Chapter 20, Problem 27P

To determine

The thickness of the insulation that needed to be used.

The time taken for the insulation to pay for itself from the energy it saves.

Explanation of Solution

Given:

The height of the side surface of the cubic industrial surface $\left(h\right)$ is $3\text{m}$.

The outer surface temperature of the section $\left(T_s\right)$ is $110°\text{C}$.

The temperature of the furnace room including the surface $\left(T_{\infty }\right)$ is $30°\text{C}$.

The effectiveness of the outer surface of furnace $\left(\epsilon \right)$ is $0.7$.

The thermal conductivity of the glass wool insulation $\left(k_{ins}\right)$ is $0.038\text{W}/\text{m}\cdot \text{K}$.

The effectiveness of reflective sheet $\left({\epsilon }_r\right)$ is $0.2$.

The reduction in heat loss $\left(R\right)$ is $90\%$.

The efficiency of the furnace $\left(\eta \right)$ is $78\%$.

The price of natural gas $\left(P\right)$ is $\$1.10/\text{therms}$.

The continuous operation time of furnace $\left(\Delta t\right)$ is $1\text{year}$.

The cost of the installation of insulation $\left(C_{insu}\right)$ is $\$550$.

Calculation:

Calculate the film temperature $\left(T_f\right)$ using the relation.

    $\begin{array}{c}{T}_f=\frac{T_s+T_{\infty }}{2}\\ =\frac{110°\text{C}+30°\text{C}}{2}\\ =70°\text{C}\end{array}$

Refer Table A-22 “Properties of air at 1 atm pressure”.

Obtain the following values of different properties corresponding film temperature $70°\text{C}$ to the as follows:

$\begin{array}{l}k=0.02881\text{W}/\text{m}\cdot \text{K}\\ \mathrm{Pr}=0.7177\\ v=1.995\times {10}^{-5}{\text{m}}^{\text{2}}/\text{s}\end{array}$

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

    $\begin{array}{c}\beta =\frac{1}{T}\\ =\frac{1}{\left(70°\text{C}+273\right)\text{K}}\\ =\frac{1}{343\text{K}}\\ =0.002915{\text{K}}^{-1}\end{array}$

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

    $\begin{array}{c}{L}_c=h\\ =3\text{m}\end{array}$

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

    $\begin{array}{c}\text{Ra}=\frac{g\beta \left(T_s-T_{\infty }\right){\left(L_c\right)}^3}{v^2}\mathrm{Pr}\\ =\frac{\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(0.002915{\text{K}}^{-1}\right)\left(\begin{array}{l}110°\text{C}\\ -30°\text{C}\end{array}\right){\left(3\text{m}\right)}^3}{{\left(1.995\times {10}^{-5}{\text{m}}^{\text{2}}/\text{s}\right)}^2}\left(0.7177\right)\\ =\frac{\left(0.02859\text{m}/{\text{s}}^2\cdot \text{K}\right)\left[\begin{array}{r}\left(110°\text{C}+273\right)\text{K}\\ -\left(30°\text{C}+273\right)\text{K}\end{array}\right]\left(27{\text{m}}^{\text{3}}\right)}{{\left(1.995\times {10}^{-5}{\text{m}}^{\text{2}}/\text{s}\right)}^2}\left(0.7177\right)\\ =1.114\times {10}^{11}\end{array}$

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

    $\begin{array}{c}\text{Nu}={\left[0.825+\frac{0.389{\left(\text{Ra}\right)}^{1/6}}{{\left[1+{\left(0.492/\mathrm{Pr}\right)}^{9/16}\right]}^{8/27}}\right]}^2\\ ={\left[0.825+\frac{0.389{\left(1.114\times {10}^{11}\right)}^{1/6}}{{\left[1+{\left(0.492/0.7177\right)}^{9/16}\right]}^{8/27}}\right]}^2\\ ={\left[0.825+\frac{26.983}{1.19}\right]}^2\\ =552.24\end{array}$

Calculate the surface area of heat transfer $\left(A\right)$ using the relation.

    $\begin{array}{c}A=4{\left(L\right)}^2\\ =4{\left(3\text{m}\right)}^2\\ =4\left(9{\text{m}}^2\right)\\ =36{\text{m}}^{\text{2}}\end{array}$

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

    $\begin{array}{c}h=\frac{k}{L_c}Nu\\ =\frac{0.02881\text{W}/\text{m}\cdot \text{K}}{3\text{m}}\left(552.24\right)\\ =5.303\text{W}/{\text{m}}^2\cdot \text{K}\end{array}$

Calculate the rate of heat transfer $\left(\dot{Q}\right)$ using the relation.

