Chapter 10, Problem 10.1P

In a heat exchanger, as shown in the accompanying figure, air flows over brass tubes of 1.8-cm 1D and 2.1-cm OD containing steam. The convection heat transfer coefficients on the air and steam sides of the tubes are ${\text{70 W/m}}^{\text{2}}{\text{K and 210 W/m}}^{\text{2}}\text{K}$, respectively. Calculate the overall heal transfer coefficient for the heal exchanger (a) based on the inner tube area and (b) based on the outer tube area.

To determine

The overall heat transfer coefficient based on the inner tube area.

Answer to Problem 10.1P

Over all heat transfer co-efficient based on inner surface area is $58.75\text{ W}/{\text{m}}^2\text{K}$

Explanation of Solution

Given information:

Inner diameter of the tube $\text{Di}=1.8\text{ cm}=0.018\text{ m}$

Outer diameter of the tube ${\text{D}}_{\text{o}}=2.1\text{ cm}=0.021\text{m}$

Air flows outside the tubes and steam flows inside the tubes.

Convective heat-transfer coefficients onair side and steam side are ${\text{h}}_{\text{o}}=70\text{ W}/{\text{m}}^2\text{K}$ and

${\text{h}}_{\text{i}}=210\text{ W}/{\text{m}}^2\text{K}$

From Appendix 2 (Table 10),

Thermal conductivity of brass $\text{k}=111\text{ W}/\text{mK}$

Part a:

Over all hear transfer co-efficient based on inner surface area.

Over all heat transfer co-efficient based on inner surface area is given by (refer to Equation 10.4 in text book)

${\text{U}}_{\text{i}}=\frac{1}{\frac{1}{{\text{h}}_{\text{i}}}+\left[\frac{{\text{A}}_{\text{i}}\ln \left({\text{R}}_{\text{o}}/{\text{R}}_{\text{i}}\right)}{2\text{πkL}}\right]}+\frac{{\text{A}}_{\text{i}}}{{\text{A}}_{\text{o}}{\text{h}}_{\text{o}}}$

${\text{A}}_{\text{i}}=$ Inside surface area of the tube $={\text{πD}}_{\text{i}}\text{L}$

${\text{A}}_{\text{o}}=$ Outside surface area of the tube $={\text{πD}}_{\text{o}}\text{L}$

${\text{U}}_{\text{i}}=\frac{1}{\frac{1}{{\text{h}}_{\text{i}}}+\left[\frac{{\text{πD}}_{\text{i}}\text{L}\ln \left({\text{R}}_{\text{o}}/{\text{R}}_{\text{i}}\right)}{2\text{πkL}}\right]+\frac{{\text{πD}}_{\text{i}}\text{L}}{{\text{πD}}_{\text{o}}{\text{Lh}}_{\text{o}}}}$

${\text{U}}_{\text{i}}=\frac{1}{\frac{1}{{\text{h}}_{\text{i}}}+\left[\frac{{\text{D}}_{\text{i}}\ln \left({\text{D}}_{\text{o}}/{\text{D}}_{\text{i}}\right)}{2\text{k}}\right]+\frac{{\text{D}}_{\text{i}}}{{\text{D}}_{\text{o}}{\text{h}}_{\text{o}}}}$

${\text{U}}_{\text{i}}=\frac{1}{\left[\frac{1}{210}\right]+\left[\frac{0.018\ln \left(\frac{0.021}{0.018}\right)}{2\times 111}\right]+\left[\frac{0.018}{0.021\times 70}\right]}$

${\text{U}}_{\text{i}}=58.75\text{ W}/{\text{m}}^2\text{K}$

To determine

Overall heat transfer coefficient based on outer surface area of the tube

Answer to Problem 10.1P

Overall heat transfer coefficient based on outer surface area of the tube is $50.36\text{ W}/{\text{m}}^2\text{K}$

Explanation of Solution

Overall heat transfer coefficient based on outer surface area of the tube is given by (refer to equation 10.3 in text book)

${\text{U}}_{\text{o}}=\frac{1}{\left[\frac{{\text{A}}_{\text{o}}}{{\text{A}}_{\text{i}}{\text{h}}_{\text{i}}}\right]+\left[\frac{{\text{A}}_{\text{o}}*\ln \left({\text{R}}_{\text{o}}/{\text{R}}_{\text{i}}\right)}{2\text{πkL}}\right]+\left[\frac{1}{{\text{h}}_{\text{o}}}\right]}$

${\text{U}}_{\text{o}}=\frac{1}{\left[\frac{{\text{πD}}_{\text{o}}\text{L}}{{\text{πD}}_{\text{i}}{\text{Lh}}_{\text{i}}}\right]+\left[\frac{{\text{πD}}_{\text{o}}\text{L*ln}\left({\text{D}}_{\text{o}}/{\text{D}}_{\text{i}}\right)}{2\text{πkL}}\right]+\left[\frac{1}{{\text{h}}_{\text{o}}}\right]}$

${\text{U}}_{\text{o}}=\frac{1}{\left[\frac{{\text{D}}_{\text{o}}}{{\text{D}}_{\text{i}}{\text{h}}_{\text{i}}}\right]+\left[\frac{{\text{D}}_{\text{o}}\ln \left({\text{D}}_{\text{o}}/{\text{D}}_{\text{i}}\right)}{2\text{k}}\right]+\left[\frac{1}{{\text{h}}_{\text{o}}}\right]}$

${\text{U}}_{\text{o}}=\frac{1}{\left[\frac{0.021}{0.018\times 210}\right]+\left[\frac{0.021\text{*ln}\left(\frac{0.021}{0.018}\right)}{2\times 111}\right]+\left[\frac{1}{70}\right]}$

${\text{U}}_{\text{o}}=50.36\text{ W}/{\text{m}}^2\text{K}$

Overall heat transfer coefficient based on inner surface area is 50.36 $\text{W}/{\text{m}}^2\text{K}.$