Question

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

Short answer

a) 10°C b) 20°C c) 30°C d) 40°C e) 50°C Correct answer: b) 20°C

Step-by-step solution

Temperature Difference

The temperature difference, \(\Delta T\), between the steam and outer surface of the tube is given by: \(\Delta T = T_{steam} - T_{outer}\) where \(T_{steam}\) is the steam temperature and \(T_{outer}\) is the temperature of the outer surface. \(\Delta T = 50^{\circ}\mathrm{C} - 30^{\circ}\mathrm{C} = 20^{\circ}\mathrm{C}\) #Step 2: Calculate the Nusselt number for condensation on the outer surface of the tube#

Nusselt Number

The Nusselt number for condensation on the outer surface of a horizontal tube is given by the following equation: \(Nu = 0.54 \left(\frac{k_{l}}{6\cdot10^{-2}\mathrm{m}}\left[\frac{g \rho_{l}^{2}(h_{fg} + 0.68c_{p l}\Delta T)\Delta T}{\mu_{l}k_{l}}\right]^{1/4}\right) \cdot \mathrm{m}\) Using the given properties of water and the temperature difference, we can calculate the Nusselt number. #Step 3: Calculate the heat transfer coefficient using the Nusselt number#

Heat Transfer Coefficient

The heat transfer coefficient \(h\) is given by the formula: \(h = \frac{Nu \times k_{l}}{6\cdot10^{-2}\mathrm{m}}\) By plugging in the Nusselt number from Step 2, we obtain the heat transfer coefficient, which will be one of the given options (a)-(e). #Step 4: Calculate the area of the tube's outer surface#

Surface Area

The surface area of a tube can be calculated as: \(A = n \times \pi D L\) where \(n\) is the number of tubes, \(D\) is the diameter of each tube, and \(L\) is the length of each tube. In this case, \(A = 20 \times \pi (6\cdot10^{-2}\mathrm{m})(3\mathrm{m})\) #Step 5: Calculate the rate of heat transfer#

Rate of Heat Transfer

The rate of heat transfer \(q\) for condensation on the outer surface of the tube is given by the formula: \(q = h \times A \times \Delta T\) Using the area calculated in Step 4 and heat transfer coefficient from Step 3, we can calculate the rate of heat transfer. #Step 6: Calculate the rate of condensation of steam#

Rate of Condensation

The rate of condensation, \(m'\), can be found using the formula: \(m' = \frac{q}{h_{f g}}\) where \(h_{f g}\) is the enthalpy of vaporization at the saturation temperature. By plugging in the values for \(q\) and \(h_{fg}\), we can obtain the rate of condensation, which is one of the given options (a)-(e).

Key concepts

Nusselt number

Understanding the Nusselt number is key to analyzing heat transfer in fluid systems. It compares the rate of heat transfer through a fluid by conduction with the rate of heat transfer by convection. In essence, it indicates how effective convection is compared to conduction. For condensation on the outer surface of a horizontal tube, the Nusselt number can be calculated using the formula:

• \[Nu = 0.54 \left(\frac{k_{l}}{D}\left[\frac{g \rho_{l}^{2}(h_{fg} + 0.68c_{p l}\Delta T)\Delta T}{\mu_{l}k_{l}}\right]^{1/4}\right)\cdot D\]

Here, fluids' properties like density \(\rho_{l}\), dynamic viscosity \(\mu_{l}\), thermal conductivity \(k_{l}\), specific heat \(c_{p l}\), and enthalpy of vaporization \(h_{fg}\) play vital roles in determining the number. The Nusselt number impacts how heat transfers across the fluid layer and helps in designing efficient heat transfer systems.

Heat transfer coefficient

The heat transfer coefficient \(h\) quantifies how well heat is conducted from the steam to the tube wall during condensation. Once the Nusselt number is known, you can find \(h\) using:

• \[h = \frac{Nu \times k_{l}}{D}\]

A higher heat transfer coefficient indicates a more effective transfer of heat. Factors that influence \(h\) include the temperature difference between the steam and the tube surface, fluid properties, and the geometry of the system such as the diameter \(D\) of the tube. Knowing \(h\) helps in understanding the efficiency of the cooling process and making necessary adjustments in practical applications.

Rate of condensation

Rate of condensation, \(m'\), is vital in determining how quickly steam transforms into liquid. This metric is derived from the rate of heat transfer \(q\) due to condensation, using:

• \[m' = \frac{q}{h_{f g}}\]

Here, \(h_{f g}\) is the enthalpy of vaporization, denoting the energy required to convert liquid to vapor at a specific temperature. The rate of condensation tells us how effective the setup is in condensing the steam, helping optimize the sizing and design of heat exchangers. Efficient condensation means a higher \(m'\), indicative of high heat transfer and effective cooling.

Temperature difference

Temperature difference \(\Delta T\) is the driving force for heat transfer in condensation processes. It is simply the difference between the steam temperature \(T_{steam}\) and the tube's outer surface temperature \(T_{outer}\):

• \[\Delta T = T_{steam} - T_{outer}\]

For our problem, \(\Delta T = 20^{\circ}\mathrm{C}\). This difference influences both the Nusselt number and the heat transfer coefficient. A larger \(\Delta T\) usually results in enhanced heat transfer, making processes more efficient. It essentially dictates the rate at which energy transfers from the steam to the cooler surface.