Question

Inside a condenser, there is a bank of seven copper tubes with cooling water flowing in them. Steam condenses at a rate of \(0.6 \mathrm{~kg} / \mathrm{s}\) on the outer surfaces of the tubes that are at a constant temperature of \(68^{\circ} \mathrm{C}\). Each copper tube is \(5-\mathrm{m}\) long and has an inner diameter of \(25 \mathrm{~mm}\). Cooling water enters each tube at \(5^{\circ} \mathrm{C}\) and exits at \(60^{\circ} \mathrm{C}\). Determine the average heat transfer coefficient of the cooling water flowing inside each tube and the cooling water mean velocity needed to achieve the indicated heat transfer rate in the condenser.

Short answer

#tag_title#Step 2: Calculate the surface area of the tubes#tag_content#We are given the dimensions of the tubes: length \(L = 5 \mathrm{~m}\) and outer diameter \(D_o = 20 \mathrm{~mm}\). There are 1500 tubes, so the total surface area of the tubes can be calculated as: \(A = N_{tubes} \pi D_o L\) where \(N_{tubes}\): Number of tubes \(A = (1500) (\pi) (0.02 \mathrm{~m}) (5 \mathrm{~m}) \approx 471.24 \mathrm{~m^2}\) #tag_title#Step 3: Find the average temperature difference#tag_content#We are given the mean temperature of the cooling water going in and out of the tubes: \(T_i = 20^\circ \mathrm{C}\) and \(T_o = 35^\circ \mathrm{C}\). We can calculate the average temperature difference by finding the difference between the temperature of the tubes (\(T_t = 50^\circ \mathrm{C}\)) and the average temperature of the cooling water (\(\frac{T_i + T_o}{2}\)): \(\Delta T = T_t - \frac{T_i + T_o}{2}\) \(\Delta T = 50^\circ \mathrm{C} - \frac{20^\circ \mathrm{C} + 35^\circ \mathrm{C}}{2} = 22.5^\circ \mathrm{C}\) #tag_title#Step 4: Calculate the average heat transfer coefficient#tag_content#Now that we have the heat transfer rate, surface area of the tubes, and average temperature difference, we can calculate the average heat transfer coefficient: \(h = \frac{Q}{A \Delta T}\) \(h = \frac{1356 \mathrm{~kJ/s}}{(471.24 \mathrm{~m^2})(22.5 \mathrm{~K})} \approx 12.16 \mathrm{~W/(m^2 K)}\) #tag_title#Step 5: Calculate the cooling water flow rate#tag_content#We can now use the heat transfer rate equation to find the cooling water flow rate: \(\dot{m} = \frac{Q}{c_p \Delta T_{water}}\) where \(c_p\) for water is approximately \(4.18 \mathrm{~kJ/(kg K)}\), and \(\Delta T_{water} = 15^\circ \mathrm{C}\): \(\dot{m} = \frac{1356 \mathrm{~kJ/s}}{(4.18 \mathrm{~kJ/(kg K)})(15 \mathrm{~K})} \approx 21.74 \mathrm{~kg/s}\) #tag_title#Step 6: Calculate the cooling water mean velocity#tag_content#Finally, we can calculate the cooling water mean velocity by dividing the flow rate by the total cross-sectional area of the tubes: \(v = \frac{\dot{m}}{N_{tubes} \pi \frac{D_i^2}{4}}\) where \(D_i\) is the inner diameter of the tubes, which can be calculated as \(D_i = D_o - 2t\), where \(t = 2 \mathrm{~mm}\) is the thickness of the tube. This gives us \(D_i = 16 \mathrm{~mm}\). \(v = \frac{21.74 \mathrm{~kg/s}}{(1500) (\pi) (\frac{0.016 \mathrm{~m}}{2})^2} \approx 2.16 \mathrm{~m/s}\) Therefore, the average heat transfer coefficient of the cooling water flowing inside each tube is approximately \(12.16 \mathrm{~W/(m^2 K)}\), and the cooling water mean velocity needed to achieve the given heat transfer rate in the condenser is approximately \(2.16 \mathrm{~m/s}\).

Step-by-step solution

Find the heat transfer rate

First, we need to find the total heat transfer rate from the steam condensing on the outer surfaces of the tubes. We're given the steam condensation rate as \(0.6 \mathrm{~kg/s}\), and the latent heat of steam is approximately \(2260 \mathrm{~kJ/kg}\). Therefore, the total heat transfer rate is: \(Q = \dot{m}_{steam} L_{steam}\) where \(\dot{m}_{steam}\): Steam condensation rate \(L_{steam}\): Latent heat of steam \(Q = (0.6 \mathrm{~kg/s}) (2260 \mathrm{~kJ/kg}) = 1356 \mathrm{~kJ/s}\)

Key concepts

Condenser

A condenser is a device commonly used in industrial and residential settings to condense steam, often back into water. By doing so, it allows for recovery of the latent heat of vaporization, which is crucial for efficient thermal energy use. In the context of this problem, the condenser consists of copper tubes through which cooling water flows. As steam comes into contact with the cooler surface of these tubes, it transfers heat, causing the steam to condense.

Condensers are pivotal in various applications such as power plants, refrigerators, and air conditioning systems. They play an essential role in increasing the efficiency of a thermal cycle by reclaiming and reusing the energy from the steam. By converting steam back into liquid form, condensers reduce energy waste and improve overall system performance.

Copper Tubes

Copper tubes are widely used in heat exchange applications due to their excellent thermal conductivity. In this exercise, they form the backbone of the condenser structure. The copper material allows efficient transfer of heat from the steam to the cooling water within the tubes.

Each copper tube here is designed with a length of 5 meters and an inner diameter of 25 mm. Such dimensions are crucial because they influence the heat transfer rate and the flow characteristics of the cooling water.

Using copper tubes in condensers ensures that they can handle substantial thermal loads while minimizing potential losses associated with lower conductivity materials. This makes copper an ideal choice in situations where heat must be efficiently transferred from one medium to another.

Latent Heat

Latent heat is the amount of heat required to change the state of a substance without changing its temperature. In this problem, we focus on the latent heat of steam, which is the heat needed to transform steam into water.

Given as approximately 2260 kJ/kg, latent heat is a critical concept in thermodynamics. It describes the energy exchange involved in phase changes, such as the steam condensation occurring within the condenser.

The understanding of latent heat is essential when calculating the heat transfer rate, as it determines the total amount of energy released during condensation. In the exercise, knowing this value helps to compute the heat transfer rate, which is necessary to maintain the condenser’s efficient operation.

Cooling Water Flow

The flow of cooling water through the copper tubes is fundamental to the functionality of the condenser. It facilitates the removal of heat from the condensing steam, aiding in the transformation back to a liquid state.

In this scenario, the cooling water enters the tubes at a low temperature of 5°C and exits at a higher temperature of 60°C. This process indicates the movement and absorption of energy, as cooler water absorbs heat from the steam and carries it away.

Determining the average heat transfer coefficient and the necessary mean velocity of the cooling water are key to ensuring optimal heat removal. By understanding these parameters, we can design a system that effectively dissipates heat, maintaining desired operational and safety standards.