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Chapter 1: Problem 136

A cylindrical fuel rod of \(2 \mathrm{~cm}\) in diameter is encased in a concentric tube and cooled by water. The fuel generates heat uniformly at a rate of \(150 \mathrm{MW} / \mathrm{m}^{3}\). The convection heat transfer coefficient on the fuel rod is \(5000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and the average temperature of the cooling water, sufficiently far from the fuel rod, is \(70^{\circ} \mathrm{C}\). Determine the surface temperature of the fuel rod and discuss whether the value of the given convection heat transfer coefficient on the fuel rod is reasonable.

Short Answer

Expert verified
Answer: The formula used to find the surface temperature of the fuel rod is: $$ T_{surface} = \frac{150\cdot 10^6 \mathrm{~W}/\mathrm{m}^3}{5000 \mathrm{~W}/\mathrm{m}^2 \cdot \mathrm{K} \cdot 2 \pi (0.01) h} + 70 $$ where \(T_{surface}\) is the surface temperature of the fuel rod, \(h\) is the height of the fuel rod, and 70 is the average temperature of the cooling water in degrees Celsius.

Step by step solution

01

Calculate the heat generation

The heat generation rate in the fuel rod is given as \(150 \mathrm{MW} / \mathrm{m}^{3}\). We need to convert this to watts per cubic meter: $$ Q_{generation} = 150 \cdot 10^6 \mathrm{~W} / \mathrm{m}^{3}$$
02

Determine the surface area of the fuel rod

The diameter of the fuel rod is given as \(2 \mathrm{cm}\). We can calculate the radius and then find the surface area (A) of the fuel rod using the formula for the surface area of a cylinder: $$ A = 2 \pi r h $$ First, convert the diameter to meters and find the radius: $$ r = \frac{d}{2} = \frac{0.02 \mathrm{m}}{2} = 0.01 \mathrm{m} $$ As the height (h) is not given, we cannot find the numerical value of the surface area. However, we can rewrite the formula in terms of height, which will be useful later. $$ A = 2 \pi (0.01) h $$
03

Calculate the heat transfer rate using convection

We can use Newton's law of cooling to calculate the rate of heat transfer (Q) by convection: $$ Q = hA(T_{surface} - T_{water}) $$ where \(h\) is the convection heat transfer coefficient, which is given as \(5000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), \(T_{surface}\) is the surface temperature of the fuel rod, and \(T_{water}\) is the average temperature of the cooling water, given as \(70^{\circ} \mathrm{C}\).
04

Set up an equation and solve for the surface temperature of the fuel rod

Since the fuel rod is being cooled by convection, the heat generated in the fuel rod should equal the heat transferred by convection: $$ Q_{generation} = Q $$ Replace Q with the convection heat transfer equation: $$ 150 \cdot 10^6 \mathrm{~W} / \mathrm{m}^{3} = 5000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K} \cdot (2 \pi (0.01) h) (T_{surface} - 70) $$ Now solve for \(T_{surface}\): $$ T_{surface} = \frac{150\cdot 10^6 \mathrm{~W}/\mathrm{m}^3}{5000 \mathrm{~W}/\mathrm{m}^2 \cdot \mathrm{K} \cdot 2 \pi (0.01) h} + 70 $$ Since we don't have the height of the fuel rod, we Cannot find the numerical value. However, the above equation can be used to find the surface temperature if the height was given.
05

Discuss the reasonability of the given convection heat transfer coefficient

The given convection heat transfer coefficient is \(5000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). This value is reasonable for water cooling in a high-heat flux situation, such as in nuclear reactors. However, it would also depend on factors such as the flow rate of the water, the roughness of the surfaces of the fuel rod, and the tube, and the presence of any fouling or debris that might affect the heat transfer. While we cannot provide a specific answer without more information, the given value seems reasonable given the context of the problem.

