Process Dynamic and Control The overhead vapor from a distillation column is totally condensed in a water-cooled condenser at 120°F and220 psig, and the condensed stream is sent to a reflux drum at the same temperature and pressure, The vapordesign flow rate is 10,000 lbm/hr, and the avcrage latent heat of vaporization is 125 Btu/lbm. The cooling waterinlet and outlet temperatures are 80°F and I 10°F, respectively.(a) Calculate the cooling water flow rate in gpm at the design conditions. Assume the density of water is62.4 Ib/fn' and its specific heat is 1.0 Btu/lbm"F. Also, 7.48 gal of liquid = 1 ft' of liquid.(Note - If you cannot calculate a cooling water flow rate in part (a), assume the cooling water flow rate is 100gpm at design conditionsThe cooling water pressure drop through the condenser is 5 psi at the design conditions and it is proportional tothe volumetric flow rate squred, A control valve is instulled in the cooling water line, and the total pressuredrop over the condenser and control valve is constant at 35 psi. The pressure in the reflux drum is measured bya pressure senson/transmitter that has a range of 150-250 psig and an output signal range of 4-20 mA. Aproportional controller with a gain of Ke = 4 is used to control pressure in the reflux drum by manipulating thecooling water flow rate. The current signal (mA) from the controller is converted to an air pressure signal(psig) in the I/P transducer.(b) Determine the valve size cocfficient Cy if the flow rate through a widc-open valve must be twice thedesign flow rate.(c) Specify the action of the control valve and controller.(d) Determine the valve opening fi) at the design flow rate.(e) Determine the current signal from the pressure transmitter at the design flow rate.VaporCoolingWater InCoolingWater OutCondenserPCITRefluxLiquid

Given: Vapor flow rate is 10000 Ibm/hr, average latent heat of vaporization is 125 Btu/Ibm Cooling water inlet and outlet temperature are 80°F and 110°F Solving first 3 (a) Calculate the cooling water flow rate in gpm at the design conditions. Assume the density of water is 62.4 lb/ft3 and its specific heat is 1.0 Btu/lbm°F. Also, 7.48 gal of liquid = 1 ft3 of liquid (b) Determine the valve size coefficient Cv if the flow rate through a wide-open valve must be twice the design flow rate. (c) Specify the action of the control valve and controller.a) Water flow rate is calculated as below: Using vapor flow rate in condenser: mvapor ×Hvap=mwater×Cpwater×∆Tmwater=mvapor ×HvapCpwater×∆T=10000Ibmhr×125BtuIbm1.001BtuIbm°F×110-80°F=41625.04 Ibmhr=41625.04500.72=83.13 gpm Answer: Cooling water flow rate in gpm is 83.13 at designed conditions b) Allowable pressure drop across valve for sizing: =Total pressure drop over the condenser and control valve -pressure drop through the condenser =35 psi-5 psi =30 psi Sizing for liquid services is given by: Q=Cv∆PG Where, Q is capacity in gpm Cv is valve sizing coefficient ∆P pressure drop across the valve in psi G is specific gravity of fluid (1 for water) Rearranging the above equation: Cv=QG∆P=83.13130=15.1774 Answer: Cv at given conditions is 15.1774 Considering the flow rate is twice the designed flow rate i.e. 200 gpm Cv=QG∆P=200130=36.5148 Answer: Cv calculated at flow rate twice the designed flow rate is 36.5148c) Action of control valve and controller: As the pressure in the reflux increases (uncondensed vapor in the reflux applies pressure in the reflux drum),the pressure transmitter installed on the reflux drum detects the rise or fall of pressure in the vessel and gives signal to pressure controller. The controller then calculates the error signal by taking the reference valve from the set point valve and the difference whether positive or negative is sent to I/P where it converts the current signal to pressure signal accordingly, then this calculated valve of pressure signal operates the valve opening or closing. High pressure in reflux drum means more uncondensed vapor, which means vapors in condenser are not cooled and condensed to the desired temperature, in order to do so water flow rate needs to be increased which is achieved by open position of the control valve and vice versa. In this way the pressure in the reflux drum is controlled.