Question

Problem 3 (40pts) A shell and tube heat exchanger is being designed for an application of cooling hot oil. When new, the heat exchanger has an overall heat transfer coefficient, U = 300 W/m2- K. The cooling water is available at 4.5 kg/sec and 20°C. (Cpoil = 2.3 kJ/kg-K & Cpwater = 4.18 kJ/kg-K) Based on experience with this water supply, the heat exchanger will experience fouling due to mineral deposits in the water. The fouling factor is estimated to be, R =.0004 m2 K/W. Without fouling, the heat exchanger surface area was computed to be 30 m². In order to still achieve the same cooling load for the oil, estimate the new surface area with fouling present. Useful equations: 0 = UA LMTD R = ün where R = thermal resistance The thermal resistance increases with fouling. Ryout Rnew + L Where Riou is thermal resistance after fouling, Rnew is thermal resistance for new condition (ie, before fouling), and R is fouling factor, R = .0004 m2 KW

Answer 1

we know ä= Ux Asx LMTD LMTD = 300x30 (for new ease on without fouling) Now after fouling the overall heat transfer Co-efficient and surface abea (As) will be Changed but cooling load (à) and LMTD will be same as new ease.. So, for the fouled tube, us = ť + RA Zoo + 0.0004 229/W Uf = 267.86 m² K/w 300 x 30 2 267.86 x Ag & al Asp = 33.6 so new surface abea will be 33.6 m².