Question

Water \(\left(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) enters a 12 -cm-diameter and \(8.5-\mathrm{m}\)-long tube at \(75^{\circ} \mathrm{C}\) at a rate of \(0.35 \mathrm{~kg} / \mathrm{s}\), and is cooled by a refrigerant evaporating outside at \(-10^{\circ} \mathrm{C}\). If the average heat transfer coefficient on the inner surface is \(500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), the exit temperature of water is (a) \(18.4^{\circ} \mathrm{C}\) (b) \(25.0^{\circ} \mathrm{C}\) (c) \(33.8^{\circ} \mathrm{C}\) (d) \(46.5^{\circ} \mathrm{C}\) (e) \(60.2^{\circ} \mathrm{C}\)

Short answer

Answer: (d) 46.5°C

Step-by-step solution

Calculate the surface area of the tube

The surface area A of a tube is given by \(A = 2\pi rL\), where r is the radius and L is the length of the tube. Given the diameter of \(12\text{ cm}\), we have a radius \(r=6\text{ cm}=0.06\text{ m}\). And the length of the tube \(L=8.5\text{ m}\). Now we can calculate the surface area of the tube: \(A = 2\pi(0.06)(8.5) = 3.198\mathrm{~m}^2\).

Calculate the rate of heat transfer

The rate of heat transfer \(q\) is given by \(q= hA \Delta T\), where \(h = 500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) is the average heat transfer coefficient, and \(\Delta T = (75-(-10)) = 85\mathrm{~K}\) is the temperature difference. Now we can calculate the rate of heat transfer: \(q = (500)(3.198)(85) = 135950 \mathrm{~W}\).

Calculate the heat loss by the water to find the exit temperature

The heat loss by the water equals the rate of heat transfer and is given by the formula: \(q=mc_p\Delta T_{exit}\), where \(m = 0.35 \mathrm{~kg} / \mathrm{s}\) is the mass flow rate, \(c_p=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) is the specific heat of water, and \(\Delta T_{exit}\) is the temperature difference between the entrance and exit of the water. Now we can solve for \(\Delta T_{exit}\): \(\Delta T_{exit} = \frac{q}{mc_p} = \frac{135950}{(0.35)(4180)}= 29.2\)°C$.

Determine the exit temperature of water

To find the exit temperature of the water, subtract the temperature difference, \(\Delta T_{exit}\), from the entrance temperature: \(T_{exit}=T_{entrance} - \Delta T_{exit} = 75°\text{C} - 29.2°\text{C} = 45.8°\text{C}\). Now, we need to find which option from the given choices matches with our calculated exit temperature. Comparing our answer to the given choices, the closest option is: (d) \(46.5^{\circ} \mathrm{C}\). Therefore, the exit temperature of the water is approximately \(46.5^{\circ} \mathrm{C}\).

Key concepts

Convective Heat Transfer

Convective heat transfer is an essential concept when analyzing how heat moves through fluids such as water or air. This process is driven by the movement of the fluid itself—when warmer parts of a fluid flow move to colder regions, heat is transferred. This can occur naturally, as in the case of buoyancy-driven movement, or be assisted by external forces like pumps or fans, which is known as forced convection.

In the exercise, water flows through a pipe and is subject to cooling by a refrigerant. The heat transfer coefficient, represented by the symbol 'h', reflects the efficiency of heat being transferred from the water inside the tube to the surrounding refrigerant. A higher 'h' value means that the fluid is effectively losing heat. The specific formula used here, \(q= hA \Delta T\), encapsulates the core idea of convective heat transfer by relating the heat transfer rate (q), the heat transfer coefficient (h), the contact area (A), and the temperature difference (\(\Delta T\)) between the tube's water and the refrigerant.

Thermal Analysis

In the realm of physics and engineering, thermal analysis is a branch that deals with how heat moves within different materials and across interfaces. It incorporates assessing temperature variations, heat transfer rates, and energy change within a system to predict how temperature changes over time.

To carry out a thermal analysis, one must understand the properties of the materials involved, such as specific heat capacity, which is a measure of how much energy is needed to raise the temperature of a unit mass of a substance by one degree. In the given exercise, the water's specific heat capacity \(c_p\) plays a pivotal role in determining how much heat is lost, which in turn affects the water exit temperature. After calculating the rate of heat transfer using convective heat transfer principles, we apply the concept to quantify the water's heat loss through the formula \(q = mc_p \Delta T_{exit}\). This approach is a classic example of applying thermal analysis to solve real-world problems.

Heat Exchanger Design

Heat exchangers are devices designed to efficiently transfer heat from one medium to another. They are widely used in various industries such as power generation, chemical processing, and HVAC systems. Design considerations include the type of heat exchanger, the materials used, surface area for heat transfer, and the overall thermal performance.

The exercise demonstrates a simple heat exchanger where water transfers its heat to a refrigerant through the pipe's wall. The effectiveness of this process depends on the pipe's dimensions (which influence the surface area available for heat transfer), the properties of the flowing fluid, and operational conditions such as mass flow rate and inlet temperature. The solution process involves determining the exit temperature, it implicitly touches upon the critical aspects of heat exchanger design—calculating the necessary surface area and understanding how the flow rate and temperature difference drive the thermal performance of the system.