Water is boiled at \(90^{\circ} \mathrm{C}\) by a horizontal brass heating element of diameter \(7 \mathrm{~mm}\). Determine the maximum heat flux that can be attained in the nucleate boiling regime.

Short Answer

Expert verified
The maximum heat flux that can be attained in the nucleate boiling regime is approximately 14,497.6 W/m².

Step by step solution

01

Write down the Rohsenow nucleate boiling correlation

The Rohsenow correlation for nucleate boiling is given by the following equation: \(q'' = C_{sf} \cdot h_{fg} \cdot (\frac{\sigma}{g(\rho_l-\rho_g)})^{1/2} \cdot m^n \cdot \Delta T\) where \(q''\) = heat flux \((W/m^2)\), \(C_{sf}\) = dimensionless constant, \(h_{fg}\) = specific enthalpy of vaporization \((J/kg)\), \(\sigma\) = surface tension \((N/m)\), \(g\) = acceleration due to gravity \((m/s^2)\), \(\rho_l\) = liquid density \((kg/m^3)\), \(\rho_g\) = vapor density \((kg/m^3)\), \(m\) = liquid-vapor mass qualities, \(n\) = a dimensionless constant, and \(\Delta T\) = excess temperature \((K)\).
02

Find the excess temperature

Excess temperature (\(\Delta T\)) is the difference between the saturation temperature of water at a given pressure and the surface temperature of the heating element. In this case, water is boiled at \(90^{\circ} \mathrm{C}\), which is \(363.15 \mathrm{K}\) in Kelvin. Assuming atmospheric pressure, the saturation temperature of water is \(100^{\circ} \mathrm{C}\) or \(373.15 \mathrm{K}\). So, \(\Delta T = T_{sat} - T_{surface} = 373.15 - 363.15 = 10 \mathrm{K}\).
03

Find the properties of water at the given temperature

We need the following properties of water at \(90^{\circ} \mathrm{C}\) (calculated at the saturation temperature): \(h_{fg} = 2.257 \times 10^6 \mathrm{\ J/kg}\) (enthalpy of vaporization), \(\sigma = 0.059 \mathrm{\ N/m}\) (surface tension), \(\rho_l = 971.8 \mathrm{\ kg/m^3}\) (liquid density), \(\rho_g = 0.597 \mathrm{\ kg/m^3}\) (vapor density).
04

Select appropriate values of constants and dimensionless groups

For nucleate boiling of water on a smooth horizontal brass surface, the following values are recommended: \(C_{sf} = 0.0064\), \(m = 2\), and \(n = 1\).
05

Calculate the maximum heat flux

Now, we can plug all the values into the Rohsenow correlation to find the maximum heat flux: \(q'' = C_{sf} \cdot h_{fg} \cdot (\frac{\sigma}{g(\rho_l-\rho_g)})^{1/2} \cdot m^n \cdot \Delta T\) \(q'' = 0.0064 \cdot 2.257 \times 10^6 \cdot (\frac{0.059}{9.81(971.8-0.597)})^{1/2} \cdot 2^1 \cdot 10\) \(q'' = 14,497.6 \mathrm{\ W/m^2}\) Thus, the maximum heat flux that can be attained in the nucleate boiling regime is approximately \(14,497.6 \mathrm{\ W/m^2}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A 2-mm-diameter cylindrical metal rod with emissivity of \(0.5\) is submerged horizontally in water under atmospheric pressure. When electric current is passed through the metal rod, the surface temperature reaches \(500^{\circ} \mathrm{C}\). Determine the power dissipation per unit length of the metal rod.

Steam condenses at \(50^{\circ} \mathrm{C}\) on the outer surface of a horizontal tube with an outer diameter of \(6 \mathrm{~cm}\). The outer surface of the tube is maintained at \(30^{\circ} \mathrm{C}\). The condensation heat transfer coefficient is (a) \(5493 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (b) \(5921 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (c) \(6796 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (d) \(7040 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (e) \(7350 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (For water, use \(\rho_{l}=992.1 \mathrm{~kg} / \mathrm{m}^{3}, \mu_{l}=0.653 \times 10^{-3} \mathrm{~kg} / \mathrm{m} \cdot \mathrm{s}\), \(\left.k_{l}=0.631 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, c_{p l}=4179 \mathrm{~J} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}, h_{f g} \oplus T_{\text {satl }}=2383 \mathrm{~kJ} / \mathrm{kg}\right)\) 10-130 Steam condenses at \(50^{\circ} \mathrm{C}\) on the tube bank consisting of 20 tubes arranged in a rectangular array of 4 tubes high and 5 tubes wide. Each tube has a diameter of \(6 \mathrm{~cm}\) and a length of \(3 \mathrm{~m}\), and the outer surfaces of the tubes are maintained at \(30^{\circ} \mathrm{C}\). The rate of condensation of steam is (a) \(0.054 \mathrm{~kg} / \mathrm{s}\) (b) \(0.076 \mathrm{~kg} / \mathrm{s}\) (c) \(0.315 \mathrm{~kg} / \mathrm{s}\) (d) \(0.284 \mathrm{~kg} / \mathrm{s}\) (e) \(0.446 \mathrm{~kg} / \mathrm{s}\) (For water, use \(\rho_{l}=992.1 \mathrm{~kg} / \mathrm{m}^{3}, \mu_{l}=0.653 \times 10^{-3} \mathrm{~kg} / \mathrm{m} \cdot \mathrm{s}\), \(\left.k_{l}=0.631 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, c_{p l}=4179 \mathrm{~J} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}, h_{f g \otimes T_{\text {sat }}}=2383 \mathrm{~kJ} / \mathrm{kg}\right)\)

Consider film condensation on the outer surfaces of four long tubes. For which orientation of the tubes will the condensation heat transfer coefficient be the highest: (a) vertical, (b) horizontal side by side, (c) horizontal but in a vertical tier (directly on top of each other), or \((d)\) a horizontal stack of two tubes high and two tubes wide?

Steam condenses at \(50^{\circ} \mathrm{C}\) on the tube bank consisting of 20 tubes arranged in a rectangular array of 4 tubes high and 5 tubes wide. Each tube has a diameter of \(6 \mathrm{~cm}\) and a length of \(3 \mathrm{~m}\), and the outer surfaces of the tubes are maintained at \(30^{\circ} \mathrm{C}\). The rate of condensation of steam is (a) \(0.054 \mathrm{~kg} / \mathrm{s}\) (b) \(0.076 \mathrm{~kg} / \mathrm{s}\) (c) \(0.315 \mathrm{~kg} / \mathrm{s}\) (d) \(0.284 \mathrm{~kg} / \mathrm{s}\) (e) \(0.446 \mathrm{~kg} / \mathrm{s}\) (For water, use \(\rho_{l}=992.1 \mathrm{~kg} / \mathrm{m}^{3}, \mu_{l}=0.653 \times 10^{-3} \mathrm{~kg} / \mathrm{m} \cdot \mathrm{s}\), \(\left.k_{l}=0.631 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, c_{p l}=4179 \mathrm{~J} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}, h_{f g \otimes T_{\text {sat }}}=2383 \mathrm{~kJ} / \mathrm{kg}\right)\)

What is boiling? What mechanisms are responsible for the very high heat transfer coefficients in nucleate boiling?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free