Workshop 5: Steady heat conduction

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1 Insulated steam pipe

Heat loss through an insulated pipe

Saturated steam at 130 ∘C$130\,{}^\circ\mathrm{C}$ is flowing inside a steel pipe having an ID of 2.09 cm and an OD of 2.67 cm. The pipe is insulated with 3.8 cm of insulation on the outside. The convective coefficient for the inside steam surface of the pipe is estimated to be ℎ𝑖$h_i$ = 5680 W/m2.K$\mathrm{W/m^2.K}$, while on the outside of the insulation it is estimated to be ℎ𝑜$h_o$ = 11.4 W/m2.K$\mathrm{W/m^2.K}$. The mean thermal conductivities of the metal and the insulation material are 45 W/m.K$\mathrm{W/m.K}$ and 0.064 W/m.K$\mathrm{W/m.K}$, respectively. Calculate the heat loss for 1 m of pipe using the method of resistances if the surrounding air is at 26 ∘C$26\,{}^\circ\mathrm{C}$.

Worked solution

Heat flows radially from the steam, through the steel and insulation, and then into the air. The same heat rate passes through four resistances.

Figure 1: Cross-section of an insulated pipe. The inner and outer steel radii are r1 and r2; insulation extends to r3. Heat crosses the inner film, steel, insulation and outer film in series.

Data

Quantity	Symbol	Value
Pipe length	𝐿$L$	1.0 m$1.0\ \mathrm m$
Inside diameter	𝐷1$D_1$	0.0209 m$0.0209\ \mathrm m$
Outside steel diameter	𝐷2$D_2$	0.0267 m$0.0267\ \mathrm m$
Radial insulation thickness	𝑡$t$	0.0380 m$0.0380\ \mathrm m$
Steel conductivity	𝑘𝑠$k_s$	45 W m−1 K−1$45\ \mathrm{W\,m^{-1}\,K^{-1}}$
Insulation conductivity	𝑘𝑖$k_i$	0.064 W m−1 K−1$0.064\ \mathrm{W\,m^{-1}\,K^{-1}}$
Steam and air temperatures	𝑇𝑠𝑡𝑒𝑎𝑚,𝑇𝑎𝑖𝑟$T_{steam},T_{air}$	130,26 ∘C$130,26\,{}^\circ\mathrm C$
Inside and outside coefficients	ℎ𝑖,ℎ𝑜$h_i,h_o$	5680,11.4 W m−2 K−1$5680,11.4\ \mathrm{W\,m^{-2}\,K^{-1}}$

We need the heat loss from the specified one-metre length.

Assumptions

• Steady radial heat flow with negligible end losses.
• Constant conductivities and heat transfer coefficients.
• Perfect thermal contact between steel and insulation.
• No separate radiation or fouling resistance.

Calculations

First convert diameters to radii. Insulation thickness is already a radial length, so it is added directly to the outside steel radius:

𝑟1=𝐷12=0.01045 m,𝑟2=𝐷22=0.01335 m$$r_1=\frac{D_1}{2}=0.01045\ \mathrm m,\qquad r_2=\frac{D_2}{2}=0.01335\ \mathrm m$$𝑟3=𝑟2+𝑡=0.01335+0.0380=0.05135 m$$r_3=r_2+t=0.01335+0.0380=0.05135\ \mathrm m$$

Each film uses its own cylindrical area, 2𝜋𝑟𝐿$2\pi rL$. Conduction through a cylindrical layer uses both bounding radii. The four resistances are

Element	Resistance relation	Value, K W−1$\mathrm{K\,W^{-1}}$
Inside steam film	𝑅𝑐𝑖 =1/(ℎ𝑖2𝜋𝑟1𝐿)$R_{ci}=1/(h_i2\pi r_1L)$	0.00268$0.00268$
Steel wall	𝑅𝑠 =ln⁡(𝑟2/𝑟1)/(2𝜋𝑘𝑠𝐿)$R_s=\ln(r_2/r_1)/(2\pi k_sL)$	0.000866$0.000866$
Insulation	𝑅𝑖 =ln⁡(𝑟3/𝑟2)/(2𝜋𝑘𝑖𝐿)$R_i=\ln(r_3/r_2)/(2\pi k_iL)$	3.35$3.35$
Outside air film	𝑅𝑐𝑜 =1/(ℎ𝑜2𝜋𝑟3𝐿)$R_{co}=1/(h_o2\pi r_3L)$	0.272$0.272$

For example, the largest resistance is obtained from

𝑅𝑖=ln⁡(0.051350.01335)2𝜋(0.064)(1.0)≃3.35 K W−1$$R_i=\frac{\ln\left(\frac{0.05135}{0.01335}\right)}{2\pi(0.064)(1.0)} \simeq3.35\ \mathrm{K\,W^{-1}}$$

