You have been assigned to design a heat exchanger to cool a gaseous process stream in a chemical plant. The stream (20 kg/s) enters at 431 K, has a heat capacity of 3.45 J/gC and needs to be cooled to 402 K. Cooling water is available at 85F and has a specified maximum temperature of 120F. The over all heat transfer coefficient is approximately 570 W/m^2K. How much area must the heat exchanger have?
In every second, the gas stream must lose (29K)(20000 g)(3.45 J/gK) = 2001 kJ. The maximum water temperature is 322K, so the temperature difference is always at least 80K. (2001 kW)/[(570 W/m^2K)*(80K)] = 43.9 m^2 should be more than sufficient, as the temperature difference near the entry point would be greater than 80K.
energy transferred = mass of hot gas * specific heat * temperature change calculate the log mean temperature difference for the heat exchanger Q = U * A * dTlogmean Substitute and find the area.
Hello Fatima : Get the heat exchanger duty: ---------------------------------------... Let H indicate the hot gaseous process stream. QH = ( mdotH ) ( CpH ) ( TH1 - TH2 ) QH = ( 20 kg/s ) ( ( 1000 g / 1 kg ) ( 3.45 J / g - C ) ( 431 K - 402 K ) QH = 2.001 x 10^6 J QHX = Heat Exchanger Duty = QH = QCW Assume counter-rcurrent flow and get the log mean temperature difference ( LMTD ) : ---------------------------------------... TH1 = 431 K = 157.8 C TCW2 = 120 F = 48.9 C TH2 = 402 K = 128.8 C TCW1 = 85 F = 29.4 C Delta T sub II = TH1 - TCW2 = ( 157.8 C - 48.9 C ) = 108.9 C Delta T sub I = TH2 - TCW1 = ( 128.8 C - 29.4 C ) = 99.4 C LMTD = ( 108.9 C - 99.4 C ) / ln ( 108.9 / 99.4 ) = 104.1 C Now apply the Heat Exchanger Design Equation : ---------------------------------------... QHX = ( U ) ( Ahx ) ( LMTD ) Ahx = ( QHX ) / ( U ) ( LMTD ) Ahx = ( 2.001 x 10^6 J ) / ( 570 J/s - sq m - C ) ( 104.1 C ) Ahx = required heat transfer area for the heat exchanger = 34 sq m <------------------------------------
