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Design a heat exchanger for a specific process that involves heating a liquid from 20°C to 60°C. The liquid has a flow rate of 500 L/hr, and it will be heated by steam at a pressure of 3 bar. The heat transfer coefficient for the steam is 3000 W/(m²·K), and the overall heat transfer coefficient for the heat exchanger is 500 W/(m²·K). Determine the surface area requ

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To design the heat exchanger, we first need to determine the amount of heat required to heat the liquid from 20C to 60C. We can use the formula:Q = m * Cp * Twhere Q is the heat required, m is the mass flow rate of the liquid, Cp is the specific heat capacity of the liquid, and T is the temperature difference.Assuming the liquid is water, the specific heat capacity  Cp  is approximately 4.18 kJ/ kgK . We also need to convert the flow rate from L/hr to kg/s:Flow rate = 500 L/hr *  1 m/1000 L  *  1000 kg/m  *  1 hr/3600 s  = 0.139 kg/sNow we can calculate the heat required:Q = 0.139 kg/s * 4.18 kJ/ kgK  *  60C - 20C  = 23.2 kWNext, we can determine the required surface area of the heat exchanger using the overall heat transfer coefficient  U  and the temperature difference between the steam and the liquid  T :A = Q /  U * T Assuming the steam temperature is constant at the saturation temperature corresponding to 3 bar pressure, which is approximately 134C, the temperature difference between the steam and the liquid is:T = 134C -  60C + 20C /2 = 89CNow we can calculate the required surface area:A = 23.2 kW /  0.5 kW/ mK  * 89 K  = 1.04 mFinally, we can determine the flow rate of steam required to heat the liquid. We can use the heat transfer coefficient for the steam  h  and the surface area  A  to calculate the steam flow rate:Q = m_steam * h * A * TRearranging the equation to solve for the steam flow rate:m_steam = Q /  h * A * T m_steam = 23.2 kW /  3 kW/ mK  * 1.04 m * 89 K  = 0.092 kg/sIn conclusion, the required surface area for the heat exchanger is 1.04 m, and the flow rate of steam required to heat the liquid is 0.092 kg/s.

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