Chapter 19, Problem 86P

To determine

The outlet mean temperature of water.

The surface temperature of the parallel plates.

The total rate of heat transfer.

Explanation of Solution

Given:

The value of width of plate is $\left(a\right)$ is $1\text{m}$.

The value of length of plate is $\left(b\right)$ is $10\text{m}$.

The mass flow rate $\left(\dot{m}\right)$ is $0.58\text{kg}/\text{s}$

The free stream velocity $\left(V\right)$ is $5\text{m}/\text{s}$.

Calculation:

Refer to table A-15, “Properties of saturated water”.

Obtain the following properties of water corresponding to temperature $30°\text{C}$ as follows:

$\begin{array}{l}\rho =997\text{kg}/{\text{m}}^3\\ k=0.6129\text{W}/\text{m}\cdot \text{K}\\ c_p=4178\text{J}/\text{kg}\cdot \text{K}\\ \mathrm{Pr}=5.68\\ \nu =0.8315\times {10}^{-6}{\text{m}}^2/\text{s}\end{array}$

Calculate the average velocity of flow $\left(V_{avg}\right)$ using the relation.

    $\begin{array}{c}{V}_{avg}=\frac{\dot{m}}{\rho A_c}\\ =\frac{0.58\text{kg}/\text{s}}{\left(997\text{kg}/{\text{m}}^3\right)\left(1\text{m}\times \left(1\text{2}\text{.5}\text{mm}\frac{{10}^{-3}\text{m}}{1\text{mm}}\right)\right)}\\ =0.0465\text{m}/\text{s}\end{array}$

Calculate the perimeter $\left(P\right)$ using the relation.

    $\begin{array}{c}P=2\left(a+b\right)\\ =2\left(1\text{m}+\left(12.5\text{mm}\frac{{10}^{-3}\text{m}}{1\text{mm}}\right)\right)\\ =2.025\text{m}\end{array}$

Calculate the hydraulic diameter of the tube $\left(D_h\right)$ using the relation.

    $\begin{array}{c}{D}_h=\frac{4A_c}{P}\\ =\frac{4\left(ab\right)}{2.025\text{m}}\\ =\frac{4\left(1\text{m}\times \left(12.5\text{mm}\frac{{10}^{-3}\text{m}}{1\text{mm}}\right)\right)}{2.025\text{m}}\\ =0.0246\text{m}\end{array}$

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

    $\begin{array}{c}\mathrm{Re}=\frac{V_{avg}{D}_h}{\nu }\\ =\frac{\left(0.0465\text{m}/\text{s}\right)\times \left(0.0246\text{m}\right)}{\left(0.8315\times {10}^{-6}{\text{m}}^2/\text{s}\right)}\\ =1376\end{array}$

The value of Reynolds number is less than $2300$. Therefore the flow is laminar flow.

Calculate the ratio of dimensions $a$ and $b$ using the relation.

    $\begin{array}{c}\frac{a}{b}=\frac{1\text{m}}{0.0125\text{m}}\\ =80\end{array}$

The Nusselt number value corresponding to $\frac{a}{b}=80$ for fully developed laminar flow is $8.24$.

    $Nu=8.24$

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

    $\begin{array}{c}h=Nu\left(\frac{k}{D}\right)\\ =8.24\left(\frac{0.6129\text{W}/\text{m}\cdot \text{K}}{\text{0}\text{.0246}\text{m}}\right)\\ =205.29\text{W}/{\text{m}}^2\cdot \text{K}\end{array}$

Calculate the total surface area $\left(A_s\right)$ using the relation.

    $\begin{array}{c}{A}_s=L\times 2\left(a+b\right)\\ =10\text{m}\times 2\left(1\text{m}+\left(12.5\text{mm}\frac{{10}^{-3}\text{m}}{1\text{mm}}\right)\right)\\ =20.25{\text{m}}^2\end{array}$

Calculate the value of $\frac{h{A}_s}{\dot{m}{c}_p}$ using the relation.

