## Related questions with answers

A device for performing boiling experiments consists of a copper bar $(k=400 \mathrm{W} / \mathrm{m} \cdot \mathrm{K})$, which is exposed to a boiling liquid at one end, encapsulates an electrical heater at the other end, and is well insulated from its surroundings at all but the exposed surface. Thermocouples inserted in the bar are used to measure temperatures at distances of $x_{1}=10 \mathrm{mm}$ and $x_{2}=25 \mathrm{mm}$ from the surface. (a) An experiment is performed to determine the boiling characteristics of a special coating applied to the exposed surface. Under steady-state conditions, nucleate boiling is maintained in saturated water at atmospheric pressure, and values of $T_{1}=133.7^{\circ} \mathrm{C}$ and $T_{2}=158.6^{\circ} \mathrm{C}$ are recorded. If n=1, what value of the coefficient $C_{s, f}$ is associated with the Rohsenow correlation? (b) Assuming applicability of the Rohsenow correlation with the value $C_{s, f}$ determined from part (a), compute and plot the excess temperature $\Delta T_{e}$ as a function of the boiling heat flux for $10^{5} \leq q_{s}^{\prime \prime} \leq 10^{6} \mathrm{W} / \mathrm{m}^{2}$. What are the corresponding values of $T_{1}$ and $T_{2}$ for $q_{s}^{\prime \prime}=10^{6} \mathrm{W} / \mathrm{m}^{2} ?$ If $q_{s}^{\prime \prime}$ were increaset to $1.5 \times 10^{6} \mathrm{W} / \mathrm{m}^{2}$, could the foregoing results be extrapolated to infer the corresponding values of $\Delta T_{e}, T_{1}$, and $T_{2} ?$

Solution

VerifiedGiven$\Rightarrow$ Thermal conductivity of copper, $k=400\ \frac{\text{W}}{\text{m.K}}$; $x_1=10\ \text{mm}$; $x_2=25\ \text{mm}$; $T_1=133.7^{o} \ \text{C}$; $T_2=158.6^{o} \ \text{C}$; $n=1$

Thermophysical Properties of $\textbf{saturated boiling water}$:

$T_{sat} =323\ \text{K}$; $h_{fg} =2257\ \frac{\text{kJ}}{\text{kg}}$; $\rho_{l}=957.9\ \frac{\text{kg}}{\text{m}^3}$; $\rho_{v}=0.5955\ \frac{\text{kg}}{\text{m}^3}$; $c_{p,l}=4.217\ \frac{\text{kJ}}{\text{kg} \cdot K}$; $\mu_{l}=279 \times 10^{-6} \ \frac{\text{N} \cdot s}{\text{m}^2}$; $Pr=1.76$, $\sigma=0.0589\ \frac{\text{N}}{\text{m}}$

$\textbf{(a)}$ Heat flux between points $1$ and $2$ can be given by Fourier’s law:

$q''=k\dfrac{dT}{dx} \ \ \ \ \ (1)$

Substituting values in above expression:

$\begin{align*} q'' &=400 \left(\dfrac{T_2-T_1}{x_2-x_1}\right)\\ &=400\left(\dfrac{158.6-133.7}{0.025-0.01}\right)\\ &=664\ \frac{\text{kW}}{\text{m}^2} \end{align*}$

Heat flux or heat transfer per unit area can be calculated by using Rohsenow equation:

$q'' = \mu_{l}\cdot h_{fg} \left [\frac{g(\rho_{l} - \rho_{v})}{\sigma}\right]^{\frac{1}{2}} \left( \frac{C_{p,l} \Delta T_{e}}{C_{s,f}h_{fg}Pr_{l}^n} \right)^{3}$

where $\Delta T_e$ denotes excess temperature given by $\Delta T_e=T_{s}-T_{sat}$

For steady state conditions, the temperature distribution in the bar is given as:

$\begin{align*} T_s &=T_1 - \dfrac{dT}{dx} x_1 \ \ \ \ \ \ (2) \\ &=T_1 - \left(\dfrac{T_2 - T_1}{x_2 - x_1} \right)x_1\\ &=133.7 - \left(\dfrac{158.6 - 133.7}{0.025 - 0.01} \right)0.01\\ &=117.1^{o} \ \text{C} \end{align*}$

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