Question

at Your objective in this problem is to design a "simple" annular flow heat exchanger (see the "simple" description the beginning of this HW). A heat exchanger of this kind can easily be built in your garage with two different size pper pipes along with some solder. Assume both fluids are water with the cold fluid flowing in the outer tube. The pipe diameters are based in Type L copper tube dimensions: outside diameter 1/2 -inch nominal tube is 0.625 inch, with a wall thickness of 0.04 inches and the outside diameter of t thickness of 0.05 inches. The mass flow-rate of water through both tubes should be based on 10 ftu/s inlet temperature to the cold side is 40°F and the inlet to the hot side is 90°F. What length of h required to achieve a hot-side outlet temperature of 60°F if the overall heat transfer coefficier exchanger is 400 Btu/hr ft2°F. Also, calculate the log mean temperature difference and effective not have to calculate convective heat transfer coefficients because the value of U is given. he one-inch nominal tube is 1.125 inches with a wall of water eat exchanger is e heat ness. (NOTE: you do

Answer 1

T_h1 = 90 degree F rightarrow inlet of hot fluid T_h0 = 60 degree F rightarrow outlet of hot fluid T_c1 = 40 degree F rightarrow inlet of cold fluid T_c0 = ? u = 400 B + 4/hr ft^2 degree F rightarrow overall heat transfer m = 10 ft/sec times 60 sec/hr = 600 ft/hr q = UA (LMTD) = mc_p (T_hi - T - h0) = mc_c (T_c0 - T_ci) downdoublearrow Energy balance mc_p (T_hi - T - h0) = mc_c (T_c0 - T_ci) specific heat capacities are equal so it is a Balanced counter flaw heat exchanger \r\n T_hi - T_h0 = T_c0 - T_ci L = length of heat exchanger T_c0 = 30 + 40 degree = 70 degree F LMTD = (increment T)_i = (increment T)_0 = (T_hi - T_c0) = (T_h0 - T_ci) = 20 degree F For length of a tube, the heat has to be from innertube innerside to outerside. Q = UA_i LMTD = mc_p (T_hi - T_h0) = 400 times pi D_i L times LMTD