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A 8390 g/hr stream of liquid methyl alcohol, also called methanol, (CH_3OH) at 6.20 atm and 24.0 degree C was held at constant pressure, vaporized and brought to 337.0 degree C. At what rate must heat be supplied to this system? Assume that methyl alcohol vapor behaves ideally for the temperature range and pressure given.

Respuesta :

Answer:

heat must be supplied to the system at a rate of 6670.22 KJ/h

Explanation:

m = 8390 g/h * ( Kg / 1000g )

⇒ m = 8.39 Kg/h

P1 = P2 = 6,20 atm.......... constant pressure

T1 = 24°C = 297 K

T2 = 337.0°C = 610 K

⇒ ΔT = 610 - 297 = 313 K

Q = ?

starting from first law at constant pressure:

  • Qp = ΔH = mCpΔT

∴ Cp (24°C) = 2,54 KJ / Kg°K.......engineering toolbox CH3OH

⇒ Q = ( 8.39 Kg/h ) * ( 2.54 KJ / Kg K ) * ( 313 K )

⇒ Q = 6670.22 KJ/h

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