Insulation thickness to stop a pipe freezing: 22-Mec-A1 Applied Thermodynamics and Heat Transfer, December 2017, Q5

Question 5 — Insulation thickness to stop a pipe freezing (Part B, equal value)

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. Stationary water (initially 15 °C) in a plastic pipe must stay above 0 °C for 60 h with ambient at −10 °C and $h=30\ \text{W/m}^2$·°C. Pipe: $r_1=0.03$ m, $r_2=0.033$ m, $k_p=0.16$; insulation $k_i=0.0105\ \text{W/m}$·°C. Find. the insulation thickness $t=r_3-r_2$ that just prevents freezing (per unit length).

Quantity	Value
Pipe inner / outer radius	0.030 / 0.033 m
Pipe / insulation conductivity	0.16 / 0.0105 W/m·°C
Outside convection coefficient, $h$	30 W/m²·°C
Ambient / initial / freeze temperatures	−10 / 15 / 0 °C
Closed period, $t$	60 h = 216,000 s

Figure 5 — Cross-section: pipe wall (thin) plus weather-jacketed insulation of outer radius $r_3$; the required insulation makes the total conduction+convection resistance large enough that the water cannot cool to 0 °C within 60 h.

Approach. Treat the water as a lumped mass (no internal resistance) losing heat to the ambient through the series pipe-wall, insulation and outside-convection resistances; require that its lumped-capacitance temperature not fall to 0 °C in 60 h, which sets a minimum total resistance and hence the insulation radius.

Thermal capacity of the water per unit length. With $\rho=1000\ \text{kg/m}^3$, $c_p=4186\ \text{J/kg}\cdot\text{K}$: $$(\rho c_p)' = \rho c_p \pi r_1^2 = (1000)(4186)\pi(0.03)^2 = 1.184\times10^{4}\ \text{J/(m}\cdot\text{K)}$$
Lumped-capacitance cooling requirement. The water temperature decays as $T(t)=T_\infty+(T_0-T_\infty)e^{-t/(\,(\rho c_p)'R'\,)}$. Setting $T=0$ °C at $t=60$ h and solving for the required resistance: $$R' = \frac{t}{(\rho c_p)'\ln\!\frac{T_0-T_\infty}{0-T_\infty}} = \frac{216{,}000}{(1.184\times10^{4})\ln\!\frac{25}{10}} = 19.92\ \text{m}\cdot\text{°C/W}$$ Required total resistance $R' = 19.92$ m·°C/W
Resistance network (per unit length). The pipe wall and outside film contribute $$R'_p = \frac{\ln(r_2/r_1)}{2\pi k_p} = \frac{\ln(0.033/0.030)}{2\pi(0.16)} = 0.0948,\qquad R'_\text{conv} = \frac{1}{h\,2\pi r_3}$$ so the insulation must supply $R'_i = R' - R'_p - R'_\text{conv}$ with $R'_i = \dfrac{\ln(r_3/r_2)}{2\pi k_i}$.
Solve for the insulation radius. Iterating the implicit equation $$\frac{\ln(r_3/0.033)}{2\pi(0.0105)} + \frac{1}{(30)2\pi r_3} + 0.0948 = 19.92$$ gives $r_3 = 0.1217$ m (outside-film term $\approx 0.0436$ m·°C/W). The insulation thickness is $$t = r_3 - r_2 = 0.1217 - 0.033 = 0.0887\ \text{m}$$ $t \approx 8.9$ cm of insulation

Check — modelling and properties.

The water is modelled as a single well-mixed lump (the problem explicitly disregards internal water resistance), cooling exponentially toward −10 °C. Using $c_p=4186$ J/kg·K and $\rho=1000\ \text{kg/m}^3$; taking $c_p=4200$ shifts the required thickness by under 1 mm. Only sensible cooling (15 → 0 °C) is counted — latent heat is not, since we require the water not to reach the freezing point.

Quantity	Result
Water heat capacity per length	1.184 × 10⁴ J/(m·K)
Required total resistance	19.92 m·°C/W
Outer insulation radius $r_3$	0.122 m
Insulation thickness $t$	≈ 8.9 cm

Concept & Theory

Topic. Transient lumped-capacitance cooling of a stationary fluid through a composite cylindrical wall — an insulation-sizing (freeze-protection) problem.

Key relations. Lumped decay $\frac{T-T_\infty}{T_0-T_\infty}=e^{-t/\tau}$ with $\tau=(\rho c_p)'R'$; cylindrical conduction $R'=\ln(r_o/r_i)/(2\pi k)$; convection $R'=1/(h\,2\pi r_o)$; resistances in series.

Why this works. Because the water is stagnant and the problem waives its internal resistance, the whole slug sits at one (falling) temperature, so a first-order lumped model applies. Requiring the time constant to be long enough that the exponential has not driven the temperature to 0 °C after 60 h converts directly into a minimum series resistance, and the insulation layer is the only adjustable resistance. The small pipe radius means the outside film and pipe wall are almost negligible next to the thick low-$k$ insulation.

Common pitfall. Treating the loss as steady-state at the initial 25 °C driving difference (over-conservative), or including latent heat of fusion (the water must not even begin to freeze). Also forgetting that the convective resistance changes as $r_3$ grows.

Source. Çengel & Ghajar, Heat and Mass Transfer §4 (transient/lumped systems) and §3 (steady conduction in cylindrical layers; critical radius of insulation).

Question 5 of 8: Insulation thickness to stop a pipe freezing

Question 5 — Insulation thickness to stop a pipe freezing (Part B, equal value)