Chapter 19, Problem 121RQ

To determine

The rate of heat loss from the steam per unit length of the pipe when air is flowing across the pipe.

Explanation of Solution

Given:

The inlet temperature of steam $\left(T_{1\infty }\right)$ is $250°\text{C}$.

The thermal conductivity of the stainless steel pipe $\left(k\right)$ is $15\text{W}/\text{m}\cdot \text{K}$.

The inner diameter of the pipe $\left(d_i\right)$ is $4\text{cm}$.

The outer diameter of the pipe $\left(d_o\right)$ is $4.6\text{cm}$.

The thickness of the insulation $\left(t\right)$ is $3.5\text{cm}$.

The thermal conductivity of the insulation $\left(k_1\right)$ is $0.038\text{W}/\text{m}\cdot \text{K}$.

The emissivity of outer surface is $\left(\epsilon \right)$ is $0.3$.

The heat transfer coefficient inside the pipe $\left(h\right)$ is $80\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}$.

The flow rate of air $\left(V\right)$ is $4\text{m}/\text{s}$.

Calculation:

Calculate the outer diameter of the insulated pipe $\left(d_{io}\right)$ using the relation.

  $\begin{array}{c}{d}_{io}=d_o+2t\\ =4.6\text{cm}+2\left(3.5\text{cm}\right)\\ =11.6\text{cm}\end{array}$

Calculate the Reynolds number $\left(\mathrm{Re}\right)$ by using the relation.

  $\begin{array}{c}\mathrm{Re}=\frac{V{d}_{io}}{\nu }\\ =\frac{\left(4\text{m}/\text{s}\right)\left(11.6\text{cm}\right)}{\left(1.426\times {10}^{-5}{\text{m}}^{\text{2}}/\text{s}\right)}\\ =\frac{\left(4\text{m}/\text{s}\right)\left(11.6\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)}{\left(1.426\times {10}^{-5}{\text{m}}^{\text{2}}/\text{s}\right)}\\ =3253856942\end{array}$

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

Obtain the following properties corresponding to $10°\text{C}$ as follows:

$k_2=0.02439\text{W}/\text{m}\cdot \text{K}$

$\nu =1.426\times {10}^{-5}{\text{m}}^{\text{2}}/\text{s}$

$\mathrm{Pr}=0.7336$

Calculate the Nusselt number for flow across a cylinder $\left(Nu\right)$ using the relation.

  $\begin{array}{c}Nu=0.3+\frac{0.62\sqrt{\mathrm{Re}}{\mathrm{Pr}}^{1/3}}{{\left[1+{\left(\frac{0.4}{\mathrm{Pr}}\right)}^{2/3}\right]}^{1/4}}{\left[1+{\left(\frac{\mathrm{Re}}{282000}\right)}^{5/8}\right]}^{4/5}\\ =0.3+\frac{0.62\sqrt{\left(3253856942\right)}{\left(0.7336\right)}^{1/3}}{{\left[1+{\left(\frac{0.4}{0.7336}\right)}^{2/3}\right]}^{1/4}}{\left[1+{\left(\frac{3253856942}{282000}\right)}^{5/8}\right]}^{4/5}\\ =107\end{array}$

Calculate the heat transfer coefficient through the air $\left(h_o\right)$ using the relation.

  $\begin{array}{c}Nu=\frac{h_o{d}_{io}}{k_2}\\ \left(107\right)=\frac{h_o\left(11.6\text{cm}\right)}{\left(0.02439\text{W}/\text{m}\cdot \text{K}\right)}\\ h_o=\frac{\left(107\right)\left(0.02439\text{W}/\text{m}\cdot \text{K}\right)}{\left(11.6\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)}\\ h_o=22.5\text{W}/{\text{m}}^2\cdot \text{K}\end{array}$

Calculate the area of the outer surface of the pipe $\left(A_o\right)$ using the relation.

  $\begin{array}{c}{A}_o=\pi d_{io}L\\ =\pi \left(11.6\text{cm}\right)\left(1\text{m}\right)\\ =\pi \left(11.6\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)\left(1\text{m}\right)\\ =0.364{\text{m}}^{\text{2}}\end{array}$

Calculate the area of the inner surface of the pipe $\left(A_i\right)$ using the relation.

  $\begin{array}{c}{A}_i=\pi d_{i}L\\ =\pi \left(4\text{cm}\right)\left(1\text{m}\right)\\ =\pi \left(4\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)\left(1\text{m}\right)\\ =0.125{\text{m}}^{\text{2}}\end{array}$

Calculate the convection thermal resistence $\left(R_{\text{conv}}\right)$ using the relation.