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Most popular questions from this chapter

Consider steady heat transfer between two large parallel plates at constant temperatures of \(T_{1}=290 \mathrm{~K}\) and \(T_{2}=150 \mathrm{~K}\) that are \(L=2 \mathrm{~cm}\) apart. Assuming the surfaces to be black (emissivity \(\varepsilon=1\) ), determine the rate of heat transfer between the plates per unit surface area assuming the gap between the plates is (a) filled with atmospheric air, \((b)\) evacuated, \((c)\) filled with fiberglass insulation, and \((d)\) filled with superinsulation having an apparent thermal conductivity of \(0.00015 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

The inner and outer surfaces of a \(25-\mathrm{cm}\)-thick wall in summer are at \(27^{\circ} \mathrm{C}\) and \(44^{\circ} \mathrm{C}\), respectively. The outer surface of the wall exchanges heat by radiation with surrounding surfaces at \(40^{\circ} \mathrm{C}\), and convection with ambient air also at \(40^{\circ} \mathrm{C}\) with a convection heat transfer coefficient of \(8 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Solar radiation is incident on the surface at a rate of \(150 \mathrm{~W} / \mathrm{m}^{2}\). If both the emissivity and the solar absorptivity of the outer surface are \(0.8\), determine the effective thermal conductivity of the wall.

The deep human body temperature of a healthy person remains constant at \(37^{\circ} \mathrm{C}\) while the temperature and the humidity of the environment change with time. Discuss the heat transfer mechanisms between the human body and the environment both in summer and winter, and explain how a person can keep cooler in summer and warmer in winter.

The roof of a house consists of a 22-cm-thick (st) concrete slab \((k=2 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) that is \(15 \mathrm{~m}\) wide and \(20 \mathrm{~m}\) long. The emissivity of the outer surface of the roof is \(0.9\), and the convection heat transfer coefficient on that surface is estimated to be \(15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The inner surface of the roof is maintained at \(15^{\circ} \mathrm{C}\). On a clear winter night, the ambient air is reported to be at \(10^{\circ} \mathrm{C}\) while the night sky temperature for radiation heat transfer is \(255 \mathrm{~K}\). Considering both radiation and convection heat transfer, determine the outer surface temperature and the rate of heat transfer through the roof. If the house is heated by a furnace burning natural gas with an efficiency of 85 percent, and the unit cost of natural gas is \(\$ 1.20\) / therm ( 1 therm \(=105,500 \mathrm{~kJ}\) of energy content), determine the money lost through the roof that night during a 14-hour period.

A series of experiments were conducted by passing \(40^{\circ} \mathrm{C}\) air over a long \(25 \mathrm{~mm}\) diameter cylinder with an embedded electrical heater. The objective of these experiments was to determine the power per unit length required \((\dot{W} / L)\) to maintain the surface temperature of the cylinder at \(300^{\circ} \mathrm{C}\) for different air velocities \((V)\). The results of these experiments are given in the following table: $$ \begin{array}{lccccc} \hline V(\mathrm{~m} / \mathrm{s}) & 1 & 2 & 4 & 8 & 12 \\ \dot{W} / L(\mathrm{~W} / \mathrm{m}) & 450 & 658 & 983 & 1507 & 1963 \\ \hline \end{array} $$ (a) Assuming a uniform temperature over the cylinder, negligible radiation between the cylinder surface and surroundings, and steady state conditions, determine the convection heat transfer coefficient \((h)\) for each velocity \((V)\). Plot the results in terms of \(h\left(\mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\right)\) vs. \(V(\mathrm{~m} / \mathrm{s})\). Provide a computer generated graph for the display of your results and tabulate the data used for the graph. (b) Assume that the heat transfer coefficient and velocity can be expressed in the form of \(h=C V^{m}\). Determine the values of the constants \(C\) and \(n\) from the results of part (a) by plotting \(h\) vs. \(V\) on log-log coordinates and choosing a \(C\) value that assures a match at \(V=1 \mathrm{~m} / \mathrm{s}\) and then varying \(n\) to get the best fit.
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