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Question

A device for performing boiling experiments consists of a copper bar (k=400W/mK)(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 x1=10mmx_{1}=10 \mathrm{mm} and x2=25mmx_{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 T1=133.7CT_{1}=133.7^{\circ} \mathrm{C} and T2=158.6CT_{2}=158.6^{\circ} \mathrm{C} are recorded. If n=1, what value of the coefficient Cs,fC_{s, f} is associated with the Rohsenow correlation? (b) Assuming applicability of the Rohsenow correlation with the value Cs,fC_{s, f} determined from part (a), compute and plot the excess temperature ΔTe\Delta T_{e} as a function of the boiling heat flux for 105qs106W/m210^{5} \leq q_{s}^{\prime \prime} \leq 10^{6} \mathrm{W} / \mathrm{m}^{2}. What are the corresponding values of T1T_{1} and T2T_{2} for qs=106W/m2?q_{s}^{\prime \prime}=10^{6} \mathrm{W} / \mathrm{m}^{2} ? If qsq_{s}^{\prime \prime} were increaset to 1.5×106W/m21.5 \times 10^{6} \mathrm{W} / \mathrm{m}^{2}, could the foregoing results be extrapolated to infer the corresponding values of ΔTe,T1\Delta T_{e}, T_{1}, and T2?T_{2} ?

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Given\Rightarrow Thermal conductivity of copper, k=400 Wm.Kk=400\ \frac{\text{W}}{\text{m.K}}; x1=10 mmx_1=10\ \text{mm}; x2=25 mmx_2=25\ \text{mm}; T1=133.7o CT_1=133.7^{o} \ \text{C}; T2=158.6o CT_2=158.6^{o} \ \text{C}; n=1n=1

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

Tsat=323 KT_{sat} =323\ \text{K}; hfg=2257 kJkgh_{fg} =2257\ \frac{\text{kJ}}{\text{kg}}; ρl=957.9 kgm3\rho_{l}=957.9\ \frac{\text{kg}}{\text{m}^3}; ρv=0.5955 kgm3\rho_{v}=0.5955\ \frac{\text{kg}}{\text{m}^3}; cp,l=4.217 kJkgKc_{p,l}=4.217\ \frac{\text{kJ}}{\text{kg} \cdot K}; μl=279×106 Nsm2\mu_{l}=279 \times 10^{-6} \ \frac{\text{N} \cdot s}{\text{m}^2}; Pr=1.76Pr=1.76, σ=0.0589 Nm\sigma=0.0589\ \frac{\text{N}}{\text{m}}

(a)\textbf{(a)} Heat flux between points 11 and 22 can be given by Fourier’s law:

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

Substituting values in above expression:

q=400(T2T1x2x1)=400(158.6133.70.0250.01)=664 kWm2\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=μlhfg[g(ρlρv)σ]12(Cp,lΔTeCs,fhfgPrln)3q'' = \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 ΔTe\Delta T_e denotes excess temperature given by ΔTe=TsTsat\Delta T_e=T_{s}-T_{sat}

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

Ts=T1dTdxx1      (2)=T1(T2T1x2x1)x1=133.7(158.6133.70.0250.01)0.01=117.1o C\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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