Question #169269

An organic liquid leaves a reactor at 70ºC and it is necessary to reduce its temperature to 45ºC before the next unit operation. A shell-and-tube heat exchanger is to be used, and the organic liquid will flow on the shell side. The exchanger has 110 tubes, and the coolant in the tubes will be water entering at 15ºC. The tubes have a single pass – i.e. the water is divided across all tubes equally and flows just once through the exchanger. The total flow rate of water is of 75,000 kg h^{−1} , and the water exits the exchanger at 28 ºC. Using the Information Sheet where appropriate, determine:

a) the flow rate of the process liquid (in kg s−1 ), [5 marks]

b) the log mean temperature difference if the streams are in countercurrent flow, [5 marks]

c) the tube-side film heat transfer coefficient if the tubes have an inside diameter of 14.8 mm, [10 marks]

d) the heat transfer area if the overall heat transfer coefficient is 1100 W m−2K−1 . [5 marks]

INFORMATION

Specific heat capacity of process liquid 2200 J kg^{-1}K^{-1}

Specific heat capacity of water 4190 J kg^{−1}K^{-1}

Viscosity of water 0.758 × 10^{−3} N s m^{−2 }

Thermal conductivity of water 0.617 W m^{−1}K^{−1}

Expert's answer

The rate of heat transfer is determined from

Q = m_{c}c_{pc}(T_{c,out} −T_{c,in} ) = (20.83 kg/s)(4.19 kJ/kg°C)(28°C − 15°C) = 1134.6 kW

c_{pc} = 75000 kg/h x 1h/3600 s = 20.83 kg/s

The mass flow rate of the process liquid is

m_{h} = Q /c_{ph}(T_{h,out} −T_{h,in} ) = 1134.6 kW /(2.20 kJ/kg°C)(70°C − 45°C) = 20.63 kg/s

The logarithmic mean temperature difference for counterflow arrangement is

ΔT_{1} = T_{h, in} - T_{c, out} = 70°C − 45°C = 25°C

ΔT_{2} = T_{h, out} - T_{c, in} = 28°C − 15°C = 13°C

ΔT_{lm, CF} = (ΔT_{1} - ΔT_{2}) / ln(ΔT_{1}/ΔT_{2}) = (25 - 13) / ln(25/13) = 18.35°C

A = Q/UΔT_{lm, CF} = 1134600 W / (1100 W/m^{2}K)(18.35 + 273.15 K) = 3.54 m^{2}

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