Chapter 6, Problem 6.1P

Determine the heat transfer coefficient at the stagnation point and the average value of the heat transfer coefficient for a single 5-cm-OD, 60-cm-long tube in cross-flow. The temperature of the tube surface is $\text{260°C}$, the velocity of the fluid flowing perpendicular to the tube axis is 6 m/s, and the temperature of the fluid is $\text{38°C}$. Consider the following fluids: (a) air, (b) hydrogen, and (c) water.

To determine

Heat transfer co-efficient at stagnation point and average heat transfer co-efficient with air as working fluid.

Answer to Problem 6.1P

Heat transfer co-efficient at stagnation point $\text{h}=68.9\text{ W}/{\text{m}}^2\text{K}$.

Average heat-transfer co-efficient $\overline{\text{h}}=41.9\text{ W}/{\text{m}}^2\text{K}$.

Explanation of Solution

Given Information:

Outer diameter of the tube (D) = 5 cm = 0.05 m

Length of the tube (L) = 60 cm = 0.6 m

Temperature of the tube surface (T_s) = 260⁰C

Velocity of the fluid flowing perpendicular to the tube axis is (V) = 6 m/s

Temperature of the fluid is (T_f) = 38⁰C

Explanation:

For cylindrical surfaces having cross flow, local heat-transfer co-efficient at any angular portion $\text{θ}$, is given by,

$\text{Nu}\left(\text{θ}\right)=1.14{\left[\text{Re}\right]}^{0.5}{\left[\text{Pr}\right]}^{0.4}\left[1-{\left(\frac{\text{θ}}{90}\right)}^3\right]$

Heat transfer co-efficient at stagnation point ($\text{θ}=0),\text{ is given by },$

$\text{Nu}\left(\text{θ}\right)=1.14{\left[\text{Re}\right]}^{0.5}{\left[\text{Pr}\right]}^{0.4}$

Reynolds number $\left(\text{Re}\right)=\frac{\text{VD}}{\vartheta }$

$\text{Nusselt number }\left(\text{Nu}\right)=\frac{\text{hD}}{\text{k}}$

$\frac{\text{hD}}{\text{k}}=1.14{\left[\frac{\text{VD}}{\text{ν}}\right]}^{0.5}{\left[\text{Pr}\right]}^{0.4}$

Heat transfer -coefficient at stagnation point is given by,

$\text{h}=\frac{\text{k}}{\text{D}}\times 1.14{\left[\frac{\text{VD}}{\text{ν}}\right]}^{0.5}{\left[\text{Pr}\right]}^{0.4}$ —— Equation(1)

For cross flow over cylindrical surfaces, average Nusselt number is given by:

$\overline{\text{Nu}}=\text{C }{\left[\text{Re}\right]}^{\text{m}}{\left[\text{Pr}\right]}^{\text{n }}{\left[\frac{\text{Pr}}{{\text{Pr}}_{\text{s}}}\right]}^{0.25}$

$\frac{\overline{\text{h}}\text{D}}{\text{k}}=\text{C }{\left[\frac{\text{VD}}{\text{ν}}\right]}^{\text{m}}{\left[\text{Pr}\right]}^{\text{n }}{\left[\frac{\text{Pr}}{{\text{Pr}}_{\text{s}}}\right]}^{0.25}$

Average convective heat-transfer co-efficient is given by,

$\overline{\text{h}}=\frac{\text{k}}{\text{D}}\times \text{C }{\left[\frac{\text{VD}}{\text{ν}}\right]}^{\text{m}}{\left[\text{Pr}\right]}^{\text{n }}{\left[\frac{\text{Pr}}{{\text{Pr}}_{\text{s}}}\right]}^{0.25}$———-Equation (2)

From Appendix-2, Table -28, Properties of dry air at bulk temperature of 38⁰C are,

Prandtl number (Pr) = 0.71

Thermal conductivity (k) = 0.0264 W/mK

Kinematic viscosity $\left(\text{ν}\right)=17.4\times {10}^{-6}{\text{ m}}^2/\text{s}$

At surface temperature Prandtl number, ${\text{Pr}}_{\text{s}}=0.71$

Form Equation(1), Heat transfer co-efficient at stagnation point :

$\text{h}=\frac{0.0264}{0.05}\times 1.14{\left[\frac{6\times 0.05}{17.4\times {10}^{-6}}\right]}^{0.5}{\left[0.71\right]}^{0.4}$

$\text{h}=68.9\text{ W}/{\text{m}}^2\text{K}$

From Equation(2), average heat transfer co-efficient is,

$\text{Re}=\frac{6\times 0.05}{17.4\times {10}^{-6}}=17241.37$

For Re =17241.37, C = 0.26, m = 0.6

For Pr <10, n = 0.37

$\overline{\text{h}}=\frac{\text{k}}{\text{D}}\times \text{C }{\left[\frac{\text{VD}}{\text{ν}}\right]}^{\text{m}}{\left[\text{Pr}\right]}^{\text{n }}{\left[\frac{\text{Pr}}{{\text{Pr}}_{\text{s}}}\right]}^{0.25}$

$\overline{\text{h}}=\frac{0.0264}{0.05}\times 0.26\times {\left[\frac{6\times 0.05}{17.4\times {10}^{-6}}\right]}^{0.6}{\left[0.71\right]}^{0.37}{\left[\frac{0.71}{0.71}\right]}^{0.25}$

$\overline{\text{h}}=41.9\text{ W}/{\text{m}}^2\text{K}$

Conclusion:

Heat transfer co-efficient at stagnation point $\text{h}=68.9\text{ W}/{\text{m}}^2\text{K}$.

