Chapter 13: Problem 4
What is the crossed-strings method? For what kinds of geometries is the crossed-strings method applicable?
Chapter 13: Problem 4
What is the crossed-strings method? For what kinds of geometries is the crossed-strings method applicable?
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Get started for freeA furnace is shaped like a long semicylindrical duct of diameter $D=15 \mathrm{ft}\(. The base and the dome of the furnace have emissivities of \)0.5$ and \(0.9\) and are maintained at uniform temperatures of 550 and $1800 \mathrm{R}$, respectively. Determine the net rate of radiation heat transfer from the dome to the base surface per unit length during steady operation.
Two square plates, with the sides \(a\) and \(b\) (and \(b>a\) ), are coaxial and parallel to each other, as shown in Fig. P13-142, and they are separated by a center-to-center distance of \(L\). The radiation view factor from the smaller to the larger plate, \(F_{a b+}\) is given by $$ F_{a b}=\frac{1}{2 A}\left\\{\left[(B+A)^{2}+4\right]^{0.5}-\left[(B-A)^{2}+4\right]^{0.5}\right\\} $$ where, \(A=a / L\) and \(B=b / L\). (a) Calculate the view factors \(F_{a b}\) and \(F_{b a}\) for $a=20 \mathrm{~cm}\(, \)b=60 \mathrm{~cm}\(, and \)L=40 \mathrm{~cm}$. (b) Calculate the net rate of radiation heat exchange between the two plates described above if \(T_{a}=800^{\circ} \mathrm{C}\), $T_{b}=200^{\circ} \mathrm{C}, \varepsilon_{a}=0.8\(, and \)\varepsilon_{b}=0.4$. (c) A large, square plate (with the side $c=2.0 \mathrm{~m}, \varepsilon_{c}=0.1$, and negligible thickness) is inserted symmetrically between the two plates such that it is parallel to and equidistant from them. For the data given above, calculate the temperature of this third plate when steady operating conditions are established.
Consider an enclosure consisting of eight surfaces. How many view factors does this geometry involve? How many of these view factors can be determined by the application of the reciprocity and the summation rules?
Two parallel disks of diameter \(D=0.6 \mathrm{~m}\) separated by $L=0.4 \mathrm{~m}$ are located directly on top of each other. Both disks are black and are maintained at a temperature of \(450 \mathrm{~K}\). The back sides of the disks are insulated, and the environment that the disks are in can be considered to be blackbodies at \(300 \mathrm{~K}\). Determine the net rate of radiation heat transfer from the disks to the environment. Answer: $781 \mathrm{~W}$
The convection heat transfer coefficient for a clothed person while walking in still air at a velocity of \(0.5\) to \(2 \mathrm{~m} / \mathrm{s}\) is given by \(h=8.6 V^{0.53}\), where \(V\) is in \(\mathrm{m} / \mathrm{s}\) and \(h\) is in \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Plot the convection coefficient against the walking velocity, and compare the convection coefficients in that range to the average radiation coefficient of about $5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$.
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