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In: Chemistry

QUESTION 1. Cool and condense benzene for storage (F&R 8.38) At OzChem plant, benzene vapour is...

QUESTION 1. Cool and condense benzene for storage (F&R 8.38)

At OzChem plant, benzene vapour is processed as a reagent. The plant produces benzene as a vapour (at 5780C, 1 atm), which is cooled and converted to a liquid at 250C in a continuous condenser that operates at 1 atm pressure (the pressure drop is negligible). The liquid condensate from the condenser is transferred at a rate of 150 mol/s into storage tanks. Calculate the rate (in kW) at which heat must be transferred from the benzene in the condenser.

Expert Solution

To find rate of heat transfer from benzene in the condenser,

we have benzene vapour temp. = 578C Cooled to 25C

T₁ = 25C & T₂ = 578C

Molar flowrate of condensate is 150 mol/s

Here benzene vapor is condensed to liquid so phase change must be considered.

Boiling point of Benzene is 80C at 1 atm so From 578C to 80C Specific Heat capacity will be considered and no phase will change at 80C, At 80C latent heat of phase change will be considered and from 80C to 25C cooling no phase will be change.

Q1 = From 578C to 80C

Cp = 135.69 J/mol*K

Q1 = m Cp (T2-T1)

Q1 = 150 (135.69) (578-80)

= 1013.6043 kW

Q2 = m

= 390 kJ/kg

Let convert flowrate from mol/s to kg/s

m=150 mol/s = 150* 78 g/s = 11.7 kg/s

Q2 = m = 11.7 * 390 = 4563 kJ/s = 4563 kW

Q3 = m Cp (T2-T1)

= 150 * (135.69) (80-25)

= 1119.4425 kW

Total Heat Transferred during condensation,

Q = Q1+Q2+Q3

=1013.6043+4563+1119.425

= 6696.0293 kW

queen_honey_blossom answered 2 years ago

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