    $\begin{array}{c}\dot{Q}=hA\left(T_s-T_{\infty }\right)+\epsilon A\sigma \left(T_s^4-T_{surr}^4\right)\\ =\left[\begin{array}{l}\left(5.303\text{W}/{\text{m}}^2\cdot \text{K}\right)\left(36{\text{m}}^2\right)\left(\begin{array}{l}\left(110°\text{C}+273\right)\text{K}\\ -\left(25°\text{C}+273\right)\text{K}\end{array}\right)\\ +\left(0.7\right)\left(36{\text{m}}^{\text{2}}\right)\left(5.67\times {10}^{-8}\text{W}/{\text{m}}^{\text{2}}\cdot {\text{K}}^{\text{4}}\right)\left[\begin{array}{l}{\left(\left(110°\text{C}+273\right)\text{K}\right)}^4\\ -{\left(\left(25°\text{C}+273\right)\text{K}\right)}^4\end{array}\right]\end{array}\right]\\ =\left(190.908\text{W}/\text{K}\right)\left(85\text{K}\right)+\left(1.428\times {10}^{-6}\text{W}/{\text{K}}^{\text{4}}\right)\left(1.363\times {10}^{10}{\text{K}}^4\right)\\ =35690.8\text{W}\end{array}$

Calculate the rate of heat transfer after insulation $\left({\dot{Q}}_{saved}\right)$ using the relation.

    $\begin{array}{c}{\dot{Q}}_{saved}=R\dot{Q}\\ =\left(90\%\right)\left(35690.8\text{W}\right)\\ =\left(0.9\right)\left(35690.8\text{W}\right)\\ =32121.72\text{W}\end{array}$

Calculate the rate of heat loss after insulation $\left({\dot{Q}}_{loss}\right)$ using the relation.

    $\begin{array}{c}{\dot{Q}}_{loss}=\left(1-R\right)\dot{Q}\\ =\left(1-90\%\right)\left(35690.8\text{W}\right)\\ =\left(0.1\right)\left(35690.8\text{W}\right)\\ =3569.08\text{W}\end{array}$

Calculate the energy saved by installing the insulation $\left(E_{saved}\right)$ using the relation.

  $\begin{array}{c}{E}_{saved}={\dot{Q}}_{saved}\left(\Delta t\right)\\ =\left(32121.72\text{W}\times \frac{{10}^{-3}\text{kJ}/\text{s}}{1\text{W}}\right)\left(1\text{year}\times \frac{365\text{day}}{1\text{year}}\right)\\ =\left(32.12172\text{kJ}/\text{s}\right)\left(365\text{day}\times \frac{86400\mathrm{s}}{1\text{days}}\right)\\ =1.013\times {10}^9\text{kJ}\end{array}$

Calculate the total money saved $\left(M_{saved}\right)$ using the relation.

    $\begin{array}{c}{M}_{saved}=\left(E_{saved}\right)\left(P\right)\\ =\left(1.013\times {10}^9\text{kJ}\times \frac{1\text{therms}}{105500\text{kJ}}\right)\left(\$1.10/\text{therms}\right)\\ =\$10562.08\end{array}$

Means the money saved due to insulation in a year is $\$10562.08$.

Calculate the time taken by the insulation to pay for the cost $\left(t_{ins}\right)$ using the relation.

    $\begin{array}{c}{t}_{ins}=\frac{C_{ins}}{M_{saved}}\\ =\frac{\$550}{\$10562.081/\text{year}\times \frac{1\text{year}}{365\text{days}}}\\ =19\text{days}\end{array}$

Thus, the time taken for the insulation to pay for itself from the energy it saves is $19\text{days}$.

Insulation will reduce the temperature of the surface which reduce the Rayleigh number, Nusselt number, hence for the convection coefficient the temperature of the surface is consider as $50°\text{C}$.

Calculate the film temperature $\left(T_f\right)$ using the relation.

    $\begin{array}{c}{T}_f=\frac{T_s+T_{\infty }}{2}\\ =\frac{50°\text{C}+30°\text{C}}{2}\\ =\frac{80°\text{C}}{2}\\ =40°\text{C}\end{array}$

Refer Table A-22 “Properties of air at 1 atm pressure”.

