Chapter 20, Problem 48P

To determine

The thickness of the insulation.

The money saved by the insulation during the time period.

Explanation of Solution

Given:

The temperature $\left(T_s\right)$ of the $73°\text{C}$.

The length $\left(L\right)$ of the pipe is $100\text{m}$.

The diameter $\left(D\right)$ of the pipe is $30\text{cm}$.

The ambient air temperature $\left(T_a\right)$ is $0°\text{C}$.

The pipe temperature $\left(T_s\right)$ is $25°\text{C}$.

The effective sky temperature $\left(T\right)$ is $0°\text{C}$.

The time period $\left(t\right)$ is $15\text{h}$.

The emissivity $\left(\epsilon \right)$ of the pipe is $0.1$.

The thermal conductivity $\left(k\right)$ of the insulation is $0.035\text{W}/\text{m}\cdot \text{K}$.

Calculation:

Refer table $\text{A-22}$ “Properties of air at $1\text{atm}$ pressure”.

Obtain the following properties of air corresponding to the temperature of $5°\text{C}$.

$\begin{array}{c}k=0.024\text{W}/\text{m}\cdot \text{K}\\ v=1.382\times {10}^{-5}{\text{m}}^2/\text{s}\\ \mathrm{Pr}=0.735\end{array}$

Calculate the volume expansion coefficient $\left(\beta \right)$ using the relation.

    $\begin{array}{c}\beta =\frac{1}{T_m}\\ =\left[\frac{1}{\left(5°\text{C}+\text{273}\right)\text{K}}\right]\\ =0.003597{\text{K}}^{-1}\end{array}$

Calculate the heat loss $\left(\dot{Q}\right)$ from the pipe using the relation.

    $\begin{array}{c}{\dot{Q}}_{nat}=h{A}_s(T_s-T_a)\\ =4.366\text{W}/{\text{m}}^2\cdot \text{K}\times \pi \left(30\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)\left(100\text{m}\right)\left(\begin{array}{l}\left(25°\text{C}+273\right)\text{K}-\\ \left(0°\text{C}+273\right)\text{K}\end{array}\right)\\ =10287\text{W}\end{array}$

Calculate the heat loss $\left(\dot{Q}\right)$ by radiation from the pipe using the relation.

  $\begin{array}{c}{\dot{Q}}_{rad}=\epsilon A_s\sigma (T_s{}^4-T_a{}^4)\\ =\left[\begin{array}{l}\left(0.8\right)\left(\pi \left(30\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)\left(100\text{m}\right)\right)\times \\ \left(5.67\times {10}^{-8}\text{W}/{\text{m}}^2\cdot {\text{K}}^4\right)\end{array}\right]\left[\begin{array}{l}{\left(25°\text{C}+273\right)}^4{\text{K}}^4-\\ {\left(-30°\text{C}+273\right)}^4{\text{K}}^4\end{array}\right]\\ =18808\text{W}\end{array}$

Calculate the total amount of heat loss using the relation.

    $\begin{array}{c}{\dot{Q}}_{total}={\dot{Q}}_{natural}+{\dot{Q}}_{radiation}\\ =10287\text{W}+18808\text{W}\\ =29094\text{W}\end{array}$

Calculate the reduction in heat loss due to insulation.

    $\begin{array}{c}\dot{Q}=\left(1-0.85\right){\dot{Q}}_{total}\\ =0.15\times 29094\\ =4364\text{W}\end{array}$

Calculate the energy $\left(Q\right)$ saved using the relation.

    $\begin{array}{c}{Q}_{savedtotal}={\dot{Q}}_{saved}\Delta t\\ =\left(0.85\times 29094\text{W}\times \frac{1\text{kW}}{1000\text{W}}\right)\left(10\text{h}\right)\\ =247.3\text{kWh}\end{array}$

Calculate the money saved using the relation.

    $\begin{array}{c}\text{Money}\text{saved}=\left(\text{energy}\text{saved}\right)\left(\text{unit}\text{of}\text{cost}\text{energy}\right)\\ =\left(243.7\text{kWh}\right)\left(\$0.09/\text{kWh}\right)\\ =\$22.3\end{array}$

Calculate the characteristics length $\left(L_c\right)$ using the relation.

