675\mathrm{\small~W/m\cdot K})(1\,m)}

=\;3.176\;\mathrm{K/W}\qquad\left(1.675\;\frac{\mathrm{h\;R}}{\mathrm{Btu}}\right)

and for the outside surface of the insulation,

\begin{array}{l}{{R_{5}=1/[(22.7\,\mathrm{W/m^{2}}\cdot{ K})(\pi)(0.1027\,\mathrm{m})(1\,\mathrm{m})]}}\end{array}

=~0.1365~\mathrm{K/W}~~~~~~~\left(0.0720~{\frac{\mathrm{hR}}{\mathrm{Btu}}}\right)

Thus, the heat loss for the insulated pipe becomes

q\ ={\frac{\Delta T}{\sum R}}={\frac{\Delta T}{R_{1}+R_{3}+R_{4}+R_{5}}}={\frac{404-294\ \mathrm{K}}{3.316\ \mathrm{K/W}}}

=~33.2~\mathrm{W}~~~~~~~\left(113~{\frac{\mathrm{Btu}}{\mathrm{h}}}\right)

This is a reduction of approximately 85%!