A This is a RETROFIT problem All of the design information needed is given U

A this is a retrofit problem all of the design

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A. This is a RETROFIT problem. All of the design information needed is given (U and area) B. We must use the ε -NTU method here. C h = m h c ph = 2.2 kg/s * 4.18 kJ/kg-K = 9.196 kW/K C c = m c c pc = 1.475 kg/s * 3.35 kJ/kg-K = 4.94 kW/K The cold fluid (solvent) has the lower heat capacity rate, so it will heat more than the hot fluid (water) cools. q max = c min (T h,in -T c,in ) = 4.94 kW/K * (95 – 28 o C) = 331 kW Need ε to find q. A o = N P * N T * pi * d o * L NTU = UA/c min = U o A o /C min = 475 W/m2-K * 4 * 88 * pi * (0.0072 m) * 1.72 m / (4940 W/K) NTU = 1.317 Need ε from NTU. There is an equation in Table 11.4, but it’s much easier to read off of Figure 11.13 (matches: 2 shells, 4 tube passes) C r = C min /C max = 4.94/9.196 = 0.537 Reading from Figure 11.13, ε is approximately 0.61 (somewhere in that area 0.59 to 0.63 is a reasonable estimate) Thus, q = 0.61 * q max = 0.61 * 331 kW = 201.9 kW q = dU/dt = m c c pc (T co – T ci ) 201900 W = 4940 (T co – 28 C) T co = 68.9 C (67 – 70, depending on effectiveness value) C. If saturated steam at 95 o C were used instead of liquid water, the heat capacity ratio C min /C max would drop to zero since C max for steam is infinity. NTU would not change, since U, A and C min do not change compared to part B. So using Figure 11.13, with C r = 0, and NTU = 1.317, the effectiveness would increase to approximately 0.73 This can also be easily calculated from Table 11.3- the bottom row has the equation for all heat exchangers where C r = 0 (i.e., when one fluid is changing phases). Here, ε = 1 – exp (- NTU) = 1 – exp (- 1.317) = 0.732 (about the same as we got by reading the Figure). So, using steam at 95 C would cause q = 0.732 * qmax = 0.732 * 331 kW = 242.3 kW 242300 W = 4940 (T co – 28 C) T co = 77.0 o C. This is somewhat higher than using liquid water. The main difference here is that the fluid temperature on the shell side retains a higher temperature throughout the exchanger, giving a higher dTlm driving force.
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7. Open heart surgery is normally done in hypothermic conditions, where the patient’s blood is cooled before the surgery and re-warmed afterward. A proposed concentric tube counterflow heat exchanger is designed to have a length of 0.50 m and inside tube diameter of 55 mm (thin-walled), and has an overall heat transfer coefficient, U, of 500 W/m 2 -K. Given that the blood enters the heat exchanger at 18 o C and 0.050 kg/s, and water enters the annulus at 60 o C at 0.10 kg/s: A. What is the temperature of the blood exiting the exchanger? B. How long would the exchanger need to be to achieve a blood outlet temperature of 37 o C? Note: You may assume the mean temperature of water to be 55 o C (for determining fluid properties ONLY), and that the heat capacity of the blood is 3500 J/kg-K. Part A is a RETROFIT problem, since neither outlet temperature is known. Solve using ε -NTU. Part B is a DESIGN problem, so you can use either ε -NTU or LMTD to solve.
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8. A large shell and tube heat exchanger has U = 640 W/m 2 -K. There are 200 thin-walled tubes with diameter 2.7 cm and 5.5 m length. They make 4 total passes through the exchanger. There are two shell passes.
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