The elements are in series, so

𝑅𝑡𝑜𝑡𝑎𝑙=𝑅𝑐𝑖+𝑅𝑠+𝑅𝑖+𝑅𝑐𝑜≃3.626 K W−1$$R_{total}=R_{ci}+R_s+R_i+R_{co}\simeq3.626\ \mathrm{K\,W^{-1}}$$

Use the steam-to-air temperature difference across the complete network:

˙𝑄=𝑇𝑠𝑡𝑒𝑎𝑚−𝑇𝑎𝑖𝑟𝑅𝑡𝑜𝑡𝑎𝑙=130−263.626≃28.7 W$$\dot Q=\frac{T_{steam}-T_{air}}{R_{total}} =\frac{130-26}{3.626}\simeq28.7\ \mathrm W$$

The outer insulation temperature provides a useful direction check:

𝑇𝑠,𝑜=𝑇𝑎𝑖𝑟+˙𝑄𝑅𝑐𝑜≃26+28.7(0.272)=33.8 ∘C$$T_{s,o}=T_{air}+\dot QR_{co} \simeq26+28.7(0.272)=33.8\,{}^\circ\mathrm C$$

It lies above the air temperature and below the steam temperature. The insulation supplies about 92 percent of the total resistance. Its large temperature drop is consistent with the small heat loss.

Answer

The heat loss is approximately 28.7 W$28.7\ \mathrm W$ from one metre of pipe, or 28.7 W m−1$28.7\ \mathrm{W\,m^{-1}}$. This is a heat rate per pipe length, not a heat flux per surface area.

2 Electrically heated slab

Insulation surface temperature

In a laboratory experiment, an electric current is passed through a slab of thickness 2𝛿$2\delta$. As a result, heat generation 𝑆𝑒$S_e$ takes place in the slab. The slab is insulated on both sides (thickness = t) to satisfy the safety considerations related to the danger stemming from the fact that an electric current is passing. In addition, the insulated slab is cooled on both sides by a fluid of temperature 𝑇𝑐$T_c$, with convective coefficient of h. Determine the temperature at the surface of the insulation, assuming that the thermal conductivities for the slab and the insulation material are constants, denoted by 𝑘𝑠$k_s$ and 𝑘𝑖$k_i$, respectively.

Worked solution

Choose 𝑥 =0$x=0$ at the slab centre and 𝑥 =𝛿$x=\delta$ at its right face.

Symmetry gives 𝑑𝑇/𝑑𝑥 =0$dT/dx=0$ at the centre.

The slab energy balance is

𝑘𝑠𝑑2𝑇𝑑𝑥2+𝑆𝑒=0.$$k_s\frac{d^2T}{dx^2}+S_e=0.$$

Integrating with the symmetry condition gives

𝑑𝑇𝑑𝑥=−𝑆𝑒𝑥𝑘𝑠.$$\frac{dT}{dx}=-\frac{S_ex}{k_s}.$$

The heat flux leaving either slab face, in W m−2$\mathrm{W\,m^{-2}}$, is therefore

𝑞′′=−𝑘𝑠[𝑑𝑇𝑑𝑥]𝛿=𝑆𝑒𝛿.$$q''=-k_s\left[\frac{dT}{dx}\right]_{\delta}=S_e\delta.$$

Each face removes half the generation per unit area of the whole slab.

The insulation carries this constant flux, so

𝑆𝑒𝛿=ℎ(𝑇𝑜−𝑇𝑐)=𝑘𝑖𝑡(𝑇𝑖−𝑇𝑜).$$S_e\delta=h(T_o-T_c) =\frac{k_i}{t}(T_i-T_o).$$

Hence the outer surface and interface temperatures are

𝑇𝑜=𝑇𝑐+𝑆𝑒𝛿ℎ,$$T_o=T_c+\frac{S_e\delta}{h},$$𝑇𝑖=𝑇𝑐+𝑆𝑒𝛿ℎ+𝑆𝑒𝛿𝑡𝑘𝑖.$$T_i=T_c+\frac{S_e\delta}{h} +\frac{S_e\delta t}{k_i}.$$

Both expressions return kelvin.

The overall generation fixes the surface heat flux, so 𝑘𝑠$k_s$ affects the temperature gradient inside the slab rather than 𝑇𝑜$T_o$.

For example the centre temperature is

𝑇𝑖+𝑆𝑒𝛿22𝑘𝑠,$$T_i+\frac{S_e\delta^2}{2k_s},$$

also in kelvin.

