Question

Air enters a 25-cm-diameter 12 -m-long underwater duct at \(50^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) at a mean velocity of \(7 \mathrm{~m} / \mathrm{s}\), and is cooled by the water outside. If the average heat transfer coefficient is \(85 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and the tube temperature is nearly equal to the water temperature of \(10^{\circ} \mathrm{C}\), determine the exit temperature of air and the rate of heat transfer. Evaluate air properties at a bulk mean temperature of \(30^{\circ} \mathrm{C}\). Is this a good assumption?

Short answer

Answer: The exit temperature of the air is approximately \(20^{\circ}\mathrm{C}\), and the rate of heat transfer is approximately \(-32016\ \mathrm{W}\).

Step-by-step solution

Calculate the surface area of the duct

The surface area of the duct is essential in determining the total heat transfer. First, we need to find the surface area of the duct \((A)\) using the formula: \(A = \pi D L\), where \(D\) is the duct diameter and \(L\) is the length of the duct. Given, \(D = 0.25 \ \mathrm{m}\) and \(L = 12 \ \mathrm{m}\), we can calculate the surface area as: \(A = \pi (0.25\ \mathrm{m}) (12\ \mathrm{m}) = 9.42\ \mathrm{m^2}\)

Calculate the mass flow rate of air

To calculate the exit temperature of the air, we need to find the mass flow rate of air \((\dot{m})\). We can use the formula: \(\dot{m} = \rho V A\), where \(\rho\) is the density of air, \(V\) is the mean velocity, and \(A\) is the cross-sectional area of the duct. First, we need to find the cross-sectional area: \(A_c = \frac{\pi D^2}{4}\) Given, \(D = 0.25 \ \mathrm{m}\), the cross-sectional area is: \(A_c = \frac{\pi (0.25\ \mathrm{m})^2}{4} = 0.0491\ \mathrm{m^2}\) At \(30^{\circ}\mathrm{C}\) and \(1\ \mathrm{atm}\), the density of air is approximately \(1.164\ \mathrm{kg/m^3}\), and the mean velocity is \(7\ \mathrm{m/s}\). We can now calculate the mass flow rate: \(\dot{m} = (1.164\ \mathrm{kg/m^3})(7\ \mathrm{m/s})(0.0491\ \mathrm{m^2}) = 0.399\ \mathrm{kg/s}\)

Calculate the heat transfer

Using the heat transfer coefficient \((h)\) and the temperature difference between the water and air, we can calculate the heat transfer \((\dot{Q})\): \(\dot{Q} = hA(T_w - T_i)\) Given, \(h = 85\ \mathrm{W/m^2·K}\), \(T_w = 10^{\circ}\mathrm{C}\), and \(T_i = 50^{\circ}\mathrm{C}\), we can find the heat transfer: \(\dot{Q} = (85\ \mathrm{W/m^2·K})(9.42\ \mathrm{m^2})(10^{\circ}\mathrm{C} - 50^{\circ}\mathrm{C}) = -32016\ \mathrm{W}\) Since the heat transfer is negative, it confirms that the air is losing heat to the water.

Calculate the exit temperature of the air

We can find the exit temperature of the air \((T_e)\) using the formula: \(\dot{Q} = \dot{m}c_p(T_e - T_i)\), where \(c_p\) is the specific heat capacity of air at constant pressure. At a bulk mean temperature of \(30^{\circ}\mathrm{C}\), the specific heat capacity of air is approximately \(1005\ \mathrm{J/kg·K}\). We can rearrange the formula to find \(T_e\): \(T_e = T_i + \frac{\dot{Q}}{\dot{m}c_p}\) Given, \(T_i = 50^{\circ}\mathrm{C}\), \(\dot{Q} = -32016\ \mathrm{W} = -32016\ \mathrm{J/s}\), \(\dot{m} = 0.399\ \mathrm{kg/s}\), and \(c_p = 1005\ \mathrm{J/kg·K}\), we can find the exit temperature: \(T_e = 50^{\circ}\mathrm{C} + \frac{-32016\ \mathrm{J/s}}{(0.399\ \mathrm{kg/s})(1005\ \mathrm{J/kg·K})} \approx 20^{\circ}\mathrm{C}\)

Verify the assumption

Our initial assumption was that the air properties could be evaluated at a bulk mean temperature of \(30^{\circ}\mathrm{C}\). Since the exit temperature of the air is approximately \(20^{\circ}\mathrm{C}\) and the initial temperature is \(50^{\circ}\mathrm{C}\), the bulk mean temperature is: \(T_{bulk} = \frac{T_i + T_e}{2} = \frac{50^{\circ}\mathrm{C} + 20^{\circ}\mathrm{C}}{2} = 35^{\circ}\mathrm{C}\) As the difference between the assumed bulk mean temperature and the actual bulk mean temperature is only \(5^{\circ}\mathrm{C}\), this assumption can be considered reasonable. In summary, the exit temperature of the air is approximately \(20^{\circ}\mathrm{C}\), and the rate of heat transfer is approximately \(-32016\ \mathrm{W}\). The assumption of air properties being evaluated at \(30^{\circ}\mathrm{C}\) is reasonable.

Key concepts

Understanding Heat Transfer Coefficient

The heat transfer coefficient is a measure of a material's ability to conduct heat. In the context of our exercise, it's used to quantify how effectively the underwater duct can transfer heat between the air inside and the surrounding water.

The given coefficient is 85 W/m²·K, signifying that for every square meter of the duct's surface and for every degree of temperature difference between the air and water, 85 watts of heat are transferred. Higher values indicate more efficient heat transfer. Knowing this coefficient allows us to calculate the rate of heat transfer using the formula:
\[\begin{equation}\dot{Q} = hA(T_w - T_i),\end{equation}\]where \(\dot{Q}\) is the heat transfer per unit time, h is the heat transfer coefficient, A is the duct surface area, \(T_w\) is the water temperature, and \(T_i\) is the initial air temperature.

Calculating Mass Flow Rate

Mass flow rate is a critical concept in fluid mechanics and thermodynamics, representing the amount of mass passing through a given surface per unit time. It's calculated with the formula:
\[\begin{equation}\dot{m} = \rho V A_c,\end{equation}\]where \(\dot{m}\) is the mass flow rate, \(\rho\) is the density of the air, V is the velocity of the air, and \(A_c\) is the cross-sectional area of the duct.

In the exercise, we used this concept to determine how much air is cooled as it passes through the duct. It's important to get this step right because an incorrect mass flow rate will lead to errors in the calculation of the exit temperature and the total heat transfer.

Exit Temperature of Air

The exit temperature of the air is the final temperature of the air after it has passed through the duct and has been subject to heat transfer. In our problem, we use the following formula to determine the exit temperature (\(T_e\)):
\[\begin{equation}T_e = T_i + \frac{\dot{Q}}{\dot{m}c_p},\end{equation}\]where \(T_i\) is the initial air temperature, \(\dot{Q}\) is the rate of heat transfer calculated earlier, \(\dot{m}\) is the mass flow rate, and c_p is the specific heat capacity of the air.

This value is crucial for systems involving thermal regulation, like heating and cooling ducts, because it indicates whether the ducting system effectively brings the air to the desired temperature.