Question 15.1: A steel pipe having an inside diameter of 1.88 cm and a wall......

The first law of thermodynamics applied to this problem will reduce to the form \delta Q/d t=0, indicating that the rate of heat transfer into the control volume is equal to the rate leaving—that is, Q=q=\mathrm{constant}.

As the heat flow will be in the radial direction, the independent variable is r, and the proper form for the Fourier rate equation is

q_{r}=-k A{\frac{d T}{d r}}

Writing A=2\pi r L, we have

q_{r}=-k(2\pi r L){\frac{d T}{d r}}

where q_{r} is constant, which may be separated and solved as follows:

q_{r}\,\int_{r_{i}}^{r_{o}}{\frac{d r}{r}}=\,-2\pi k L\int_{T_{i}}^{T_{o}}d T=2\pi k L\int_{T_{o}}^{T_{i}}d T

\begin{array}{r l}{q_{r}\ln{\frac{r_{o}}{r_{i}}}}&{=\ 2\pi k L(T_{i}-T_{o})}\end{array}             (15-9)

q_{r}\ \ =\ \ \frac{2\pi k L}{\ln{r_{o}}/r_{i}}(T_{i}-T_{o})

Substituting the given numerical values, we obtain

q_{r}={\frac{2\pi(42.90\ \mathrm{W/m}\cdot\mathrm{K})(367-344)\mathrm{K}}{\ln(2.66/1.88)}}

=17.860\,\mathrm{W/m}\,(18,600\,\mathrm{Btu}/\mathrm{h}\cdot\mathrm{ft})

The inside and outside surface areas per unit length of pipe are

A_{i}\,=\,\pi(1.88)(10^{-2})(1)=0.059\,\mathrm{m^{2}/m}\,(0.194\,\mathrm{ft^{2}/f t})

A_{o}=\pi(2.662)(10^{-2})(1)=0.084\,{\mathrm{m}}^{2}/{\mathrm{m}}\,(0.275\,{\mathrm{ft}}^{2}/{\mathrm{ft}})

giving

{\frac{q_{r}}{A_{i}}}={\frac{17,860}{0.059}}=302.7{\mathrm{~kW/m^{2}~(95,900~B t u/h\cdot f t^{2})}}

\frac{q_{o}}{A_{i}}=\frac{17,860}{0.084}=212.6\;\mathrm{kW/m^{2}}\;(67,400\;\mathrm{Btu/h}\cdot\mathrm{ft}^{2})