Question

Jakob (1949) suggests the following correlation be used for square tubes in a liquid crossflow situation: $$ \mathrm{Nu}=0.102 \mathrm{Re}^{0.675} \mathrm{Pr}^{1 / 3} $$ Water $(k=0.61 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=6)\( at \)50^{\circ} \mathrm{C}$ flows across a 1 -cm-square tube with a Reynolds number of 10,000 and surface temperature of $75^{\circ} \mathrm{C}\(. If the tube is \)3 \mathrm{~m}$ long, the rate of heat transfer between the tube and water is (a) \(9.8 \mathrm{~kW}\) (b) \(12.4 \mathrm{~kW}\) (c) \(17.0 \mathrm{~kW}\) (d) \(19.6 \mathrm{~kW}\) (e) \(24.0 \mathrm{~kW}\)

Short answer

Question: In a liquid crossflow situation for square tubes, find the rate of heat transfer between the tube and the water using the given correlation: Nu = 0.102 Re^0.675 Pr^(1/3). The Reynolds number, Re, is 10000, the Prandtl number, Pr, is 6, the thermal conductivity (k) of water is 0.61 W/mK, the surface temperature of the tube is 75°C, the water temperature is 50°C, the length of the square tube is 3 m, and the diameter of the tube is 1 cm. Answer: The rate of heat transfer is approximately 18.3 kW.

Step-by-step solution

Calculate the Nusselt number

Using the correlation, we can find Nu as follows: $$ \mathrm{Nu}=0.102 \mathrm{Re}^{0.675} \mathrm{Pr}^{1 / 3} = 0.102 (10000)^{0.675}(6)^{1/3} $$ Compute the value of Nu: $$ \mathrm{Nu} \approx 100 $$

Calculate the heat transfer coefficient

Using the Nusselt number, we can calculate the heat transfer coefficient (h) with: $$ h = \frac{\mathrm{Nu} \cdot k}{d} $$ We know k = 0.61 W/mK, Nu = 100, and the diameter d = 0.01m (1 cm). Compute the value of h: $$ h = \frac{100 \cdot 0.61}{0.01} = 6100 \, \mathrm{W} / \mathrm{m}^{2} \mathrm{K} $$

Calculate the surface area

Calculate the surface area (A) of the square tube using the length (3 m) and the perimeter (4 sides of 1 cm): $$ A = 4 \cdot ( 0.01 \, \mathrm{m} ) \cdot ( 3 \, \mathrm{m} ) = 0.12 \, \mathrm{m}^{2} $$

Use Newton's law of cooling to find the rate of heat transfer

Finally, we use Newton's law of cooling, which states that the rate of heat transfer (Q) is proportional to the temperature difference, the surface area, and the heat transfer coefficient: $$ Q = h \cdot A \cdot \Delta T $$ We are given the surface temperature (75°C) and the water temperature (50°C), so the temperature difference is: $$ \Delta T = 75 - 50 = 25 \, \mathrm{K} $$ Compute the value of Q: $$ Q = 6100 \, \mathrm{W} / \mathrm{m}^{2} \mathrm{K} \cdot 0.12 \, \mathrm{m}^{2} \cdot 25 \, \mathrm{K} = 18300 \, \mathrm{W} = 18.3 \, \mathrm{kW} $$ The rate of heat transfer is approximately 18.3 kW, which is closest to choice (d), 19.6 kW. Therefore, the answer is (d).