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A benzene/toluene liquid mixture consisting of 31 mole% benzene is fed to a reboiler, where 1250...

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:

Solutions

Expert Solution

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

Amanda K answered 1 year ago
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