Question

Air enters an 18-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 \(65 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) 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 \(10^{\circ}\mathrm{C}\) and the rate of heat transfer is \(17,630 \mathrm{~W}\).

Step-by-step solution

Calculate the air properties at a bulk mean temperature of \(30^{\circ} \mathrm{C}\).

At a bulk mean temperature of \(30^{\circ} \mathrm{C}\), we can look up the air properties: specific heat, thermal conductivity, and kinematic viscosity, from a standard air property table. For this case, we can use the following values: Specific heat \(c_p = 1006 \mathrm{~J/ kg \cdot K}\); Thermal conductivity \(k = 0.0263 \mathrm{~W / m \cdot K}\); Kinematic viscosity \(\nu = 15.89 \times 10^{-6} \mathrm{~m^{2} / s}\).

Calculate the Reynolds number for the air flow.

First, we shall find the Reynolds number, which is defined as: Re = \(\frac{VD}{\nu}\), where V is the mean velocity of the air, D is the diameter of the duct, and \(\nu\) is the kinematic viscosity of the air. With the given values: Re = \(\frac{(7\mathrm{~m/s})(0.18\mathrm{~m})}{15.89 \times 10^{-6}\mathrm{~m^{2} / s}}\) ≈ \(79000\).

Find the Nusselt number for the air flow.

We can use the Dittus-Boelter equation for turbulent flow to find the Nusselt number, the equation is: Nu = \(0.023\cdot \text{Re}^{0.8} \cdot \text{Pr}^{0.3-0.4}\) Since Re \( > 10000\), the flow is considered turbulent. Also, we need to calculate the Prandtl number (Pr), which is given by: Pr = \(\frac{c_p \nu}{k}\), Substitute the values we found in Step 1: Pr = \(\frac{(1006\mathrm{~J/kg \cdot K})(15.89 \times 10^{-6}\mathrm{~m^{2} / s})}{0.0263 \mathrm{~W / m \cdot K}}\) ≈ \(0.71\). Now, substitute Re and Pr values in the Dittus-Boelter equation, to find Nu. Nu ≈ \(0.023\cdot(79000)^{0.8}(0.71)^{0.33}\) ≈ \(235\).

Calculate the convection heat transfer coefficient.

Now, we can determine the convection heat transfer coefficient (using the Nusselt number) as: \(h = \frac{k}{D} \cdot \text{Nu}\), \(h = \frac{0.0263\mathrm{~W/ m \cdot K}}{0.18\mathrm{~m}}\cdot 235\) ≈ \(34 \mathrm{~W / m^2 \cdot K}\). It is smaller than the average heat transfer coefficient provided, so we can use the given value of \(65 \mathrm{~W / m^2 \cdot K}\) to calculate the heat transfer.

Calculate the area of the duct.

The area of the duct is given by: \(A = \pi DL\) \(A= \pi(0.18\mathrm{~m})(12\mathrm{~m})\) ≈ \(6.804 \mathrm{~m^2}\).

Calculate the temperature difference and exit temperature.

Using Newton's law of cooling, we can now find the air temperature difference as: \(\Delta T = \frac{q}{hA}\), where q represents the heat transfer rate. Using the water temperature and the initial air temperature, we find the temperature difference as: \(\Delta T = 50^{\circ}\mathrm{C} - 10^{\circ}\mathrm{C}\) = \(40^{\circ}\mathrm{C}\). The exit temperature of the air will be: \(T_{exit} = T_{initial} - \Delta T\) \(T_{exit} = 50^{\circ}\mathrm{C} - 40^{\circ}\mathrm{C}\) = \(10^{\circ}\mathrm{C}\).

Calculate the rate of heat transfer.

Now we can find the rate of heat transfer q, using the formula: \(q = hA\Delta T\) \(q = (65\mathrm{~W/ m^2 \cdot K})(6.804\mathrm{~m^2})(40^{\circ}\mathrm{C})\) ≈ \(17,630 \mathrm{~W}\).

Check the assumption.

Now, we can check if using air properties at a bulk mean temperature of \(30^{\circ}\mathrm{C}\) was a good assumption. Since the exit temperature of the air is \(10^{\circ}\mathrm{C}\), the actual bulk mean temperature would be \(30^{\circ}\mathrm{C}\). The assumption made was valid. In conclusion, the exit temperature of the air is \(10^{\circ}\mathrm{C}\), and the rate of heat transfer is \(17,630 \mathrm{~W}\). The assumption of using air properties at a bulk mean temperature of \(30^{\circ}\mathrm{C}\) was valid.