    $\begin{array}{c}\frac{h{A}_s}{\dot{m}{c}_p}=\left[\frac{-\left(205.29\text{W}/{\text{m}}^2\cdot \text{K}\right)\times \left(20.25{\text{m}}^2\right)}{\left(0.58\text{kg}/\text{s}\right)\left(4178\text{J}/\text{kg}\cdot \text{K}\right)}\right]\\ =-1.71\end{array}$

Calculate the outlet temperature using interpolation from Figure 19-27 of book corresponding to $\frac{h{A}_s}{\dot{m}{c}_p}$ value.

    $\begin{array}{c}\frac{5-\frac{h{A}_s}{\dot{m}{c}_p}}{5-1}=\frac{99.5-T_e}{99.5-70.6}\\ \frac{5-\left[-1.71\right]}{5-1}=\frac{99.5-T_e}{99.5-70.6}\\ T_e=75.766°\text{C}\end{array}$

Thus, the outlet mean temperature of water is $75.766°\text{C}$.

Calculate the surface temperature of fluid $\left(T_s\right)$ using the relation.

    $\begin{array}{c}{T}_s=\frac{T_e-T_i\text{exp}\left[\frac{-h{A}_s}{\dot{m}{c}_p}\right]}{1-\text{exp}\left[\frac{-h{A}_s}{\dot{m}{c}_p}\right]}\\ =\frac{75.766°\text{C}-\left(20°\text{C}\right)\text{exp}\left[-1.71\right]}{1-\text{exp}\left[-1.71\right]}\\ =88°\text{C}\end{array}$

Thus, the surface temperature of the parallel plates is $88°\text{C}$.

Refer to table A-23, “Properties of gases at $1\text{atm}$ pressure”.

Obtain the following properties of $H_2$ gas corresponding to temperature $100°\text{C}$:

$\begin{array}{l}\rho =0.06584\text{kg}/{\text{m}}^3\\ k=0.2095\text{W}/\text{m}\cdot \text{K}\\ c_p=14473\text{J}/\text{kg}\cdot \text{K}\\ \mathrm{Pr}=0.7196\\ \nu =1.582\times {10}^{-4}{\text{m}}^2/\text{s}\end{array}$

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

    $\begin{array}{c}\mathrm{Re}=\frac{V_{avg}L}{\nu }\\ =\frac{\left(5\text{m}/\text{s}\right)\times \left(10\text{m}\right)}{\left(1.582\times {10}^{-4}{\text{m}}^2/\text{s}\right)}\\ =316055.62\end{array}$

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

    $\begin{array}{c}Nu=0.332{\mathrm{Re}}^{0.5}{\mathrm{Pr}}^{0.33}\\ =0.332{\left(316055.62\right)}^{0.5}{\left(0.677\right)}^{0.33}\\ =164.10\end{array}$

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

    $\begin{array}{c}h=Nu\left(\frac{k}{D}\right)\\ =164.10\left(\frac{0.21981\text{W}/\text{m}\cdot \text{K}}{10\text{m}}\right)\\ =3.60\text{W}/{\text{m}}^2\cdot \text{K}\end{array}$

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

    $\begin{array}{c}Q=h{A}_s\Delta T_m\\ =h{A}_s\frac{\Delta T_e-\Delta T_i}{\ln \frac{\Delta T}{\Delta T_i}}\\ =\left(3.60\text{W}/{\text{m}}^2\cdot \text{K}\right)\left(20.25{\text{m}}^2\right)\frac{\left(88°\text{C}-\text{75}\text{.766}°\text{C}\right)-\left(88°\text{C}-\text{20}°\text{C}\right)}{\ln \frac{\left(88°\text{C}-\text{75}\text{.766}°\text{C}\right)}{\left(88°\text{C}-\text{20}°\text{C}\right)}}\\ =72.9\text{W}/\text{K}\times \frac{\left(12.234°\text{C}+273\right)\text{K}-\left(68°\text{C}+273\right)\text{K}}{-1.7153}\\ =2370.06\text{W}\end{array}$

Thus, the uniform heat flux on tube surface is $2370.06\text{W}$.