Question

A concentric annulus tube has inner and outer diameters of \(25 \mathrm{~mm}\) and \(100 \mathrm{~mm}\), respectively. Liquid water flows at a mass flow rate of \(0.05 \mathrm{~kg} / \mathrm{s}\) through the annulus with the inlet and outlet mean temperatures of \(20^{\circ} \mathrm{C}\) and \(80^{\circ} \mathrm{C}\), respectively. The inner tube wall is maintained with a constant surface temperature of \(120^{\circ} \mathrm{C}\), while the outer tube surface is insulated. Determine the length of the concentric annulus tube. Assume flow is fully developed.

Short answer

In this exercise, we determined the length of a concentric annulus tube through which liquid water flows. We were given the inner and outer diameters of the tubes, mass flow rate of water, and inlet and outlet mean temperatures of the liquid. Using principles of heat transfer, we found the heat transfer rate, annulus area, heat flux, thermal resistance of the liquid, and the overall heat transfer coefficient. Finally, we calculated the length of the concentric annulus tube to be approximately 0.383 meters.

Step-by-step solution

Find the heat transfer rate between the liquid and the inner tube.

As the outer tube surface is insulated and the flow is fully developed, we can say that all heat transfer occurs between liquid water and the inner tube, and we can use the simple energy balance to find the heat transfer rate \(Q\), using the mass flow rate \(m\) and the specific heat capacity \(c_p\) of water: $$Q = m\cdot c_p \cdot (T_{out} - T_{in})$$ We know the mass flow rate, as well as inlet and outlet temperatures, but we need the specific heat capacity of water. It is approximately \(4.18\mathrm{~kJ/(kg\cdot K)}\) or \(4180\mathrm{~J/(kg\cdot K)}\) at near room and at moderate temperatures.

Compute the heat transfer rate

Now we can calculate the heat transfer rate using the given data: $$Q = 0.05\,\mathrm{kg/s} \cdot 4180\,\mathrm{J/(kg\cdot K)} \cdot (80-20)\,\mathrm{K} = 12540\,\mathrm{W}$$

Determine the annulus area.

Next, we need to find the annulus area, \(A_{annulus}\), where the heat transfer occurs between the liquid and the inner tube wall. This area can be calculated by subtracting the area of the inner tube from the area of the outer tube: $$A_{annulus} = \pi D_o D_i$$ where \(D_o\) is the outer diameter of the annulus and \(D_i\) is the inner diameter of the annulus. Given values are in millimeters, so we should convert them to meters: $$D_i = 25\,\mathrm{mm} = 0.025\,\mathrm{m}$$ $$D_o = 100\,\mathrm{mm} = 0.1\,\mathrm{m}$$ Now we can calculate the annulus area: $$A_{annulus} = \pi \cdot 0.1\,\mathrm{m} \cdot 0.025\,\mathrm{m} = 0.007854\,\mathrm{m^2}$$

Calculate the heat flux.

We will use the heat transfer rate that we found in Step 2, together with the annulus area found in Step 3, to calculate the heat flux, \(q^{\prime\prime}\): $$q^{\prime\prime} = \frac{Q}{A_{annulus}}$$ $$q^{\prime\prime} = \frac{12540\,\mathrm{W}}{0.007854\,\mathrm{m^2}} = 1595643\,\mathrm{W/m^2}$$

Calculate the thermal resistance of the liquid.

Now we need to calculate the thermal resistance, \(R_{th}\), of the liquid flowing through the annulus. This can be found using the temperature difference between the inner tube wall and the inlet temperature of the liquid, and the heat flux: $$R_{th} = \frac{T_{wall} - T_{in}}{q^{\prime\prime}}$$ $$R_{th} = \frac{120\,\mathrm{^\circ C} - 20\,\mathrm{^\circ C}}{1595643\,\mathrm{W/m^2}} = 6.273 \times 10^{-5}\,\mathrm{m^2K/W}$$

Determine the overall heat transfer coefficient, U.

