I am solving the following problem:

A Pipe 200 mm outside diameter and 20 m length is covered with a layer of 70 mm thick insulation having thermal conductivity of 0.05 W/m·K and a thermal conductance of 10 W/m2·K at the outer surface. If the temperature of the pipe is 350 °C and the ambient temperature is 15 °C, calculate the external surface temperature of the lagging.

I have calculated the thermal resistances as:

$${ R_{pipe} = \frac{1}{A \times h_o} = \frac{1}{ (\pi \times 200^2 \times (1/10^3)^2 \times ()} = 3.183098862 }$$

$${ R_{insulation} = \frac{1}{2\pi KL} = \frac{1}{ (2\pi \times 0.05 \times 20)} = 0.04776312933 }$$

The heat transfer is supposed to be:

$${ Q = \frac{\triangle T}{ R_{total}} }$$

What does it mean by "temperature of the lagging"? The change in temperature from the outside and inside already given.

  • 1
    $\begingroup$ You really need to double-check those calculations. Your R_pipe is missing a term and none of the arithmetic is correct. $\endgroup$
    – Air
    Jul 15 '15 at 15:54

What does it mean by "temperature of the lagging" ?  

Air's comment:

The lagging is not really the insulation layer itself but the cladding around the insulation layer. The insulation has no real structural properties to speak of; lagging not only holds the insulation in place and protects it from the environment (including insects and rodents), it can also provide a lower-emissivity surface for the insulation layer, further reducing heat transfer.

however, since there is no given $k$ (thermal conductivity) value for the lagging I'll assume that the required temperature is the outer surface temperature of the insulation layer.

  Your $R_{pipe}$ term should be the resistance due to convection heat transfer from insulation surface to the ambient air:   $$ R_{pipe} = \frac{1}{A_o  h_o} =  \frac{1}{\pi D_{ins} L h_o} = \frac{1}{\pi *  0.17 * 2 * 20 *  10} = 0.00468\ \text{°C/Watt}$$ and since the thermal insulation is a cylinder with inner and outer radius of $r_1$ and $r_2$ respectively, its thermal resistance should be calculated as follows:   $${ R_{ins} = \frac{ln(\frac{r_2}{r_1})}{2\pi kL} = \frac{ln(\frac{170\ mm}{100\ mm})}{ (2\pi \times 0.05 \times 20)}   = 0.0844\ \text{°C/Watt} }$$

$$Q = \frac{\triangle T}{ R_{total}} = \frac{350 - 15}{ 0.00468+0.0844} = 3760.664 \text{ Watt}$$   finally temperature of outer surface of insulation:   $$ \triangle T = Q\times  R_{pipe} = 3760.664* 0.00468=17.6\ \mathrm{°C} = T_{outer} - T_{ambient}$$   $$T_{outer} = T_{ambient} + \triangle T = 15 + 17.6 =  32.6\ \mathrm{°C}$$

  • 1
    $\begingroup$ @Air I made the assumption that they are both the same since there is no separate $k$ given and thanks for spotting the surface area error. All is fixed now $\endgroup$
    – Algo
    Jul 15 '15 at 17:48
  • 1
    $\begingroup$ Readers less comfortable with these equations should note that the effective thickness of the cylindrical insulation is $r_2\ln(r_2/r_1)$ but the $r_2$ outside the logarithm cancels in the denominator. $\endgroup$
    – Air
    Jul 15 '15 at 17:53
  • $\begingroup$ Regarding the separate $k$, it's a perfectly reasonable assumption to make; the contribution of the cladding toward thermal resistance is typically negligible. It could be nothing more than aluminum foil. $\endgroup$
    – Air
    Jul 15 '15 at 17:55
  • $\begingroup$ @algo why is D_ins 0.17∗2 ? $\endgroup$
    – james
    Jul 16 '15 at 2:12
  • $\begingroup$ @james pipe outer radius is 100 mm adding a 70 mm thickness of insulation layer makes total of 170 mm radius to the outer surface of insulation and so the diameter is 2 * 170 mm = 2 * 0.17 meters $\endgroup$
    – Algo
    Jul 16 '15 at 10:46

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