Help with Heat Transfer Paper?

Help! Heat transfer problem?

• You have to calculate and construct a tube-in-tube heat exchanger. The inner tube for your design is 20 mm ID (inner diameter) and has 2.0 mm thick walls. The tube is made from stainless steel with a thermal conductivity of 17 W•m-1•K-1. Calculate the inner diameter for the outer tube if you want the cross sectional areas of both flow channels to be identical. You intend to use the heat exchanger in counter-current flow, and you will pass equal amounts of water through both sides. The hot water enters at 70 oC, and the cold water at 20 oC. Assume that the heat capacity and the density are independent of temperature. In this case, the temperature change in the ‘hot’ stream is equal (but opposite in sign) to the temperature change in the ‘cold’ stream, and the temperature difference along the heat exchanger is always constant. Calculate the Reynolds number for the inside and outside flow. The linear velocity of the water in the exchanger is in both cases 0.3 m/s. Take other properties of water at 45 oC. Use the Dittus-Boelter formula to calculate the heat transfer coefficients hi and ho and the overall heat transfer coefficient U. Again, take all property values at 45 oC. At which side do you get the higher heat transfer coefficient h? Explain!

• Answer:

  A1 = A2 pi*di^2/4 = pi*Di^2/4 - pi*(di+2t)^2/4 di^2 = Di^2 - (di+2t)^2 20^2 = Di^2 - (24)^2 Di = 31.241 mm Thi = 70 deg. C Tci = 20 deg. C Energy balance, m_dot*c*(Thi - Tho) = m_dot*c*(Tco-Tci) (Thi - Tho) = (Tco-Tci) Reynolds number, Re = ρVD/μ Solve for diameter. Use hydraulic diameter definition, dh = 4A/p, p = wetted perimeter dh for inside pipe dh = 4*(pi*di^2/4) / pi*D dh = di Dh for outside pipe Evaluate properties at T = 45 deg. C. Dh = 4A/p Dh = 4*(pi*Di^2/4 - pi*(di+2t)^2/4) / [pi*Di + pi*(di+2t)], Do = di+2t Dh = (Di^2 - Do^2) / [Di + Do] Dh = Di - Do Reynolds for both cases can be solved, evaluate properties at T = 45 deg. C Dittus-Boelter formula, Nu_D = 0.023*Re^(4/5)*Pr^n n = 0.4 for heating and n = 0.3 for cooling Find coefficient of convection using Nusselt definition, Nu_D = hD/k h = NU_d*k / D Again use diameter, D, as Hydraulic diameter, Dh(outer pipe) and dh (inside pipe). U = 1 / [(1/hi + di/2k*ln(Do/di) + 1/ho)]