D The heat exchanger is operated for a few years so that the lake water has

D the heat exchanger is operated for a few years so

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D. The heat exchanger is operated for a few years so that the lake water has generated a fouling factor R f,i ” = 0.00070 m 2 -K/W. Knowing that the thermal resistance of the tubes can be ignored and that there is no fouling on the glycerin side, determine the three thermal resistance terms in the heat exchanger (in K/W), and the value of U o . E. Compare q and dU/dt to determine if the heat exchanger is big enough to achieve the internal energy changes for the two fluids? (or is it oversized or undersized?) Notes: Use the LMTD method to find q. You will need to refer to Chapter 9 in this problem. The heat capacity of liquid water changes very little over the entire temperature range for water; use a reasonable guess for T mean to look up c p,L . A. B. Free convection over horizontal tubes. Nu D = CRa D n where the value of Ra yields C = 0.850 and n = 0.188, then Nu = 2.924 = hD/kf Look up kf at Tmean = 310 K. kf = 0.286 W/m-K h = 2.924 * 0.286 W/m-K / 0.0107 m = 78.1 W/m2-K C. dU/dt for glycerin = m cp (Tout – Tin) = 0.82 kg/s * 2490 J/kg-K * (-10 K) = -20400 W must guess cp for water since Tout is unknown = 4180 J/kg-K (very little change with temperature) dU dt for water = 20400 W = m cp (Tout – Tin) = 1.33 kg/s * 4180 J/kg-K * (Tout – 295K) Tout = 298.7 K D. UA = 1/sum R = 1 / {1/h i A i + R fi /A i + 1/h o A o } A i = N T N P pi * Di * L = 54 * 2 * pi * 0.0092 m * 4.6 m = 14.36 m 2 A o = N T N P pi * Di * L = 54 * 2 * pi * 0.0107 m * 4.6 m = 16.70m 2 UA = 1/ {1/(630 W/m 2 -K * 14.36 m 2 + .0007 m 2 K/W/14.36 m 2 + 1/(78.1 W/m 2 -K)(16.7 m 2 )} = 1/ {1.105 * 10 -4 + 4.87 * 10 -5 + 8.1 * 10 -4 }
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R water convection = 1.105 * 10 -4 K/W R water fouling = 4.87 * 10 -5 K/W R glycerin convection = 7.66 * 10 -4 K/W UA = U o A o = 1080 W/K U o = 64.7 W/m2-K E. q = UAF dTlm,CF dTlm,CF = (315 – 298.7) – (305 – 295) / ln((315 – 298.7) – (305 – 295)) = 12.9 K F comes from the handout in class. eta h = (298.7 – 295) / (315 – 295) = 0.185 Z = (315 – 305)/(298.7 – 295) = 2.7 From the figure for 1 shell, 2 tube passes, F is approximately 0.96 (difficult to read, around 0.95- 0.97) Thus, q = 1080 W/K * 0.96 * 12.9 K = 13,370 W Compared with the answer we found in part B where dU/dt is 20,400, the heat exchanger is too small to achieve the desired outlet temperature.
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6. A shell and tube heat exchanger is used to heat a solvent stream flowing at 1.475 kg/s through 88 steel tubes. The solvent has a heat capacity of 3.35 kJ/kg-K and an inlet temperature of 28 o C. Hot water enters the shell (which has a total of four tube passes inside two shell passes) at 95 o C with a mass flow rate of 2.20 kg/s and heat capacity of 4.18 kJ/kg-K. The tubes are 1.72 m long, and have inner and outer diameters of d i = 0.55 cm and d o = 0.72 cm. The outside overall heat transfer coefficient for the heat exchanger, U o , is 475 W/m 2 -K. A. Is this a RETROFIT ( ε -NTU) or DESIGN (LMTD) problem? B. What is the temperature of the solvent stream when it exits the heat exchanger? C. What would the outlet temperature of the solvent be if saturated steam at 95 o C were used in place of liquid water? (Note: the heat capacity rate for a fluid changing phases is infinity; also, saturated steam at 95 o C simply means that the steam is under a slight vacuum pressure).
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