q k A dTdx 237 Wm K 128 m 2 35 50012 m q 758 10 7 W or 758 MW from outside to

Q k a dtdx 237 wm k 128 m 2 35 50012 m q 758 10 7 w

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q = -k A dT/dx = 237 W/m-K *(128 m 2 ) * (35 -5)/(0.012 m) q = 7.58 * 10 7 W, or 75.8 MW from outside to inside the reactor. B. q” = q/A = 75.8 MW / 128 m 2 q” = 592,000 W/m 2
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2. How thick would the materials listed below need to be for the same heat transfer rate as a 6.5 cm thick common brick: (assume same surface temperatures, and look up k at 300 K) A. hardboard siding B. cement plaster (sand aggregate) C. tin D. cork board Note: use Appendix A.1-A.3 to find k values. You may assume a temperature of 300 K for looking up the properties. This involves looking up k values again. Brick, k = 0.72 W/m-K Hardboard siding k = 0.094 W/m-K cement plaster k = 0.72 tin k = 66.6 cork board, k = 0.039 Starting with q = -k A (T2-T1)/dx, in this problem q is held constant, as is the area, T2, and T1, and we are to find new values for dx based on different k values. dx and k must be directly proportionate, so a smaller k means that a smaller thickness of the material is needed; conversely, if the material is more conducting, that material must be thicker. So kA/kbrick = dxA/dxbrick, of dxA (the thickness of material A) = kA/kbrick * dxbrick A. Hardboard siding: dx = (0.094 / 0.72) * 6.5 cm = 0.84 cm (thinner!) B. cement plaster (k is the same!) = 6.5 cm (same thickness) C. tin = (66.6 / 0.72) * 6.5 cm = 601 cm (over 6 meters!) D. cork board = 0.039 / 0.72 * 6.5 cm = 0.35 cm (only 3.5 millimeters! that’s because it is highly insulating!)
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3. Pressurized water at 68 o C flows inside a 7.5-cm inner diameter, 10 m long cylindrical tube with the inside surface temperature maintained at 30 o C. The convective heat transfer coefficient between the water and the tube surface is 814 W/m 2 - o C. A. Find the convective heat transfer coefficient in metric (W/m 2 -K)and Engineering (English) units (Btu/h-ft 2 - o F). B. What is the heat transfer rate, q, from the tube to the water (in kW), and heat flux from the tube to the water at the inside surface (in kW/m 2 )? C. Does the water exit the tube at 68 o C? If so, explain where the thermal energy input from the tube ended up. If not, estimate the new exit temperature for a flow rate of 1.7 kg/s. D. Based on your answer to part C, would you expect the local heat flux near the entrance to the tube to be _the same as_, _greater than_, or _less than_ the local heat flux at the exit of the tube? Why? (assume h = constant) Note: There is a conversion table on the back inside cover of the book to help with English/SI conversions. Water properties are in Table A.6. The subscript f stands for liquid, and g for gas. Using the handy table in the back of your textbook (inside back cover), you will find that the conversion is 1 W/m 2 -K (which is the same as W/m 2 - o C) = 0.17611 Btu/h-ft 2 - o F, so h = 814 W/m 2 - o C = 143 Btu/h-ft 2 - o F B. What is the heat transfer rate, q, from the tube to the water (in kW), and heat flux from the tube to the water at the inside surface (in kW/m 2 )? (surface area of a cylinder = pi * D * L... you’ll need to know this for 306) q = h A (Ts - Tinf) = 814 W/m 2 -C * 3.14159 * 0.075 m * 10 m * (30 - 68 o C) * 1 kW/1000W q = -72.9 kW (note, here the negative means the heat is transferred from the water to the tube) q” = q / A = -30.9 kW/m 2 C. Does the water exit the tube at 80 o C? If not, estimate the new exit temperature for a flow rate of 1.7 kg/s.
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