The overall heat transfer coefficient, \(U\), can be related to the thermal resistance of the annulus using this equation: $$R_{th} = \frac{1}{U \cdot A_{annulus}}$$ Now we can solve for \(U\): $$U = \frac{1}{R_{th} \cdot A_{annulus}}$$ $$U = \frac{1}{6.273 \times 10^{-5}\,\mathrm{m^2K/W} \cdot 0.007854\,\mathrm{m^2}} = 2025.4\,\mathrm{W/(m^2K)}$$

Calculate the length of the concentric annulus tube.

Finally, we can find the length of the concentric annulus tube, \(L\), using the heat transfer rate, the overall heat transfer coefficient, and the temperature difference between the inner tube wall and the inlet temperature of the liquid: $$Q = U \cdot A_{annulus} \cdot L \cdot (T_{wall} - T_{in})$$ $$L = \frac{Q}{U \cdot A_{annulus} \cdot (T_{wall} - T_{in})}$$ $$L = \frac{12540\,\mathrm{W}}{2025.4\,\mathrm{W/(m^2K)} \cdot 0.007854\,\mathrm{m^2} \cdot (120\,\mathrm{^\circ C} - 20\,\mathrm{^\circ C})} = 0.383\,\mathrm{m}$$ The length of the concentric annulus tube is approximately \(0.383\,\mathrm{m}\).

Key concepts

Concentric Annulus

In heat transfer problems involving pipes, a concentric annulus refers to an arrangement where there is an inner tube and an outer tube. The fluid flows through the space or gap, known as the annular region, between these two tubes. It plays a crucial role in various thermal applications, as it allows for effective heat exchange between the fluid and the tube walls. In this setup:

• The inner diameter corresponds to the diameter of the inner tube.
• The outer diameter corresponds to the diameter of the outer tube.
• The annular area is the region available for fluid flow and heat transfer.

Calculating key parameters like the annulus area involves simple geometric formulas, subtracting the inner circle's area from the larger circle's area in the case of cylinders. Understanding the physical layout of a concentric annulus is fundamental as it impacts the distribution of heat and the overall efficiency of the heat exchange process.

Thermal Resistance

Thermal resistance is a critical concept in analyzing heat transfer within a system. It quantifies the resistance of a material or a flow path to heat flow. In the context of fluid flowing through a concentric annulus, the thermal resistance is influenced by:

• The temperature difference between the inner tube's wall and the fluid.
• The heat flux, which is the rate of heat transfer per unit area.

Thermal resistance is given by the formula:\[ R_{th} = \frac{T_{wall} - T_{in}}{q''} \]where \(T_{wall}\) is the wall temperature, \(T_{in}\) is the inlet fluid temperature, and \(q''\) is the heat flux. A low thermal resistance suggests efficient heat transfer. Conversely, a high thermal resistance indicates that the material or configuration resists heat flow, leading to less efficient heat transfer.

Overall Heat Transfer Coefficient

The overall heat transfer coefficient, often denoted as \(U\), combines all modes of heat transfer (conduction, convection) into a single value. This coefficient is crucial in thermal calculations as it provides a measure of how effectively heat is transferred from one medium to another over a certain area.

• It is inversely related to thermal resistance, meaning a low thermal resistance translates to a high overall heat transfer coefficient.
• \(U\) takes into account the material properties and conditions (e.g., fluid velocity, temperature) of both the inner and outer surfaces of the heat transfer system.

The formula to find the overall heat transfer coefficient is:\[ U = \frac{1}{R_{th} \cdot A_{annulus}} \]where \(R_{th}\) is the thermal resistance and \(A_{annulus}\) is the area through which the heat is being transferred. A high value of \(U\) indicates that the system is efficient in transferring heat.

Heat Flux Calculation

Heat flux is the rate of heat energy transfer per unit area, commonly expressed in \( \mathrm{W/m^2} \). In concentric annulus systems, calculating the heat flux is a key step in understanding how heat moves from the heated inner wall to the cooler fluid.To calculate heat flux \(q''\), we use the formula:\[ q'' = \frac{Q}{A_{annulus}} \]where:

• \(Q\) is the total heat transfer rate.
• \(A_{annulus}\) is the annulus area, the space between the inner and the outer tube.

This calculation shows how spread out the heat transfer is across the area. It helps in assessing the efficiency and safety of the heat transfer system, as high heat flux could indicate potential overheating or material failure. Keeping heat flux at optimal levels is essential for the longevity and performance of any system involving heat transfer.