# Quiz3s - of energy equation to solve for it 29 29 kg/s...

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Fall 2007 1 ME 416 Computer Assisted Design of Thermal Systems Quiz #3 Solution A concentric tube (double pipe), counterflow heat exchanger is used to heat kerosene flowing at 0.03 kg/s from 260 K to 330 K with air whose temperature changes from 380 K to 350 K. The kerosene has a c P of 2500 J/(kg K) and the air has a c P of 1000 J/(kg K). The overall heat transfer coefficient is 15 W/(m 2 K). Determine the required mass flow rate of air and the surface area of the heat exchanger. Solution We recognize this as a sizing problem, so we will proceed with that methodology 1. Our c P 's are provided in the problem statement. Note that kerosene is the cold fluid and air is the hot fluid. 2. We see that we do not know the air mass flow rate, so we will use our conservation
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Unformatted text preview: of energy equation to solve for it. ( 29 ( 29 kg/s 0.175 350)-(1000)(380 260)-0)(330 (0.03)(250 T-T c T-T ) c m ( = m out H, in H, H P, in c, out C, C P H = = & & 3. min C P, C C C W/K 75 0) (0.03)(250 c m C = = = = & max H P, H H C W/K 175 00) (0.175)(10 c m C = = = = & 0.429 175 75 C C = C max min R = = 4. U = 15 W/(m 2 ⋅ K) 5. For the effectiveness, we calculate ( 29 ( 29 ( 29 ( 29 0.583 260-380 260-330 T-T T-T = T-T C T-T C = in C, in H, in C, out C, in C, in H, min in C, out C, C = = ε 6. Our NTU relationship is 1.028 1-29) 0.583)(0.4 ( 1-83 .5 ln 1-0.429 1 1-C 1-ln 1-C 1 = NTU R R = = ε ε 7. Our surface area is then 2 min m 5.14 15 ) (75)(1.228 U NTU C = A = =...
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