    $\begin{array}{c}\left(R_{\text{conv}}\right)=\frac{1}{h{A}_i}\\ =\frac{1}{\left(80\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}\right)\left(0.125{\text{m}}^{\text{2}}\right)}\\ =0.0995\text{K}/\text{W}\end{array}$

Calculate the pipe thermal resistence $\left(R_{\text{pipe}}\right)$ using the relation.

    $\begin{array}{c}{R}_{\text{pipe}}=\frac{\ln \left(d_0/d\right)}{2\pi kL}\\ =\frac{\ln \left(\frac{4.6\text{cm}}{4\text{cm}}\right)}{2\pi \left(15\text{W}/\text{m}\cdot \text{K}\right)\left(1\text{m}\right)}\\ =0.0015\text{K}/\text{W}\end{array}$

Calculate the insulation thermal resistence $\left(R_{\text{insulation}}\right)$ using the relation.

    $\begin{array}{c}{R}_{\text{insulation}}=\frac{\ln \left(d_{i0}/d_0\right)}{2\pi kL}\\ =\frac{\ln \left(\frac{11.6\text{cm}}{4.6\text{cm}}\right)}{2\pi \left(0.038\text{W}/\text{m}\cdot \text{K}\right)\left(1\text{m}\right)}\\ =3.874\text{K}/\text{W}\end{array}$

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

  $\begin{array}{c}Q=\frac{T-T_s}{R_{\text{conv}}+R_{\text{pipe}}+R_{\text{insulation}}}\\ =\frac{250°\text{C}-T_s}{\left(0.0995\text{K}/\text{W}\right)+\left(0.0015\text{K}/\text{W}\right)+\left(3.874\text{K}/\text{W}\right)}\\ =\frac{\left[\left(250°\text{C}+273\right)\text{K}\right]-T_s}{3.975\text{K}/\text{W}}\\ =\frac{523\text{K}-T_s}{3.975\text{K}/\text{W}}\end{array}$        (I)

Calculate the surface temperature $\left(T{s}_{}\right)$ by using the relation.

  $\begin{array}{c}Q=h_0{A}_0\left(T_s-T_{suur}\right)+\epsilon A_0\sigma \left(T_s{}^4-T_{suur}^4\right)\\ \frac{523\text{K}-T_s}{3.975\text{K}/\text{W}}=\left[\begin{array}{l}\left(22.5\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}\right)\left(0.364{\text{m}}^{\text{2}}\right)\left(T_s-3\text{°C}\right)+\\ \left(0.3\right)\left(0.364{\text{m}}^{\text{2}}\right)\left(5.67\times {10}^{-8}\text{W}/{\text{m}}^{\text{2}}{\text{×K}}^{\text{4}}\right)\left[T_s{}^4-{\left(\text{3°C}\right)}^4\right]\end{array}\right]\\ \frac{523\text{K}-T_s}{3.975\text{K}/\text{W}}=\left[\begin{array}{l}\left(22.5\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}\right)\left(0.364{\text{m}}^{\text{2}}\right)\left\{\left(T_s-3\text{°C}+273\right)\text{K}\right\}+\\ \left(0.3\right)\left(0.364{\text{m}}^{\text{2}}\right)\left(5.67\times {10}^{-8}\text{W}/{\text{m}}^{\text{2}}\cdot {\text{K}}^{\text{4}}\right)\left[T_s{}^4-{\left\{\left(\text{3°C}+273\right)\text{K}\right\}}^4\right]\end{array}\right]\\ \frac{523\text{K}-T_s}{3.975\text{K}/\text{W}}=\left(8.19\text{W}/\text{K}\right)\left(T_s-278\text{K}\right)+0.619\text{W}/{\text{K}}^4\left(T_s{}^4-5802782976{\text{K}}^{\text{4}}\right)\\ T_s=276\text{K}\end{array}$

Calculate the heat transfer by using equation (I).

  $\begin{array}{c}Q=\frac{523\text{K}-\left(276\text{K}\right)}{3.975\text{K}/\text{W}}\\ =\frac{247\text{K}}{3.975\text{K}/\text{W}}\\ =62.14\text{W}\end{array}$

Thus, the rate of heat loss from the steam per unit length of the pipe when air is flowing across the pipe is $62.14\text{W}$.