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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?

Short Answer

Expert verified
Answer: To calculate the heat transfer rate, first calculate the temperature difference (∆T) between the steam and the tube surface. Then, use the Nusselt number relation for laminar film condensation to determine the heat transfer coefficient (h), Reynolds and Prandtl numbers. Use Newton's Law of Cooling to find the heat transfer rate (Q). To find the required tube diameter (D), calculate the mass flow rate of steam and the condensation rate (G). Finally, use the Reynolds and Prandtl numbers to solve for the tube diameter.

Step by step solution

01

Calculate the temperature difference

We know the surface temperature of the tube is \(40^{\circ} \mathrm{C}\) and the steam temperature is \(60^{\circ} \mathrm{C}\). The temperature difference (∆T) is: ∆T = \(60^{\circ} \mathrm{C}\) - \(40^{\circ} \mathrm{C}\) = \(20^{\circ} \mathrm{C}\)
02

Calculate the heat transfer coefficient (h)

Given that the tube diameter is large relative to the thickness of the liquid film, we can use the following Nusselt number relation for laminar film condensation: Nu = \(0.943 \times (RePr)^{\frac{1}{3}}\) Now, find the Reynolds and Prandtl numbers: Re = \(\frac{4G}{\pi \mu D}\) Pr = \(\frac{c_p \mu}{k}\)
03

Calculate heat transfer rate (Q)

Using Newton's Law of Cooling and knowing the area of the tube's outer surface (A) and the temperature difference (∆T), we can calculate the heat transfer rate (Q) as follows: Q = h A ∆T 2. Finding the tube diameter
04

Calculate mass flow rate of steam

We are given that \(12 \mathrm{~kg} / \mathrm{h}\) of steam must be condensed. Calculate the mass flow rate in \( \mathrm{kg} / \mathrm{s}\): \(\dot{m} = \frac{12 \mathrm{~kg}}{3600 \mathrm{~s}}\)
05

Calculate condensation rate

Now, calculate the condensation rate (G) in \(\mathrm{kg} / \mathrm{m^2\cdot s}\): G = \(\frac{\dot{m}}{A}\)
06

Solve for tube diameter (D)

With G known, Re and Pr can be calculated. Then h can be determined using the Nusselt number relation, and finally, the tube diameter (D) can be calculated: D = \(\frac{4G}{\pi \mu Re}\) Now we have found the heat transfer rate to the tube and the required tube diameter to condense the desired amount of steam. To assess the quality of the assumptions, we can observe the chosen Nusselt number relation and the given tube diameter with respect to the liquid film thickness. If the tube diameter is indeed large relative to the liquid film thickness, then the wavy-laminar flow assumption is reasonable.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Film Condensation
Film condensation is a heat transfer process where a vapor comes into contact with a cooler surface and condenses to form a liquid film. This liquid film then flows down the surface under the influence of gravity in a laminar or turbulent manner depending on various conditions like surface temperature and vapor flow. In the example above, a vertical tube is cooling saturated steam, leading to the formation of a liquid film on its surface.

The effectiveness of condensation is influenced by the mode of condensation (wavy or laminar flow), thickness of the condensate film, temperature difference between the vapor and the surface, and properties of the condensate. Typically, a thinner film provides better heat transfer due to less thermal resistance. When assuming wavy-laminar flow, it indicates a mix of laminar flow with some disturbances that do not fully transition into turbulent flow, which can be a reasonable approximation when the liquid film is relatively thin compared to the tube's diameter.
Nusselt Number
The Nusselt number (Nu) is a dimensionless quantity used in heat transfer calculations and indicates the ratio of convective to conductive heat transfer across a fluid boundary. In the context of condensation, it's a measure of the efficiency of heat transfer from the condensing steam to the cooling surface.

To calculate the Nusselt number, a relation that considers both the Reynolds (Re) and Prandtl (Pr) numbers is used. The formula provided in the original problem, Nu = 0.943×(RePr)^(1/3), takes into account the flow conditions of the condensate film. It assumes a certain correlation between the heat transfer, the momentum transfer (Reynolds number), and the thermal properties of the fluid (Prandtl number), which are crucial to estimating the heat transfer coefficient (h) that is applied to determine the heat transfer rate.
Reynolds Number
The Reynolds number (Re) is a dimensionless value that represents the ratio of inertial forces to viscous forces within a fluid flow. It helps determine the flow regime, characterizing the flow as laminar, transitional, or turbulent. Calculated as Re = (4G)/(πμD) in the textbook solution, where G is the mass flux, μ is the dynamic viscosity, and D is the tube diameter, it is essential for evaluating the behavior of the fluid on the surface while condensing.

A lower Reynolds number corresponds to a smoother, more orderly flow, which suits the film condensation process. In applications like the vertical tube in the given problem, maintaining a certain range for Reynolds number ensures that the wavy-laminar flow assumption remains valid, and thereby the heat transfer coefficient can be calculated more precisely.
Prandtl Number
The Prandtl number (Pr) is another dimensionless number in fluid mechanics that expresses the ratio of momentum diffusivity (kinematic viscosity) to thermal diffusivity. It is given by Pr = (cpμ)/k, where cp is the specific heat at constant pressure, μ is the fluid's dynamic viscosity, and k is the thermal conductivity. The Prandtl number gives insight into the relative thickness of the velocity boundary layer to the thermal boundary layer.

In the process of condensation, a higher Prandtl number would mean the thermal boundary layer is thinner compared to the velocity boundary layer, which can affect the rate of heat transfer. For the steam condensing on the tube surface, the Prandtl number helps ascertain the effect of thermal properties on the condensation process and, along with the Nusselt and Reynolds numbers, determines the heat transfer rate during the condensation.

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Most popular questions from this chapter

A non-boiling two-phase flow of air and engine oil in a 25 -mm diameter tube has a bulk mean temperature of \(140^{\circ} \mathrm{C}\). If the flow quality is \(2.1 \times 10^{-3}\) and the mass flow rate of the engine oil is \(0.9 \mathrm{~kg} / \mathrm{s}\), determine the mass flow rate of air and the superficial velocities of air and engine oil.

Consider film condensation on a vertical plate. Will the heat flux be higher at the top or at the bottom of the plate? Why?

Saturated steam at \(30^{\circ} \mathrm{C}\) condenses on the outside of a 4-cm- outer-diameter, 2-m-long vertical tube. The temperature of the tube is maintained at \(20^{\circ} \mathrm{C}\) by the cooling water. Determine \((a)\) the rate of heat transfer from the steam to the cooling water, \((b)\) the rate of condensation of steam, and \((c)\) the approximate thickness of the liquid film at the bottom of the tube. 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?

Water is boiled at atmospheric pressure by a horizontal polished copper heating element of diameter \(D=0.5\) in and emissivity \(\varepsilon=0.05\) immersed in water. If the surface temperature of the heating element is \(788^{\circ} \mathrm{F}\), determine the rate of heat transfer to the water per unit length of the heating element.

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

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