Question

A \(10-\mathrm{cm} \times 10\)-cm horizontal flat heater is used for vaporizing refrigerant- \(134 \mathrm{a}\) at \(350 \mathrm{kPa}\). The heater is supplied with \(0.35 \mathrm{MW} / \mathrm{m}^{2}\) of heat flux, and the surface temperature of the heater is \(25^{\circ} \mathrm{C}\). If the experimental constant in the Rohsenow correlation is \(n=1.7\), determine the value of the coefficient \(C_{s f}\). Discuss whether or not the Rohsenow correlation is applicable in this analysis.

Short answer

Answer: The dimensionless Rohsenow correlation constant, \(C_{sf}\), for this case is approximately 0.00674. The Rohsenow correlation seems to be applicable for this scenario, as the heat flux is moderate, and the physical properties of refrigerant-134a are typical of many common refrigerants. However, it's important to compare the predictions of the Rohsenow correlation with experimental data or more detailed modeling techniques for a more accurate analysis and evaluation.

Step-by-step solution

Find the properties of refrigerant-134a at the given pressure

Using a refrigeration property table such as the one found in the ASHRAE Handbook of Fundamentals or an online calculator, we can find the thermo-physical properties of the refrigerant-134a at the given pressure of 350 kPa. For refrigerant-134a at 350 kPa, we have: The saturation temperature: \(T_{sat} = 32.12^{\circ}C\) The liquid density: \(\rho_l = 1208 kg/m^3\) The vapor density: \(\rho_v = 20.6 kg/m^3\) The fluid viscosity: \(\mu _f = 1.53 \times 10^{-4} kg/m·s\) The enthalpy of vaporization: \(h_{fg} = 1.92 \times 10^5 J/kg\)

Calculate the temperature difference ΔT_sat

Now, we can calculate the temperature difference between the saturation temperature and the surface temperature of the heater: \(\Delta T_{sat} = T_{sat} - T_{surface} = 32.12^{\circ}C - 25^{\circ}C = 7.12 K\)

Plug values into Rohsenow correlation equation and solve for C_sf

Using the Rohsenow correlation formula, we can solve for the coefficient \(C_{sf}\): \(q''= C_{sf} \cdot \mu _f \cdot h_{fg} \cdot g \cdot \Delta T_sat^{n} \cdot (\rho_l - \rho_v) / \rho_v\) Substituting the given values and rearranging the equation to solve for \(C_{sf}\): \(C_{sf} = q'' / (\mu _f \cdot h_{fg} \cdot g \cdot \Delta T_{sat}^n \cdot (\rho_l - \rho_v) / \rho_v)\) \(C_{sf} = (0.35 \times 10^6 W/m^2) / ((1.53 \times 10^{-4} kg/m·s)(1.92 \times 10^5 J/kg)(9.81 m/s^2)(7.12K ^{1.7})((1208 kg/m^3)-(20.6 kg/m^3))/(20.6 kg/m^3))\) After calculating the above expression, we get: \(C_{sf} \approx 0.00674\)

Discuss the applicability of the Rohsenow correlation

The Rohsenow correlation is generally applicable for pool boiling, where the heat transfer is dominated by buoyancy-induced natural convection and nucleate boiling. It may not be applicable for very low or very high heat fluxes, for fluids with a high Prandtl number, and for situations where flow boiling or liquid agitation is significant. In our case, the heat flux is moderate, and the physical properties of refrigerant-134a are typical of many common refrigerants. Therefore, the Rohsenow correlation seems to be applicable for this scenario. However, it's essential to compare the predictions of the Rohsenow correlation with experimental data or more detailed modeling techniques for more accurate analysis and evaluation.