Average heat-transfer co-efficient $\overline{\text{h}}=41.9\text{ W}/{\text{m}}^2\text{K}$.

To determine

Heat transfer co-efficient at stagnation point and average heat transfer co-efficient with hydrogen as working fluid.

Answer to Problem 6.1P

Heat transfer co-efficient at stagnation point $\text{h}=184.36\text{ W}/{\text{m}}^2\text{K }$

Average heat-transfer co-efficient $\overline{\text{h}}=96.14\text{ W}/{\text{m}}^2\text{K}$

Explanation of Solution

From Appendix – 2, table-32, Properties of hydrogen at bulk temperature of 38⁰C are,

Prandtl number (Pr) = 0.704

Thermal conductivity (k) = 0.187 W/mK

Kinematic viscosity $\left(\text{ν}\right)=116.6\times {10}^{-6}{\text{ m}}^2/\text{s}$

At surface temperature Prandtl number, ${\text{Pr}}_{\text{s}}=0.671$

Form Equation(1), Heat transfer co-efficient at stagnation point :

$\text{h}=\frac{\text{k}}{\text{D}}\times 1.14{\left[\frac{\text{VD}}{\text{ν}}\right]}^{0.5}{\left[\text{Pr}\right]}^{0.4}$

$\text{h}=\frac{0.187}{0.05}\times 1.14{\left[\frac{6\times 0.05}{116.6\times {10}^{-6}}\right]}^{0.5}{\left[0.671\right]}^{0.4}=184.36\text{ W}/{\text{m}}^2\text{K}$

From Equation(2), average heat transfer co-efficient is,

$\overline{\text{h}}=\frac{\text{k}}{\text{D}}\times \text{C }{\left[\frac{\text{VD}}{\text{ν}}\right]}^{\text{m}}{\left[\text{Pr}\right]}^{\text{n }}{\left[\frac{\text{Pr}}{{\text{Pr}}_{\text{s}}}\right]}^{0.25}$

$\text{Re}=\frac{6\times 0.05}{116.6\times {10}^{-6}}=2573$

From table 6.1, For Re =2573, C = 0.26, m = 0.6

For Pr < 10, n = 0.37

$\overline{\text{h}}=\frac{0.187}{0.05}\times \left(0.26\right){\left[2573\right]}^{0.6}{\left[0.704\right]}^{0.37}{\left[\frac{0.704}{0.671}\right]}^{0.25}$

$\overline{\text{h}}=96.14\text{ W}/{\text{m}}^2\text{K}$

Conclusion:

Heat transfer co-efficient at stagnation point $\text{h}=184.36\text{ W}/{\text{m}}^2\text{K }$

Average heat-transfer co-efficient $\overline{\text{h}}=96.14\text{ W}/{\text{m}}^2\text{K}$

To determine

Heat transfer co-efficient at stagnation point and average heat transfer co-efficient with water as working fluid.

Answer to Problem 6.1P

Heat transfer co-efficient at stagnation point $\text{h}=17231.51\text{ W}/{\text{m}}^2\text{K}$

Average heat-transfer co-efficient $\overline{\text{h}}=22432\text{ W}/{\text{m}}^2\text{K}$

Explanation of Solution

From Appendix – 2, table-13, Properties of water at bulk temperature of 38⁰C are,

Prandtl number (Pr) = 4.5

Thermal conductivity (k) = 0.629 W/mK

Kinematic viscosity $\left(\text{ν}\right)=0.685\times {10}^{-6}{\text{ m}}^2/\text{s}$

At surface temperature Prandtl number, ${\text{Pr}}_{\text{s}}=0.86$

Form Equation(1), Heat transfer co-efficient at stagnation point :

$\text{h}=\frac{\text{k}}{\text{D}}\times 1.14{\left[\frac{\text{VD}}{\text{ν}}\right]}^{0.5}{\left[\text{Pr}\right]}^{0.4}$

$\text{h}=\frac{0.629}{0.05}\times 1.14{\left[\frac{6\times 0.05}{0.685\times {10}^{-6}}\right]}^{0.5}{\left[4.5\right]}^{0.4}=17231.51\text{ W}/{\text{m}}^2\text{K}$

From Equation(2), average heat transfer co-efficient is,

$\overline{\text{h}}=\frac{\text{k}}{\text{D}}\times \text{C }{\left[\frac{\text{VD}}{\text{ν}}\right]}^{\text{m}}{\left[\text{Pr}\right]}^{\text{n }}{\left[\frac{\text{Pr}}{{\text{Pr}}_{\text{s}}}\right]}^{0.25}$

$\text{Re}=\frac{6\times 0.05}{0.685\times {10}^{-6}}=437956.2$

From table 6.1, For Re = $437956.2$, C = 0.076, m = 0.7

For Pr < 10, n = 0.37

$\overline{\text{h}}=\frac{0.629}{0.05}\times \left(0.076\right){\left[437956.2\right]}^{0.7}{\left[4.5\right]}^{0.37}{\left[\frac{4.5}{0.86}\right]}^{0.25}$

$\overline{\text{h}}=22432\text{ W}/{\text{m}}^2\text{K}$

Conclusion:

Heat transfer co-efficient at stagnation point $\text{h}=17231.51\text{ W}/{\text{m}}^2\text{K}$

Average heat-transfer co-efficient $\overline{\text{h}}=22432\text{ W}/{\text{m}}^2\text{K}$