To design the heat exchanger, we first need to calculate the heat duty Q required to cool the liquid from 80C to 40C. We will assume that the specific heat capacity Cp of the liquid is constant and equal to 4 kJ/kgK. Q = mass flow rate * Cp * TWhere:Q = heat duty kW mass flow rate = 5 kg/min 0.0833 kg/s Cp = 4 kJ/kgK 4000 J/kgK T = 80 - 40 C = 40CQ = 0.0833 kg/s * 4000 J/kgK * 40 K = 13,328 W 13.328 kW Next, we need to calculate the overall heat transfer coefficient U using the heat transfer coefficients of the liquid and water. We can use the formula:1/U = 1/h1 + 1/h2Where:U = overall heat transfer coefficient W/mK h1 = heat transfer coefficient of the liquid 1000 W/mK h2 = heat transfer coefficient of the water 5000 W/mK 1/U = 1/1000 + 1/5000U = 833.33 W/mKNow, we can calculate the required heat transfer area A using the formula:A = Q / U * Tm Where:A = heat transfer area m Tm = log mean temperature difference LMTD For the LMTD, we need to determine the temperature difference at both ends of the heat exchanger. Assuming the water enters the heat exchanger at 20C and exits at 60C, we have:T1 = 80 - 20 C = 60CT2 = 40 - 60 C = -20CTm = T1 - T2 / ln T1/T2 = 60 - -20 / ln 60/ -20 = 34.9CNow we can calculate the required heat transfer area:A = 13,328 W / 833.33 W/mK * 34.9 K = 0.453 mSince the maximum allowable area is 20 m, the required area of 0.453 m is well within the limit.For the optimal flow configuration, a counterflow configuration would be more efficient than a parallel flow configuration, as it provides a higher temperature gradient between the two fluids throughout the heat exchanger. This results in better heat transfer and a smaller heat exchanger surface area requirement.In summary, the required surface area of the heat exchanger is 0.453 m, and the optimal flow configuration is counterflow.