Question 9.34: A pipeline of 100 mm outside diameter, carrying steam at 420......
A pipeline of 100 mm outside diameter, carrying steam at 420 K, is to be insulated with a lagging material which costs £10/m³ and which has a thermal conductivity of 0.1 W/m K. The ambient temperature may be taken as 285 K, and the coefficient of heat transfer from the outside of the lagging to the surroundings as 10 W/m² K. If the value of heat energy is 7.5 x 10^{-4} £/MJ and the capital cost of the lagging is to be depreciated over 5 years with an effective simple interest rate of 10 per cent per annum based on the initial investment, what is the economic thickness of the lagging?
Is there any possibility that the heat loss could actually be increased by the application of too thin a layer of lagging?
For a thick-walled cylinder, the rate of conduction of heat through lagging is given by equation 9.21:
Q=\frac{2\pi l k(T_{1}-T_{2})}{\ln(r_{2}/r_{1})} (9.21)
Q={\frac{2\pi l k(T_{i}-T_{o})}{\ln(d_{o}/d_{i})}} W
where d_{o} and d_{i} are the external and internal diameters of the lagging and T_{o} and T_{i} the corresponding temperatures.
Substituting k = 0.1 W/mK, T_{o} = 420K (neglecting temperature drop across pipe wall), and d_{i} = 0.1 m, then:
{\frac{Q}{l}}={\frac{2\pi\times0.1(420-T_{o})}{\ln(d_{o}/0.1)}} W/m
The term Q/l must also equal the heat loss from the outside of the lagging,
or: \frac{Q}{l}=h_{o}(T_{o}-285)\pi d_{o}=10(T_{o}-285)\pi d_{o}\mathrm{{\bf~W/m}}
Thus: T_{o}=\left\{{\frac{Q}{l}}{\frac{1}{10\pi d_{o}}}+285\right\} K
Substituting: \frac{Q}{l}=\frac{2\pi\times0.1\left[135-\frac{Q}{l}\frac{1}{10\pi d_{o}}\right]}{\ln(d_{o}/0.1)}
or: \frac{Q}{l}=\frac{2\pi\times0.1\times135}{\ln(d_{o}/0.1)+2\pi\times0.1\times\frac{1}{10\pi d_{o}}}=\frac{84.82}{\ln(d_{o}/0.1)+(0.02/d_{o})} W/m
Value of heat lost = £7.5 x 10^{-4}/MJ
or: \frac{84.82}{\ln(d_{o}/0.1)+(0.02/d_{o})}\times7.5\times10^{-4}\times10^{-6}=\frac{6.36\times10^{-8}}{\ln(d_{o}/2.1)+(0.02/d_{o})} £/m s
Volume of lagging per unit pipe length = \frac{\pi}{4}[d_{o}^{2}-(0.1)^{2}]\ m^{3}/\mathrm{m}
Capital cost of lagging = £10/m³ or \frac{\pi}{4}[d_{o}^{2}-0.01]10=7.85(d_{o}^{2}-0.01) £/m
Noting that 1 year = 31.5 M s, then :
Depreciation = 7.85[d_{o}^{2}-0.01]/(5\times31.5\times10^{6})=4.98\times10^{-8}(d_{o}^{2}-0.01) £/Ms
Interest charges = (0.1\times7.85)(d_{o}^{2}-0.01)/(31.5\times10^{6})=2.49\times10^{-8}(d_{o}^{2}-0.01) £/Ms
Total capital charges = 7.47\times10^{-8}(d_{o}^{2}-0.01) £/Ms
Total cost (capital charges + value of heat lost) is given by:
C=\left\{\frac{6.36}{\ln(d_{o}/0.1)+(0.02/d_{o})}+7.47(d_{o}^{2}-0.01)\right\}10^{-8} £/Ms
Differentiating with respect to d_{o}:
10^{8}\frac{\mathrm{d}C}{\mathrm{d}d_{o}}=6.36\left[\frac{-1}{\mathrm{[ln}(d_{o}/0.1)+(0.02/d_{o})]^{2}}\right]\left[\frac{1}{d_{o}}-\frac{0.02}{d_{o}^{2}}\right]+7.47(2d_{o})In order to obtain the minimum value of C, dC/dd_{o} must be put equal to zero.
Then: \frac{1}{[\ln(d_{o}/0.1)+(0.02/d_{o})]^{2}}=\frac{(7.47\times2)}{6.36}\left[\frac{d_{o}}{(1/d_{o})-(0.02/d_{o}^{2})}\right]
that is: \frac{1}{[\ln(d_{o}/0.1)+(0.02/d_{o})]^{2}}=2.35\frac{d_{o}^{3}}{(d_{o}-0.02)}
A trial and error solution gives d_{o} = 0.426 m or 426 mm
Thus, the economic thickness of lagging = (426 — 100)/2 = 163 mm
For this pipeline: {\frac{h r}{k}}={\frac{10\times(50\times10^{-3})}{0.1}}=5
From equation 9.267, the critical value of hr/k, below which the heat loss may be increased by a thin layer of lagging, is 1. For hr/k > 1, as in this problem, the situation will not arise.
{\frac{h r}{k}}=1 (9.267)