Obtain the following values of different properties corresponding film temperature $40°\text{C}$ to the as follows:

$\begin{array}{l}k=0.02662\text{W}/\text{m}\cdot \text{K}\\ \mathrm{Pr}=0.7255\\ v=1.702\times {10}^{-5}{\text{m}}^{\text{2}}/\text{s}\end{array}$

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

    $\begin{array}{c}\beta =\frac{1}{T}\\ =\frac{1}{\left(40°\text{C}+273\right)\text{K}}\\ =\frac{1}{313\text{K}}\\ =0.003195{\text{K}}^{-1}\end{array}$

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

    $\begin{array}{c}\text{Ra}=\frac{g\beta \left(T_s-T_{\infty }\right){\left(L_c\right)}^3}{v^2}\mathrm{Pr}\\ =\frac{\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(0.003195{\text{K}}^{-1}\right)\left(\begin{array}{l}50°\text{C}\\ -30°\text{C}\end{array}\right){\left(3\text{m}\right)}^3}{{\left(1.702\times {10}^{-5}{\text{m}}^{\text{2}}/\text{s}\right)}^2}\left(0.7255\right)\\ =\frac{\left(0.03134\text{m}/{\text{s}}^2\cdot \text{K}\right)\left[\begin{array}{r}\left(50°\text{C}+273\right)\text{K}\\ -\left(30°\text{C}+273\right)\text{K}\end{array}\right]\left(27{\text{m}}^{\text{3}}\right)}{{\left(1.702\times {10}^{-5}{\text{m}}^{\text{2}}/\text{s}\right)}^2}\left(0.7255\right)\\ =4.238\times {10}^{10}\end{array}$

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

    $\begin{array}{c}\text{Nu}={\left[0.825+\frac{0.389{\left(\text{Ra}\right)}^{1/6}}{{\left[1+{\left(0.492/\mathrm{Pr}\right)}^{9/16}\right]}^{8/27}}\right]}^2\\ ={\left[0.825+\frac{0.389{\left(4.238\times {10}^{10}\right)}^{1/6}}{{\left[1+{\left(0.492/0.7255\right)}^{9/16}\right]}^{8/27}}\right]}^2\\ ={\left[0.825+\frac{22.969}{1.19}\right]}^2\\ =405.08\end{array}$

Calculate the surface area of heat transfer $\left(A\right)$ using the relation.

    $\begin{array}{c}A=4\left(L\right)\left(L+2t\right)\\ =4\left(3\text{m}\right)\left(3\text{m}+2t\right)\\ =12\text{m}\left(3\text{m}+2t\right)\end{array}$

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

    $\begin{array}{c}h=\frac{k}{L_c}Nu\\ =\frac{0.02662\text{W}/\text{m}\cdot \text{K}}{3\text{m}}\left(405.08\right)\\ =3.594\text{W}/{\text{m}}^2\cdot \text{K}\end{array}$

Calculate the rate of heat transfer $\left(\dot{Q}\right)$ using the relation.

    $\begin{array}{c}\dot{Q}=hA\left(T_s-T_{\infty }\right)+\epsilon A\sigma \left(T_s^4-T_{surr}^4\right)\\ 35690.8\text{W}=\left[\begin{array}{l}\left(3.594\text{W}/{\text{m}}^2\cdot \text{K}\right)A\left(\begin{array}{l}{T}_s\\ -\left(30°\text{C}+273\right)\text{K}\end{array}\right)\\ +\left(0.7\right)A\left(5.67\times {10}^{-8}\text{W}/{\text{m}}^{\text{2}}\cdot {\text{K}}^{\text{4}}\right)\left[\begin{array}{l}{T}_s{}^4\\ -{\left(\left(30°\text{C}+273\right)\text{K}\right)}^4\end{array}\right]\end{array}\right]\\ 35690.8\text{W=}\left[\begin{array}{l}\left(3.594\text{W}/{\text{m}}^2\cdot \text{K}\right)A\left(\begin{array}{l}{T}_s\\ -\left(30°\text{C}+273\right)\text{K}\end{array}\right)\\ +\left(3.969\times {10}^{-8}\text{W}/{\text{m}}^2\cdot \text{K}\right)A\left[\begin{array}{l}{T}_s{}^4\\ -{\left(\left(30°\text{C}+273\right)\text{K}\right)}^4\end{array}\right]\end{array}\right]\end{array}$

The heat loss from the insulation by radiation and convection is equal to heat conduction through insulation.

Calculate the rate of heat transfer through insulation $\left({\dot{Q}}_{cond}\right)$ using the relation.

    $\begin{array}{c}{\dot{Q}}_{cond}=k_{ins}A\frac{\left(T_{s\left(\text{furnace}\right)}-T_s\right)}{t}\\ 35690.8\text{W}=\left(0.038\text{W}/\text{m}\cdot \text{K}\right)A\left(\frac{\left(110°\text{C}+273\right)\text{K}-T_s}{t}\right)\end{array}$

By use of trial and error method the surface temperature of insulation obtained is $41.2°\text{C}$ and thickness of the insulation is $2.84\text{cm}$.

Thus, the time taken for the insulation to pay for itself from the energy it saves is $19\text{days}$ and thickness of the insulation is $2.84\text{cm}$.