Question

1) A 2-inch Schedule 160 horizontal steel pipe delivers hexane at 81.2 °C at a velocity of 5.22 m/min. The wall temperature inside the pipe is 78.3 °C. The outside air temperature is -5 °C. For a basis of 40 mm of length: a) Determine the temperature outside the pipe that is exposed to air. b) What is the heat transfer rate of the system? c) What are the values of the heat transfer coefficient for hexane and air? d) What is the overall heat-transfer coefficient of the system based on the outside area​

Answer 1

Answer:

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Answer 2

Given:

$T_{1}=81.2^{\circ} \mathrm{C},T_{4}=-5^{\circ} \mathrm{C},V=5.22 \mathrm{~m} / \mathrm{min},\\T_{2}=78.3^{\circ} \mathrm{C},\mathrm{L}=40 \mathrm{~mm}$

We have to obtain the data for the Hexane.

$\text { Hexane @ } 81.2^{\circ} \mathrm{C} \text { : exists as gas state }\\\mu=7.9 \mu P a \cdot S=7.9 \times 10^{-6} P a \cdot S\\\rho=0.04287 \mathrm{~mol} / \mathrm{dm}^{3}\left(\frac{86.1 \mathrm{~g}}{\mathrm{~mol}}\right)\left(\frac{1 \mathrm{~kg}}{1000 \mathrm{~g}}\right)\left(\frac{10 \mathrm{dm}}{1 \mathrm{~m}}\right)^{3}=3.69511 \mathrm{~kg} / \mathrm{m}^{3}\\C_{p}=164,507.474 \mathrm{~J} / \mathrm{kmol} \cdot \mathrm{K}=164.5075 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}$

$k=18.5109 \mathrm{~mW} / \mathrm{m} \cdot K=0.0185 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}$

$\text { Now we have to obtain necessary data for the pipe }\\\text { From Table 10-22 of Perry (2008) }\\\begin{array}{l}d_{o}=2.375 \mathrm{in}\left(\frac{0.0254 \mathrm{~m}}{i n}\right)=0.0603 \mathrm{~m} \\d_{i}=1.687 \mathrm{in}\left(\frac{0.0254 \mathrm{~m}}{i n}\right)=0.0428 \mathrm{~m}\end{array}\\\\\text { From A.3-16 of Geankoplis et al. (2018) }$

$k_{\text {steel }}=45 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\\\begin{array}{l}r_{o}=\frac{0.0603 \mathrm{~m}}{2}=0.0302 \mathrm{~m} \\r_{i}=\frac{0.0428 \mathrm{~m}}{2}=0.0214 \mathrm{~m}\end{array}\\\text { for the Reynolds number }\\\\R e=\frac{d_{i} V \rho}{\mu}=\frac{(0.0428 \mathrm{~m})(0.087 \mathrm{~m} / \mathrm{s})\left(3.69511 \mathrm{~kg} / \mathrm{m}^{3}\right)}{7.9 \times 10^{-6} \mathrm{~kg} / \mathrm{m} \cdot \mathrm{s}} \approx 1742$

$\text { Since the fluid flow is laminar, we can use equation } 2.17 \text { of module 2 }$

$\text { for the } D / L\\D / L=0.0428 \mathrm{~m} / 0.04 \mathrm{~m}=1.07\\\text { for the Prandtl number using equation } 2.14 \text { of Module } 2 .$

$\begin{array}{l}P r=\frac{\mu C_{p}}{k}=\frac{\left(7.9 \times 10^{-6} \mathrm{~kg} / \mathrm{m} \cdot \mathrm{s}\right)\left(164.5075 \mathrm{~kg} \cdot \mathrm{m}^{2} \cdot \mathrm{s}^{-2} / \mathrm{mol} \cdot \mathrm{K}\right)\left({ }^{1 \mathrm{~mol} / 86.1 \mathrm{~g}}\right)\left({ }^{1000 \mathrm{~g}} / \mathrm{kg}\right)}{0.0185 \mathrm{~kg} \cdot \mathrm{m}^{2} \cdot \mathrm{s}^{-3} / \mathrm{m} \cdot \mathrm{K}} \\P r=0.8159\end{array}$

$\text { the viscosity at the inner wall }\\\mu_{w}=7.8359 \times 10^{-6} \mathrm{~Pa} \cdot \mathrm{s}\\\left.\frac{h_{a}(0.0428 \mathrm{~m})}{\left(0.0185^{W} / \mathrm{m} \cdot \mathrm{K}\right.}\right)=1.86(1521)^{1 / 3}\left(\frac{7.9}{7.8359}\right)^{0.14}\\h_{a}=9.2565^{W} / m^{2} \cdot K\\\text { for the surface area inside the pipe. }\\A_{1}=2 \pi r_{i} L=2 \pi(0.0214 \mathrm{~m})(0.04 \mathrm{~m})=5.37 \times 10^{-3} \mathrm{~m}^{2}$

$\begin{array}{l}q=h_{a} A_{1}\left(T_{1}-T_{2}\right)=\left(9.2565^{W} /_{m^{2} \cdot K}\right)\left(5.37 \times 10^{-3} m^{2}\right)[81.2-78.3] K \\q=\mathbf{0 . 1 4 4 2} W\end{array}\\q=0.1442 W, A_{1}=5.37 \times 10^{-3} \mathrm{~m}^{2}, k_{\text {steel }}=45 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\\\text { the area log mean of the pipe. }$

$\begin{array}{l}A_{2}=2 \pi r_{o} L=2 \pi(0.0302 \mathrm{~m})(0.04 \mathrm{~m})=7.59 \times 10^{-3} \mathrm{~m}^{2} \\A_{l m, p i p e}=\frac{A_{2}-A_{1}}{\ln \left(A_{2} / A_{1}\right)}=6.41 \times 10^{-3} \mathrm{~m}^{2}\end{array}$

For the outside temp. of the pipe:

$\begin{array}{l}q=\frac{k_{\text {steel }}}{A_{\text {lmpipe }}} \frac{T_{2}-T_{3}}{r_{2}-r_{1}} ; 0.1442 \mathrm{~W}=\frac{45 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}}{6.41 \times 10^{-3} \mathrm{~m}^{2}} \frac{(78.3+273.15) \mathrm{K}-T_{3}}{(0.0302-0.0214) \mathrm{m}} \\T_{3}=351.45 \mathrm{~K}=78.3{ }^{\circ} \mathrm{C}\end{array}$

$\begin{array}{l}q=h_{a i r} A_{2}\left(T_{3}-T_{4}\right) \\0.1442 W=h_{a i r}\left(7.59 \times 10^{-3} \mathrm{~m}^{2}\right)[78.3-(-5)] K \\h_{a i r}=0.2281 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\end{array}$

$\begin{array}{l}q=U_{o} A_{o}\left(T_{1}-T_{4}\right) \\0.1442 \mathrm{~W}=U_{o}\left(7.59 \times 10^{-3} \mathrm{~m}^{2}\right)[81.2-(-5)] \mathrm{K} \\U_{o}=0.2204 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\end{array}$

Hence,

$a) \text { Ans: } 78.3^{\circ} \mathrm{C} \text { }\\b)\text { Ans: } 0.1442 \mathrm{~W}\\c) \text { Ans: } \quad h_{\text {hexane }}= \\9.2565 & w / m^{2, K}, h_{\text {air }}=0.2281 w / m^{2}, K\end{array}$