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Chapter 10: Problem 35

A \(10 \mathrm{~cm} \times 10 \mathrm{~cm}\) horizontal flat heater is used for vaporizing refrigerant-134a 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

Expert verified
Based on the Rohsenow correlation, the coefficient \(C_{sf}\) for the given scenario with a horizontal flat heater and refrigerant-134a is approximately \(0.0042\). The correlation appears to be reasonably applicable in this case due to the surface temperature of the heater being not far from the saturation temperature and vaporization of the refrigerant-134a. However, it is important to verify the applicability for such heating conditions using more specific correlations for refrigerants or conducting experiments.

Step by step solution

01

Calculate the temperature difference

First, let's find the temperature difference across the interface. The surface temperature of the heater is \(25^{\circ} \mathrm{C}\). To find the interface temperature, we need to find the saturation temperature at the given pressure of \(350 \mathrm{kPa}\). Using refrigerant-134a property tables or steam tables, we can find the saturation temperature to be approximately \(38^{\circ} \mathrm{C}\). Thus, the temperature difference is: $$\Delta T = T_{sat} - T_s = 38 - 25 = 13^{\circ} \mathrm{C}$$
02

Find the latent heat of vaporization for refrigerant-134a

Now we need to find the latent heat of vaporization (\(h_{fg}\)) for refrigerant-134a at the given pressure. Using refrigerant-134a property tables or steam tables, we can find that: $$h_{fg} \approx 240 \mathrm{~kJ/kg}$$
03

Convert the heat flux to the appropriate unit

We need to convert the heat flux from \(\mathrm{MW/m^2}\) to \(\mathrm{kJ/(m^2s)}\). Given the heat flux of \(0.35 \mathrm{MW}/\mathrm{m^2}\): $$q = 0.35 \times 10^3 \mathrm{~kJ/(m^2s)} = 350 \mathrm{~kJ/(m^2s)}$$
04

Solve for the coefficient \(C_{sf}\)

Now, we can use the Rohsenow correlation to solve for the coefficient \(C_{sf}\): $$q = C_{sf}\Delta T^n h_{fg}$$ Rearranging the equation to solve for \(C_{sf}\): $$C_{sf} = \frac{q}{\Delta T^n h_{fg}}$$ Plugging in the values from the previous steps: $$C_{sf} = \frac{350}{(13)^{1.7} \times 240} \approx 0.0042$$ So, the value of the coefficient \(C_{sf}\) is approximately \(0.0042\).
05

Discuss the applicability of Rohsenow correlation

The Rohsenow correlation is an empirical correlation that is widely used to predict the rate of heat transfer during boiling, particularly for pool boiling. However, it may not be applicable in all scenarios, especially when the system deviates significantly from the original experimental conditions. In this case, it seems reasonable to apply Rohsenow correlation since the surface temperature of the heater is not far from the saturation temperature and the refrigerant-134a experiences vaporization. However, it's important to check with more specific correlations for refrigerants or conduct experiments to verify the applicability for such heating conditions.

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Most popular questions from this chapter

Saturated water vapor at atmospheric pressure condenses on the outer surface of a \(0.1\)-m-diameter vertical pipe. The pipe is \(1 \mathrm{~m}\) long and has a uniform surface temperature of \(80^{\circ} \mathrm{C}\). Determine the rate of condensation and the heat transfer rate by condensation. Discuss whether the pipe can be treated as a vertical plate. Assume wavy-laminar flow and that the tube diameter is large relative to the thickness of the liquid film at the bottom of the tube. Are these good assumptions?

Does the amount of heat absorbed as \(1 \mathrm{~kg}\) of saturated liquid water boils at \(100^{\circ} \mathrm{C}\) have to be equal to the amount of heat released as \(1 \mathrm{~kg}\) of saturated water vapor condenses at \(100^{\circ} \mathrm{C}\) ?

A \(1.5-\mathrm{m}\)-long vertical tube is used for condensing saturated steam at \(60^{\circ} \mathrm{C}\). The surface temperature of the tube is maintained at a uniform temperature of \(40^{\circ} \mathrm{C}\) by flowing coolant inside the tube. Determine the heat transfer rate to the tube and the required tube diameter to condense \(12 \mathrm{~kg} / \mathrm{h}\) of steam during the condensation process. Assume wavy-laminar flow and that the tube diameter is large relative to the thickness of the liquid film at the bottom of the tube. Are these good assumptions?

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Saturated water vapor at a pressure of \(12.4 \mathrm{kPa}\) is condensed over 100 horizontal tubes in a rectangular array of 5 tubes high and 20 tubes wide, each with a diameter of \(8 \mathrm{~mm}\). If the tube surfaces are maintained with a uniform temperature of \(30^{\circ} \mathrm{C}\), determine the condensation rate per unit length (in \(\mathrm{kg} / \mathrm{s} \cdot \mathrm{m}\) ) of the tubes.
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