Problem 148
Water enters a circular tube whose walls are maintained at constant temperature at a specified flow rate and temperature. For fully developed turbulent flow, the Nusselt number can be determined from $\mathrm{Nu}=0.023 \mathrm{Re}^{0.8} \operatorname{Pr}^{0.4}$. The correct temperature difference to use in Newton's law of cooling in this case is (a) The difference between the inlet and outlet water bulk temperatures. (b) The difference between the inlet water bulk temperature and the tube wall temperature. (c) The log mean temperature difference. (d) The difference between the average water bulk temperature and the tube temperature. (e) None of the above.
Problem 149
Air \(\left(c_{p}=1007 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) enters a \(17-\mathrm{cm}\)-diameter and 4-m-long tube at $65^{\circ} \mathrm{C}\( at a rate of \)0.08 \mathrm{~kg} / \mathrm{s}$ and leaves at \(15^{\circ} \mathrm{C}\). The tube is observed to be nearly isothermal at \(5^{\circ} \mathrm{C}\). The average convection heat transfer coefficient is (a) \(24.5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (b) \(46.2 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (c) \(53.9 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (d) \(67.6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (e) \(90.7 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\)
Problem 150
Water \(\left(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) enters a \(12-\mathrm{cm}\)-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 the 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) \(60.2^{\circ} \mathrm{C}\)
Problem 153
Engine oil flows in a 15 -cm-diameter horizontal tube with a velocity of $1.3 \mathrm{~m} / \mathrm{s}\(, experiencing a pressure drop of \)12 \mathrm{kPa}$. The pumping power requirement to overcome this pressure drop is (a) \(190 \mathrm{~W}\) (b) \(276 \mathrm{~W}\) (c) \(407 \mathrm{~W}\) (d) \(655 \mathrm{~W}\) (e) \(900 \mathrm{~W}\)
Problem 154
Water enters a 5-mm-diameter and 13 -m-long tube at \(15^{\circ} \mathrm{C}\) with a velocity of \(0.3 \mathrm{~m} / \mathrm{s}\) and leaves at $45^{\circ} \mathrm{C}\(. The tube is subjected to a uniform heat flux of \)2000 \mathrm{~W} / \mathrm{m}^{2}$ on its surface. The temperature of the tube surface at the exit is (a) \(48.7^{\circ} \mathrm{C}\) (b) \(49.4^{\circ} \mathrm{C}\) (c) \(51.1^{\circ} \mathrm{C}\) (d) \(53.7^{\circ} \mathrm{C}\) (e) \(55.2^{\circ} \mathrm{C}\) (For water, use $k=0.615 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=5.42, \nu=0.801 \times\( \)10^{-6} \mathrm{~m}^{2} / \mathrm{s}$.)
Problem 155
Water enters a 5-mm-diameter and 13-m-long tube at \(45^{\circ} \mathrm{C}\) with a velocity of \(0.3 \mathrm{~m} / \mathrm{s}\). The tube is maintained at a constant temperature of \(8^{\circ} \mathrm{C}\). The exit temperature of the water is (a) \(4.4^{\circ} \mathrm{C}\) (b) \(8.9^{\circ} \mathrm{C}\) (c) \(10.6^{\circ} \mathrm{C}\) (d) \(12.0^{\circ} \mathrm{C}\) (e) \(14.1^{\circ} \mathrm{C}\) (For water, use $k=0.607 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=6.14, \nu=0.894 \times\( \)10^{-6} \mathrm{~m}^{2} / \mathrm{s}, c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, \rho=997 \mathrm{~kg} / \mathrm{m}^{3}$.)
Problem 157
Air enters a 7-cm-diameter and 4-m-long tube at \(65^{\circ} \mathrm{C}\) and leaves at \(15^{\circ} \mathrm{C}\). The tube is observed to be nearly isothermal at \(5^{\circ} \mathrm{C}\). If the average convection heat transfer coefficient is \(20 \mathrm{~W} / \mathrm{m}^{2},{ }^{\circ} \mathrm{C}\), the rate of heat transfer from the air is (a) \(491 \mathrm{~W}\) (b) \(616 \mathrm{~W}\) (c) \(810 \mathrm{~W}\) (d) \(907 \mathrm{~W}\) (e) \(975 \mathrm{~W}\)
Problem 158
Air \(\left(c_{p}=1000 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) enters a \(16-\mathrm{cm}\)-diameter and \(19-\mathrm{m}\)-long underwater duct at \(50^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) at an average velocity of $7 \mathrm{~m} / \mathrm{s}$ and is cooled by the water outside. If the average heat transfer coefficient is $35 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$ and the tube temperature is nearly equal to the water temperature of \(5^{\circ} \mathrm{C}\), the exit temperature of the air is (a) \(6^{\circ} \mathrm{C}\) (b) \(10^{\circ} \mathrm{C}\) (c) \(18^{\circ} \mathrm{C}\) (d) \(25^{\circ} \mathrm{C}\) (e) \(36^{\circ} \mathrm{C}\)
Problem 159
Water enters a 2-cm-diameter and 3-m-long tube whose walls are maintained at \(100^{\circ} \mathrm{C}\) with a bulk temperature of \(25^{\circ} \mathrm{C}\) and a volume flow rate of \(3 \mathrm{~m}^{3} / \mathrm{h}\). Neglecting the entrance effects and assuming turbulent flow, the Nusselt number can be determined from \(\mathrm{Nu}=0.023 \mathrm{Re}^{0.8} \mathrm{Pr}^{0.4}\). The convection heat transfer coefficient in this case is (a) \(4140 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (b) \(6160 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (c) \(8180 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (d) \(9410 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (e) \(2870 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (For water, use $k=0.610 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=6.0, \mu=9.0 \times\( \)10^{-4} \mathrm{~kg} / \mathrm{m} \cdot \mathrm{s}, \rho=1000 \mathrm{~kg} / \mathrm{m}^{3}$.)
Problem 160
Air at \(110^{\circ} \mathrm{C}\) enters an \(18-\mathrm{cm}\)-diameter and \(9-\mathrm{m}\)-long duct at a velocity of \(4.5 \mathrm{~m} / \mathrm{s}\). The duct is observed to be nearly isothermal at \(85^{\circ} \mathrm{C}\). The rate of heat loss from the air in the duct is (a) \(760 \mathrm{~W}\) (b) \(890 \mathrm{~W}\) (c) \(1210 \mathrm{~W}\) (d) \(1370 \mathrm{~W}\) (e) \(1400 \mathrm{~W}\) (For air, use $k=0.03095 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=0.7111, \nu=2.306 \times\( \)10^{-5} \mathrm{~m}^{2} / \mathrm{s}, c_{p}=1009 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}$.)
