Additional Magnesia Insulation on a Hot Pipe: 23-Chem-A2 Unit Operations and Separation Processes, May 2016, Q6

Question B3: Additional Magnesia Insulation on a Hot Pipe (25 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Composite cylindrical wall.

Quantity	Value
Pipe outer radius $r_1$	25 mm (surface at 1100 K)
First insulation to $r_2$	+50 mm ⇒ $r_2=75$ mm, $k_1=0.17$
Magnesia layer	$k_2=0.09$ W/m·K, $T_{max}=615$ K
Target outer surface T	370 K
Air temperature	280 K
Surface coefficient $h$	10 W/m²·K

Find. The magnesia thickness $t=r_3-r_2$, and a check that the magnesia stays below 615 K.

Figure B3 — Two conduction resistances in series (existing insulation then magnesia) discharge to air through a combined convection/radiation coefficient; the added magnesia must drop the outer face to 370 K.

Approach. At steady state the same heat rate per metre passes through both conduction layers and off the outer surface; equating the conduction expression (1100 K to 370 K) with the surface loss ($h$, air) gives one equation for the outer radius $r_3$, hence the thickness — then check the magnesia hot-face temperature.

Heat loss set by the target surface. With the outer face fixed at $T_3=370$ K, the surface loss per metre is $$q=h(2\pi r_3)(T_3-T_{air})=10(2\pi r_3)(370-280).$$ This grows with $r_3$.
Same $q$ through the two conduction layers. From 1100 K to 370 K across the series resistance, $$q=\frac{1100-370}{\dfrac{\ln(r_2/r_1)}{2\pi k_1}+\dfrac{\ln(r_3/r_2)}{2\pi k_2}}.$$ This falls as $r_3$ grows (thicker magnesia).
Solve for the outer radius. Equating the two expressions gives a single transcendental equation in $r_3$; numerically $$r_3=0.0924\ \mathrm{m}\;\Rightarrow\; t=r_3-r_2=0.0924-0.075=\;\boxed{17.4\ \mathrm{mm}}.$$ The corresponding loss is $q=522$ W per metre of pipe.
Check the magnesia service temperature. The temperature at the magnesia hot face (inner radius $r_2$) is $$T_{r_2}=1100-q\,\frac{\ln(r_2/r_1)}{2\pi k_1}=1100-522\frac{\ln(3)}{2\pi(0.17)}=563\ \mathrm{K}.$$ Since $563 \mathrm{K}<615 \mathrm{K}$, the magnesia operates safely. Magnesia thickness ≈ 17.4 mm; hot face 563 K < 615 K limit

Quantity	Result
Outer radius $r_3$	92.4 mm
Magnesia thickness $t$	≈ 17.4 mm
Heat loss $q$	522 W/m
Magnesia hot-face T	563 K (< 615 K ✓)

Check (interpretation): the 50 mm pipe "outer diameter" is taken as $r_1=25$ mm with the pipe metal at 1100 K, and the first 50 mm insulation therefore extends to $r_2=75$ mm; the single coefficient $h=10$ W/m²·K lumps convection and radiation at the outer face as the exam states.

Concept & Theory

Topic. Steady radial conduction through composite cylinders with a surface (convection + radiation) resistance.

Key relations. Cylindrical conduction resistance $R=\ln(r_o/r_i)/2\pi kL$; series addition; surface loss $q=h\,A_o(T_s-T_\infty)$ with $A_o=2\pi r_3 L$; energy continuity $q_{cond}=q_{surf}$.

Why this works. In steady state the heat rate is identical through every series element, so fixing the outer surface temperature ties the surface-loss equation (increasing in $r_3$) to the conduction equation (decreasing in $r_3$); their intersection is the required outer radius. The intermediate-face temperature then follows from the temperature drop across the first layer alone.

Common pitfall. Using a plane-wall ($\Delta x/k$) resistance instead of the logarithmic cylindrical form; and forgetting to verify the material temperature limit — magnesia would fail if placed where the interface exceeds 615 K.

Source. Incropera & DeWitt, Fundamentals of Heat and Mass Transfer ch. 3 (one-dimensional conduction, composite cylinders); Coulson & Richardson Vol. 1 ch. 9.

Question 6 of 6: Additional Magnesia Insulation on a Hot Pipe

Question B3: Additional Magnesia Insulation on a Hot Pipe (25 marks)