Question

A shell-and-tube heat exchanger is to be designed to cool down the petroleum- based organic vapor available at a flow rate of \(5 \mathrm{~kg} / \mathrm{s}\) and at a saturation temperature of \(75^{\circ} \mathrm{C}\). The cold water \(\left(c_{p}=4187 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) used for its condensation is supplied at a rate of \(25 \mathrm{~kg} / \mathrm{s}\) and a temperature of \(15^{\circ} \mathrm{C}\). The cold water flows through copper tubes with an outside diameter of \(20 \mathrm{~mm}\), a thickness of \(2 \mathrm{~mm}\), and a length of \(5 \mathrm{~m}\). The overall heat transfer coefficient is assumed to be \(550 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and the latent heat of vaporization of the organic vapor may be taken to be \(580 \mathrm{~kJ} / \mathrm{kg}\). Assuming negligible thermal resistance due to pipe wall thickness, determine the number of tubes required.

Short answer

Answer: 17 tubes are required for the shell-and-tube heat exchanger.

Step-by-step solution

Calculate the heat transfer needed for condensation

First, we need to calculate the amount of heat transfer required to condense the organic vapor. We'll use the flow rate and latent heat of vaporization values for this calculation: $$Q = \dot{m}_{vapor} \cdot L_{vapor}$$ where: \(Q\) - Heat transfer (W) \(\dot{m}_{vapor}\) - Flow rate of organic vapor (5 kg/s) \(L_{vapor}\) - Latent heat of vaporization of the organic vapor (580 kJ/kg) $$Q = 5 \frac{kg}{s} \cdot 580000 \frac{J}{kg} = 2900000 \mathrm{~W}$$

Calculate the heat transfer rate on the water side

Next, we need to determine the heat transfer rate on the water side. We'll use the flow rate, specific heat capacity, and temperature difference values for this calculation: $$\dot{Q} = \dot{m}_{water} \cdot c_{p} \cdot \Delta T$$ where: \(\dot{Q}\) - Heat transfer rate (W) \(\dot{m}_{water}\) - Flow rate of cold water (25 kg/s) \(c_{p}\) - Specific heat capacity of water (4187 J/kg·K) \(\Delta T\) - Temperature difference (75°C - 15°C = 60°C) $$\dot{Q} = 25 \frac{kg}{s} \cdot 4187 \frac{J}{kg \cdot K} \cdot 60 K = 6304500 \mathrm{~W}$$

Calculate the heat transfer coefficient for the system

Now, we need to determine the heat transfer coefficient for the system. The overall heat transfer coefficient was given in the problem statement: $$U = 550 \frac{W}{m^{2} \cdot K}$$

Calculate the convective heat transfer surface area

We need to calculate the convective heat transfer surface area for the pipes with the given diameter and length: $$A = N \cdot \pi d \cdot L$$ where: \(A\) - Convective heat transfer surface area (m²) \(N\) - Number of tubes \(d\) - Diameter of the tubes (20 mm = 0.02 m) \(L\) - Length of the tubes (5 m)

Calculate the number of tubes needed

The heat transfer equation, assuming negligible thermal resistance due to pipe wall thickness, is: $$Q = U \cdot A \cdot \Delta T$$ Rearrange the equation to find the number of tubes: $$N = \frac{Q}{\pi d \cdot L \cdot U \cdot \Delta T}$$ Now substitute the values: $$N = \frac{2900000 \mathrm{~W}}{\pi (0.02 \mathrm{~m}) (5 \mathrm{~m}) (550 \frac{W}{m^{2} \cdot K}) (60 \mathrm{~K})} = 16.65$$ Since we can't have a fraction of a tube, we'll round up to the next whole number: $$N = 17$$ There are 17 tubes required for the shell-and-tube heat exchanger.

Key concepts

Shell-and-Tube Heat Exchanger

A shell-and-tube heat exchanger is a type of heat exchanger widely used in various industrial applications. It employs a collection of tubes housed within a shell. One fluid flows through the tubes, and a different fluid passes around the tubes within the shell, allowing heat exchange without the fluids mixing. This design is particularly efficient for large volume heat transfer.
In practice, shell-and-tube heat exchangers are favored for their robust structure and effectiveness in handling high-pressure fluids. They come in various configurations, allowing specific designs to be applied according to the heat transfer requirements of a particular process. This versatility explains why they are commonly used in industries such as oil refining, chemical processing, and power generation.
Such systems can have several passes on either the shell side or tube side. This flexibility in design allows for optimizing heat transfer and maintaining fluid flow pressures.

Heat Transfer Calculations

When designing a heat exchanger, precise heat transfer calculations are critical to ensuring the system's efficiency and effectiveness. The fundamental equation used in these calculations is based on the principle of energy conservation. For the system in the exercise, two main equations were derived to represent the heat transfer in both the condensation of organic vapor and the heating of water.

• For the vapor condensation: \[ Q = \dot{m}_{vapor} \times L_{vapor} \]where \( Q \) represents the heat transfer needed, \( \dot{m}_{vapor} \) the mass flow rate, and \( L_{vapor} \) the latent heat.


• For the heating of water:\[ \dot{Q} = \dot{m}_{water} \times c_{p} \times \Delta T \]This uses the mass flow rate of the water, its specific heat capacity \( c_{p} \), and the temperature difference \( \Delta T \).

The balance between these two equations ensures that the heat exchanger performs optimally, with both sides exchanging the necessary heat at rates that are manageable given the system constraints.

Thermodynamics Education

Understanding thermodynamics is fundamental to learning how heat exchangers work. This branch of physics focuses on energy transfer, particularly heat, and its effects on matter. In the context of a heat exchanger, the main focus is on the laws of thermodynamics that govern energy interactions.
Key concepts include the first law of thermodynamics, which deals with the conservation of energy, and how it applies to both closed and open systems. For heat exchangers, this implies that the energy lost by one fluid must be gained by the other, barring any heat losses to the environment. The second law, which introduces the principle of entropy, helps to understand why these processes aren't perfectly reversible and always involve some loss of usable energy.
By integrating these principles, students can appreciate how systems like heat exchangers operate effectively within the bounds of natural laws, leading to more informed and efficient designs.

Engineering Problem Solving

Engineering problem-solving involves applying scientific and mathematical principles to design effective solutions for complex real-world issues. With heat exchangers, this means not only performing calculations but also optimally considering materials, fluid dynamics, and system configuration.
In the given exercise, solving for the number of tubes required an understanding of how shell-and-tube exchangers function and involved multiple steps such as calculating the heat to be transferred, evaluating the water's ability to absorb heat, and applying the overall heat transfer coefficient. This structured approach showcases how engineers break down a larger problem into fundamental parts.

• Identify the problem: Define the parameters such as the type of exchanger needed, the fluids involved, and their respective properties.
• Calculate heat transfer needs: Use heat transfer equations to establish thermal requirements.
• Determine system requirements: Calculate physical specifications for effective heat transfer, such as the number of tubes.

Such methodical problem-solving requires understanding all aspects involved and an ability to apply theoretical knowledge to practical applications, which is crucial in engineering design processes.