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 \(\rho_{l}=992.1 \mathrm{~kg} / \mathrm{m}^{3}, \mu_{l}=0.653 \times 10^{-3} \mathrm{~kg} / \mathrm{m} \cdot \mathrm{s}\), \(\left.k_{l}=0.631 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, c_{p l}=4179 \mathrm{~J} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}, h_{f g \otimes T_{\text {sat }}}=2383 \mathrm{~kJ} / \mathrm{kg}\right)\)

Step by step solution

01

Find the total surface area of the tubes

First, we need to find the total surface area of the 20 tubes. Each tube has a diameter (D) of \(6 \mathrm{~cm}\) and a length (L) of \(3 \mathrm{~m}\). To find the surface area of a single tube, use the formula: \(A = \pi \times D \times L\) To find the total surface area, multiply by the number of tubes (N = 20). \(A_{total} = N \times \pi \times D \times L\)

02

Calculate the total surface area

Substituting the values we have: D = 0.06 m (converted from cm) L = 3 m N = 20 tubes \(A_{total} = 20 \times \pi \times 0.06 \times 3\) \(A_{total} \approx 11.3097 \mathrm{m}^{2}\)

03

Find the formula for the rate of condensation

Based on the concept of heat transfer by condensation, the rate of condensation is given by the formula: \(q' = N_u \times k_l \times \frac{A_{total}}{D} \times \frac{T_{sat} - T_{surface}}{h_{fg}}\) Where: - \(N_u\) is the Nusselt number - A critical concept for this exercise is that for laminar film condensation on horizontal tubes, the Nusselt number is constant and equal to 0.725. - \(T_{sat}\) is the saturated temperature of steam which is given as \(50^{\circ} \mathrm{C}\) - \(T_{surface}\) is the temperature of the outer surface of the tube, given as \(30^{\circ} \mathrm{C}\).

04

Calculate the rate of condensation

Substituting the values: \(Nu = 0.725\) \(k_l = 0.631 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) \(T_{sat} = 50^{\circ} \mathrm{C}\) \(T_{surface} = 30^{\circ} \mathrm{C}\) \(h_{fg} = 2383 \mathrm{~kJ} / \mathrm{kg}\) (converted to J/kg: \(2383000 \mathrm{~J} / \mathrm{kg}\)) \(q' = 0.725 \times 0.631 \times \frac{11.3097}{0.06} \times \frac{50-30}{2383000}\) \(q' \approx 0.0537 \mathrm{~kg} / \mathrm{s}\)

05

Choose the closest value provided

Comparing the calculated value \(0.0537 \mathrm{~kg} / \mathrm{s}\) to the given options: (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) \(0.446 \mathrm{~kg} / \mathrm{s}\) The closest value to our calculated result is (a) \(0.054 \mathrm{~kg} / \mathrm{s}\).

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