Question

A benzene/toluene liquid mixture consisting of 31 mole% benzene is fed to a reboiler, where 1250 KW of heat is provided causing the mixture to partially vaporize. The vapor output has a flowrate of 80 kmol/hr and consists of 40 mole% benzene. The liquid output has a flowrate of 100 kmol/hr. The two output streams are in equilibrium and exit the reboiler at 130 oC. Determine the temperature of the feed stream. Neglect the effect of pressure and consider the following additional data:

Answer 1

	Cp(g)	Cp(l)	delta H (vap) kJ/mol
			@ BP
Benzene	82.44	134.8	30.77
Toluene	103.7	155.96	38.06

	Cp(g)	Cp(l)	delta H (vap) kJ/mol	BP
			@ BP
Benzene	82.44	134.8	30.77	80.1
Toluene	103.7	155.96	38.06	110.6
Mass Balance,				gas Out	80 mol/hr			Q (BP to 130 deg C)		Q (T to BP deg C)
				Benzene (40%)	32	mol/hr		131904		=32*134.8*(80-T)
				Toluene	48			99552		=48*155.96*(110-T)
Input = (Gas + Liq Outlet)	180 mol/hr	Reboiler						231456	= Q (gas) Total
Benzene (31 %)	55.8
Toluene	124.2
				Liq out	100 mol/hr
				Benzene	23.8
				Toluene	76.2
Heat required for vaporisation
= [gas moles Benzene x H (vap)] + [gas moles Toluene x H (vap)]
=	2811.52	kJ
1250 kW in 1 hr is provided
= 3600 x 1250		kJ
=	4500000	kJ
By heat balance, (temperatures in K) Heat for the gas moles to go from temp T to respective boiling points + Heat of vaporisation and then the heat for the gas moles to go from their Boiling points to 130 deg C
=> [32*134.8*(353-T)] + [48*155.96*(393-T)] + 380784 + 2811.52 = 4500000
=> [4313.6* (353-T)] + [7486*(393-T)] = 4116404.48
=> 1522700 + 2941998 - 11800 T = 4116404.48
=> T = 29.51 deg C