    $\begin{array}{c}{L}_c=D+2t_{insulation}\\ =0.3+2t_{insulation}\end{array}$

Calculate the Rayleigh number $\left({\text{Ra}}_D\right)$ using the relation.

  $\begin{array}{c}\text{Ra}=\left(\frac{g\beta \left(T_s-T_a\right)L^3}{v^2}\mathrm{Pr}\right)\\ =\left[\frac{\begin{array}{l}9.81\text{m}/{\text{s}}^2\left(0.003597{\text{K}}^{-1}\right)\times \\ \left(T_s-273\text{K)}\right){\left(30\text{cm}\times \frac{1\text{m}}{100\text{cm}}+2t_{insulation}\right)}^3\end{array}}{{\left(1.382\times {10}^{-5}{\text{m}}^2/\text{s}\right)}^2}\left(0.735\right)\right]\end{array}$        (I)

Calculate the Nusselt number $\left(\text{Nu}\right)$ using the relation.

    $\begin{array}{c}\text{Nu}={\left[0.6+\frac{0.387{\text{Ra}}^{1/6}}{{\left[1+{\left(\frac{0.559}{\mathrm{Pr}}\right)}^{9/16}\right]}^{8/27}}\right]}^2\\ ={\left[0.6+\frac{0.387R{a}^{1/6}}{{\left[1+{\left(\frac{0.559}{0.735}\right)}^{9/16}\right]}^{8/27}}\right]}^2\end{array}$        (II)

Calculate the heat transfer coefficient $\left(h\right)$ using the relation.

  $\begin{array}{c}h=\frac{k}{L_c}\text{Nu}\\ \text{=}\frac{0.024\text{W}/\text{m}\cdot \text{K}}{\left(30\text{cm}\times \frac{1\text{m}}{100\text{cm}}+2t_{insulation}\right)}\left(\text{Nu}\right)\end{array}$        (III)

Calculate the area $\left(A_s\right)$ using the relation.

    $\begin{array}{c}{A}_s=\pi DL\\ =\pi \left(30\text{cm}\times \frac{1\text{m}}{100\text{cm}}+2t_{insulation}\right)(100\text{m)}\end{array}$        (IV)

Calculate the heat loss $\left(\dot{Q}\right)$ from the insulated pipe by convection using the relation.

    $\begin{array}{c}Q={\dot{Q}}_{natural}+{\dot{Q}}_{radiation}\\ =h{A}_s\left(T_s-T_a\right)+\epsilon A_s\sigma (T_s{}^4-T_a{}^4)\\ 4364\text{W}=\left[\begin{array}{l}h{A}_s\left(T_s-273\right)\text{K}+\left(0.1\right)A_s\times \\ \left(5.67\times {10}^{-8}\text{W}/{\text{m}}^2\cdot {\text{K}}^4\right)\end{array}\right]\left[{\left(T_s\right)}^4-{\left(\left(-30°\text{C}+273\right)\text{K}\right)}^4\right]\end{array}$        (V)

Calculate the heat loss $\left(\dot{Q}\right)$ by radiation from the pipe using the relation.

  $\begin{array}{c}\dot{Q}=Q_{insulation}\\ =\frac{2\pi kL\left(T_{\text{tank}}-T_s\right)}{\text{ln}\left(L_c/D\right)}\\ 4364\text{W}=\frac{2\pi \left(0.035\text{W}/\text{m}\cdot \text{K}\right)\left(100\text{m}\right)\left(\left(20+273\right)\text{K}-T_s\right)}{\text{ln}\frac{\left(30\text{cm}\times \frac{1\text{m}}{100\text{cm}}+2t_{insulation}\right)}{\left(30\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}}\end{array}$        (VI)

Solve all the above Equations to obtain the values.

    $\begin{array}{c}{T}_s=281.5\text{K}\\ t_{insulation}=0.013\text{m}\end{array}$

Thus, thickness of the insulation in pipe is $0.013\text{m}$ and the total amount of money saved by the insulation is $\$22.3$.