Question 9.33: A steam pipe, 150 mm i.d. and 168 mm o.d., is carrying steam......

In this case:

d = 150 mm or 0.150 m
d_{o} = 168 mm or 0.168 m
d_{w} = 159 mm or 0.159 m
d_{s} = 268 mm or 0.268 m
d_{m}, the log mean of d_{o} and d_{s} = 215 mm or 0.215 m.

The coefficient for condensing steam together with that for any scale will be taken as 8500 W/m² K, k_{w} as 45 W/m K, and k_{l} as 0.073 W/m K.

The temperature on the outside of the lagging is estimated at 314 K and (h_{r} + h_{c}) will be taken as 10 W/m² K.

The thermal resistances are therefore:

\frac{1}{h_{i}\pi d}=\frac{1}{8500\times\pi\times0.150} = 0.00025

\frac{x_{w}}{k_{w}\pi d_{w}}=\frac{0.009}{45\times\pi\times0.159} = 0.00040

{\frac{x_{l}}{k_{l}\pi d_{m}}}={\frac{0.050}{0.073\times\pi\times0.215}} = 1.013

{\frac{1}{(h_{r}+h_{c})\pi d_{s}}}={\frac{1}{10\times\pi\times0.268}} = 0.119

The first two terms may be neglected and hence the total thermal resistance is 1.132 m K/W.

The heat loss per metre length = (444 – 294)/1.132 = \underline{\underline{132.5\ W/m}} (from equation 9.261).

\frac{Q}{l}=\Sigma\,\Delta T\Big/\left[\frac{1}{h_{i}\pi d}+\frac{x_{w}}{k_{w}\pi d_{w}}+\frac{x_{l}}{k_{r}\pi d_{m}}+\frac{1}{(h_{r}+h_{c})\pi d_{s}}\right]                         (9.261)

The temperature on the outside of the lagging may now be checked as follows:

ΔT(lagging) = 0.895(444 – 294) = 134 deg K
Thus the temperature on the outside of the lagging is (444 — 134) = 310 K, which approximates to the assumed value.

Taking an emissivity of 0.9, from equation 9.119:

h_{r}=q/(T_{1}-T_{2})=\frac{\mathrm{e}\sigma(T_{1}^{4}-T_{2}^{4})}{T_{1}-T_{2}}=\mathrm{e}\sigma(T_{1}^{3}+T_{1}^{2}T_{2}+T_{1}T_{2}^{2}+T_{3}^{3})                     (9.119)

From Table 9.5 for air (Gr Pr = 10^{4} – 10^{9}), n = 0.25 and C^{\prime \prime} = 1.32.

Substituting in equation 9.105 (putting l = diameter = 0.268 m):

h=C^{\prime\prime}(\Delta T)^{n}l^{3n-1}\;\mathrm{(W/m^{2}K)}                  (9.105)

and: h_{c}=C^{\prime\prime}(\Delta T)^{n}l^{3n-1}=1.32\left[\frac{310-294}{0.268}\right]^{0.25} = 3.67 W/m² K

Thus (h_{r} + h_{c}) = 11.1 W/m² K, which is close to the assumed value. In practice it is rare for forced convection currents to be absent, and the heat loss is probably higher than this value.

If the pipe were unlagged, (h_{r} + h_{c}) for ΔT = 150 K would be about 20 W/m² K and the heat loss would then be:

= (20 x π x 0.168 x 150) = 1584 W/m
or: \underline{\underline{1.58\ kW/m}}

Under these conditions it is seen that the heat loss has been reduced by more than 90 per cent by the addition of a 50 mm thickness of lagging.