3 Pipe location in a hot wall

Temperature-dependent conductivity

A pipe designed to transport liquid metal is to be embedded in a 1.2 m thick wall. The inside surface of the wall has a constant surface temperature of 925 K, whereas its external surface is exposed to air with a convective heat transfer coefficient of 23 W/m2.K$\mathrm{W/m^2.K}$ and 300 K. The thermal conductivity of the wall varies with temperature according to 𝑘 =0.073 +3.942 ×10−4𝑇$k = 0.073 + 3.942\times10^{-4}T$, where T is temperature in Kelvin, and k is in W/m.K$\mathrm{W/m.K}$. It is desired that the pipe is located at a point in the wall where the temperature is < 650 K. Calculate how far from the hot surface should the pipe be located and the heat flux for the wall.

Worked solution

Data

Quantity	Symbol	Value
Wall thickness	𝐿$L$	1.2 m$1.2\ \mathrm m$
Hot-face temperature	𝑇ℎ$T_h$	925 K$925\ \mathrm K$
Outside air	𝑇∞$T_\infty$	300 K$300\ \mathrm K$
Outside coefficient	ℎ$h$	23 W m−2 K−1$23\ \mathrm{W\,m^{-2}\,K^{-1}}$
Conductivity	𝑘(𝑇)$k(T)$	0.073 +3.942 ×10−4𝑇 W m−1 K−1$0.073+3.942\times10^{-4}T\ \mathrm{W\,m^{-1}\,K^{-1}}$
Pipe temperature limit	𝑇𝑙𝑖𝑚𝑖𝑡$T_{limit}$	650 K$650\ \mathrm K$

We need the outward heat flux and the depth at which the wall reaches the pipe-temperature limit.

Assumptions

• Steady one-dimensional conduction with no heat generation.
• Negligible disturbance of the wall temperature field by the embedded pipe.
• The given conductivity relation applies throughout the temperature range.

Calculations

Write 𝑘(𝑇) =𝑎 +𝑏𝑇$k(T)=a+bT$, where 𝑎 =0.073 W m−1 K−1$a=0.073\,\mathrm{W\,m^{-1}\,K^{-1}}$ and 𝑏 =3.942 ×10−4 W m−1 K−2$b=3.942\times10^{-4}\,\mathrm{W\,m^{-1}\,K^{-2}}$.

Let 𝑥$x$ increase from the hot face towards the air, 𝐿 =1.2 m$L=1.2\,\mathrm{m}$, and 𝑇𝑜 =𝑇(𝐿)$T_o=T(L)$.

With no generation the positive heat flux is constant:

𝑞′′=−𝑘(𝑇)𝑑𝑇𝑑𝑥.$$q''=-k(T)\frac{dT}{dx}.$$

Integrating across the wall and applying convection gives

𝑞′′𝐿=𝑎(925−𝑇𝑜)+𝑏2(9252−𝑇2𝑜),𝑞′′=23(𝑇𝑜−300).$$\begin{aligned} q''L&=a(925-T_o) +\frac b2(925^2-T_o^2),\\ q''&=23(T_o-300). \end{aligned}$$(1)

Substituting the second relation into the first produces

1.971×10−4𝑇2𝑜+27.673𝑇𝑜−8516.17=0.$$1.971\times10^{-4}T_o^2+27.673T_o -8516.17=0.$$

The physical root is 𝑇𝑜 ≈307.07 K$T_o\approx307.07\,\mathrm{K}$.

The other root is negative and lies outside the specified temperature interval.

From Eq. 1, the outward heat flux is

𝑞′′=23(𝑇𝑜−300)≈162.6 W m−2.$$q''=23(T_o-300) \approx162.6\,\mathrm{W\,m^{-2}}.$$

Integrating from 925 K$925\,\mathrm{K}$ to 650 K$650\,\mathrm{K}$ gives

𝑞′′𝑥650=𝑎(925−650)+𝑏2(9252−6502),$$q''x_{650}=a(925-650) +\frac b2(925^2-650^2),$$𝑥650≈0.648 m.$$x_{650}\approx0.648\,\mathrm{m}.$$

Since 𝑘(𝑇) >0$k(T)>0$ throughout the wall, the temperature decreases with 𝑥$x$.

The pipe must lie farther than 0.648 m$0.648\,\mathrm{m}$ from the hot face towards the cold face.

Equality gives 650 K$650\,\mathrm{K}$ and does not meet the strict limit.

A finite pipe must have its entire occupied region on the cold side if every part must remain below the limit.

Answer

The outward heat flux is 163 W m−2$163\ \mathrm{W\,m^{-2}}$. The pipe must be placed more than 0.648 m$0.648\ \mathrm m$ from the hot face towards the cold face to remain below 650 K$650\ \mathrm K$.