Question

Two liquid streams of carbon tetrachloride are to be evaporated to produce a vapor at 200 C and 1 atm. If the first feed stream is at 1 atm and 30 C with a flow rate of 1000 kg/h and the second feed stream is at 70 C and 1 atm and has a flow rate of 500 kg/h, calculate the heat which must be supplied to the evaporator. Cp for CCl4 (gas phase in cal/gmol K) = 9.725 + 4.893 x 10^-2(T) - 5.421 x 10^-5(T²) + 2.112 x 10^-8(T³)

Answer 1

Boiling point of CCl4= 76.7 deg.c

Up to 76.7 deg.c, CCl4 needs to be supplied with sensible heat and then latent heat

Molecular weight of CCl4= 154 g/mole

Moles of CCl4 in 1000 kg/hr (Stream-1)= 1000*1000/154 gmoles/hr=6493.5 gmoles/hr

Moles of CCl4 in 500 kg/hr (Stream-2) = 500*1000/154= 3246.7 gmoles/hr

Specific heat of liquid CCl4=131.3 J/mol.K

1. Energy required to take the stream 1 up to boiling point= 6493.5*131.3*(76.7-30)= 39816259 joules=39816.259 Kj
2. Energy required to heat the stream 2 up to boiling point= 3246.7*131.3*(76.7-70)= 2856154 joules=2856.154 Kj
3. Now latent heat needs to be supplied to stream-1 and 2 for converting them into vapor. Total of stream-1+2= 6493.5+3246.7=9740.26 gmoles/hr. Enthalpy change of vaporization= 32.54 Kj/mol. Hence heat to be supplied for vaporizing the stream of 1 and 2= 9740.26*32.54Kj=316948.1 Kj
4. Sensible heat of vapor =

Cp= 9.725+4.893*10-3T- 5.421*10-5T²

Sensible heat= 9740.26*4816.978/(473.15-350)= 380987.6 Cal=1592528 joules= 1592.53 Kj

Total heat to be supplied= 39816.259Kj ( from 1)+ 2856.154( from 2)+ 316948.1 Kj ( from 3)+ 1592.53 Kj ( from 4)= 3213746 KJ