Pipe Insulation Thickness to Prevent Freezing During Shutdown: 17-Phys-B6 Applied Thermodynamics and Heat Transfer, December 2017, Q5

Question 5: Pipe Insulation Thickness to Prevent Freezing During Shutdown

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. A 60.0 mm ID / 66.0 mm OD plastic pipe, water stagnant inside starting at $15\,{}^{\circ}\text{C}$, wrapped in an unknown thickness of low-conductivity insulation and exposed to the worst-case ambient (coldest air, highest wind-driven $h$) for the full 60-hour shutdown; the internal convective resistance is explicitly neglected.

Given data
Quantity	Symbol	Value
Pipe inside radius	$r_i$	3.0 cm
Pipe outside radius	$r_o$	3.3 cm
Pipe thermal conductivity	$k_{pipe}$	0.16 W/m·K
Insulation thermal conductivity	$k_{ins}$	0.0105 W/m·K
Worst-case outside heat-transfer coefficient	$h$	30 W/m²·K
Worst-case ambient temperature	$T_\infty$	$-10\,{}^{\circ}\text{C}$
Initial water temperature	$T_{w,i}$	$15\,{}^{\circ}\text{C}$
Freezing threshold	$T_{w,f}$	$0\,{}^{\circ}\text{C}$
Shutdown duration	$t$	60 h

Find. The insulation thickness $t_{ins}$ (per unit pipe length) that prevents the water from reaching $0\,{}^{\circ}\text{C}$ within 60 hours under the worst conditions.

Composite radial wall: water core, pipe wall, insulation of unknown thickness, losing heat by convection to worst-case ambient air.

Approach. The true problem is transient (the water cools as it loses heat, which would shrink the driving $\Delta T$ over the 60 hours). A defensible bound — and the natural reading of the exam's own "worst conditions" instruction — is to hold the driving temperature difference at its LARGEST value, $\Delta T=T_{w,i}-T_\infty=25\,{}^{\circ}\text{C}$, constant for the entire 60 hours; since the real $\Delta T$ can only be smaller than this at every later instant, any insulation sized against this bound keeps the water at or above $0\,{}^{\circ}\text{C}$ under the actual (milder) transient history. Set the resulting bounding steady-state heat-loss rate, integrated over 60 hours, equal to the water's own sensible heat capacity between $15\,{}^{\circ}\text{C}$ and $0\,{}^{\circ}\text{C}$, and solve for the insulation outer radius.

Available sensible energy per metre of pipe. Water cross-section area $A_w=\pi r_i^2=\pi(0.030)^2=0.002827\text{ m}^2$; mass per metre $m'=\rho A_w=1000\times0.002827=2.827\text{ kg/m}$. Using $c_p=4.19\text{ kJ/kg}\cdot\text{K}$, $$E'=m'c_p(T_{w,i}-T_{w,f})=2.827\times4.19\times15=177.7\text{ kJ/m}$$
Allowable steady heat-loss rate. Over $t=60\text{ h}=216{,}000\text{ s}$, $$Q'_{allow}=\frac{E'}{t}=\frac{177{,}700}{216{,}000}$$ $$\boxed{Q'_{allow}=0.823\text{ W/m}}$$
Resistance network and root-find for the insulation radius. Per unit length, the pipe-wall resistance is $$R'_{pipe}=\frac{\ln(r_o/r_i)}{2\pi k_{pipe}}=\frac{\ln(3.3/3.0)}{2\pi\times0.16}=0.0948\text{ m}\cdot\text{K/W}$$ and for an insulation outer radius $r_3$, $$R'_{tot}(r_3)=R'_{pipe}+\frac{\ln(r_3/r_o)}{2\pi k_{ins}}+\frac{1}{h\,(2\pi r_3)}$$ Solving $\Delta T/R'_{tot}(r_3)=Q'_{allow}$ numerically (the insulation term dominates because $k_{ins}$ is so low) gives $r_3=0.243\text{ m}$, i.e. $$\boxed{t_{ins}=r_3-r_o=0.243-0.033=0.210\text{ m}\approx21.0\text{ cm}}$$ At this radius $R'_{tot}=30.4\text{ m}\cdot\text{K/W}$, giving $Q'=\Delta T/R'_{tot}=25/30.4=0.823\text{ W/m}$, matching $Q'_{allow}$ exactly by construction.

Check: 21 cm is a large thickness for pipe insulation, a direct consequence of the unusually low quoted insulation conductivity (0.0105 W/m·K, more typical of a high-performance foam than ordinary weather-jacketed fibreglass) combined with the deliberately conservative constant-$\Delta T$ "worst conditions" reading; a genuinely transient (exponential cooling) analysis would allow a somewhat thinner layer.

Question 5 — results
Quantity	Value
Allowable steady heat-loss rate $Q'_{allow}$	0.823 W/m
Pipe-wall resistance $R'_{pipe}$	0.0948 m·K/W
Total resistance at solution $R'_{tot}$	30.4 m·K/W
Insulation outer radius $r_3$	0.243 m
Insulation thickness $t_{ins}$	21.0 cm

Topic. Worst-case, steady-conduction-bounded freeze-protection sizing for an insulated pipe.

Key relations. Cylindrical conduction resistance $R=\ln(r_2/r_1)/(2\pi kL)$; convection resistance $R=1/(hA)$; series composite radial-wall resistance; sensible heat $Q=mc_p\Delta T$.

Why this works. An exact solution needs a transient lumped-capacitance ODE (the water's own $\Delta T$ shrinks as it cools). Substituting the WORST-CASE constant $\Delta T$ — the water's initial temperature held against the coldest ambient for the whole 60 hours — turns this into an ordinary steady-conduction sizing problem while still guaranteeing safety, because the real instantaneous heat-loss rate can only be smaller than this bound at every later time. This is precisely the reading the exam's own "under the worst conditions" phrase is asking for.

Common pitfall. Solving the honest transient decay (which gives a smaller, more optimistic thickness than the exam's own worst-case framing intends) or including the water's own internal convective resistance, which the question explicitly instructs to disregard.

Source. F. P. Incropera and D. P. DeWitt, Fundamentals of Heat and Mass Transfer, 7th ed., Ch. 3 (composite plane/cylindrical walls, critical radius of insulation).

Question 5 of 8: Pipe Insulation Thickness to Prevent Freezing During Shutdown

Question 5: Pipe Insulation Thickness to Prevent Freezing During Shutdown