P6.46 A hot benzene stream (1000 gmol/h) at 500°C and 1 bar pressure is rap-idly cooled by mixing with cold liquid benzene at 25°C and 1 bar pres-sure. The final temperature of the combined benzene streams must be200°C. How much liquid benzene must be added?uolding torches but it is much more expensive than

The system here we have to consider is mixer Here we will consider mass balance and energy balance . In energy balance there are no heat and work terms in this system. There are no change in system energy as it is in steady state conditions.There is only one component in the system which is benzene. The units are all consistent in the system. Higher temperature streams are hot streams and lower temperatures are cold streams Mass balance Bhot+Bcold=Bout1000 gmol/h+Bcold=Bout we will choose reference stream the liquid benzene stream. The enthalpy of cold stream is 0 kJ/gmol The enthalpy of hot stream =∫298K353K(-6.2106+0.565T-3.141×10-4T2)dT+30700+∫353K773K(-33.92+0.4739T-3.017×10-4T2+7.13×10-8T3)×dTJ/gmol=7939+30700+61872 J/gmol=100511J/gmol=100.511kJ/gmol The enthalpy of out stream =∫298K353K(-6.2106+0.565T-3.141×10-4T2)dT+30700+∫353K573K(-33.92+0.4739T-3.017×10-4T2+7.13×10-8T3)×dTJ/gmol=7939+30700+27958 J/gmol=66597J/gmol=66.597kJ/gmol Energy balance and we will get the mass flow rate of liquid benzene . Hout×Bhot+Hcold×Bcold=Hout×Bout1000 gmol/h×100.511 kJ/gmol+0×Bcold=66.597 kJ/gmol×Bout100511kJ/h=66.597 kJ/gmol×BoutBout=1509.24gmol/hBhot+Bcold=1509.24gmol/hBcold=1509.24gmol/h-BhotBcold=1509.24gmol/h-1000 gmol/hBcold=509.24 gmol/h Final answer The liquid benzene to be added is 509.24 gmol/h